   # 6.23 through 6.30 Determine the maximum deflection for beam shown by the moment-area method. FIG. P6.25, P6.51

#### Solutions

Chapter
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Chapter 6, Problem 25P
Textbook Problem
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## 6.23 through 6.30 Determine the maximum deflection for beam shown by the moment-area method.FIG. P6.25, P6.51 To determine

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

### Explanation of Solution

Given information:

The Young’s modulus (E) is 200 GPa.

The moment of inertia (I) is 600×106mm4.

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 (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×24(80×12)=0RA=96024RA=40kN

Determine the reaction at support C;

V=0RA+RC80=0RC=8040RC=40kN

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

Refer Figure (2),

Determine the bending moment at A;

MA=(40×24)(80×12)=0

Determine the bending moment at B;

MB=(40×12)=480kNm

The bending moment at B is 480kNm, 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 240EI.

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

Show the deflected shape of the given beam as in Figure (3).

Refer Figure (3),

Determine the deflection between A and C using the relation;

Divide the M/EI into triangle and rectangle to evaluate the deflection.

ΔCA=[MEI(Areaoftriangle)+MEI(Areaoftriangle)]

Substitute 480EI for MEI, (12×12×(12+13×12)) for area of triangle, (480EI240EI) for MEI, and (12×12×(23×12)) for area of triangle.

ΔCA=[480EI(12×12×(12+13×12))+(480EI240EI)(12×12×(23×12))]=1EI[46,080+11,520]=57,600kNm3EI

Determine the slope at A using the relation;

θA=ΔCAL

Substitute 57,600kNm3EI for ΔDA and 24 m for L.

θA=57,600kNm3EI24=2,400kNm2EI

The maximum deflection will occur for the given beam at point M and take a distance of xM from point A then the slope at point M is equal to the slope at point A

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