   # 6.23 through 6.30 Determine the maximum deflection for the beam shown by the moment-area method. FIG. P6.29, P6.55

#### Solutions

Chapter
Section
Chapter 6, Problem 29P
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. P6.29, P6.55

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 350in.4.

Calculation:

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

To draw the M/EI the given beam can be divided into two segments such as AB and AC.

Consider span AB:

Show the free body diagram for span AB as in Figure (1).

Determine the support reaction at A;

MB=0RA×15(2×15×152)=0RA=22515RA=15kips

Determine the support reaction at B;

V=0RA+RB(2×15)=0RB=3015RB=15kips

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

Determine the bending moment at A using the relation;

MA=15×15(2×15×152)=225225=0

Determine the bending moment at centre of the beam AB;

Mcenter=15×7.5(2×7.5×7.52)=112.556.25=56.25kips-ft

Determine the bending moment at B;

MB=15×15+(2×15×152)=225+225=0

Consider span AC:

The uniformly distributed load is acting in span BC.

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

Determine the support reaction at B;

MA=0RB×15+(2×21×212)=0RB=44115RB=29.4kips()

Determine the support reaction at A;

V=0RA+RB(2×21)=0RA=4229.4RA=12.6kips()

Show the reactions of the beam as in Figure (4).

Determine the bending moment at A;

MA=29.4×15(2×21×212)=405405=0

Determine the bending moment at B;

MB=12.6×15(2×15×152)=189225=36kips-ft

Determine the bending moment at C;

MC=12.6×21(2×21×212)+(29.4×6)=264.6441+176.4=0

Show the M/EI diagram of the given beam as in Figure (5).

Show the elastic curve of the given beam as in Figure (6).

Determine the between point B and A using the relation;

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

Substitute 56.25EI for MEI, (23×15×(152)) for area of parabola, (36EI) for MEI, and (12×15×(13×15)) for area of triangle.

ΔBA=[56.25EI(23×15×(152))+(36EI)(12×15×(13×15))]=1EI(4,218.751,350)=2,868.75kips-ft3EI

Determine the slope at A using the relation;

θA=ΔBAL

Substitute 2,868.75kips-ft3EI for ΔBA and 15 ft for L.

θA=2,868

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