# 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.

### Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931

Chapter
Section

### Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931
Chapter 7, Problem 61P
Textbook Problem
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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 250 GPa and I is 600×106mm4.

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 load P at point C in the desired direction to find the deflection.

The value of load P is 100 kN.

Sketch the beam with 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(20)P(15)200(5)=020Dy=1,000+15PDy=50+3P4

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

+Fy=0Ay+Dy200P=0Ay+50+3P4200P=0Ay=150+P4

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

 Segment x-coordinate M ∂M∂P Origin Limits (m) AB A 0−5 (150+P4)x 0.25x BC A 5−15 (150+P4)x−200(x−5) 0.25x DC D 0−5 (50+3P4)x 0.75x

Refer to Table 1.

The value of load P is 100 kN.

For span AB,

M=(150+1004)x=175x

For span BC,

M=(150+1004)x200(x5)=175x200x+1,000=25x+1,000

For span DC,

M=(50+3(100)4)x=125x

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.

Substitute I for span AB, 2I for span BC, and I for span CD.

Rearrange Equation (1) for the limits 05, 515, and 05 as follows.

ΔC=1EI[05(MP)Mdx+12515(MP)Mdx+05(MP)Mdx]

Substitute 0

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