   # Determine the absolute maximum bending moment in a 60 ft long simply supported beam due to the series of four moving concentrated loads shown in Fig. P9.14. FIG. P9.14, P9.16, P9.19, P9.23

#### Solutions

Chapter
Section
Chapter 9, Problem 23P
Textbook Problem
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## Determine the absolute maximum bending moment in a 60 ft long simply supported beam due to the series of four moving concentrated loads shown in Fig. P9.14. FIG. P9.14, P9.16, P9.19, P9.23

To determine

Find the absolute maximum bending moment in a 60 ft long simply supported beam.

### Explanation of Solution

Calculation:

Sketch the simply supported beam as shown in Figure 1.

Refer Figure 2.

PR=10+20+20+5=55k

Find the location of the resultant (x¯).

Consider moment of the resultant about a point equals the sum of the moments of the individual loads about the same point.

PR(x¯)=20(15)+20(25)+5(35)PR(x¯)=975

Substitute 55 k for PR.

(55)x¯=975x¯=17.72ft

Find the position of loads on the beam.

We observe that the resultant is closest to the second load. Hence, the maximum bending moment occurs under the second load. The series of the load is positioned on the beam so that the midspan of the beam is located halfway between the load 2 and the resultant.

Find the distance between the resultant and second load.

x2=x¯15

Substitute 17.72ft for x¯.

x2=17.7215=2.72ft

Find the position of second load and resultant from center of the beam.

Location of resultant=x22

Substitute 2

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