   # 7.34 and 7.35 Use the virtual work method to determine the slope and deflection at point D of the beam shown. FIG. P7.35, P7.64

#### Solutions

Chapter
Section
Chapter 7, Problem 35P
Textbook Problem
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## 7.34 and 7.35 Use the virtual work method to determine the slope and deflection at point D of the beam shown. FIG. P7.35, P7.64

To determine

Find the slope and deflection at point D of the beam using virtual work method.

### Explanation of Solution

Given information:

The beam is given in the Figure.

Value of E is 30,000 ksi, I is 4,000in.4 for span AB, and 3,000in.4 for span BD.

Calculation:

Consider the real system.

Draw a diagram showing all the given real loads acting on it.

Let an equation expressing the variation of bending moment due to real loading be M.

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

Find the reactions and moment at the supports:

Consider portion BCD, Summation of moments about B is equal to 0.

MB=08Cy35(16)=0Cy=70k

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

+Fy=0Ay+Cy2.5(16)35=0Ay+702.5(16)35=0Ay=5k

Summation of moments about A is equal to 0.

MA=0MA35(32)+70(24)2.5(16)(162)=0MA=240k-ft

Consider the virtual system.

Draw a diagram of beam without the given real loads. For deflection apply unit load at the point and in the desired direction.

For slope calculation, apply a unit couple at the point on the beam where the slope is desired.

Sketch the virtual system of the beam with unit couple at point D as shown in Figure 2.

Let an equation expressing the variation of bending moment due to virtual couple be Mv1.

Consider portion BCD, Summation of moments about B is equal to 0.

MB=08Cy1=0Cy=18k

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

+Fy=0Ay+Cy=0Ay+18Ay=18k

Summation of moments about A is equal to 0.

MA=0MA1+18(24)=0MA=2k-ft

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

Let an equation expressing the variation of bending moment due to virtual load be Mv2.

Consider portion BCD, Summation of moments about B is equal to 0.

MB=08Cy1(16)=0Cy=2k

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

+Fy=0Ay+Cy1=0Ay+21=0Ay=1k

Summation of moments about A is equal to 0.

MA=0MA1(32)+2(24)=0MA=16k-ft

Find the equations for M, Mv1, and Mv2 for the 3 segments of the beam as shown in Table 1.

 Segment x-coordinate M (kN⋅m) Mv1 (kN⋅m) Mv2 (kN⋅m) Origin Limits (m) DC D 0−8 −35x −1 −1x CB D 8−16 −35x+70(x−8) −1+18(x−8) x−16 AB A 0−16 240+5x−1.25x2 2−x8 16−x

Find the slope at D using the virtual work expression:

1(θD)=0LMv1MEIdx (1)

Here, L is the length of the beam, E is the young’s modulus, and I is the moment of inertia.

Rearrange Equation (1) for the limits 08, 816, and 016 as follows.

θD=1E[1I08(Mv1M)dx+1I816(Mv1M)dx+1I016(Mv1M)dx]

Substitute 1 for Mv1, 35x for M for the limit 08, 1+18(x8) for Mv1, 35x+70(x8) for M for the limit 816, 2x8 for Mv1, 4I3 for I, and 240+5x1.25x2 for M for the limit 016.

θD=1E[1I08(1)(35x)dx+1I816(1+18(x8))(35x+70(x8))dx+143I016(2x8)(240+5x1.25x2)dx]=1E[1I08(35x)dx+1I816(70x140x+1,12035x28+70x28560x8)dx+34I016(480+10x2

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