   # Using the conjugate-beam method, determine the smallest moments of inertia I required for the beams shown in Figs. P6.18 through P6.22, so that the maximum beam deflection does not exceed the limit of 1/360 of the span length (i.e., Δ max ≤ L / 360). FIG. P6.22, P6.48

#### Solutions

Chapter
Section
Chapter 6, Problem 48P
Textbook Problem
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## Using the conjugate-beam method, determine the smallest moments of inertia I required for the beams shown in Figs. P6.18 through P6.22, so that the maximum beam deflection does not exceed the limit of 1/360 of the span length (i.e., Δmax ≤ L/ 360). FIG. P6.22, P6.48

To determine

Find the smallest moment of inertia (I) required for the beam if the deflection does not exceed the limit l/360 of the span length by using conjugate-beam method.

### Explanation of Solution

Given information:

The Young’s modulus (E) is 29,000 ksi.

Calculation:

Consider the flexural rigidity EI of the beam is constant.

Show the given beam as in Figure (1).

Refer Figure (1),

Consider upward is positive and downward is negative.

Consider clockwise is negative and counterclockwise is positive.

Determine the support reaction at A using the Equation of equilibrium;

ME=0RA×40+(60×30)+(60×20)600=0RA=2,40040RA=60k

Determine the reaction at support E;

V=0RA+RE6060=0RE=12060RE=60k

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

Refer Figure (2),

Determine the bending moment at E;

ME=(60×40)(60×30)(60×20)=2,4003,000=600kips-ft

Determine the bending moment at D;

MD=600(60×10)=600600=0

Determine the bending moment at C;

MC=(60×20)(60×10)=1,200600=600kips-ft

Determine the bending moment at B;

MB=(60×10)=600kips-ft

Determine the bending moment at A;

MA=(60×40)600(60×20)(60×10)=2,4002,400=0

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

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

Determine the reaction at support A using the relation;

MD=0RA(30)[12(600EI)(10)(13×10+20)+(10)(600EI)(10+102)+12(600EI)(10)(23×10)+12(600EI)(10)(23×10)]=0RA=130EI(70,000+90,000+20,000+20,000)RA=6,666.67kips-ft2EI

Consider that the maximum bending moment in the conjugate beam occur at point M, at a distance of xM from point B.

Determine the sear force at point M using the relation;

SM=0[RA+12×b1×h1+(b2×h2)]=0

Here, b is the width and h is the height of the respective triangle and rectangle.

Substitute 6,666.67kips-ft2EI for RA, 10 ft for b1, 600EI for h1, xM for b2, 600EI for h2

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