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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 56P
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 56P, 6.49 through 6.56 Determine the maximum deflection for the beams shown in Figs. P6.23 through P6.30

FIG. P6.30, P6.56

To determine

Find the maximum deflection for the beam using the conjugate-beam method.

Explanation of Solution

Given information:

The Young’s modulus is 200 GPa.

The moment of inertia is 500×106mm4.

Calculation:

Consider flexural rigidity EI of the beam is constant.

The given beam can be taken into two segments to draw the M/EI diagram. First segment is A to D; the reactions are calculated by considering all the forces acting throughout the span. Second segment is a simply supported beam of span B to C and the reaction at each support can be calculated by considering the load acting between B and C.

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

Refer Figure (1),

Determine the support reaction at A;

MB=0RC×10(25×14×142)+(25×3×32)=0RC=2,4501,12510RC=233.75kN

Determine the support reaction at B;

V=0RB+RC(25×17)=0RB=425233.75RB=191.25kN

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

Refer Figure (2),

Determine the bending moment at A;

MA=191.25×3+233.75×13(25×17×172)=3,612.53,612.5=0

Determine the bending moment at B;

MB=(25×3×32)=112.5kNm

Determine the bending moment at D;

MD=(25×4×42)=200kNm

Consider span BC:

Show the free body diagram of the beam as in Figure (3).

Refer Figure (3),

Determine the support reaction at B;

MC=0RB×10(25×10×102)=0RB=1,25010RB=125kN

Determine the support reaction at C;

V=0RB+RC(25×10)=0RC=250125RB=125kN

Determine the bending at the centre of beam BC;

M5ft=25×1028=312.5kN-m

Show the reaction and moment as in Figure (4).

Show the conjugate beam diagram for the whole span as in Figure (5).

Refer Figure (5),

The end condition of the original beam is free but in the conjugate-beam method the ends are changed to fixed end and the reaction at point B and C are calculated by considered the intermediate supports B and C as hinge.

Show the conjugate-beam diagram for the beam portion BC as in Figure (6).

Determine the conjugate beam reaction at B;

MB=0RB(10)1EI[23×10×312.5×10210×112.5×10212×10×(200112.5)×(13×10)]=0RB=3,333.33(10)EIRB=333.33kN-m2EI

Determine the shear force at M using the relation;

SM=1EI[RB+12×wM×xM+23(25xM28)xM112.5xM+12×(8.75xM)×xM]

Substitute 333.33kN-m2EI for RB and 125xM12.5xM2 for wM.

SM=1EI[333.33+12×(125xM12.5xM2)×xM+23(25xM28)xM112.5xM+12×(8.75xM)×xM]=1EI[333

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Chapter 6 Solutions

Structural Analysis
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