# Finite Element Method Questions and Answers – Beams and Frames – Boundary Conditions

«
»

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

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

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 (K) of an element given value of force (F) and displacement (Q)?
a) FQ=K
b) KQ=F
c) KF=Q
d) KFQ=1

Explanation: The correct equation is given by KQ=F. The value of force (F) is the product of stiffness (K) and displacement (Q). The value of stiffness (K) 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

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

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

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

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 (E), modulus of rigidity (G), and Poisson’s ratio (u)?
a) E = 2*G*(1+u)
b) E = 3*G*(1+u)
c) E = 2*G*(1-u)
d) E = 3*G*(1-u)

Explanation: The relation between the material properties is given by
E = 2*G*(1+u)
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.

Sanfoundry Global Education & Learning Series – Finite Element Method. 