Finite Element Method Questions and Answers – Beams and Frames – Finite Element Formulation

«
»

This set of Finite Element Method Multiple Choice Questions & Answers (MCQs) 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
View Answer

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.
advertisement

2. Beams are horizontal members used for supporting transverse loading.
a) True
b) False
View Answer

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
View Answer

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
advertisement
advertisement

4. The displacements in beam elements are interpolated using _______
a) shape elements
b) shape functions
c) shape parameters
d) shape factors
View Answer

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
View Answer

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.
advertisement

6. In beam elements the cross section of the element is assumed.
a) True
b) False
View Answer

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
View Answer

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.
advertisement

8. The element stiffness matrix (k) for beam element is given by which of the following expressions?
a)\(\frac{EI}{l^3}\begin{bmatrix}12&6l&-12&6l\\6l&4l^2&-6l&2l^2\\-12&-6l&12&-6l\\6l&2l^2&-6l&4l^2\end{bmatrix}\)
b) \(\frac{EI}{l^2}\begin{bmatrix}12&-6l&-12&6l\\-6l&4l^2&6l&2l^2\\-12&6l&12&6l\\6l&2l^2&6l&4l^2\end{bmatrix}\)
c)\(\frac{EI}{l^3}\begin{bmatrix}12&-6l&-12&6l\\-6l&4l^2&6l&2l^2\\-12&6l&12&6l\\6l&2l^2&6l&4l^2\end{bmatrix}\)
d)\(\frac{EI}{l^2}\begin{bmatrix}12&6l&-12&6l\\6l&4l^2&-6l&2l^2\\-12&-6l&12&-6l\\6l&2l^2&-6l&4l^2\end{bmatrix}\)
View Answer

Answer: a
Explanation: The value of stiffness matrix is given by
k=\(\frac{EI}{l^3}\begin{bmatrix}12&6l&-12&6l\\6l&4l^2&-6l&2l^2\\-12&-6l&12&-6l\\6l&2l^2&-6l&4l^2\end{bmatrix}\)
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.

Sanfoundry Global Education & Learning Series – Finite Element Method.

advertisement

To practice all areas of Finite Element Method, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement
Manish Bhojasia - Founder & CTO at Sanfoundry
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