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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 54P
Textbook Problem
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6.49 through 6.56 Determine the maximum deflection for the beams shown in Figs. P6.23 through P6.30 by the conjugate-beam method.

Chapter 6, Problem 54P, 6.49 through 6.56 Determine the maximum deflection for the beams shown in Figs. P6.23 through P6.30

FIG. P6.29, P6.55

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 29,000 ksi.

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

Calculation:

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

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

Consider upward is positive and downward is negative.

Consider clockwise is negative and counterclowise is positive.

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 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 (due to uniformly distributed load) 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;

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 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 for the given beam as in Figure (6).

Show the conjugate beam M/EI (uniformly distributed load) diagram as in Figure (7).

Determine the conjugate beam reaction at support A and C;

RA=RC=13×(486EI)×36=5,832kips-ftEI=5,832kips-ftEI()

Show the conjugate beam M/EI (concentrated load) diagram as in Figure (8).

Determine the conjugate support reaction at support A;

MC=0RA×36+(12×36×560EI×23×24)=0RA=161,280kips-ft36EIRA=4,480kips-ftEI()

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.

Show the section xM (uniformly distributed load) and distance xM as in Figure (9).

Refer Figure (9),

Determine the intensity of load at xM using the relation;

wM=54xM3×xM×xM2=54xM1.5xM2

Show point M (Concentrated load) and distance xM as in Figure (10).

Refer Figure (10),

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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Chapter 6 Solutions

Structural Analysis
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