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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 64P
Textbook Problem
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7.63 and 7.64 Use Castigliano’s second theorem to determine the slope and deflection at point D of the beam shown in Figs. P7.34 and P7.35.

Chapter 7, Problem 64P, 7.63 and 7.64 Use Castiglianos second theorem to determine the slope and deflection at point D of

FIG. P7.35, P7.64

To determine

Find the slope and deflection at point D of the beam using Castigliano’s second theorem.

Explanation of Solution

Given information:

The beam is given in the Figure.

Value of E is 30,000 ksi, I is 4,000in.4 for span AB, and 3,000in.4 for span BD.

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

The value of load P is 35 k and couple M¯ is zero.

Sketch the beam with load P and couple M¯ 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.

The derivative of M with respect to M¯ is MM¯.

Find the reactions and moment at the supports:

Consider portion BCD, Summation of moments about B is equal to 0.

MB=08CyP(16)M¯=08Cy=16P+M¯Cy=2P+M¯8

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

+Fy=0Ay+Cy2.5(16)P=0Ay+2P+M¯82.5(16)P=0Ay=40PM¯8

Summation of moments about A is equal to 0.

MA=0MAP(32)M¯+Cy(24)2.5(16)(162)=0MA32PM¯+48P+3M¯320=0MA=32016P2M¯

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

Segmentx-coordinateMMM¯MP
OriginLimits (ft)
DCD08PxM¯1x
CBD816PxM¯+(2P+M¯8)(x8)1+18(x8)x16
ABA016(32016P2M¯)+(40PM¯8)x1.25x22x816x

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

θD=0L(MM¯)MEIdx (1)

Here, L is the length of the beam.

Rearrange Equation (1) for the limits 08, 816, and 016 as follows.

θD=1E[1I08(MM¯)Mdx+1I816(MM¯)Mdx+1I016(MM¯)Mdx]

Substitute the value of load P as 35 k and couple M¯ as 0 in the column 4 of Table 1.

Substitute 1 for MM¯, 35x for M for the limit 08, 1+18(x8) for MM¯, 35x+70(x8) for M for the limit 816, 2x8 for MM¯, 4I3 for I, and 240+5x1.25x2 for M for the limit 016.

θD=1E[1I08(1)(35x)dx+1I816(1+18(x8))(35x+70(x8))dx+143I016(2x8)(240+5x1.25x2)dx]=1E[1I08(35x)dx+1I816(70x140x+1,12035x28+70x28560x8)dx+34I016(480+10x2

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Chapter 7 Solutions

Structural Analysis
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