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Civil Engineering (CE) Exam  >  Civil Engineering (CE) Tests  >  Test: Method of Virtual Work: Beams & Frames - Civil Engineering (CE) MCQ

Test: Method of Virtual Work: Beams & Frames - Civil Engineering (CE) MCQ


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14 Questions MCQ Test - Test: Method of Virtual Work: Beams & Frames

Test: Method of Virtual Work: Beams & Frames for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Test: Method of Virtual Work: Beams & Frames questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Method of Virtual Work: Beams & Frames MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Method of Virtual Work: Beams & Frames below.
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Test: Method of Virtual Work: Beams & Frames - Question 1
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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?

Detailed Solution for Test: Method of Virtual Work: Beams & Frames - Question 1

Answer: d
Explanation: On substituting value of internal energy in earlier theorem, we can get this.

Test: Method of Virtual Work: Beams & Frames - Question 2
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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.

 P is treated here as:-

Detailed Solution for Test: Method of Virtual Work: Beams & Frames - Question 2

Answer: b
Explanation: P is treated as variable and N is expressed in its term for partial differentiation.

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Test: Method of Virtual Work: Beams & Frames - Question 3
Save

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.

Detailed Solution for Test: Method of Virtual Work: Beams & Frames - Question 3

Answer: a
Explanation: P is applied in above said direction. That is how we have been calculating the work done till now.

Test: Method of Virtual Work: Beams & Frames - Question 4
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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.

 N is caused by:- 
Detailed Solution for Test: Method of Virtual Work: Beams & Frames - Question 4

Answer: c
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.

Test: Method of Virtual Work: Beams & Frames - Question 5
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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 the external work performed during application of load?
Detailed Solution for Test: Method of Virtual Work: Beams & Frames - Question 5

Answer: a
Explanation: Since loads are gradually applied, work done will be average load times deflection. We can also find by integration.

Test: Method of Virtual Work: Beams & Frames - Question 6
Save

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?
Detailed Solution for Test: Method of Virtual Work: Beams & Frames - Question 6

Answer: b
Explanation: At this time p1 and p2 are already applied, only dp1 is gradually applied.

Test: Method of Virtual Work: Beams & Frames - Question 7
Save

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.
Detailed Solution for Test: Method of Virtual Work: Beams & Frames - Question 7

Answer: a
Explanation: It is true as the third term can be ignored as it is very small.

Test: Method of Virtual Work: Beams & Frames - Question 8
Save

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?
Detailed Solution for Test: Method of Virtual Work: Beams & Frames - Question 8

Answer: d
Explanation: Now, since all the loads are gradually applied, all will have a factor of half.

Test: Method of Virtual Work: Beams & Frames - Question 9
Save

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?
Detailed Solution for Test: Method of Virtual Work: Beams & Frames - Question 9

Answer: d
Explanation: We will get this by just subtracting two works done. This will be termed as dw.

Test: Method of Virtual Work: Beams & Frames - Question 10
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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.

Which of the following is equal to Δ1?
Detailed Solution for Test: Method of Virtual Work: Beams & Frames - Question 10

Answer: d
Explanation: Just substitute value of p2d Δ2 in dw using one of the earlier equation

Test: Method of Virtual Work: Beams & Frames - Question 11
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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.
Detailed Solution for Test: Method of Virtual Work: Beams & Frames - Question 11

Answer: a
Explanation: To calculate Δ we equate work done on both side which will mean m multiplied by angular displacement which is M/EI.

Test: Method of Virtual Work: Beams & Frames - Question 12
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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?
Detailed Solution for Test: Method of Virtual Work: Beams & Frames - Question 12

Answer: c
Explanation: Integration is done all over the beam, as it will give the work done

Test: Method of Virtual Work: Beams & Frames - Question 13
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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?
Detailed Solution for Test: Method of Virtual Work: Beams & Frames - Question 13

Answer: d
Explanation: Since we have to integrate wrt x, we express m in terms of x.

Test: Method of Virtual Work: Beams & Frames - Question 14
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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?
Detailed Solution for Test: Method of Virtual Work: Beams & Frames - Question 14

Answer: b
Explanation: Term shown above basically reprents virual load multiplied by displacement.

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