   # 7.57 and 7.58 Use Castigliano’s second theorem to determine the slope and deflection at point B of the beam shown in Figs. P7.20 and P7.21.

#### Solutions

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Chapter 7, Problem 58P
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## 7.57 and 7.58 Use Castigliano’s second theorem to determine the slope and deflection at point B of the beam shown in Figs. P7.20 and P7.21. To determine

Find the slope and deflection at point B 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 70 GPa and I is 164(106)mm4.

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 and couple M¯ at point B in the desired direction to find the deflection and slope.

The value of fictitious load P is zero.

Sketch the beam with fictitious load P and couple M¯ as shown in Figure 1.

Let the equation for bending moment at distance x in terms of load P and couple M¯ be M.

The couple M¯ is 50kNm.

Refer to Figure 1.

The equation for bending moment is,

M=M¯Px (1)

Substitute 0 for P and 50kNm for M¯.

M=50

Find the derivative MM¯ using Equation (1).

MM¯=1

The expression for slope at B using Castigliano’s second theorem (θB) is shown as follows:

θB=0L(MM¯)MEIdx

Here, L is the length of the beam.

Substitute 4 m for L, 1 for MM¯, and 50 for M.

θB=1EI04(1)(50)dx=1EI04(50)dx=1EI[50x]04=1EI(50×4)

θB=200EI

Substitute 70 GPa for E and 164(106)mm4 for I.

θB=200kNm2(70GPa)(164(106)mm4)=200kNm2(103N1kN)(70GPa×109N/m21GPa)(164(106)mm4×1016m41mm4)=0

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