   # 7.68 and 7.69 Use Castigliano’s second theorem to determine the horizontal deflection at joint C of the frames shown in Figs. P7.45 and P7.46. FIG. P7.45, FIG. P7.68

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Chapter 7, Problem 68P
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## 7.68 and 7.69 Use Castigliano’s second theorem to determine the horizontal deflection at joint C of the frames shown in Figs. P7.45 and P7.46. FIG. P7.45, FIG. P7.68

To determine

Find the horizontal deflection at joint C of the frame using Castigliano’s second theorem.

### Explanation of Solution

Given information:

The frame is given in the Figure.

The value of E is 200 GPa and I is 400×106mm4.

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:

Let apply a fictitious load P at joint C in the horizontal direction to find the deflection.

The value of fictitious load P is zero.

Sketch the frame 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 D:

Summation of moments about A is equal to 0.

MA=0Dy(10)3(10)(102)20(10)P(10)=0Dy=35+P

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

+Fy=0Ay+Dy3(10)=0Ay+35+P3(10)=0Ay=5+P

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

+Fx=0Ax+20+P=0Ax=20+P

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

 Segment x-coordinate M ∂M∂P Origin Limits (m) AB A 0−10 (20+P)x x CB C 0−10 (35+P)x−1.5x2 x

The expression for horizontal 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 010 and 010 as follows.

ΔC=1EI[010(MP)Mdx+010(MP)Mdx]

Substitute the value of fictitious load P as zero in the Column 4 of Table 1

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