Category: Finite Element Objective Questions

Finite Element Method Questions 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 c₀ should be EA and ρA, respectively.
(-frac{d}{dx}(afrac{dU}{dx}))=λc₀ U
a) True
b) False
Answer: a
Clarification: For the given eigenvalue equation, the values of the parameters a and c₀ depends upon the physical properties and phenomena involved in the problem. For a bar problem, a=EA and c₀=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 c₀=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)(frac{h}{k})
b)(T_∞-T)(frac{1}{L} )
c) T_∞-T
d) T-T_∞
Answer: a
Clarification: 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(frac{W}{mK}) was initially maintained 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 surface of the wall is 9(frac{W}{m^2K}), then what is the temperature gradient developed at the surface?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: 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(35-45)
T_x = (frac{9}{-45}) (35-45)
=(frac{1}{-5}) (-10)
=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
Clarification: 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
Clarification: 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_∞, 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{partial^2 T}{partial x^2} = frac{1}{alpha} frac{partial T}{partial t})
a) (frac{k}{pc})
b) (frac{pc}{k})
c) (frac{c}{kp})
d) (frac{c}{k})
Answer: a
Clarification: 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(x,t) = U(x)exp(λt). What is λ appropriately called?
(frac{-partial}{partial x} (a frac{partial u}{partial x}) + b frac{partial u}{partial t}) + cu = 0 for 0a) Natural frequency
b) Eigenvalue
c) Thermal diffusivity
d) Thermal flux
Answer: b
Clarification: The governing equation of an unsteady one-dimensional heat transfer problem is a parabolic equation. Hence, its solution is given by u(x,t) = U(x)exp(λt), where u represents temperature along a direction x at any time t, U(x) 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(x, t) = U(x) e^-iwt. Which option is not correct about the solution equation?
a) w denotes the natural frequency
b) w² denotes eigenvalue
c) U(x) denotes mode shape
d) u(x, t) denotes transverse displacements
Answer: d
Clarification: The unsteady natural axial oscillations of a bar are periodic. They are measured by assuming a solution u(x, t) = U(x) e^-iwt, where w is natural frequency, w² is an eigenvalue, U(x) 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
Clarification: 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(ddot{x}) + K(dot{x}) + Cx = F
b) M(ddot{x}) is time-dependent
c) All the forces are time-independent
d) The equation is of 3^rd order
Answer: b
Clarification: 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 (begin{pmatrix} 1&0&0 \ 0&1&0 \ 0&0&3 end{pmatrix}). What is the value of a, if K⁷ = (begin{pmatrix} c&0&0\ 0&b&0 \ 0&0&a end{pmatrix})?
a) 2187
b) 729
c) 6561
d) 5⁷
Answer: a
Clarification: Since K is a diagonal matrix, its higher powers are obtained by raising its diagonal elements to the same power. If K=(begin{pmatrix} 1&0&0 \ 0&1&0 \ 0&0&3 end{pmatrix}) then K⁷=(begin{pmatrix} 1^7&0&0 \ 0&1^7&0 \ 0&0&3^7end{pmatrix}). Equating the corresponding elements of (begin{pmatrix} c&0&0\ 0&b&0 \ 0&0&a end{pmatrix}) and (begin{pmatrix} 1^7&0&0 \ 0&1^7&0 \ 0&0&3^7end{pmatrix}) we get
a=3⁷
a=2187.

12. In matrix algebra, what is the eigenvalue of the matrix (begin{pmatrix} 1&1&1 \ 1&1&1 \ 1&1&1 end{pmatrix})?
a) 1
b) 2
c) 3
d) 4
Answer: c
Clarification: The eigenvalue, L of a matrix is equal to the root (factor) of the equation |K-LI|=0.
Let the given matrix be denoted by K then K-LI = (begin{pmatrix} 1&1&1 \ 1&1&1 \ 1&1&1 end{pmatrix} – L begin{pmatrix}1&0&0 \ 0&1&0 \0&0&1 end{pmatrix})
= (begin{pmatrix} 1-L&1&1 \ 1&1-L&1 \ 1&1&1-L end{pmatrix})
|K-LI| = (begin{vmatrix} 1-L&1&1 \ 1&1-L&1 \ 1&1&1-L end{vmatrix})
= (1-L)((1-L)²-1)-1(-L)+1(L)
= (1-L)(L²-2L) + 2L
= -L³ + 3L²
= -L² (L-3).
Given -L² (L-3) = 0
On simplification L = 0, 0 and 3.

13. In matrix algebra, what is the value of a-b if the eigenvector of (begin{pmatrix}1&1&1 \ 1&1&1 \ 1&1&1 end{pmatrix}) corresponding to eigenvalue three is (begin{pmatrix}a \ b \ a end{pmatrix})?
a) 0
b) 1
c) 2
d) 3
Answer: a
Clarification: If X is an eigenvector corresponding to an eigenvalue L of a matrix K, then KX=LX. The eigenvector of (begin{pmatrix}1&1&1 \ 1&1&1 \ 1&1&1 end{pmatrix}) corresponding to eigenvalue three is (begin{pmatrix}1 \ 1 \ 1 end{pmatrix}). Equating the corresponding elements of (begin{pmatrix}a \ b \ a end{pmatrix}) and (begin{pmatrix}1 \ 1 \ 1 end{pmatrix})
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{12EI}{h^3})
b) (frac{12EA}{h^3})
c) (frac{12EA}{h})
d) (frac{12AI}{h^3})
Answer: a
Clarification: 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) (frac{rho A}{3})
b) (frac{rho A}{6})
c) 0
d) (frac{rho I}{3})
Answer: a
Clarification: 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.

Finite Element Method for Entrance exams,

Finite Element Method Multiple Choice Questions 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
Clarification: 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, (int_{L} sigma^T epsilon(phi)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
Clarification: 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=[ψ₁, ψ₂]ⁿ for element n, then the variational form is ______________
a) ψ^T(KQ–F)=0
b) ψ(KQ-F)=0
c) ψ(KQ)=F
d) ψ(F)=0
Answer: a
Clarification: 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=(frac{2EI}{l})
b) K=(frac{2EI}{l}begin{bmatrix}2 & 1 \ 1 & 2 end{bmatrix})
c) K=(frac{2E}{l}begin{bmatrix}2 \ 1 end{bmatrix})
d) K=(frac{2E}{l}begin{bmatrix}1 & 1 \ 1 & 1 end{bmatrix})
Answer: b
Clarification: 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
Clarification: 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=(frac{EA}{l})
b) F=ma
c) f(x)=y
d) ф=ф(x)
Answer: d
Clarification: 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) ε(ф)=(frac{dx}{dphi})
b) ε(ф)=(frac{dphi}{dx})
c) ε(ф)=(frac{dx}{dvarepsilon})
d) ф(ε)=(frac{dvarepsilon}{dphi})
Answer: b
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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.

Finite Element Method Multiple Choice Questions 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
Clarification: The term isoparametric is derived from the use of the same shape functions (or interpolation functions) [N] to define the element’s geometric shape as are used to define the displacements within the element.

2. The vector q=[q₁,q₂………q₈]^T of a four noded quadrilateral denotes ____
a) Load vector
b) Transition matrix
c) Element displacement vector
d) Constant matrix
Answer: c
Clarification: 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
Clarification: Master Element (ME) 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
Clarification: Master Element (ME) 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.

5. Shape function can be written as _____
a) N_t=(1-ξ)(1-η)
b) N_t=(1-ξ)
c) N_t=(1-η)
d) N_t=(frac{1}{4})(1-ξ)(1-η)
Answer: d
Clarification: 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=(frac{1}{4})(1-ξ)(1-η)
b) N_t=(1-ξ)(1-η)
c) N_t=(frac{1}{4})(1+ξξ_i)(1+ηη_i)
d) Undefined
Answer: c
Clarification: 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): DOF 1 and 2 for solid displacement (u) and DOF 3 for fluid pressure (p).

7. For a four noded quadrilateral elements, In u^T=[u.v]^T the displacement elements can be represented as u=N₁q₁+N₂q₃+ N₃q₅+ N₄q₇
v= N₁q₂+N₂q₄+ N₃q₆+ N₄q₈
then the shape function can be represented as _____
a) (N=left[begin{array}{ |c c c c}q_1 & q_5 \ q_2 &q_6\q_3 &q_7\q_4 & q_8end{array}right])
b) (N=begin{bmatrix}q_1 &q_3 &q_5 &q_7 \ q_2 &q_4&q_6&q_8end{bmatrix})
c) (N=begin{bmatrix}q_1 \ q_2end{bmatrix})
d) (N=begin{bmatrix}N_1 & 0 & N_3 & 0&N_5&0&N_7 & 0 \ 0 & N_2 &0 &N_4&0&N_6&0&N_8end{bmatrix})
Answer: d
Clarification: 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
Clarification: 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
Clarification: A body force that is distributed force per unit volume, a vector, many people probably call up Vector’s definition (from Despicable Me). 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 (more than one piece of information).

10. Shape functions are linear functions along the _____
a) Surfaces
b) Edges
c) Elements
d) Planes
Answer: b
Clarification: 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.

Finite Element Method Multiple Choice Questions on “Library of Elements and Interpolation Functions – 1”.

1. Which option is not correct about the three-noded triangular plane stress (linear) 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
Clarification: 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 (i.e. triangular elements with interpolation functions of higher degree) systematically?
a) Pascal’s triangle
b) Galerkin method
c) Jacobi method
d) Delaunaytriangulation
Answer: a
Clarification: The Higher-order triangular elements (also the Lagrange family of 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
Clarification: 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
Clarification: 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 (i.e., the degree of the polynomial is 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
Clarification: 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.
(begin{matrix}
& & & 1 & & &\
& & x & & y & & \
& x^2 & & xy & & y^2 & \
x^3 & & x^2y & & xy^2 & & y^3 \
end{matrix}
)

a) True
b) False
Answer: a
Clarification: 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(s), and it corresponds to all triangular elements with at least 0.5n(n+1) 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
Clarification: 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?

a) 1
b) xy
c) x²y
d) xy²
Answer: b
Clarification: 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. x², xy and y² in the same order, the interior node corresponds to the term xy. An n^th row of Pascal’s triangle contains n term(s) and it corresponds to all triangular elements with at least 0.5n(n+1) 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
Clarification: 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^nddegree polynomial corresponds to 6 noded triangle
Answer: c
Clarification: 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 N₁ and N₂?
a) N₁u₁+N₂u₂
b) N₁u₂+N₂u₁
c) N₁u₁-N₂u₂
d) N₁u₂-N₂u₁
Answer: a
Clarification: 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 u₁, u₂ and corresponding shape functions N₁, N₂. The displacement function is given by N₁u₁+N₂u₂.

12. For the following element, what is the value of the distance variable s at the 1^st row?

a) 1
b) 2
c) 0.5
d) 0
Answer: c
Clarification: 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}).

a) True
b) False
Answer: b
Clarification: 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}).

14. In the FEM element library, what is the correct name for a six noded triangular element?
a) Linear strain triangular element
b) Constant strain triangular element
c) Variable strain triangular element
d) Higher-order triangular element
Answer: a
Clarification: A Constant strain triangular (CST) element is the simplest triangular element with three end nodes. A Linear strain triangular element (LST) 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.