Finite Element Method Pune University MCQs

Finite Element Method Pune University MCQs

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Matrix Algebra”.

1. What is a matrix?

a) Group of elements

b) Array of elements

c) Group of columns and rows

d) Array of numbers

Answer: b

Explanation: A matrix is an array of elements. The matrix A is denoted as [A]. An element located in the ith row and j th column is denoted as aij. A matrix is a collection of numbers arranged into a fixed number of rows and columns.

2. Which of the following is a row vector?

a) \

\(\left[

\right]\)

Answer: a

Explanation: A matrix of dimension is called row vector. A matrix of dimension is called column vector.

For example

d=[ 1 2 3 4] is a row vector.

c = \(\left[

\right]\) is a column vector.

3. T = _______

a) T

b) B T C T A T

c) C T B T A T

d) A T B T C T

Answer: c

Explanation: A matrix which is formed by turning all the rows of given matrix into columns and vice versa is called a transpose of matrix. The transpose of a product is given as the product of the transposes in the reverse order.

T = C T B T A T .

4. The derivative of Ax with respect to variable x p is given by __________

a) \

=x p

b) \(\frac{d}{dx}\)(x p )=A x

c) ∫ A x=x p

d) ∫x p =Ax

Answer: a

Explanation: Let A be an matrix of constants and x = [x 1 x 2 x 3 …… x n ] T be column vector of n variables. Then, derivative of A x with respect to variable x p is given by

=x p .

5. A symmetric matrix is called ____________, if all its Eigen values are strictly positive i.e., greater than zero.

a) Negative definite

b) Positive definite

c) Co- definite

d) Alternative definite

Answer: b

Explanation: If all Eigen values of symmetric matrix are positive then the matrix is called as positive definite matrix. A symmetric matrix A of dimension is positive definite if, for any non zero vector x = [x 1 x 2 x 3 …… x n ] T . That is x T Ax > 0.

6. A A -1 =A -1 A is a condition for ________

a) Singular matrix

b) Nonsingular matrix

c) Matrix inversion

d) Ad joint of matrix

Answer: c

Explanation: If det A not equal to zero, then A has an inverse, denoted by A -1 . The inverse satisfies the relation

A A -1 =A -1 A= I

7. A positive definite symmetric matrix A can be decomposed into form A=LL T this decomposition is called ________

a) Cholesky

b) Rayleighs

c) Galerkins

d) Potential energy

Answer: a

Explanation: L is the lower triangular matrix, and its transpose L T is upper triangular matrix. This is called Cholesky decomposition. It is a decomposition of a positive definite matrix into a product of lower triangular matrix and its conjugate transpose.

8. Det=0 is a ________

a) Characteristic equation

b) Matrix equation

c) Inversion of matrix

d) Cholesky’s equation

Answer: a

Explanation: A non zero solution will occurs when Ʌ is a singular matrix or detɅ=0 it is a characteristic equation. A characteristic equation is the equation which is solved to find the Eigen values, also called the characteristic polynomials.

9. \Missing open brace for subscript Principle diagonal matrix

b) Upper triangular matrix

c) Lower triangular matrix

d) Singular matrix

Answer: b

Explanation: An upper triangular or right triangular matrix is one whose elements below the principal diagonal elements are zero. The sum or product or inverse of any two upper triangular matrixes is an upper triangular matrix.

10. A=\

a) 120

b) -80

c) -175

d) 0

Answer: c

Explanation: det\(

= a 11 (a 22 a 33 -a 32 a 23 )-a 12 (a 21 a 33 -a 31 a 23 )+a 13 (a 21 a 32 -a 31 a 22 )

= 3*-)-2*-)+1*-)

=-175.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Gaussian Elimination”.

1. Gaussian elimination is a name given to a well known method of solving simultaneous equation by successively eliminating _________

a) Variables

b) Equations

c) Unknown

d) Algorithms

Answer: c

Explanation: Gaussian elimination is an approach for solving equations type of Ax=B in matrix form. Gaussian elimination is a name given to a well known method of solving simultaneous equation by successively eliminating Unknowns.

2. Step number in Gaussian elimination is denoted as ___________

a) Superscript

b) Subscript

c) Unknown

d) Elimination

Answer: a

Explanation: Gaussian elimination is an algorithm for solving systems of linear equations. The idea at step 1 is to use equation 1 in eliminating x 1 from remaining equations. We know the step numbers as superscript set in parentheses.

3. In Gaussian elimination, A is defined as symmetric matrix then its multiplier is defined as ____

a) C = a kk /a ik

b) C = a ki /a kk

c) C = a ik /a kk

d) C = a kk /a ki

Answer: b

Explanation: In a Gaussian elimination, If A is a symmetric matrix then its algorithm can be modified in two methods, one method is its multiplier is defined as C = a ki /a kk . 2 nd modification is related to DO LOOP.

4. A banded matrix is defined as ____________

a) Non zero elements are contained in band

b) Zero elements are contained in a band

c) Non zero elements are contained out of a band

d) Both Non zero elements and Zero elements

Answer: a

Explanation: A band matrix is a sparse matrix whose non zero entries are confined to a diagonal band. In a banded matrix, all of the non zero elements are contained within a band; outside of the band all elements are zero.

5. In a symmetric banded matrix __________

a) a ij =a ji

b) a ji *a ij

c) a ij ≠a ji

d) a ii =a jj

Answer: a

Explanation: For a symmetric banded matrix a ij =a ji . A symmetric banded matrix is a symmetric matrix whose nonzero elements are arranged uniformly near the diagonal.

6. Consider a nxn symmetric matrix: \Missing open brace for subscript Full band width

b) Half band width

c) Semi band width

d) Co band width

Answer: b

Explanation: In semi banded matrix of nxn matrix nbw is denoted as Half band width of the matrix. By this we can easily be solved further. The term band or banded matrix is used for a matrix whose band width is reasonably small.

7. The line separating from the top zeros from the first non-zero element is called ____

a) Equation

b) Gaussian solution

c) Skyline solution

d) Both Gaussian and skyline solutions

Answer: c

Explanation: If there are zeros at the top of the column, only the elements starting from the first non zero value need be stored. The line separating from the top zeroes from the first non- zero element is called Skyline solution.

8. Frontal method is a _______ of Gaussian elimination method that uses the structure of finite element problem.

a) Structure

b) Variation

c) Algorithm

d) Data

Answer: b

Explanation: Frontal method is a variation of Gaussian elimination method that uses the structure of finite element problem. Elements can be stored in-core in a clique sequence as recently proposed by areas, this subset is called front and it is essentially the transition region between the part of the system already finished.

9. Frontal method is implemented for ________

a) Hexahedral element

b) Polyhedral element

c) Octahedral element

d) Both Hexahedral and Polyhedral

Answer: a

Explanation: The frontal method is implemented for the hexahedral element. By this method we can recombine tetrahedral element to hexahedral element. However, non conformal quadrilateral faces adjacent to triangular faces.

10. Frontal method involves __________

a) Computer programming

b) Manual programming

c) C- programming

d) Computing

Answer: a

Explanation: The elimination process is handled by writing the eliminated equation to the computer hard disk. Processing the front involves dense matrix operations, which use the CPU efficiently. In a typical implementation, only the front is in memory while factors in decomposition are written into files.

This set of Finite Element Method online test focuses on “Conjugate Gradient Method for Equation Solving”.

1. Conjugate gradient method is a ________

a) Standard method

b) Equation method

c) Iterative method

d) Elimination method

Answer: c

Explanation: The conjugate gradient method is an iterative method for the solution of equations. And it is implemented in several computer codes. This method is used to solve un-constrained optimization problems such as energy minimization.

2. Conjugate gradient method is for only _________

a) Non symmetric matrix

b) Symmetric matrix

c) Symmetric and Non symmetric matrix

d) Identity matrix

Answer: b

Explanation: Conjugate gradient method generally used to solve symmetric matrices. This method is often implemented as an iterative algorithm. Large sparse systems often arise when numerically solving partial differential equations or optimization problems.

3. Which version is used to solve conjugate gradient method of the algorithm of symmetric matrices?

a) Rayleighs version

b) Fletcher-Reeves version

c) Frontal method

d) Galerkins method

Answer: b

Explanation: We generally present Fletcher-Reeves version of the algorithm of the symmetric matrices in Conjugate gradient method. This method is of algorithm model and also as an iterative method. Conjugate gradient method is unstable with respect to even small perturbations.

4. The iterations are continued until g k T g x reaches a ________

a) Small value

b) Infinite value

c) Large value

d) Negative value

Answer: a

Explanation: When k=0 1 2 3…… The iterations are continued until g k T g x reaches small value in conjugate gradient method. By getting small values we can easily conclude the problems. And also can be easily iterated to next easiest method like Gaussian elimination.

5. This method is robust and has ________ n iterations.

a) Diverges

b) Converges

c) Converge and diverge

d) Symmetric

Answer: b

Explanation: When we solve a symmetric matrix in conjugate gradient method, it can be simplified in n iterations and also this method is robust and converges “n” no of iterations. Conjugate gradient method can be implemented to optimization of solution.

6. Conjugate gradient method is implemented in _________

a) Algorithm solving

b) Iterative solving

c) Program solving

d) Program CG solving

Answer: d

Explanation: Conjugate gradient algorithm procedure is implemented in program CG solving, which is included on disk. This method is also easily to solve in MAT Lab also. Several algorithms have been proposed e.g., CGLS, LSQR.

7. Conjugate gradient method includes several ______

a) Computer codes

b) Programs

c) Algorithms

d) Matrices

Answer: a

Explanation: Conjugate gradient method is an iterative method for the solution of equations. This method is become increasingly popular and is implemented in several computer codes. Because, it is mainly done on computer software such as MAT lab.

8. Conjugate gradient algorithm can be accelerated by using _________

a) Algorithm strategies

b) Preconditioning strategies

c) Conditioning strategies

d) Computer strategies

Answer: b

Explanation: Conjugate gradient algorithm can be accelerated by using preconditioning strategies. Preconditioner of a matrix has to be symmetric positive definite and fixed and that cannot be changed from iteration to iteration. The behavior of the preconditioned conjugate gradient method may become unpredictable.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “One Dimensional Problems – Finite Element Modelling”.

1. If the structure is divided into discrete areas or volumes then it is called an _______

a) Structure

b) Element

c) Matrix

d) Boundaries

Answer: b

Explanation: An element is a basic building block of finite element analysis. An element is a mathematical relation that defines how the degrees of freedom of node relate to next. The structure is divided into discrete areas or volumes known as elements.

2. In finite element modeling nodal points are connected by unique ________

a) Surface

b) Shape

c) Eigen values

d) Matrix

Answer: a

Explanation: A node is a co-ordinate location in a space where the degrees of freedom can be defined. A node may be limited in calculated motions for a variety of reasons. Element boundaries are defined when nodal points are connected by unique polynomial curve or surface.

3. In finite element modeling every element connects to _______

a) 4 nodes

b) 3 nodes

c) 2 nodes

d) Infinite no of nodes

Answer: c

Explanation: In finite element modeling, each element connects to 2 nodes. Better approximations are obtained by increasing the number of elements. It is convenient to define a node at each location where the point load is applied.

4. In one dimensional problem, each node has _________ degrees of freedom.

a) 2 degrees of freedom

b) 3 degrees of freedom

c) No degrees of freedom

d) 1 degree of freedom

Answer: d

Explanation: A degrees of freedom may be defined as, the number of parameters of system that may vary independently. It is the number of parameters that determines the state of a physical system. In one dimensional problem, every node is permitted to displace only in the direction. Thus each node has only one degree of freedom.

5. Which relations are used in one dimensional finite element modeling?

a) Stress-strain relation

b) Strain-displacement relation

c) Total potential energy

d) Total potential energy; Stress-strain relation; Strain-displacement relation.

Answer: d

Explanation: The basic procedure for a one dimensional problem depends upon total potential energy, stress-strain relation and strain-displacement relation are used in developing the finite element modeling.

6. One dimensional element is the linear segments which are used to model ________

a) Bars and trusses

b) Plates and beams

c) Structures

d) Solids

Answer: a

Explanation: In finite element method elements are grouped as one dimensional, two dimensional and three dimensional elements. One dimensional element is the linesegment which is used to model bars and trusses.

7. Modeling is defined as ________________

a) Elemental area with uniform cross section

b) Elemental area with non uniform cross section

c) Structural area with uniform cross section

d) Non structural area with non uniform cross section

Answer: a

Explanation: Modeling is one of the basic steps in finite element method. Let us model a stepped shaft consists of discrete no of elements each having a uniform cross sectional area. Average cross section area within each region is evaluated and used to define elemental area with uniform cross sectional area.

Stepped Shaft Model consisting discrete number of elements with uniform cross sectional area

A1=A 1 ’+A 2 ’/2.

8. Discretization includes __________ numbering.

a) Element and node

b) Only nodal

c) Only elemental

d) Either nodal or elemental

Answer: a

Explanation: The process of dividing a body into equivalent number of finite elements associated with nodes is called discretization. Discretization includes both node and element numbering, in this model every element connects two nodes.

9. The loading on an element includes _______

a) Body force

b) Traction force

c) Point load

d) Body force, Traction force & Point load

Answer: d

Explanation: The loading on an element includes body force; traction force & point load. Body force is distributed force acting on every elemental volume. Traction force is a distributed load along the surface of a body.

10. Global nodes corresponds to _______

a) Entire body

b) On surface

c) On interface

d) On element

Answer: a

Explanation: Global coordinate system corresponds to the entire body. It is used to define nodes in the entire body.

11. Local node number corresponds to ______________

a) Entire body

b) On element

c) On interface

d) On surface

Answer: b

Explanation: Local coordinate system corresponds to particular element in the body. The numbering is done to that particular element neglecting the entire body.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “One Dimensional Problems – Co-ordinates and Shape Functions”.

1. Natural or intrinsic coordinate system is used to define ___________

a) Co-ordinates

b) Shape functions

c) Displacement functions

d) Both shape functions and co-ordinate functions

Answer: b

Explanation: Natural coordinate system is another way of representing direction. It is based on the relative motion of the object. We use this system of coordinates in defining shape functions, which are used in interpolating the displacement field.

2. In q=[q 1 ,q 2 ] T is defined as __________

a) Element displacement vector

b) Element vector

c) Displacement vector

d) Shape function vector

Answer: a

Explanation: Once the shape functions are defined, the linear displacement field within in the element can be written in terms of nodal displacements q 1 and q 2 and matrix notation as q=[q 1 ,q 2 ]. Here q is referred as element displacement function.

3. Shape function is just a ___________

a) Displacement function

b) Equation

c) Interpolation function

d) Matrix function

Answer: c

Explanation: The shape function is the function which interpolates the solution between the discrete values obtained at the mesh nodes. Low order polynomials are typically chosen as shape functions. Interpolation within the shape functions is achieved through shape functions.

4. Isoparametric formula is ______________

a) x=N 1 x 1 +N 2 x 2

b) x=N 2 x 1 +N 1 x 2

c) x=N 1 x 1 -N 2 x 2

d) x=N 2 x 1 -N 1 x 2

Answer: a

Explanation: From nodal displacement equation we can write that isoparametric equation as

x=N 1 x 1 +N 2 x 2

Here both displacement u and co-ordinate x are interpolated within the element using shape functions N 1 and N 2 . This is called isoparametric formulation in literature.

5. B=\Missing open brace for subscript Strain matrix

b) Element-strain displacement matrix

c) Displacement matrix

d) Elemental matrix

Answer: b

Explanation: ε= Bq

Here B is element strain displacement matrix. Use of linear shape functions results in a constant B matrix. Hence, in a constant strain within the element. The stress from Hooke’s law is

σ= EBq .

6. Deformation at the end of elements are called _____________

a) Load

b) Displacement functions

c) Co-ordinates

d) Nodes

Answer: d

Explanation: Nodes are the points where displacement, reaction force, deformation etc.., can be calculated. Corner of each element is called a node. A node is a co-ordinate location in space where degrees of freedom are defined.

7. Write the shape function of the given element.

Find the Shape Function of the element u= N 1 u 1 +N 2 u 2 . Here N 1 & N 2 are

a) N 1 =1-x/l e &N 2 =x/l e

b) N 1 =x/l e &N 2 =1-x/l e

c) N 1 =0 & N 2 =x

d) N 1 =x & N 2 =0

Answer: a

Explanation:

Shape Function of the element

1 2 --- local variables

I j --- global variables

u1(e) u2(e)

x1=0 x2=0

Then matrix notation form is

u=\(

u 1 =c 1 +c 2 =c 1

u 2 = c 1 +c 2 (l e )

In matrix equation

By solving we get

N 1 =1-x/l e & N 2 =x/l e .

8. In shape functions, first derivatives must be _______ within an element.

a) Infinite

b) Finite

c) Natural

d) Integer

Answer: b

Explanation: In general shape functions need to satisfy that, first derivatives must be finite within element. Shape functions are interpolation functions. First derivatives are finite within element because for easy calculations.

9. In shape functions, _________ must be continuous across the element boundary.

a) Derivatives

b) Nodes

c) Displacement

d) Shape function

Answer: c

Explanation: Shape functions are interpolation functions. In general shape functions need to satisfy that, displacements must be continuous across the element boundary.

10. Stresses due to rigid body motion are _______________

a) Zero

b) Considered

c) Not considered

d) Infinite

Answer: c

Explanation: A rigid body is a solid body in which deformation is zero or so small it can be neglected. A rigid body is usually considered as a continuous distribution of mass. By rigid body deformation is neglected so stresses are not considered.

11. The expressions u=Nq; ε= Bq; σ= EBq relate ____________

a) Displacement, Strain and Stress

b) Strain and stress

c) Strain and displacement

d) Stress and displacement

Answer: a

Explanation: Stress is defined as force per unit area. Strain is defined as the amount of deformation in the direction of applied force. Displacement is the difference between the final and initial position of a point. The given expressions show the relationship between stress, strain and displacement of a body.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “One Dimensional Problems – Potential Energy Approach”.

1. Continuum is discretized into_______ elements.

a) Infinite

b) Finite

c) Unique

d) Equal

Answer: b

Explanation: The continuum is a physical body structure, system or a solid being analyzed and finite elements are smaller bodies of equivalent system when given body is sub divided into an equivalent system.

2. U e =\(\frac{1}{2}\int\) σ T εA dx is a _____________

a) Potential equation

b) Element strain energy

c) Load

d) Element equation

Answer: b

Explanation: The given equation is Element strain energy equation. The strain energy is the elastic energy stored in a deformed structure. It is computed by integrating the strain energy density over the entire volume of the structure.

3. Which is the correct option for the following equation?

K e =\

Load vector

b) Energy matrix

c) Node matrix

d) Element stiffness matrix

Answer: d

Explanation: The given matrix is element stiffness matrix. A stiffness matrix represents the system of linear equations that must be solved in order to as certain an approximate solution to the differential equation. The stiffness matrix is a inherent property of a structure. Stiffness matrix is positive definite. K e is linearly proportional to the product E e A e and inversely proportional to length l e .

4. Body force vector f e = _____________

a) \

A e l e \

A e l e f \(

Answer: a

Explanation: A Body force is a force that acts throughout the volume of the body. Forces due to gravity, electric and magnetic fields are examples of body forces.

5. Between wheel and ground how much of traction force is required?

a) High traction force

b) Low traction force

c) Infinite traction force

d) No traction force

Answer: a

Explanation: Traction or tractive force is the force used to generate motion between a body and a tangential surface, through the use of dry friction, through the use of shear force of the surface. In the design of wheeled or tracked vehicles, high traction between wheel and ground should be more desirable.

6. Element traction force is given by ___

a) T e =Tl e \

T e =Tl e

c) T e =\

Undefined

Answer: c

Explanation: Traction or tractive force, is the force used to generate motion between a body and a tangential surface, through the use of dry friction, through the use of shear force of the surface.

7. ∏ = \(\frac{1}{2}\) Q T KQ-Q T F In this equation F is defined as _________

a) Global displacement vector

b) Global load vector

c) Global stiffness matrix

d) Local displacement vector

Answer: b

Explanation: Global load vector is assembly of all local load vectors. This load vector is obtained by due to given load. In the given equation F is defined as global load vector.

8. What are the basic unknowns on stiffness matrix method?

a) Nodal displacements

b) Vector displacements

c) Load displacements

d) Stress displacements

Answer: a

Explanation: Stiffness matrix represents systems of linear equations that must be solved in order to as certain an approximate solution to the differential equation. In stiffness matrix nodal displacements are treated as basic unknowns for the solution of indeterminate structures. The external loads and the internal member forces must be in equilibrium at the nodal points.

9. Write the element stiffness for a truss element.

a) K=\

K=\

K=AE

Answer: b

Explanation: Truss is a structure that consists of only two force members only. Where the members are organized so that the assemblage as a whole behaves as a single object.

10. Formula for global stiffness matrix is ____________

a) No. of nodes*Degrees of freedom per node

b) No. of nodes

c) Degrees of freedom per node

d) No. of elements

Answer: a

Explanation: Generally global stiffness matrix is used to complex systems. Stiffness matrix method is used for structures such as simply supported, fixed beams and portal frames. Size of stiffness matrix is defined as:

Size of global stiffness matrix=No. of nodes*Degrees of freedom per node.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “One Dimensional Problems – Galerkin Approach”.

1. Galerkin technique is also called as _____________

a) Variational functional approach

b) Direct approach

c) Weighted residual technique

d) Variational technique

Answer: c

Explanation: The equivalent of applying the variation of parameters to a function space, by converting the equation into weak formulation. Galerkin’s method provide powerful numerical solution to differential equations and modal analysis. The Galerkin method of weighted residuals, the most common method of calculating the global stiffness matrix in the finite element method.

2. In the equation, \

Adx -\int_{L} \phi^T f Adx -\int_{L}\phi^Tdx – \sum_{i}\phi_i P_i=0\) First term represents _______

a) External virtual work

b) Virtual work

c) Internal virtual work

d) Total virtual work

Answer: c

Explanation: In the given equation first term represents internal virtual work. Virtual work means the work done by the virtual displacements. The principle of virtual work is equivalent to the conditions for static equilibrium of a rigid body expressed in terms of total forces and torques. The virtual work done by internal forces is called internal virtual work.

3. Considering element connectivity, for example for element ψ=[ψ 1 , ψ 2 ] n for element n, then the variational form is ______________

a) ψ T =0

b) ψ=0

c) ψ=F

d) ψ=0

Answer: a

Explanation: Element connectivity means Assemble the element equations. To find the global equation system for the whole solution region we must assemble all the element equations. For formulation of a variational form for a system of differential equations. First method treats each equation independently as a scalar equation, while the other method views the total system as a vector equation with a vector function as a unknown.

4. Write the element stiffness matrix for a beam element.

a) K=\

K=\

K=\(\frac{2E}{l}

Answer: b

Explanation: Element stiffness matrix means it is a matrix method that makes use of the members stiffness relations for computing member forces and displacements in the structures.

5. Element connectivities are used for _____

a) Traction force

b) Assembling

c) Stiffness matrix

d) Virtual work

Answer: b

Explanation: Element connectivity means “Assemble the element equations. To find the global equation system for the whole solution region we must assemble all the element equations. In other words we must combine local element equations for all the elements used for discretization.

6. Virtual displacement field is _____________

a) K=\

F=ma

c) f=y

d) &fcy;=&fcy;

Answer: d

Explanation: Virtual work is defined as work done by a real force acting through a virtual displacement. Virtual displacement is an assumed infinitesimal change of system coordinates occurring while time is held constant.

7. Virtual strain is ____________

a) εMisplaced &=\

εMisplaced &=\

&fcy;=\(\frac{d\varepsilon}{d\phi}\)

Answer: b

Explanation: Virtual work is defined as the work done by a real force acting through a virtual displacement. A virtual displacement is any displacement is any displacement consistent with the constraints of the structure.

8. To solve a galerkin method of approach equation must be in ___________

a) Equation

b) Vector equation

c) Matrix equation

d) Differential equation

Answer: d

Explanation: Galerkin method of approach is also called as weighted residual technique. This method of approach can be used for irregular geometry with a regular pattern of nodes. The solution function is substituted in a differential equation, this differential equation will not be satisfied and will give a residue.

9. By the Galerkin approach equation can be written as __________

a) {P}-{K}{Δ}=0

b) {K}-{P}{Δ}=0

c) {Δ}-{p}{K}=0

d) Undefined

Answer: a

Explanation: Galerkin’s method of weighted residuals, the most common method of calculating the global stiffness matrix in fem. This requires the boundary element for solving integral equations.

10. In basic equation Lu=f, L is a ____________

a) Matrix function

b) Differential operator

c) Degrees of freedom

d) No. of elements

Answer: b

Explanation: The method of weighted residual technique uses the weak form of physical problem or the direct differential equation. The basic equation Lu=f in that L is an differential operator. It uses the principle of orthogonality between Residual function and basis function.

This set of Finite Element Method Interview Questions and Answers focuses on “One Dimensional Problems – Assembly of the Global Stiffness Matrix and Load Vector”.

1. How is Assembly of stiffness matrix symbolically denoted?

a) K={k} e

b) K←∑ e K e

c) K←∑K e

d) Undefined

Answer: b

Explanation: The stiffness matrix represents the system of linear equations that must be solved in order to ascertain an approximate solution to differential equation.

2. What is the Strain energy equation?

a) U e =\(\frac{1}{2}\)q T k e q

b) U e =\(\frac{1}{2}\)q e k e q

c) U e =\(\frac{1}{2}\)qk e

d) U e =\(\frac{1}{2}\)q T k e

Answer: a

Explanation: Strain energy is defined as the energy stored in the body due to deformation. The strain energy per unit volume is known as strain energy density and the area under stress-strain curve towards the point of deformation. When the applied force is released, the system returns to its original shape.

3. What is the actual equation of stiffness matrix?

a) K=\

K=\

K=\(\frac{AE}{l}

Answer: d

Explanation: Stiffness matrix represents the system of linear equations that must be solved in order to ascertain an approximate solution to the differential equation. The stiffness matrix is an inherent property of the structure. A stiffness matrix is a positive definite.

4. From where does the global load vector F is assembled?

a) Element force vectors only

b) Point loads only

c) Both element force vectors and point loads

d) Undefined

Answer: c

Explanation: Global load vector is assembling of all local load variables. This global load vector is get from assembling of both element force vectors and point loads.

5. For an element as given below, what will be the 1 ST element stiffness matrix?

Find the 1st element stiffness matrix for the given element

a) \

\(\frac{EA_1}{l_1}

Answer: a

Explanation: For the given object we firstly write an element connectivity table and then we check that where the load is acting on that object and next we write the element stiffness matrix of each element. For this object first element stiffness matrix is as given.

6. Principal of minimum potential energy follows directly from the principal of ________

a) Elastic energy

b) Virtual work energy

c) Kinetic energy

d) Potential energy

Answer: b

Explanation: The total potential energy of an elastic body is defined as sum of total strain energy and the work potential energy. Therefore the principal of minimum potential energy follows directly the principal of virtual work energy.

7. The points at where kinetic energy increases dramatically then those points are called _______

a) Stable equilibrium points

b) Unstable equilibrium points

c) Equilibrium points

d) Unique points

Answer: b

Explanation: If an external force acts to give the particles of the system some small initial velocity and kinetic energy will developed in that body then the point where kinetic energy decreased that point is Stable equilibrium point and the point where the kinetic energy dramatically increased then the point is called Unstable equilibrium points.

8. We can obtain same assembly procedure by Stiffness matrix method and _______

a) Potential energy method

b) Rayleigh method

c) Galerkin approach

d) Vector method

Answer: c

Explanation: Galerkin method provides powerful numerical solution to differential equations and modal analysis. Assembling procedure is same for both stiffness matrix method and galerkin approach method in Finite element modeling.

9. By element stiffness matrix we can get relation of members in an object in _____

a) Different matrices

b) One matrix

c) Identity matrix

d) Singular matrix

Answer: b

Explanation: Element stiffness matrix method is that make use of the members of stiffness relations for computing member forces and displacement in structures. So by this element stiffness matrix method we can get relation of members in an object in one matrix.

10. What is the Global stiffness method called?

a) Multiple matrix

b) Direct stiffness matrix

c) Unique matrix

d) Vector matrix

Answer: b

Explanation: Global stiffness matrix method makes use of the members stiffness relations for computing member forces and displacements in structures. Hence Global stiffness matrix or Direct stiffness matrix or Element stiffness matrix can be called as one.

11. Which technique do traditional workloads use?

a) Scale out technique

b) Scale up technique

c) Building technique

d) Shrinking technique

Answer: b

Explanation: When the workload increases on the system, the machine scales up by adding more RAM, CPU and storage spaces.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “One Dimensional Problems – Properties of K”.

1. Dimension of global stiffness matrix is _______

a) N X N , where N is no of nodes

b) M X N , where M is no of rows and N is no of columns

c) Linear

d) Eliminated

Answer: a

Explanation: A global stiffness matrix is a method that makes use of members stiffness relation for computing member forces and displacements in structures. The dimension of global stiffness matrix K is N X N where N is no of nodes.

2. Each node has only _______

a) Two degrees of freedom

b) One degree of freedom

c) Six degrees of freedom

d) Three degrees of freedom

Answer: b

Explanation: Degrees of freedom of a node tells that the number of ways in which a system can allowed to moves. In a stiffness matrix each node can have one degree of freedom.

3. Global stiffness K is a______ matrix.

a) Identity matrix

b) Upper triangular matrix

c) Lower triangular matrix

d) Banded matrix

Answer: d

Explanation: A banded matrix is a sparse matrix whose non zero entities are confined to a diagonal band, comprising the main diagonal and zero or more diagonals on either side. A global stiffness matrix K is a banded matrix. That is, all the elements outside the band are zero.

4. The dimension of K banded is _____

a) [N X NBW ]

b) [NBW X N]

c) [N X N]

d) [NBW X NBW]

Answer: a

Explanation: K can be compactly represented in banded form. As K banded is of dimension [N X NBW] where NBW is the half band width.

5. In many one-dimensional problems, the banded matrix has only two columns. Here NBW=____

a) 6

b) 3

c) 7

d) 2

Answer: d

Explanation: NBW means half bandwidth. Many of the One- dimensional problems banded matrix has only 2 columns then NBW=2. We know that

NBW =max +1

6. Stiffness matrix represents a system of ________

a) Programming equations

b) Iterative equations

c) Linear equations

d) Program CG SOLVING equations

Answer: c

Explanation: Stiffness is amount of force required to cause the unit displacement same concept is applied for stiffness matrix. The stiffness matrix represents a system of linear equations that must be solved in order to ascertain an approximate solution to differential equation.

7. Stiffness matrix is _____

a) Non symmetric and square

b) Symmetric and square

c) Non symmetric and rectangular

d) Symmetric and rectangular

Answer: b

Explanation: Stiffness matrix is a inherent property of the structure. The property of a stiffness matrix, as the stiffness matrix is square and symmetric.

8. In stiffness matrix, all the _____ elements are positive.

a) Linear

b) Zigzag

c) Diagonal

d) Rectangular

Answer: c

Explanation: Stiffness matrix represents system of linear equations that must be solved in order to ascertain an approximate solution to the differential equation. The stiffness matrix is an inherent property of a structure. In stiffness matrix all the diagonal elements are positive.

9. The size of global stiffness matrix will be equal to the total ______ of the structure.

a) Nodes

b) Degrees of freedom

c) Elements

d) Structure

Answer: b

Explanation: For a global stiffness matrix, a structural system is an assemblage of number of elements. These elements are interconnected to form the whole structure. The size of global stiffness matrix will be equal to the total degrees of freedom of the structure.

10. Element stiffness is obtained with respect to its ___

a) Degrees of freedom

b) Nodes

c) Axes

d) Elements

Answer: c

Explanation: A stiffness matrix represents system of linear equations that must be solved in order to ascertain an approximate solution to the differential equation. Element stiffness is obtained with respect to its axes.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Finite Element Equations – Treatment of Boundary Conditions”.

1. Types of Boundary conditions are ______

a) Potential- Energy approach

b) Penalty approach

c) Elimination approach

d) Both penalty approach and elimination approach

Answer: d

Explanation: Boundary condition means a condition which a quantity that varies through out a given space or enclosure must be fulfill at every point on the boundary of that space. In fem, Boundary conditions are basically two types they are Penalty approach and elimination approach.

2. Potential energy, π = _________

a) \(\frac{1}{2}\)Q T KQ-Q T F

b) QKQ-QF

c) \

\(\frac{1}{2}\)QF

Answer: a

Explanation: Minimum potential energy theorem states that “Of all possible displacements that satisfy the boundary conditions of a structural system, those corresponding to equilibrium configurations make the total potential energy assume a minimum value.”

Potential energy π=\(\frac{1}{2}\)Q T KQ-Q T F

3. Equilibrium conditions are obtained by minimizing ______

a) Kinetic energy

b) Force

c) Potential energy

d) Load

Answer: c

Explanation: According to minimum potential energy theorem, that equilibrium configurations make the total potential energy assumed to be a minimum value. Therefore, Equilibrium conditions are obtained by minimizing Potential energy.

4. In elimination approach, which elements are eliminated from a matrix ____

a) Force

b) Load

c) Rows and columns

d) Undefined

Answer: c

Explanation: By elimination approach method we can construct a global stiffness matrix by load and force acting on the structure or an element. Then reduced stiffness matrix can be obtained by eliminating no of rows and columns of a global stiffness matrix of an element.

5. In elimination approach method, extract the displacement vector q from the Q vector. By using ___

a) Potential energy

b) Load

c) Force

d) Element connectivity

Answer: d

Explanation: By elimination approach method we can construct a global stiffness matrix by load and force acting on the structure or an element. Then we extract the displacement vector q from the Q vector. By using Element connectivity, and determine the element stresses.

6. Penalty approach method is easy to implement in a ______

a) Stiffness matrix

b) Iterative equations

c) Computer program

d) Cg solving

Answer: c

Explanation: Penalty approach is the second approach for handling boundary conditions. This method is used to derive boundary conditions. This approach is easy to implement in a computer program and retains it simplicity even when considering general boundary conditions.

7. If Q 1 =a 1 then a 1 is _________

a) Displacement

b) Symmetric

c) Non symmetric

d) Specified displacement

Answer: d

Explanation: In penalty approach method a 1 is known as specified displacement of 1. This is used to model the boundary conditions.

8. The first step of penalty approach is, adding a number C to the diagonal elements of the stiffness matrix. Here C is a __________

a) Large number

b) Positive number

c) Real number

d) Zero

Answer: a

Explanation: Penalty approach is one of the method to derive boundary conditions of an element or a structure. The first step is adding a large number C to the diagonal elements of the stiffness matrix. Here C is a large number.

9. In penalty approach evaluate _______ at each support.

a) Load vector

b) Degrees of freedom

c) Force vector

d) Reaction force

Answer: d

Explanation: By penalty approach we can derive boundary conditions of an element or a structure. The first step of this approach is to add a large number to the diagonal elements. Second step is to extract element displacement vector. Third step is to evaluate reaction force at each point.

10. For modeling of inclined roller or rigid connections, the method used is ___

a) Elimination approach

b) Multiple constraints

c) Penalty approach

d) Minimum potential energy theorem

Answer: b

Explanation: Multiple constraints is one of the method for boundary conditions it is generally used in problems for modeling inclined rollers or rigid connections.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “One Dimensional Problems – Quadratic Shape Function”.

1. What is a shape function?

a) Interpolation function

b) Displacement function

c) Iterative function

d) Both interpolation and displacement function

Answer: d

Explanation: The shape function is a function which interpolates the solution between discrete values obtained at the mesh nodes. Lower order polynomials are chosen as shape functions. Shape function is a displacement function as well as interpolation function.

2. Quadratic shape functions give much more _______

a) Precision

b) Accuracy

c) Both Precision and accuracy

d) Identity

Answer: b

Explanation: The shape function is function which interpolates the solution between discrete values obtained at the mesh nodes. The unknown displacement field was interpolated by linear shape functions within each element. Use of quadratic interpolation leads to more accurate results.

3. Strain displacement relation ______

a) ε=\

ε=\

x=\

Cannot be determined

Answer: a

Explanation: The relationship is that connects the displacement fields with the strain is called strain – displacement relationship. If strain is ε then strain – displacement relation is

ε=\(\frac{du}{dx}\)

4. The _____ and ______ can vary linearly.

a) Force and load

b) Precision and accuracy

c) Strain and stress

d) Distance and displacement

Answer: c

Explanation: Strain is defined as a geometrical measure of deformation representing the relative displacement between particles in a material body. Stress is a physical quantity that expresses the internal forces that neighboring particles of a continuous material exert on each other. In quadratic shape functions strain and stress can vary linearly.

5. By Hooke’s law, stress is ______

a) σ=Bq

b) σ=EB

c) B=σq

d) σ=EBq

Answer: d

Explanation: Hooke’s law states that the strain in a solid is proportional to the applied stress within the elastic limit of that solid.

6. Nodal displacement as _____

a) u=Nq

b) N=uq

c) q=Nu

d) Program SOLVING

Answer: a

Explanation: Nodes will have nodal displacements or degrees of freedom which may include translations, rotations and for special applications, higher order derivatives of displacements.

7. At the condition, at , N 1 =1 at ξ=-1 which yields c=\Missing open brace for subscript Computer functions

b) Programming functions

c) Galerkin function

d) Lagrange shape functions

Answer: d

Explanation: The lagrange shape function sum to unity everywhere. At the given condition the shape functions are named as Lagrange shape functions.

8. Element body force vector is given by ______

a) f e =\(\frac{A_el_ef}{2} \int_{-1}^{1}\) N T dξ

b) f e = \(\frac{A_el_e}{2}\int_{}^{1}\) N T dξ

c) Conditioning strategies

d) f e =∫ N T dξ

Answer: a

Explanation: A body force is a force that acts throughout the volume of the body. In FEM, Element body force vector is given by

f e =\(\frac{A_el_ef}{2} \int_{-1}^{1}\) N T dξ

9. Element traction force is given by ___

a) T e =\(\frac{l_e}{2}\int_{-1}^{1}\)N T dξ

b) T e =\(\frac{l_eT}{2}\int_{-1}^{1}\)N T dξ

c) T e =\(\frac{l_eT}{2}\int_{-1}^{1}\)N T

d) Undefined

Answer: b

Explanation: The term traction force can either refer to the total traction a vehicle exerts on a surface or the amount of total traction that is parallel to the direction of motion.

Element traction force is given by

T e =\(\frac{l_eT}{2}\int_{-1}^{1}\)N T dξ

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “One Dimensional Problems – Temperature Effects”.

1. With temperature effect which will vary linearly?

a) Horizontal stress load

b) Potential energy

c) Vertical stress load

d) Kinematic energy

Answer: c

Explanation: Temperature is a variant which varies from one point to another point. It has adverse effects on different structures. By temperature effect Vertical stress load vary linearly.

2. α means ____

a) Co-efficient of thermal expansion

b) Co-efficient of linear expansion

c) Thermal expansion

d) Thermal effect

Answer: a

Explanation: The co-efficient of thermal expansion describes how the size of an object changes with a change in temperature. Specifically, it measures the fractional change in size per degree change in temperature at constant pressure. It is denoted by symbol α.

3. In temperature effect, initial strain, ε 0 = ____

a) α ΔT

b) α+ΔT

c) α-ΔT

d) Load

Answer: a

Explanation: Strain is relative change in shape or size of an object due to externally applied forces. Temperature is a variant which varies from one point to another point. In temperature effect of FEM, Initial strain ε 0 =α ΔT.

4. In a structure, a crack is formed as a result of ______

a) Thermal expansion

b) Thermo couple

c) Thermal strain

d) Thermal stress

Answer: d

Explanation: Thermal stress is caused by differences in temperature or by differences in thermal expansion. A crack formed as a result of Thermal stress produced by rapid cooling from a high temperature.

5. In the given diagram, the line indicates ____________

The line indicating stress-strain relation in the diagram

a) Stress relation

b) Strain relation

c) Stress – strain relation

d) Undefined

Answer: c

Explanation: Stress is physical quantity that expresses the internal forces that neighboring particles of a continues material exert on each other. While, strain is the measure of deformation of the material. In the given diagram, the line indicates Stress-strain relation.

6. Stress – strain relation is given as _____

a) σ=E

b) σ=Eε

c) σ=E(ε-ε 0 )

d) Undefined

Answer: c

Explanation: Stress is a physical quantity that expresses the internal forces that neighboring particles of a continuous material exert on each other, while strain is the measure of the deformation of the material. Stress- strain relation is given as

σ=E(ε-ε 0 ).

7. Strain energy per unit volume is ___

a) u 0 =E(ε-ε 0 )

b) u 0 =\(\frac{1}{2}\)E(ε-ε 0 )

c) Non symmetric

d) Specified displacement

Answer: b

Explanation: Energy stored in a body due to deformation is called Strain energy. The strain energy per unit volume is called strain energy density and the area under the stress strain curve towards the point is deformation. Strain energy per unit volume is

u 0 =\(\frac{1}{2}\)E(ε-ε 0 )

8. Temperature change is denoted as_____

a) ΔT=(T 2 -T 1 )

b) θ e

c) l e

d) A e

Answer: b

Explanation: Temperature is the measure of average kinetic energy of the particles in a system. Adding heat to system causes temperature rise. In Finite element analysis temperature is denoted as θ e .

9. θ e =\

True

b) False

Answer: a

Explanation: Temperature is the measure of average kinetic energy of the particles in a system. Adding heat to system causes its temperature to rise. Temperature effect describes that how much of temperature is rised in body when load is applied. Temperature effect formula is as shown

θ e =\(\frac{E_eA_el_e\alpha\Delta T}{x_2-x_1}

\).

10. Stress in each element is ____

a) Eliminated

b) σ=EBq

c) σ=αΔT

d) σ=E

Answer: d

Explanation: Stress is a physical quantity that expresses the internal forces that neighboring particles of a continuous material exert on each other. After solving Finite element equations KQ=F for the displacements Q, the stress in each element can be obtained from equation.

σ=E.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Plane Trusses”.

1. Plane trusses are also known as _____

a) One–dimensional trusses

b) Two-dimensional trusses

c) Three-dimensional trusses

d) Poly dimensional trusses

Answer: b

Explanation: Truss elements are two- node members which allow arbitrary orientation in XYZ co-ordinate system. Truss transmits axial force only. Planar truss is one where all members and nodes lie within Two dimensional plane.

2. A truss structure consists only ___ force members.

a) Only one

b) Two

c) Three

d) Poly

Answer: b

Explanation: Truss elements are two node members which allow arbitrary orientation in XYZ co-ordinate system. Truss transmits axial force only, in general, three degree of freedom element. A truss structure consists only 2 truss members.

3. Plane truss element can be shown in _____

a) Local coordinate system

b) Global coordinate system

c) Local and global coordinate systems

d) Dimensional structure

Answer: c

Explanation: Truss is one where all members and nodes lie within two dimensional plane. Truss is that elements of a truss have various orientation. These different orientations can be shown in local and global coordinate system.

4. The truss element is a _____ when we see it in a local co-ordinate system.

a) Three dimensional

b) One dimensional

c) Two dimensional

d) Thermal component

Answer: b

Explanation: Truss is one where all members and nodes lie within two dimensional plane. Truss is that elements of a truss have various orientations. When we see a truss in local co-ordinate system, the element of a truss can be seen as one dimensional element.

5. Where l and m are direct cosines, then transformation matrix L is given by ___

a) \

Answer: d

Explanation: The direct cosines l and m are introduced as l = cos θ and m=cos &fcy;. These direction cosines are the cosines of the angles that the local x’-axis makes with the global x-, y- axes.

6. Formula for direct cosine l=cosθ= ______

a) x 2 -x 1

b) \

\(\frac{x_2-x_1}{l_e}\)

Answer: d

Explanation: A truss is a two node element which allows arbitrary orientation in XYZ co-ordinate system. Planar truss is one where all the members and nodes lie within two dimensional plane.

7. The direct cosine m is given by formula cos&fcy;= ____

a) \

\(\frac{x_2-y_2}{l_e}\)

Answer: b

Explanation: The direct cosines l and m are introduced as l=cosθ and m=cos&fcy;. These direction cosines are cosines of the angles of that local x’- axis makes with global x- and y- axes. Let (x 1 ,y 1 ) and (x 2 ,y 2 ) be the co-ordinates of nodes 1 and 2.

8. Strain energy in global co-ordinates can be written as ____

a) q ’ T k ’ q ’

b) q ’ T k ’

c) k ’ q ’

d) \(\frac{1}{2}\) q ’ T k ’ q ’

Answer: d

Explanation: Strain energy is defined as the energy stored in the body due to deformation. Strain energy per unit volume is known as strain energy density and the area under stress- strain curve towards the point of deformation.

9. The stress σ in a truss element is given by ____

a) σ=E

b) σ=ε

c) σ=E e ε

d) σ=α

Answer: c

Explanation: Stress is a physical quantity that expresses the internal forces that neighboring the particles of a continuous material exert on each other. Expression for element stresses can be obtained by noting that a truss element in local co-ordinates is simple two- force member.

10. In a truss element, element temperature load is ___

a) θ e =E e A e

b) θ e =E e A e ε 0

c) θ e =E e A e ε 0 \

θ e =E e A e ε 0 \(

Answer: d

Explanation: In a truss, temperature effect is arised then thermal stress problem is considered here. Since the element is simply a one dimensional element when viewed in local co-ordinate system, the element temperature load in the local co-ordinate system.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Three Dimensional Trusses”.

1. In a 3-D truss element nodal displacement vector in a local co-ordinate is q’= ___

a) [q 1 ‘ ]

b) [q 2 ‘ ]

c) [q 1 ‘ ,q 2 ‘ ,q 3 ‘ ] T

d) [q 1 ‘ ,q 2 ‘ ] T

Answer: d

Explanation: Local co-ordinates are measurement indices into a local co-ordinate system or local co-ordinate space. Nodes will have nodal displacements or degrees of freedom which may include translations, rotations, and for special applications, higher order derivatives of displacements.

2. Transformation between local and global co-ordinates is _____

a) q ‘ =Lq

b) q ‘ =L

c) q ‘ =L+q

d) q ‘ =\(\frac{L}{q}\)

Answer: a

Explanation: Transformations are frequently used in linear algebra and computer graphics, since transformations can easily be represented, combined and computed.

3. For a 3D truss element, transformation matrix L between local and global co-ordinates is given by ___

a) \

Answer: d

Explanation: A transformation matrix is a special matrix that can describe 2D and 3D transformations. Transformations are frequently used in linear algebra and computer graphics. Since transformations can be easily represented, combined and computed.

4. The direct cosines l, m, n are of local _____

a) z’- axes

b) x’- axes

c) y’- axes

d) None of the above

Answer: b

Explanation: Direction cosine refers to the cosine of the angle between any two vectors and three co-ordinate axes. Direction cosines are analogous extension of the usual notion of the slope to higher dimensions. In a truss element the direction cosines l, m, n are of local co-ordinate of x’- axes.

5. The direction cosines l, m, n with respect to ____ x-, y-, z- axes respectively.

a) Local

b) Local and global

c) Global

d) Transformation

Answer: c

Explanation: The direction cosines of a vector are the cosines of the angles between the vector and three co-ordinate axes. Direction cosines are analogous extension of the usual notion of the slope and higher dimensions. The direction cosines l, m, n are with respect to global x-,y-, z- axes .

6. In 3D truss, formula for direct cosine l = _____

a) x 2 -x 1

b) \

\(\frac{x_2-x_1}{l_e}\)

Answer: d

Explanation: A truss is a two node element which allows arbitrary orientation in XYZ co-ordinate system. The 3D truss element can be treated as straight forward generalization of the 2D truss element.

7. The direct cosine m is given by formula ____

a) \

\(\frac{x_2-y_2}{l_e}\)

Answer: b

8. The direction cosine n is given by formula _____

a) \

Undefined

Answer: c

Explanation: A truss is a two node element. The three D truss element can be treated as straight forward generalization of the 2D truss element.

9. In 3D truss element, the nodal displacement vector in global co-ordinates is _____

a) q ‘ = [q 1 ,q 2 ]

b) q ‘ = [q 1 ,q 2 ,q 3 ]

c) q ‘ = [q 1 ,q 2 ,q 3 ,q 4 ,q 5 ,q 6 ]

d) Undefined

Answer: c

Explanation: Nodes will have nodal displacements or degrees of freedom which may include translations, rotations, and for special applications, higher order derivatives of displacements.

10. In a truss element, element temperature load is ___

a) θ e =E e A e

b) θ e =E e A e ε 0

c) θ e =E e A e ε 0 \

θ e =E e A e ε 0 \(

Answer: d

This set of Finite Element Method Questions and Answers for Freshers focuses on “Assembly of Global Stiffness Matrix for the Banded & Skyline Solutions”.

1. What is a banded matrix?

a) Sparse matrix

b) Rectangular matrix

c) Unit matrix

d) Square matrix

Answer: a

Explanation: In matrix theory band matrix is a sparse matrix, whose non-zero entities are confined to a diagonal band. Comprising the main diagonal and zero are more diagonals on either side.

2. Skyline matrix storage is in the form of ______

a) Banded matrix

b) Sparse matrix

c) Singular matrix

d) Identity matrix

Answer: b

Explanation: In scientific computing, skyline matrix storage or SKS or a variable banded matrix storage or envelope storage scheme is form of a sparse matrix storage format matrix that reduces the storage requirement of matrix more than banded storage.

3. Symmetry and sparsity of the global stiffness matrix can be approached by _____ methods.

a) One

b) Three

c) Two

d) Four

Answer: c

Explanation: In assembly of global stiffness matrix, the solution for finite element equations can take advantage of symmetry and sparsity of global stiffness matrix. There are two methods to identify them. They are banded approach and skyline approach methods.

4. Which of these was one of the methods for determining assembly of global stiffness matrix?

a) Galerkin approach

b) Skyline approach

c) Rayleigh method

d) Assembly method

Answer: b

Explanation: In assembly of global stiffness matrix, there are two methods to determine the global stiffness matrices. They are banded approach and skyline approach. In which the assembly procedure of the matrix was easy.

5. In banded matrix, elements are _____ placed in stiffness matrix.

a) Singular

b) Determinant values

c) Directly

d) Indirectly

Answer: c

Explanation: A band matrix is a sparse matrix whose non zero entities are confined to a diagonal band comprising the main diagonal and zero or more diagonals on either side. In the banded approach, the elements of each element stiffness matrix K e are directly placed in banded matrix S.

6. In Skyline matrix, the elements in a stiffness matrix can be placed in _______

a) Direct values

b) Determinant values

c) Load values

d) Vector form

Answer: d

Explanation: In skyline matrix storage, or SKS or, variable band matrix storage or, envelope storage scheme is a form of a sparse matrix that reduces the storage requirement of the matrix more than banded storage. In skyline approach, the elements of K e are placed in a vector form with certain identification pointers.

7. Formula for maximum span or half band width in banded approach is _____

a) m e =[|i-j|+1]

b) \

q ‘ =lq

d) m e =[2|i-j|+1]

Answer: d

Explanation: The matrix has a given band width what we mean is that the band width is at most the quantity, not that it is necessarily exactly equal to that quantity. In a banded approach, half band width is as given

m e =[2|i-j|+1]

8. The first step of skyline assembly matrix involves evaluation of ____

a) Column height

b) Row height

c) Matrix height

d) Undefined

Answer: a

Explanation: Skyline assembly matrix scheme of form of a sparse matrix that reduces the storage requirement of a matrix than banded approach. The first step of skyline assembly involves the evaluation of the skyline height or the column height for each diagonal location.

9. The second step in skyline approach is assembling the element stiffness values into _____

a) Row vector

b) Identity vector

c) Column vector

d) Determinant vector

Answer: c

Explanation: Skyline assembly matrix scheme of form of sparse matrix that reduces the storage requirement of a matrix than banded approach. The second step in skyline approach is assembling the element stiffness values into column vector.

10. The details of a skyline assembly matrix are implemented in a program called ____

a) Boolean program

b) Cholesky program

c) Truss program

d) Trussky program

Answer: d

Explanation: As this assembly was done to trusses by default all the steps applied in skyline approach were implemented in program TRUSSKY.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “ Two Dimensional Problems – Finite Element Modelling”.

1. In 2D elements. Discretization can be done. The points where triangular elements meet are called ____

a) Displacement

b) Nodes

c) Vector displacements

d) Co-ordinates

Answer: b

Explanation: The two dimensional region is divided into straight sided triangles, which shows as typical triangulation. The points where the corners of the triangles meet are called nodes.

2. Each triangle formed by three nodes and three sides is called a ______

a) Node

b) Force matrix

c) Displacement vector

d) Element

Answer: d

Explanation: An element is a basic building block of finite element analysis. An element is a mathematical relation that defines how the degrees of freedom of a node relate to next. In discretization of 2D element each triangle is called element.

3. The finite element method is used to solve the problem ______

a) Uniformly

b) Vigorously

c) Approximately

d) Identically

Answer: c

Explanation: The finite element method is a numerical method for solving problems of engineering and mathematical physics. Typical problems areas of interest include structure analysis, heat transfer, fluid flow, mass transport and electromagnetic potential etc..,. The method yields approximate values of the unknowns at discrete number of points.

4. In two dimensional modeling each node has ____ degrees of freedom.

a) One

b) Infinity

c) Finite

d) Two

Answer: d

Explanation: In two dimensional problem, each node is permitted to displace in the two directions x and y. Thus each node has two degrees of freedom.

5. For a triangular element,element displacement vector can be denoted as ___

a) q=[q 1 ,q 2 ,q 3 ] T

b) q=[q 1 ,q 2 ] T

c) q=[q 1 ,q 2 ,……q 6 ] T

d) Load vector

Answer: c

Explanation: The displacement components of a local node is represented in x and y directions, respectively. For that we denote element displacement vector as

q=[q 1 ,q 2 ,……q 6 ] T .

6. In two dimensional analysis, stresses and strains are related as ___

a) σ=Dε

b) σ=ε

c) Load values

d) ε=Dσ

Answer: a

Explanation: When a material is loaded with force, it produces stress. Which then cause material to deform. Strain is response of a system t an applied stress.

7. In two dimensional modeling, body force is denoted as ___

a) f=[f x ,f y ] T

b) σ=Dε

c) q ‘ =lq

d) f=[2|i-j|+1]

Answer: a

Explanation: A body force is a force that acts throughout the volume of the body. Body forces contrast with contact forces or the classical definition of surface forces which are exerted to the surface of the object. Body force is denoted as

f=[f x ,f y ] T .

8. The information of array of size and number of elements and nodes per element can be seen in ___

a) Column height

b) Element connectivity table

c) Matrix form

d) Undefined

Answer: b

Explanation: An element connectivity table specifies global node number corresponding to the local node element. Element connectivity is the nodal information for the individual element with details how to fit together to form the complete original system.

9. In two dimensional modeling, traction force is denoted as ____

a) Row vector

b) T=[T x ,T y ] T

c) f=[f x ,f y ] T

d) σ=Dε

Answer: b

Explanation: Traction or tractive force is the force used to generate motion between body and a tangential surface, through the use of dry friction, through the use of hear force. Tractive force is defined as

T=[T x ,T y ] T

10. In two dimensional modeling, elemental volume is given by ____

a) dV=tdA

b) dV=dA

c) f=[f x ,f y ] T

d) Trussky program

Answer: a

Explanation: In mathematics, a volume element provides a means for integrating a function with respect to volume in various co-ordinate systems such as spherical co-ordinates and cylindrical co-ordinates. Then elemental volume is given by

dV=tdA.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Two Dimensional Problems – Constant Strain Triangle”.

1. Finite element method uses the concept of _____

a) Nodes and elements

b) Nodal displacement

c) Shape functions

d) Assembling

Answer: c

Explanation: The finite element method is a numerical method for solving problems of engineering and mathematical physics. Finite element method uses the concept of shape functions in systematically developing the interpolations.

2. For constant strain elements the shape functions are ____

a) Spherical

b) Quadratical

c) Polynomial

d) Linear

Answer: d

Explanation: The constant strain triangle element is a type of element used in finite element analysis which is used to provide an approximate solution in a 2D domain to the exact solution of a given differential equation. For CST shape functions are linear over the elements.

3. Linear combination of these shape functions represents a ______

a) Square surface

b) Linear surface

c) Plane surface

d) Combinational surface

Answer: c

Explanation: A constant strain element is used to provide an approximate solution to the 2D domain to the exact solution of the given differential equation. The shape function is a function which interpolates the solution between the discrete values obtained at the mesh nodes.

4. In particular, N 1 +N 2 +N 3 represent a plane at a height of one at nodes ______

a) One

b) Two

c) Three

d) One, two and three

Answer: d

Explanation: Any linear combination of these shape functions also represents a plane surface. In particular, N 1 +N 2 +N 3 represents a plane height of one at nodes one, two, and, three and thus it is parallel to the triangle 123.

5. If N 3 is dependent shape function, It is represented as ____

a) N 3 =ξ

b) N 3 =1-ξ

c) N 3 =1-η

d) N 3 =1-ξ-η

Answer: d

Explanation: The shape function is a function which interpolates the solution between the discrete values obtained at the mesh nodes. N 1 , N 2 , N 3 are not linearly independent only one of two of these are independent.

6. In two dimensional problems x-, y- co-ordinates are mapped onto ____

a) x-, y- co-ordinates

b) x-, ξ – co-ordinates

c) η-, y- co-ordinates

d) ξ-η-Co-ordinates

Answer: d

Explanation: The similarity with one dimensional element should be noted ; in one dimensional problem the x- co-ordinates were mapped onto ξ- co-ordinates and the shape functions were defined as functions of ξ. In the two dimensional elements the x-, y-, co-ordinates are mapped onto ξ-,,η – co-ordinates.

7. The shape functions are physically represented by _____

a) Triangular co-ordinates

b) ξ-,η-Co-ordinates

c) Area co-ordinates

d) Surface co-ordinates

Answer: c

Explanation: The shape function is a function which interpolates the solution between discrete values obtained at the mesh nodes. Therefore appropriate functions have to be used and as already mentioned; low order typical polynomials are used in shape functions. The shape functions are physically represented by area co-ordinates.

8. A 1 is the first area and N 1 is its shape function then shape function N 1 = ___

a) A 1 /A

b) A-A 1

c) A 1 +A

d) A 1

Answer: a

Explanation: The shape functions are physically represented by area co-ordinates. A point in a triangle divides into three areas. The shape functions are precisely represented as

N 1 =A 1 /A .

9. The equation u=Nq is a _____ representation.

a) Nodal

b) Isoparametric

c) Uniparametric

d) Co-ordinate

Answer: b

Explanation: The isoparametric representation of finite elements is defined as element geometry and displacements are represented by same set of shape functions.

10. For plane stress or plane strain, the element stiffness matrix can be obtained by taking _____

a) Shape functions, N

b) Material property matrix, D

c) Iso parametric representation, u

d) Degrees of freedom, DoF

Answer: b

Explanation: The material property matrix is represented as ratio of stress to strain that is σ=Dε . Therefore by this relation element stiffness matrix can be obtained by material property matrix.

11. In a constant strain triangle, element body force is given as ____

a) f e =[f x ,f y ,f x ,f y ,f x ,f y ] T

b) f e =\

f e =\(\frac{t_eA_e}{3}\)[f x ,f y ,f x ,f y ,f x ,f y ] T

d) f e =\(\frac{t_eA_e}{3}\)[f x ,f y ] T

Answer: c

Explanation: A body force is a force which acts through the volume of the body. Body forces contrast with the contact forces or the classical definition of the surface forces which are exerted to the surface of the body.

12. Traction force term represented as ___

a) ∫u T Tl

b) ∫u T T

c) ∫u T

d) u T Tl

Answer: a

Explanation: Traction force or tractive force are used to generate a motion between a body and a tangential surface, through the use of dry friction, through the use of shear force of the surface is also commonly used.

13. In the equation KQ=F, K is called as ____

a) Stiffness matrix

b) Modified stiffness matrix

c) Singular stiffness matrix

d) Uniform stiffness matrix

Answer: b

Explanation: The stiffness matrix represents system of linear equations that must be solved in order to ascertain an approximate solution to differential equation. The stiffness and force modifications are made to account for the boundary conditions.

14. Principal stresses and their directions are calculated by using ____

a) Galerkin approach

b) Rayleigh method

c) Potential energy method

d) Mohr’s circle method

Answer: d

Explanation: Mohr’s circle is two dimensional graphical representation of the transformation law. While considering longitudinal stresses and vertical stresses in a horizontal beam during bending.

15. I the distribution of the change in temperature ΔT, the strain due to this change is ____

a) Constant strain

b) Stress

c) Initial strain

d) Uniform strain

Answer: c

Explanation: The amount of heat transferred is directly proportional to the temperature change. The distribution of change in temperature, the strain due to this change is initial strain.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Two Dimensional Problem Modelling and Boundary Conditions”.

1. Finite element method is used for computing _____ and _____

a) Stress and strain

b) Nodes and displacement

c) Nodes and elements

d) Displacement and strain

Answer: d

Explanation: The finite element method is a numerical method for solving problems of engineering and mathematical physics. To solve the problem it subdivides a larger problem into smaller, simpler parts that are called finite elements.

2. In deformation of the body, the symmetry of ______ and symmetry of ____ can be used effectively.

a) Stress and strain

b) Nodes and displacement

c) Geometry and strain

d) Geometry and loading

Answer: d

Explanation: Deformation changes in an object’s shape or form due to the application of a force or forces. Deformation proportional to the stress applied within the elastic limits of the material.

3. For a circular pipe under internal or external pressure, by symmetry all points move _____

a) Radially

b) Linearly

c) Circularly

d) Along the pipe

Answer: a

Explanation: The boundary conditions require that points along x and n are constrained normal to the two lines respectively. If a circular pipe under internal or external pressure, by symmetry all the points move radially.

4. Boundary conditions can be easily considered by using _______

a) Rayleigh method

b) Penalty approach method

c) Galerkin approach

d) Potential energy approach

Answer: b

Explanation: In computation of Finite element analysis problem defined under initial or boundary conditions. For implementation of boundary conditions we need a staggered grid.

5. When dividing an area into triangles, avoid large _____

a) Dimensions

b) Loading

c) Aspect ratios

d) Boundary conditions

Answer: c

Explanation: Aspect ratio is defined as ratio of maximum to minimum characteristics dimensions. For this reason we can avoid large aspect ratios when dividing an area into triangles.

6. In dividing the elements a good practice may be to choose corner angles in the range of ____

a) 30-120°

b) 90-180°

c) 25-75°

d) 45-180°

Answer: a

Explanation: The best elements are those that approach an equilateral triangular configuration. Such configurations are usually not possible. A good practice is to choose corner angle in the range of 30-120°.

7. Stresses can be change widely at ____

a) Large circular sections

b) Notches and fillets

c) Corners

d) Crystals

Answer: b

Explanation: In a structure geometrical notches, such as holes cannot be avoided. The notches are causing in a homogeneous stress distribution, as notches fillets are also a cause for in homogenous stress distribution.

8. The Constant strain triangle can give____ stresses on elements.

a) Linear

b) Constant

c) Uniform

d) Parallel

Answer: b

Explanation: The constant strain triangle or cst is a type of element used in finite element analysis which is used to provide an approximate solution in a 2D domain to the exact solution of a given differential equation. By this we get constant stresses on elements.

9. The _____ can be obtained even with coarser meshes by plotting and extrapolating.

a) Minimum stresses

b) Minimum strain

c) Maximum stresses

d) Maximum strain

Answer: c

Explanation: Coarse mesh is more accurate in getting values. The smaller elements will better represent the distribution. Better estimates of maximum stress may be obtained even with coarser meshes. Coarse meshes are recommended for initial trails.

10. Coarser meshes are recommended for _____

a) Loading

b) Notches and fillets

c) Crystals

d) Initial trails

Answer: d

Explanation: The smaller elements will better represent the distribution. Coarse mesh is more accurate in getting values. Better estimates of maximum stress may obtained even with the coarse meshes.

11. Increasing the number of nodes in coarse mesh regions where stress variations are high, should give better results. This method is called _____

a) Divergence

b) Convergence

c) Convergent- divergent

d) Un defined

Answer: b

Explanation: At the initial trails, errors may be fixed, before running large number of elements. The convergence is successively increasing the number of elements in finite element meshes.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Two Dimensional Problems – Orthotropic Materials”.

1. Give an example of orthotropic material?

a) Topaz

b) Aluminum

c) Barium

d) Sodium

Answer: a

Explanation: Orthotropic materials have material properties that differ along three mutually orthogonal two fold axis of rotational symmetry. They are a subset of anisotropic materials, because their properties change when measured from different directions.

2. Unidirectional fiber- reinforced composites also exhibit _______ behavior.

a) Isotropic

b) Orthotropic

c) Material

d) Unidirection composite

Answer: b

3. Orthotropic planes have ____ mutually perpendicular planes of elastic symmetry.

a) One

b) Two

c) Three

d) Four

Answer: c

Explanation: Orthotropic materials have material properties that differ along three mutually orthogonal two fold axis of rotational symmetry. Wood may also consider to be orthotropic. These materials have three mutually perpendicular planes.

4. The principal material axes that are normal to the _______

a) Co-ordinates

b) Number of nodes

c) Principal axes

d) Plane of symmetry

Answer: d

Explanation: Orthotropic materials are a subset of anisotropic; their properties depend upon the direction in which they are measured. Orthotropic materials have three planes of symmetry. That is normal to principal material axes.

5. v 12 indicates that the poisson’s ratio that characterizes the decrease in ______ during tension applied in ______

a) 2- direction and 1- direction

b) 2- direction and 3- direction

c) 1- direction and 2- direction

d) 2- direction and 4- direction

Answer: a

Explanation: Poisson’s ratio is the ratio of transverse contraction strain to longitudinal extension strain in the direction of stretching force. Tensile deformation is considered positive and compressive deformation is considered negative.

6. Unidirectional composites are stacked at different fiber orientations to form a ______

a) Laminate

b) Orthotropic material

c) Isotropic material

d) Anisotropic material

Answer: a

Explanation: A unidirectional fabric is one in which the majority of fibers run in one direction only. A laminate is a tough material that is made by sticking together two or more layers of a particular substance.

7. When an orthotropic plate is loaded parallel to its material axes, it results only _____

a) Shear strains

b) Normal strains

c) Parallel strains

d) Uniform strains

Answer: b

Explanation: Orthotropic materials are a subset of anisotropic; their properties depend upon the direction in which they are measured. When an orthotropic plate is loaded parallel to its material axes, it results normal strains.

8. When the stresses are determined in an orthotropic material, then they are used to determine ____

a) Strains

b) Deformation

c) Factor of safety

d) Loads

Answer: c

Explanation: Factors of safety , is also known as safety factor , is a term describing the load carrying capacity of a system beyond the expected or actual loads. Essentially, the factor of safety is how much stronger the system is than it needs to be for an intended load. After determining the stresses in orthotropic materials by using an appropriate failure theory we can find factor of safety.

9. Stress- strain law defined as ______

a) σ=D(ε-ε 0 )

b) σ=D

c) σ=Dε

d) σ=Dε 0

Answer: a

Explanation: The relationship between the stress and strain that a particular material displays is known as that particular material’s stress–strain curve. It is unique for each material and is found by recording the amount of deformation at distinct intervals of tensile or compressive loading .

10. E 1 value of Balsa wood is ___

a) 0.125*10 6 psi

b) 12.04*10 6 psi

c) 23.06*10 6 psi

d) 7.50*10 6 psi

Answer: a

Explanation: A material’s property is an intensive, often quantitative, property of some material. Quantitative properties may be used as a metric by which the benefits of one material versus another can be assessed, thereby aiding in materials selection.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Axis Symmetric Formulation”.

1. Consider an Axisymmetric problem of which is having a radius of r, rotational angle θ and length l. Then r dl dθ is known as ____

a) Elemental volume

b) Element

c) Elemental surface area

d) Elemental surface

Answer: c

Explanation: Axial symmetry is symmetry around an axis; an object is axially symmetric if its appearance is unchanged if rotated around an axis. Surface element may refer to an infinitesimal portion of a 2D surface, as used in a surface integral in a 3D space.

2. The stress vector is correspondingly defined as ___________

a) σ=[σ y ,σ z ,τ yz ,σ θ ] T

b) σ=[σ y ,σ z ] T

c) u=[u, w] T

d) T=[T y ,T z ] T

Answer: a

Explanation: The stress is expressed by the Cauchy traction vector T defined as the traction force F between adjacent parts of the material across an imaginary separating surface S, divided by the area of S.

3. Problems involving three- dimensional axisymmetric solids or solids of revolution, subjected to _____ loading.

a) Rotational

b) Two-dimensional

c) Three-dimensional

d) Axisymmetric loading

Answer: d

4. All deformations and stresses are independent of _______

a) Co-ordinates

b) Number of nodes

c) Rotational angle, θ

d) Area

Answer: c

Explanation: The point around which you rotate is called the center of rotation, and the smallest angle you need to turn is called the angle of rotation. The angle of rotation is a measurement of the amount, the angle, by which a figure is rotated counterclockwise about a fixed point, often the center of a circle.

5. Revolving bodies like fly wheels can be analyzed by introducing _______ in body force term.

a) Gravitational force

b) Revolving force

c) Centrifugal force

d) Centripetal force

Answer: c

Explanation: The centrifugal force is an inertial force directed away from the axis of rotation that appears to act on all objects when viewed in a rotating frame of reference.

6. The stress- strain relation is given as ___________

a) σ=D

b) σ=Dε

c) σ=ε

d) ε=Dσ

Answer: b

Explanation: The Hook’s law, states that within the elastic limits the stress is proportional to the strain since for most materials it is impossible to describe the entire stress – strain curve with simple mathematical expression, in any given problem the behavior of the materials is represented by an idealized stress – strain curve, which emphasizes those aspects of the behaviors which are most important is that particular problem.

7. Axisymmetric problems are totally defined in ______

a) xy planes

b) yz planes

c) rz planes

d) rθ planes

Answer: c

8. For axisymmetry solids gravity forces can be considered if deformation and stresses act on _____

a) X direction

b) Z direction

c) Y direction

d) Parallel to plane

Answer: b

Explanation: Gravity, or gravitation, is a natural phenomenon by which all things with mass are brought toward one another, including objects ranging from atoms and photons, to planets and stars. Gravity forces can be considered if acting in z direction.

9. In axisymmetric solids, stress- strain law can be defined as ______

a) σ=D(ε-ε 0 )

b) σ=D

c) σ=Dε

d) σ=Dε 0

Answer: c

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Finite Element Modelling – Triangular Element”.

1. The transformation relationships into strain displacement relations. Then the equation can be written as ____

a) ε=Bq

b) ε=Dq

c) ε=q

d) Elemental surface

Answer: a

Explanation: For a triangular element, it can be modeled by using isoparametric formulation and then by chain rule, a jacobian matrix can be formed and then by transforming the matrix into simple form it is represented as

ε=Bq.

2. In the equation U e =\(\frac{1}{2}\)2q T (2∏ ∫ B T DBrdA)q the quantity inside the paranthesis is _____

a) Axisymmentric

b) Strain displacement relationships

c) Stiffness matrix

d) Symmetric matrix

Answer: c

Explanation: In the finite element method for the numerical solution of elliptic partial differential equations, the stiffness matrix represents the system of linear equations that must be solved in order to ascertain an approximate solution to the differential equation.

3. The volume of ring shaped element is _____

a) A e =\

A e =detJ

c) 2πr

d) 4πr 2

Answer: a

Explanation: A ring-shaped object, a region bounded by two concentric circles. … Informally, it has the shape of a hardware washer. The volume of the ring-shaped element is A e =\(\frac{1}{2}\mid det J \mid\).

4. The element body force vector f e is given by _____

a) Co-ordinates

b) f e =\(\frac{2Πr̅A_e}{3}\)[f̅ r ,f̅ z ,f̅ r ,f̅ z ,f̅ r ,f̅ z ] T

c) f e =\(\frac{2Πr̅A_e}{3}\)[f̅ x ,f̅ y ] T

d) f e =\(\frac{2Πr̅A_e}{3}\)

Answer: b

Explanation: A body force is a force that acts throughout the volume of a body. Forces due to gravity, electric fields and magnetic fields are examples of body forces. Body forces contrast with contact forces or the classical definition of surface forces which are exerted to the surface of an object.

5. A rotating flywheel with its axis in the z direction. We consider the flywheel to be stationary and apply the equivalent radial centrifugal force per unit volume is _____

a) 2Πr

b) 4Πr 2

c) ρrω 2

d) ρω 2

Answer: c

Explanation: The centrifugal force is an inertial force directed away from the axis of rotation that appears to act on all objects when viewed in a rotating frame of reference.

6. Surface traction of a uniformly distributed load with components T 1 and T 2 is _____

a) q T T e =2Π∫ e u T Trdl

b) q T T e =2Π

c) σ=ε

d) ε=Dσ

Answer: a

Explanation: Traction, or tractive force, is the force used to generate motion between a body and a tangential surface, through the use of dry friction, though the use of shear force of the surface. Traction can also refer to the maximum tractive force between a body and a surface, as limited by available friction.

7. On summing up the strain energy and force terms over all the elements and modifying for the boundary conditions while minimizing the total potential energy. We get ______

a) σ=D

b) Kinematic energy

c) ε=Dσ

d) KQ=F

Answer: c

Explanation: KQ=F by this we can obtain unknown displacement vectors. A displacement is a vector whose length is the shortest distance from the initial to the final position of a point P. It quantifies both the distance and direction of an imaginary motion along a straight line from the initial position to the final position of the point.

8. Using the connectivity of the elements, the internal virtual work can be expressed in the form _____

a) Ψ T =KQ

b) Ψ T =K

c) KQ=F

d) σ=Dε

Answer: a

Explanation: The principle of virtual work states that in equilibrium the virtual work of the forces applied to a system is zero. Newton’s laws state that at equilibrium the applied forces are equal and opposite of the reaction, or constraint forces.

9. In axisymmetric problems, by using stress strain relation and strain displacement relation we can obtain an equation that is ____

a) σ=DB̅q

b) σ=D

c) σ=Dε

d) σ=Dε 0

Answer: c

Explanation: The relationship between the stress and strain that a particular material displays is known as that particular material’s stress–strain curve. The strains give information about the deformation of material particles but, since they do not encompass translations and rotations, they do not give information about the precise location in space of particles.

10. The temperature effect on axisymmetric formulation. The vector ε̅ 0 is the initial strain evaluated at the centroid, representing the average temperature rise of the element is _____

a) θ e =2Πr̅

b) θ e =2Πr̅A e B̅ T Dε̅ 0

c) K=QF

d) σ=Dε 0

Answer: b

Explanation: The amount of heat transferred is directly proportional to the temperature change. Temperature is a proportional measure of the average kinetic energy of the random motions of the constituent microscopic particles in a system ; but more rigorous definitions include all quantum states of matter.

11. Uniform increase in temperature of, ΔT introduces initial _____

a) Normal strain

b) Strain

c) Stresses

d) Kinetic energy

Answer: a

Explanation: Strain, is a term used to measure the deformation or extension of a body that is subjected to a force or set of forces. The strain of a body is generally defined as the change in length divided by the initial length. The elongation of the bar is assumed normal, or perpendicular, to the cross section. Therefore, like stress, the strain is called a normal strain.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Axis Symmetric Problem Modelling and Boundary Conditions”.

1. Axisymmetry implies that points lying on the z- axis remains _____ fixed.

a) Tangentially

b) Spherically

c) Radially

d) Circularly

Answer: c

2. Modeling of a cylinder of infinite length subjected to external pressure. The length dimensions are assumed to be _____

a) Finite

b) Non uniform

c) Perpendicular

d) Constant

Answer: d

Explanation: The traditional view is still used in elementary treatments of geometry, but the advanced mathematical viewpoint has shifted to the infinite curvilinear surface and this is how a cylinder is now defined in various modern branches of geometry and topology.

3. Press fit of a ring of length L and internal radius r j onto a rigid shaft of radius r 1 +δ is considered. When symmetry is assumed about the mid plane, this plane is restrained in the _____

a) X direction

b) Y direction

c) Z direction

d) Undefined

Answer: c

Explanation: A drive shaft, driveshaft, driving shaft, propeller shaft , or Cardan shaft is a mechanical component for transmitting torque and rotation, usually used to connect other components of a drive train that cannot be connected directly because of distance or the need to allow for relative movement between them.

4. The condition that nodes at the internal radius have to displace radially by δ , a large stiffness C is added to the _____

a) Co-ordinates

b) Length

c) Diagonal locations

d) Radius

Answer: c

Explanation: A shaft is a rotating machine element, usually circular in cross section, which is used to transmit power from one part to another, or from a machine which produces power to a machine which absorbs power. The various members such as pulleys and gears are mounted on it.

5. Press fit on elastic shaft, may define pairs of nodes on the contacting boundary, each pair consisting of one node on the _____ and one on the ______

a) Shaft and couple

b) Sleeve and shaft

c) Shaft and sleeve

d) Sleeve and couple

Answer: b

Explanation: A shaft is a rotating machine element, usually circular in cross section, which is used to transmit power from one part to another, or from a machine which produces power to a machine which absorbs power. A flexible shaft or an elastic shaft is a device for transmitting rotary motion between two objects which are not fixed relative to one another.

6. For a Belleville spring the load is applied on _____

a) Shaft

b) Hole

c) Periphery of the circle

d) Coupling

Answer: c

Explanation: The Belleville spring, also called the Belleville washer, is a conical disk spring. The load is applied on the periphery of the circle and supported at the bottom.

7. On Belleville spring the load is applied in ______

a) X direction

b) Z direction

c) Y direction

d) Axial direction

Answer: d

Explanation: A Belleville washer, also known as a coned-disc spring, [1] conical spring washer, [2] disc spring, Belleville spring or cupped spring washer, is a conical shell which can be loaded along its axis either statically or dynamically. A Belleville washer is a type of spring shaped like a washer. It is the frusto-conical shape that gives the washer a spring characteristic.

8. In the Belleville spring, the load-deflection curve is _____

a) Linear

b) Curved

c) Non linear

d) Parabolic

Answer: c

9. A steel sleeve inserted into a rigid insulated wall. The sleeve fits snugly, and then the temperature is raised by _____

a) Uniform

b) Non uniform

c) σ

d) ΔT

Answer: d

Explanation: A sleeve is a tube of material that is put into a cylindrical bore, for example to reduce the diameter of the bore or to line it with a different material. Sometimes there is a metal sleeve in the bore to give it more strength. The pistons run directly in the bores without using cast iron sleeves.

10. In a Belleville spring, load-deflection characteristics and stress distribution can be obtained by dividing the area into ____

a) Surfaces

b) Nodes

c) Elements

d) Loads

Answer: c

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Two Dimensional Isoparametric Elements – Four Node Quadrilateral”.

1. In two dimensional isoparametric elements, we can generate element stiffness matrix by using ____

a) Numerical integration

b) Differential equations

c) Partial derivatives

d) Undefined

Answer: a

Explanation: The term isoparametric is derived from the use of the same shape functions [N] to define the element’s geometric shape as are used to define the displacements within the element.

2. The vector q=[q 1 ,q 2 ………q 8 ] T of a four noded quadrilateral denotes ____

a) Load vector

b) Transition matrix

c) Element displacement vector

d) Constant matrix

Answer: c

Explanation: A displacement is a vector whose length is the shortest distance from the initial to the final position of a point P. It quantifies both the distance and direction of an imaginary motion along a straight line from the initial position to the final position of the point.

3. For a four noded quadrilateral, we define shape functions on _____

a) X direction

b) Y direction

c) Load vector

d) Master element

Answer: d

Explanation: Master Element is the main point of reference in our analysis. The ME represents the person itself, and it gives us a primary layer of our personality. To determine the quality of ME, and overall chart, we have to analyze what kind of connection and access ME has to other Elements. The shape function is the function which interpolates the solution between the discrete values obtained at the mesh nodes.

4. The master element is defined in ______

a) Co-ordinates

b) Natural co-ordinates

c) Universal co-ordinates

d) Radius

Answer: b

5. Shape function can be written as _____

a) N t =

b) N t =

c) N t =

d) N t =\

Answer: d

Explanation: The shape function is the function which interpolates the solution between the discrete values obtained at the mesh nodes. Therefore, appropriate functions have to be used and, as already mentioned, low order polynomials are typically chosen as shape functions.

6. For a four noded element while implementing a computer program, the compact representation of shape function is ____

a) N t =\

b) N t =

c) N t =\(\frac{1}{4}\)(1+ξξ i )(1+ηη i )

d) Undefined

Answer: c

Explanation: FourNodeQuad is a four-node plane-strain element using bilinear isoparametric formulation. This element is implemented for simulating dynamic response of solid-fluid fully coupled material, based on Biot’s theory of porous medium. Each element node has 3 degrees-of-freedom : DOF 1 and 2 for solid displacement and DOF 3 for fluid pressure .

7. For a four noded quadrilateral elements, In u T =[u.v] T the displacement elements can be represented as u=N 1 q 1 +N 2 q 3 + N 3 q 5 + N 4 q 7

v= N 1 q 2 +N 2 q 4 + N 3 q 6 + N 4 q 8

then the shape function can be represented as _____

a) \

\(N=

Answer: d

Explanation: Displacement function in FEM. When the nodes displace, they will drag the elements along in a certain manner dictated by the element formulation. In other words, displacements of any points in the element will be interpolated from the nodal displacements , and this is the main reason for the approximate nature of the solution.

8. The stiffness matrix from the quadrilateral element can be derived from _____

a) Uniform energy

b) Strain energy

c) Stress

d) Displacement

Answer: b

Explanation: In the finite element method for the numerical solution of elliptic partial differential equations, the stiffness matrix represents the system of linear equations that must be solved in order to as certain an approximate solution to the differential equation.

9. For four noded quadrilateral element, the global load vector can be determined by considering the body force term in _____

a) Kinetic energy

b) Potential energy

c) Kinematic energy

d) Temperature

Answer: b

Explanation: A body force that is distributed force per unit volume, a vector, many people probably call up Vector’s definition . He says: It’s a mathematical term. A quantity represented by an arrow with both direction and magnitude. … Vector: a quantity with more than one element .

10. Shape functions are linear functions along the _____

a) Surfaces

b) Edges

c) Elements

d) Planes

Answer: b

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Numerical Integration”.

1. Which method of approach is useful for evaluating four noded quadratic elements?

a) Numerical integration

b) Penality approach method

c) Gaussian quadrature approach

d) Rayleighs method

Answer: c

Explanation: Gaussian quadrature is to select the n Gauss points and n weights such that provides an exact answer for polynomials f of as large ∼ degree as possible. In other words, the Idea is that if the n-point integration formula is exact for all polynomials up to as high a degree as possible, then the formula will work well even if f is not a polynomial.

2. One point formula in quadratic approach is ____

a) w 1 f(ξ 1 )

b) σ=εD

c) N t =

d) Constant matrix

Answer: a

Explanation: In elementary algebra, the quadratic formula is the solution of the quadratic equation. There are other ways to solve the quadratic equation instead of using the quadratic formula, such as factoring, completing the square, or graphing. This is seen to be the familiar midpoint rule.

3. Two point formula of a quadratic approach is _____

a) X direction

b) w 1 f(ξ 1 )+w 2 f(ξ 2 )

c) N t =

d) σ=D

Answer: b

4. The extension of Gaussian quadrature to two-dimensional integrals of the form of _____

a) I≈\(\sum_{i=1}^{n}\sum_{j=1}^{n}\)w i w j f(ξ i ,η j )

b) Natural co-ordinates

c) w 1 f(ξ 1 )+w 2 f(ξ 2 )

d) w 1 f(ξ 1 )

Answer: a

Explanation: In numerical analysis, a quadrature rule is an approximation of the definite integral of a function, usually stated as a weighted sum of function values at specified points within the domain of integration. An n-point Gaussian quadrature rule, is a quadrature rule constructed to yield an exact result for polynomials of degree 2n − 1 or less by a suitable choice of the points xi and weights wi for i=1,…, n. The domain of integration for such a rule is conventionally taken as [−1, 1].

5. Stiffness integration for quadratic element for 2*2 matrix is ____

a) N t =

b) k ij =\(\sum_{IP=1}^{4}\)W IP ∅ IP

c) N t =

d) N t =\

Answer: b

Explanation: Stiffness is the rigidity of an object, the extent to which it resists deformation in response to an applied force. The complementary concept is flexibility or pliability: the more flexible an object is, the less stiff it is. A stiff equation is a differential equation for which certain numerical methods for solving the equation are numerically unstable, unless the step size is taken to be extremely small.

6. The stresses in the quadratic element are not ______

a) Linear

b) Uniform

c) Constant

d) Undefined

Answer: c

7. The stresses are evaluated at the __________

a) Nodal points

b) Nodal displacements

c) Gauss points

d) Elements

Answer: c

Explanation: The stress applied to a material is the force per unit area applied to the material. The maximum stress a material can stand before it breaks is called the breaking stress or ultimate tensile stress. The forces acting on it are trying to stretch the material. In numerical analysis, a quadrature rule is an approximation of the definite integral of a function, usually stated as a weighted sum of function values at specified points within the domain of integration.

8. For quadrilateral with 2X2 integration gives _____ sets of stress values.

a) One

b) Two

c) Three

d) Four

Answer: d

9. For degenerate four noded quadrilateral element the errors are _____

a) Constant

b) Uniform

c) Higher

d) Lesser

Answer: c

Explanation: A degenerated element is an element whose characteristic face shape is quadrilateral, but is modeled with at least one triangular face. Degenerated elements are often used for modeling transition regions between fine and coarse meshes, or for modeling irregular and warped surfaces.

10. Gauss points are also the points used for numerical evaluation of _____

a) Surfaces

b) k e

c) Elements

d) Planes

Answer: b

Explanation: Stiffness is the rigidity of an object, the extent to which it resists deformation in response to an applied force. The complementary concept is flexibility or pliability: the more flexible an object is the less stiff it is. A stiff equation is a differential equation for which certain numerical methods for solving the equation are numerically unstable, unless the step size is taken to be extremely small.

This set of Finite Element Method Questions and Answers for Experienced people focuses on “Four Node Quadrilateral for Axis Symmetric Problems”.

1. In the four-node quadrilateral element, the shape functions contained terms _________

a) ξ

b) σ

c) ∅

d) Undefined

Answer: a

2. A _________ element by using nine-node shape function.

a) Load vector

b) Sub parametric

c) Element displacement vector

d) Constant matrix

Answer: b

Explanation: The Nine-Node Biquadratic Quadrilateral This element is often abbreviated to Quad9 in the FEM literature. This element has three types of shape functions, which are associated with corner nodes, midside nodes and center node, respectively.

3. Eight-Node Quadrilateral. This element belongs to the ________ family of elements.

a) Serendipity

b) Constant matrix

c) Load vector

d) Master element

Answer: a

Explanation: The Eight-Node “Serendipity” Quadrilateral. This element is often abbreviated to Quad8 in the FEM literature. It is an eight-node quadrilateral element that results when the center node 9 of the biquadratic quadrilateral is eliminated by kinematic constraints.

4. N 1 , is of the form ____

a) Co-ordinates

b) N 1 =c

c) N 1 =

d) N 1 =

Answer: b

5. Six node triangular elements is also known as _____

a) Triangle

b) Quadratic triangle

c) Interpolation

d) Shape function

Answer: b

Explanation: The six-node triangle is shown in Figs. 7.8a and b. By referring to the master element where ϛ=1-ξ-η. Because of terms ξ 2 ,η 2 etc. in the shape functions, this element is also called a quadratic triangle. The isoparametric representation is

u= Nq .

6. In six node triangular element, the gauss points of a triangular element can be defined by ____

a) Two point rule

b) Three point rule

c) One point rule

d) Undefined

Answer: c

7. The mid node should not be outside of the triangular element this condition should ensures that det J does not attain a value ____

a) Constant

b) Zero

c) Unity

d) Infinite

Answer: b

Explanation: The Mid-Node Admissible Spaces [1,2] for two-dimensional quadratic triangular finite elements are extended to three-dimensional quadratic tetrahedral finite elements . The MAS concept for 3DQTE postulates a bounded region within which a mid-side node of a curved edge of the 3DQTE can be placed to ensure maintaining a specified minimum and maximum Jacobian determinant value at any point of the element.

8. The nodal temperature load can be evaluated by using _____

a) Uniform energy

b) Strain energy

c) Numerical integration

d) Displacement

Answer: c

Explanation: A temperature can be applied to nodes, surfaces, or parts in a model. A surface temperature applies nodal temperatures to each node on the surface, and a part temperature applies nodal temperatures to each node in the part. A temperature is used for a thermal stress analysis.

9. The gauss points for a triangular region differ from the _____ region.

a) Rectangular

b) Triangular

c) Square

d) Temperature

Answer: c

Answer: b

This set of Finite Element Method Interview Questions and Answers for freshers focuses on “Conjugate Gradient Implementation of Quadrilateral Element”.

1. Which of the following elements are used for structural and fatigue analysis?

a) Triangular Elements

b) Quadrilateral Elements

c) Shell type Elements

d) Pentagonal Elements

Answer: b

Explanation: Quadrilateral Elements are used for structural and fatigue analysis. Shell type Elements are used for dynamic analysis. Triangular Elements are used for mold flow analysis. Pentagonal elements can be used in CFD analysis.

2. In conjugate gradient implementation of quadrilateral elements the stiffness of each element is stored in three dimensional arrays.

a) True

b) False

Answer: a

Explanation: The stiffness of each quadrilateral element is stored in three dimensional arrays. Thus stiffness of each element can be recalled without recalculating for iterations carried out.

3. The total number of dofs in quadrilateral element is ______

a) thirty six

b) eighteen

c) twenty four

d) forty two

Answer: c

Explanation: Number of dofs per node=6

Number of nodes in quadrilateral element=4

Total number of dofs=6*4=24

4. Which of the following statement is false?

a) Linear quad element is more accurate than linear tria element

b) Linear tria element is more accurate than linear quad element

c) Parabolic elements are better than linear elements

d) Parabolic quad element has eight codes

Answer: b

Explanation: As quad element has one more term in the displacement function than tria elements it is more accurate. The displacement function is u=a 0 +a 1 x+a 2 y+a 3 xy.

5. Which of the following statements about warp angle is not true?

a) It is a not applicable for tria elements

b) It is an angle formed between two normal of planes formed by splitting quad element along diagonal

c) Its value should be less than 10°

d) It is the minimum angle between the planes of diagonally split quad element

Answer: d

Explanation: Warp angle is a quality check parameter mostly used in quad elements. It is the maximum value between the normal of planes formed by diagonally splitting quad elements.

6. Acceptable included angle for quad elements lies between 45° and 135°.

a) True

b) False

Answer: a

Explanation: Included angle check is applied for individual angles and. Skew is based on overall shape and doesn’t take in account individual angles.

7. Which of the following is the correct expression for calculating skew of quadrilateral element?

a) Skew=90°+minimum angle formed by joining opposite mid sides of quad element

b) Skew=90°-minimum angle formed by joining opposite mid sides of quad element

c) Skew=90°*minimum angle formed by joining opposite mid sides of quad element

d) Skew=90°/minimum angle formed by joining opposite mid sides of quad element

Answer: b

Explanation: Skew is a quality check parameter for quadrilateral elements. The ideal value of skew should be 0°. The value of skew is acceptable till 45°.

8. Which of the following is the correct expression for calculating skew of quadrilateral element?

Find the value of stretch for Quadrilateral element

a) Stretch=L min *(2 0.5 /d max )

b) Stretch=L min +(2 0.5 /d max )

c) Stretch=L min -(2 0.5 /d max )

d) Stretch=L min *(2 0.5 +d max )

Answer: a

Explanation: The value of stretch is given by:

Stretch=L min *(2 0.5 /d max )

Here L min is length of minimum side of quad element, and d max is the length of maximum diagonal of quad element. Ideally its value should be 1, but values greater than 0.2 are acceptable.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Beams and Frames – Finite Element Formulation”.

1. During finite element formulation of beam each node has _______ degrees of freedom.

a) three

b) two

c) one

d) six

Answer: b

Explanation: During finite element formulation of beam, each node has two degrees of freedom. The two degrees of freedom on each node represent the transverse displacement and the slope or rotation of the node.

2. Beams are horizontal members used for supporting transverse loading.

a) True

b) False

Answer: a

Explanation: Beams are horizontal members supporting transverse loads acting on it. Some examples of beams include shafts, bridges, and members in buildings.

3. The total number of degrees of freedom in a beam with four nodes is ______

a) four

b) eight

c) sixteen

d) thirty two

Answer: b

Explanation: Number of degrees of freedom per node in a beam element=2

Number of nodes in beam element=4

Total number of degrees of freedom=2*4=8

4. The displacements in beam elements are interpolated using _______

a) shape elements

b) shape functions

c) shape parameters

d) shape factors

Answer: b

Explanation: Shape functions are used in beam elements to interpolate the displacements. The shape functions along with nodal displacements can be used to interpolate displacements in beam elements.

5. The shape functions for interpolation on beam elements are defined on the range of ________

a) 0 to +1

b) -1 to 0

c) 0 to +2

d) -1 to +1

Answer: d

Explanation: The shape function for interpolation is defined in the range of -1 to +1. The value varies between -1 to +1 where value on one node is -1 and the value on other node is +1.

6. In beam elements the cross section of the element is assumed.

a) True

b) False

Answer: a

Explanation: Beam elements are a type of one dimensional element. The cross section of the geometry is assumed where the beam element is used.

7. Which of the following statements is correct?

a) Stiffness coefficient value for numerical solution is less than the value of exact solution

b) Stiffness coefficient value for numerical solution is greater than the value of exact solution

c) Stiffness coefficient value for numerical solutions is equal to the value of exact solution

d) Stiffness coefficient value for numerical solutions is twice the value of exact solution

Answer: b

Explanation: The stiffness coefficient value for numerical solution is greater than the value of exact solutions. This yields the displacement value in numerical solution lower as compared to the value of exact solution.

8. The element stiffness matrix for beam element is given by which of the following expressions?

a)\

\(\frac{EI}{l^2}

Answer: a

Explanation: The value of stiffness matrix is given by

k=\(\frac{EI}{l^3}

Here l is length of beam element, and E is the modulus of elasticity, and I moment of inertia of the beam element. The stiffness matrix of the beam element affects the displacement of the nodes and their interpolation in the beam element. The beam element stiffness matrix is a symmetric matrix.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Beams and Frames – Load Vector”.

1. On a simply supported beam with uniformly distributed load over length the value of reaction force at one support is _______

a) p*l

b) /2

c) 2

d) /3

Answer: b

Explanation: Overall force acting on the beam=p*l

The reaction force at one of the two supports=/2.

2. In Galerkin method we convert a continuous operator problem into a discrete problem.

a) True

b) False

Answer: a

Explanation: In Galerkin method we convert a continuous operator problem into a discrete problem. In beams the problem is formulated into finite elements using Galerkin Method.

3. Which of the following is not a one dimensional element?

a) bar

b) brick

c) beam

d) rod

Answer: b

Explanation: Brick is not a one dimensional element. One dimensional element is used when one dimension of the model geometry is significantly greater than the other two dimensions.

4. Three geometrically identical beams made out of steel, aluminum, and titanium are axially loaded. Which of the following statements is correct?

a) Stress in titanium is the least

b) Stress in cast iron is the highest

c) Stress in steel is the least

d) Stress in all three beams is the same

Answer: d

Explanation: The stress in all three beams will be induced equally. Stress in not dependent on the material, but the geometry and cross section of the element.

5. The application of force at which of the following point on a beam will nullify the effect of torsion?

a) Centroid

b) Center of mass

c) Extreme fiber

d) Shear centre

Answer: d

Explanation: The application of force on shear centre nullifies the effect of torsion on a beam element. This is useful when the cross section of beam is asymmetric and creates a twisting effect on application of force.

6. For a taper beam element two cross sections are necessary to define the geometry.

a) True

b) False

Answer: a

Explanation: A taper beam element requires two cross sections to define the geometry. Regular beam element cannot take into account the variation in cross section required to define the geometry.

7. Which of the following statements is correct?

a) Beam elements are recommended for unsymmetrical cross sections

b) Bar elements are recommended for unsymmetrical cross sections

c) Beam and bar elements can both be used for unsymmetrical cross sections

d) Neither beam nor bar elements can be used for unsymmetrical cross sections

Answer: a

Explanation: Beam elements are recommended for unsymmetrical cross sections due to their ability to take into account the shear centre. Thus shear centre can nullify effects of torsion acting on unsymmetrical cross section which is limited in case of bar elements.

8. The equivalent load acting on a simply supported beam element loaded with uniformly distributed load over an element of length is given by which of the following expressions?

a) f=\

f=\

f=\([\frac{pl}{2},\frac{pl^2}{12},\frac{-pl}{2},-\frac{pl^2}{12}]\)

Answer: a

Explanation: The value of equivalent load is given by

f=\([\frac{pl}{2},\frac{pl^2}{12},\frac{pl}{2},-\frac{pl^2}{12}]\)

Here l is length of beam element, and p is the uniformly distributed load. Here \(\frac{pl}{2}\) represents the reaction force on the supports, and \(\frac{pl^2}{12}\) represents the moment acting on the supports.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “ Beams and Frames – Boundary Conditions”.

1. Symmetry in application of boundary conditions should be avoided in which of the following type of analysis?

a) Linear static analysis

b) Modal analysis

c) Thermal analysis

d) Nonlinear static analysis

Answer: b

Explanation: Symmetric boundary conditions should not be used for modal analysis. Symmetric model would miss some of the modes of modal analysis or out of phase modes.

2. Boundary conditions are applied to simulate the physical constraints on the finite element model.

a) True

b) False

Answer: a

Explanation: Boundary conditions simulate the physical constraints on the finite element model. Application of boundary conditions is a crucial preprocessing step to yield accurate solution.

3. Which of the following is the correct equation for stiffness of an element given value of force and displacement ?

a) FQ=K

b) KQ=F

c) KF=Q

d) KFQ=1

Answer: b

Explanation: The correct equation is given by KQ=F. The value of force is the product of stiffness and displacement . The value of stiffness of the element determines the displacement of the node.

4. Which of the following conditions must be fulfilled to apply symmetry in a finite element model?

a) Geometry of the model is symmetric

b) Boundary conditions to be applied are symmetric

c) Geometry model has large number of nodes

d) Geometry of the model is symmetric and boundary conditions to be applied are symmetric

Answer: d

Explanation: The geometry and boundary conditions both have to be symmetric to apply any kind of symmetry. Half or quarter portions of a model can be used to reduce computational cost.

5. Which of the following boundary conditions cannot be directly applied on solid elements?

a) Force

b) Pressure

c) Support

d) Torque

Answer: d

Explanation: Torque cannot be directly applied on solid element in finite element model. Since solid elements have three translational degrees of freedom and no rotational degrees of freedom torque cannot be directly applied on solid elements.

6. Traction is force acting on an area in any direction other than normal.

a) True

b) False

Answer: a

Explanation: Traction is force acting on an area in any direction other than normal. The force acting on an area in normal direction is called as pressure. Traction is boundary condition applied where force acting on a surface is not normal to the surface such as friction and drag.

7. Which of the following statements is correct?

a) Reaction force at supports is equal to the sum of the product of stiffness and displacement

b) Reaction force at supports is equal to the product of sum of stiffness and displacement

c) Reaction force at supports is equal to the sum of stiffness’s

d) Reaction force at supports is equal to the sum of displacements

Answer: a

Explanation: Reaction force at supports is equal to the sum of the product of stiffness and displacement. The stiffness and displacement matrices are multiplied for each element and then cumulated to find the reaction force at supports.

8. Which of the following equations give the relation between material properties like modulus of elasticity , modulus of rigidity , and Poisson’s ratio ?

a) E = 2*G*

b) E = 3*G*

c) E = 2*G*

d) E = 3*G*

Answer: a

Explanation: The relation between the material properties is given by

E = 2*G*

Here E is the ratio of normal stress to normal strain. G is the ratio of shear stress to shear strain and u is the ratio of lateral strain to longitudinal strain.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Shear Force & Bending Moment”.

1. Which of the following statements are correct about a cantilevered beam with point load acting on the extreme end of the beam?

a) Bending stresses induced in the beam are constant throughout the length of the beam

b) Bending stresses induced in the beam decreases linearly from fixed end to free end

c) Bending stresses induced in the beam increases linearly from fixed end to free end

d) Bending stresses induced in the beam decreases exponentially from fixed end to free end

Answer: b

Explanation: Bending stresses induced in the beam decreases linearly from fixed end to free end. The point load acting induces normal as well as shear stresses, but when length beam is large the shear stresses are negligible.

2. Shear stress acts parallel to the cross section of the beam.

a) True

b) False

Answer: a

Explanation: Shear stress acts parallel to the cross section causing distortion. Shear stress results in angular deformation which is measured in terms of angle.

3. Which of the following is the correct equation for bending moment of an element given value of young’s modulus , moment of inertia , and radius of curvature ?

a) M= EIR

b) M= EI/R

c) M= 2 R

d) M= 2 /R

Answer: b

Explanation: The correct equation is given by M= EI/R. The value of bending moment is directly proportional to young’s modulus and moment of inertia . The value of bending moment is inversely proportional to radius of curvature .

4. Which of the following statements are correct about a cantilevered beam with point load acting on the extreme end of the beam?

a) Shear stress along the length of the beam increases linearly

b) Shear stress along the length of the beam deceases linearly

c) Shear stress along the length of the beam decreases exponentially

d) Shear stress along the length of the beam remains constant

Answer: d

Explanation: Shear stress along the length of the beam remains constant. There is no change in the magnitude of the shear stress acting on the beam along its length.

5. Which of the following statements are correct regarding shear stress distribution across the cross section in a cantilevered beam with point load acting on the extreme end of the beam?

a) Shear stress distribution across the cross section of beam is constant

b) Shear stress distribution across the cross section of beam is zero

c) Shear stress distribution across the cross section of beam is zero at center and maximum at extreme ends

d) Shear stress distribution across the cross section of beam is zero at extreme ends and maximum at center

Answer: d

Explanation: The shear stress distribution across the cross section of beam is zero at extreme ends and maximum at center.The variation in the magnitude of shear stress from center to the extreme ends follows a circular curve.

6. Torque acting on the face of a cylindrical body induces bending moment in the body.

a) True

b) False

Answer: b

Explanation: Torque acting on the face of a cylindrical body does not induce bending moment in the body. The torque acting on the body induces torsional shear stress in the body.

7. Which of the following equations is the correct expression for shear force in an element, given the modulus of elasticity , moment of inertia , element length (l e ), displacement in a uniformly distributed load on a simply supported beam?

a) V=\(\frac{EI}{^3}\)(2q 1 +l e q 2 -2q 3 +l e q 4 )

b) V=\(\frac{EI}{^2}\)(2q 1 +l e q 2 -2q 3 +l e q 4 )

c) V=\(\frac{EI}{^2}\)(2q 1 +l e q 2 +2q 3 +l e q 4 )

d) V=\(\frac{EI}{^3}\)(2q 1 +l e q 2 +2q 3 +l e q 4 )

Answer: a

Explanation: The correct expression is given by

V=\(\frac{EI}{^3}\)(2q 1 +l e q 2 -2q 3 +l e q 4 )

Here E is the ratio of normal stress to normal strain. Here q1, q2, q3, q4 are the four displacements at the two supported nodes of the simply supported beam.

8. Which of the following equations is the correct expression for bending moment in an element, given the modulus of elasticity , moment of inertia , element length (l e ), shape function and displacement in a uniformly distributed load on a simply supported beam?

a) M=\(\frac{EI}{^2}\)[6ξq 1 +l e q 2 -6ξq 3 +l e q 4 ]

b) M=\(\frac{EI}{^3}\)[6ξq 1 -l e q 2 -6ξq 3 -l e q 4 ]

c) M=\(\frac{EI}{^2}\)[6ξq 1 +l e q 2 -6ξq 3 -l e q 4 ]

d) M=\(\frac{EI}{^3}\)[6ξq 1 +l e q 2 -6ξq 3 +l e q 4 ]

Answer: a

Explanation: The correct expression is given by

M=\(\frac{EI}{^2}\)[6ξq 1 +l e q 2 -6ξq 3 +l e q 4 ]

Here E is the ratio of normal stress to normal strain. Here q1, q2, q3, q4 are the four displacements at the two supported nodes of the simply supported beam. The value of shape function varies between -1 to +1.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Beams on Elastic Support”.

1. Which of the following is a class of applications formed by beams supported on soil?

a) Soil foundation

b) Winkler foundation

c) Elastic Beam foundation

d) Reinforced foundation

Answer: b

Explanation: Winkler foundation is a class of applications formed by beams supported on soil. There are many engineering applications where the beams are supported on elastic members.

2. Single row bearings can be considered by a node at each bearing location.

a) True

b) False

Answer: a

Explanation: Single row bearings can be considered by a node at each bearing location. Also for single row bearing, bearing stiffness can be added to the vertical degree of freedom.

3. Which of the following statements is not correct about beams on elastic supports?

a) Rotational stiffness has to be considered for journal bearings

b) Single row bearings can be considered by a node at each bearing location

c) In wide journal bearing stiffness per unit length has to be considered

d) Rotational stiffness need not be considered for roller bearings

Answer: d

Explanation: Rotational stiffness has to be considered for roller bearings. Also in wide journal bearings stiffness per unit length has to be considered.

4. Which of the following statements are correct regarding beam elements?

a) Beam elements can resist force and moment about x and y axis

b) Beam elements can resist only force about x and y axis

c) Beam elements can resist only force about x, y, and z axis

d) Beam elements can resist force and moment about x, y, and z axis

Answer: d

Explanation: Beam elements can resist force and moment about x, y, and z axis.

5. Rod elements cannot resist shear force applied on the element.

a) True

b) False

Answer: a

Explanation: Rod elements cannot resist shear force applied on the element. Rod element does not have the required vertical degree of freedom to resist shear force.

6. Which of the following expressions must be added to the element stiffness matrix for beams supported on elastic supports, where l is length of beam element, and s is the stiffness per unit length of element?

a) \

\(\frac{EI}{420}

Answer: a

Explanation: The value of stiffness matrix to be added for beams with elastic support is

\(\frac{sl}{420}

Here l is length of beam element, and s is the stiffness per unit length of element. The stiffness matrix of the beam element affects the displacement of the nodes and their interpolation in the beam element.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Boundary Value Problems – 1”.

1. In Finite Element Methods , in two-dimensional problems, we approximate solution on a domain but not the domain itself.

a) True

b) False

Answer: b

Explanation: In two-dimensional problems, the finite elements are simple geometric shapes. These shapes can be used to approximate a given domain as well as the solution over it. Because of double approximation the accuracy of solution depends on both the approximations and is errors in two approximations gets multiplied.

2. In Finite Element Methods , a boundary value problem is a set of differential equations with a solution, which also satisfies some additional constraints, known as ___

a) boundary conditions

b) nodal values

c) equilibrium equations

d) energy minimum

Answer: a

Explanation: A boundary value problem is a set of differential equation with a solution, which also satisfies some additional constraints, called the boundary conditions. Nodal Values are the values of a variable at each node obtained as a result of finite element analysis. Equilibrium equations are used in free body diagrams.

3. Which of the following expression is the correct solution for finite element approximation over the 3 noded element shown?

Find the expression u for finite element approximation over the 3 noded element in diagram

a) u=c1+x*c2+y*c3

b) u=c1+x*c2+y*c1

c) u=c1+x*c2+y*c3+x*y

d) u=x*c1+y*c2

Answer: a

Explanation: The polynomial u=c1+x*c2+y*c3 is the correct solution for finite element approximations over the 3 noded element. It satisfies the conditions in order for the approximate solution to be convergent to the true solution. Here c1, c2 and c3 are shape functions and the set of coefficients {1,x,y} is continuous, linearly independent and complete.

4. For A1, A2, and A3 as area coordinates and s1, s2 and s3 as shape functions for the element shown, which relation is correct?

Find the relation of A1, A2 & A3 area coordinates & s1, s2 & s3 shape functions of the element

a) s1=A1/

b) s1=/A1

c) s1=A1+A2+A3

d) s1=A1

Answer: a

Explanation: For A1, A2, and A3 as area coordinates and s1, s2 and s3 as shape functions for the element shown, the total area is equal to the sum of A1, A2 and A3 and s1 is calculated by the relation s1=A1/total area

=A1/.

5. In Finite Element Methods , In two-dimensional problems, we shall have two types of errors, one due to the approximation of the solution and the other due to approximation of the domain.

a) True

b) False

Answer: a

Explanation: In two-dimensional problems, the finite elements are simple geometric shapes that can be used to approximate a given domain as well as the solution over it. We not only approximate solution on a domain but also the domain itself by using a suitable mesh. As a result, we shall have two types of errors, one due to the approximation of the solution and the other due to approximation of the domain.

6. For A1=5, A2=10, A3=5, what is the value of the shape function at node 1 of the triangular element shown?

Find the value of the shape function at node 1 of the triangular element

a) 0.15

b) 0.25

c) 0.35

d) 0.45

Answer: b

Explanation: Total area is A.

A=A1+A2+A3

A=5+10+5

=20.

The shape function at node 1 is given by

=5/20

=0.25.

7. If x1, x2 and x3 are displacements of nodes 1, 2 and 3 respectively and e, n are shape functions then which expression correctly describes displacement of any point on the element along x direction?

Find the displacement of any point (x,y) on the element along x direction

a) x=e+n+x3

b) x=n*x1+e*x2+x3

c) x=e+n+x3

d) x=e+n+x3

Answer: a

Explanation: A triangular element has three nodes and hence three shape functions. Here, the three shape functions are shown as e,n and 1-e-n. Thus, displacement of any point on element along x direction is x=++x3* or x=e+n+x3.

8. u=u[x,y] and v=v[x,y] Using the chain rule of partial derivatives, we get Jacobian of the transformation, J. The relation between area of 2D three noded triangular element and Jacobian is given by _____

a) A=1*|detJ|

b) A=*|detJ|

c) A=0.5*|detJ|

d) A=2*|detJ|

Answer: c

Explanation: For a 2D three noded triangular element area A=*|*-*|. Using the chain rule of partial derivatives, we get Jacobian of the transformation, J=\

*-* and in terms of detJ

A=A=0.5*|detJ|

9. In a two-dimensional static structural problem, the various non-zero stresses are ____

a) σ=[σxσyσz]

b) σ=[σxσyσx+σy]

c) σ=[σxσyσx-σy]

d) σ=[σxσyτxy]

Answer: d

Explanation: In a two-dimensional static structural problem, the stresses are in a single plane.They are the two normal stresses σx, σy and one shear stress τxy.

10. If 1, 2 and 3 is the order of sequence of elemental nodes of e=1 then to make elemental connectivity, which of the following is the correct order for e=2?

Find order of elemental nodes of e=2 to make elemental connectivity

a) 2 3 1

b) 1 3 2

c) 3 2 1

d) 2 1 3

Answer: a

Explanation: Since, the order of nodes for e=1 is in counter clockwisesense, the same has to be followed for ordering the nodes of e=2. Thus, 2 3 1 is the correct option.

11. If 1, 2; 3, 4; 5, 6 and 7, 8 are the degrees of freedom at nodes 1,2,3 and 4 respectively then the common degrees of freedom for e=1 and e=2 are ____

Find the common degrees of freedom for e=1 and e=2

a) 3 4, 7 8

b) 1 2, 3 4

c) 5 6, 7 8

d) 1 2, 5 6

Answer: a

Explanation: The common degrees of freedom for e=1 and e=2 are the one corresponding to common nodes between e=1 and e=2. The common nodes are 2 and 3. Thus, the common degrees of freedom are 3, 4 and 7, 8.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Boundary Value Problems – 2”.

1. For A1=5, A2=10, A3=5, what is the value of the shape function at node 1 of the element shown?

Find the value of the shape function at node 1 of the element

a) 0.15

b) 0.5

c) 0.35

d) 0.25

Answer: b

Explanation:

Total area, A=A1+A2+A3

A=5+10+5

=20.

The shape function at node 2 is given by

=10/20

=0.5.

2. In a solid of revolution, if the geometry, support conditions, loads, and material properties are all symmetric about the axis and are independent of θ, then the problem can be treated as a ____

a) two-dimensional one

b) one-dimensional one

c) three-dimensional one

d) plane strain

Answer: a

Explanation: In a solid of revolution, if the geometry, support conditions, loads, and material properties are all symmetric about the axis and are independent of θ, then the problem can be treated as a two-dimensional problem. Moreover, due to the absence of stress variation in the third dimension, such a problem is treated as a plain stress problem.

3. A function Q is evaluated at boundary 1-2 by boundary integral Q=∮q*Sds where q=q 0 and shape functions S are S 1 , S 2 .S 1 =1- and S 2 =1-S 1 then Q 1 is given by expression ____

Find expression of Q1 if function Q is evaluated at boundary 1-2 by boundary integral

a) \

\)*q 0 *l

b) q 0 *l

c) \

\)*q 0 *l

d) \

\)*q 0 *l

Answer: a

Explanation: Given Q=∮q*Sds

Q 1 =\

*S1*ds

=\(\int_{0}^l\)q 0 *

)\)*ds

=[q 0 *s*

\))]

Putting limits of s from zero to l

Q 1 =[

*q₀*l]-0

\)*q₀*l.

4. In a static structural type Boundary Value Problem, at any fixed support, How many non-zero Degrees Of Freedom exist?

a) 0

b) 1

c) 2

d) 3

Answer: a

Explanation: In a static structural type Boundary Value Problem, three types of supports exist. They are roller, fixed and hinged support. A fixed support has zero degrees of freedom where as a roller and a hinged support have two and one degree of freedom respectively.

5. A function Q is evaluated at boundary 1-2 by boundary integral Q=∮q*Sds where q=q 0 and shape functions S are S 1 , S 2 .S 1 =1- and S 2 =1-S 1 then Q 3 is given by the value ____

Find expression of Q3 if function Q is evaluated at boundary 1-2 by boundary integral

a) \

b) 1

c) \

d) 0

Answer: d

Explanation: Given Q=∮q*Sds

Since there is no q defined on sides 2-3 and 3-1 we take q=0.

Q 3 =\(\int_{0}^l\)0*S1*ds

=0.

6. In a static structural type Boundary Value Problem, at any roller support, How many non-zero Degrees Of Freedom exist?

a) 0

b) 1

c) 2

d) 3

Answer: c

7. A function Q is evaluated at boundary 1-2 by boundary integral Q=∮q*Sds where q=q 0 * and shape functions S are S 1 , S 2 .S 1 =1- and S 2 =1-S 1 then Q 1 is given by expression ___

Find Q1 if function Q is evaluated at boundary 1-2 & shape function is S2.S1=1-(s/l) & S2=1-S1

a) \

\)*q₀*l

b) q₀*l

c) \

\)*q₀*l

d) \

\)*q₀*l

Answer: d

Explanation: Given Q=∮q*Sds

Along line 1-2, Q 1 =\

*S1*ds

=\(\int_{0}^l\)q 0 *\

\)*

)\)*ds

\(\int_{0}^l\)q 0 *\

*ds-\int_{0}^l\)q 0 \

^2)\)*ds

Putting limits of s from zero to l

Q 1 =\

\)*q 0 *l-\

\)*q 0 *l

=q₀*l*\

)\)

\)*q₀*l.

8. In a static structural type Boundary Value Problem, at any hinged support, How many non-zero Degrees Of Freedom exist?

a) 0

b) 1

c) 2

d) 3

Answer: b

9. A function Q is evaluated at boundary 1-2 by boundary integral Q=∮q*Sds where q=q₀* and shape functions S are S 1 , S 2 .S 1 =1- and S 2 =1-S 1 then Q 2 is given by expression ___

Find expression of Q2 if function Q is evaluated at boundary 1-2 by boundary integral

a) \

\)*q₀*l

b) q₀*l

c) \

\)*q₀*l

d) \

\)*q₀*l

Answer: c

Explanation: Given Q=∮q*Sds

Along line 1-2, Q 2 =\

*S2*ds

*ds\)

^2)*ds\)

Putting limits of s from zero to l

= \

\)*q₀*l.

10. For a linear triangular element with as the coordinates of the ith node of the element, which option denotes twice the Area of the triangle?

a) + +

b) + +

c) +

d) + +

Answer: a

Explanation: A linear triangular element has 3 nodes. With as coordinates of i th node, the twice of area is given by determinant of the matrix \

+ + .

11. For a linear triangular element with as the coordinates of the i th node of the element the area=10units, the value of ∑αi from the standard relation αi+βiX+γiY=*Area where X=∑xi, Y=∑yi is ___

a) 10

b) 20

c) 30

d) 40

Answer: b

Explanation: A linear triangular element has 3 nodes. With as coordinates of ith node, the twice of area is given by determinant of the matrix \

+ + . Then from the standard relation we have ∑αi = + +

=2*Area

=2*10

=20.

12. For a linear triangular element with as the coordinates of the ith node of the element the area=10units, the value of ∑βi from the standard relation αi+βiX+γiY=*Area where X=∑xi, Y=∑yi is ___

a) 0

b) 10

c) 20

d) 30

Answer: a

Explanation: A linear triangular element has 3 nodes. With as coordinates of ith node, the twice of area is given by determinant of the matrix \

++. Then from the standard relation we have ∑βi=++

=y2−y3+y3−y1+y1–y2

=0.

13. In a 3D axisymmetric solid, because of symmetry about the longitudinal axis, the stresses do not vary along ___ coordinate.

a) x

b) y

c) z

d) θ

Answer: d

Explanation: In a 3D axisymmetric solid, because of the symmetry about the longitudinal z-axis, the stresses does not vary along circumferential direction i.e. along θ coordinate and such a problem can be treated as a two-dimensional problem.

14. For a linear triangular element with as the coordinates of the ith node of the element the area=10units, the value of ∑γi from the standard relation αi+βiX+γiY=*Area where X=∑xi, Y=∑yi is ___

a) 0

b) 10

c) 20

d) 30

Answer: a

Explanation: A linear triangular element has 3 nodes. With as coordinates of ith node, the twice of area is given by determinant of the matrix\

++.Then from the standard relation we have ∑γi=−−−

=−x2+x3−x3+x1−x1+x2.

=0.

This set of Finite Element Method Questions & Answers for Exams focuses on “Some Comments on Mesh Generation and Imposition of Boundary Conditions”.

1. In Finite Element Method , which option is not correct with respect to discretization or mesh generation?

a) Density of elements depends on the degree of accuracy desired

b) Choice of element type may depend on the geometry of domain

c) For better accuracy correct formulae has to be applied

d) Some general rules govern mesh generation

Answer: c

Explanation: In Finite Element Method , for discretization purpose there are no specific formulae to obtain mesh generation as it depends on the expertise of the analyst. However, the density of elements,choice of element type may depend on the degree of accuracy desired and the geometry of domain.

2. In Finite Element Method , Mesh convergence is a measure of the number of _____ required in a model to ensure that the results of an analysis are not affected by changing the size of the _____

a) elements, mesh

b) nodes, mesh

c) nodes, element

d) degrees of freedom, element

Answer: a

Explanation: Mesh convergence determines how many elements are required in a FEM model to ensure that the results of an analysis are not affected by varying the size of the mesh. Once the mesh is converged, no change is observed in the results even after changing the density of the mesh.

3. In Finite Element Method , While domain discretization, at which of the following geometrical cases a node is not required?

a) Change in cross-section area only

b) Change in cross-section only

c) Change in cross section area as well as change in cross-section

d) No change

Answer: d

Explanation: While domain discretization, a node is necessary to be placed whenever there is a geometric discontinuity. For example, a change in cross section area or change in cross-section or both.

4. In Finite Element Method , In order to increase accuracy of a solution, while domain discretization, more than one node is to be placed at a point source.

a) True

b) False

Answer: b

Explanation: While domain discretization, a single node is placed at a point source and the point source is assumed to lumped at that node. Since there is no division of a point load, its discretization plays no role in determining accuracy of the solution.

5. For an inviscid flow of a fluid around a cylinder in an open channel as shown, the flow in the surroundings of the cylinder gets _____ and thus, the flow becomes ___

Inviscid flow of a fluid around a cylinder in an open channel

a) accelerated, non-uniform

b) decelerated, unsteady

c) accelerated, unsteady

d) compressed, non-uniform

Answer: a

Explanation: Since the section of the cylinder is smaller than the inlet section of the channel, for an inviscid flow around the cylinder in an open channel as shown, the flow in the surroundings of the cylinder gets accelerated and thus, the flow becomes non-uniform. This phenomenon is a result of conservation of mass.

6. For an inviscid flow of a fluid around the cylinder in an open channel as shown, the flow in the surroundings of the cylinder gets accelerated. At which points, A, B, C and D the mesh is created finer than the other points?

Inviscid flow of a fluid around the cylinder in an open channel

a) Both A and B

b) Either A or B

c) Both C and D

d) Only at D

Answer: c

Explanation: For an inviscid flow around the cylinder in an open channel as shown, the flow in the surroundings of the cylinder gets accelerated and away from cylinder flow is uniform. Knowledge of qualitative behavior of the flow helps one to use a coarse mesh at far from the cylinder, and a fine one at closer distances. Another purpose of using a refined mesh near the cylinder is to exactly represent the curved boundary of the domain there.

7. In Finite Element Method , for discretization purpose in general, a refined mesh is employed in areas where changes in geometry, boundary conditions, loading, material properties or solution are present.

a) True

b) False

Answer: a

Explanation: For an inviscid flow around the cylinder in an open channel, the flow in the surroundings of the cylinder gets accelerated and away from cylinder flow is uniform. Knowledge of qualitative behavior of the flow helps one to use a coarse mesh at far from the cylinder, and a fine one at closer distances. Another purpose of using a refined mesh near the cylinder is to exactly represent the curved boundary of the domain there.

8. In Finite Element Method , for discretization purpose in general, which 2D element is an undesired one?

a) An equilateral triangle of side 5 units

b) A quadrilateral

c) A rhombus of side 10 units

d) A rectangle of dimensions 100×1

Answer: d

Explanation: During discretization, when a mesh is made, care should be taken to avoid elements with very large aspect ratios or small angles. In finite element equations the coefficient matrices depend on the aspect ratios. If the aspect ratios is very large, the resulting coefficient matrices are ill—conditioned .

9. In Finite Element Method , for discretization purpose in general, which optionis not true for mesh refinement?

a) Previous meshes should be contained in the refined mesh

b) Any point in the body can be included within an arbitrarily small element at any stage of the mesh refinement

c) The same order of approximation for the solution may be preserved through all stages of the refinement process

d) Elements with very large aspect ratios are desired

Answer: d

Explanation: A mesh refinement should satisfy three conditions, 1) previous meshes should be contained in the refined mesh, 2) Any point in the body can be included within an arbitrarily small element at any stage of the mesh refinement 3) The same order of approximation for the solution may be preserved through all stages of the refinement process.

10. In Finite Element Method , for discretization purpose in general, which of the following mesh cannot be a part of mesh refinement process?

Mesh refinement process for discretization purpose in Finite Element Method (FEM)

a) A

b) B

c) C

d) D

Answer: c

Explanation: A mesh refinement should satisfy three conditions, i) previous meshes should be contained in the refined mesh otherwise mesh refinement is not correct, ii) Any point in the body can be included within an arbitrarily small element at any stage of the mesh refinement iii) The same order of approximation for the solution may be preserved through all stages of the refinement process.

11. To discretize a domain using rectangular elements, if the aspect ratio is to be limited to 10, which dimension of rectangular element gives finest mesh?

a) 1×0.5

b) 1.1×0.1

c) 0.11×0.01

d) 0.1×0.05

Answer: d

Explanation: During discretization, when a mesh is made, care should be taken to avoid elements with very large aspect ratios . For a>b, if axb are dimension of rectangular element then it is given that a/b should not be more than 10. The smaller the element the finer is the mesh.

12. In Finite Element Method , choose the incorrect option regarding to the dimensions of elements in a finite element mesh?

a) Elements with very large aspect ratios are avoided

b) Elements with small angles are undesired

c) The coefficient matrices in finite element equations depend on dimensions of elements

d) Rectangular elements with high aspect ratio are preferred

Answer: d

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Single Variable Problems – Applications – 1”.

1. If the finite element model shown below represents heat conduction in axisymmetric or plane geometries then which option is not true?

-\frac{\partial}{\partial y}

\) +a 00 u-f=0

a) u is temperature

b) a 11 , a 22 are conductivities in x, y directions

c) f is internal heat generation

d) a 00 =1

Answer: d

Explanation: Given model equation is a 2 nd order partial differential equation. For heat conduction in axisymmetric or plane geometries, the given model equation is applicable for u as temperature, a 11 and a 22 as conductivities in x and y directions respectively, f as internal heat generation and a 00 =0.

2. In steady state heat transfer finite element model [K+H]*{T}={Q}*{P} if convective boundary conditions are neglected then which option is applicable?

a) K=0

b) H=0

c) Q=0

d) T=0

Answer: b

Explanation: When convective boundary conditions are neglected in the steady state heat transfer finite element model [K+H]*{T}={Q}*{P}, it reduces to [K]*{T}={Q} form. The terms H and P which are related to convection become zero.

3. For a convective boundary, the natural boundary condition is a balance of energy transfer across the boundary due to conduction and/or convection.

a) True

b) False

Answer: a

Explanation: The existence of convection heating or cooling leads to the convection boundary condition, known as Newton boundary condition. For a convective boundary, the natural boundary condition is a balance of energy transfer across the boundary due to conduction and/or convection.

4. In the below equation for steady-state heat transfer in plane systems, what does β stands for?

\(k_x\frac{\partial T}{\partial x}n_x+k_y\frac{\partial T}{\partial y}n_y\)+β(T-T ∞ )=\(\hat{q}\) n

a) Convective heat transfer coefficient

b) Thermal expansion

c) Thermal conductivity

d) Diffusivity

Answer: a

5. In a steady state heat conduction problem with a thermal conductivity of 22), for a typical element of mesh of linear triangular elements, what is the value of a in stiffness matrix,

K=a*\

b) 12

c) 10

d) 5.5

Answer: a

Explanation: For a typical element of mesh of linear triangular elements K=*\

is 22 then

a=k/2

a=22/2

=11.

6. Heat conduction problem over a rectangular domain is shown in figure. Which nodes have temperature value of zero units?

Heat conduction problem over a rectangular domain

a) Only1

b) Only 7

c) Only 2 and 3

d) 1, 2 and 3

Answer: d

Explanation: All the nodes present at the boundary along which temperature value is zero will have T=0. In the given domain, the nodes 1, 2 and 3 are on boundary with temperature value zero. The node 7 is along an insulated boundary where temperature value depends on the temperature of other boundaries.

7. If q denotes the amount of heat flow through any boundary then what is the value of q at node7?

Find the value of q (amount of heat flow through any boundary) at insulated node

a) Dependent on a,b

b) =0

c) >0

d) <0

Answer: b

Explanation: A boundary which is insulated will have no heat flow. Thus All the nodes present at that boundary will have q=0, irrespective of the value of T at other boundaries. In the given domain, the nodes 6, 7 and 8 are on insulated boundary, thus no heat flow.

8. What is the value of T at node 3 if the length of each element is 1/a units?

Find the value of T at node 3 if the length of each element is 1/a units

a) 0.5*b

b) b

c) 2*b

d) b*cos

Answer: a

Explanation: Given that the temperature distribution at boundary follows T=bcos\(\frac{\pi ax}{6}\).

At node 3, x=2/a units,

ax=2

T=bcos\(\frac{\pi2}{6}\)

T=bcos\(\frac{\pi}{3}\)

=b*0.5.

9. The velocity field of a fluid flow is denoted by v=2xi+3j. Which option exactly describes the flow?

a) Ideal

b) Compressible

c) Irrational

d) Inviscid

Answer: b

Explanation: A fluid is said to be incompressible if ∇.V=0.

∇.V=\(\frac{d}{dx}+\frac{d}{dy}\)

=2+0

=2.

Thus, the flow is compressible.

10. For a two dimensional problem, the evaluation of boundary integrals amounts to evaluation of line integrals.

a) True

b) False

Answer: a

Explanation: The boundary of a 2D element consists of line segments, which are one dimensional elements. Thus, for a two dimensional problem, the evaluation of boundary integrals amounts to evaluation of line integrals .

11. A river flows along node 1 through node 4, infiltrating an aquifer at constant rate of 5(m 3 /day*m 2 ). The length of elements is 0.2m. What is the value of global force due to infiltration of the river at node 1?

Find the value of global force due to infiltration of the river at node 1

a) 0.5

b) 1.0

c) 1.5

d) 2.0

Answer: a

Explanation: The value of global force due to infiltration of the river at node 1 is given by 0.5*q*h, where 0.5 denotes the value of the shape function, q is rate of infiltration and h is length of the element.

=0.5*q*h

Given q=5, h=0.2

=0.5*5*0.2

=0.5.

12. In a Heat Transfer problem, which option is used for interpolation of temperature inside a finite element?

a) Natural coordinates

b) Global coordinates

c) Temperature gradients

d) Shape functions

Answer: d

Explanation: Shape Functions are used for interpolation of temperature inside a finite element. Global coordinates are used to apply boundary conditions. Natural coordinates are used to derive shape functions. Temperature gradients can’t be used for interpolation.

13. Consider a system where a wall of a tank containing a hot liquid at a temperature T 0 , with an air stream of temperature T∞ flows on the outside of the tank, maintaining the outside wall temperature of TL. What is the expression for boundary condition of the system?

a) q=h

b) T=T∞

c) T=TL

d) TL-T∞

Answer: a

Explanation: The boundary conditions are mainly of three kinds: constant temperature, constant heat flux , and convection. In this problem, the temperature difference between wall and surroundings creates the constant heat flux of q=h. This heat flux is taken as a boundary condition.

This set of Finite Element Method Question Bank focuses on “Single Variable Problems – Applications – 2”.

1. A river flows along node 1 through node 5, infiltrating an aquifer at a constant rate. What is the value of s1a+s2a+s3c+s4c if s denotes a linear interpolation function in FEM?

Find the value of s1a+s2a+s3c+s4c if s denotes a linear interpolation function in FEM

a) 1

b) 2

c) 3

d) 4

Answer: b

Explanation: Since s denotes a linear interpolation function in FEM, the sum of the two linear interpolation functions corresponding to one element must be equal to unity. Thus, s1a+s2a =1 and s3c+s4c=1.

→s1a+s2a+s3c+s4c =2.

2. In a thermodynamics process, what is the correct term for the series of states through which a system passes?

a) Path

b) Phase

c) Cycle

d) Direction

Answer: a

Explanation: A thermodynamic path is series of states through which a system passes from an initial state of equilibrium to a final state of equilibrium and can be plotted on a pressure-volume , pressure-temperature , and temperature-entropy diagrams to view graphically. Phase is a quantity of matter that can be separated mechanically from a heterogeneous mixture.

3. A river flows along node 1 through node 4, infiltrating an aquifer at constant rate of 5(m 3 /day*m 2 ). The length of elements shown is 2m. What is the value of global force due to infiltration of the river at node 2?

Find the value of global force due to infiltration of the river at node 2

a) 5

b) 10

c) 15

d) 20

Answer: b

Explanation: The value of global force due to infiltration of the river at node 2 is given by 0.5*q*h+0.5*q*h, where 0.5 denotes the value of the two shape functions, q is rate of infiltration and h is length of the element.

=0.5*q*h+0.5*q*h

=q*h

Given q=5, h=2

=5*2

=10.

4. In which thermodynamics process the temperature remains constant?

a) Isobaric

b) Isentropic

c) Isothermal

d) Isochoric

Answer: c

Explanation: An Isobaric process takes place at constant pressure an Isentropic process takes place at constant entropy. An Isothermal process takes place at constant temperature whereas an isochoric process takes place at constant volume.

5. If ψ=2x+3y represents stream function then what is the magnitude of velocity?

a) 3

b) 5

c) 6.3

d) 3.6

Answer: d

Explanation: The magnitude of velocity is (V 2 +U 2 ) .

V=\(\frac{\partial \psi}{\partial y}\)=3 and U=\(\frac{-\partial \psi}{\partial x}\)=-2

v=(3 2 +2 2 )

=3.6.

6. In which process does a thermodynamic system remains infinitesimally closed to an equilibrium state at all times?

a) Path equilibrium process

b) Cycle equilibrium process

c) Phase equilibrium process

d) Quasi-state or quasi-equilibrium process

Answer: d

Explanation: Quasi-static Process, in thermodynamics, is a process that happens infinitely slowly. However, it is very important to note that no real process is Quasistatic. A Quasistatic process ensures that the system will go through a sequence of states that are infinitesimally close to equilibrium.

7. In analysis of mechanical stresses, which of the following is the correct form of the 3D equation of stress equilibrium?

a) \

No such equation exists

Answer: a

Explanation: For Cartesian problems in three dimensions, the stresses σx, τxy, and τxz act on the face normal to X-axis and the body is said to be in equilibrium if\(\frac{\partial \sigma x}{\partial x}+\frac{\partial \tau xy}{\partial y}+\frac{\partial \tau xz}{\partial z}\)=0 on that face.

8. In thermodynamics, what does the term open system refer to?

a) Control mass

b) Control volume

c) Control energy

d) Control temperature

Answer: b

Explanation: In thermodynamics, an open system freely exchanges matter and energy with its surroundings. For instance, when you are boiling water in an open bowl on a stove, energy and matter are being transferred to the surroundings through steam. In this case, the volume remains constant making it a control volume process.

9. Consider the wall of a tank containing a hot liquid at a temperature T 0 , with an airstream passed on the outside, maintaining a wall temperature of T L at the boundary. From the figure shown,what is the boundary condition at X=0?

Find boundary condition at X=0 for wall of a tank containing a hot liquid at temperature T0

a) T=T 0

b) T=T∞

c) T=T L

d) Data insufficient

Answer: a

Explanation: For heat transfer problems, the boundary conditions are mainly of three types. They are specified temperature, specified heat flux and convection. From the figure shown, at X=0, the boundary condition specified is T=T 0 .

10. In thermodynamics, what is the name of a process with identical end states?

a) Cycle

b) Path

c) Phase

d) Either Path or phase

Answer: a

Explanation: A thermodynamic cycle is a process with identical end states. A thermodynamic path is series of states or the path through which a system passes from an initial state of equilibrium to a final state of equilibrium. Phase is a quantity of matter that can be separated mechanically from a heterogeneous mixture.

11. For the given heat radiation equation, which of the following terminology is not correct?

q {x} n {x} +q {y} n {y} +q {z} n {z} =σε\(T_{\{s\}}^{\{4\}}\)-αq {r}

a) α is Stefan-Boltzmann constant

b) σ is the surface emission coefficient

c) ε is surface absorption coefficient

d) q r is heat generated per unit surface area

Answer: d

Explanation: In the given equation,the first term in the RHS denotes radiated power and it has no sign before it. The second term has negative sign before it. Thus, q r denotes negative heat flow per unit surface area. q r does not denote heat generated per unit surface area.

12. A system is said to be in thermodynamic equilibrium if it maintains which of the following equilibriums?

a) Mechanical and phase

b) Thermal and chemical

c) Thermal, mechanical and chemical

d) Thermal, phase, mechanical and chemical

Answer: c

Explanation: Thermodynamic equilibrium describes a system whose properties will not change without some sort of outside interference. In other words, a system in thermodynamic equilibrium will not change unless something is added or subtracted from it. Thus, it should be under Thermal, mechanical and chemical equilibrium.

13. Which method is used to write the basic heat transfer equation in the following form using weighted residuals?

\Missing or unrecognized delimiter for \right\) N i dV = 0

a) Galerkin method

b) Jacobi method

c) Rayleigh Ritz method

d) Delaunay method

Answer: a

Explanation: Finite element equations are obtained using Galerkin method. Rayleigh Ritz method does not use interpolation functions, N i whereas Galerkin method uses interpolation functions, N i and thus, is used in FEM. Jacobi is used for Eigen value problems. Delaunay method is used to generate mesh for triangular elements.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Eigen Value and Time Dependent Problems – 1”.

1. The simultaneous linear equations used in FEM for solution of static problems are KX=F, the methods available for solving these equations are divided into two types: direct and iterative.

a) True

b) False

Answer: a

Explanation: When FEM is used for solution of static problems, we deal with a set of simultaneous Linear Equations of the form KX=F, where K is stiffness Matrix, X is displacement matrix and F is load vector. The order of matrix K is very large and the methods available for solving the equation are divided into two types: direct and iterative. Direct methods are used for equations without any round of error and iterative methods are used for the equations which start with an initial approximation.

2. For the following equations, what is the value of x 2 using Gaussian elimination method?

x 1 -x 2 +3x 3 =10——–

5x 2 -5x 3 =-5————

-7x 3 =-28 —————-

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: From the 3 rd equation it’s seen that x 3 =\(\frac{-28}{-7}\)=4. Using x 3 in 2 nd equation we get

5*x 2 -5*4=-5

5*x 2 =15

x 2 =3

3. For the following equations, what is the value of x 1 using Gaussian elimination method?

x 1 -x 2 +3x 3 =10——–

5x 2 -5x 3 =-5————

-7x 3 =-28—————-

a) 1

b) 2

c) 3

d) 4

Answer: a

Explanation: From the 3 rd equation it’s seen that x 3 =\(\frac{-28}{-7}\)=4. Using x 3 in 2 nd equation we get

5*x 2 -5*4=-5

5*x 2 =15

x 2 =3

Using x 3 , x 2 in 1 st equation

x 1 -3+3*4=10

x 1 -3+12=10

x 1 =1

4. Which option is not correct about direct methods for solving system of linear equations?

a) In the absence of errors it yields exact solution

b) Errors arising from round off and truncation may give useless results

c) Gaussian elimination method is an example

d) Starts with an initial approximation

Answer: d

Explanation: The methods used for solving a system of linear equations are classified as: direct and iterative. Direct methods are those, which, in the absence of round-off and other errors, yield an exact solution in a finite number of elementary arithmetic operations. Indeed the errors arising from round-off and truncation may lead to extremely poor or even useless results. The fundamental method used for direct solutions is Gaussian elimination.

5. Which option is not correct about iterative methods for solving system of linear equations?

a) Convergence yields a good approximate solution

b) Insensitive to the growth of round-off errors

c) Gaussian elimination method is an example

d) Starts with an initial approximation

Answer: c

Explanation: The methods available for solving a system of linear equations can be divided into two types: direct and iterative. Iterative methods are those, which start with an initial approximation. When the process converges, we can expect to get a good approximate solution. The main advantages of iterative methods are the simplicity and uniformity of the operations to be performed, which make them well suited for use on computers and their relative insensitivity to growth of round-off errors.

6. In structural mechanics, which option is not correct about linear analysis?

a) Displacements are infinitesimally small

b) Material is linearly elastic

c) Externally applied loads are a function of time

d) Applied loads are not a function of time

Answer: c

Explanation: In a linear analysis, we assume that the displacements of a finite element assemblage are infinitesimally small and the material is linearly elastic. In addition, we also assume that a nature of boundary conditions remains unchanged during application of loads on Finite element assemblage. Loads are constant with respect to time.

7. A generalized Eigen value problem [K- ω 2 M]X=0 has a non-zero solution for X. What can be the value of determinant of the matrix [K- ω 2 M]?

a) Any integer

b) 0

c) +1

d) Positive integer

Answer: b

Explanation: A generalized Eigen value problem is represented by homogeneous matrix equation, [K- ω 2 M]X=0. From matrix equations methods, the equation has a non-zero solution for X if the determinant of the matrix [K-ω 2 M] equals to zero.

8. In FEM, the forced vibrations equation after Finite Element discretization of a structure can be expressed as which option?

a) Mẍ+Kẋ=F

b) Mẍ+Kẋ=0

c) Mẍ+Kx=F

d) Mẍ+Kx=0

Answer: c

Explanation: The forced vibrations equation after Finite Element discretization of a structure can be expressed as Mẍ+Kx=F where M and K are the mass and stiffness matrices of the structure, F is the external load vector; x and ẍ are the displacement and acceleration vectors. In the forced vibration equation the force vector is non-zero.

9. The free vibrations equation after Finite Element discretization of a structure is expressed as Mẍ+Kx=0. Which option is not correct about the free vibration case?

a) Displacements are harmonic

b) x=Xe iωt where X is amplitude

c) [K-ω 2 M]X=0

d) KX=Mω 2

Answer: d

Explanation: In a free vibration analysis, the external load vector is zero and the displacements, x are harmonic x=Xe iωt where X is amplitude, on substituting x in governing equation we get [K-ω 2 M]X=0 or KX=Mω 2 X.

10. The generalized Eigen value problem [K-ω 2 M]X=0 has a non-zero solution for X. What is the value of natural frequency, ω if K=\

b) 1/9

c) 9

d) 1/3

Answer: d

Explanation: The generalized Eigen value problem [K-ω 2 M]=0 has a non-zero solution for X if the determinant of the matrix [K-ω 2 M] equals zero,

K=ω 2 M

\)=ω 2 *\(

\) Equating corresponding elements, we get 9*ω 2 =1

ω 2 =1/9

Natural frequency, ω =1/3.

11. Which option is not correct about free vibration analysis problem KX= λMX, where X represents the amplitude of displacement x?

a) The displacements are harmonic

b) X represent mode shapes or Eigen vectors

c) λ represent Eigen value

d) ω represents Eigen value

Answer: d

Explanation: In a free vibration analysis KX= λMX, the external load vector is zero and the displacements are harmonic x=Xe iωt where X represents the amplitude of displacement x called Eigen vectors and λ= ω 2 represent Eigen value.

12. After Finite Element discretization of a structure, which option expresses the free vibrations equation?

a) Mẍ+Kẋ=F

b) Mẍ+Kẋ=0

c) Mẍ+Kx=F

d) Mẍ+Kx=0

Answer: d

Explanation: After Finite Element discretization of a structure, the free vibrations equation can be expressed as Mẍ+Kx=0 where M and K are the mass and stiffness matrices of the structure; x and ẍ are the displacement and acceleration vectors respectively. In a free vibration analysis, the external load vector is zero.

13. For the eigenvalue problem of the form A = λB, which option is not correct about the parameters used in the equation below?

A=\

B=1

c) B=0

d) λ is called eigenvalue

Answer: c

Explanation: For the eigenvalue problem of the form A = λB, A and B denote linear differential operators, has nontrivial solutions u. The values of λ are called eigenvalues and the associated functions U are called Eigen functions. For example, the given equation has A=\(\frac{d^2x}{dx^2}\) and B=1.

14. The generalized Eigen value problem [K- λM]X=0 has a non-zero solution for X. What is the value of λ if K=\

b) \

d) \(\frac{1}{4}\)

Answer: d

Explanation: The generalized Eigen value problem [K- λM]X=0 has a non-zero solution for X if the determinant of the matrix [K- λM] equals zero or K=λM

\)=λ\(

Equating corresponding elements, we get 1=4*λ

λ=\(\frac{1}{4}\).

15. For the following eigenvalue equation to represent a heat transfer problem, a=kA and C=ρcA.

\)=λCU

a) True

b) False

Answer: a

Explanation: For the given eigenvalue equation the quantities a and C depend on the physics of problem. For a heat transfer problem, a=kA and C=ρcA where k is thermal conductivity, A is cross-sectional area and c is specific heat.

This set of Finite Element Method Questions and Answers for Entrance exams focuses on “Eigen Value and Time Dependent Problems – 2”.

1. Suppose the following eigenvalue equation represents a bar problem, then the value of the parameters a and c 0 should be EA and ρA , respectively.

\)=λc 0 U

a) True

b) False

Answer: a

Explanation: For the given eigenvalue equation, the values of the parameters a and c 0 depends upon the physical properties and phenomena involved in the problem. For a bar problem, a=EA and c 0 =ρA, where E is Young’s modulus, A is cross-sectional area and m is the mass density. For a heat transfer problem, a=kA and c 0 =ρcA.

2. A plane wall of length L units and Cross-section area A units was initially maintained at a temperature of T units. It is subjected to an ambient temperature of T ∞ units at one surface. If the heat transfer coefficient at the surface of the wall is assumed to be h units, then what is the temperature gradient developed at the surface?

a)(T ∞ -T)\

(T ∞ -T)\

T ∞ -T

d) T-T ∞

Answer: a

Explanation: Let T x be temperature gradient developed at the surface. If the heat transfer coefficient at the surfaces of a wall is assumed to be h units, then the heat interaction at the surfaces of the wall is evaluated by Equating the conduction heat transfer to the convection heat transfer, i.e.,

kAT x = hA(T-T ∞ )

T x =(T ∞ -T)\(\frac{h}{k}\).

3. A plane wall of thermal conductivity of 45\

b) 2

c) 3

d) 4

Answer: b

Explanation: Let T x be temperature gradient developed at the surface. If the heat transfer coefficient at the surface of a wall is is 9\(\frac{W}{m^2K}\) then the heat interaction at the surface of the wall is evaluated by equating the conduction heat transfer to the convection heat transfer, i.e.,

45T x = 9

T x = \

=2.

4. A plane wall was maintained initially at a temperature of T units. It is subjected to an ambient temperature of T ∞ units at one surface. If the heat transfer coefficient at the surfaces of the wall is assumed to be infinite, then what is the new temperature at the wall?

a) T

b) T ∞

c) T ∞ -T

d) T-T ∞

Answer: b

Explanation: Let X be the unknown new temperature at the wall surface. If the heat transfer coefficient at the surfaces of a wall is assumed to be h, then the heat interaction at the surfaces of the wall is evaluated by equating the conduction heat transfer to the convection heat transfer, i.e.,

kAT x = hA(X – T ∞ )

\(\frac{-kT_x}{h}\) = X – T ∞

Given h = ∞

\(\frac{-kT_x}{\infty}\) = X – T ∞

0 = X – T ∞

X = T ∞ .

5. A plane wall was maintained initially at a temperature of 35°C. It is subjected to an ambient temperature of 45°C at one surface. If the heat transfer coefficient at the surfaces of the wall is assumed to be infinite, then what is the new temperature at the wall surface?

a) 35°C

b) 45°C

c) 40°C

d) 50°C

Answer: b

kAT x =hA(X-T ∞ )

\(\frac{-kT_x}{h}\)=X-T ∞

Given h = ∞

\(\frac{-kT_x}{\infty}\)=X-T ∞

0=X-T ∞

X=T ∞ , given T ∞ =45°C

X=45°C.

6. In thermodynamics, the following equation represents a diffusion process. If k is thermal conductivity, p is density, and c is the specific heat at constant pressure, then what is α?

\(\frac{c}{k}\)

Answer: a

Explanation: The term α is called diffusion coefficient, and it is equal to \(\frac{k}{pc}\). This equation governs one-dimensional temperature distribution in a plain wall. A one-dimensional problem is solved using bar elements with one degree of freedom at each node.

7. The governing equation of an unsteady one-dimensional heat transfer problem is given below. It has a solution u = Uexp. What is λ appropriately called?

+ b \frac{\partial u}{\partial t}\) + cu = 0 for 0<x<L

a) Natural frequency

b) Eigenvalue

c) Thermal diffusivity

d) Thermal flux

Answer: b

Explanation: The governing equation of an unsteady one-dimensional heat transfer problem is a parabolic equation. Hence, its solution is given by u = Uexp, where u represents temperature along a direction x at any time t, U is the corresponding mode shape, and λ is the eigenvalue of the equation. The solution is periodic.

8. The unsteady natural axial oscillations of a bar are periodic, and they are determined by assuming a solution u = U e -iwt . Which option is not correct about the solution equation?

a) w denotes the natural frequency

b) w 2 denotes eigenvalue

c) U denotes mode shape

d) u denotes transverse displacements

Answer: d

Explanation: The unsteady natural axial oscillations of a bar are periodic. They are measured by assuming a solution u = U e -iwt , where w is natural frequency, w 2 is an eigenvalue, U is mode shape, and u is instantaneous axial displacement. The problem is solved in FEM by employing bar elements and appropriate shape functions.

9. In matrix algebra, which option is not correct about an eigenvalue problem of the type Ax = Lx?

a) It has a discrete solution

b) It has solution only if A non-singular

c) x is called eigenvector

d) L is called eigenvalue

Answer: b

Explanation: An eigenvalue problem of the type Ax = Lx looks as if it should have a continuous solution, but instead, it has discrete ones. The problem is to find the numbers denoted by L, called eigenvalues, and their matching vectors denoted by x, called eigenvectors. It may have a solution irrespective of whether the matrix A is singular or not.

10. The dynamic equation of motion of a structure contains M, C and K as mass, damping and stiffness matrices of the structure, respectively. If F is an external load vector, then which option is correct about the equation?

a) M\

All the forces are time-independent

d) The equation is of 3 rd order

Answer: b

Explanation: The dynamic equation of motion of a structure is a 2 nd order equation. It is written as M\(\ddot{x}\) + C\(\dot{x}\) + Kx = F, where M, C and K are the mass, damping and stiffness matrices of structure, respectively. All the forces in the equation are time-dependent. M\(\ddot{x}\) is inertia force, Kx is spring force, and C\(\ddot{x}\) is damping force.

11. In matrix algebra, a matrix K equals \(

\). What is the value of a, if K 7 = \

2187

b) 729

c) 6561

d) 5 7

Answer: a

Explanation: Since K is a diagonal matrix, its higher powers are obtained by raising its diagonal elements to the same power. If K=\(

\) then K 7 =\(

\). Equating the corresponding elements of \(

\) and \(

\) we get

a=3 7

a=2187.

12. In matrix algebra, what is the eigenvalue of the matrix \

b) 2

c) 3

d) 4

Answer: c

Explanation: The eigenvalue, L of a matrix is equal to the root of the equation |K-LI|=0.

Let the given matrix be denoted by K then K-LI = \

2 -1)-1+1

= (L 2 -2L) + 2L

= -L 3 + 3L 2

= -L 2 .

Given -L 2 = 0

On simplification L = 0, 0 and 3.

13. In matrix algebra, what is the value of a-b if the eigenvector of \

b) 1

c) 2

d) 3

Answer: a

Explanation: If X is an eigenvector corresponding to an eigenvalue L of a matrix K, then KX=LX. The eigenvector of \(

\) corresponding to eigenvalue three is \(

\). Equating the corresponding elements of \(

\) and \(

a=b=1

a-b=0.

14. From the Euler-Bernoulli beam theory of natural vibrations, using cubic Hermite polynomials approximation, what is the 1 st element of the stiffness matrix?

a) \

\(\frac{12AI}{h^3}\)

Answer: a

Explanation: In the formulation of the Euler-Bernoulli beam theory, there are two degrees of freedom at a point, w and \(\frac{dw}{dx}\). Typically, the finite element model of this theory uses cubic polynomial. The first element of the stiffness matrix is \(\frac{12EI}{h^3}\), where E is Young’s modulus, I is the area moment of inertia and h is the length of the element.

15. From the Timoshenko beam theory of natural vibrations, using cubic Hermite polynomials approximation, what is the 1 st element of the mass matrix?

a) \

d) \(\frac{\rho I}{3}\)

Answer: a

Explanation: Using the Timoshenko beam theory applied to natural vibrations, mode shape is approximated using the cubic Hermite polynomials \(\psi_i^e\) and \(\psi_j^e\). The first element of a mass matrix is \(M_{ij}^{11} = \int_{x_a}^{x_b} \rho A \psi_i^e \psi_j^e\) dx, where x is the length of the element. For the 1 st element, using appropriate values of \(\psi_i^e\) and \(\psi_j^e\), the term \(M_{ij}^{11}\) reduces to \(\frac{\rho A}{3}\), where ρ is the density of the beam material, and A is the cross-section area of the beam.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Library of Elements and Interpolation Functions – 1”.

1. Which option is not correct about the three-noded triangular plane stress element used in FEM?

a) It has six degrees of freedom

b) It belongs to both the isoparametric and superparametric element families

c) It can be improved by the addition of internal degrees of freedom

d) Delaunay triangulation can be used for its mesh generation

Answer: c

Explanation: The three-node triangular element with linear displacements for the plane stress problem is simply called a linear triangle. It has six degrees of freedom and it belongs to both the isoparametric and superparametric element families. A mesh of linear trianglecan be easily generated using Delaunay triangulation, but the element cannot be improved by the addition of internal degrees of freedom; rather, it can be improved by increasing the number of nodes.

2. In FEM, which option is used to develop the Higher-order triangular elements systematically?

a) Pascal’s triangle

b) Galerkin method

c) Jacobi method

d) Delaunaytriangulation

Answer: a

Explanation: The Higher-order triangular elements can be systematically developed with the help of Pascal’s triangle. Finite element equations are obtained using the Galerkin method. Jacobi is used for eigenvalue problems. The Delaunay method is used to generate mesh for triangular elements.

3. In FEM, What is the number of displacement polynomials necessary for finding displacements in a linear triangular element?

a) 1

b) 2

c) 3

d) 4

Answer: b

Explanation: The number of displacement polynomials for an element is equal to the degrees of freedom of each node of the element. A linear triangular element has three nodes and two degrees of freedom at each node. Thus, the total number of displacement polynomials necessary for finding displacements is two.

4. Concerning triangular elements in FEM, which option is not correct about the mathematical formula of Pascal’s triangle?

a) It contains the terms in two coordinates only

b) The position of the terms can be viewed as the nodes of a triangular element

c) The position of the first and last terms of a row is at the vertices of a triangular element

d) A triangular element of order 2 corresponds to the second row

Answer: d

Explanation: A Pascal’s triangle contains the terms of polynomials of various degrees in two coordinates. We can view the positions of the terms as nodes of a triangular element, with the constant term and the first and last terms of a given row being the vertices of the triangle. A triangular element of order 2 contains six nodes and corresponds to the third row of Pascal’s triangle.

5. Which option is not correct about the four-noded rectangular plane stress element used in FEM?

a) It has eight degrees of freedom

b) Shape functions N1, N2, N3 and N4 are bilinear functions of x and y

c) The displacement field is continuous across elements

d) Its Delaunay triangulation is unique

Answer: d

Explanation: The four-node quadrilateral element with linear displacements for a plane stress problem has two degrees of freedom at each node. The total degrees of freedom of the element is eight. The displacement field is continuous across elements connected at nodes and the shape functions N1, N2, N3 and N4 are bilinear functions of x and y. Its Delaunay triangulationis not unique, but it has two solutions.

6. In FEM, if x, y represents Cartesian coordinates, then the following triangular array of binomial coefficients forms Pascal’s triangle.

True

b) False

Answer: a

Explanation: In mathematics, Pascal’s triangle is a triangular array of binomial coefficients. It contains the terms of polynomials of various degrees in two coordinates x and y. An n th row of Pascal’s triangle contains n term, and it corresponds to all triangular elements with at least 0.5n number of nodes. The 1st row contains number one only.

7. In the FEM element library, what is the other name of a higher-order element?

a) Complex element

b) Simplex element

c) Linear element

d) Nonlinear element

Answer: a

Explanation: Simplex and linear elements contain nodes only at endpoints but not at the interior. They have linear polynomials as interpolation functions. Higher-order elements can be created easily from simplex elements by adding additional intermediate nodes to each element. They are also called complex elements.

8. In FEM, if x, y represents Cartesian coordinates, then which term of Pascal’s triangle corresponds to the position of the interior node of the following element?

The position of interior node of 10 noded triangle corresponding to the Pascal’s triangle term

a) 1

b) xy

c) x 2 y

d) xy 2

Answer: b

Explanation: The interior node of the ten noded triangular elements can be viewed as the middle term of the third row of Pascal’s triangle. Since the 3 rd row contains three terms viz. x 2 , xy and y 2 in the same order, the interior node corresponds to the term xy. An n th row of Pascal’s triangle contains n term and it corresponds to all triangular elements with at least 0.5n number of nodes.

9. In the FEM element library, an eight noded quadrilateral element belongs to which family?

a) Serendipity

b) Linear

c) Simplex

d) Quadratic

Answer: a

Explanation: The Serendipity elements are the rectangular elements with intermediate nodes but no interior nodes, i.e., all nodes lie on boundary. Since four nodes of an eight noded quadrilateral element are intermediate nodes, it belongs to the Serendipity family. Simplex and linear elements contain nodes only at endpoints but not at intermediate points. They have linear polynomials as interpolation functions. A quadratic element contains interior nodes.

10. In FEM, which option is not correct about the Lagrange family of triangular elements?

a) The nodes are uniformly spaced

b) Pascal’s triangle can be viewed as a triangular element

c) Dependent variables and their derivatives are continuous at inter-element boundaries

d) 2 nd degree polynomial corresponds to 6 noded triangle

Answer: c

Explanation: In Lagrange family elements the nodes are regularly placed everywhere on the grid i.e., they are uniformly spaced. The location of the terms in Pascal’s triangle gives the location of nodes in elements. Thus, Pascal’s triangle can be viewed as a triangular element. The derivatives of dependent variables are not continuous at inter-element boundaries. 2 nd -degree polynomial corresponds to 6 noded triangles.

11. What is the displacement function for one-dimensional, two noded linear elements in terms of its shape functions N 1 and N 2 ?

a) N 1 u 1 +N 2 u 2

b) N 1 u 2 +N 2 u 1

c) N 1 u 1 -N 2 u 2

d) N 1 u 2 -N 2 u 1

Answer: a

Explanation: For a linear element, the displacement function is a linear polynomial of nodal displacements. A one-dimensional, two noded linear elements have two nodes with corresponding displacements u 1 , u 2 and corresponding shape functions N 1 , N 2 . The displacement function is given by N 1 u 1 +N 2 u 2 .

12. For the following element, what is the value of the distance variable s at the 1 st row?

Find the value of the distance variable s at 1st row for a quadratic element if k = 3

a) 1

b) 2

c) 0.5

d) 0

Answer: c

Explanation: The value of the distance variable s is \(\frac{p}{k-1}\); where p is the row at which s is calculated, k is the number of uniformly spaced nodes per side of the element. For a quadratic element, we have k = 3.

s=\(\frac{p}{3-1}\)

=\(\frac{p}{2}\)

At first row, p=1

s=\(\frac{1}{2}\)

=0.5.

13. For the following element, if s is the distance variable as shown, then its value at the 2 nd row is \(\frac{1}{3}\).

Find the value of the distance variable s at 2nd row for an element if k = 2

a) True

b) False

Answer: b

Explanation: The value of the distance variables s is \(\frac{p}{k-1}\); where p is the row at which s is calculated, k is the number of uniformly spaced nodes per side of the element. For a given element, we have k = 4.

s=\(\frac{p}{4-1}\)

=\(\frac{p}{3}\)

At second row, p=2

s=\(\frac{2}{3}\).

Answer: a

Explanation: A Constant strain triangular element is the simplest triangular element with three end nodes. A Linear strain triangular element is a six-noded triangular element with three intermediate nodes in addition to three end nodes. For plane stress applications, LST gives an accurate result compare to the three-noded CST element. The variable strain triangular element is a higher-order triangular element with more than six nodes.

Sanfoundry Global Education & Learning Series – Finite Element Method.

This set of Finite Element Method Questions and Answers for Campus interviews focuses on “Library of Elements and Interpolation Functions – 2”.

1. In the FEM element library, what is the correct name for a three noded triangular element?

a) Linear strain triangular element

b) Constant strain triangular element

c) Variable strain triangular element

d) Higher-order triangular element

Answer: b

Explanation: A Constant Strain Triangular element is the simplest triangular element with only three nodes that are located at its ends. A Linear Strain Triangular element is a six-noded triangular element with three intermediate nodes in addition to three end nodes. For plane stress applications, LST gives an accurate result compare to the three-noded CST element. The variable strain triangular element is a higher-order triangular element with more than six nodes.

2. If the geometry and other parameters of an element are defined in terms of only one spatial coordinate, then the element is a one-dimensional element?

a) True

b) False

Answer: a

Explanation: A one-dimensional element possesses one degree of freedom at each node. It is also known as a bar element or line element. Geometry and other parameters of a bar element are defined in terms of one spatial coordinate only. If a one-dimensional element has two nodes with corresponding displacements u 1 , u 2 and corresponding shape functions N 1 , N 2 , then the displacement function is given by N 1 u 1 +N 2 u 2 .

3. In FEM, what is the name of the element specified by a polynomial of order two or more?

a) Nonlinear element

b) Higher-order element

c) Linear element

d) Master element

Answer: b

4. In FEM, what is the name of the shape function of an Euler-Bernoulli beam element?

a) Hermite cubic interpolation function

b) Lagrange cubic interpolation function

c) Consistent element functions

d) Quadratic interpolation functions

Answer: a

Explanation: Interpolation function of a beam element is continuous with nonzero derivatives up to order two. It is derived by interpolating the displacement polynomial as well as its derivative at the nodes. Such interpolation functions are called as Hermite cubic interpolation function. The Lagrange cubic interpolation Functions are derived by interpolating the displacement polynomial but not its derivatives.

5. In FEM, which option is used to develop the Higher-order rectangular elements systematically?

a) A rectangular array of binomial coefficients

b) Galerkin method

c) Jacobi method

d) Delaunay triangulation

Answer: a

Explanation: Analogous to the Lagrange family of triangular elements, the Lagrange family of rectangular elements can be developed from a rectangular array of binomial coefficients. Since a linear rectangular element has four corners , the polynomial should have the first four terms 1, x, y, and xy

6. In FEM, what are the elements in which the same shape functions describe the geometry and field displacement variables?

a) Iso-parametric

b) Axi-Symmetric

c) Super-parametric

d) Sub-parametric

Answer: a

Explanation: In sub-parametric formulations, the geometry is represented by elements of a lower order than those used to approximate the dependent variable. An example of this category is provided by the beam element. In iso-parametric formulations , the same element is used to approximate the geometry as well as the dependent unknowns. In super-parametric formulations, the geometry is represented by elements of a higher order than those used to approximate the dependent variables.

7. In the Finite Element Method , if the geometry is represented by elements of a higher order than those used to approximate the field displacement variables, then it is called super-parametric formulation.

a) True

b) False

Answer: a

Explanation: In super-parametric formulations, the geometry is represented by elements of a higher order than those used to approximate the dependent variables. In sub-parametric formulations, the geometry is represented by elements of a lower order than those used to approximate the dependent variable. The beam element provides an example of this category. In iso-parametric formulations , the same element is used to approximate the geometry as well as the dependent unknowns.

8. Which option is not correct about the Lagrange rectangular element used in FEM?

a) Second-order Lagrange element has nine nodes

b) Zero-order Lagrange element has one node

c) First-order Lagrange element has four nodes

d) Third-order Lagrange element has fifteen nodes

Answer: d

Explanation: In general a p th -order Lagrange rectangular element has n nodes with n= 2 , where p=0,1,2 …

For p=0,

n=1.

For p=1,

n=2 2 ,

=4.

For p=2,

n=3 2

=9.

For p=3,

n=4 2

=16.

9. Which option is not correct about the Lagrange family of triangular elements used in FEM?

a) 2 nd -degree polynomial corresponds to 6 noded triangle

b) 0 th -degree polynomial corresponds to 1 noded triangle

c) 1 st -degree polynomial corresponds to 3 noded triangle

d) 3 rd -degree polynomial corresponds to 9 noded triangle

Answer: d

Explanation: A p th degree polynomial corresponds to n noded triangular element with n=0.5, where n=0, 1, 2 …

For p=0,

n=1.

For p=1,

n=0.5*3*2,

=3.

For p=2,

n=0.5*3*4

=6.

For p=3,

n=0.5*4*5

=10.

10. In FEM, if x, y represents Cartesian coordinates, then which term of Pascal’s triangle corresponds to the position of the interior node of the following element?

Pascal's triangle corresponding to position of interior node of 9 noded Lagrange rectangle

a) 1

b) xy

c) x 2 y

d) xy 2

Answer: b

Explanation: The interior node of the nine noded Lagrange rectangular element is the central term in the corresponding parallelogram on Pascal’s triangle. The central term is the middle term of the 3 rd row. As the 3 rd row contains three terms viz. x 2 , xy and y 2 in the same order, the interior node corresponds to the term xy.

11. What is the number of nodes present on the boundary of the Lagrange quadratic rectangular element used in FEM?

a) 1

b) 4

c) 8

d) 9

Answer: c

Explanation: The Lagrange quadratic rectangular element has nine regularly spaced nodes. Four nodes are paced at the four corners, four at midpoints of the sides, and one at the center of the element. Thus, a total of 8 boundary nodes are present. Its associated polynomial has a total of nine terms, including the second degree and third-degree terms.

12. Which nodes of the higher-order elements of the Lagrange family do not contribute to the inter-element connectivity?

a) Interior nodes

b) All nodes

c) Corner nodes

d) Intermediate nodes

Answer: a

Explanation: Since the Interior nodes of the higher-order elements of the Lagrange family do not contribute to the inter-element connectivity, they can be condensed out at the element level so that the size of the element matrices is reduced. The elements formed after removing the Interior nodes are called serendipity elements.

13. What is the reason for an element in the Serendipity family to have a smaller size of stiffness matrix compare to a similar element in the Lagrange family?

a) Absence of interior nodes

b) Modified element connectivity

c) Lesser interpolation functions

d) Irregular arrangements of nodes

Answer: a

Explanation: The internal nodes of the higher-order elements of the Lagrange family do not contribute to inter-element connectivity, and hence they are condensed out at element level; as a result, the size of the element matrices is reduced. However, the element connectivity remains unaffected. The elements formed after removing the internal nodes are called serendipity elements.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “ Modelling Considerations”.

1. In mathematical modeling of a process, which option is not a characteristic of an analytical solution?

a) Mathematical equations are used to describe a process

b) Most practical problems cannot be solved

c) Exact information on the quantities of interest is obtained

d) Finite element method is used

Answer: d

Explanation: In the development of a mathematical model, we derive the mathematical relationships governing a system. The mathematical model is often in the form of differential equations. If the relationships are simple, it is possible to obtain exact information on the quantities of interest, this is known as the analytical solution. The finite element method is not used in an analytical solution. Most practical problems are too complicated to allow analytical solutions.

2. In mathematical modeling of a process, which option is not a characteristic of a numerical solution method?

a) It does not give exact information on the quantities of interest

b) A set of assumptions are made about the process

c) Applicable to simple problems only

d) Finite element method is used

Answer: c

Explanation: In the development of a mathematical model, we often make a set of assumptions about the process to derive the mathematical relationships governing the system. A numerical method, for example, the finite element method, gives an approximate solution to a given problem. Most practical problems are solved using the numerical solution method.

3. In FEM, which option is not correct with respect to the generation of element geometries?

a) It involves coordinate transformation

b) It is based on a one-to-many mapping

c) A point in actual element maps uniquely to master element

d) Tiny interior angles are avoided

Answer: b

Explanation: The numerical evaluation of integrals over actual elements involves a coordinate transformation from the actual element to a master element. The transformation is acceptable if and only if every point in the actual element is mapped uniquely into a point in the master element, and vice versa. Such mappings are termed one-to-one. To avoid numerical failure of matrices, the interior angle at each vertex of a triangular element should be reasonably larger than 0°.

4. In the generation of element geometries, if dx dy represents an area element in the real element and dεdη represents the corresponding area element in the master element, then what is the expression for Jacobian j e ?

b) \

\(\frac{1}{ }\)

Answer: b

Explanation: The numerical evaluation of integrals over actual elements involves a one-to-one mapping between the actual element and the master element. This requirement can be expressed as j e >0, where j e is the Jacobian matrix. J e represents the ratio of an area element in the real element to the corresponding area element in the master element.

5. In the generation of element geometries, for what value of j e , the element geometries lie within the limits of acceptable distortion?

a) j e >0

b) j e <0

c) j e =0

d) Any real value of j e

Answer: a

Explanation: If j e is zero, then a nonzero area element in the real element is mapped into zero areas in the master element, which is unacceptable. Also, if j e <0, a right-handed coordinate system is mapped into a left-handed coordinate system. In order to keep the element geometries within acceptable distortion, the j e must be greater than zero, and excessive distortion is avoided.

6. Which of the following finite element geometries contains unacceptable vertex angles, in practice?

Find the finite element geometries containing unacceptable vertex angles in practice

a) 1 and 2

b) 2 and 4

c) Only 3

d) 1, 3 and 5

Answer: d

Explanation: The element geometries are kept within acceptable distortion and hence, excessive distortion is avoided. Some geometric shapes of real elements are also avoided. For example, the interior angle at each vertex of a triangular element should not be equal to either 0° or 180°. Indeed, in practice, the angle should reasonably be larger than 0 0 and smaller than 180° to avoid numerical ill-conditioning of element matrices. 1 and 5 contain too large angles, whereas 3 contains a too small angle.

7. For a finite element mesh to be valid, which option is not true regarding the elements used?

a) The number of elements used is not exact

b) Elements can be of different orders

c) Elements can be of different types

d) The choice of elements and mesh is problem-independent

Answer: d

Explanation: A valid mesh can be coarse or refined, and may consist of one or more orders and types of elements. A judicious choice of element order and type could save computational cost while giving accurate results. It should be noted that the choice of elements and mesh is problem-dependent.

8. A finite element mesh is refined by subdividing existing elements into two or more elements of the same type, what is the name of such a refinement?

a) h-version mesh refinement

b) p-version mesh refinement

c) h, p-version mesh refinement

d) p, h-version mesh refinement

Answer: a

Explanation: Refining a mesh by subdividing its existing elements into two or more elements of the same type is called the h-version mesh refinement. Alternatively, in p-version mesh refinement, the existing elements can be replaced by elements of a higher order. The h, p-version mesh refinement, in which the elements are subdivided into two or more elements in some places and replaced with higher-order elements in other places.

9. In local mesh refinement, if very small elements are placed adjacent to very large ones, then the mesh becomes unacceptable.

a) True

b) False

Answer: a

Explanation: Generally, local mesh refinements should be such that very small elements are not placed adjacent to very large ones. A mesh can be coarse or refined , and may consist of one or more orders and types of elements .

10. In FEM, which option is the most realistic representation of the actual force between deformable bodies?

a) Point load

b) Uniformly varying load

c) Sine distribution

d) Uniformly distributed load

Answer: c

Explanation: A situation where the representation of boundary forces is subject to different interpretations is found when the force is due to contact between two bodies. For example, a solid plate in contact with a circular disc generates a reactive force that can be represented either as a point load or as a locally distributed force. A sine distribution might be a more realistic representation of the actual force between deformable bodies.

11.For the following element, what is the value of a+b such that the Jacobian J=\

+b-2] is positive, where are natural coordinates?

Find the value of a+b such that the Jacobian J is positive

a) >2

b) <2

c) =2

d) 0

Answer: a

Explanation: The point corresponds to in natural coordinates. Thus = .

J=\

+b-2]

-4]

-1

Given j>0,

-1>0

>2.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Plane Elasticity – Governing Equations”.

1. In solid mechanics, what does linearized elasticity deal with?

a) Small deformations in linear elastic solids

b) Large deformations in linear elastic solids

c) Large deformations in non-Hookean solids

d) Small deformations in non-Hookean solids

Answer: a

Explanation: The part of solid mechanics that deals with stress and deformation of solid continua is called Elasticity. Linearized elasticity is concerned with small deformations in linear elastic solids or Hookean solids

2. For plane elasticity problems in three dimensions, which option is not responsible for making the solutions independent of one of the dimensions?

a) Geometry

b) Boundary conditions

c) Externally applied loads

d) Material

Answer: d

Explanation: Elasticity is the part of solid mechanics that deals with stress and deformation of solid continua. There is a class of problems in elasticity whose solution is not dependent on one of the coordinates because of their geometry, boundary conditions, and externally applied loads. Such problems are called plane elasticity problems.

3. For a plane strain problem, which strain value is correct if the problem is characterized by the displacement field u x =u x , u y =u y and u z =0?

a) ε xy =0

b) ε xz =0

c) ε yz ≠0

d) ε xz ≠0

Answer: b

Explanation: The plane strain problems are characterized by the displacement field u x =u x , u y =u y and u z =0, where (u x , u y , u z ) denote the components of this displacement vector u in the coordinate system. The displacement field results in the following strain field:

\(\epsilon_{xz}=\epsilon_{yz}=\epsilon_{zz}=0, \epsilon_{xx}=\frac{\partial u_x}{\partial x}, 2\epsilon_{xy}=\frac{\partial u_x}{\partial y}+\frac{\partial u_y}{\partial x}\) and \(\epsilon_{yy}=\frac{\partial u_y}{\partial y}\).

4. Underplane strain condition, what is the value of ε yy if the problem is characterized by the displacement field u x =2x+3y, u y =5y 2 , and u z =0?

a) 10y

b) 5y

c) 3

d) 0

Answer: a

Explanation: The plane strain problems are characterized by the displacement field u x =u x , u y =u y and u z =0, where (u x , u y , u z ) denote the components of the displacement vector u in the coordinatesystem. The displacement field results in the following strain field:

Thus, \(\epsilon_{yy}=\frac{\partial 5y^2}{\partial y}\)

=5*\(\frac{\partial y^2}{\partial y}\)

=5*2y

=10y.

5.For a plane strain problem, the relation between stress and strain components for an orthotropic material is σ=Cε. Which option is the correct structure of the matrix C?

a) \

Answer: a

Explanation: For an orthotropic material under plane strain, with principal material axes (x 1 , x 2 , x 3 ) coinciding with the coordinates, the relation between stress and strain components is \(

\) where C is the elastic stiffness matrix. The state of stress is σ xz =σ yz =0 and \

\).

6. For an orthotropic material, if E and v represent Young’s modulus and the poisons ratio, respectively, then what is the value of v 12 if E 1 =200 Gpa, E 2 =160 Gpa and v 21 =0.25?

a) 0.3125

b) 0.05

c) 0.2125

d) 0.3

Answer: a

Explanation: For an orthotropic material, E 1 and E 2 are the principal

moduli in the x and y directions, respectively. The poisons ratio and Young’s moduli are related by the equation

v 12 =v 21 \(\frac{E_1}{E_2}\).

v 12 =0.25*\(\frac{200}{160}\)

=0.25*1.25

=0.3125.

7. Under plane stress condition in the XYZ Cartesian system, which stress value is correct if a problem is characterized by the stress field σ xx =σ xx , σ yy =σ yy and σ zz =0?

a) σ xy =0

b) σ yx ≠0

c) σ zx ≠0

d) σ yz ≠0

Answer: b

Explanation: A state of plane stress in XYZ Cartesian system is defined as one in which the following stress field exists:

σ xz =σ yz =σ zz =0, σ xx , σ xy =σ xy and σ yy =σ yy .

Thus, σ xx , σ xy and σ yy are non-zero stresses. Such a problem in three dimensions can be dealt with as a two-dimensional problem.

8. For theplane stress problem in XYZ Cartesian system, σ xx =σ xx , σ yy =σ yy and σ zz =0, which option is correct regarding the associated strain field?

a) ε xx =0

b) ε yx =0

c) ε zx =0

d) ε yy =0

Answer: c

Explanation: The strain field associated with the given stress field has the form ε=Sσ, where the matrix S is a symmetric matrix, and it is called elastic compliances matrix. In the XYZ Cartesian system, all the strain components except ε yz and ε zx are non-zero. Thus, ε xx ≠0, ε yy ≠0, ε zz ≠0, ε xy ≠0, where as ε yz =0 and ε zx =0.

9. For any two cases of plane elasticity problems, if the constitutive equations are different, then their final equations of motion are also different.

a) True

b) False

Answer: a

Explanation: The equations of motion for plane elasticity problems are given by D*σ+f=ρü in the vector form, where f denotes body force vector, σ is the stress vector, u is displacement vector, D is a matrix of the differential operator, and ρ is the density. Note that the equations of motion of plane stress and plane strain cases differ from each other only on account of the difference in their constitutive equations.

10. In solid mechanics, which option is not a characteristic of a plane stress problem in the XYZ Cartesian system?

a) One dimension is very small compared to the other two dimensions

b) All external loads are coplanar

c) Strain along any one direction is zero

d) Stress along any one direction is zero

Answer: c

Explanation: An example of a plane stress problem is provided by a plate in the XYZ Cartesian system that is thin along the Z-axis. It is acted upon by external loads lying in the xy plane that are independent of the Z coordinate. Thus, stresses and strains are observed in all directions except that the stress is zero along the Z-axis.

11. In solid mechanics, what is the correct vector form of the equations of motion for a plane elasticity problem?

a) D*σ+f=ρü

b) D*σ+f=ρu̇

c) D 2 *σ+f=ρü

d) D*σ+f=ρu

Answer: a

Explanation: For plane elasticity problems, the equations of motion are one of the governing equations. The vector form of equations of motion is D*σ+f=ρü, where f denotes body force vector, σ is the stress vector, u is the displacement vector, D is a matrix of differential operator and ρ is the density.

Answer: b

Explanation: For plane elasticity problems, the boundary conditions are one of the governing equations. There are two types of boundary conditions, namely, essential boundary conditions and natural boundary conditions. The equation t x ≡σ xx n x +σ xy n y represents natural boundary condition or Neumann boundary condition.

Sanfoundry Global Education & Learning Series – Finite Element Method.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Plane Elasticity – Weak Formulations”.

1. In FEM, which method is not used to construct the weak forms and associated finite element model of the plane elasticity equations?

a) Principle of virtual displacements

b) The principle of minimum total potential energy

c) Weak form of governing differential equations

d) Hamiltonian principle

Answer: d

Explanation: There are two different ways of constructing the weak forms and associated finite element model of the plane elasticity equations. The first one uses the principle of virtual displacements , while the second approach follows a three-step procedure to obtain a weak form of governing differential equations.

2. In constructing the weak forms of plane elasticity problems, which option is not related to the principle of virtual displacements?

a) Displacements to strains

b) Strains to stresses

c) The equations of motion

d) Body forces

Answer: d

Explanation: Among the ways of constructing the weak forms and associated finite element model of the plane elasticity equations, the principle of virtual displacements is expressed in terms of matrices relating displacements to strains, strains to stresses, and the equations of motion. This approach is used in most finite element texts on solid mechanics.

3. What is the correct form of the principle of virtual displacements applied to plane finite elastic element If V e is the volume of element and s e is its surface?

a) 0=\(\int_{V_e}\)(σ ij δε ij +ρü i δu i )dV-\(\int_{V_e}\)f i δu i dV-∮ s e \(\hat{t_i}\)δu i ds

b) 0=\(\int_{V_e}\)(σ ij δε ij +ρu̇ i δu i )dV-\(\int_{V_e}\)f i δu i dV-∮ s e \(\hat{t_i}\)δu i ds

c) 0=\(\int_{V_e}\)(σ ij δε ij +ρü i δu i )dV+\(\int_{V_e}\)f i δu i dV-∮ s e \(\hat{t_i}\)δu i ds

d) 0=\(\int_{V_e}\)(σ ij δε ij +ρu̇ i δu i )dV\(\int_{V_e}\)f i δu i dV+∮ s e \(\hat{t_i}\)δu i ds

Answer: a

Explanation:The vector form of the principle of virtual displacements applied to plane finite elastic element with volume, V e and surface, s e is 0=\(\int_{V_e}\)(σ ij δε ij +ρü i δu i )dV-\(\int_{V_e}\)f i δu i dV-∮ s e \(\hat{t_i}\) δu i ds, where “δ” denotes the variational operator, (σ ij and ε ij are the components of stress and strain tensors, respectively, and f i and t i are the components of the body force and boundary stress vectors, respectively.

4. In the weak formulation of the plane elasticity equations, even though the methods, the principle of virtual displacements and the three-step weak formulation, give, mathematically different finite element models, they are the same in their algebraic forms.

a) True

b) False

Answer: b

Explanation: There are two different ways of constructing the weak forms and associated finite element model of the plane elasticity equations. The first one is the principle of virtual displacements, while the second is a three-step procedure to obtain a weak form of governing differential equations. Of course, both methods give, mathematically, the same finite element model, but differ in their algebraic forms.

5. In the weak form of the principle of virtual displacements, 0=\(\int_{V_e}\)(σ ij δε ij +ρü i δu i )dV-\(\int_{V_e}\)f i δu i dV-∮ s e t̂ i δu i ds , applied to plane finite elastic element,which term corresponds to virtual strain energy stored in the body?

a) \(\int_{V_e}\)(σ ij δε ij )dV

b) ∮ s e t̂ i δu i ds

c) \(\int_{V_e}\)(ρü i δu i )dV

d) \(\int_{V_e}\)f i δu i dV

Answer: a

Explanation: There are four terms in the vector form of the principle of virtual displacements applied to plane finite elastic element, 0=\(\int_{V_e}\)(σ ij δε ij +ρü i δu i )dV-\(\int_{V_e}\)f i δu i dV-∮ s e t̂ i δu i ds. The first term in the equation corresponds to the virtual strain energy stored in the body.The second term deals with the kinetic energy stored in the body; the third term represents the virtual work done by the body force, and the fourth term represents the virtual work done by the surface traction.

6. In the weak form of the principle of virtual displacements applied to a plane elastic finite element, what does the term \(\int_{V_e}\)(ρü i δu i )dV correspond to?

a) Virtual strain energy

b) Kinetic energy

c) Virtual work done by the body force

d) Virtual work done by the surface traction

Answer: b

Explanation: Thefour terms in the vector form of the principle of virtual displacements are present in the equation, 0=\(\int_{V_e}\)(σ ij δε ij +ρü i δu i )dV-\(\int_{V_e}\)f i δu i dV-∮ s e t̂ i δu i ds. The given term corresponds to the kinetic energy stored in the body, the third term represents the virtual work done by the body force, and the fourth term represents the virtual work done by the surface traction.

7. Under plane elasticity, which force is responsible for doing the virtual work \(\int_{V_e}\)f i δu i dV in the weak form of the principle of virtual displacements?

a) Body force

b) Concentrated loads

c) Surface traction force

d) Pressure force

Answer: a

Explanation:Of the four terms present in the vector form of the principle of virtual displacements applied to plane finite elastic element, 0=\(\int_{V_e}\)(σ ij δε ij +ρü i δu i )dV-\(\int_{V_e}\)f i δu i dV-∮ s e t̂ i δu i ds, the first term one corresponds to the virtual strain energy stored in the body. The second term corresponds to the kinetic energy stored in the body. The given term is the third term in the equation, and that represents the virtual work done by the body force.

8. For a plane elasticity problem, which term in the weak form of the principle of virtual displacements is affected by a change in the applied surface traction forces?

a) \(\int_{V_e}\)(σ ij δε ij )dV

b) ∮ s e t̂ i δu i ds

c) \(\int_{V_e}\)(ρü i δu i )dV

d) \(\int_{V_e}\)f i δu i dV

Answer: b

Explanation:The vector form of the principle of virtual displacements applied to the plane finite elastic element consists of four terms. Two terms correspond to energy stored in the body. The term, ∮ s e t̂ i δu i ds represents the virtual work done by the surface traction forces, and thus, it is affected by a change in the applied surface traction forces.

9. If U,V denotes the components of the displacement vector, then which option is the correct primary nodal degrees of freedom present in the following figure of a plane elasticity problem?

Find the primary nodal degrees of freedom present in the figure of a plane elasticity problem

a) U 1 =V 1 =0

b) U 2 ≠0, V 2 =0

c) U 2 =0, V 2 ≠0

d) U 1 =V 1 ≠0

Answer: a

Explanation: A fixed connection implies that all components of displacement are zero, whereas roller support indicates that the displacement normal to the wall is zero. In the given problem, U and V denote the horizontal and vertical displacements, respectively, at the global i th node of the mesh. The specified primary degrees of freedom include U 1 =V 1 = U 10 =V 10 =0, U 2 ≠0 and V 2 ≠0.

10. If F x , F y denotes the components of the force vector, then which option is the correct secondary degrees of freedom present in the following figure of a plane elasticity problem?

Find the secondary degrees of freedom present in the figure of a plane elasticity problem

a) \

\(F_7^x\ne F_7^y\)=0

Answer: d

Explanation: The procedure to calculate the nodal forces is the same as that used for the calculation of nodal sources single-variable problems, except that the nodal values must be decomposed into the x and y components. Since the distributed force is along the x coordinate, all nodal computed nodal forces are along the x coordinate. The specified secondary degrees of freedom include \(F_6^x=F_6^y=F_7^y=0\) But

\(F_7^x\)≠0.

11. For plane elasticity problems, which option represents the essential boundary conditions among the governing equations?

a) Displacements, u x and u y at the boundary

b) Surface traction at the boundary

c) The displacements (u x and u y ) and surface traction at the boundary

d) Stresses in the element

Answer: a

Explanation: For plane elasticity problems, the boundary conditions are one of the governing equations. There are two types of boundary conditions, namely, essential boundary conditions and natural boundary conditions. The displacements specified in the problem are the essential boundary condition or Dirichlet boundary condition.

12. If the equation ∫ Ω c h e \((\frac{\partial w_1}{\partial x}\sigma_{xx}+\frac{\partial w_1}{\partial y}\)σ xy -w 1 f x +ρw 1 \

\)dxdy-&oint; Γ c h e w 1 (σ xx n x +σ xy n y )ds=0 represents the weak form of plane elasticity equations, then the weight functions w 1 and w 2 are the first variations of u x and u y , respectively.

a) True

b) False

Answer: a

Explanation: The equations ∫ Ω c h e \((\frac{\partial w_1}{\partial x}\sigma_{xx}+\frac{\partial w_1}{\partial y}\)σ xy -w 1 f x +ρw 1 \

\)dxdy-&oint; Γ c h e w 1 (σ xx n x +σ xy n y )ds=0 and ∫ Ω c h e \((\frac{\partial w_2}{\partial x}\sigma_{xy}+\frac{\partial w_2}{\partial y}\)σ yy -w 2 f y +ρw 2 \

\)dxdy-&oint; Γ c h e w 2 (σ xy n x +σ yy n y )ds=0 represents the weak forms of plane elasticity equations, where Ω is the area of cross-section of the domain, ┌ is a portion of element boundary, and the weight functions w1 and w2 are the first variations of u x and u y respectively.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Plane Elasticity – Finite Element Model”.

1. If only the first derivatives of u x and u y appear in the weak forms, then their interpolation must be at least bilinear.

a) True

b) False

Answer: a

Explanation: An examination of the weak form, 0=∫ Ω e h e \

+ c_{66}\frac{\partial w_1}{\partial y}

\)+ρw 1 u x ]dxdy-∫ Ω e h e w 1 f x dxdy-∮ ┌ d h e w 1 t x ds reveals that only first derivatives of u x and u y with respect to x and y appear respectively. Therefore, u x and u y must be approximated by the Lagrange family of interpolation functions, and at least a bilinear interpolation is required.

2. In FEM, if two independent variables are components of the same vector, then they can be approximated by two different types of interpolations.

a) True

b) False

Answer: b

Explanation: u x and u y are the primary variables in the expanded weak forms of the plane elasticity problems. Although u x and u y are independent of each other, they are the components of the displacement vector. Therefore, both components should be approximated using the same type and degree of interpolation.

3. What can one conclude about the displacement components u x and u y in the finite element model of the plane elasticity equations?

a) They are primary variables and must be carried as primary nodal degrees of freedom

b) They are secondary variables and must be carried as primary nodal degrees of freedom

c) They are primary variables and must be carried as secondary nodal degrees of freedom

d) They are secondary variables and must be carried as secondary nodal degrees of freedom

Answer: b

Explanation: An examination of the weak form, 0=∫ Ω e h e \

+ c_{66}\frac{\partial w_1}{\partial y}

\)+ρw 1 u x ]dxdy-∫ Ω e h e w 1 f x dxdy-∮ ┌ d h e w 1 t x ds reveals the following:

u x and u y are the primary variables, which must be carried as the primary nodal degrees of freedom.

only first derivatives of u x and u y with respect to x and y, respectively, appear.

4. Which interpolation functions must be used for the primary variables in weak forms of plane elasticity equations?

a) Hermite family interpolation function

b) Lagrange family interpolation function

c) Hierarchical interpolation function

d) Quadratic interpolation function

Answer: b

Explanation: An examination of expanded weak forms of the plane elasticity problems reveals that the variables u x and u y are the primary variables, and only first derivatives of u x and u y with respect to x and y appear, respectively. Therefore, u x and u y must be approximated by the Lagrange family of interpolation functions, and at least bilinear interpolation is required.

5. What are the simplest elements that fit for the finite element model of the plane elasticity equations?

a) Linear triangular and quadratic quadrilateral elements

b) Quadratic triangular and linear quadrilateral elements

c) Linear triangular and linear quadrilateral elements

d) Quadratic triangular and quadratic quadrilateral elements

Answer: c

Explanation: Because the first-derivatives of the primary variables, u x and u y with respect to x and y, respectively, appear in the expanded weak forms of the plane elasticity problems, they must be approximated by the Lagrange family of interpolation functions, with at least a bilinear interpolation. The simplest elements that satisfy those requirements are the linear triangular and linear quadrilateral elements.

6. What is the correct statement regarding the shape function S of a linear triangular element?

a) The first derivatives of S and hence, all the strains are element-wise constant

b) The first derivatives of S are linear, but the strains are element-wise constant

c) The first derivatives of S and hence the strains are linear functions

d) The first derivatives of S are element-wise constant, but the strains are linear

Answer: a

Explanation: A linear triangular element has two degrees of freedom, u x and u y per node and a total of six nodal displacements per element. Since the shape functions are linear, their first derivatives are element-wise constant, and hence, all the strains computed for the linear triangular element are element-wise constant.

7. What is the other name of a linear triangular element for plane elasticity problems?

a) Constant-strain triangle

b) Linear-strain triangle

c) Quadratic-strain triangle

d) Variable-strain triangle

Answer: a

Explanation: In a linear triangular element , there are two degrees of freedom, u x and u y per node and a total of six nodal displacements per element. Since the first derivatives of shape functions for a triangular element are element-wise constant, all the strains computed for the linear triangular element are element-wise constant. Therefore, the linear triangular element for a plane elasticity problem is known as the constant-strain triangular element.

8. What is the function on \

denotes the shape function of a linear quadrilateral element?

a) Linear in e and constant in n

b) Constant in e and Linear in n

c) Linear in both e and n

d) Constant in both e and n

Answer: a

Explanation: For a quadrilateral element, the first derivatives of the shape function are not constant. \(\frac{\partial s}{\partial n}\) is linear in e and constant in n whereas \(\frac{\partial s}{\partial e}\) is linear in n and constant in e. Also, the first derivatives of shape functions for a triangular element are constant element wise, and hence all the strains computed are constant element wise.

9. For a linear triangular element, what is the order of matrix B in the strain-displacement relation ε=BD, where D denotes the displacement matrix?

a) 6×3

b) 3×6

c) 3×8

d) 8×3

Answer: b

Explanation: For plane elasticity problems the strain-displacement relation is given by ε=BD, where ε=[ε xx ε yy ε xy ] T , B=\(

\) and D=\(

\) T . The order of matrix B is 3x2n, where n is the number of nodes in the element. A linear triangular element has three nodes, thus n=3.

Order of B is 3×2*3

=3×6.

10. For a linear quadrilateral element, what is the order of matrix B in the strain-displacement relation ε=BD, where D denotes the displacement matrix?

a) 6×3

b) 3×6

c) 3×8

d) 8×3

Answer: c

Explanation: For plane elasticity problems the strain-displacement relation is given by ε=BD, where ε=[ε xx ε yy ε xy ] T , B=\(

\) and D=\(

\) T . The order of matrix B is 3x2n, where n is the number of nodes in the element. A linear quadrilateral element has four nodes, thus n=4.

Order of B is 3×2*4

=3×8.

11. Which option is not correct with respect to the orders of the matrices in the following finite element model of plane elastic equations?

M e \(\ddot{\Delta}^e\)+K e Δ e =F e +Q e

a) Mass matrix M has order 2n x 2n

b) Stiffness matrix K has order 2n x n

c) The element load vector F has order 2n x 1

d) The vector of internal forces Q has order 2n x 1

Answer: b

Explanation: For the given vector form of finite element model of plane elastic equations, the element mass matrix M and stiffness matrix K are of the order 2n x 2n, the element load vector F and the vector of internal forces Q are of the order 2n x 1, where n is the number of nodes in a Lagrange finite element.

12. Which form of a periodic solution is sought for the natural vibration study of plane elastic bodies?

a) {Δ}={Δ 0 }e iωt

b) {Δ}={Δ 0 }e -iω

c) {Δ}={Δ 0 }e -iωt

d) {Δ}={Δ 0 }e -iω/t

Answer: c

Explanation: For natural vibration study of plane elastic bodies, we seek a periodic solution of the form {Δ}={Δ 0 }e -iωt , where Δ denotes the displacements, ω is the frequency of natural vibration and i=\(\sqrt{-1}\). With this, the finite element models of plane elastic problems reduce to an eigen value problem (-ω 2 M e +K e )\(\Delta_0^e\)=Q e .

13. Which option is correct for the first derivative of the shape function S in the study of plane elasticity problems?

a) It is element-wise constant for triangular element whereas not a constant for quadrilateral element

b) It is element-wise constant for both the triangular element as well as a quadrilateral element

c) It is not a constant for triangular element whereas element-wise constant for quadrilateral element

d) It is a combination of a linear function and constant for both the triangular element and a quadrilateral element

Answer: a

Explanation: In the study of plane elasticity problems, the first derivatives of the shape function S for a triangular element are element-wise constant, whereas, for a quadrilateral element, they are not constant. Notably, \(\frac{\partial s }{\partial n}\) is linear in e and constant in n whereas \(\frac{\partial s }{\partial e}\) is linear in n and constant in e.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Plane Elasticity – Evaluation of Integrals”.

1. In the Finite Element Method, which expression is correct for a linear triangular element if S is the shape function, A e is its area, and K is a constant?

a) \

\(\frac{\partial S}{\partial x}\)=KA e

d) \(\frac{\partial S}{\partial y}\)=KA e 2

Answer: a

Explanation: For a linear triangular element, the shape function \

\), \(\frac{\partial \psi_i^e}{\partial x}=\frac{\beta_i^e}{2A_e}\) and \(\frac{\partial \psi_i^e}{\partial y}=\frac{\gamma_i^e}{2A_e}\)where A e is the area of the element, α, β and γ are constants. Note that the derivatives of the shape function are constants.

2. In Finite Element Analysis, what is the correct load vector for a linear triangular element with area A e , thickness h e and uniform body force vector f?

a) \

\(\frac{h_e}{4A_e}\)f

Answer: b

Explanation: For a linear triangular element, for the case in which the body force is uniform and thus the body force components fx and fy are element-wise constant

, the load vector F has the form F=∫ Ω c h e (ψ e ) T \(f_0^e\)dx

=\(\frac{A_e h_e}{4}

\). The internal load vector Q is computed only when the element falls on the boundary of the domain on which tractions are specified.

3. In Finite Element Analysis, what is the correct load vector for the linear quadrilateral element with area A e , thickness h e and uniform body force vector f?

a) \

\(\frac{h_e}{4A_e}\)f

Answer: a

Explanation: For a linear quadrilateral element,for the case in which the body force is uniform and thus the body force components are element-wise constant

, the load vector F has the form F=\

^T f_0^edx\) =\(\frac{A_e h_e}{4}

\). The internal load vector Q is computed only when the element falls on the boundary of the domain on which tractions are specified.

4. In the Finite Element Method, the vector of internal forces is computed only when the element falls on the boundary of the domain on which tractions are absent.

a) True

b) False

Answer: b

Explanation: In Finite Element Analysis, internal load vector Q is computed only when the element falls on the boundary of the domain on which tractions are specified . Computation of Q involves the evaluation of line integrals . In practice, it is convenient to express the surface traction t in the element coordinates. In that case, Q can be evaluated in the element coordinates and then transformed to the global coordinates for assembly.

5. Which option is not correct concerning the internal load vector in the finite element model of plane elasticity problems?

a) It is computed at all the nodes interior of the element

b) It is computed only when the element falls on the boundary of the domain on which tractions are known

c) Its computation doesn’t involve evaluation of line integrals for any type of element

d) It is evaluated in global coordinates but not in element coordinates

Answer: b

6. In transformations, what is the transformation matrix R in the relation F=RQ if the load vector in global coordinates is F and the load vector in element coordinates is Q?

a) \

Answer: a

Explanation: In practice, it is convenient to express the surface traction T in the element coordinates. In that case, the element load vector can be evaluated in the element coordinates and then transformed to the global coordinates for assembly. If Q denotes the element load vector referred to the element coordinates, then the corresponding load vector referred to the global coordinates is given by

F=R T Q, where R is the transformation matrix R=\(

\) and α is the angle between the global x-axis and the traction vector T.

7. What is the global load vector in Finite Element Analysis of the following structure if the local load vector is \(

\) and θ=0?

Find the global load vector in Finite Element Analysis of the structure

a) \

Answer: a

Explanation: If Q denotes the element load vector referred to the element coordinates then the corresponding load vector referred to the global coordinates is given by F=R T Q where R is the transformation matrix R=\(

\) and α is the angle between the global x-axis and the traction vector T.

Here α=90-θ, given θ=0

α=90-0

=90.

Cos90=0 and sin90=1.

R=\(

F=\(

^T\)x\(

=\(

\).

8. What is the expression for the traction term t n in the element load vector Q e =&oint; ┌ c h e ψ T tds of the following figure where L 23 is the length of the line 2-3?

Find the expression for the traction term in the element load vector of the figure

a) t n =-T

b) t n =T

c) t n =-T

d) t n =T

Answer: b

Explanation: For the element shown in the given figure, the side 2-3 is subjected to linearly varying normal force t n . The traction term on the side 2-3 of the element is t n =-T

\), where the minus sign for T accounts for the direction of the applied traction. Traction is acting towards the body in the present case. The local coordinate system s used in the above expression is chosen along the side connecting node 2 to node 3, with its origin at node 2. We are not restricted to this choice.

9. In Finite Element Analysis, what are the values of nodal forces in the following element if the line 2-4 is 160 in long?

Find the values of nodal forces in element - Finite Element Analysis

a) 1600 along both the DOF 3 and 7

b) 800 and 0 along the DOF 3 and 4 respectively

c) 0 and 800 along the DOF 7 and 8 respectively

d) 0 and 800 along the DOF 3 and 4 respectively

Answer: b

Explanation: Consider the thin elastic plate subjected to a uniformly distributed edge load, as shown in the given figure. The specified displacement degrees of freedom for the problem are U 1 =U 2 =0, and the known forces are F 3 = \(\frac{pbh}{2}\), F 4 =0, F 7 =\(\frac{pbh}{2}\) and F 8 =0.

Given p*h=10 and b=160 thus \(\frac{pbh}{2}\)

=\(\frac{10×160}{2}\)

=800.

Thus, the forces along the DOF 3 and 4 are 800 and 0, respectively.

10. In vibration and transient analysis of beams, if the linear acceleration scheme predicts the solution,then it is unstable for the first several time steps, but it eventually becomes stable.

a) True

b) False

Answer: b

Explanation: In the determination of natural frequencies and transient response using plane elements, the time approximation scheme is used. Note that the solution predicted by the linear acceleration scheme is stable for the first several time steps, but it eventually becomes unstable.

11. In Finite Element Analysis, which option is correct for computation of load due to specified boundary stress?

a) Can be computed using a local coordinate system and one-dimensional interpolation functions

b) Can be computed using a local coordinate system but not one-dimensional interpolation functions

c) Cannot be computed using a local coordinate system but one-dimensional interpolation functions can be used

d) Neither a local coordinate system nor one-dimensional interpolation functions can be used

Answer: a

Explanation: In general, the loads due to specified boundary stresses can be computed using an appropriate local coordinate system and one-dimensional interpolation functions. When higher-order elements are involved, the corresponding order of one-dimensional interpolation functions must be used.

12. In the Finite Element Method, which element is known for the slowest convergence?

a) Linear triangular element

b) Quadratic triangular element

c) Linear rectangular elements

d) Quadratic rectangular elements

Answer: a

Explanation: Mesh convergence determines how many elements are required in a finite element model to ensure that the results of an analysis are not affected by varying the size of the mesh. Once a mesh is converged, no change is observed in the results even after changing its density. The linear triangular element mesh has the slowest convergence compared to the quadratic triangular element, linear and quadratic rectangular elements.

This set of Finite Element Method Question Paper focuses on “Plane Elasticity – Assembly, Boundary and Initial Conditions”.

1. From solid mechanics, what is the correct displacement boundary condition for the following plane stress problem of the beam?

Find displacement (u) boundary condition for plane stress problem of beam

a) uy

b) uy

c) ux

≠0

d) ux

=≠0

Answer: b

Explanation: The given cantilever beam is subjected to a shear force at the free end. The other end is supported by roller and hinge support. Because the hinge support restrains translation, ux

=uy

=0. As the roller support restrains translation perpendicular to the base, only ux=0 which implies ux

≠0 and ux

≠0 are incorrect.

2. From solid mechanics, which traction boundary condition is not correct for the following beam of thickness h?

Find traction (t) boundary condition for beam of thickness h

a) ty

b) ty

c) tx=0

d) ty=-hT

Answer: a

Explanation: The given cantilever beam is subjected to a shear force at the free end, thus tx=0 and ty=-hT. The other end is supported by both roller and hinge support. Because the hinge support restrains translation by offering a reactive force along the directions x and y, ty

≠0 and ty

≠0 whereas the roller support restrains translation perpendicular to the base; thus, only tx≠0. The option ty

=0 is incorrect.

3. In Finite Element Analysis of the beam, which primary variable does not belong to the following mesh?

Find primary variable which does not belong to the mesh in Finite Element Analysis of the beam

a) U 9 =0

b) U 19 =0

c) U 10 =0

d) U 20 =0

Answer: c

Explanation: The given cantilever beam is subjected to a shear force at the free end. The other end is supported by roller and hinge support. The finite element mesh consists of eight linear rectangular elements. The node 1, 2, 3… represents the DOF , , … respectively. Since the translation along x is constrained, U 9 =U 19 =U 29 =0. Because of the hinge at node 10, U 20 =0. The roller support doesn’t restrain vertical movement, thus U 10 ≠0.

4. What is the total size of the assembled stiffness matrix of a plane elastic structure such that its finite element mesh has eight nodes and two degrees of freedom at each node?

a) 16×16

b) 8×8

c) 2×2

d) 4×4

Answer: a

Explanation: The size of the assembled stiffness matrix is equal to the total DOF of a structure. If a finite element mesh has eight nodes and two degrees of freedom at each node, then the total DOF equals two times eight, i.e., sixteen. Thus the order of the assembled stiffness matrix is 16×16.

5. What is the element at the index position 3×3 of the assembled stiffness matrix of the following mesh if K 1 =\(

\) and K 2 =\(

\)?

Find the element at the index position 3×3 of the assembled stiffness matrix of the mesh

a) 9

b) 11

c) 13

d) 4

Answer: a

Explanation: The assembled stiff matrix, K of the given mesh, is of the order 8×8. K is a combination of K 1 and K 2 such that K 1 corresponds to the DOF in order 1,2,3,4, 7 and 8. K 2 corresponds to the DOF in the order 3, 4, 5, 6, 7 and 8. Thus, K=\(

\) and the elementin K at the index 3×3 is 6+3

=9.

6. In the Finite Element Method, if two different values of the same degree of freedom are specified at a point, then such point is called as a singular point.

a) True

b) False

Answer: a

Explanation: The points at which both displacement and force degrees of freedom are known or when two different values of the same degree of freedom are specified are called as singular points. In problems with multiple DOF, we are required to decide as to which degree of freedom is known when singular points are encountered.

7. For time-dependent problems in FEA, which variables must be specified for each component of the displacement field problems?

a) The initial displacement and velocity

b) The initial displacement only

c) The final velocity

d) The initial displacement and final velocity

Answer: a

Explanation: Concerning the specification of the displacements and forces in a finite element mesh, in general, only one of the quantities of each of the pairs (u x , t x ) and (u y , t y ) is known at a nodal point in the mesh. For time-dependent problems, the initial displacement and velocity must be specified for each component of the displacement field.

8. What is the magnitude of the force at node 22 if the moment M is replaced by an equivalent distributed force at x=acm?

Find magnitude of force at node 22 if moment M is replaced by equivalent distributed force

a) \

Always zero

c) \

\(\frac{-M}{b}\)

Answer: b

Explanation: To calculate the magnitude, assume that the force causing the moment is linear with y. At node 11, the beam is pushed towards negative x; thus, the effective force at 11 is negative. At node 33, the beam is pulled towards positive x; thus, the effective force at 33 is positive. As node 22 is located at the center, it is neither pushed nor pulled; thus, the effective force at node 22 is always zero.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Governing Equations”.

1. For fluid flows, which expression does not characterize the slow flow of a viscous and incompressible fluid in a closed domain?

a) v.∇v=0

b) μ≠0

c) =0

d) =0

Answer: d

Explanation: The slow flow of a viscous and incompressible fluid in a closed domain is characterized by the following expressions:

Slow : v.∇v=0.

Viscous: μ≠0.

Incompressible: =0.

It is not a uniform flow; thus, ≠0.

2. In mathematical modeling, for which option, a fluid flow can be approximated by a two-dimensional model?

a) One of the dimensions is very small, and there is some flow along with it

b) One of the dimensions is very long, and there is no flow along with it

c) One of the dimensions is very long, and there is some flow along with it

d) The velocity components in two directions vary along the third direction

Answer: b

Explanation: For a fluid flow, assuming that one of the dimensions say, along the z-direction of the domain is very long, and there is no flow along that direction. The velocity components in the other two directions (v x and v y ) do not vary with the z-direction. Under these conditions, the flow can be approximated by a two-dimensional model.

3. For fluid flows obeying conservation of mass, what is the value of k if v=4x+ky denotes the velocity at any point in the flow?

a) -4

b) 4

c) -2

d) 2

Answer: a

Explanation: A flow obeys conservation of mass if \(\frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}\)=0. Comparing v=4x+ky with v=v x +v y , we get v x =4x and v y =ky.

Using the conservation of mass, we get \(\frac{\partial }{\partial x}+\frac{\partial }{\partial y}\)=0

4+k=0

K=-4.

4. The following equation represents the momentum equation for a fluid flow that is approximated by a two-dimensional model. What does k stand for?

ρ\

-\frac{\partial}{\partial y}[k

]+\frac{\partial P}{\partial x}\)-f x =0

a) Thermal conductivity

b) Fluid viscosity

c) Density

d) Pressure

Answer: b

Explanation: By using constitutive relations the momentum equation is expressed as ρ\

-\frac{\partial}{\partial y}[k

]+\frac{\partial P}{\partial x}\)-f x =0, where v x , v y are the velocity components, P is the pressure, k is the viscosity, f x is the component of the body force vector, and ρ is the density.

5. If a finite element model involves the natural and direct formulation of momentum and continuity equations, then it is known as the velocity-pressure formulation or mixed formulation.

a) True

b) False

Answer: a

Explanation: There are two different finite element models of momentum and continuity equations. The first one is a natural and direct formulation of momentum and continuity equations. It is known as the velocity-pressure formulation or mixed formulation. The second model is based on the interpretation that the continuity equation is an additional relation among the velocity components.

6. In the formulation of the finite element model for fluids flows, which option is not correct about the penalty formulation?

a) The continuity equation is an additional relation among the velocity components

b) The constraint is satisfied in an approximate sense

c) Uses the penalty function method

d) Involves natural and direct formulation of momentum and continuity equations

Answer: d

Explanation: There are two different finite element models of momentum and continuity equations. The penalty formulation is based on the interpretation that the continuity equation is an additional relation among the velocity components. The constraint is satisfied in a least-squares sense. This particular method of including the constraint in the formulation is known as the penalty function method. It does not involve the natural and direct formulation of momentum and continuity equations.

7. Consider the unsteady flow of a viscous fluid squeezed between two horizontal parallel plates, as shown in the following figure. The flow is induced by the uniform motion of the plates toward each other. What are the boundary conditions at the boundary ‘b’ of the domain if v x , v y are the velocity components?

The unsteady flow of a viscous fluid squeezed between two horizontal parallel plates

a) v y ≠0 and v x ≠0

b) v y =0 and v x ≠0

c) v y ≠0 and v x =0

d) v y =0 and v x =0

Answer: b

Explanation: For the unsteady flow of a viscous fluid squeezed between two horizontal parallel plates, because of the symmetry, the computational domain is taken to be one-quarter of the flow field. Since the boundary ‘b’ is located equidistant from the plates, the pushing effect of one plate balances the pushing effect from the opposite one and results in a vertical constraint (v y =0) for the plate. The fluid is free to move in a horizontal direction, thus, v x ≠0.

8. Consider the unsteady flow of a viscous fluid squeezed between two horizontal parallel plates, as shown in the following figure. The flow is induced by the uniform motion of the plates toward each other. What are the boundary conditions at the boundary ‘a’ of the domain if v x , v y are the velocity components?

Unsteady flow of a viscous fluid squeezed between two horizontal parallel plates

a) v y =-V 0 and v x =0

b) v y =-V 0 and v x ≠0

c) v y =V 0 and v x =0

d) v y =V 0 and v x ≠0

Answer: a

Explanation: For the unsteady flow of a viscous fluid squeezed between two horizontal parallel plates, because of the symmetry, the computational domain is taken to be one-quarter of the flow field. The top plate has a velocity –V 0 . The fluid along the top plate is imparted –V 0 along the vertical direction and zero velocity in the horizontal direction, thus v y =-V 0 and v x =0.

9. Which statement is not true regarding the variables present in the viscous ﬂow problems?

a) Velocity components are zero on ﬁxed walls

b) Shear stress is zero along the line of symmetry

c) Vertical velocity component v y and horizontal stress t x must be zero along the horizontal line of symmetry

d) Shear stress is non zero along the line of symmetry

Answer: d

Explanation: In general, the primary variables are obtained using FEM and secondary variables are calculated using the primary variables. In the viscous ﬂow, both the velocity components are zero on ﬁxed walls, and shear stress is zero along the line of symmetry. Vertical velocity component v y and horizontal stress t x must be zero along the horizontal line of symmetry and v x , t y must be zero along the vertical line of symmetry.

10. Considering the problem of bending of beams according to the Euler-Bernoulli beam theory, if the beam is in equilibrium, then solving the equations governing the equilibrium of the Euler-Bernoulli beam is equivalent to minimizing the total potential energy.

a) True

b) False

Answer: a

Explanation: Consider the problem of bending of beams according to the Euler-Bernoulli beam theory. The principle of minimum total potential energy states that if the beam is in equilibrium, then the total potential energy associated with the equilibrium conﬁguration is the minimum; i.e., the equilibrium displacements make the total potential energy a minimum. Thus, solving the equations governing the equilibrium of the Euler-Bernoulli beam is equivalent to minimizing the total potential energy.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Velocity – Pressure Finite Element Model”.

1. In velocity-pressure formulation in FEM, which step is not used in the development of a weak form?

a) Multiply governing equations with weight functions

b) Integrating over the element domain

c) Integrating by parts

d) Performing coordinate transformation

Answer: d

Explanation: In the development of weak form, we consider a typical element and develop the finite element model over it by following the three-step procedure. Firstly, multiply governing equations with weight functions and Integrate over the element domain. Secondly, perform integration by parts and, thirdly, define the coefficient of weight function.

2. In the formulation of governing equations, which option does not signify the characteristics of a weight function?

a) Weight functions are multiplied to governing equations to obtain weak forms

b) Weight functions are interpreted from the physical setup of the problem

c) Weight function must denote a non-dimensional quantity

d) Weight function can be interpreted as a velocity

Answer: c

Explanation: In the development of the weak form, we consider a typical element and multiply its governing equations with weight functions. They can be interpreted physically for a given equation; for example, in the momentum equation, the weight function must be interpreted as velocity. For the representing volume change, weight function is like pressure. Thus, it can be a dimensional quantity.

3. For a governing equation, what does one conclude from the weak formulation if it does not contain boundary integral involving weight function?

a) Integration by parts is used

b) Integration by parts is not used

c) The weight function is as a primary variable

d) The weight function is to be made continuous across inter-element boundaries

Answer: b

Explanation: For a governing equation, the weak formulation does not contain boundary integral if there is no integration by parts used, this implies that the weight function is not a primary variable but a part of the secondary variables, this, in turn, requires that the weight function is not to be made continuous across inter-element boundary.

4. In FEM, which option is the correct weak form of the following momentum equation?

ρ\

l\(\int_{\Omega_e}[\rho w_1\frac{\partial v_x}{\partial t}+\frac{\partial w_1}{\partial x}\sigma_{xx}+\frac{\partial w_1}{\partial y}\sigma_{xy}-w_1 f_x]\)dxdy+∮ ┏ c w 1 (σ xx n x +σ xy n y )ds=0

b) l\

l\(\int_{\Omega_e}[\rho w_1\frac{\partial v_x}{\partial t}+\frac{\partial w_1}{\partial x}\sigma_{xx}+\frac{\partial w_1}{\partial y}\sigma_{xy}-w_1 f_x]\)dxdy+∮ ┏ c w 1 (σ xx +σ xy )ds=0

d) l\(\int_{\Omega_e} w_1[\rho\frac{\partial v_x}{\partial t}-\frac{\partial \sigma_{xx}}{\partial x}-\frac{\partial \sigma_{xy}}{\partial y}-f_x]\)dxdy++∮ ┏ c w 1 (σ xx n x +σ xy n y )ds=0

Answer: a

Explanation: The three-step procedure obtains the weak forms of the momentum equation over an element. Firstly, multiply the governing equation with weight function w 1 and integrate over the element domain Ω e . Secondly, perform integration by parts and thirdly define the coefficient of weight function. Thus, the weak form must contain two separate integrals, the terms \(\frac{\partial w_1}{\partial x}\) and σ xx n x .

5. Using constitutive relations, what is the value of τ xx if μ=0.3 and v=4x?

a) 0.24

b) 2.4

c) 0.12

d) 1.2

Answer: b

Explanation: From constitutive relations, τ xx =2μ\(\frac{\partial v_x}{\partial x}\), where μ is the coefficient of viscosity.

\(\frac{\partial v_x}{\partial x}\)

= \(\frac{\partial }{\partial x}\)

=4.

Thus τ xx =2*μ*4.

Given μ=0.3

τ xx =2*0.3*4

=2.4.

6. Using constitutive relations, what is the value of τ xy if μ=0.3 and v=4xy-6y?

a) 0.24x

b) 2.4x

c) 0.12x

d) 1.2x

Answer: d

Explanation: From constitutive relations, τ xy =μ\

\), where μ is the coefficient of viscosity. On comparing v=4xy+6y with v=vx+vy, we get vx=4xy and vy =6y.

\(\frac{\partial v_x}{\partial y}+\frac{\partial v_y}{\partial x}\)

=4x+0

=4x.

Thus τ xy =μ*.

Given μ=0.3

τ xx =0.3*4x

=1.2x.

7. If a governing equation represents volume change in an element, then the weight function in its weak form must be like a force that causes the volume change.

a) True

b) False

Answer: a

Explanation: The governing equation \

in its weak form must be like a force that causes the volume change. Volume changes occur under the action of hydrostatic pressure, hence w=-P.

8. Which option is not correct concerning the pressure variable, P in the weak form of the momentum and continuity equation?

a) P is a primary variable

b) P is a part of the secondary variables

c) P=constant is the minimum continuity requirement for interpolation

d) It is discontinuous across inter-element boundaries

Answer: a

Explanation: In the weak form of the continuity equation, there is no boundary integral involving weight function because no integration by parts is used, this implies that P is not a primary variable; it is a part of the secondary variables, this, in turn, requires that P not be made continuous across inter-element boundaries. The minimum continuity requirement for interpolation of P is that P=constant.

9. Which option is not correct concerning the velocity variables, v x and v y in the weak form of the momentum and continuity equation?

a) They are primary variables

b) The minimum continuity requirement for interpolation is that they are linear in x and y

c) The minimum continuity requirement for interpolation is that they are constant

d) They are continuous across the inter-element boundary

Answer: c

Explanation: In the weak form of the continuity equation, the boundary integral involving weight function is present by the use of integration by parts. It implies that v x and v y are primary variables; this, in turn, requires that v x and v y to be continuous across inter-element boundaries. The minimum continuity requirement for interpolation of v x and v y is that they are linear in x and y.

10. Which equation is the correct vector form of the finite element model of momentum and continuity equations in the flow domain?

a) MΔ+K 11 Δ+K 12 P=0

b) MΔ+K 11 Δ+K 22 P=F 1

c) MΔ+K 22 Δ+K 12 P=F 1

d) MΔ+K 11 Δ+K 12 P=F 1

Answer: d

Explanation: The equation MΔ+K 11 Δ+K 12 P=F 1 gives the vector form of the finite element model of momentum and continuity equations in the flow domain, where F is force matrix, M is a mass matrix, K 11 is related to velocity term, K 12 is related to the pressure term, P is the pressure matrix, and Δ is the velocity matrix.

11. For the vector form of the finite element model of momentum and continuity equations MΔ+K 11 Δ+K 12 P=0, what is the correct expression for mass matrix M?

a) M=∫ Ω c ρψ T ψdx

b) M=∫ Ω c fψ T ψdx

c) M=∫ Ω c tψ T ψdx

d) M=∫ Ω c Pψ T ψdx

Answer: a

Explanation: The vector form of the finite element model in the flow domain is , MΔ+K 11 Δ+K 12 P=F 1 where F is force matrix, M is a mass matrix, K 11 is related to velocity term, K 12 is related to the pressure term, P is the pressure matrix, and Δ is the velocity matrix. M is a mass matrix as it contains the mass density values of elements, thus M=∫ Ω c ρψ T ψdx, where ρ denotes the mass density.

12. For the vector form of the finite element model of momentum and continuity equations MΔ+K 11 Δ+K 12 P=0 what is the order of matrix F if the order of M is 2n x 2n?

a) 2n x 1

b) 1 x 2n

c) m x 2n

d) 2n x m

Answer: a

Explanation: In the vector form, the terms M and K 11 are of the order 2n x 2n, K 12 is of the order 2n x m, K 21 is of the order m x 2n. The equation MΔ+K 11 Δ+K 12 P=F 1 gives the vector form, where F is force matrix, M is a mass matrix, K 11 is related to velocity term, K 12 is related to the pressure term, P is the pressure matrix, and Δ is the velocity matrix. Note that

13. In matrix algebra, if a matrix is positive definite, then all its eigenvalues are greater than zero.

a) True

b) False

Answer: a

Explanation: A positive definite matrix is a symmetric matrix with all eigenvalues greater than zero. The FEM model of the continuity equation does not contain the pressure term, P. Therefore, the assembled equations also have zero in diagonal elements corresponding to the nodal values of P .

14. What is the physical interpretation of the weight function w3 in the following weak form of the continuity equation?

-∫ Ω e w 3 \

\)dxdy=0

a) Hydrostatic pressure

b) Axial force

c) Surface traction

d) Body force

Answer: a

Explanation: The weak form of the continuity equation represents the volume change in an element of dimensions dx and dy. Therefore, the weight function must be like a force that causes the volume change. Volume changes occur under the action of hydrostatic pressure. Hence, w3 is like -P.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Penalty – Finite Element Model – 1”.

1. Which type of problem can be obtained by reformulating a problem with differential constraints by using the penalty method?

a) A problem with no constraints

b) A problem with variable constraints

c) A problem with fixed constraints

d) A problem with structural constraints

Answer: a

Explanation: Penalty method is introduced in connection with constraint equations. It can also be used to reformulate a problem with differential constraints as one without constraints. In the viscous fluid flow problem, the constraints are created on velocity components by the elimination of pressure term, and thus, the Penalty method is applied.

2. In the interest of the simple formulation of viscous flows, which case does not involve time derivative terms?

a) Static case

b) Transient case

c) Unsteady case

d) Non-periodic

Answer: a

Explanation: The equations governing flows of viscous incompressible fluids can be viewed as equivalent to minimizing a quadratic functional with a constraint. In the penalty method, in the interest of simplicity, for the static case, the constraint condition does not involve time derivative terms. Then, we add time derivative terms to study transient problems.

3. In the penalty formulation of the fluid flow model, if the velocity field (v x , v y ) satisfies the continuity equation, then the weight functions also satisfy the continuity equation.

a) True

b) False

Answer: b

Explanation: In the penalty formulation, we begin with the unconstrained problem described by the weak forms of the mixed model, without the time—derivative terms. Now, suppose that the velocity field (v x , v y ) is such that the continuity equation is satisfied identically. Then the weight functions being variations of the velocity components, also satisfy the continuity equation.

4. In the weak forms of the fluid flow model, as the weight functions are virtual variations of the velocity components(v x , v y ),respectively, which relation is satisfied by the weight functions?

a) \

\(\frac{\partial w2}{\partial x}-\frac{\partial w1}{\partial y}\)=0

Answer: a

Explanation: In the weak forms of the fluid flow model, suppose that the velocity field (v x , v y ) is such that the continuity equation \

being variations of the velocity components, also satisfy the continuity equation. Thus the relation possessed by the weight functions is \(\frac{\partial w1}{\partial x}+\frac{\partial w2}{\partial y}=0\).

5. In finite element modeling, which formulation introduces constraints on variables and satisfies them in an approximate sense?

a) Velocity-pressure formulation

b) Penalty formulation

c) Mixed formulation

d) Lagrange multiplier formulation

Answer: b

Explanation: There are two different finite element models of momentum and continuity equations. The first one is a natural and direct formulation of momentum and continuity equations. The second model, penalty formulation, is based on the interpretation that the continuity equation is a new relationship among the velocity components, and it eliminates the pressure term leading to a constraint on the velocity components. The constraint is satisfied in a least-squares sense.

6. In the formulation of the finite element model, which option is the complete restated form of the weak forms of viscous fluids flow equations?

a) Only B t +B v -B̅ p =l

b) B t +B v -B̅ p =l and –B p (w 3 , v)=0

c) Only –B p (w 3 , v)=0

d) B t +B v -B̅ p =0

Answer: b

, \)v=\(

,\) f=\(

\) and t=\(

\).

7. In the following variational problem of finding velocity components and pressure, which bilinear form includes time-derivative terms?

B t +B v -B̅ p =l; –B p (w 3 , v)=0

a) B t

b) B v

c) B̅ p

d) B p (w 3 ,v)

Answer: a

Explanation: The problem described by weak forms of viscous fluids flow equations can be restated as a variational problem of finding (v x , v y , P) such that B t +B v -B̅ p =l and –B p (w 3 , v)=0 holds for all weight functions and t>0. The term B t =∫ Ω e ρw T vdx contains the time derivative term v. Further more, in the penalty method, we add time derivative terms to study transient problems.

8. In the weak forms of the fluid flow model, since the weight functions are linearly dependent on each other, the sum of the three weak forms is the same as the three individual equations.

a) True

b) False

Answer: b

Explanation: if x1, x2 and x3 represent three weak forms, then x1=ax2+bx3 is true only if the weight functions are linearly dependent. Since the weight functions are linearly independent of each other, the sum of the three weak forms is the same as the three individual equations i.e, the sum must be the expression x1+x2+x3 only.

9. Which option is not correct concerning the bilinear term B in the variational problem of the viscous fluid flow equation?

a) It is symmetric

b) It contains the viscosity matrix

c) B v = B v

d) It is bilinear in both w and v

Answer: d

Explanation: The complete variational problem of the viscous fluid flow equation is B t +B v -B̅ p =l and –B p (w 3 , v)=0. The term, B v is linear in both v and w, it is symmetric because it contains the symmetric matrix, C. Thus B v = B . The variational problem B v =l is a constrained problem.

10. In the variational problem of fluid flow, what is the correct matrix form of C in the bilinear form B v =∫ Ω e T Cdx?

a) \

μ\

μ\(

Answer: c

Explanation: The complete restated form of the weak forms of viscous fluids flow equations is given by B t +B v -B̅ p =l and –B p (w 3 ,v)=0. The bilinear form, B v =∫ Ω e T Cdx is symmetric because it contains the symmetric matrix μ \(

\), denoted by the letter C. C is the viscosity matrix since it contains the viscosity term μ.

Answer: c

Explanation: Since the bilinear term has the form B̅ p =∫ Ω e \

^T\)Pdx; K=\

^T\)p.

We have D 1 =\(

\) and w=\(

\(D_1^T w=

= \(\frac{\partial }{\partial y}+\frac{\partial }{\partial x}\)

=2+3

=5.

K=(D 1 T w) T p

=5*3

=15.

Sanfoundry Global Education & Learning Series – Finite Element Method.

This set of Finite Element Method Assessment Questions and Answers focuses on “Penalty – Finite Element Model – 2”.

1. Which option is correct regarding constraints in the viscous flow problems governed by the continuity and momentum equation?

a) Constrains are readily available

b) Constraints are created by the elimination of pressure

c) Constraints are created by the elimination of velocity

d) No constraints are dealt with in penalty formulation

Answer: b

Explanation: In viscous flow problems, constrains are not readily available from the presented equations. Since pressure is not present in all the governing equations, we eliminate it from the given other equations. The constraints are created on velocity by the elimination of pressure, and thus, the penalty method is applied.

2. In the finite element method, which option is not a natural and direct formulation of momentum and continuity equations?

a) Velocity-pressure formulation

b) Penalty formulation

c) Mixed formulation

d) Lagrange multiplier formulation

Answer: b

Explanation: There are two different finite element models of momentum and continuity equations. The first one is a natural and direct formulation of momentum and continuity equations. It is known as the velocity-pressure formulation or mixed formulation. Lagrange multiplier formulation is similar to the velocity-pressure formulation. The penalty formulation is not natural and direct.

3. For the functional I v =\(\frac{1}{2}\)B v -l, which option is equivalent to the equations governing steady flows of viscous incompressible fluids?

a) Minimize l v

b) Maximize l v

c) Stationary l v

d) Derivative of l v

Answer: a

Explanation: For the quadratic functional given by the equation I v =\(\frac{1}{2}\)B v -l, one can state that the equations governing steady flows of viscous incompressible fluids are equivalent to minimize I v , subject to the constraint \(\frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}\)=0. It is a constrained problem of finding velocity.

4. In FEM, what is the characteristic of the problem with the functional I v representing governing equations of steady viscous incompressible flows?

a) It is subjected to constraint \

It is an unconstrained problem

c) It can be reformulated as a constrained one, by using the penalty method

d) It cannot be reformulated as an unconstrained one, by using the penalty method

Answer: a

Explanation: For the functional I v , one can state that the equations governing steady flows of viscous incompressible fluids are equivalent to minimize I v , subject to the constraint \(\frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}\)=0. It is a constrained problem that can be reformulated as an unconstrained one by using the penalty method.

5. In the Lagrange multiplier method, a constrained problem is reformulated as one of finding the stationary points of an unconstrained function, whereas in the penalty method, a problem with differential constraints is reformulated to one without constraints.

a) True

b) False

Answer: a

Explanation: In the Lagrange multiplier method, a constrained problem is reformulated as one of finding the stationary points of the unconstrained functional I L ≡I v +∫ Ω c λGdxdy where λ is the Lagrange multiplier. In the penalty method, a problem with differential constraints is reformulated to one without constraints.

6. For the functional I L ≡I v +∫ Ω c λGdxdy, what is the necessary condition for I L to have a stationary value?

a) δ v x I L +δ v y I L +δ λ I L =0

b) δ v x I L -δ v y I L -δ λ I L =0

c) δ v x I L -δ v y I L +δ λ I L =0

d) δ v x I L +δ v y I L -δ λ I L =0

Answer: a

Explanation: In the Lagrange multiplier method the constrained problem is reformulated as one of finding the stationary points of the unconstrained functional I L ≡I v +∫ Ω c λGdxdy where λ is the Lagrange multiplier. The necessary condition for I L to have stationary value is δ v x I L +δ v y I L +δ λ I L =0.

7. In FEM, which option is a negative aspect of the mixed formulation?

a) Presence of zeros on matrix diagonals corresponding to pressure variable

b) Involves a natural and direct formulation

c) It is a velocity-pressure formulation

d) It is similar to the Lagrange multiplier method

Answer: a

Explanation: A negative aspect of the mixed finite element model is the presence of zeroes on the matrix diagonals corresponding to the pressure variables. Direct equation-solving methods must use some type of pivoting strategy, while the use of iterative solvers is severely handicapped by poor convergence behavior.

8. In the computation of mixed formulation in FEM, which option is correct for the nodal DOF of the following element?

Find nodal DOF of element in computation of mixed formulation in FEM

a) o nodes with u,v and P

b) * nodes with u,v and P

c) o nodes with P only

d) *nodes with u and v

Answer: b

Explanation: Commonly used elements for two-dimensional flows of viscous incompressible fluids are the triangular and quadrilateral elements. In the case of linear elements, the pressure is treated as discontinuous between elements. In quadratic elements, the DOF at intermediate and interior nodes is velocity, i.e., u and v, whereas DOF at corner nodes are u, v and P.

9. For the following slider bearing, the upper pad is inclined to the base pad. Which option is correct for a load applied normal to the base pad?

Slider or slipper bearing with the upper pad inclined to the base pad

a) A pressure gradient develops in the lubricant, and the bearing can support some transverse load

b) The pressure remains uniform in the lubricant, and the bearing can support some transverse load

c) A pressure gradient develops in the lubricant, and the bearing cannot support any transverse load

d) The pressure remains uniform in the lubricant, and the bearing cannot support any transverse load

Answer: a

Explanation: The slider bearing consists of a short sliding pad over a stationary pad, and the small gap between the two pads is filled with a lubricant. If the upper pad is inclined to the base plate, then a pressure gradient develops in the lubricant, and the bearing can support some transverse load. The penalty formulation in FEM is used to solve for pressure and velocities of flow.

10. For the following slider bearing, the upper pad is parallel to the base pad. Which option is correct for a load applied normal to the base pad?

Slider or slipper bearing with the upper pad is parallel to the base pad

a) The pressure in the lubricant is atmospheric, and the bearing cannot support any load

b) The pressure in the lubricant is greater than atmospheric, and the bearing can support any load

c) The pressure in the lubricant is less than atmospheric, and the bearing cannot support any load

d) The pressure in the lubricant is atmospheric, and the bearing can support some load

Answer: a

Explanation: The slider bearing consists of a short sliding pad over a stationary pad, and the small gap between the two pads is filled with a lubricant. If the upper pad is parallel to the base plate, then the pressure everywhere in the gap is atmospheric , and the bearing cannot support any transverse load. The penalty formulation in FEM is used to solve for pressure and velocities of flow.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “ Classical Plate Model”.

1. In FEM, what does the simple two-dimensional plate theories account for?

a) The kinematics of bending deformation of thin bodies without to transverse loads

b) The kinematics of bending deformation of thin bodies subjected to transverse loads

c) Thin bodies subjected to transverse loads without bending deformation

d) Thin plates subjected to transverse loads with bending deformation

Answer: b

Explanation: The term “plate” refers to solid bodies that are bounded by two parallel planes whose lateral dimensions are large compared with the separation between them . In most cases, the thickness is no greater than one-tenth of the smallest in-plane dimension. Geometrically, plate problems are similar to the plane stress problems except that plates are also subjected to transverse loads. Thus, the simple two-dimensional plate theories account for the kinematics of bending deformation of thin bodies subjected to transverse loads.

2. In the bending of elastic plates, although the plate problems are similar to the plane stress problems, plates are also subjected to transverse loads that cause bending about axes normal to the plane of the plate.

a) True

b) False

Answer: b

Explanation: A plate is a solid body that is bounded by two parallel planes whose lateral dimensions are large compared with the separation between them . Mostly, the thickness is lesser than one-tenth of the smallest in-plane dimension. Geometrically, plate problems are similar to the plane stress problems except that plates are also subjected to transverse loads that cause bending about axes in the plane of the plate.

3. Which option is correct about an analog of a plate in simple plate theory used in FEM?

a) A plate is a one-dimensional analog of a beam

b) A plate is a two-dimensional analog of a beam

c) A plate is a three-dimensional analog of a beam

d) A plate is not analogous to beam

Answer: b

Explanation: Geometrically, plate problems are similar to the plane stress problems except that plates are also subjected to transverse loads that cause bending about axes in the plane of the plate. In other words, a plate is a two-dimensional analog of a beam. Because of the smallness of the thickness dimension, it is often not necessary to model plates using three-dimensional elasticity theory. Simple two-dimensional theories that account for the kinematics of bending deformation of thin bodies subjected to transverse loads have been developed, and they are known as plate theories.

4. In FEM, which principle is used to derive the governing equations of displacement-based plate theories?

a) The principle of virtual forces

b) The principle of virtual work

c) The principle of virtual displacements

d) The principle of least energy

Answer: c

Explanation: Governing equations of displacement-based plate theories are derived using the principle of virtual displacements, the principle of virtual displacements directly yields the weak forms of the governing equations. The starting point in the development of the governing equations of a plate theory is to choose a displacement field. Typically, the displacement components are selected in the form of a linear combination of unknown functions and powers of the thickness coordinate z so that certain kinematics of the plate are represented.

5. In FEM, which theory is an extension of the Euler-Bernoulli beam theory?

a) Classical Plate Theory

b) Shear Deformation Theory

c) Hencky-Mindlin plate theory

d) Shell theory

Answer: a

Explanation: The two most commonly used displacement-based plate theories are the Classical Plate Theory and first-order Shear Deformation Theory . CPT is an extension of the Euler-Bernoulli beam theory from one dimension to two dimensions and is also known as the Kirchhoff plate theory. Shear Deformation Theory is an extension of the Timoshenko beam theory.

6. In displacement-based plate theories, which option is not an assumption of Classical Plate Theory ?

a) A straight line perpendicular to the plane of the plate is inextensible

b) A straight line perpendicular to the plane of the plate remains straight

c) A straight line perpendicular to the plane of the plate remains straight and does not rotate

d) A straight line perpendicular to the plane of the plate rotates such that it remains perpendicular to the tangent to the deformed surface

Answer: c

7. Which option specifiesan assumption made in Classical Plate Theory for a plate lying in the plane XY?

a) ε zz =0

b) ε xz ≠0

c) ε yz ≠0

d) ε xy =0

Answer: a

Explanation: The assumption made in Classical Plate Theory is that a straight line perpendicular to the plane of the plate is inextensible, remains straight, and rotates such that it remains perpendicular to the tangent to the deformed surface. These assumptions are equivalent to specifying ε zz =0, ε yz =0, ε xz =0 for a plate in XY plane.

8. The following diagram is the mid-plane of an elastic plate. Which option shows the correct moments Mxx and Mxy if the plate is loaded as per the Classical Plate Theory?

The mid-plane of an elastic plate loaded as per the Classical Plate Theory

a) Find the boundary conditions in Classical Plate Theory

b) Bending moment and shear force in a plate lying in XY plane

c) Find Mxx and Mxy in mid-plane of an elastic plate if it is loaded as per Classical Plate Theory

d) Concentrated transverse load & shear force in the mid-plane of an elastic plate

Answer: b

Explanation: The given diagram shows bending moment and shear force in a plate lying in XY plane. The term Q is concentrated transverse load, N is shear force, and Mxx is edge bending moment whereas Mxy is the twisting moment. The option ‘a’ is correct for the sense of the moments but not for their positions.

9. The following figure shows the mid-plane of an elastic plate. What are the boundary conditions in Classical Plate Theory if the plate is clamped and the displacement along Z is w?

Find boundary conditions in clamped plate

a) w=0,\

w=0,V n =0

c) w=0,\

w=0,M nn =0

Answer: a

Explanation: Geometrically, plate problems are similar to the plane stress problems except that plates are also subjected to transverse loads that cause bending about axes in the plane of the plate. The boundary condition for a clamped plate is the absence of deflection and normal derivative of deflection , i.e., w=\(\frac{\partial w}{\partial n}\)=0. Because a simply supported end does not restrict rotation, the reactive moment is zero, i.e., w=M nn =0. For a free end, both, the reactive moment and the shear force are absent, i.e., M nn = V n =0.

10. The following figure shows the mid-plane of an elastic plate. In Classical Plate Theory, what are the boundary conditions ifthe plate is simply supported with deflection w along Z?

Find boundary conditions if plate is supported with deflection in Classical Plate Theory

a) w=0,\

w=0,V n =0

c) w=0,\

w=0,M nn =0

Answer: d

Explanation: Plate problems are geometrically similar to the plane stress problems except that plates are also subjected to transverse loads. A clamped plate has no deflection and normal derivative of deflection , i.e., w=\(\frac{\partial w}{\partial n}\)=0. In a simply supported end, rotation is not restricted; thus, the reactive moment is zero, i.e., w= M nn =0. A free end has no reactive moment and the shear force, i.e., M nn = V n =0.

11. In FEM, what are the primary variables in the Classical Plate Theory of plate deformation ?

a) The transverse deflection w only

b) The transverse deflection w and the normal derivative of w

c) The normal derivative of w only

d) The normal and the time derivative of w

Answer: b

Explanation: An examination of the boundary terms in the weak form of Classical Plate Theory suggests that the essential boundary conditions involve specifying the transverse deflection w and the normal derivative of w, which constitute the primary variables of the problem . Hence, the finite element interpolation of w must be such that w and \(\frac{dw}{dn}\) are continuous across the inter-element boundaries in CPT elements.

12.How many termpolynomial is used in the approximation of w for a three noded triangular element with three DOF at each node?

a) 9-term

b) 18-term

c) 6-term

d) 12-term

Answer: a

Explanation: Several C1 rectangular and triangular plate bending elements with

or with

as the degrees of freedom at each node exists in the literature. A triangular element with three nodes, with

at each node, requires the 9-term polynomial approximation of w, i.e.,w=a 1 +a 2 x+a 3 y+a 4 xy+a 5 x 2 +a 6 y 2 +a 7 (x 2 y+xy 2 )+a 8 x 3 +a 9 y 3 .

13. In Classical Plate Theory, what is the correct comment for the polynomial w=a 1 +a 2 x+a 3 y+a 4 xy+a 5 x 2 +a 6 y 2 +a 7 (x 2 y+xy 2 )+a 8 x 3 +a 9 y 3 ?

a) It is used in the approximation of deflection for a rectangular element

b) It is a complete third order polynomial

c) It’s an incomplete third order polynomial

d) w varies as a quadratic along any line x=constant

Answer: c

Explanation: Several C1 rectangular and triangular plate bending elements with

or with

as the degrees of freedom at each node exists in the literature. A triangular element with three nodes, with

at each node, requires the 9-term polynomial approximation of w, i.e., w=a 1 +a 2 x+a 3 y+a 4 xy+a 5 x 2 +a 6 y 2 +a 7 (x 2 y+xy 2 )+a 8 x 3 +a 9 y 3 . This is an incomplete third-order polynomial because x 2 y and y 2 x do not vary independently. We note from the equation that w varies as a cubic along with any line x=constant or y=constant.

14. In Classical Plate Theory, because the 4-noded rectangular element is non-conforming, it gives poor results in approximation.

a) True

b) False

Answer: b

Explanation: Elements that violate any of the continuity conditions are known as non-conforming elements. Thus, the four-node rectangular element with w approximated by a 12-term polynomial is non-conforming. Despite this deficiency, the element is known to give good results. A similar discussion leads to the conclusion that the three-node triangular element is the non-conforming type and found to have convergence problems and singular behavior for certain meshes.

15. In Classical Plate Theory, how many triangles assemble to give a conforming triangular element with DOF w, \

b) 3

c) 4

d) 12

Answer: b

Explanation: A non-conforming element is the one that violates any of the continuity conditions. The three-node triangular element is non-conforming and found to have convergence problems and singular behavior for certain meshes. A conforming triangular element is an assemblage of three triangles. The interpolation functions for the triangular element can be expressed in terms of the area coordinates. A confirming rectangular element is formed as an assembly of twelve triangular elements.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Shear Deformable Plate Model”.

1. In FEM, which theory is an extension of the Timoshenko beam theory?

a) Classical Plate Theory

b) Hencky-Mindlin plate theory

c) Kirchhoff plate theory

d) Shell theory

Answer: b

2. In displacement-based plate theories, if a linear theory based on infinitesimal strains and orthotropic material properties is used, then the in-plane displacements are coupled with the transverse deflection.

a) True

b) False

Answer: b

Explanation: For a linear theory based on infinitesimal strains and orthotropic material properties, the in-plane displacements are uncoupled from the transverse deflection uz=w. The plane elasticity equations govern the in-plane displacements . The in-plane displacements are zeroin the absence of in-plane forces, andhence, we discuss only the equations governing the bending deformation.

3. In FEM, what are the primary variables in the Shear Deformation Theory of plate deformation ?

a) The transverse deflection w only

b) The transverse deflection w and the normal derivative of w

c) The transverse deflection w and the angles of rotation of the transverse normal about in-plane axes

d) The angles of rotation of the transverse normal about in-plane axes only

Answer: c

Explanation: An examination of the boundary terms in the weak form of Shear Deformation Theory suggests that the essential boundary conditions involve specifying the transverse deflection w and the angles of rotation of the transverse normal about in-plane axes (φ x , φ y ), which constitute the primary variables of the problem . Hence, the finite element interpolation of w must be such that w, (φ x and φ y are continuous across the inter-element boundaries in SDT elements.

4. Which equation correctly describes Hamilton’s principle used in FEM?

a) 0=\

]dt

b) 0=\

]dt

c) 0=\

]dt

d) 0=\

]dt

Answer: a

Explanation: Governing equations of displacement-based plate theories are derived using the principle of virtual displacements. The principle of virtual displacements or Hamilton’s principle requires that 0=\

]dt where δU, δV and δK denote the virtual strain energy, virtual work done by externally applied forces, and virtual kinetic energy, respectively. These quantities are expressed in terms of actual stresses and virtual strains, which depend on the assumed displacement functions and their variations.

5. In displacement-based plate theories, which option is correct about Shear Deformation Theory ?

a) It is also called Kirchhoff plate theory

b) It is an extension of Euler-Bernoulli beam theory from one dimension to two dimensions

c) It does not involve Timoshenko beam theory

d) It is often known as Hencky-Mindlin plate theory

Answer: d

6. In displacement-based plate theories, which assumption of Classical Plate Theory is relaxed in Shear Deformation Theory?

a) A straight-line perpendicular to the plane of the plate is inextensible

b) A straight line perpendicular to the plane of the plate remains straight

c) A straight line perpendicular to the plane of the plate rotates such that it remains perpendicular to the tangent to the deformed surface

d) A straight line perpendicular to the plane of the plate rotates

Answer: c

Explanation: In the SDT, we relax the normality assumption of CPT, i.e., transverse normal may rotate without remaining perpendicular to the mid-plane. The Classical Plate Theory is based on the assumption that a straight line perpendicular to the plane of the plate is inextensible, remains straight, and rotates such that it remains perpendicular to the tangent to the deformed surface.

7. Which option specifies an assumption made in Shear Deformation Theory for a plate lying in the plane XY?

a) ε zz ≠0

b) ε xz =0

c) ε yz ≠0

d) ε xy =0

Answer: c

Explanation: The Classical Plate Theory is based on the assumption that a straight line perpendicular to the plane of the plate is inextensible, remains straight, and rotates such that it remains perpendicular to the tangent to the deformed surface, but In the SDT, we relax the normality assumption of CPT, i.e., transverse normal may rotate without remaining normal to the mid-plane. Thus, for a plate in the XY plane, the assumptions are equivalent to specifying ε zz =0 only whereas ε yz and ε xz are non-zero.

8. In SDT, what are the boundary conditions for a plate that is clamped if &fcy; represents the rotation of the transverse normal about an in-plane axis and w is the transverse deflection?

a) w=0,\

w=0,&fcy;=0

c) w=0,\

w=0, M nn =0

Answer: b

Explanation: Geometrically, plate problems are similar to the plane stress problems except that plates are also subjected to transverse loads that cause bending about axes in the plane of the plate. In SDT, the boundary condition for a clamped plate is the absence of deflection and rotation of the transverse normal about any in-plane axis, i.e., w=&fcy;=0. Because a simply supported end does not restrict rotation, the reactive moment is zero, i.e., w=M nn =0. For a free end, both, the reactive moment and the shear force are absent, i.e., M nn =Q n =0.

9. In SDT, what are the boundary conditions for a plate that is simply supported if &fcy; represents the rotation of the transverse normal about an in-plane axis and w is the transverse deflection?

a) w=0,\

w=0,&fcy;=0

c) w=0,\

w=0, M nn =0

Answer: d

Explanation: Plate problems are geometrically similar to the plane stress problems except that plates are also subjected to transverse loads. In SDT, a clamped plate has no deflection and rotation of the transverse normal about any in-plane axis, i.e., w=&fcy;=0. In a simply supported end, rotation is not restricted; thusthe reactive moment is zero, i.e., w= M nn =0. A free end does not have reactive moment and the shear force, i.e., M nn =Q n =0.

10. In FEM, which option is correct for a linear plate theory based on infinitesimal strains and orthotropic material properties?

a) The plane elasticity equations govern the transverse deflections

b) The transverse deflections are coupled with in-plane displacements

c) The in-plane displacements are zero in the absence of in-plane forces

d) The transverse deflections are zero in the absence of in-plane forces

Answer: c

Explanation: For a linear theory based on infinitesimal strains and orthotropic material properties, the in-plane displacements are uncoupled from the transverse deflection uz=w. The plane elasticity equations govern the in-plane displacements . The in-plane displacements are zero if there are no in-plane forces and hence, we discuss only the equations governing the bending deformation and the associated finite element models.

11. In the Shear Deformation plate theory, when does the transverse shear strains in the element equations present computational difficulties?

a) If the plate is thick

b) If the side to thickness ratio of the plate is large

c) If the side to thickness ratio of the plate is small

d) If higher-order finite elements are used

Answer: b

Explanation: The transverse shear strains in the element equations of Shear Deformation Theory present computational difficulties when the side-to-thickness ratio of the plateis large . For thin plates, the transverse shear strains are negligible, and consequently, the element stiffness matrix becomes stiff and yields erroneous results for the generalized displacements. This phenomenon is known as shear locking.

12. In the Shear Deformation plate theory, what characteristic contributes to shear locking?

a) Transverse shear strains in thick plates present computational difficulties

b) Transverse shear strains in thin plates present computational efficiency

c) For thick plates, the element stiffness matrix yields erroneous results for the generalized displacements

d) For thin plates, the element stiffness matrix becomes stiff and yields erroneous results

Answer: d

Explanation: The transverse shear strains in the element equations of Shear Deformation Theory cause computational difficulties when the side-to-thickness ratio of the plate is large. Shear locking is observed when the transverse shear strains in thin plates are negligible, and consequently, the element stiffness matrix becomes stiff and yields erroneous results for the generalized displacements.This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Matrix Algebra”.

1. What is a matrix?

a) Group of elements

b) Array of elements

c) Group of columns and rows

d) Array of numbers

Answer: b

2. Which of the following is a row vector?

a) \

\(\left[

\right]\)

Answer: a

Explanation: A matrix of dimension is called row vector. A matrix of dimension is called column vector.

For example

d=[ 1 2 3 4] is a row vector.

c = \(\left[

\right]\) is a column vector.

3. T = _______

a) T

b) B T C T A T

c) C T B T A T

d) A T B T C T

Answer: c

T = C T B T A T .

4. The derivative of Ax with respect to variable x p is given by __________

a) \

=x p

b) \(\frac{d}{dx}\)(x p )=A x

c) ∫ A x=x p

d) ∫x p =Ax

Answer: a

Explanation: Let A be an matrix of constants and x = [x 1 x 2 x 3 …… x n ] T be column vector of n variables. Then, derivative of A x with respect to variable x p is given by

=x p .

5. A symmetric matrix is called ____________, if all its Eigen values are strictly positive i.e., greater than zero.

a) Negative definite

b) Positive definite

c) Co- definite

d) Alternative definite

Answer: b

6. A A -1 =A -1 A is a condition for ________

a) Singular matrix

b) Nonsingular matrix

c) Matrix inversion

d) Ad joint of matrix

Answer: c

Explanation: If det A not equal to zero, then A has an inverse, denoted by A -1 . The inverse satisfies the relation

A A -1 =A -1 A= I

7. A positive definite symmetric matrix A can be decomposed into form A=LL T this decomposition is called ________

a) Cholesky

b) Rayleighs

c) Galerkins

d) Potential energy

Answer: a

8. Det=0 is a ________

a) Characteristic equation

b) Matrix equation

c) Inversion of matrix

d) Cholesky’s equation

Answer: a

9. \Missing open brace for subscript Principle diagonal matrix

b) Upper triangular matrix

c) Lower triangular matrix

d) Singular matrix

Answer: b

10. A=\

a) 120

b) -80

c) -175

d) 0

Answer: c

Explanation: det\(

= a 11 (a 22 a 33 -a 32 a 23 )-a 12 (a 21 a 33 -a 31 a 23 )+a 13 (a 21 a 32 -a 31 a 22 )

= 3*-)-2*-)+1*-)

=-175.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Gaussian Elimination”.

1. Gaussian elimination is a name given to a well known method of solving simultaneous equation by successively eliminating _________

a) Variables

b) Equations

c) Unknown

d) Algorithms

Answer: c

2. Step number in Gaussian elimination is denoted as ___________

a) Superscript

b) Subscript

c) Unknown

d) Elimination

Answer: a

3. In Gaussian elimination, A is defined as symmetric matrix then its multiplier is defined as ____

a) C = a kk /a ik

b) C = a ki /a kk

c) C = a ik /a kk

d) C = a kk /a ki

Answer: b

4. A banded matrix is defined as ____________

a) Non zero elements are contained in band

b) Zero elements are contained in a band

c) Non zero elements are contained out of a band

d) Both Non zero elements and Zero elements

Answer: a

5. In a symmetric banded matrix __________

a) a ij =a ji

b) a ji *a ij

c) a ij ≠a ji

d) a ii =a jj

Answer: a

Explanation: For a symmetric banded matrix a ij =a ji . A symmetric banded matrix is a symmetric matrix whose nonzero elements are arranged uniformly near the diagonal.

6. Consider a nxn symmetric matrix: \Missing open brace for subscript Full band width

b) Half band width

c) Semi band width

d) Co band width

Answer: b

7. The line separating from the top zeros from the first non-zero element is called ____

a) Equation

b) Gaussian solution

c) Skyline solution

d) Both Gaussian and skyline solutions

Answer: c

8. Frontal method is a _______ of Gaussian elimination method that uses the structure of finite element problem.

a) Structure

b) Variation

c) Algorithm

d) Data

Answer: b

9. Frontal method is implemented for ________

a) Hexahedral element

b) Polyhedral element

c) Octahedral element

d) Both Hexahedral and Polyhedral

Answer: a

10. Frontal method involves __________

a) Computer programming

b) Manual programming

c) C- programming

d) Computing

Answer: a

This set of Finite Element Method online test focuses on “Conjugate Gradient Method for Equation Solving”.

1. Conjugate gradient method is a ________

a) Standard method

b) Equation method

c) Iterative method

d) Elimination method

Answer: c

2. Conjugate gradient method is for only _________

a) Non symmetric matrix

b) Symmetric matrix

c) Symmetric and Non symmetric matrix

d) Identity matrix

Answer: b

3. Which version is used to solve conjugate gradient method of the algorithm of symmetric matrices?

a) Rayleighs version

b) Fletcher-Reeves version

c) Frontal method

d) Galerkins method

Answer: b

4. The iterations are continued until g k T g x reaches a ________

a) Small value

b) Infinite value

c) Large value

d) Negative value

Answer: a

5. This method is robust and has ________ n iterations.

a) Diverges

b) Converges

c) Converge and diverge

d) Symmetric

Answer: b

6. Conjugate gradient method is implemented in _________

a) Algorithm solving

b) Iterative solving

c) Program solving

d) Program CG solving

Answer: d

7. Conjugate gradient method includes several ______

a) Computer codes

b) Programs

c) Algorithms

d) Matrices

Answer: a

8. Conjugate gradient algorithm can be accelerated by using _________

a) Algorithm strategies

b) Preconditioning strategies

c) Conditioning strategies

d) Computer strategies

Answer: b

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “One Dimensional Problems – Finite Element Modelling”.

1. If the structure is divided into discrete areas or volumes then it is called an _______

a) Structure

b) Element

c) Matrix

d) Boundaries

Answer: b

2. In finite element modeling nodal points are connected by unique ________

a) Surface

b) Shape

c) Eigen values

d) Matrix

Answer: a

3. In finite element modeling every element connects to _______

a) 4 nodes

b) 3 nodes

c) 2 nodes

d) Infinite no of nodes

Answer: c

4. In one dimensional problem, each node has _________ degrees of freedom.

a) 2 degrees of freedom

b) 3 degrees of freedom

c) No degrees of freedom

d) 1 degree of freedom

Answer: d

5. Which relations are used in one dimensional finite element modeling?

a) Stress-strain relation

b) Strain-displacement relation

c) Total potential energy

d) Total potential energy; Stress-strain relation; Strain-displacement relation.

Answer: d

6. One dimensional element is the linear segments which are used to model ________

a) Bars and trusses

b) Plates and beams

c) Structures

d) Solids

Answer: a

7. Modeling is defined as ________________

a) Elemental area with uniform cross section

b) Elemental area with non uniform cross section

c) Structural area with uniform cross section

d) Non structural area with non uniform cross section

Answer: a

Stepped Shaft Model consisting discrete number of elements with uniform cross sectional area

A1=A 1 ’+A 2 ’/2.

8. Discretization includes __________ numbering.

a) Element and node

b) Only nodal

c) Only elemental

d) Either nodal or elemental

Answer: a

9. The loading on an element includes _______

a) Body force

b) Traction force

c) Point load

d) Body force, Traction force & Point load

Answer: d

10. Global nodes corresponds to _______

a) Entire body

b) On surface

c) On interface

d) On element

Answer: a

Explanation: Global coordinate system corresponds to the entire body. It is used to define nodes in the entire body.

11. Local node number corresponds to ______________

a) Entire body

b) On element

c) On interface

d) On surface

Answer: b

Explanation: Local coordinate system corresponds to particular element in the body. The numbering is done to that particular element neglecting the entire body.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “One Dimensional Problems – Co-ordinates and Shape Functions”.

1. Natural or intrinsic coordinate system is used to define ___________

a) Co-ordinates

b) Shape functions

c) Displacement functions

d) Both shape functions and co-ordinate functions

Answer: b

2. In q=[q 1 ,q 2 ] T is defined as __________

a) Element displacement vector

b) Element vector

c) Displacement vector

d) Shape function vector

Answer: a

3. Shape function is just a ___________

a) Displacement function

b) Equation

c) Interpolation function

d) Matrix function

Answer: c

4. Isoparametric formula is ______________

a) x=N 1 x 1 +N 2 x 2

b) x=N 2 x 1 +N 1 x 2

c) x=N 1 x 1 -N 2 x 2

d) x=N 2 x 1 -N 1 x 2

Answer: a

Explanation: From nodal displacement equation we can write that isoparametric equation as

x=N 1 x 1 +N 2 x 2

Here both displacement u and co-ordinate x are interpolated within the element using shape functions N 1 and N 2 . This is called isoparametric formulation in literature.

5. B=\Missing open brace for subscript Strain matrix

b) Element-strain displacement matrix

c) Displacement matrix

d) Elemental matrix

Answer: b

Explanation: ε= Bq

Here B is element strain displacement matrix. Use of linear shape functions results in a constant B matrix. Hence, in a constant strain within the element. The stress from Hooke’s law is

σ= EBq .

6. Deformation at the end of elements are called _____________

a) Load

b) Displacement functions

c) Co-ordinates

d) Nodes

Answer: d

7. Write the shape function of the given element.

Find the Shape Function of the element u= N 1 u 1 +N 2 u 2 . Here N 1 & N 2 are

a) N 1 =1-x/l e &N 2 =x/l e

b) N 1 =x/l e &N 2 =1-x/l e

c) N 1 =0 & N 2 =x

d) N 1 =x & N 2 =0

Answer: a

Explanation:

Shape Function of the element

1 2 --- local variables

I j --- global variables

u1(e) u2(e)

x1=0 x2=0

Then matrix notation form is

u=\(

u 1 =c 1 +c 2 =c 1

u 2 = c 1 +c 2 (l e )

In matrix equation

By solving we get

N 1 =1-x/l e & N 2 =x/l e .

8. In shape functions, first derivatives must be _______ within an element.

a) Infinite

b) Finite

c) Natural

d) Integer

Answer: b

9. In shape functions, _________ must be continuous across the element boundary.

a) Derivatives

b) Nodes

c) Displacement

d) Shape function

Answer: c

Explanation: Shape functions are interpolation functions. In general shape functions need to satisfy that, displacements must be continuous across the element boundary.

10. Stresses due to rigid body motion are _______________

a) Zero

b) Considered

c) Not considered

d) Infinite

Answer: c

11. The expressions u=Nq; ε= Bq; σ= EBq relate ____________

a) Displacement, Strain and Stress

b) Strain and stress

c) Strain and displacement

d) Stress and displacement

Answer: a

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “One Dimensional Problems – Potential Energy Approach”.

1. Continuum is discretized into_______ elements.

a) Infinite

b) Finite

c) Unique

d) Equal

Answer: b

2. U e =\(\frac{1}{2}\int\) σ T εA dx is a _____________

a) Potential equation

b) Element strain energy

c) Load

d) Element equation

Answer: b

3. Which is the correct option for the following equation?

K e =\

Load vector

b) Energy matrix

c) Node matrix

d) Element stiffness matrix

Answer: d

4. Body force vector f e = _____________

a) \

A e l e \

A e l e f \(

Answer: a

Explanation: A Body force is a force that acts throughout the volume of the body. Forces due to gravity, electric and magnetic fields are examples of body forces.

5. Between wheel and ground how much of traction force is required?

a) High traction force

b) Low traction force

c) Infinite traction force

d) No traction force

Answer: a

6. Element traction force is given by ___

a) T e =Tl e \

T e =Tl e

c) T e =\

Undefined

Answer: c

Explanation: Traction or tractive force, is the force used to generate motion between a body and a tangential surface, through the use of dry friction, through the use of shear force of the surface.

7. ∏ = \(\frac{1}{2}\) Q T KQ-Q T F In this equation F is defined as _________

a) Global displacement vector

b) Global load vector

c) Global stiffness matrix

d) Local displacement vector

Answer: b

Explanation: Global load vector is assembly of all local load vectors. This load vector is obtained by due to given load. In the given equation F is defined as global load vector.

8. What are the basic unknowns on stiffness matrix method?

a) Nodal displacements

b) Vector displacements

c) Load displacements

d) Stress displacements

Answer: a

9. Write the element stiffness for a truss element.

a) K=\

K=\

K=AE

Answer: b

Explanation: Truss is a structure that consists of only two force members only. Where the members are organized so that the assemblage as a whole behaves as a single object.

10. Formula for global stiffness matrix is ____________

a) No. of nodes*Degrees of freedom per node

b) No. of nodes

c) Degrees of freedom per node

d) No. of elements

Answer: a

Size of global stiffness matrix=No. of nodes*Degrees of freedom per node.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “One Dimensional Problems – Galerkin Approach”.

1. Galerkin technique is also called as _____________

a) Variational functional approach

b) Direct approach

c) Weighted residual technique

d) Variational technique

Answer: c

2. In the equation, \

Adx -\int_{L} \phi^T f Adx -\int_{L}\phi^Tdx – \sum_{i}\phi_i P_i=0\) First term represents _______

a) External virtual work

b) Virtual work

c) Internal virtual work

d) Total virtual work

Answer: c

3. Considering element connectivity, for example for element ψ=[ψ 1 , ψ 2 ] n for element n, then the variational form is ______________

a) ψ T =0

b) ψ=0

c) ψ=F

d) ψ=0

Answer: a

4. Write the element stiffness matrix for a beam element.

a) K=\

K=\

K=\(\frac{2E}{l}

Answer: b

Explanation: Element stiffness matrix means it is a matrix method that makes use of the members stiffness relations for computing member forces and displacements in the structures.

5. Element connectivities are used for _____

a) Traction force

b) Assembling

c) Stiffness matrix

d) Virtual work

Answer: b

6. Virtual displacement field is _____________

a) K=\

F=ma

c) f=y

d) &fcy;=&fcy;

Answer: d

7. Virtual strain is ____________

a) εMisplaced &=\

εMisplaced &=\

&fcy;=\(\frac{d\varepsilon}{d\phi}\)

Answer: b

8. To solve a galerkin method of approach equation must be in ___________

a) Equation

b) Vector equation

c) Matrix equation

d) Differential equation

Answer: d

9. By the Galerkin approach equation can be written as __________

a) {P}-{K}{Δ}=0

b) {K}-{P}{Δ}=0

c) {Δ}-{p}{K}=0

d) Undefined

Answer: a

Explanation: Galerkin’s method of weighted residuals, the most common method of calculating the global stiffness matrix in fem. This requires the boundary element for solving integral equations.

10. In basic equation Lu=f, L is a ____________

a) Matrix function

b) Differential operator

c) Degrees of freedom

d) No. of elements

Answer: b

This set of Finite Element Method Interview Questions and Answers focuses on “One Dimensional Problems – Assembly of the Global Stiffness Matrix and Load Vector”.

1. How is Assembly of stiffness matrix symbolically denoted?

a) K={k} e

b) K←∑ e K e

c) K←∑K e

d) Undefined

Answer: b

Explanation: The stiffness matrix represents the system of linear equations that must be solved in order to ascertain an approximate solution to differential equation.

2. What is the Strain energy equation?

a) U e =\(\frac{1}{2}\)q T k e q

b) U e =\(\frac{1}{2}\)q e k e q

c) U e =\(\frac{1}{2}\)qk e

d) U e =\(\frac{1}{2}\)q T k e

Answer: a

3. What is the actual equation of stiffness matrix?

a) K=\

K=\

K=\(\frac{AE}{l}

Answer: d

4. From where does the global load vector F is assembled?

a) Element force vectors only

b) Point loads only

c) Both element force vectors and point loads

d) Undefined

Answer: c

Explanation: Global load vector is assembling of all local load variables. This global load vector is get from assembling of both element force vectors and point loads.

5. For an element as given below, what will be the 1 ST element stiffness matrix?

Find the 1st element stiffness matrix for the given element

a) \

\(\frac{EA_1}{l_1}

Answer: a

6. Principal of minimum potential energy follows directly from the principal of ________

a) Elastic energy

b) Virtual work energy

c) Kinetic energy

d) Potential energy

Answer: b

7. The points at where kinetic energy increases dramatically then those points are called _______

a) Stable equilibrium points

b) Unstable equilibrium points

c) Equilibrium points

d) Unique points

Answer: b

8. We can obtain same assembly procedure by Stiffness matrix method and _______

a) Potential energy method

b) Rayleigh method

c) Galerkin approach

d) Vector method

Answer: c

9. By element stiffness matrix we can get relation of members in an object in _____

a) Different matrices

b) One matrix

c) Identity matrix

d) Singular matrix

Answer: b

10. What is the Global stiffness method called?

a) Multiple matrix

b) Direct stiffness matrix

c) Unique matrix

d) Vector matrix

Answer: b

11. Which technique do traditional workloads use?

a) Scale out technique

b) Scale up technique

c) Building technique

d) Shrinking technique

Answer: b

Explanation: When the workload increases on the system, the machine scales up by adding more RAM, CPU and storage spaces.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “One Dimensional Problems – Properties of K”.

1. Dimension of global stiffness matrix is _______

a) N X N , where N is no of nodes

b) M X N , where M is no of rows and N is no of columns

c) Linear

d) Eliminated

Answer: a

2. Each node has only _______

a) Two degrees of freedom

b) One degree of freedom

c) Six degrees of freedom

d) Three degrees of freedom

Answer: b

Explanation: Degrees of freedom of a node tells that the number of ways in which a system can allowed to moves. In a stiffness matrix each node can have one degree of freedom.

3. Global stiffness K is a______ matrix.

a) Identity matrix

b) Upper triangular matrix

c) Lower triangular matrix

d) Banded matrix

Answer: d

4. The dimension of K banded is _____

a) [N X NBW ]

b) [NBW X N]

c) [N X N]

d) [NBW X NBW]

Answer: a

Explanation: K can be compactly represented in banded form. As K banded is of dimension [N X NBW] where NBW is the half band width.

5. In many one-dimensional problems, the banded matrix has only two columns. Here NBW=____

a) 6

b) 3

c) 7

d) 2

Answer: d

Explanation: NBW means half bandwidth. Many of the One- dimensional problems banded matrix has only 2 columns then NBW=2. We know that

NBW =max +1

6. Stiffness matrix represents a system of ________

a) Programming equations

b) Iterative equations

c) Linear equations

d) Program CG SOLVING equations

Answer: c

7. Stiffness matrix is _____

a) Non symmetric and square

b) Symmetric and square

c) Non symmetric and rectangular

d) Symmetric and rectangular

Answer: b

Explanation: Stiffness matrix is a inherent property of the structure. The property of a stiffness matrix, as the stiffness matrix is square and symmetric.

8. In stiffness matrix, all the _____ elements are positive.

a) Linear

b) Zigzag

c) Diagonal

d) Rectangular

Answer: c

9. The size of global stiffness matrix will be equal to the total ______ of the structure.

a) Nodes

b) Degrees of freedom

c) Elements

d) Structure

Answer: b

10. Element stiffness is obtained with respect to its ___

a) Degrees of freedom

b) Nodes

c) Axes

d) Elements

Answer: c

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Finite Element Equations – Treatment of Boundary Conditions”.

1. Types of Boundary conditions are ______

a) Potential- Energy approach

b) Penalty approach

c) Elimination approach

d) Both penalty approach and elimination approach

Answer: d

2. Potential energy, π = _________

a) \(\frac{1}{2}\)Q T KQ-Q T F

b) QKQ-QF

c) \

\(\frac{1}{2}\)QF

Answer: a

Potential energy π=\(\frac{1}{2}\)Q T KQ-Q T F

3. Equilibrium conditions are obtained by minimizing ______

a) Kinetic energy

b) Force

c) Potential energy

d) Load

Answer: c

4. In elimination approach, which elements are eliminated from a matrix ____

a) Force

b) Load

c) Rows and columns

d) Undefined

Answer: c

5. In elimination approach method, extract the displacement vector q from the Q vector. By using ___

a) Potential energy

b) Load

c) Force

d) Element connectivity

Answer: d

6. Penalty approach method is easy to implement in a ______

a) Stiffness matrix

b) Iterative equations

c) Computer program

d) Cg solving

Answer: c

7. If Q 1 =a 1 then a 1 is _________

a) Displacement

b) Symmetric

c) Non symmetric

d) Specified displacement

Answer: d

Explanation: In penalty approach method a 1 is known as specified displacement of 1. This is used to model the boundary conditions.

8. The first step of penalty approach is, adding a number C to the diagonal elements of the stiffness matrix. Here C is a __________

a) Large number

b) Positive number

c) Real number

d) Zero

Answer: a

9. In penalty approach evaluate _______ at each support.

a) Load vector

b) Degrees of freedom

c) Force vector

d) Reaction force

Answer: d

10. For modeling of inclined roller or rigid connections, the method used is ___

a) Elimination approach

b) Multiple constraints

c) Penalty approach

d) Minimum potential energy theorem

Answer: b

Explanation: Multiple constraints is one of the method for boundary conditions it is generally used in problems for modeling inclined rollers or rigid connections.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “One Dimensional Problems – Quadratic Shape Function”.

1. What is a shape function?

a) Interpolation function

b) Displacement function

c) Iterative function

d) Both interpolation and displacement function

Answer: d

2. Quadratic shape functions give much more _______

a) Precision

b) Accuracy

c) Both Precision and accuracy

d) Identity

Answer: b

3. Strain displacement relation ______

a) ε=\

ε=\

x=\

Cannot be determined

Answer: a

Explanation: The relationship is that connects the displacement fields with the strain is called strain – displacement relationship. If strain is ε then strain – displacement relation is

ε=\(\frac{du}{dx}\)

4. The _____ and ______ can vary linearly.

a) Force and load

b) Precision and accuracy

c) Strain and stress

d) Distance and displacement

Answer: c

5. By Hooke’s law, stress is ______

a) σ=Bq

b) σ=EB

c) B=σq

d) σ=EBq

Answer: d

Explanation: Hooke’s law states that the strain in a solid is proportional to the applied stress within the elastic limit of that solid.

6. Nodal displacement as _____

a) u=Nq

b) N=uq

c) q=Nu

d) Program SOLVING

Answer: a

Explanation: Nodes will have nodal displacements or degrees of freedom which may include translations, rotations and for special applications, higher order derivatives of displacements.

7. At the condition, at , N 1 =1 at ξ=-1 which yields c=\Missing open brace for subscript Computer functions

b) Programming functions

c) Galerkin function

d) Lagrange shape functions

Answer: d

Explanation: The lagrange shape function sum to unity everywhere. At the given condition the shape functions are named as Lagrange shape functions.

8. Element body force vector is given by ______

a) f e =\(\frac{A_el_ef}{2} \int_{-1}^{1}\) N T dξ

b) f e = \(\frac{A_el_e}{2}\int_{}^{1}\) N T dξ

c) Conditioning strategies

d) f e =∫ N T dξ

Answer: a

Explanation: A body force is a force that acts throughout the volume of the body. In FEM, Element body force vector is given by

f e =\(\frac{A_el_ef}{2} \int_{-1}^{1}\) N T dξ

9. Element traction force is given by ___

a) T e =\(\frac{l_e}{2}\int_{-1}^{1}\)N T dξ

b) T e =\(\frac{l_eT}{2}\int_{-1}^{1}\)N T dξ

c) T e =\(\frac{l_eT}{2}\int_{-1}^{1}\)N T

d) Undefined

Answer: b

Explanation: The term traction force can either refer to the total traction a vehicle exerts on a surface or the amount of total traction that is parallel to the direction of motion.

Element traction force is given by

T e =\(\frac{l_eT}{2}\int_{-1}^{1}\)N T dξ

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “One Dimensional Problems – Temperature Effects”.

1. With temperature effect which will vary linearly?

a) Horizontal stress load

b) Potential energy

c) Vertical stress load

d) Kinematic energy

Answer: c

Explanation: Temperature is a variant which varies from one point to another point. It has adverse effects on different structures. By temperature effect Vertical stress load vary linearly.

2. α means ____

a) Co-efficient of thermal expansion

b) Co-efficient of linear expansion

c) Thermal expansion

d) Thermal effect

Answer: a

3. In temperature effect, initial strain, ε 0 = ____

a) α ΔT

b) α+ΔT

c) α-ΔT

d) Load

Answer: a

4. In a structure, a crack is formed as a result of ______

a) Thermal expansion

b) Thermo couple

c) Thermal strain

d) Thermal stress

Answer: d

5. In the given diagram, the line indicates ____________

The line indicating stress-strain relation in the diagram

a) Stress relation

b) Strain relation

c) Stress – strain relation

d) Undefined

Answer: c

6. Stress – strain relation is given as _____

a) σ=E

b) σ=Eε

c) σ=E(ε-ε 0 )

d) Undefined

Answer: c

σ=E(ε-ε 0 ).

7. Strain energy per unit volume is ___

a) u 0 =E(ε-ε 0 )

b) u 0 =\(\frac{1}{2}\)E(ε-ε 0 )

c) Non symmetric

d) Specified displacement

Answer: b

u 0 =\(\frac{1}{2}\)E(ε-ε 0 )

8. Temperature change is denoted as_____

a) ΔT=(T 2 -T 1 )

b) θ e

c) l e

d) A e

Answer: b

9. θ e =\

True

b) False

Answer: a

θ e =\(\frac{E_eA_el_e\alpha\Delta T}{x_2-x_1}

\).

10. Stress in each element is ____

a) Eliminated

b) σ=EBq

c) σ=αΔT

d) σ=E

Answer: d

σ=E.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Plane Trusses”.

1. Plane trusses are also known as _____

a) One–dimensional trusses

b) Two-dimensional trusses

c) Three-dimensional trusses

d) Poly dimensional trusses

Answer: b

2. A truss structure consists only ___ force members.

a) Only one

b) Two

c) Three

d) Poly

Answer: b

3. Plane truss element can be shown in _____

a) Local coordinate system

b) Global coordinate system

c) Local and global coordinate systems

d) Dimensional structure

Answer: c

4. The truss element is a _____ when we see it in a local co-ordinate system.

a) Three dimensional

b) One dimensional

c) Two dimensional

d) Thermal component

Answer: b

5. Where l and m are direct cosines, then transformation matrix L is given by ___

a) \

Answer: d

Explanation: The direct cosines l and m are introduced as l = cos θ and m=cos &fcy;. These direction cosines are the cosines of the angles that the local x’-axis makes with the global x-, y- axes.

6. Formula for direct cosine l=cosθ= ______

a) x 2 -x 1

b) \

\(\frac{x_2-x_1}{l_e}\)

Answer: d

Explanation: A truss is a two node element which allows arbitrary orientation in XYZ co-ordinate system. Planar truss is one where all the members and nodes lie within two dimensional plane.

7. The direct cosine m is given by formula cos&fcy;= ____

a) \

\(\frac{x_2-y_2}{l_e}\)

Answer: b

8. Strain energy in global co-ordinates can be written as ____

a) q ’ T k ’ q ’

b) q ’ T k ’

c) k ’ q ’

d) \(\frac{1}{2}\) q ’ T k ’ q ’

Answer: d

9. The stress σ in a truss element is given by ____

a) σ=E

b) σ=ε

c) σ=E e ε

d) σ=α

Answer: c

10. In a truss element, element temperature load is ___

a) θ e =E e A e

b) θ e =E e A e ε 0

c) θ e =E e A e ε 0 \

θ e =E e A e ε 0 \(

Answer: d

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Three Dimensional Trusses”.

1. In a 3-D truss element nodal displacement vector in a local co-ordinate is q’= ___

a) [q 1 ‘ ]

b) [q 2 ‘ ]

c) [q 1 ‘ ,q 2 ‘ ,q 3 ‘ ] T

d) [q 1 ‘ ,q 2 ‘ ] T

Answer: d

2. Transformation between local and global co-ordinates is _____

a) q ‘ =Lq

b) q ‘ =L

c) q ‘ =L+q

d) q ‘ =\(\frac{L}{q}\)

Answer: a

Explanation: Transformations are frequently used in linear algebra and computer graphics, since transformations can easily be represented, combined and computed.

3. For a 3D truss element, transformation matrix L between local and global co-ordinates is given by ___

a) \

Answer: d

4. The direct cosines l, m, n are of local _____

a) z’- axes

b) x’- axes

c) y’- axes

d) None of the above

Answer: b

5. The direction cosines l, m, n with respect to ____ x-, y-, z- axes respectively.

a) Local

b) Local and global

c) Global

d) Transformation

Answer: c

6. In 3D truss, formula for direct cosine l = _____

a) x 2 -x 1

b) \

\(\frac{x_2-x_1}{l_e}\)

Answer: d

7. The direct cosine m is given by formula ____

a) \

\(\frac{x_2-y_2}{l_e}\)

Answer: b

8. The direction cosine n is given by formula _____

a) \

Undefined

Answer: c

Explanation: A truss is a two node element. The three D truss element can be treated as straight forward generalization of the 2D truss element.

9. In 3D truss element, the nodal displacement vector in global co-ordinates is _____

a) q ‘ = [q 1 ,q 2 ]

b) q ‘ = [q 1 ,q 2 ,q 3 ]

c) q ‘ = [q 1 ,q 2 ,q 3 ,q 4 ,q 5 ,q 6 ]

d) Undefined

Answer: c

Explanation: Nodes will have nodal displacements or degrees of freedom which may include translations, rotations, and for special applications, higher order derivatives of displacements.

10. In a truss element, element temperature load is ___

a) θ e =E e A e

b) θ e =E e A e ε 0

c) θ e =E e A e ε 0 \

θ e =E e A e ε 0 \(

Answer: d

This set of Finite Element Method Questions and Answers for Freshers focuses on “Assembly of Global Stiffness Matrix for the Banded & Skyline Solutions”.

1. What is a banded matrix?

a) Sparse matrix

b) Rectangular matrix

c) Unit matrix

d) Square matrix

Answer: a

Explanation: In matrix theory band matrix is a sparse matrix, whose non-zero entities are confined to a diagonal band. Comprising the main diagonal and zero are more diagonals on either side.

2. Skyline matrix storage is in the form of ______

a) Banded matrix

b) Sparse matrix

c) Singular matrix

d) Identity matrix

Answer: b

3. Symmetry and sparsity of the global stiffness matrix can be approached by _____ methods.

a) One

b) Three

c) Two

d) Four

Answer: c

4. Which of these was one of the methods for determining assembly of global stiffness matrix?

a) Galerkin approach

b) Skyline approach

c) Rayleigh method

d) Assembly method

Answer: b

5. In banded matrix, elements are _____ placed in stiffness matrix.

a) Singular

b) Determinant values

c) Directly

d) Indirectly

Answer: c

6. In Skyline matrix, the elements in a stiffness matrix can be placed in _______

a) Direct values

b) Determinant values

c) Load values

d) Vector form

Answer: d

7. Formula for maximum span or half band width in banded approach is _____

a) m e =[|i-j|+1]

b) \

q ‘ =lq

d) m e =[2|i-j|+1]

Answer: d

m e =[2|i-j|+1]

8. The first step of skyline assembly matrix involves evaluation of ____

a) Column height

b) Row height

c) Matrix height

d) Undefined

Answer: a

9. The second step in skyline approach is assembling the element stiffness values into _____

a) Row vector

b) Identity vector

c) Column vector

d) Determinant vector

Answer: c

10. The details of a skyline assembly matrix are implemented in a program called ____

a) Boolean program

b) Cholesky program

c) Truss program

d) Trussky program

Answer: d

Explanation: As this assembly was done to trusses by default all the steps applied in skyline approach were implemented in program TRUSSKY.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “ Two Dimensional Problems – Finite Element Modelling”.

1. In 2D elements. Discretization can be done. The points where triangular elements meet are called ____

a) Displacement

b) Nodes

c) Vector displacements

d) Co-ordinates

Answer: b

Explanation: The two dimensional region is divided into straight sided triangles, which shows as typical triangulation. The points where the corners of the triangles meet are called nodes.

2. Each triangle formed by three nodes and three sides is called a ______

a) Node

b) Force matrix

c) Displacement vector

d) Element

Answer: d

3. The finite element method is used to solve the problem ______

a) Uniformly

b) Vigorously

c) Approximately

d) Identically

Answer: c

4. In two dimensional modeling each node has ____ degrees of freedom.

a) One

b) Infinity

c) Finite

d) Two

Answer: d

Explanation: In two dimensional problem, each node is permitted to displace in the two directions x and y. Thus each node has two degrees of freedom.

5. For a triangular element,element displacement vector can be denoted as ___

a) q=[q 1 ,q 2 ,q 3 ] T

b) q=[q 1 ,q 2 ] T

c) q=[q 1 ,q 2 ,……q 6 ] T

d) Load vector

Answer: c

Explanation: The displacement components of a local node is represented in x and y directions, respectively. For that we denote element displacement vector as

q=[q 1 ,q 2 ,……q 6 ] T .

6. In two dimensional analysis, stresses and strains are related as ___

a) σ=Dε

b) σ=ε

c) Load values

d) ε=Dσ

Answer: a

Explanation: When a material is loaded with force, it produces stress. Which then cause material to deform. Strain is response of a system t an applied stress.

7. In two dimensional modeling, body force is denoted as ___

a) f=[f x ,f y ] T

b) σ=Dε

c) q ‘ =lq

d) f=[2|i-j|+1]

Answer: a

f=[f x ,f y ] T .

8. The information of array of size and number of elements and nodes per element can be seen in ___

a) Column height

b) Element connectivity table

c) Matrix form

d) Undefined

Answer: b

9. In two dimensional modeling, traction force is denoted as ____

a) Row vector

b) T=[T x ,T y ] T

c) f=[f x ,f y ] T

d) σ=Dε

Answer: b

T=[T x ,T y ] T

10. In two dimensional modeling, elemental volume is given by ____

a) dV=tdA

b) dV=dA

c) f=[f x ,f y ] T

d) Trussky program

Answer: a

dV=tdA.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Two Dimensional Problems – Constant Strain Triangle”.

1. Finite element method uses the concept of _____

a) Nodes and elements

b) Nodal displacement

c) Shape functions

d) Assembling

Answer: c

2. For constant strain elements the shape functions are ____

a) Spherical

b) Quadratical

c) Polynomial

d) Linear

Answer: d

3. Linear combination of these shape functions represents a ______

a) Square surface

b) Linear surface

c) Plane surface

d) Combinational surface

Answer: c

4. In particular, N 1 +N 2 +N 3 represent a plane at a height of one at nodes ______

a) One

b) Two

c) Three

d) One, two and three

Answer: d

5. If N 3 is dependent shape function, It is represented as ____

a) N 3 =ξ

b) N 3 =1-ξ

c) N 3 =1-η

d) N 3 =1-ξ-η

Answer: d

6. In two dimensional problems x-, y- co-ordinates are mapped onto ____

a) x-, y- co-ordinates

b) x-, ξ – co-ordinates

c) η-, y- co-ordinates

d) ξ-η-Co-ordinates

Answer: d

7. The shape functions are physically represented by _____

a) Triangular co-ordinates

b) ξ-,η-Co-ordinates

c) Area co-ordinates

d) Surface co-ordinates

Answer: c

8. A 1 is the first area and N 1 is its shape function then shape function N 1 = ___

a) A 1 /A

b) A-A 1

c) A 1 +A

d) A 1

Answer: a

Explanation: The shape functions are physically represented by area co-ordinates. A point in a triangle divides into three areas. The shape functions are precisely represented as

N 1 =A 1 /A .

9. The equation u=Nq is a _____ representation.

a) Nodal

b) Isoparametric

c) Uniparametric

d) Co-ordinate

Answer: b

Explanation: The isoparametric representation of finite elements is defined as element geometry and displacements are represented by same set of shape functions.

10. For plane stress or plane strain, the element stiffness matrix can be obtained by taking _____

a) Shape functions, N

b) Material property matrix, D

c) Iso parametric representation, u

d) Degrees of freedom, DoF

Answer: b

Explanation: The material property matrix is represented as ratio of stress to strain that is σ=Dε . Therefore by this relation element stiffness matrix can be obtained by material property matrix.

11. In a constant strain triangle, element body force is given as ____

a) f e =[f x ,f y ,f x ,f y ,f x ,f y ] T

b) f e =\

f e =\(\frac{t_eA_e}{3}\)[f x ,f y ,f x ,f y ,f x ,f y ] T

d) f e =\(\frac{t_eA_e}{3}\)[f x ,f y ] T

Answer: c

12. Traction force term represented as ___

a) ∫u T Tl

b) ∫u T T

c) ∫u T

d) u T Tl

Answer: a

13. In the equation KQ=F, K is called as ____

a) Stiffness matrix

b) Modified stiffness matrix

c) Singular stiffness matrix

d) Uniform stiffness matrix

Answer: b

14. Principal stresses and their directions are calculated by using ____

a) Galerkin approach

b) Rayleigh method

c) Potential energy method

d) Mohr’s circle method

Answer: d

Explanation: Mohr’s circle is two dimensional graphical representation of the transformation law. While considering longitudinal stresses and vertical stresses in a horizontal beam during bending.

15. I the distribution of the change in temperature ΔT, the strain due to this change is ____

a) Constant strain

b) Stress

c) Initial strain

d) Uniform strain

Answer: c

Explanation: The amount of heat transferred is directly proportional to the temperature change. The distribution of change in temperature, the strain due to this change is initial strain.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Two Dimensional Problem Modelling and Boundary Conditions”.

1. Finite element method is used for computing _____ and _____

a) Stress and strain

b) Nodes and displacement

c) Nodes and elements

d) Displacement and strain

Answer: d

2. In deformation of the body, the symmetry of ______ and symmetry of ____ can be used effectively.

a) Stress and strain

b) Nodes and displacement

c) Geometry and strain

d) Geometry and loading

Answer: d

Explanation: Deformation changes in an object’s shape or form due to the application of a force or forces. Deformation proportional to the stress applied within the elastic limits of the material.

3. For a circular pipe under internal or external pressure, by symmetry all points move _____

a) Radially

b) Linearly

c) Circularly

d) Along the pipe

Answer: a

4. Boundary conditions can be easily considered by using _______

a) Rayleigh method

b) Penalty approach method

c) Galerkin approach

d) Potential energy approach

Answer: b

Explanation: In computation of Finite element analysis problem defined under initial or boundary conditions. For implementation of boundary conditions we need a staggered grid.

5. When dividing an area into triangles, avoid large _____

a) Dimensions

b) Loading

c) Aspect ratios

d) Boundary conditions

Answer: c

Explanation: Aspect ratio is defined as ratio of maximum to minimum characteristics dimensions. For this reason we can avoid large aspect ratios when dividing an area into triangles.

6. In dividing the elements a good practice may be to choose corner angles in the range of ____

a) 30-120°

b) 90-180°

c) 25-75°

d) 45-180°

Answer: a

7. Stresses can be change widely at ____

a) Large circular sections

b) Notches and fillets

c) Corners

d) Crystals

Answer: b

8. The Constant strain triangle can give____ stresses on elements.

a) Linear

b) Constant

c) Uniform

d) Parallel

Answer: b

9. The _____ can be obtained even with coarser meshes by plotting and extrapolating.

a) Minimum stresses

b) Minimum strain

c) Maximum stresses

d) Maximum strain

Answer: c

10. Coarser meshes are recommended for _____

a) Loading

b) Notches and fillets

c) Crystals

d) Initial trails

Answer: d

Explanation: The smaller elements will better represent the distribution. Coarse mesh is more accurate in getting values. Better estimates of maximum stress may obtained even with the coarse meshes.

11. Increasing the number of nodes in coarse mesh regions where stress variations are high, should give better results. This method is called _____

a) Divergence

b) Convergence

c) Convergent- divergent

d) Un defined

Answer: b

Explanation: At the initial trails, errors may be fixed, before running large number of elements. The convergence is successively increasing the number of elements in finite element meshes.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Two Dimensional Problems – Orthotropic Materials”.

1. Give an example of orthotropic material?

a) Topaz

b) Aluminum

c) Barium

d) Sodium

Answer: a

2. Unidirectional fiber- reinforced composites also exhibit _______ behavior.

a) Isotropic

b) Orthotropic

c) Material

d) Unidirection composite

Answer: b

3. Orthotropic planes have ____ mutually perpendicular planes of elastic symmetry.

a) One

b) Two

c) Three

d) Four

Answer: c

4. The principal material axes that are normal to the _______

a) Co-ordinates

b) Number of nodes

c) Principal axes

d) Plane of symmetry

Answer: d

5. v 12 indicates that the poisson’s ratio that characterizes the decrease in ______ during tension applied in ______

a) 2- direction and 1- direction

b) 2- direction and 3- direction

c) 1- direction and 2- direction

d) 2- direction and 4- direction

Answer: a

6. Unidirectional composites are stacked at different fiber orientations to form a ______

a) Laminate

b) Orthotropic material

c) Isotropic material

d) Anisotropic material

Answer: a

7. When an orthotropic plate is loaded parallel to its material axes, it results only _____

a) Shear strains

b) Normal strains

c) Parallel strains

d) Uniform strains

Answer: b

8. When the stresses are determined in an orthotropic material, then they are used to determine ____

a) Strains

b) Deformation

c) Factor of safety

d) Loads

Answer: c

9. Stress- strain law defined as ______

a) σ=D(ε-ε 0 )

b) σ=D

c) σ=Dε

d) σ=Dε 0

Answer: a

10. E 1 value of Balsa wood is ___

a) 0.125*10 6 psi

b) 12.04*10 6 psi

c) 23.06*10 6 psi

d) 7.50*10 6 psi

Answer: a

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Axis Symmetric Formulation”.

1. Consider an Axisymmetric problem of which is having a radius of r, rotational angle θ and length l. Then r dl dθ is known as ____

a) Elemental volume

b) Element

c) Elemental surface area

d) Elemental surface

Answer: c

2. The stress vector is correspondingly defined as ___________

a) σ=[σ y ,σ z ,τ yz ,σ θ ] T

b) σ=[σ y ,σ z ] T

c) u=[u, w] T

d) T=[T y ,T z ] T

Answer: a

3. Problems involving three- dimensional axisymmetric solids or solids of revolution, subjected to _____ loading.

a) Rotational

b) Two-dimensional

c) Three-dimensional

d) Axisymmetric loading

Answer: d

4. All deformations and stresses are independent of _______

a) Co-ordinates

b) Number of nodes

c) Rotational angle, θ

d) Area

Answer: c

5. Revolving bodies like fly wheels can be analyzed by introducing _______ in body force term.

a) Gravitational force

b) Revolving force

c) Centrifugal force

d) Centripetal force

Answer: c

Explanation: The centrifugal force is an inertial force directed away from the axis of rotation that appears to act on all objects when viewed in a rotating frame of reference.

6. The stress- strain relation is given as ___________

a) σ=D

b) σ=Dε

c) σ=ε

d) ε=Dσ

Answer: b

7. Axisymmetric problems are totally defined in ______

a) xy planes

b) yz planes

c) rz planes

d) rθ planes

Answer: c

8. For axisymmetry solids gravity forces can be considered if deformation and stresses act on _____

a) X direction

b) Z direction

c) Y direction

d) Parallel to plane

Answer: b

9. In axisymmetric solids, stress- strain law can be defined as ______

a) σ=D(ε-ε 0 )

b) σ=D

c) σ=Dε

d) σ=Dε 0

Answer: c

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Finite Element Modelling – Triangular Element”.

1. The transformation relationships into strain displacement relations. Then the equation can be written as ____

a) ε=Bq

b) ε=Dq

c) ε=q

d) Elemental surface

Answer: a

ε=Bq.

2. In the equation U e =\(\frac{1}{2}\)2q T (2∏ ∫ B T DBrdA)q the quantity inside the paranthesis is _____

a) Axisymmentric

b) Strain displacement relationships

c) Stiffness matrix

d) Symmetric matrix

Answer: c

3. The volume of ring shaped element is _____

a) A e =\

A e =detJ

c) 2πr

d) 4πr 2

Answer: a

4. The element body force vector f e is given by _____

a) Co-ordinates

b) f e =\(\frac{2Πr̅A_e}{3}\)[f̅ r ,f̅ z ,f̅ r ,f̅ z ,f̅ r ,f̅ z ] T

c) f e =\(\frac{2Πr̅A_e}{3}\)[f̅ x ,f̅ y ] T

d) f e =\(\frac{2Πr̅A_e}{3}\)

Answer: b

5. A rotating flywheel with its axis in the z direction. We consider the flywheel to be stationary and apply the equivalent radial centrifugal force per unit volume is _____

a) 2Πr

b) 4Πr 2

c) ρrω 2

d) ρω 2

Answer: c

Explanation: The centrifugal force is an inertial force directed away from the axis of rotation that appears to act on all objects when viewed in a rotating frame of reference.

6. Surface traction of a uniformly distributed load with components T 1 and T 2 is _____

a) q T T e =2Π∫ e u T Trdl

b) q T T e =2Π

c) σ=ε

d) ε=Dσ

Answer: a

7. On summing up the strain energy and force terms over all the elements and modifying for the boundary conditions while minimizing the total potential energy. We get ______

a) σ=D

b) Kinematic energy

c) ε=Dσ

d) KQ=F

Answer: c

8. Using the connectivity of the elements, the internal virtual work can be expressed in the form _____

a) Ψ T =KQ

b) Ψ T =K

c) KQ=F

d) σ=Dε

Answer: a

9. In axisymmetric problems, by using stress strain relation and strain displacement relation we can obtain an equation that is ____

a) σ=DB̅q

b) σ=D

c) σ=Dε

d) σ=Dε 0

Answer: c

10. The temperature effect on axisymmetric formulation. The vector ε̅ 0 is the initial strain evaluated at the centroid, representing the average temperature rise of the element is _____

a) θ e =2Πr̅

b) θ e =2Πr̅A e B̅ T Dε̅ 0

c) K=QF

d) σ=Dε 0

Answer: b

11. Uniform increase in temperature of, ΔT introduces initial _____

a) Normal strain

b) Strain

c) Stresses

d) Kinetic energy

Answer: a

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Axis Symmetric Problem Modelling and Boundary Conditions”.

1. Axisymmetry implies that points lying on the z- axis remains _____ fixed.

a) Tangentially

b) Spherically

c) Radially

d) Circularly

Answer: c

2. Modeling of a cylinder of infinite length subjected to external pressure. The length dimensions are assumed to be _____

a) Finite

b) Non uniform

c) Perpendicular

d) Constant

Answer: d

3. Press fit of a ring of length L and internal radius r j onto a rigid shaft of radius r 1 +δ is considered. When symmetry is assumed about the mid plane, this plane is restrained in the _____

a) X direction

b) Y direction

c) Z direction

d) Undefined

Answer: c

4. The condition that nodes at the internal radius have to displace radially by δ , a large stiffness C is added to the _____

a) Co-ordinates

b) Length

c) Diagonal locations

d) Radius

Answer: c

5. Press fit on elastic shaft, may define pairs of nodes on the contacting boundary, each pair consisting of one node on the _____ and one on the ______

a) Shaft and couple

b) Sleeve and shaft

c) Shaft and sleeve

d) Sleeve and couple

Answer: b

Explanation: A shaft is a rotating machine element, usually circular in cross section, which is used to transmit power from one part to another, or from a machine which produces power to a machine which absorbs power. A flexible shaft or an elastic shaft is a device for transmitting rotary motion between two objects which are not fixed relative to one another.

6. For a Belleville spring the load is applied on _____

a) Shaft

b) Hole

c) Periphery of the circle

d) Coupling

Answer: c

Explanation: The Belleville spring, also called the Belleville washer, is a conical disk spring. The load is applied on the periphery of the circle and supported at the bottom.

7. On Belleville spring the load is applied in ______

a) X direction

b) Z direction

c) Y direction

d) Axial direction

Answer: d

8. In the Belleville spring, the load-deflection curve is _____

a) Linear

b) Curved

c) Non linear

d) Parabolic

Answer: c

9. A steel sleeve inserted into a rigid insulated wall. The sleeve fits snugly, and then the temperature is raised by _____

a) Uniform

b) Non uniform

c) σ

d) ΔT

Answer: d

10. In a Belleville spring, load-deflection characteristics and stress distribution can be obtained by dividing the area into ____

a) Surfaces

b) Nodes

c) Elements

d) Loads

Answer: c

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Two Dimensional Isoparametric Elements – Four Node Quadrilateral”.

1. In two dimensional isoparametric elements, we can generate element stiffness matrix by using ____

a) Numerical integration

b) Differential equations

c) Partial derivatives

d) Undefined

Answer: a

Explanation: The term isoparametric is derived from the use of the same shape functions [N] to define the element’s geometric shape as are used to define the displacements within the element.

2. The vector q=[q 1 ,q 2 ………q 8 ] T of a four noded quadrilateral denotes ____

a) Load vector

b) Transition matrix

c) Element displacement vector

d) Constant matrix

Answer: c

3. For a four noded quadrilateral, we define shape functions on _____

a) X direction

b) Y direction

c) Load vector

d) Master element

Answer: d

4. The master element is defined in ______

a) Co-ordinates

b) Natural co-ordinates

c) Universal co-ordinates

d) Radius

Answer: b

5. Shape function can be written as _____

a) N t =

b) N t =

c) N t =

d) N t =\

Answer: d

6. For a four noded element while implementing a computer program, the compact representation of shape function is ____

a) N t =\

b) N t =

c) N t =\(\frac{1}{4}\)(1+ξξ i )(1+ηη i )

d) Undefined

Answer: c

7. For a four noded quadrilateral elements, In u T =[u.v] T the displacement elements can be represented as u=N 1 q 1 +N 2 q 3 + N 3 q 5 + N 4 q 7

v= N 1 q 2 +N 2 q 4 + N 3 q 6 + N 4 q 8

then the shape function can be represented as _____

a) \

\(N=

Answer: d

8. The stiffness matrix from the quadrilateral element can be derived from _____

a) Uniform energy

b) Strain energy

c) Stress

d) Displacement

Answer: b

Explanation: In the finite element method for the numerical solution of elliptic partial differential equations, the stiffness matrix represents the system of linear equations that must be solved in order to as certain an approximate solution to the differential equation.

9. For four noded quadrilateral element, the global load vector can be determined by considering the body force term in _____

a) Kinetic energy

b) Potential energy

c) Kinematic energy

d) Temperature

Answer: b

10. Shape functions are linear functions along the _____

a) Surfaces

b) Edges

c) Elements

d) Planes

Answer: b

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Numerical Integration”.

1. Which method of approach is useful for evaluating four noded quadratic elements?

a) Numerical integration

b) Penality approach method

c) Gaussian quadrature approach

d) Rayleighs method

Answer: c

2. One point formula in quadratic approach is ____

a) w 1 f(ξ 1 )

b) σ=εD

c) N t =

d) Constant matrix

Answer: a

3. Two point formula of a quadratic approach is _____

a) X direction

b) w 1 f(ξ 1 )+w 2 f(ξ 2 )

c) N t =

d) σ=D

Answer: b

4. The extension of Gaussian quadrature to two-dimensional integrals of the form of _____

a) I≈\(\sum_{i=1}^{n}\sum_{j=1}^{n}\)w i w j f(ξ i ,η j )

b) Natural co-ordinates

c) w 1 f(ξ 1 )+w 2 f(ξ 2 )

d) w 1 f(ξ 1 )

Answer: a

5. Stiffness integration for quadratic element for 2*2 matrix is ____

a) N t =

b) k ij =\(\sum_{IP=1}^{4}\)W IP ∅ IP

c) N t =

d) N t =\

Answer: b

6. The stresses in the quadratic element are not ______

a) Linear

b) Uniform

c) Constant

d) Undefined

Answer: c

7. The stresses are evaluated at the __________

a) Nodal points

b) Nodal displacements

c) Gauss points

d) Elements

Answer: c

8. For quadrilateral with 2X2 integration gives _____ sets of stress values.

a) One

b) Two

c) Three

d) Four

Answer: d

9. For degenerate four noded quadrilateral element the errors are _____

a) Constant

b) Uniform

c) Higher

d) Lesser

Answer: c

10. Gauss points are also the points used for numerical evaluation of _____

a) Surfaces

b) k e

c) Elements

d) Planes

Answer: b

Explanation: Stiffness is the rigidity of an object, the extent to which it resists deformation in response to an applied force. The complementary concept is flexibility or pliability: the more flexible an object is the less stiff it is. A stiff equation is a differential equation for which certain numerical methods for solving the equation are numerically unstable, unless the step size is taken to be extremely small.

This set of Finite Element Method Questions and Answers for Experienced people focuses on “Four Node Quadrilateral for Axis Symmetric Problems”.

1. In the four-node quadrilateral element, the shape functions contained terms _________

a) ξ

b) σ

c) ∅

d) Undefined

Answer: a

2. A _________ element by using nine-node shape function.

a) Load vector

b) Sub parametric

c) Element displacement vector

d) Constant matrix

Answer: b

3. Eight-Node Quadrilateral. This element belongs to the ________ family of elements.

a) Serendipity

b) Constant matrix

c) Load vector

d) Master element

Answer: a

4. N 1 , is of the form ____

a) Co-ordinates

b) N 1 =c

c) N 1 =

d) N 1 =

Answer: b

5. Six node triangular elements is also known as _____

a) Triangle

b) Quadratic triangle

c) Interpolation

d) Shape function

Answer: b

u= Nq .

6. In six node triangular element, the gauss points of a triangular element can be defined by ____

a) Two point rule

b) Three point rule

c) One point rule

d) Undefined

Answer: c

7. The mid node should not be outside of the triangular element this condition should ensures that det J does not attain a value ____

a) Constant

b) Zero

c) Unity

d) Infinite

Answer: b

8. The nodal temperature load can be evaluated by using _____

a) Uniform energy

b) Strain energy

c) Numerical integration

d) Displacement

Answer: c

9. The gauss points for a triangular region differ from the _____ region.

a) Rectangular

b) Triangular

c) Square

d) Temperature

Answer: c

Answer: b

This set of Finite Element Method Interview Questions and Answers for freshers focuses on “Conjugate Gradient Implementation of Quadrilateral Element”.

1. Which of the following elements are used for structural and fatigue analysis?

a) Triangular Elements

b) Quadrilateral Elements

c) Shell type Elements

d) Pentagonal Elements

Answer: b

2. In conjugate gradient implementation of quadrilateral elements the stiffness of each element is stored in three dimensional arrays.

a) True

b) False

Answer: a

Explanation: The stiffness of each quadrilateral element is stored in three dimensional arrays. Thus stiffness of each element can be recalled without recalculating for iterations carried out.

3. The total number of dofs in quadrilateral element is ______

a) thirty six

b) eighteen

c) twenty four

d) forty two

Answer: c

Explanation: Number of dofs per node=6

Number of nodes in quadrilateral element=4

Total number of dofs=6*4=24

4. Which of the following statement is false?

a) Linear quad element is more accurate than linear tria element

b) Linear tria element is more accurate than linear quad element

c) Parabolic elements are better than linear elements

d) Parabolic quad element has eight codes

Answer: b

Explanation: As quad element has one more term in the displacement function than tria elements it is more accurate. The displacement function is u=a 0 +a 1 x+a 2 y+a 3 xy.

5. Which of the following statements about warp angle is not true?

a) It is a not applicable for tria elements

b) It is an angle formed between two normal of planes formed by splitting quad element along diagonal

c) Its value should be less than 10°

d) It is the minimum angle between the planes of diagonally split quad element

Answer: d

Explanation: Warp angle is a quality check parameter mostly used in quad elements. It is the maximum value between the normal of planes formed by diagonally splitting quad elements.

6. Acceptable included angle for quad elements lies between 45° and 135°.

a) True

b) False

Answer: a

Explanation: Included angle check is applied for individual angles and. Skew is based on overall shape and doesn’t take in account individual angles.

7. Which of the following is the correct expression for calculating skew of quadrilateral element?

a) Skew=90°+minimum angle formed by joining opposite mid sides of quad element

b) Skew=90°-minimum angle formed by joining opposite mid sides of quad element

c) Skew=90°*minimum angle formed by joining opposite mid sides of quad element

d) Skew=90°/minimum angle formed by joining opposite mid sides of quad element

Answer: b

Explanation: Skew is a quality check parameter for quadrilateral elements. The ideal value of skew should be 0°. The value of skew is acceptable till 45°.

8. Which of the following is the correct expression for calculating skew of quadrilateral element?

Find the value of stretch for Quadrilateral element

a) Stretch=L min *(2 0.5 /d max )

b) Stretch=L min +(2 0.5 /d max )

c) Stretch=L min -(2 0.5 /d max )

d) Stretch=L min *(2 0.5 +d max )

Answer: a

Explanation: The value of stretch is given by:

Stretch=L min *(2 0.5 /d max )

Here L min is length of minimum side of quad element, and d max is the length of maximum diagonal of quad element. Ideally its value should be 1, but values greater than 0.2 are acceptable.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Beams and Frames – Finite Element Formulation”.

1. During finite element formulation of beam each node has _______ degrees of freedom.

a) three

b) two

c) one

d) six

Answer: b

2. Beams are horizontal members used for supporting transverse loading.

a) True

b) False

Answer: a

Explanation: Beams are horizontal members supporting transverse loads acting on it. Some examples of beams include shafts, bridges, and members in buildings.

3. The total number of degrees of freedom in a beam with four nodes is ______

a) four

b) eight

c) sixteen

d) thirty two

Answer: b

Explanation: Number of degrees of freedom per node in a beam element=2

Number of nodes in beam element=4

Total number of degrees of freedom=2*4=8

4. The displacements in beam elements are interpolated using _______

a) shape elements

b) shape functions

c) shape parameters

d) shape factors

Answer: b

Explanation: Shape functions are used in beam elements to interpolate the displacements. The shape functions along with nodal displacements can be used to interpolate displacements in beam elements.

5. The shape functions for interpolation on beam elements are defined on the range of ________

a) 0 to +1

b) -1 to 0

c) 0 to +2

d) -1 to +1

Answer: d

Explanation: The shape function for interpolation is defined in the range of -1 to +1. The value varies between -1 to +1 where value on one node is -1 and the value on other node is +1.

6. In beam elements the cross section of the element is assumed.

a) True

b) False

Answer: a

Explanation: Beam elements are a type of one dimensional element. The cross section of the geometry is assumed where the beam element is used.

7. Which of the following statements is correct?

a) Stiffness coefficient value for numerical solution is less than the value of exact solution

b) Stiffness coefficient value for numerical solution is greater than the value of exact solution

c) Stiffness coefficient value for numerical solutions is equal to the value of exact solution

d) Stiffness coefficient value for numerical solutions is twice the value of exact solution

Answer: b

8. The element stiffness matrix for beam element is given by which of the following expressions?

a)\

\(\frac{EI}{l^2}

Answer: a

Explanation: The value of stiffness matrix is given by

k=\(\frac{EI}{l^3}

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Beams and Frames – Load Vector”.

1. On a simply supported beam with uniformly distributed load over length the value of reaction force at one support is _______

a) p*l

b) /2

c) 2

d) /3

Answer: b

Explanation: Overall force acting on the beam=p*l

The reaction force at one of the two supports=/2.

2. In Galerkin method we convert a continuous operator problem into a discrete problem.

a) True

b) False

Answer: a

Explanation: In Galerkin method we convert a continuous operator problem into a discrete problem. In beams the problem is formulated into finite elements using Galerkin Method.

3. Which of the following is not a one dimensional element?

a) bar

b) brick

c) beam

d) rod

Answer: b

Explanation: Brick is not a one dimensional element. One dimensional element is used when one dimension of the model geometry is significantly greater than the other two dimensions.

4. Three geometrically identical beams made out of steel, aluminum, and titanium are axially loaded. Which of the following statements is correct?

a) Stress in titanium is the least

b) Stress in cast iron is the highest

c) Stress in steel is the least

d) Stress in all three beams is the same

Answer: d

Explanation: The stress in all three beams will be induced equally. Stress in not dependent on the material, but the geometry and cross section of the element.

5. The application of force at which of the following point on a beam will nullify the effect of torsion?

a) Centroid

b) Center of mass

c) Extreme fiber

d) Shear centre

Answer: d

6. For a taper beam element two cross sections are necessary to define the geometry.

a) True

b) False

Answer: a

Explanation: A taper beam element requires two cross sections to define the geometry. Regular beam element cannot take into account the variation in cross section required to define the geometry.

7. Which of the following statements is correct?

a) Beam elements are recommended for unsymmetrical cross sections

b) Bar elements are recommended for unsymmetrical cross sections

c) Beam and bar elements can both be used for unsymmetrical cross sections

d) Neither beam nor bar elements can be used for unsymmetrical cross sections

Answer: a

8. The equivalent load acting on a simply supported beam element loaded with uniformly distributed load over an element of length is given by which of the following expressions?

a) f=\

f=\

f=\([\frac{pl}{2},\frac{pl^2}{12},\frac{-pl}{2},-\frac{pl^2}{12}]\)

Answer: a

Explanation: The value of equivalent load is given by

f=\([\frac{pl}{2},\frac{pl^2}{12},\frac{pl}{2},-\frac{pl^2}{12}]\)

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “ Beams and Frames – Boundary Conditions”.

1. Symmetry in application of boundary conditions should be avoided in which of the following type of analysis?

a) Linear static analysis

b) Modal analysis

c) Thermal analysis

d) Nonlinear static analysis

Answer: b

Explanation: Symmetric boundary conditions should not be used for modal analysis. Symmetric model would miss some of the modes of modal analysis or out of phase modes.

2. Boundary conditions are applied to simulate the physical constraints on the finite element model.

a) True

b) False

Answer: a

Explanation: Boundary conditions simulate the physical constraints on the finite element model. Application of boundary conditions is a crucial preprocessing step to yield accurate solution.

3. Which of the following is the correct equation for stiffness of an element given value of force and displacement ?

a) FQ=K

b) KQ=F

c) KF=Q

d) KFQ=1

Answer: b

Explanation: The correct equation is given by KQ=F. The value of force is the product of stiffness and displacement . The value of stiffness of the element determines the displacement of the node.

4. Which of the following conditions must be fulfilled to apply symmetry in a finite element model?

a) Geometry of the model is symmetric

b) Boundary conditions to be applied are symmetric

c) Geometry model has large number of nodes

d) Geometry of the model is symmetric and boundary conditions to be applied are symmetric

Answer: d

Explanation: The geometry and boundary conditions both have to be symmetric to apply any kind of symmetry. Half or quarter portions of a model can be used to reduce computational cost.

5. Which of the following boundary conditions cannot be directly applied on solid elements?

a) Force

b) Pressure

c) Support

d) Torque

Answer: d

6. Traction is force acting on an area in any direction other than normal.

a) True

b) False

Answer: a

7. Which of the following statements is correct?

a) Reaction force at supports is equal to the sum of the product of stiffness and displacement

b) Reaction force at supports is equal to the product of sum of stiffness and displacement

c) Reaction force at supports is equal to the sum of stiffness’s

d) Reaction force at supports is equal to the sum of displacements

Answer: a

8. Which of the following equations give the relation between material properties like modulus of elasticity , modulus of rigidity , and Poisson’s ratio ?

a) E = 2*G*

b) E = 3*G*

c) E = 2*G*

d) E = 3*G*

Answer: a

Explanation: The relation between the material properties is given by

E = 2*G*

Here E is the ratio of normal stress to normal strain. G is the ratio of shear stress to shear strain and u is the ratio of lateral strain to longitudinal strain.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Shear Force & Bending Moment”.

1. Which of the following statements are correct about a cantilevered beam with point load acting on the extreme end of the beam?

a) Bending stresses induced in the beam are constant throughout the length of the beam

b) Bending stresses induced in the beam decreases linearly from fixed end to free end

c) Bending stresses induced in the beam increases linearly from fixed end to free end

d) Bending stresses induced in the beam decreases exponentially from fixed end to free end

Answer: b

2. Shear stress acts parallel to the cross section of the beam.

a) True

b) False

Answer: a

Explanation: Shear stress acts parallel to the cross section causing distortion. Shear stress results in angular deformation which is measured in terms of angle.

3. Which of the following is the correct equation for bending moment of an element given value of young’s modulus , moment of inertia , and radius of curvature ?

a) M= EIR

b) M= EI/R

c) M= 2 R

d) M= 2 /R

Answer: b

4. Which of the following statements are correct about a cantilevered beam with point load acting on the extreme end of the beam?

a) Shear stress along the length of the beam increases linearly

b) Shear stress along the length of the beam deceases linearly

c) Shear stress along the length of the beam decreases exponentially

d) Shear stress along the length of the beam remains constant

Answer: d

Explanation: Shear stress along the length of the beam remains constant. There is no change in the magnitude of the shear stress acting on the beam along its length.

5. Which of the following statements are correct regarding shear stress distribution across the cross section in a cantilevered beam with point load acting on the extreme end of the beam?

a) Shear stress distribution across the cross section of beam is constant

b) Shear stress distribution across the cross section of beam is zero

c) Shear stress distribution across the cross section of beam is zero at center and maximum at extreme ends

d) Shear stress distribution across the cross section of beam is zero at extreme ends and maximum at center

Answer: d

6. Torque acting on the face of a cylindrical body induces bending moment in the body.

a) True

b) False

Answer: b

Explanation: Torque acting on the face of a cylindrical body does not induce bending moment in the body. The torque acting on the body induces torsional shear stress in the body.

a) V=\(\frac{EI}{^3}\)(2q 1 +l e q 2 -2q 3 +l e q 4 )

b) V=\(\frac{EI}{^2}\)(2q 1 +l e q 2 -2q 3 +l e q 4 )

c) V=\(\frac{EI}{^2}\)(2q 1 +l e q 2 +2q 3 +l e q 4 )

d) V=\(\frac{EI}{^3}\)(2q 1 +l e q 2 +2q 3 +l e q 4 )

Answer: a

Explanation: The correct expression is given by

V=\(\frac{EI}{^3}\)(2q 1 +l e q 2 -2q 3 +l e q 4 )

Here E is the ratio of normal stress to normal strain. Here q1, q2, q3, q4 are the four displacements at the two supported nodes of the simply supported beam.

a) M=\(\frac{EI}{^2}\)[6ξq 1 +l e q 2 -6ξq 3 +l e q 4 ]

b) M=\(\frac{EI}{^3}\)[6ξq 1 -l e q 2 -6ξq 3 -l e q 4 ]

c) M=\(\frac{EI}{^2}\)[6ξq 1 +l e q 2 -6ξq 3 -l e q 4 ]

d) M=\(\frac{EI}{^3}\)[6ξq 1 +l e q 2 -6ξq 3 +l e q 4 ]

Answer: a

Explanation: The correct expression is given by

M=\(\frac{EI}{^2}\)[6ξq 1 +l e q 2 -6ξq 3 +l e q 4 ]

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Beams on Elastic Support”.

1. Which of the following is a class of applications formed by beams supported on soil?

a) Soil foundation

b) Winkler foundation

c) Elastic Beam foundation

d) Reinforced foundation

Answer: b

Explanation: Winkler foundation is a class of applications formed by beams supported on soil. There are many engineering applications where the beams are supported on elastic members.

2. Single row bearings can be considered by a node at each bearing location.

a) True

b) False

Answer: a

Explanation: Single row bearings can be considered by a node at each bearing location. Also for single row bearing, bearing stiffness can be added to the vertical degree of freedom.

3. Which of the following statements is not correct about beams on elastic supports?

a) Rotational stiffness has to be considered for journal bearings

b) Single row bearings can be considered by a node at each bearing location

c) In wide journal bearing stiffness per unit length has to be considered

d) Rotational stiffness need not be considered for roller bearings

Answer: d

Explanation: Rotational stiffness has to be considered for roller bearings. Also in wide journal bearings stiffness per unit length has to be considered.

4. Which of the following statements are correct regarding beam elements?

a) Beam elements can resist force and moment about x and y axis

b) Beam elements can resist only force about x and y axis

c) Beam elements can resist only force about x, y, and z axis

d) Beam elements can resist force and moment about x, y, and z axis

Answer: d

Explanation: Beam elements can resist force and moment about x, y, and z axis.

5. Rod elements cannot resist shear force applied on the element.

a) True

b) False

Answer: a

Explanation: Rod elements cannot resist shear force applied on the element. Rod element does not have the required vertical degree of freedom to resist shear force.

a) \

\(\frac{EI}{420}

Answer: a

Explanation: The value of stiffness matrix to be added for beams with elastic support is

\(\frac{sl}{420}

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Boundary Value Problems – 1”.

1. In Finite Element Methods , in two-dimensional problems, we approximate solution on a domain but not the domain itself.

a) True

b) False

Answer: b

2. In Finite Element Methods , a boundary value problem is a set of differential equations with a solution, which also satisfies some additional constraints, known as ___

a) boundary conditions

b) nodal values

c) equilibrium equations

d) energy minimum

Answer: a

3. Which of the following expression is the correct solution for finite element approximation over the 3 noded element shown?

Find the expression u for finite element approximation over the 3 noded element in diagram

a) u=c1+x*c2+y*c3

b) u=c1+x*c2+y*c1

c) u=c1+x*c2+y*c3+x*y

d) u=x*c1+y*c2

Answer: a

4. For A1, A2, and A3 as area coordinates and s1, s2 and s3 as shape functions for the element shown, which relation is correct?

Find the relation of A1, A2 & A3 area coordinates & s1, s2 & s3 shape functions of the element

a) s1=A1/

b) s1=/A1

c) s1=A1+A2+A3

d) s1=A1

Answer: a

=A1/.

5. In Finite Element Methods , In two-dimensional problems, we shall have two types of errors, one due to the approximation of the solution and the other due to approximation of the domain.

a) True

b) False

Answer: a

6. For A1=5, A2=10, A3=5, what is the value of the shape function at node 1 of the triangular element shown?

Find the value of the shape function at node 1 of the triangular element

a) 0.15

b) 0.25

c) 0.35

d) 0.45

Answer: b

Explanation: Total area is A.

A=A1+A2+A3

A=5+10+5

=20.

The shape function at node 1 is given by

=5/20

=0.25.

Find the displacement of any point (x,y) on the element along x direction

a) x=e+n+x3

b) x=n*x1+e*x2+x3

c) x=e+n+x3

d) x=e+n+x3

Answer: a

a) A=1*|detJ|

b) A=*|detJ|

c) A=0.5*|detJ|

d) A=2*|detJ|

Answer: c

Explanation: For a 2D three noded triangular element area A=*|*-*|. Using the chain rule of partial derivatives, we get Jacobian of the transformation, J=\

*-* and in terms of detJ

A=A=0.5*|detJ|

9. In a two-dimensional static structural problem, the various non-zero stresses are ____

a) σ=[σxσyσz]

b) σ=[σxσyσx+σy]

c) σ=[σxσyσx-σy]

d) σ=[σxσyτxy]

Answer: d

Explanation: In a two-dimensional static structural problem, the stresses are in a single plane.They are the two normal stresses σx, σy and one shear stress τxy.

10. If 1, 2 and 3 is the order of sequence of elemental nodes of e=1 then to make elemental connectivity, which of the following is the correct order for e=2?

Find order of elemental nodes of e=2 to make elemental connectivity

a) 2 3 1

b) 1 3 2

c) 3 2 1

d) 2 1 3

Answer: a

Explanation: Since, the order of nodes for e=1 is in counter clockwisesense, the same has to be followed for ordering the nodes of e=2. Thus, 2 3 1 is the correct option.

11. If 1, 2; 3, 4; 5, 6 and 7, 8 are the degrees of freedom at nodes 1,2,3 and 4 respectively then the common degrees of freedom for e=1 and e=2 are ____

Find the common degrees of freedom for e=1 and e=2

a) 3 4, 7 8

b) 1 2, 3 4

c) 5 6, 7 8

d) 1 2, 5 6

Answer: a

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Boundary Value Problems – 2”.

1. For A1=5, A2=10, A3=5, what is the value of the shape function at node 1 of the element shown?

Find the value of the shape function at node 1 of the element

a) 0.15

b) 0.5

c) 0.35

d) 0.25

Answer: b

Explanation:

Total area, A=A1+A2+A3

A=5+10+5

=20.

The shape function at node 2 is given by

=10/20

=0.5.

2. In a solid of revolution, if the geometry, support conditions, loads, and material properties are all symmetric about the axis and are independent of θ, then the problem can be treated as a ____

a) two-dimensional one

b) one-dimensional one

c) three-dimensional one

d) plane strain

Answer: a

3. A function Q is evaluated at boundary 1-2 by boundary integral Q=∮q*Sds where q=q 0 and shape functions S are S 1 , S 2 .S 1 =1- and S 2 =1-S 1 then Q 1 is given by expression ____

Find expression of Q1 if function Q is evaluated at boundary 1-2 by boundary integral

a) \

\)*q 0 *l

b) q 0 *l

c) \

\)*q 0 *l

d) \

\)*q 0 *l

Answer: a

Explanation: Given Q=∮q*Sds

Q 1 =\

*S1*ds

=\(\int_{0}^l\)q 0 *

)\)*ds

=[q 0 *s*

\))]

Putting limits of s from zero to l

Q 1 =[

*q₀*l]-0

\)*q₀*l.

4. In a static structural type Boundary Value Problem, at any fixed support, How many non-zero Degrees Of Freedom exist?

a) 0

b) 1

c) 2

d) 3

Answer: a

5. A function Q is evaluated at boundary 1-2 by boundary integral Q=∮q*Sds where q=q 0 and shape functions S are S 1 , S 2 .S 1 =1- and S 2 =1-S 1 then Q 3 is given by the value ____

Find expression of Q3 if function Q is evaluated at boundary 1-2 by boundary integral

a) \

b) 1

c) \

d) 0

Answer: d

Explanation: Given Q=∮q*Sds

Since there is no q defined on sides 2-3 and 3-1 we take q=0.

Q 3 =\(\int_{0}^l\)0*S1*ds

=0.

6. In a static structural type Boundary Value Problem, at any roller support, How many non-zero Degrees Of Freedom exist?

a) 0

b) 1

c) 2

d) 3

Answer: c

7. A function Q is evaluated at boundary 1-2 by boundary integral Q=∮q*Sds where q=q 0 * and shape functions S are S 1 , S 2 .S 1 =1- and S 2 =1-S 1 then Q 1 is given by expression ___

Find Q1 if function Q is evaluated at boundary 1-2 & shape function is S2.S1=1-(s/l) & S2=1-S1

a) \

\)*q₀*l

b) q₀*l

c) \

\)*q₀*l

d) \

\)*q₀*l

Answer: d

Explanation: Given Q=∮q*Sds

Along line 1-2, Q 1 =\

*S1*ds

=\(\int_{0}^l\)q 0 *\

\)*

)\)*ds

\(\int_{0}^l\)q 0 *\

*ds-\int_{0}^l\)q 0 \

^2)\)*ds

Putting limits of s from zero to l

Q 1 =\

\)*q 0 *l-\

\)*q 0 *l

=q₀*l*\

)\)

\)*q₀*l.

8. In a static structural type Boundary Value Problem, at any hinged support, How many non-zero Degrees Of Freedom exist?

a) 0

b) 1

c) 2

d) 3

Answer: b

9. A function Q is evaluated at boundary 1-2 by boundary integral Q=∮q*Sds where q=q₀* and shape functions S are S 1 , S 2 .S 1 =1- and S 2 =1-S 1 then Q 2 is given by expression ___

Find expression of Q2 if function Q is evaluated at boundary 1-2 by boundary integral

a) \

\)*q₀*l

b) q₀*l

c) \

\)*q₀*l

d) \

\)*q₀*l

Answer: c

Explanation: Given Q=∮q*Sds

Along line 1-2, Q 2 =\

*S2*ds

*ds\)

^2)*ds\)

Putting limits of s from zero to l

= \

\)*q₀*l.

10. For a linear triangular element with as the coordinates of the ith node of the element, which option denotes twice the Area of the triangle?

a) + +

b) + +

c) +

d) + +

Answer: a

Explanation: A linear triangular element has 3 nodes. With as coordinates of i th node, the twice of area is given by determinant of the matrix \

+ + .

a) 10

b) 20

c) 30

d) 40

Answer: b

Explanation: A linear triangular element has 3 nodes. With as coordinates of ith node, the twice of area is given by determinant of the matrix \

+ + . Then from the standard relation we have ∑αi = + +

=2*Area

=2*10

=20.

a) 0

b) 10

c) 20

d) 30

Answer: a

Explanation: A linear triangular element has 3 nodes. With as coordinates of ith node, the twice of area is given by determinant of the matrix \

++. Then from the standard relation we have ∑βi=++

=y2−y3+y3−y1+y1–y2

=0.

13. In a 3D axisymmetric solid, because of symmetry about the longitudinal axis, the stresses do not vary along ___ coordinate.

a) x

b) y

c) z

d) θ

Answer: d

a) 0

b) 10

c) 20

d) 30

Answer: a

Explanation: A linear triangular element has 3 nodes. With as coordinates of ith node, the twice of area is given by determinant of the matrix\

++.Then from the standard relation we have ∑γi=−−−

=−x2+x3−x3+x1−x1+x2.

=0.

This set of Finite Element Method Questions & Answers for Exams focuses on “Some Comments on Mesh Generation and Imposition of Boundary Conditions”.

1. In Finite Element Method , which option is not correct with respect to discretization or mesh generation?

a) Density of elements depends on the degree of accuracy desired

b) Choice of element type may depend on the geometry of domain

c) For better accuracy correct formulae has to be applied

d) Some general rules govern mesh generation

Answer: c

2. In Finite Element Method , Mesh convergence is a measure of the number of _____ required in a model to ensure that the results of an analysis are not affected by changing the size of the _____

a) elements, mesh

b) nodes, mesh

c) nodes, element

d) degrees of freedom, element

Answer: a

3. In Finite Element Method , While domain discretization, at which of the following geometrical cases a node is not required?

a) Change in cross-section area only

b) Change in cross-section only

c) Change in cross section area as well as change in cross-section

d) No change

Answer: d

Explanation: While domain discretization, a node is necessary to be placed whenever there is a geometric discontinuity. For example, a change in cross section area or change in cross-section or both.

4. In Finite Element Method , In order to increase accuracy of a solution, while domain discretization, more than one node is to be placed at a point source.

a) True

b) False

Answer: b

5. For an inviscid flow of a fluid around a cylinder in an open channel as shown, the flow in the surroundings of the cylinder gets _____ and thus, the flow becomes ___

Inviscid flow of a fluid around a cylinder in an open channel

a) accelerated, non-uniform

b) decelerated, unsteady

c) accelerated, unsteady

d) compressed, non-uniform

Answer: a

Inviscid flow of a fluid around the cylinder in an open channel

a) Both A and B

b) Either A or B

c) Both C and D

d) Only at D

Answer: c

a) True

b) False

Answer: a

8. In Finite Element Method , for discretization purpose in general, which 2D element is an undesired one?

a) An equilateral triangle of side 5 units

b) A quadrilateral

c) A rhombus of side 10 units

d) A rectangle of dimensions 100×1

Answer: d

9. In Finite Element Method , for discretization purpose in general, which optionis not true for mesh refinement?

a) Previous meshes should be contained in the refined mesh

b) Any point in the body can be included within an arbitrarily small element at any stage of the mesh refinement

c) The same order of approximation for the solution may be preserved through all stages of the refinement process

d) Elements with very large aspect ratios are desired

Answer: d

10. In Finite Element Method , for discretization purpose in general, which of the following mesh cannot be a part of mesh refinement process?

Mesh refinement process for discretization purpose in Finite Element Method (FEM)

a) A

b) B

c) C

d) D

Answer: c

11. To discretize a domain using rectangular elements, if the aspect ratio is to be limited to 10, which dimension of rectangular element gives finest mesh?

a) 1×0.5

b) 1.1×0.1

c) 0.11×0.01

d) 0.1×0.05

Answer: d

12. In Finite Element Method , choose the incorrect option regarding to the dimensions of elements in a finite element mesh?

a) Elements with very large aspect ratios are avoided

b) Elements with small angles are undesired

c) The coefficient matrices in finite element equations depend on dimensions of elements

d) Rectangular elements with high aspect ratio are preferred

Answer: d

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Single Variable Problems – Applications – 1”.

1. If the finite element model shown below represents heat conduction in axisymmetric or plane geometries then which option is not true?

-\frac{\partial}{\partial y}

\) +a 00 u-f=0

a) u is temperature

b) a 11 , a 22 are conductivities in x, y directions

c) f is internal heat generation

d) a 00 =1

Answer: d

2. In steady state heat transfer finite element model [K+H]*{T}={Q}*{P} if convective boundary conditions are neglected then which option is applicable?

a) K=0

b) H=0

c) Q=0

d) T=0

Answer: b

3. For a convective boundary, the natural boundary condition is a balance of energy transfer across the boundary due to conduction and/or convection.

a) True

b) False

Answer: a

4. In the below equation for steady-state heat transfer in plane systems, what does β stands for?

\(k_x\frac{\partial T}{\partial x}n_x+k_y\frac{\partial T}{\partial y}n_y\)+β(T-T ∞ )=\(\hat{q}\) n

a) Convective heat transfer coefficient

b) Thermal expansion

c) Thermal conductivity

d) Diffusivity

Answer: a

5. In a steady state heat conduction problem with a thermal conductivity of 22), for a typical element of mesh of linear triangular elements, what is the value of a in stiffness matrix,

K=a*\

b) 12

c) 10

d) 5.5

Answer: a

Explanation: For a typical element of mesh of linear triangular elements K=*\

is 22 then

a=k/2

a=22/2

=11.

6. Heat conduction problem over a rectangular domain is shown in figure. Which nodes have temperature value of zero units?

Heat conduction problem over a rectangular domain

a) Only1

b) Only 7

c) Only 2 and 3

d) 1, 2 and 3

Answer: d

7. If q denotes the amount of heat flow through any boundary then what is the value of q at node7?

Find the value of q (amount of heat flow through any boundary) at insulated node

a) Dependent on a,b

b) =0

c) >0

d) <0

Answer: b

8. What is the value of T at node 3 if the length of each element is 1/a units?

Find the value of T at node 3 if the length of each element is 1/a units

a) 0.5*b

b) b

c) 2*b

d) b*cos

Answer: a

Explanation: Given that the temperature distribution at boundary follows T=bcos\(\frac{\pi ax}{6}\).

At node 3, x=2/a units,

ax=2

T=bcos\(\frac{\pi2}{6}\)

T=bcos\(\frac{\pi}{3}\)

=b*0.5.

9. The velocity field of a fluid flow is denoted by v=2xi+3j. Which option exactly describes the flow?

a) Ideal

b) Compressible

c) Irrational

d) Inviscid

Answer: b

Explanation: A fluid is said to be incompressible if ∇.V=0.

∇.V=\(\frac{d}{dx}+\frac{d}{dy}\)

=2+0

=2.

Thus, the flow is compressible.

10. For a two dimensional problem, the evaluation of boundary integrals amounts to evaluation of line integrals.

a) True

b) False

Answer: a

Find the value of global force due to infiltration of the river at node 1

a) 0.5

b) 1.0

c) 1.5

d) 2.0

Answer: a

=0.5*q*h

Given q=5, h=0.2

=0.5*5*0.2

=0.5.

12. In a Heat Transfer problem, which option is used for interpolation of temperature inside a finite element?

a) Natural coordinates

b) Global coordinates

c) Temperature gradients

d) Shape functions

Answer: d

a) q=h

b) T=T∞

c) T=TL

d) TL-T∞

Answer: a

This set of Finite Element Method Question Bank focuses on “Single Variable Problems – Applications – 2”.

1. A river flows along node 1 through node 5, infiltrating an aquifer at a constant rate. What is the value of s1a+s2a+s3c+s4c if s denotes a linear interpolation function in FEM?

Find the value of s1a+s2a+s3c+s4c if s denotes a linear interpolation function in FEM

a) 1

b) 2

c) 3

d) 4

Answer: b

→s1a+s2a+s3c+s4c =2.

2. In a thermodynamics process, what is the correct term for the series of states through which a system passes?

a) Path

b) Phase

c) Cycle

d) Direction

Answer: a

Find the value of global force due to infiltration of the river at node 2

a) 5

b) 10

c) 15

d) 20

Answer: b

=0.5*q*h+0.5*q*h

=q*h

Given q=5, h=2

=5*2

=10.

4. In which thermodynamics process the temperature remains constant?

a) Isobaric

b) Isentropic

c) Isothermal

d) Isochoric

Answer: c

5. If ψ=2x+3y represents stream function then what is the magnitude of velocity?

a) 3

b) 5

c) 6.3

d) 3.6

Answer: d

Explanation: The magnitude of velocity is (V 2 +U 2 ) .

V=\(\frac{\partial \psi}{\partial y}\)=3 and U=\(\frac{-\partial \psi}{\partial x}\)=-2

v=(3 2 +2 2 )

=3.6.

6. In which process does a thermodynamic system remains infinitesimally closed to an equilibrium state at all times?

a) Path equilibrium process

b) Cycle equilibrium process

c) Phase equilibrium process

d) Quasi-state or quasi-equilibrium process

Answer: d

7. In analysis of mechanical stresses, which of the following is the correct form of the 3D equation of stress equilibrium?

a) \

No such equation exists

Answer: a

8. In thermodynamics, what does the term open system refer to?

a) Control mass

b) Control volume

c) Control energy

d) Control temperature

Answer: b

Find boundary condition at X=0 for wall of a tank containing a hot liquid at temperature T0

a) T=T 0

b) T=T∞

c) T=T L

d) Data insufficient

Answer: a

10. In thermodynamics, what is the name of a process with identical end states?

a) Cycle

b) Path

c) Phase

d) Either Path or phase

Answer: a

11. For the given heat radiation equation, which of the following terminology is not correct?

q {x} n {x} +q {y} n {y} +q {z} n {z} =σε\(T_{\{s\}}^{\{4\}}\)-αq {r}

a) α is Stefan-Boltzmann constant

b) σ is the surface emission coefficient

c) ε is surface absorption coefficient

d) q r is heat generated per unit surface area

Answer: d

12. A system is said to be in thermodynamic equilibrium if it maintains which of the following equilibriums?

a) Mechanical and phase

b) Thermal and chemical

c) Thermal, mechanical and chemical

d) Thermal, phase, mechanical and chemical

Answer: c

13. Which method is used to write the basic heat transfer equation in the following form using weighted residuals?

\Missing or unrecognized delimiter for \right\) N i dV = 0

a) Galerkin method

b) Jacobi method

c) Rayleigh Ritz method

d) Delaunay method

Answer: a

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Eigen Value and Time Dependent Problems – 1”.

1. The simultaneous linear equations used in FEM for solution of static problems are KX=F, the methods available for solving these equations are divided into two types: direct and iterative.

a) True

b) False

Answer: a

2. For the following equations, what is the value of x 2 using Gaussian elimination method?

x 1 -x 2 +3x 3 =10——–

5x 2 -5x 3 =-5————

-7x 3 =-28 —————-

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: From the 3 rd equation it’s seen that x 3 =\(\frac{-28}{-7}\)=4. Using x 3 in 2 nd equation we get

5*x 2 -5*4=-5

5*x 2 =15

x 2 =3

3. For the following equations, what is the value of x 1 using Gaussian elimination method?

x 1 -x 2 +3x 3 =10——–

5x 2 -5x 3 =-5————

-7x 3 =-28—————-

a) 1

b) 2

c) 3

d) 4

Answer: a

Explanation: From the 3 rd equation it’s seen that x 3 =\(\frac{-28}{-7}\)=4. Using x 3 in 2 nd equation we get

5*x 2 -5*4=-5

5*x 2 =15

x 2 =3

Using x 3 , x 2 in 1 st equation

x 1 -3+3*4=10

x 1 -3+12=10

x 1 =1

4. Which option is not correct about direct methods for solving system of linear equations?

a) In the absence of errors it yields exact solution

b) Errors arising from round off and truncation may give useless results

c) Gaussian elimination method is an example

d) Starts with an initial approximation

Answer: d

5. Which option is not correct about iterative methods for solving system of linear equations?

a) Convergence yields a good approximate solution

b) Insensitive to the growth of round-off errors

c) Gaussian elimination method is an example

d) Starts with an initial approximation

Answer: c

6. In structural mechanics, which option is not correct about linear analysis?

a) Displacements are infinitesimally small

b) Material is linearly elastic

c) Externally applied loads are a function of time

d) Applied loads are not a function of time

Answer: c

7. A generalized Eigen value problem [K- ω 2 M]X=0 has a non-zero solution for X. What can be the value of determinant of the matrix [K- ω 2 M]?

a) Any integer

b) 0

c) +1

d) Positive integer

Answer: b

8. In FEM, the forced vibrations equation after Finite Element discretization of a structure can be expressed as which option?

a) Mẍ+Kẋ=F

b) Mẍ+Kẋ=0

c) Mẍ+Kx=F

d) Mẍ+Kx=0

Answer: c

9. The free vibrations equation after Finite Element discretization of a structure is expressed as Mẍ+Kx=0. Which option is not correct about the free vibration case?

a) Displacements are harmonic

b) x=Xe iωt where X is amplitude

c) [K-ω 2 M]X=0

d) KX=Mω 2

Answer: d

10. The generalized Eigen value problem [K-ω 2 M]X=0 has a non-zero solution for X. What is the value of natural frequency, ω if K=\

b) 1/9

c) 9

d) 1/3

Answer: d

Explanation: The generalized Eigen value problem [K-ω 2 M]=0 has a non-zero solution for X if the determinant of the matrix [K-ω 2 M] equals zero,

K=ω 2 M

\)=ω 2 *\(

\) Equating corresponding elements, we get 9*ω 2 =1

ω 2 =1/9

Natural frequency, ω =1/3.

11. Which option is not correct about free vibration analysis problem KX= λMX, where X represents the amplitude of displacement x?

a) The displacements are harmonic

b) X represent mode shapes or Eigen vectors

c) λ represent Eigen value

d) ω represents Eigen value

Answer: d

12. After Finite Element discretization of a structure, which option expresses the free vibrations equation?

a) Mẍ+Kẋ=F

b) Mẍ+Kẋ=0

c) Mẍ+Kx=F

d) Mẍ+Kx=0

Answer: d

13. For the eigenvalue problem of the form A = λB, which option is not correct about the parameters used in the equation below?

A=\

B=1

c) B=0

d) λ is called eigenvalue

Answer: c

14. The generalized Eigen value problem [K- λM]X=0 has a non-zero solution for X. What is the value of λ if K=\

b) \

d) \(\frac{1}{4}\)

Answer: d

Explanation: The generalized Eigen value problem [K- λM]X=0 has a non-zero solution for X if the determinant of the matrix [K- λM] equals zero or K=λM

\)=λ\(

Equating corresponding elements, we get 1=4*λ

λ=\(\frac{1}{4}\).

15. For the following eigenvalue equation to represent a heat transfer problem, a=kA and C=ρcA.

\)=λCU

a) True

b) False

Answer: a

This set of Finite Element Method Questions and Answers for Entrance exams focuses on “Eigen Value and Time Dependent Problems – 2”.

1. Suppose the following eigenvalue equation represents a bar problem, then the value of the parameters a and c 0 should be EA and ρA , respectively.

\)=λc 0 U

a) True

b) False

Answer: a

a)(T ∞ -T)\

(T ∞ -T)\

T ∞ -T

d) T-T ∞

Answer: a

kAT x = hA(T-T ∞ )

T x =(T ∞ -T)\(\frac{h}{k}\).

3. A plane wall of thermal conductivity of 45\

b) 2

c) 3

d) 4

Answer: b

45T x = 9

T x = \

=2.

a) T

b) T ∞

c) T ∞ -T

d) T-T ∞

Answer: b

kAT x = hA(X – T ∞ )

\(\frac{-kT_x}{h}\) = X – T ∞

Given h = ∞

\(\frac{-kT_x}{\infty}\) = X – T ∞

0 = X – T ∞

X = T ∞ .

a) 35°C

b) 45°C

c) 40°C

d) 50°C

Answer: b

kAT x =hA(X-T ∞ )

\(\frac{-kT_x}{h}\)=X-T ∞

Given h = ∞

\(\frac{-kT_x}{\infty}\)=X-T ∞

0=X-T ∞

X=T ∞ , given T ∞ =45°C

X=45°C.

6. In thermodynamics, the following equation represents a diffusion process. If k is thermal conductivity, p is density, and c is the specific heat at constant pressure, then what is α?

\(\frac{c}{k}\)

Answer: a

7. The governing equation of an unsteady one-dimensional heat transfer problem is given below. It has a solution u = Uexp. What is λ appropriately called?

+ b \frac{\partial u}{\partial t}\) + cu = 0 for 0<x<L

a) Natural frequency

b) Eigenvalue

c) Thermal diffusivity

d) Thermal flux

Answer: b

8. The unsteady natural axial oscillations of a bar are periodic, and they are determined by assuming a solution u = U e -iwt . Which option is not correct about the solution equation?

a) w denotes the natural frequency

b) w 2 denotes eigenvalue

c) U denotes mode shape

d) u denotes transverse displacements

Answer: d

9. In matrix algebra, which option is not correct about an eigenvalue problem of the type Ax = Lx?

a) It has a discrete solution

b) It has solution only if A non-singular

c) x is called eigenvector

d) L is called eigenvalue

Answer: b

a) M\

All the forces are time-independent

d) The equation is of 3 rd order

Answer: b

11. In matrix algebra, a matrix K equals \(

\). What is the value of a, if K 7 = \

2187

b) 729

c) 6561

d) 5 7

Answer: a

Explanation: Since K is a diagonal matrix, its higher powers are obtained by raising its diagonal elements to the same power. If K=\(

\) then K 7 =\(

\). Equating the corresponding elements of \(

\) and \(

\) we get

a=3 7

a=2187.

12. In matrix algebra, what is the eigenvalue of the matrix \

b) 2

c) 3

d) 4

Answer: c

Explanation: The eigenvalue, L of a matrix is equal to the root of the equation |K-LI|=0.

Let the given matrix be denoted by K then K-LI = \

2 -1)-1+1

= (L 2 -2L) + 2L

= -L 3 + 3L 2

= -L 2 .

Given -L 2 = 0

On simplification L = 0, 0 and 3.

13. In matrix algebra, what is the value of a-b if the eigenvector of \

b) 1

c) 2

d) 3

Answer: a

Explanation: If X is an eigenvector corresponding to an eigenvalue L of a matrix K, then KX=LX. The eigenvector of \(

\) corresponding to eigenvalue three is \(

\). Equating the corresponding elements of \(

\) and \(

a=b=1

a-b=0.

14. From the Euler-Bernoulli beam theory of natural vibrations, using cubic Hermite polynomials approximation, what is the 1 st element of the stiffness matrix?

a) \

\(\frac{12AI}{h^3}\)

Answer: a

15. From the Timoshenko beam theory of natural vibrations, using cubic Hermite polynomials approximation, what is the 1 st element of the mass matrix?

a) \

d) \(\frac{\rho I}{3}\)

Answer: a

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Library of Elements and Interpolation Functions – 1”.

1. Which option is not correct about the three-noded triangular plane stress element used in FEM?

a) It has six degrees of freedom

b) It belongs to both the isoparametric and superparametric element families

c) It can be improved by the addition of internal degrees of freedom

d) Delaunay triangulation can be used for its mesh generation

Answer: c

2. In FEM, which option is used to develop the Higher-order triangular elements systematically?

a) Pascal’s triangle

b) Galerkin method

c) Jacobi method

d) Delaunaytriangulation

Answer: a

3. In FEM, What is the number of displacement polynomials necessary for finding displacements in a linear triangular element?

a) 1

b) 2

c) 3

d) 4

Answer: b

4. Concerning triangular elements in FEM, which option is not correct about the mathematical formula of Pascal’s triangle?

a) It contains the terms in two coordinates only

b) The position of the terms can be viewed as the nodes of a triangular element

c) The position of the first and last terms of a row is at the vertices of a triangular element

d) A triangular element of order 2 corresponds to the second row

Answer: d

5. Which option is not correct about the four-noded rectangular plane stress element used in FEM?

a) It has eight degrees of freedom

b) Shape functions N1, N2, N3 and N4 are bilinear functions of x and y

c) The displacement field is continuous across elements

d) Its Delaunay triangulation is unique

Answer: d

6. In FEM, if x, y represents Cartesian coordinates, then the following triangular array of binomial coefficients forms Pascal’s triangle.

True

b) False

Answer: a

7. In the FEM element library, what is the other name of a higher-order element?

a) Complex element

b) Simplex element

c) Linear element

d) Nonlinear element

Answer: a

8. In FEM, if x, y represents Cartesian coordinates, then which term of Pascal’s triangle corresponds to the position of the interior node of the following element?

The position of interior node of 10 noded triangle corresponding to the Pascal’s triangle term

a) 1

b) xy

c) x 2 y

d) xy 2

Answer: b

9. In the FEM element library, an eight noded quadrilateral element belongs to which family?

a) Serendipity

b) Linear

c) Simplex

d) Quadratic

Answer: a

10. In FEM, which option is not correct about the Lagrange family of triangular elements?

a) The nodes are uniformly spaced

b) Pascal’s triangle can be viewed as a triangular element

c) Dependent variables and their derivatives are continuous at inter-element boundaries

d) 2 nd degree polynomial corresponds to 6 noded triangle

Answer: c

11. What is the displacement function for one-dimensional, two noded linear elements in terms of its shape functions N 1 and N 2 ?

a) N 1 u 1 +N 2 u 2

b) N 1 u 2 +N 2 u 1

c) N 1 u 1 -N 2 u 2

d) N 1 u 2 -N 2 u 1

Answer: a

12. For the following element, what is the value of the distance variable s at the 1 st row?

Find the value of the distance variable s at 1st row for a quadratic element if k = 3

a) 1

b) 2

c) 0.5

d) 0

Answer: c

s=\(\frac{p}{3-1}\)

=\(\frac{p}{2}\)

At first row, p=1

s=\(\frac{1}{2}\)

=0.5.

13. For the following element, if s is the distance variable as shown, then its value at the 2 nd row is \(\frac{1}{3}\).

Find the value of the distance variable s at 2nd row for an element if k = 2

a) True

b) False

Answer: b

s=\(\frac{p}{4-1}\)

=\(\frac{p}{3}\)

At second row, p=2

s=\(\frac{2}{3}\).

Answer: a

Sanfoundry Global Education & Learning Series – Finite Element Method.

This set of Finite Element Method Questions and Answers for Campus interviews focuses on “Library of Elements and Interpolation Functions – 2”.

1. In the FEM element library, what is the correct name for a three noded triangular element?

a) Linear strain triangular element

b) Constant strain triangular element

c) Variable strain triangular element

d) Higher-order triangular element

Answer: b

2. If the geometry and other parameters of an element are defined in terms of only one spatial coordinate, then the element is a one-dimensional element?

a) True

b) False

Answer: a

3. In FEM, what is the name of the element specified by a polynomial of order two or more?

a) Nonlinear element

b) Higher-order element

c) Linear element

d) Master element

Answer: b

4. In FEM, what is the name of the shape function of an Euler-Bernoulli beam element?

a) Hermite cubic interpolation function

b) Lagrange cubic interpolation function

c) Consistent element functions

d) Quadratic interpolation functions

Answer: a

5. In FEM, which option is used to develop the Higher-order rectangular elements systematically?

a) A rectangular array of binomial coefficients

b) Galerkin method

c) Jacobi method

d) Delaunay triangulation

Answer: a

6. In FEM, what are the elements in which the same shape functions describe the geometry and field displacement variables?

a) Iso-parametric

b) Axi-Symmetric

c) Super-parametric

d) Sub-parametric

Answer: a

a) True

b) False

Answer: a

8. Which option is not correct about the Lagrange rectangular element used in FEM?

a) Second-order Lagrange element has nine nodes

b) Zero-order Lagrange element has one node

c) First-order Lagrange element has four nodes

d) Third-order Lagrange element has fifteen nodes

Answer: d

Explanation: In general a p th -order Lagrange rectangular element has n nodes with n= 2 , where p=0,1,2 …

For p=0,

n=1.

For p=1,

n=2 2 ,

=4.

For p=2,

n=3 2

=9.

For p=3,

n=4 2

=16.

9. Which option is not correct about the Lagrange family of triangular elements used in FEM?

a) 2 nd -degree polynomial corresponds to 6 noded triangle

b) 0 th -degree polynomial corresponds to 1 noded triangle

c) 1 st -degree polynomial corresponds to 3 noded triangle

d) 3 rd -degree polynomial corresponds to 9 noded triangle

Answer: d

Explanation: A p th degree polynomial corresponds to n noded triangular element with n=0.5, where n=0, 1, 2 …

For p=0,

n=1.

For p=1,

n=0.5*3*2,

=3.

For p=2,

n=0.5*3*4

=6.

For p=3,

n=0.5*4*5

=10.

10. In FEM, if x, y represents Cartesian coordinates, then which term of Pascal’s triangle corresponds to the position of the interior node of the following element?

Pascal's triangle corresponding to position of interior node of 9 noded Lagrange rectangle

a) 1

b) xy

c) x 2 y

d) xy 2

Answer: b

11. What is the number of nodes present on the boundary of the Lagrange quadratic rectangular element used in FEM?

a) 1

b) 4

c) 8

d) 9

Answer: c

12. Which nodes of the higher-order elements of the Lagrange family do not contribute to the inter-element connectivity?

a) Interior nodes

b) All nodes

c) Corner nodes

d) Intermediate nodes

Answer: a

13. What is the reason for an element in the Serendipity family to have a smaller size of stiffness matrix compare to a similar element in the Lagrange family?

a) Absence of interior nodes

b) Modified element connectivity

c) Lesser interpolation functions

d) Irregular arrangements of nodes

Answer: a

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “ Modelling Considerations”.

1. In mathematical modeling of a process, which option is not a characteristic of an analytical solution?

a) Mathematical equations are used to describe a process

b) Most practical problems cannot be solved

c) Exact information on the quantities of interest is obtained

d) Finite element method is used

Answer: d

2. In mathematical modeling of a process, which option is not a characteristic of a numerical solution method?

a) It does not give exact information on the quantities of interest

b) A set of assumptions are made about the process

c) Applicable to simple problems only

d) Finite element method is used

Answer: c

3. In FEM, which option is not correct with respect to the generation of element geometries?

a) It involves coordinate transformation

b) It is based on a one-to-many mapping

c) A point in actual element maps uniquely to master element

d) Tiny interior angles are avoided

Answer: b

b) \

\(\frac{1}{ }\)

Answer: b

5. In the generation of element geometries, for what value of j e , the element geometries lie within the limits of acceptable distortion?

a) j e >0

b) j e <0

c) j e =0

d) Any real value of j e

Answer: a

6. Which of the following finite element geometries contains unacceptable vertex angles, in practice?

Find the finite element geometries containing unacceptable vertex angles in practice

a) 1 and 2

b) 2 and 4

c) Only 3

d) 1, 3 and 5

Answer: d

7. For a finite element mesh to be valid, which option is not true regarding the elements used?

a) The number of elements used is not exact

b) Elements can be of different orders

c) Elements can be of different types

d) The choice of elements and mesh is problem-independent

Answer: d

8. A finite element mesh is refined by subdividing existing elements into two or more elements of the same type, what is the name of such a refinement?

a) h-version mesh refinement

b) p-version mesh refinement

c) h, p-version mesh refinement

d) p, h-version mesh refinement

Answer: a

9. In local mesh refinement, if very small elements are placed adjacent to very large ones, then the mesh becomes unacceptable.

a) True

b) False

Answer: a

10. In FEM, which option is the most realistic representation of the actual force between deformable bodies?

a) Point load

b) Uniformly varying load

c) Sine distribution

d) Uniformly distributed load

Answer: c

11.For the following element, what is the value of a+b such that the Jacobian J=\

+b-2] is positive, where are natural coordinates?

Find the value of a+b such that the Jacobian J is positive

a) >2

b) <2

c) =2

d) 0

Answer: a

Explanation: The point corresponds to in natural coordinates. Thus = .

J=\

+b-2]

-4]

-1

Given j>0,

-1>0

>2.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Plane Elasticity – Governing Equations”.

1. In solid mechanics, what does linearized elasticity deal with?

a) Small deformations in linear elastic solids

b) Large deformations in linear elastic solids

c) Large deformations in non-Hookean solids

d) Small deformations in non-Hookean solids

Answer: a

2. For plane elasticity problems in three dimensions, which option is not responsible for making the solutions independent of one of the dimensions?

a) Geometry

b) Boundary conditions

c) Externally applied loads

d) Material

Answer: d

3. For a plane strain problem, which strain value is correct if the problem is characterized by the displacement field u x =u x , u y =u y and u z =0?

a) ε xy =0

b) ε xz =0

c) ε yz ≠0

d) ε xz ≠0

Answer: b

4. Underplane strain condition, what is the value of ε yy if the problem is characterized by the displacement field u x =2x+3y, u y =5y 2 , and u z =0?

a) 10y

b) 5y

c) 3

d) 0

Answer: a

Thus, \(\epsilon_{yy}=\frac{\partial 5y^2}{\partial y}\)

=5*\(\frac{\partial y^2}{\partial y}\)

=5*2y

=10y.

5.For a plane strain problem, the relation between stress and strain components for an orthotropic material is σ=Cε. Which option is the correct structure of the matrix C?

a) \

Answer: a

Explanation: For an orthotropic material under plane strain, with principal material axes (x 1 , x 2 , x 3 ) coinciding with the coordinates, the relation between stress and strain components is \(

\) where C is the elastic stiffness matrix. The state of stress is σ xz =σ yz =0 and \

\).

6. For an orthotropic material, if E and v represent Young’s modulus and the poisons ratio, respectively, then what is the value of v 12 if E 1 =200 Gpa, E 2 =160 Gpa and v 21 =0.25?

a) 0.3125

b) 0.05

c) 0.2125

d) 0.3

Answer: a

Explanation: For an orthotropic material, E 1 and E 2 are the principal

moduli in the x and y directions, respectively. The poisons ratio and Young’s moduli are related by the equation

v 12 =v 21 \(\frac{E_1}{E_2}\).

v 12 =0.25*\(\frac{200}{160}\)

=0.25*1.25

=0.3125.

7. Under plane stress condition in the XYZ Cartesian system, which stress value is correct if a problem is characterized by the stress field σ xx =σ xx , σ yy =σ yy and σ zz =0?

a) σ xy =0

b) σ yx ≠0

c) σ zx ≠0

d) σ yz ≠0

Answer: b

Explanation: A state of plane stress in XYZ Cartesian system is defined as one in which the following stress field exists:

σ xz =σ yz =σ zz =0, σ xx , σ xy =σ xy and σ yy =σ yy .

Thus, σ xx , σ xy and σ yy are non-zero stresses. Such a problem in three dimensions can be dealt with as a two-dimensional problem.

8. For theplane stress problem in XYZ Cartesian system, σ xx =σ xx , σ yy =σ yy and σ zz =0, which option is correct regarding the associated strain field?

a) ε xx =0

b) ε yx =0

c) ε zx =0

d) ε yy =0

Answer: c

9. For any two cases of plane elasticity problems, if the constitutive equations are different, then their final equations of motion are also different.

a) True

b) False

Answer: a

10. In solid mechanics, which option is not a characteristic of a plane stress problem in the XYZ Cartesian system?

a) One dimension is very small compared to the other two dimensions

b) All external loads are coplanar

c) Strain along any one direction is zero

d) Stress along any one direction is zero

Answer: c

11. In solid mechanics, what is the correct vector form of the equations of motion for a plane elasticity problem?

a) D*σ+f=ρü

b) D*σ+f=ρu̇

c) D 2 *σ+f=ρü

d) D*σ+f=ρu

Answer: a

Answer: b

Sanfoundry Global Education & Learning Series – Finite Element Method.

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Plane Elasticity – Weak Formulations”.

1. In FEM, which method is not used to construct the weak forms and associated finite element model of the plane elasticity equations?

a) Principle of virtual displacements

b) The principle of minimum total potential energy

c) Weak form of governing differential equations

d) Hamiltonian principle

Answer: d

2. In constructing the weak forms of plane elasticity problems, which option is not related to the principle of virtual displacements?

a) Displacements to strains

b) Strains to stresses

c) The equations of motion

d) Body forces

Answer: d

3. What is the correct form of the principle of virtual displacements applied to plane finite elastic element If V e is the volume of element and s e is its surface?

a) 0=\(\int_{V_e}\)(σ ij δε ij +ρü i δu i )dV-\(\int_{V_e}\)f i δu i dV-∮ s e \(\hat{t_i}\)δu i ds

b) 0=\(\int_{V_e}\)(σ ij δε ij +ρu̇ i δu i )dV-\(\int_{V_e}\)f i δu i dV-∮ s e \(\hat{t_i}\)δu i ds

c) 0=\(\int_{V_e}\)(σ ij δε ij +ρü i δu i )dV+\(\int_{V_e}\)f i δu i dV-∮ s e \(\hat{t_i}\)δu i ds

d) 0=\(\int_{V_e}\)(σ ij δε ij +ρu̇ i δu i )dV\(\int_{V_e}\)f i δu i dV+∮ s e \(\hat{t_i}\)δu i ds

Answer: a

a) True

b) False

Answer: b

a) \(\int_{V_e}\)(σ ij δε ij )dV

b) ∮ s e t̂ i δu i ds

c) \(\int_{V_e}\)(ρü i δu i )dV

d) \(\int_{V_e}\)f i δu i dV

Answer: a

6. In the weak form of the principle of virtual displacements applied to a plane elastic finite element, what does the term \(\int_{V_e}\)(ρü i δu i )dV correspond to?

a) Virtual strain energy

b) Kinetic energy

c) Virtual work done by the body force

d) Virtual work done by the surface traction

Answer: b

7. Under plane elasticity, which force is responsible for doing the virtual work \(\int_{V_e}\)f i δu i dV in the weak form of the principle of virtual displacements?

a) Body force

b) Concentrated loads

c) Surface traction force

d) Pressure force

Answer: a

8. For a plane elasticity problem, which term in the weak form of the principle of virtual displacements is affected by a change in the applied surface traction forces?

a) \(\int_{V_e}\)(σ ij δε ij )dV

b) ∮ s e t̂ i δu i ds

c) \(\int_{V_e}\)(ρü i δu i )dV

d) \(\int_{V_e}\)f i δu i dV

Answer: b

9. If U,V denotes the components of the displacement vector, then which option is the correct primary nodal degrees of freedom present in the following figure of a plane elasticity problem?

Find the primary nodal degrees of freedom present in the figure of a plane elasticity problem

a) U 1 =V 1 =0

b) U 2 ≠0, V 2 =0

c) U 2 =0, V 2 ≠0

d) U 1 =V 1 ≠0

Answer: a

10. If F x , F y denotes the components of the force vector, then which option is the correct secondary degrees of freedom present in the following figure of a plane elasticity problem?

Find the secondary degrees of freedom present in the figure of a plane elasticity problem

a) \

\(F_7^x\ne F_7^y\)=0

Answer: d

\(F_7^x\)≠0.

11. For plane elasticity problems, which option represents the essential boundary conditions among the governing equations?

a) Displacements, u x and u y at the boundary

b) Surface traction at the boundary

c) The displacements (u x and u y ) and surface traction at the boundary

d) Stresses in the element

Answer: a

Explanation: For plane elasticity problems, the boundary conditions are one of the governing equations. There are two types of boundary conditions, namely, essential boundary conditions and natural boundary conditions. The displacements specified in the problem are the essential boundary condition or Dirichlet boundary condition.

12. If the equation ∫ Ω c h e \((\frac{\partial w_1}{\partial x}\sigma_{xx}+\frac{\partial w_1}{\partial y}\)σ xy -w 1 f x +ρw 1 \

a) True

b) False

Answer: a

Explanation: The equations ∫ Ω c h e \((\frac{\partial w_1}{\partial x}\sigma_{xx}+\frac{\partial w_1}{\partial y}\)σ xy -w 1 f x +ρw 1 \

\)dxdy-&oint; Γ c h e w 1 (σ xx n x +σ xy n y )ds=0 and ∫ Ω c h e \((\frac{\partial w_2}{\partial x}\sigma_{xy}+\frac{\partial w_2}{\partial y}\)σ yy -w 2 f y +ρw 2 \

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Plane Elasticity – Finite Element Model”.

1. If only the first derivatives of u x and u y appear in the weak forms, then their interpolation must be at least bilinear.

a) True

b) False

Answer: a

Explanation: An examination of the weak form, 0=∫ Ω e h e \

+ c_{66}\frac{\partial w_1}{\partial y}

2. In FEM, if two independent variables are components of the same vector, then they can be approximated by two different types of interpolations.

a) True

b) False

Answer: b

3. What can one conclude about the displacement components u x and u y in the finite element model of the plane elasticity equations?

a) They are primary variables and must be carried as primary nodal degrees of freedom

b) They are secondary variables and must be carried as primary nodal degrees of freedom

c) They are primary variables and must be carried as secondary nodal degrees of freedom

d) They are secondary variables and must be carried as secondary nodal degrees of freedom

Answer: b

Explanation: An examination of the weak form, 0=∫ Ω e h e \

+ c_{66}\frac{\partial w_1}{\partial y}

\)+ρw 1 u x ]dxdy-∫ Ω e h e w 1 f x dxdy-∮ ┌ d h e w 1 t x ds reveals the following:

u x and u y are the primary variables, which must be carried as the primary nodal degrees of freedom.

only first derivatives of u x and u y with respect to x and y, respectively, appear.

4. Which interpolation functions must be used for the primary variables in weak forms of plane elasticity equations?

a) Hermite family interpolation function

b) Lagrange family interpolation function

c) Hierarchical interpolation function

d) Quadratic interpolation function

Answer: b

5. What are the simplest elements that fit for the finite element model of the plane elasticity equations?

a) Linear triangular and quadratic quadrilateral elements

b) Quadratic triangular and linear quadrilateral elements

c) Linear triangular and linear quadrilateral elements

d) Quadratic triangular and quadratic quadrilateral elements

Answer: c

6. What is the correct statement regarding the shape function S of a linear triangular element?

a) The first derivatives of S and hence, all the strains are element-wise constant

b) The first derivatives of S are linear, but the strains are element-wise constant

c) The first derivatives of S and hence the strains are linear functions

d) The first derivatives of S are element-wise constant, but the strains are linear

Answer: a

7. What is the other name of a linear triangular element for plane elasticity problems?

a) Constant-strain triangle

b) Linear-strain triangle

c) Quadratic-strain triangle

d) Variable-strain triangle

Answer: a

8. What is the function on \

denotes the shape function of a linear quadrilateral element?

a) Linear in e and constant in n

b) Constant in e and Linear in n

c) Linear in both e and n

d) Constant in both e and n

Answer: a

9. For a linear triangular element, what is the order of matrix B in the strain-displacement relation ε=BD, where D denotes the displacement matrix?

a) 6×3

b) 3×6

c) 3×8

d) 8×3

Answer: b

Explanation: For plane elasticity problems the strain-displacement relation is given by ε=BD, where ε=[ε xx ε yy ε xy ] T , B=\(

\) and D=\(

\) T . The order of matrix B is 3x2n, where n is the number of nodes in the element. A linear triangular element has three nodes, thus n=3.

Order of B is 3×2*3

=3×6.

10. For a linear quadrilateral element, what is the order of matrix B in the strain-displacement relation ε=BD, where D denotes the displacement matrix?

a) 6×3

b) 3×6

c) 3×8

d) 8×3

Answer: c

Explanation: For plane elasticity problems the strain-displacement relation is given by ε=BD, where ε=[ε xx ε yy ε xy ] T , B=\(

\) and D=\(

\) T . The order of matrix B is 3x2n, where n is the number of nodes in the element. A linear quadrilateral element has four nodes, thus n=4.

Order of B is 3×2*4

=3×8.

11. Which option is not correct with respect to the orders of the matrices in the following finite element model of plane elastic equations?

M e \(\ddot{\Delta}^e\)+K e Δ e =F e +Q e

a) Mass matrix M has order 2n x 2n

b) Stiffness matrix K has order 2n x n

c) The element load vector F has order 2n x 1

d) The vector of internal forces Q has order 2n x 1

Answer: b

12. Which form of a periodic solution is sought for the natural vibration study of plane elastic bodies?

a) {Δ}={Δ 0 }e iωt

b) {Δ}={Δ 0 }e -iω

c) {Δ}={Δ 0 }e -iωt

d) {Δ}={Δ 0 }e -iω/t

Answer: c

13. Which option is correct for the first derivative of the shape function S in the study of plane elasticity problems?

a) It is element-wise constant for triangular element whereas not a constant for quadrilateral element

b) It is element-wise constant for both the triangular element as well as a quadrilateral element

c) It is not a constant for triangular element whereas element-wise constant for quadrilateral element

d) It is a combination of a linear function and constant for both the triangular element and a quadrilateral element

Answer: a

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Plane Elasticity – Evaluation of Integrals”.

1. In the Finite Element Method, which expression is correct for a linear triangular element if S is the shape function, A e is its area, and K is a constant?

a) \

\(\frac{\partial S}{\partial x}\)=KA e

d) \(\frac{\partial S}{\partial y}\)=KA e 2

Answer: a

Explanation: For a linear triangular element, the shape function \

2. In Finite Element Analysis, what is the correct load vector for a linear triangular element with area A e , thickness h e and uniform body force vector f?

a) \

\(\frac{h_e}{4A_e}\)f

Answer: b

Explanation: For a linear triangular element, for the case in which the body force is uniform and thus the body force components fx and fy are element-wise constant

, the load vector F has the form F=∫ Ω c h e (ψ e ) T \(f_0^e\)dx

=\(\frac{A_e h_e}{4}

\). The internal load vector Q is computed only when the element falls on the boundary of the domain on which tractions are specified.

3. In Finite Element Analysis, what is the correct load vector for the linear quadrilateral element with area A e , thickness h e and uniform body force vector f?

a) \

\(\frac{h_e}{4A_e}\)f

Answer: a

Explanation: For a linear quadrilateral element,for the case in which the body force is uniform and thus the body force components are element-wise constant

, the load vector F has the form F=\

^T f_0^edx\) =\(\frac{A_e h_e}{4}

\). The internal load vector Q is computed only when the element falls on the boundary of the domain on which tractions are specified.

4. In the Finite Element Method, the vector of internal forces is computed only when the element falls on the boundary of the domain on which tractions are absent.

a) True

b) False

Answer: b

5. Which option is not correct concerning the internal load vector in the finite element model of plane elasticity problems?

a) It is computed at all the nodes interior of the element

b) It is computed only when the element falls on the boundary of the domain on which tractions are known

c) Its computation doesn’t involve evaluation of line integrals for any type of element

d) It is evaluated in global coordinates but not in element coordinates

Answer: b

6. In transformations, what is the transformation matrix R in the relation F=RQ if the load vector in global coordinates is F and the load vector in element coordinates is Q?

a) \

Answer: a

F=R T Q, where R is the transformation matrix R=\(

\) and α is the angle between the global x-axis and the traction vector T.

7. What is the global load vector in Finite Element Analysis of the following structure if the local load vector is \(

\) and θ=0?

Find the global load vector in Finite Element Analysis of the structure

a) \

Answer: a

\) and α is the angle between the global x-axis and the traction vector T.

Here α=90-θ, given θ=0

α=90-0

=90.

Cos90=0 and sin90=1.

R=\(

F=\(

^T\)x\(

=\(

\).

8. What is the expression for the traction term t n in the element load vector Q e =&oint; ┌ c h e ψ T tds of the following figure where L 23 is the length of the line 2-3?

Find the expression for the traction term in the element load vector of the figure

a) t n =-T

b) t n =T

c) t n =-T

d) t n =T

Answer: b

Explanation: For the element shown in the given figure, the side 2-3 is subjected to linearly varying normal force t n . The traction term on the side 2-3 of the element is t n =-T

9. In Finite Element Analysis, what are the values of nodal forces in the following element if the line 2-4 is 160 in long?

Find the values of nodal forces in element - Finite Element Analysis

a) 1600 along both the DOF 3 and 7

b) 800 and 0 along the DOF 3 and 4 respectively

c) 0 and 800 along the DOF 7 and 8 respectively

d) 0 and 800 along the DOF 3 and 4 respectively

Answer: b

Given p*h=10 and b=160 thus \(\frac{pbh}{2}\)

=\(\frac{10×160}{2}\)

=800.

Thus, the forces along the DOF 3 and 4 are 800 and 0, respectively.

10. In vibration and transient analysis of beams, if the linear acceleration scheme predicts the solution,then it is unstable for the first several time steps, but it eventually becomes stable.

a) True

b) False

Answer: b

11. In Finite Element Analysis, which option is correct for computation of load due to specified boundary stress?

a) Can be computed using a local coordinate system and one-dimensional interpolation functions

b) Can be computed using a local coordinate system but not one-dimensional interpolation functions

c) Cannot be computed using a local coordinate system but one-dimensional interpolation functions can be used

d) Neither a local coordinate system nor one-dimensional interpolation functions can be used

Answer: a

12. In the Finite Element Method, which element is known for the slowest convergence?

a) Linear triangular element

b) Quadratic triangular element

c) Linear rectangular elements

d) Quadratic rectangular elements

Answer: a

This set of Finite Element Method Question Paper focuses on “Plane Elasticity – Assembly, Boundary and Initial Conditions”.

1. From solid mechanics, what is the correct displacement boundary condition for the following plane stress problem of the beam?

Find displacement (u) boundary condition for plane stress problem of beam

a) uy

b) uy

c) ux

≠0

d) ux

=≠0

Answer: b

Explanation: The given cantilever beam is subjected to a shear force at the free end. The other end is supported by roller and hinge support. Because the hinge support restrains translation, ux

=uy

=0. As the roller support restrains translation perpendicular to the base, only ux=0 which implies ux

≠0 and ux

≠0 are incorrect.

2. From solid mechanics, which traction boundary condition is not correct for the following beam of thickness h?

Find traction (t) boundary condition for beam of thickness h

a) ty

b) ty

c) tx=0

d) ty=-hT

Answer: a

≠0 and ty

≠0 whereas the roller support restrains translation perpendicular to the base; thus, only tx≠0. The option ty

=0 is incorrect.

3. In Finite Element Analysis of the beam, which primary variable does not belong to the following mesh?

Find primary variable which does not belong to the mesh in Finite Element Analysis of the beam

a) U 9 =0

b) U 19 =0

c) U 10 =0

d) U 20 =0

Answer: c

4. What is the total size of the assembled stiffness matrix of a plane elastic structure such that its finite element mesh has eight nodes and two degrees of freedom at each node?

a) 16×16

b) 8×8

c) 2×2

d) 4×4

Answer: a

5. What is the element at the index position 3×3 of the assembled stiffness matrix of the following mesh if K 1 =\(

\) and K 2 =\(

\)?

Find the element at the index position 3×3 of the assembled stiffness matrix of the mesh

a) 9

b) 11

c) 13

d) 4

Answer: a

\) and the elementin K at the index 3×3 is 6+3

=9.

6. In the Finite Element Method, if two different values of the same degree of freedom are specified at a point, then such point is called as a singular point.

a) True

b) False

Answer: a

7. For time-dependent problems in FEA, which variables must be specified for each component of the displacement field problems?

a) The initial displacement and velocity

b) The initial displacement only

c) The final velocity

d) The initial displacement and final velocity

Answer: a

8. What is the magnitude of the force at node 22 if the moment M is replaced by an equivalent distributed force at x=acm?

Find magnitude of force at node 22 if moment M is replaced by equivalent distributed force

a) \

Always zero

c) \

\(\frac{-M}{b}\)

Answer: b

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Governing Equations”.

1. For fluid flows, which expression does not characterize the slow flow of a viscous and incompressible fluid in a closed domain?

a) v.∇v=0

b) μ≠0

c) =0

d) =0

Answer: d

Explanation: The slow flow of a viscous and incompressible fluid in a closed domain is characterized by the following expressions:

Slow : v.∇v=0.

Viscous: μ≠0.

Incompressible: =0.

It is not a uniform flow; thus, ≠0.

2. In mathematical modeling, for which option, a fluid flow can be approximated by a two-dimensional model?

a) One of the dimensions is very small, and there is some flow along with it

b) One of the dimensions is very long, and there is no flow along with it

c) One of the dimensions is very long, and there is some flow along with it

d) The velocity components in two directions vary along the third direction

Answer: b

3. For fluid flows obeying conservation of mass, what is the value of k if v=4x+ky denotes the velocity at any point in the flow?

a) -4

b) 4

c) -2

d) 2

Answer: a

Explanation: A flow obeys conservation of mass if \(\frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}\)=0. Comparing v=4x+ky with v=v x +v y , we get v x =4x and v y =ky.

Using the conservation of mass, we get \(\frac{\partial }{\partial x}+\frac{\partial }{\partial y}\)=0

4+k=0

K=-4.

4. The following equation represents the momentum equation for a fluid flow that is approximated by a two-dimensional model. What does k stand for?

ρ\

-\frac{\partial}{\partial y}[k

]+\frac{\partial P}{\partial x}\)-f x =0

a) Thermal conductivity

b) Fluid viscosity

c) Density

d) Pressure

Answer: b

Explanation: By using constitutive relations the momentum equation is expressed as ρ\

-\frac{\partial}{\partial y}[k

]+\frac{\partial P}{\partial x}\)-f x =0, where v x , v y are the velocity components, P is the pressure, k is the viscosity, f x is the component of the body force vector, and ρ is the density.

5. If a finite element model involves the natural and direct formulation of momentum and continuity equations, then it is known as the velocity-pressure formulation or mixed formulation.

a) True

b) False

Answer: a

6. In the formulation of the finite element model for fluids flows, which option is not correct about the penalty formulation?

a) The continuity equation is an additional relation among the velocity components

b) The constraint is satisfied in an approximate sense

c) Uses the penalty function method

d) Involves natural and direct formulation of momentum and continuity equations

Answer: d

The unsteady flow of a viscous fluid squeezed between two horizontal parallel plates

a) v y ≠0 and v x ≠0

b) v y =0 and v x ≠0

c) v y ≠0 and v x =0

d) v y =0 and v x =0

Answer: b

Unsteady flow of a viscous fluid squeezed between two horizontal parallel plates

a) v y =-V 0 and v x =0

b) v y =-V 0 and v x ≠0

c) v y =V 0 and v x =0

d) v y =V 0 and v x ≠0

Answer: a

Explanation: For the unsteady flow of a viscous fluid squeezed between two horizontal parallel plates, because of the symmetry, the computational domain is taken to be one-quarter of the flow field. The top plate has a velocity –V 0 . The fluid along the top plate is imparted –V 0 along the vertical direction and zero velocity in the horizontal direction, thus v y =-V 0 and v x =0.

9. Which statement is not true regarding the variables present in the viscous ﬂow problems?

a) Velocity components are zero on ﬁxed walls

b) Shear stress is zero along the line of symmetry

c) Vertical velocity component v y and horizontal stress t x must be zero along the horizontal line of symmetry

d) Shear stress is non zero along the line of symmetry

Answer: d

a) True

b) False

Answer: a

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Velocity – Pressure Finite Element Model”.

1. In velocity-pressure formulation in FEM, which step is not used in the development of a weak form?

a) Multiply governing equations with weight functions

b) Integrating over the element domain

c) Integrating by parts

d) Performing coordinate transformation

Answer: d

2. In the formulation of governing equations, which option does not signify the characteristics of a weight function?

a) Weight functions are multiplied to governing equations to obtain weak forms

b) Weight functions are interpreted from the physical setup of the problem

c) Weight function must denote a non-dimensional quantity

d) Weight function can be interpreted as a velocity

Answer: c

3. For a governing equation, what does one conclude from the weak formulation if it does not contain boundary integral involving weight function?

a) Integration by parts is used

b) Integration by parts is not used

c) The weight function is as a primary variable

d) The weight function is to be made continuous across inter-element boundaries

Answer: b

4. In FEM, which option is the correct weak form of the following momentum equation?

ρ\

b) l\

l\(\int_{\Omega_e}[\rho w_1\frac{\partial v_x}{\partial t}+\frac{\partial w_1}{\partial x}\sigma_{xx}+\frac{\partial w_1}{\partial y}\sigma_{xy}-w_1 f_x]\)dxdy+∮ ┏ c w 1 (σ xx +σ xy )ds=0

d) l\(\int_{\Omega_e} w_1[\rho\frac{\partial v_x}{\partial t}-\frac{\partial \sigma_{xx}}{\partial x}-\frac{\partial \sigma_{xy}}{\partial y}-f_x]\)dxdy++∮ ┏ c w 1 (σ xx n x +σ xy n y )ds=0

Answer: a

5. Using constitutive relations, what is the value of τ xx if μ=0.3 and v=4x?

a) 0.24

b) 2.4

c) 0.12

d) 1.2

Answer: b

Explanation: From constitutive relations, τ xx =2μ\(\frac{\partial v_x}{\partial x}\), where μ is the coefficient of viscosity.

\(\frac{\partial v_x}{\partial x}\)

= \(\frac{\partial }{\partial x}\)

=4.

Thus τ xx =2*μ*4.

Given μ=0.3

τ xx =2*0.3*4

=2.4.

6. Using constitutive relations, what is the value of τ xy if μ=0.3 and v=4xy-6y?

a) 0.24x

b) 2.4x

c) 0.12x

d) 1.2x

Answer: d

Explanation: From constitutive relations, τ xy =μ\

\), where μ is the coefficient of viscosity. On comparing v=4xy+6y with v=vx+vy, we get vx=4xy and vy =6y.

\(\frac{\partial v_x}{\partial y}+\frac{\partial v_y}{\partial x}\)

=4x+0

=4x.

Thus τ xy =μ*.

Given μ=0.3

τ xx =0.3*4x

=1.2x.

7. If a governing equation represents volume change in an element, then the weight function in its weak form must be like a force that causes the volume change.

a) True

b) False

Answer: a

Explanation: The governing equation \

in its weak form must be like a force that causes the volume change. Volume changes occur under the action of hydrostatic pressure, hence w=-P.

8. Which option is not correct concerning the pressure variable, P in the weak form of the momentum and continuity equation?

a) P is a primary variable

b) P is a part of the secondary variables

c) P=constant is the minimum continuity requirement for interpolation

d) It is discontinuous across inter-element boundaries

Answer: a

9. Which option is not correct concerning the velocity variables, v x and v y in the weak form of the momentum and continuity equation?

a) They are primary variables

b) The minimum continuity requirement for interpolation is that they are linear in x and y

c) The minimum continuity requirement for interpolation is that they are constant

d) They are continuous across the inter-element boundary

Answer: c

10. Which equation is the correct vector form of the finite element model of momentum and continuity equations in the flow domain?

a) MΔ+K 11 Δ+K 12 P=0

b) MΔ+K 11 Δ+K 22 P=F 1

c) MΔ+K 22 Δ+K 12 P=F 1

d) MΔ+K 11 Δ+K 12 P=F 1

Answer: d

11. For the vector form of the finite element model of momentum and continuity equations MΔ+K 11 Δ+K 12 P=0, what is the correct expression for mass matrix M?

a) M=∫ Ω c ρψ T ψdx

b) M=∫ Ω c fψ T ψdx

c) M=∫ Ω c tψ T ψdx

d) M=∫ Ω c Pψ T ψdx

Answer: a

12. For the vector form of the finite element model of momentum and continuity equations MΔ+K 11 Δ+K 12 P=0 what is the order of matrix F if the order of M is 2n x 2n?

a) 2n x 1

b) 1 x 2n

c) m x 2n

d) 2n x m

Answer: a

13. In matrix algebra, if a matrix is positive definite, then all its eigenvalues are greater than zero.

a) True

b) False

Answer: a

14. What is the physical interpretation of the weight function w3 in the following weak form of the continuity equation?

-∫ Ω e w 3 \

\)dxdy=0

a) Hydrostatic pressure

b) Axial force

c) Surface traction

d) Body force

Answer: a

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Penalty – Finite Element Model – 1”.

1. Which type of problem can be obtained by reformulating a problem with differential constraints by using the penalty method?

a) A problem with no constraints

b) A problem with variable constraints

c) A problem with fixed constraints

d) A problem with structural constraints

Answer: a

2. In the interest of the simple formulation of viscous flows, which case does not involve time derivative terms?

a) Static case

b) Transient case

c) Unsteady case

d) Non-periodic

Answer: a

3. In the penalty formulation of the fluid flow model, if the velocity field (v x , v y ) satisfies the continuity equation, then the weight functions also satisfy the continuity equation.

a) True

b) False

Answer: b

4. In the weak forms of the fluid flow model, as the weight functions are virtual variations of the velocity components(v x , v y ),respectively, which relation is satisfied by the weight functions?

a) \

\(\frac{\partial w2}{\partial x}-\frac{\partial w1}{\partial y}\)=0

Answer: a

Explanation: In the weak forms of the fluid flow model, suppose that the velocity field (v x , v y ) is such that the continuity equation \

5. In finite element modeling, which formulation introduces constraints on variables and satisfies them in an approximate sense?

a) Velocity-pressure formulation

b) Penalty formulation

c) Mixed formulation

d) Lagrange multiplier formulation

Answer: b

Explanation: There are two different finite element models of momentum and continuity equations. The first one is a natural and direct formulation of momentum and continuity equations. The second model, penalty formulation, is based on the interpretation that the continuity equation is a new relationship among the velocity components, and it eliminates the pressure term leading to a constraint on the velocity components. The constraint is satisfied in a least-squares sense.

6. In the formulation of the finite element model, which option is the complete restated form of the weak forms of viscous fluids flow equations?

a) Only B t +B v -B̅ p =l

b) B t +B v -B̅ p =l and –B p (w 3 , v)=0

c) Only –B p (w 3 , v)=0

d) B t +B v -B̅ p =0

Answer: b

, \)v=\(

,\) f=\(

\) and t=\(

\).

7. In the following variational problem of finding velocity components and pressure, which bilinear form includes time-derivative terms?

B t +B v -B̅ p =l; –B p (w 3 , v)=0

a) B t

b) B v

c) B̅ p

d) B p (w 3 ,v)

Answer: a

8. In the weak forms of the fluid flow model, since the weight functions are linearly dependent on each other, the sum of the three weak forms is the same as the three individual equations.

a) True

b) False

Answer: b

9. Which option is not correct concerning the bilinear term B in the variational problem of the viscous fluid flow equation?

a) It is symmetric

b) It contains the viscosity matrix

c) B v = B v

d) It is bilinear in both w and v

Answer: d

10. In the variational problem of fluid flow, what is the correct matrix form of C in the bilinear form B v =∫ Ω e T Cdx?

a) \

μ\

μ\(

Answer: c

\), denoted by the letter C. C is the viscosity matrix since it contains the viscosity term μ.

Answer: c

Explanation: Since the bilinear term has the form B̅ p =∫ Ω e \

^T\)Pdx; K=\

^T\)p.

We have D 1 =\(

\) and w=\(

\(D_1^T w=

= \(\frac{\partial }{\partial y}+\frac{\partial }{\partial x}\)

=2+3

=5.

K=(D 1 T w) T p

=5*3

=15.

Sanfoundry Global Education & Learning Series – Finite Element Method.

This set of Finite Element Method Assessment Questions and Answers focuses on “Penalty – Finite Element Model – 2”.

1. Which option is correct regarding constraints in the viscous flow problems governed by the continuity and momentum equation?

a) Constrains are readily available

b) Constraints are created by the elimination of pressure

c) Constraints are created by the elimination of velocity

d) No constraints are dealt with in penalty formulation

Answer: b

2. In the finite element method, which option is not a natural and direct formulation of momentum and continuity equations?

a) Velocity-pressure formulation

b) Penalty formulation

c) Mixed formulation

d) Lagrange multiplier formulation

Answer: b

Explanation: There are two different finite element models of momentum and continuity equations. The first one is a natural and direct formulation of momentum and continuity equations. It is known as the velocity-pressure formulation or mixed formulation. Lagrange multiplier formulation is similar to the velocity-pressure formulation. The penalty formulation is not natural and direct.

3. For the functional I v =\(\frac{1}{2}\)B v -l, which option is equivalent to the equations governing steady flows of viscous incompressible fluids?

a) Minimize l v

b) Maximize l v

c) Stationary l v

d) Derivative of l v

Answer: a

4. In FEM, what is the characteristic of the problem with the functional I v representing governing equations of steady viscous incompressible flows?

a) It is subjected to constraint \

It is an unconstrained problem

c) It can be reformulated as a constrained one, by using the penalty method

d) It cannot be reformulated as an unconstrained one, by using the penalty method

Answer: a

a) True

b) False

Answer: a

6. For the functional I L ≡I v +∫ Ω c λGdxdy, what is the necessary condition for I L to have a stationary value?

a) δ v x I L +δ v y I L +δ λ I L =0

b) δ v x I L -δ v y I L -δ λ I L =0

c) δ v x I L -δ v y I L +δ λ I L =0

d) δ v x I L +δ v y I L -δ λ I L =0

Answer: a

7. In FEM, which option is a negative aspect of the mixed formulation?

a) Presence of zeros on matrix diagonals corresponding to pressure variable

b) Involves a natural and direct formulation

c) It is a velocity-pressure formulation

d) It is similar to the Lagrange multiplier method

Answer: a

8. In the computation of mixed formulation in FEM, which option is correct for the nodal DOF of the following element?

Find nodal DOF of element in computation of mixed formulation in FEM

a) o nodes with u,v and P

b) * nodes with u,v and P

c) o nodes with P only

d) *nodes with u and v

Answer: b

9. For the following slider bearing, the upper pad is inclined to the base pad. Which option is correct for a load applied normal to the base pad?

Slider or slipper bearing with the upper pad inclined to the base pad

a) A pressure gradient develops in the lubricant, and the bearing can support some transverse load

b) The pressure remains uniform in the lubricant, and the bearing can support some transverse load

c) A pressure gradient develops in the lubricant, and the bearing cannot support any transverse load

d) The pressure remains uniform in the lubricant, and the bearing cannot support any transverse load

Answer: a

10. For the following slider bearing, the upper pad is parallel to the base pad. Which option is correct for a load applied normal to the base pad?

Slider or slipper bearing with the upper pad is parallel to the base pad

a) The pressure in the lubricant is atmospheric, and the bearing cannot support any load

b) The pressure in the lubricant is greater than atmospheric, and the bearing can support any load

c) The pressure in the lubricant is less than atmospheric, and the bearing cannot support any load

d) The pressure in the lubricant is atmospheric, and the bearing can support some load

Answer: a

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “ Classical Plate Model”.

1. In FEM, what does the simple two-dimensional plate theories account for?

a) The kinematics of bending deformation of thin bodies without to transverse loads

b) The kinematics of bending deformation of thin bodies subjected to transverse loads

c) Thin bodies subjected to transverse loads without bending deformation

d) Thin plates subjected to transverse loads with bending deformation

Answer: b

a) True

b) False

Answer: b

3. Which option is correct about an analog of a plate in simple plate theory used in FEM?

a) A plate is a one-dimensional analog of a beam

b) A plate is a two-dimensional analog of a beam

c) A plate is a three-dimensional analog of a beam

d) A plate is not analogous to beam

Answer: b

4. In FEM, which principle is used to derive the governing equations of displacement-based plate theories?

a) The principle of virtual forces

b) The principle of virtual work

c) The principle of virtual displacements

d) The principle of least energy

Answer: c

5. In FEM, which theory is an extension of the Euler-Bernoulli beam theory?

a) Classical Plate Theory

b) Shear Deformation Theory

c) Hencky-Mindlin plate theory

d) Shell theory

Answer: a

6. In displacement-based plate theories, which option is not an assumption of Classical Plate Theory ?

a) A straight line perpendicular to the plane of the plate is inextensible

b) A straight line perpendicular to the plane of the plate remains straight

c) A straight line perpendicular to the plane of the plate remains straight and does not rotate

d) A straight line perpendicular to the plane of the plate rotates such that it remains perpendicular to the tangent to the deformed surface

Answer: c

7. Which option specifiesan assumption made in Classical Plate Theory for a plate lying in the plane XY?

a) ε zz =0

b) ε xz ≠0

c) ε yz ≠0

d) ε xy =0

Answer: a

8. The following diagram is the mid-plane of an elastic plate. Which option shows the correct moments Mxx and Mxy if the plate is loaded as per the Classical Plate Theory?

The mid-plane of an elastic plate loaded as per the Classical Plate Theory

a) Find the boundary conditions in Classical Plate Theory

b) Bending moment and shear force in a plate lying in XY plane

c) Find Mxx and Mxy in mid-plane of an elastic plate if it is loaded as per Classical Plate Theory

d) Concentrated transverse load & shear force in the mid-plane of an elastic plate

Answer: b

9. The following figure shows the mid-plane of an elastic plate. What are the boundary conditions in Classical Plate Theory if the plate is clamped and the displacement along Z is w?

Find boundary conditions in clamped plate

a) w=0,\

w=0,V n =0

c) w=0,\

w=0,M nn =0

Answer: a

Explanation: Geometrically, plate problems are similar to the plane stress problems except that plates are also subjected to transverse loads that cause bending about axes in the plane of the plate. The boundary condition for a clamped plate is the absence of deflection and normal derivative of deflection , i.e., w=\(\frac{\partial w}{\partial n}\)=0. Because a simply supported end does not restrict rotation, the reactive moment is zero, i.e., w=M nn =0. For a free end, both, the reactive moment and the shear force are absent, i.e., M nn = V n =0.

10. The following figure shows the mid-plane of an elastic plate. In Classical Plate Theory, what are the boundary conditions ifthe plate is simply supported with deflection w along Z?

Find boundary conditions if plate is supported with deflection in Classical Plate Theory

a) w=0,\

w=0,V n =0

c) w=0,\

w=0,M nn =0

Answer: d

11. In FEM, what are the primary variables in the Classical Plate Theory of plate deformation ?

a) The transverse deflection w only

b) The transverse deflection w and the normal derivative of w

c) The normal derivative of w only

d) The normal and the time derivative of w

Answer: b

12.How many termpolynomial is used in the approximation of w for a three noded triangular element with three DOF at each node?

a) 9-term

b) 18-term

c) 6-term

d) 12-term

Answer: a

Explanation: Several C1 rectangular and triangular plate bending elements with

or with

as the degrees of freedom at each node exists in the literature. A triangular element with three nodes, with

at each node, requires the 9-term polynomial approximation of w, i.e.,w=a 1 +a 2 x+a 3 y+a 4 xy+a 5 x 2 +a 6 y 2 +a 7 (x 2 y+xy 2 )+a 8 x 3 +a 9 y 3 .

13. In Classical Plate Theory, what is the correct comment for the polynomial w=a 1 +a 2 x+a 3 y+a 4 xy+a 5 x 2 +a 6 y 2 +a 7 (x 2 y+xy 2 )+a 8 x 3 +a 9 y 3 ?

a) It is used in the approximation of deflection for a rectangular element

b) It is a complete third order polynomial

c) It’s an incomplete third order polynomial

d) w varies as a quadratic along any line x=constant

Answer: c

Explanation: Several C1 rectangular and triangular plate bending elements with

or with

as the degrees of freedom at each node exists in the literature. A triangular element with three nodes, with

14. In Classical Plate Theory, because the 4-noded rectangular element is non-conforming, it gives poor results in approximation.

a) True

b) False

Answer: b

15. In Classical Plate Theory, how many triangles assemble to give a conforming triangular element with DOF w, \

b) 3

c) 4

d) 12

Answer: b

This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Shear Deformable Plate Model”.

1. In FEM, which theory is an extension of the Timoshenko beam theory?

a) Classical Plate Theory

b) Hencky-Mindlin plate theory

c) Kirchhoff plate theory

d) Shell theory

Answer: b

a) True

b) False

Answer: b

3. In FEM, what are the primary variables in the Shear Deformation Theory of plate deformation ?

a) The transverse deflection w only

b) The transverse deflection w and the normal derivative of w

c) The transverse deflection w and the angles of rotation of the transverse normal about in-plane axes

d) The angles of rotation of the transverse normal about in-plane axes only

Answer: c

4. Which equation correctly describes Hamilton’s principle used in FEM?

a) 0=\

]dt

b) 0=\

]dt

c) 0=\

]dt

d) 0=\

]dt

Answer: a

5. In displacement-based plate theories, which option is correct about Shear Deformation Theory ?

a) It is also called Kirchhoff plate theory

b) It is an extension of Euler-Bernoulli beam theory from one dimension to two dimensions

c) It does not involve Timoshenko beam theory

d) It is often known as Hencky-Mindlin plate theory

Answer: d

6. In displacement-based plate theories, which assumption of Classical Plate Theory is relaxed in Shear Deformation Theory?

a) A straight-line perpendicular to the plane of the plate is inextensible

b) A straight line perpendicular to the plane of the plate remains straight

c) A straight line perpendicular to the plane of the plate rotates such that it remains perpendicular to the tangent to the deformed surface

d) A straight line perpendicular to the plane of the plate rotates

Answer: c

7. Which option specifies an assumption made in Shear Deformation Theory for a plate lying in the plane XY?

a) ε zz ≠0

b) ε xz =0

c) ε yz ≠0

d) ε xy =0

Answer: c

8. In SDT, what are the boundary conditions for a plate that is clamped if &fcy; represents the rotation of the transverse normal about an in-plane axis and w is the transverse deflection?

a) w=0,\

w=0,&fcy;=0

c) w=0,\

w=0, M nn =0

Answer: b

Explanation: Geometrically, plate problems are similar to the plane stress problems except that plates are also subjected to transverse loads that cause bending about axes in the plane of the plate. In SDT, the boundary condition for a clamped plate is the absence of deflection and rotation of the transverse normal about any in-plane axis, i.e., w=&fcy;=0. Because a simply supported end does not restrict rotation, the reactive moment is zero, i.e., w=M nn =0. For a free end, both, the reactive moment and the shear force are absent, i.e., M nn =Q n =0.

9. In SDT, what are the boundary conditions for a plate that is simply supported if &fcy; represents the rotation of the transverse normal about an in-plane axis and w is the transverse deflection?

a) w=0,\

w=0,&fcy;=0

c) w=0,\

w=0, M nn =0

Answer: d

10. In FEM, which option is correct for a linear plate theory based on infinitesimal strains and orthotropic material properties?

a) The plane elasticity equations govern the transverse deflections

b) The transverse deflections are coupled with in-plane displacements

c) The in-plane displacements are zero in the absence of in-plane forces

d) The transverse deflections are zero in the absence of in-plane forces

Answer: c

Explanation: For a linear theory based on infinitesimal strains and orthotropic material properties, the in-plane displacements are uncoupled from the transverse deflection uz=w. The plane elasticity equations govern the in-plane displacements . The in-plane displacements are zero if there are no in-plane forces and hence, we discuss only the equations governing the bending deformation and the associated finite element models.

11. In the Shear Deformation plate theory, when does the transverse shear strains in the element equations present computational difficulties?

a) If the plate is thick

b) If the side to thickness ratio of the plate is large

c) If the side to thickness ratio of the plate is small

d) If higher-order finite elements are used

Answer: b

12. In the Shear Deformation plate theory, what characteristic contributes to shear locking?

a) Transverse shear strains in thick plates present computational difficulties

b) Transverse shear strains in thin plates present computational efficiency

c) For thick plates, the element stiffness matrix yields erroneous results for the generalized displacements

d) For thin plates, the element stiffness matrix becomes stiff and yields erroneous results

Answer: d