   # 6.18 through 6.22 Determine the smallest moment of inertia I required for the beam shown, so that its maximum deflection does not exceed the limit of 1/360 of the span length (i.e., Δ max ≤ L /360). Use the moment-area method. FIG. P6.22, P6.48

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Chapter 6, Problem 22P
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## 6.18 through 6.22 Determine the smallest moment of inertia I required for the beam shown, so that its maximum deflection does not exceed the limit of 1/360 of the span length (i.e., Δmax ≤ L/360). Use the moment-area method.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 moment-area 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 the deflected shape of the given beam as in Figure (4).

Determine the deflection between A and D using the relation;

Substitute 600EI for MEI, (12×10×(13×10+20)) for area of triangle, (10×(102+10)) for area of rectangle, and (12×10×(23×10)) for area of triangle.

ΔDA=[600EI(12×10×(13×10+20))+(600EI)(10×(102+10))+600EI(12×10×(23×10))]=1EI[70,000+90,000+20,000]=180,000kipsft3EI

Determine the deflection between D and E using the relation;

ΔDE=[MEI(Areaoftriangle)]

Substitute 600EI for MEI and (12×10×(23×10)) for area of triangle

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