   # 6.57 and 6.58 Use the conjugate-beam method to determine the slope and deflection at point D of the beam shown in Figs. P6.31 and P6.32. FIG. P6.31, P6.57

#### Solutions

Chapter
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Chapter 6, Problem 57P
Textbook Problem
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## 6.57 and 6.58 Use the conjugate-beam method to determine the slope and deflection at point D of the beam shown in Figs. P6.31 and P6.32.FIG. P6.31, P6.57 To determine

Find the slope θD and deflection ΔD at point D of the given beam using the conjugate-beam method.

### Explanation of Solution

The Young’s modulus (E) is 200 GPa.

The moment of inertia (I) is 262×106mm4.

Calculation:

Consider elastic modulus E of the beam is constant.

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

Refer Figure (1),

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;

MC=0RA×10(120×3)=0RA=36010RA=36kN

Determine the reaction at support C using the relation;

V=0RA+RC120=0RC=12036RC=84kN

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

Refer Figure (2),

Determine the bending moment at A;

MA=(84×10)(120×7)=840840=0

Determine the bending moment at B;

MB=(36×7)=252kNm

Determine the bending moment at C;

MC=36(10)+(120×3)=360+360=0

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

Show the M/EI diagram for the conjugate-beam as in Figure (4).

Consider counterclockwise is positive and clockwise is negative.

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

RA(10)12(252EI)(7)(3+13×7)12(252EI)(3)(23×3)=0RA=5,460kNm310EIRA=546kNm2EI

Determine the shear force at D using the relation;

SD=RA+(12×b1×h1)

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

Substitute 546kNm2EI for RA, 10 m for b1, and (252EI) for h1.

SD=546kNm2EI+(12×10×252EI)=714kNm2EI

Determine the slope at D using the relation;

θD=714kNm2EI

Substitute 200 GPa for E and 262×106mm4 for I

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