Structural Analysis Multiple Choice Questions on “External Work and Strain Energy”.
Ue= work done by all external forces
Ui = internal work or strain energy
Δ = final elongation of bar
Θ = final angular deflection
1. Ui is not developed when:-
a) Structure elongates
b) Structure bends
c) Structure deforms
d) External force is zero
Answer: d
Clarification: Ui is developed when structure deforms, which is happening in other three options.
2. What will be the value of Ue if material is linear elastic? Axial force is increased from 0 to P gradually.
a) 1⁄4 P Δ
b) 1⁄3P Δ
c) 1⁄2P Δ
d) P Δ
Answer: c
Clarification: Due to linear elasticity, we can substitute force in terms of P and Δ and then integrate wrt x to get the final answer.
3. What will be the work done force P if another load external load F’ causes deflection Δ’ in the above question?
a) 1⁄4P Δ’
b) 1⁄3P Δ’
c) 1⁄2P Δ’
d) P Δ’
Answer: d
Clarification: Here, P will remain constant. So, it will be a simple integration from 0 to Δ’.
4. What will be the work done by F’?
a) 1⁄4F’ Δ’
b) 1⁄3F’ Δ’
c) 1⁄2F’ Δ’
d) F’ Δ’
Answer: c
Clarification: Here, P will remain constant. So, it will be a simple integration from 0 to Δ’.
5. What will be the value of Ue if material is linear elastic? Moment is increased from 0 to m gradually.
a) 1⁄4 M θ
b) 1⁄3 M θ
c) 1⁄2 M θ
d) M θ
Answer: c
Clarification: Due to linear elasticity, we can substitute moment in terms of M and θ and then integrate wrt x to get the final answer. Mdθ is done for moment to calculate work done.
6. What will be the work done force M if another load external load M’ causes deflection θ’ in the above question?
a) 1⁄4 M θ’
b) 1⁄3 M θ’
c) 1⁄2 M θ’
d) M θ’
Answer: d
Clarification: Here, P will remain constant. So, it will be a simple integration from 0 to Δ’.
7. What will be the work done by M’ in above question?
a) 1⁄4 M θ’
b) 1⁄3 M θ’
c) 1⁄2 M θ’
d) M θ’
Answer: c
Clarification: Due to linear elasticity, we can substitute moment in terms of M and θ and then integrate wrt x to get the final answer. Mdθ is done for moment to calculate work done.
8. If an axial force N is applied gradually to a bar which is linear elastic and has a constant cross sectional area A and length L, what will be Δ?
a) 1⁄4 NL/AE
b) 1⁄3 NL/AE
c) 1⁄2 NL/AE
d) NL/AE
Answer: d
Clarification: Hooke’s law will be valid here as material is linear elastic.
9. In the above question, what will be the value of Ui ?
a) 1⁄4 N2L/AE
b) 1⁄3 N2L/AE
c) 1⁄2 N2L/AE
d) N2L/AE
Answer: c
Clarification: Once deformation is known, we can calculate the work done using earlier equations and then Ue = Ui.
10. What will be the value of dUi in terms of E and I?
a) 1⁄4 M2dx/EI
b) 1⁄3 M2dx/EI
c) 1⁄2 M2dx/EI
d) M2dx/EI
Answer: c
Clarification: Relation between dθ and M/EI is known. So, we can use that to get results.