   # Determine the absolute maximum bending moment in a 15 m long simply supported beam due to the series of three moving concentrated loads shown in Fig. P9.13. FIG. P9.13, P9.17, P9.18, P9.22

#### Solutions

Chapter
Section
Chapter 9, Problem 22P
Textbook Problem
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## Determine the absolute maximum bending moment in a 15 m long simply supported beam due to the series of three moving concentrated loads shown in Fig. P9.13. FIG. P9.13, P9.17, P9.18, P9.22

To determine

Find the absolute maximum bending moment in a 15 m long simply supported beam.

### Explanation of Solution

Calculation:

Sketch the simply supported beam as shown in Figure 1.

Refer Figure 2.

PR=125+100+50=275kN

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¯)=100(2)+50(5)PR(x¯)=450

Substitute 275 kN for PR.

(275)(x¯)=450x¯=1.64m

Find the position of loads on the beam.

As per the observation, 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=2x¯

Substitute 1.64m for x¯.

x2=21.64=0.36m

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

Location of resultant=x22

Substitute 1.64 m for x2.

Location of resultant=1

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