This set of Finite Element Method Multiple Choice Questions & Answers (MCQs) 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) \(\frac{\partial S}{\partial x}=\frac{K}{A_e}\)

b) \(\frac{\partial S}{\partial y}=\frac{K}{A_e^2}\)

c) \(\frac{\partial S}{\partial x}\)=KA_{e}

d) \(\frac{\partial S}{\partial y}\)=KA_{e}^{2}

View Answer

Explanation: For a linear triangular (i.e., constant-strain triangle) element, the shape function \(\psi_i^e\) and its derivatives are given by \(\psi_i^e=\frac{1}{2A_e}(\alpha_i^e+\beta_i^e x+\gamma_i^e y)\), \(\frac{\partial \psi_i^e}{\partial x}=\frac{\beta_i^e}{2A_e}\) 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) \(\frac{A_e h_e}{4}\)f

b) \(\frac{A_e h_e}{3}\)f

c) \(\frac{h_e}{3A_e}\)f

d) \(\frac{h_e}{4A_e}\)f

View Answer

Explanation: For a linear triangular (i.e., constant-strain triangle) element, for the case in which the body force is uniform and thus the body force components fx and fy are element-wise constant (say, equal to, \(f_{x0}^e\) and \(f_{y0}^e\) respectively), the load vector F has the form F=∫

_{Ωc}h

_{e}(ψ

^{e})

^{T}\(f_0^e\)dx

=\(\frac{A_e h_e}{4}\begin{bmatrix}f_{x0}^e\\f_{y0}^e\\f_{x0}^e\\f_{y0}^e\\f_{x0}^e\\f_{y0}^e\end{bmatrix}\). 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) \(\frac{A_e h_e}{4} \)f

b) \(\frac{A_e h_e}{3}\)f

c) \(\frac{h_e}{3A_e}\)f

d) \(\frac{h_e}{4A_e}\)f

View Answer

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 (say, equal to, \(f_{x0}^e\) and \(f_{y0}^e\) respectively), the load vector F has the form F=\(\int_{\Omega_c}h_e(\psi^e)^T f_0^edx\) =\(\frac{A_e h_e}{4}\begin{bmatrix}f_{x0}^e\\f_{y0}^e\\f_{x0}^e\\f_{y0}^e\\f_{x0}^e\\f_{y0}^e\\f_{x0}^e\\f_{y0}^e\end{bmatrix}\). 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

View Answer

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 (i.e., known). Computation of Q involves the evaluation of line integrals (for any type of element). 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

View Answer

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 (i.e., known). Computation of Q involves the evaluation of line integrals (for any type of element). 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 using a transformation matrix.

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) \(\begin{bmatrix}

cos \alpha & sin \alpha & 0 & 0 & \\-sin \alpha & cos \alpha & 0 & 0 & \\0 & 0 & cos \alpha & sin \alpha & \\0 & 0 & -sin \alpha & cos \alpha & \\& & & & \ddots\end{bmatrix}\)

b) \(\begin{bmatrix}cos \alpha & -sin \alpha & 0 & 0 & \\

sin \alpha & cos \alpha & 0 & 0 & \\0 & 0 & cos \alpha & -sin \alpha & \\0 & 0 & sin \alpha & cos \alpha & \\& & & & \ddots\end{bmatrix}\)

c) \(\begin{bmatrix}cos \alpha & sin \alpha & 0 & 0 & \\-sin \alpha & cos \alpha & 0 & 0 & \\0 & 0 & cos \alpha & -sin \alpha & \\0 & 0 & sin \alpha & cos \alpha & \\& & & & \ddots\end{bmatrix}\)

d) \(\begin{bmatrix}cos \alpha & -sin \alpha & 0 & 0 & \\

sin \alpha & cos \alpha & 0 & 0 & \\0 & 0 & cos \alpha & sin \alpha & \\0 & 0 & -sin \alpha & cos \alpha & \\& & & & \ddots\end{bmatrix}\)

View Answer

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=\(\begin{bmatrix}cos \alpha & sin \alpha & 0 & 0 & \\-sin \alpha & cos \alpha & 0 & 0 & \\0 & 0 & cos \alpha & -sin \alpha & \\0 & 0 & sin \alpha & cos \alpha & \\& & & & \ddots\end{bmatrix}\) 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 \(\begin{bmatrix}0\\0\\2\\0\\1\\0\end{bmatrix}\) and θ=0?

a) \(\begin{bmatrix}0\\0\\0\\2\\0\\1\end{bmatrix}\)

b) \(\begin{bmatrix}0\\0\\2\\0\\0\\1\end{bmatrix}\)

c) \(\begin{bmatrix}0\\0\\2\\0\\1\\0\end{bmatrix}\)

d) \(\begin{bmatrix}0\\0\\0\\2\\1\\0\end{bmatrix}\)

View Answer

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=\(\begin{bmatrix}cos \alpha & sin \alpha & 0 & 0 & \\

-sin \alpha & cos \alpha & 0 & 0 & \\0 & 0 & cos \alpha & sin \alpha & \\0 & 0 & -sin \alpha & cos \alpha & \\& & & & \ddots\end{bmatrix}\) 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=\(\begin{bmatrix}0&1&0&0&0&0\\-1&0&0&0&0&0\\0&0&0&1&0&0\\0&0&-1&0&0&0\\0&0&0&0&0&1\\-1&0&0&0-1&0\end{bmatrix}\)

F=\(\begin{bmatrix}0&1&0&0&0&0\\ -1&0&0&0&0&0\\0&0&0&1&0&0\\0&0&-1&0&0&0\\0&0&0&0&0&1\\-1&0&0&0-1&0\end{bmatrix}^T\)x\(\begin{bmatrix}0\\0\\2\\0\\1\\0\end{bmatrix}\)

=\(\begin{bmatrix}0\\0\\0\\2\\0\\1\end{bmatrix}\).

8. What is the expression for the traction term t_{n} in the element load vector Q^{e}=∮_{┌c}h_{e}ψ^{T}tds of the following figure where L_{23} is the length of the line 2-3?

a) t_{n}=-T(-1+\(\frac{s}{L_{23}})\)

b) t_{n}=T(-1+\(\frac{s}{L_{23}})\)

c) t_{n}=-T(1+\(\frac{s}{L_{23}})\)

d) t_{n}=T(1+\(\frac{s}{L_{23}})\)

View Answer

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(1- \(\frac{s}{L_{23}})\), 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?

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

View Answer

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

View Answer

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

View Answer

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

View Answer

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.

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