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Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931

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Section
BuyFindarrow_forward

Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931
Chapter 6, Problem 28P
Textbook Problem
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6.23 through 6.30 Determine the maximum deflection for the beam shown by the moment-area method.

FIG. P6.28, P6.54

Chapter 6, Problem 28P, 6.23 through 6.30 Determine the maximum deflection for the beam shown by the moment-area method.

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 29,000 ksi.

The moment of inertia (I) is 5,000in.4.

Calculation:

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

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

The given beam carries uniformly distributed load and a point load. The loading of a beam has to be divided into carrying a point load and uniformly distributed load respectively to draw the M/EI diagram.

Consider upward is positive and downward is negative.

Consider clockwise is negative and counterclockwise is positive.

Consider uniformly distributed load:

Take moment about point C;

Determine the reaction at A using the relation;

MC=0RA×36(3×36×362)=0RA=1,94436RA=54kips

Determine the support reaction at C using the relation;

V=0RA+RC(3×36)=0RC=10854RC=54kips

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

Determine the bending moment at A;

MA=(54×36)(3×36×362)=0

Determine the bending moment at center of the beam;

Mcenter=(3×18×182)+(54×18)=486+972=486kips-ft

Determine the bending moment at C;

MC=(54×36)(3×36×362)=1,9441,944=0

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

Consider concentrated load:

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

Take moment about point C;

Determine the reaction at A using the relation;

MC=0RA×36(70×12)=0RA=84036RA=23.333kips

Determine the support reaction at C using the relation;

V=0RA+RC70=0RC=7023.333RC=46.667kips

Show the reactions as in Figure (5).

Determine the bending moment at A;

MA=(46.667×36)(70×24)=1,6801,680=0

Determine the bending moment at B;

MB=(23.333×24)=560kips-ft

Determine the bending moment at C;

MC=(23.333×36)(70×12)=840840=0

Show the M/EI diagram (point load) for the given beam as in Figure (6).

Show the elastic curve diagram as in Figure (7).

Refer Figure (3) and (6),

Determine the deflection between C and A using the relation;

ΔCA=MomentoftheareaoftheM/EIdiagrambetweenCandAaboutA

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

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

Substitute 486EI for MEI, (23×36×(362)) for area of parabola, (560EI) for MEI, (12×24×(13×24+12)) for area of triangle, 560EI for MEI, and (12×12×(23×12)) for area of triangle.

ΔCA=[486EI(23×36×(362))+(560EI)(12×24×(13×24+12))+560EI(12×12×(23×12))]=1EI[209,952+134,400+26,880]=371,232kips-ft3EI

Determine the slope at A using the relation;

θA=ΔCAL

Substitute 371,232kips-ft3EI for ΔBD and 36 ft for L

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Chapter 6 Solutions

Structural Analysis
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