   # Determine the maximum deflection for the beams shown in Figs. P6.23 through P6.30 by the conjugate-beam method. FIG. P6.27, P6.53

#### Solutions

Chapter
Section
Chapter 6, Problem 53P
Textbook Problem
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## Determine the maximum deflection for the beams shown in Figs. P6.23 through P6.30 by the conjugate-beam method. FIG. P6.27, P6.53

To determine

Find the maximum deflection Δmax for the given beam by using conjugate-beam method.

### Explanation of Solution

Given information:

The Young’s modulus (E) is 29,000 ksi.

The moment of inertia (I) is 1,000in.4.

Calculation:

Consider Young’s modulus E of the beam is constant.

Show the free body diagram of the given beam as in Figure (1).

Refer Figure (2),

Consider upward is positive and downward is negative.

Consider clockwise is negative and counterclowise is positive.

Determine the support reaction at A using the Equation of equilibrium;

MD=0RA×30(60×10)(40×20)=0RA=1,40030RA=46.67kips

Determine the reaction at support D;

V=0RA+RD4060=0RD=10046.67RD=53.33kips

Show the reactions of the given beam as in Figure (2).

Refer Figure (2),

Determine the bending moment at A;

MA=(53.33×30)(60×20)(40×10)=1,6001,600=0

Determine the bending moment at B;

MB=46.67×10=466.7kips-ft

The bending moment at B is 466.7kips-ft, the moment of inertia between A and B is I but the moment of inertia of portion B and C is 2I. Therefore, the in M/EI diagram the moment can be labeled as 233.35EI.

Determine the bending moment at C;

MC=(53.33×10)=533.3kips-ft

The bending moment at C is 533.3kips-ft, the moment of inertia between C and D is I but the moment of inertia of portion B and C is 2I. Therefore, the in M/EI diagram the moment can be labeled as 266.65EI.

Show the M/EI diagram for the given beam as in Figure (3).

Show the conjugate diagram of the given beam as in Figure (4).

Refer Figure (4),

Determine the support reaction at A of a conjugate beam using the relation;

RA×(30)[(12×10×466.7EI)(20+13×10)+(10×233.35EI)(10+102)+12×(10)(266.65EI233.35EI)(13×10+10)+12×(10)(533.3EI)(23×10)]=0RA=130EI(54,448+35,002.5+2,220+17,777)RA=3,648.25kips-ft2EI

The maximum bending moment in the conjugate beam occur at point M, at a distance of xM from point B.

Determine the intensity of load at xM using the relation;

wMxM=(266.65kips-ft233.35kips-ft)10ftwM=3.33xM

Determine the shear force at point M using the relation;

SM=0[RA+12×b1×h1+(b2×h2)+12×wM×xM]=0

Here, b is the width and h is the height of respective triangle and rectangle.

Substitute 3,648.25kips-ft2EI for RA, 10 ft for b1, 466.7EI for h1, xM for b2, 233

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