   # 7.59 through 7.62 Use Castigliano’s second theorem to determine the deflection at point C of the beams shown in Figs. P7.24−P7.27.

#### Solutions

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Chapter 7, Problem 62P
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## 7.59 through 7.62 Use Castigliano’s second theorem to determine the deflection at point C of the beams shown in Figs. P7.24−P7.27. To determine

Find the deflection at point C of the beam using Castigliano’s second theorem.

### Explanation of Solution

Given information:

The beam is given in the Figure.

The value of E is 29,000 ksi and I is 350in.4

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:

Let apply a fictitious load P at point C in the desired direction to find the deflection.

The value of fictitious load P is zero.

Sketch the beam with fictitious load P as shown in Figure 1.

Let the equation for bending moment at distance x in terms of load P be M, the derivative of M with respect to P is MP.

Find the reactions at the supports A and B:

Summation of moments about A is equal to 0.

MA=0By(15)2(21)(212)P(21)=0By(15)=441+21P=0By=29.4+1.4P

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

+Fy=0Ay+By2(21)P=0Ay=29.41.4P+42+P=0Ay=12.60.4P

Find the equations for M and MP for the 2 segments of the beam as shown in Table 1.

 Segment x-coordinate M ∂M∂P Origin Limits (ft) AB A 0−15 (12.6−0.4P)x−x2 −0.4x CB C 0−6 −Px−x2 −x

Refer to Table 1.

The value of load P is 0.

The expression for deflection at C using Castigliano’s second theorem (ΔC) is shown as follows:

ΔC=0L(MP)MEIdx (1)

Here, L is the length of the beam.

Rearrange Equation (1) for the limits 015 and 06 as follows.

ΔC=1EI[015(MP)Mdx+06(MP)Mdx]

Substitute 0

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