This set of Structural Analysis Multiple Choice Questions & Answers focuses on “Method of Virtual Work: Beams and Frames and Castigliano’s Theorem for Trusses”.
Δ = displacement caused when force is increased by a small amount
P = external force applied
N = internal force in the member force applied
L = length of member
A = cross-sectional area of member
E = Modulus of elasticity
Same symbol is used for partial and total differentiation and they are pretty obvious.
1. What will be Δ in case of straight members using theorem?
a) 1⁄4 ΣN(dN/dP)L/AE
b) 1⁄3 ΣN(dN/dP)L/AE
c) 1⁄2 ΣN(dN/dP)L/AE
Explanation: On substituting value of internal energy in earlier theorem, we can get this.
2. P is treated here as:-
c) it doesn’t matter
d) depends upon load
Explanation: P is treated as variable and N is expressed in its term for partial differentiation.
3. Force P is applied in the direction of Δ
State whether the above statement is true or false.
Explanation: P is applied in above said direction. That is how we have been calculating the work done till now.
4. N is caused by:-
a) constant forces
b) variable forces
Explanation: It is caused by both the constant external force and variable P.
A beam has been subjected to gradually applied load P1 and P2 causing deflection Δ1 and Δ2.
Gradual increase of dp1 causes subsequent deflection of dΔ1 and dΔ2.
5. What will be the external work performed during application of load?
a) 1⁄2 (p1 Δ1 + p2 Δ2)
b) 1⁄2 (p2 Δ1 + p1 Δ2)
c) p1 Δ1 + p2 Δ2
d) p2 Δ1 + p1 Δ2
Explanation: Since loads are gradually applied, work done will be average load times deflection. We can also find by integration.
6. What will be the work done during additional application of dp1?
a) p1 dΔ1 + p2 dΔ2 + dp1d Δ1
b) p1 dΔ1 + p2 dΔ2 + 1⁄2 dp1d Δ1
c) p1 dΔ1 + 1⁄2 p2 dΔ2 + dp1d Δ1
d) 1⁄2 p1 dΔ1 + p2 dΔ2 + dp1d Δ1
Explanation: At this time p1 and p2 are already applied, only dp1 is gradually applied.
7. Additional work done due to application of dp1 is p1 dΔ1 + p2 dΔ2.
Sate whether the above statement is true or false.
Explanation: It is true as the third term can be ignored as it is very small.
8. What will be the work done if all three forces are place at once on the beam?
a) (p1 + dp1)(Δ1 + dΔ1) + (p2)( Δ2 + dΔ2)
b) (p1 + dp1)(Δ1 + dΔ1) + 1⁄2 (p2)( Δ2 + dΔ2)
c) 1⁄2 (p1 + dp1)(Δ1 + dΔ1) + (p2)( Δ2 + dΔ2)
d) 1⁄2 (p1 + dp1)(Δ1 + dΔ1) + 1⁄2 (p2)( Δ2 + dΔ2)
Explanation: Now, since all the loads are gradually applied, all will have a factor of half.
9. What will be change in work done in both case on initial application of load?
a) p1dΔ1 + dp1 Δ1 + p2dΔ2
b) 1⁄2 p1dΔ1 + dp1 Δ1 + p2dΔ2
c) 1⁄2 p1dΔ1 + 1⁄2 dp1 Δ1 + p2dΔ2
d) 1⁄2 p1dΔ1 + 1⁄2 dp1 Δ1 + 1⁄2 p2dΔ2
Explanation: We will get this by just subtracting two works done. This will be termed as dw.
10. Which of the following is equal to Δ1?
Explanation: Just substitute value of p2d Δ2 in dw using one of the earlier equation.
X is taken along the axis of beam
1 = external virtual unit load acting on the beam with direction same as that of Δ.
m = internal virtual moment in beam.
Δ = external displacement of the point caused by the real loads.
M = internal moment caused by the real loads.
E = modulus of elasticity .
I = moment of inertia of cross-sectional area.
11. Which of the following term is integrated to calculate Δ.
Explanation: To calculate Δ we equate work done on both side which will mean m multiplied by angular displacement which is M/EI.
12. If L is the length of beam, then what are the upper and lower limits of the above integration?
a) –L, L
b) –L, 0
c) 0, L
d) ½ L, L
Explanation: Integration is done all over the beam, as it will give the work done.
13. Generally, in doing such integrations in which of the following’s term is m expressed?
Explanation: Since we have to integrate wrt x, we express m in terms of x.
14. Which of the following term does 1.Δ represents?
a) work done by actual forces
b) virtual strain energy stored in beam
c) real strain energy stored in beam
d) total work done by actual and virtual forces
Explanation: Term shown above basically reprents virual load multiplied by displacement.
Sanfoundry Global Education & Learning Series – Structural Analysis.
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