# Finite Element Method Questions and Answers – Plane Elasticity – Assembly, Boundary and Initial Conditions

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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(u) boundary condition for the following plane stress problem of the beam?

a) uy(a,$$\frac{b}{3}$$)=0
b) uy(a,$$\frac{b}{2}$$)=0
c) ux(a,$$\frac{b}{3}$$)≠0
d) ux(a,$$\frac{b}{2}$$)=≠0
View Answer

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(a,$$\frac{b}{2}$$)=uy(a,$$\frac{b}{2}$$)=0. As the roller support restrains translation perpendicular to the base, only ux(a, y)=0 which implies ux(a,$$\frac{b}{3}$$)≠0 and ux(a,$$\frac{b}{2}$$)≠0 are incorrect.
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2. From solid mechanics, which traction(t) boundary condition is not correct for the following beam of thickness h?

a) ty(a,$$\frac{b}{2}$$)=0
b) ty(a,$$\frac{b}{3}$$)=0
c) tx(0, y)=0
d) ty(0, y)=-hT
View Answer

Answer: a
Explanation: The given cantilever beam is subjected to a shear force at the free end, thus tx(0, y)=0 and ty(0, y)=-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(a,$$\frac{b}{2}$$)≠0 and ty(a,$$\frac{b}{2}$$)≠0 whereas the roller support restrains translation perpendicular to the base; thus, only tx(a,y)≠0. The option ty(a,$$\frac{b}{2}$$)=0 is incorrect.

3. In Finite Element Analysis of the beam, which primary variable does not belong to the following mesh?

a) U9=0
b) U19=0
c) U10=0
d) U20=0
View Answer

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 (1, 2), (3, 4), (5, 6)… respectively. Since the translation along x is constrained, U9=U19 =U29=0. Because of the hinge at node 10, U20=0. The roller support doesn’t restrain vertical movement, thus U10≠0.
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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
View Answer

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 K1=$$\begin{bmatrix}2&1&5&0&0&0\\-1&7&5&0&0&0\\4&5&6&7&8&0\\0&0&3&3&0&0\\0&0&0&1&7&1\\-1&0&1&0&-1&11\end{bmatrix}$$ and K2=$$\begin{bmatrix}3&4&5&1&0&0\\1&6&0&3&0&0\\2&0&2&1&0&1\\3&0&-1&9&0&0\\0&7&0&1&7&1\\-1&0&0&0&-1&5\end{bmatrix}$$?

a) 9
b) 11
c) 13
d) 4
View Answer

Answer: a
Explanation: The assembled stiff matrix, K of the given mesh, is of the order 8×8. K is a combination of K1and K2 such that K1corresponds to the DOF in order 1,2,3,4, 7 and 8. K2 corresponds to the DOF in the order 3, 4, 5, 6, 7 and 8. Thus, K=$$\begin{bmatrix}2&1&5&0&0&0&0&0\\-1&7&5&0&0&0&0&0\\4&5&6+3&7+4&5&1&8&0\\0&0&3+1&3+6&0&3&0&0\\0&0&4&0&2&1&0&1\\0&0&3&0&-1&9&0&0\\0&0&0&8&0&1&14&2\\-1&0&0&3&0&0&-2&16\end{bmatrix}$$ and the elementin K at the index 3×3 is 6+3
=9.
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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
View Answer

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
View Answer

Answer: a
Explanation: Concerning the specification of the displacements (the primary degrees of freedom) and forces (the secondary degrees of freedom) in a finite element mesh, in general, only one of the quantities of each of the pairs (ux, tx) and (uy, ty) 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.
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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?

a) $$\frac{2M}{b}$$
b) Always zero
c) $$\frac{M}{b}$$
d) $$\frac{-M}{b}$$
View Answer

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.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter