Method of Virtual Work: Beams & Frames - Free MCQ Practice Test with solutions,
MCQ Practice Test & Solutions: Test: Method of Virtual Work: Beams & Frames (14 Questions)
You can prepare effectively for Civil Engineering (CE) Structural Analysis with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Method of Virtual Work: Beams & Frames". These 14 questions have been designed by the experts with the latest curriculum of Civil Engineering (CE) 2026, to help you master the concept.
Test Highlights:
- - Format: Multiple Choice Questions (MCQ)
- - Duration: 30 minutes
- - Number of Questions: 14
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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.
What will be Δ in case of straight members using theorem?
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.
Detailed Solution: Question 1
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.
P is treated here as:-
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.
Detailed Solution: Question 2
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.
Force P is applied in the direction of Δ
State whether the above statement is true or false.
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.
State whether the above statement is true or false.
Detailed Solution: Question 3
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.
N is caused by:-
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.
Detailed Solution: Question 4
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.
What will be the external work performed during application of load?
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.
Detailed Solution: Question 5
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.
What will be the work done during additional application of dp1?
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.
Detailed Solution: Question 6
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.
Additional work done due to application of dp1 is p1 dΔ1 + p2 dΔ2.
Sate whether the above statement is true or false.
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.
Sate whether the above statement is true or false.
Detailed Solution: Question 7
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.
What will be the work done if all three forces are place at once on the beam?
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.
Detailed Solution: Question 8
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.
What will be change in work done in both case on initial application of load?
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.
Detailed Solution: Question 9
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.
Which of the following is equal to Δ1?
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.
Detailed Solution: Question 10
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.
Which of the following term is integrated to calculate Δ.
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.
Detailed Solution: Question 11
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.
If L is the length of beam, then what are the upper and lower limits of the above integration?
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.
Detailed Solution: Question 12
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.
Generally, in doing such integrations in which of the following’s term is m expressed?
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.
Detailed Solution: Question 13
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.
Which of the following term does 1.Δ represents?
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.
Detailed Solution: Question 14
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