   # 6.31 and 6.32 Use the moment-area method to determine the slope and deflection at point D of the beam shown. FIG. P6.31, P6.57

#### Solutions

Chapter
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Chapter 6, Problem 31P
Textbook Problem
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## 6.31 and 6.32 Use the moment-area method to determine the slope and deflection at point D of the beam shown.FIG. P6.31, P6.57 To determine

Find the slope and deflection at point D of the given beam using the moment-area method.

### Explanation of Solution

Given information:

The Young’s modulus (E) is 200 GPa.

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

Calculation:

Consider flexural rigidity EI 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;

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 elastic curve diagram as in Figure (4).

Refer Figure (3).

Determine the deflection between A and C using the relation;

Here, b1 is the width of rectangle, h1 is the height of the triangle, b2 is the width of the triangle, and h2 is the height of the triangle.

Substitute 7 m for b1, 252EI for h1, 3 m for b2, and 252EI for h2.

ΔAC=(12×7×252EI)(23×7)+(12×3×252EI)(7+13×3)=7,140kNm3EI

Determine the slope at point C using the relation;

θC=7,140kNm3(L)EI

Here, L is the length between the supports

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