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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.

What will be Δ in case of straight members using theorem?

Solution:

Answer: d

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

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.

P is treated here as:-

Solution:

Answer: b

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

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.

Force P is applied in the direction of Δ

State whether the above statement is true or false.

Solution:

Answer: a

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

QUESTION: 4

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:-

Solution:

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.

QUESTION: 5

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?

Solution:

Answer: a

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

QUESTION: 6

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?

Solution:

Answer: b

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

QUESTION: 7

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.

Solution:

Answer: a

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

QUESTION: 8

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?

Solution:

Answer: d

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

QUESTION: 9

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?

Solution:

Answer: d

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

QUESTION: 10

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?

Solution:

Answer: d

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

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.

Which of the following term is integrated to calculate Δ.

Solution:

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.

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.

If L is the length of beam, then what are the upper and lower limits of the above integration?

Solution:

Answer: c

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

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.

Generally, in doing such integrations in which of the following’s term is m expressed?

Solution:

Answer: d

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

QUESTION: 14

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?

Solution:

Answer: b

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

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