   # 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 59P
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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.

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:

Sketch the beam 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 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 CB C 0−L2 −Px −x BA C L2−L −Px −x

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.

For span AB moment of inertia is 2I and for span BC the moment of inertia is I.

Rearrange Equation (1) for the limits 0L2 and L2L as follows.

ΔC=0L/2(MP)MEIdx+12L/2L(MP)MEIdx

Substitute x for MP, Px for M for the limits 0L2, x for MP, and Px for M for the limits L2L

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