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Structural Analysis

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

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Chapter
Section
BuyFindarrow_forward

Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931
Chapter 7, Problem 65P
Textbook Problem
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Use Castigliano’s second theorem to determine the vertical deflection at joint C of the frame shown in Fig. P7.36.

Chapter 7, Problem 65P, Use Castiglianos second theorem to determine the vertical deflection at joint C of the frame shown

To determine

Find the vertical 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 29,000 ksi and I is 2,000in.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 load P at joint C in the desired direction to find the deflection.

The value of load P is 25 k.

Sketch the frame 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 and moment at the supports A:

Summation of moments about A is equal to 0.

MA=0MAP(15)2(20)(202)=0MA15P(15)400=0MA=400+15P

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

+Fy=0AyP=0Ay=P

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

+Fx=0Ax+2(20)=0Ax=40k

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

Segmentx-coordinateMMP
OriginLimits (ft)
BAA020(400+15P)40x+2(x)(x2)15
CBC015Pxx

Substitute the value of load P as 25 in the Column 4 of Table 1.

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 020 and 015 as follows.

ΔC=1EI[020(MP)Mdx+015(MP)Mdx]

Substitute 15 for MP, (400+15(25))40x+x2 for M for the limits 020, x for MP, and 25x for M for the limits 015

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Chapter 7 Solutions

Structural Analysis
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