   # Determine the maximum deflection for the beams shown in Figs. P6.23 through P6.30 by the conjugate-beam method. FIG. P6.26, P6.52

#### Solutions

Chapter
Section
Chapter 6, Problem 52P
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. P6.26, P6.52

To determine

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

### Explanation of Solution

Given information:

The Young’s modulus (E) is 70 GPa.

The moment of inertia (I) is 95×106mm4.

Calculation:

Consider flexural rigidity (EI) of the beam is constant.

Show the given beam as in Figure (1).

Consider upward is positive and downward is negative.

Consider clockwise is negative and counterclowise is positive.

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

MD=0RB×18(90×9)(50×24)=0RB=2,01018RB=111.67kN

Determine the reaction at support D using the relation;

V=0RB+RD5090=0RD=140111.67RD=28.33kN

Show the free body diagram of the given beam as in Figure (2).

Determine the bending moment at A;

MA=+(28.33×24)(90×15)+(111.67×6)=679.921,350+670.02=0

Determine the bending moment at B;

MB=+(28.33×18)(90×9)=300kNm

Determine the bending moment at C;

MC=(50×15)+(111.67×9)=255kNm

Determine the bending moment at D;

MD=(50×24)+(111.67×18)(90×9)=1,200+2,010810=0

Determine the distance (x)where the bending moment is zero using the relation;

Mx=050(6+x)+(111.67×x)=030050x+111.67x=0x=30061.67

x=4.86

Hence, the bending moment is zero at 4.86 m form support B.

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).

Determine the reaction at support D using the relation;

MBBD=0RD(18)+[12×(255EI)(9)(6+23×9)+12(255EI)(94.86)(13×4.86+6)12(300EI)(4.86)(4.863)]=0RD=(13,7704,022.217+1,180.98)18EIRD=922

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