   # 6.44 through 6.48 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. P6.20, P6.46

#### Solutions

Chapter
Section
Chapter 6, Problem 46P
Textbook Problem
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## 6.44 through 6.48 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. P6.20, P6.46 To determine

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

### Explanation of Solution

Given information:

The Young’s modulus (E) is 200 GPa.

Calculation:

Consider the flexural rigidity EI of the beam is constant.

Show the free body diagram of 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.

Since support A is a free end there will be no reaction. Therefore, the reaction at support A is automatically zero.

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

V=0RC(12×4)=0RC=48kN

Determine the moment at point C;

MC=(12×4×(42+4))=288kNm

Determine the moment at point B;

MB=(12×4×(42))=96kNm

Determine the moment at point A;

MA=48×8(12×4×(42))288=38496288=0

Show the M/EI diagram of the given beam as in Figure (2).

Conjugate-beam method:

In the given beam system, point C is a fixed end and point A is free end. But in the conjugate-beam method the fixed end of a real beam becomes free and the free end of real beam changed into the fixed end.

Show the M/EI diagram for the conjugate-beam as in Figure (3).

Consider the external forces acting (right of A) upward on the free body diagram as positive.

Determine the deflection at A using the relation;

ΔA=ΔAC=[13×b1×h1(34×b1)(b2×h2)(b1+b22)12×b3×h3(23×b3+b1)]

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

Substitute 4 m for b1, 96EI for h1, 4 m for b2, 96EI for h2, 4 m for b3, and (288EI96EI) for h3

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