   # 7.31 through 7.33 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 method of virtual work. FIG. P7.32

#### Solutions

Chapter
Section
Chapter 7, Problem 32P
Textbook Problem
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## 7.31 through 7.33 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 method of virtual work. FIG. P7.32

To determine

Find the required moment of inertia of the beam.

### Explanation of Solution

Given information:

The beam is given in the Figure.

The maximum deflection is ΔmaxL360.

The value of E is 29,000 ksi.

Apply the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium as shown below.

• For summation of forces along x-direction is equal to zero (Fx=0), consider the forces acting towards right side as positive (+) and the forces acting towards left side as negative ().
• For summation of forces along y-direction is equal to zero (Fy=0), consider the upward force as positive (+) and the downward force as negative ().
• For summation of moment about a point is equal to zero (Matapoint=0), consider the clockwise moment as negative and the counter clockwise moment as positive.

Calculation:

Consider the real system.

Let the bending moment due to real load be M.

Sketch the real system of the beam as shown in Figure 1.

Find the reactions at the supports:

Summation of moments about A is equal to 0.

MA=020By3(20)(202)=0By=30k

Summation of forces along y-direction is equal to 0.

+Fy=0Ay+By3(20)=0Ay+303(20)=0Ay=30k

Consider the virtual system.

Remove all the real loads and apply unit load where the point to find the deflection.

Let the bending moment due to virtual load be Mv.

Sketch the virtual system of the beam with unit load at point C as shown in Figure 2.

Find the reactions at the supports:

Summation of moments about A is equal to 0.

MA=020By1(10)=0By=0.5k

Summation of forces along y-direction is equal to 0.

+Fy=0Ay+By1=0Ay+0.51=0Ay=0.5k

Find the equations for M and Mv for the 2 segments of the beam as shown in Table 1.

 Segment x-coordinate M (k-ft) Mv (k-ft) Origin Limits (ft) AM A 0−10 30x−3x22 0.5x BM B 0−10 30x−3x22 0.5x

Find the maximum deflection that occurs at M using the virtual work expression:

1(Δmax)=0LMvMEIdx (1)

Rearrange Equation (1) for the limits 010 and 010 as follows

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