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# 7.28 through 7.30 Determine the deflection at point C of the beam shown in Fig. P7.24–P7.26 by the virtual work method. Use the graphical procedure (Table 7.6) to evaluate the virtual work integrals.

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### Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931

#### Solutions

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Section
BuyFindarrow_forward

### Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931
Chapter 7, Problem 30P
Textbook Problem
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## 7.28 through 7.30 Determine the deflection at point C of the beam shown in Fig. P7.24–P7.26 by the virtual work method. Use the graphical procedure (Table 7.6) to evaluate the virtual work integrals.

To determine

Find the deflection at point C of the beam using virtual work method.

### Explanation of Solution

Given information:

The beam is given in the Figure.

The value of E is 250 GPa and I is 600×106mm4.

Apply the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium as shown below.

• For summation of forces along x-direction is equal to zero (Fx=0), consider the forces acting towards right side as positive (+) and the forces acting towards left side as negative ().
• For summation of forces along y-direction is equal to zero (Fy=0), consider the upward force as positive (+) and the downward force as negative ().
• For summation of moment about a point is equal to zero (Matapoint=0), consider the clockwise moment as negative and the counter clockwise moment as positive.

Calculation:

Consider the real system.

Sketch the real system of the beam as shown in Figure 1.

Refer Figure 1 to find the reaction and moment.

Summation of moments about A is equal to 0.

MA=020Dy100(15)200(5)=0Dy=125kN

Summation of forces along y-direction is equal to 0.

+Fy=0Ay+Dy200100=0Ay+125200100=0Ay=175kN

Find the moment at point B:

MB=125(15)100(10)=875kNm

Find the moment at point C:

MC=125(5)=625kNm

Sketch the moment diagram of the real system M as shown in Figure 2.

Consider the virtual system.

Remove all the real loads and apply unit load where the point to find the deflection.

Let the bending moment due to virtual load be Mv.

Sketch the virtual system of the beam with unit load at point B as shown in Figure 3.

Refer Figure 3 to find the reaction and moment.

Summation of moments about A is equal to 0.

MA=020Dy1(15)=0Dy=0.75kN

Summation of forces along y-direction is equal to 0.

+Fy=0Ay+Dy1=0Ay+0.751=0Ay=0.25kN

Find the moment at point B:

MB=0.75(15)1(10)=1.25kNm

Find the moment at point C:

MC=0.75(5)=3.75kNm

Sketch the moment diagram of the virtual system Mv as shown in Figure 4.

Find the deflection at C using the virtual work expression:

1(ΔC)=0LMvMEIdx (1)

Substitute I for Span AB, 2I for span BC, and I for span CD.

Rearrange Equation (1) for the limits 05, 515, and 1520 as follows.

ΔC=1EI05(MvM)dx+12EI515(MvM)dx+1EI1520(MvM)dx (2)

Refer Table 7.6, “Integrals 0LMvMdx for moment diagrams of simple geometric shapes” in the textbook.

For the moment diagram in Figure 2 and Figure 4, the value of 05(MvM)dx, 515(MvM)dx, and 1520(MvM)dx are 13Mv1M1L, 16(Mv1(2M1+M2)+Mv2(M1+2M2))L, and 13Mv1M1L

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