   # A beam of length L is designed to support a uniform load of intensity q (see figure). If the supports of the beam are placed at the ends, creating a simple beam, the maximum bending moment in the beam is qL 2 /8. However, if the supports of the beam are moved symmetrically toward the middle of the beam (as shown), the maximum bending moment is reduced. Determine the distance a between the supports so that the maximum bending moment in the beam has the smallest possible numerical value. Draw the shear-force and bending-moment diagrams for this condition. Repeat part (a) if the uniform load is replaced with a triangularly distributed load with peak intensity q 0 = q at mid-span (see Fig. b). ### Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347

#### Solutions

Chapter
Section ### Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347
Chapter 4, Problem 4.5.29P
Textbook Problem
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## A beam of length L is designed to support a uniform load of intensity q (see figure). If the supports of the beam are placed at the ends, creating a simple beam, the maximum bending moment in the beam is qL2/8. However, if the supports of the beam are moved symmetrically toward the middle of the beam (as shown), the maximum bending moment is reduced. Determine the distance a between the supports so that the maximum bending moment in the beam has the smallest possible numerical value. Draw the shear-force and bending-moment diagrams for this condition. Repeat part (a) if the uniform load is replaced with a triangularly distributed load with peak intensity q0= q at mid-span (see Fig. b). a.

To determine

The distance a between the support by having the small numerical value of maximum bending moment and to draw the diagrams of shear force and bending moment.

### Explanation of Solution

Given:

The given figure

The length of the beam L supports the load which is uniform having the intensity as q. The bending moment that is maximum is given as qL2/8

Concept Used:

Resultant force,

RA=RB=qL2

Calculation:

As the forces are symmetry,

RA=RB

Fy=0RA+RBqL=0RA=RB=qL2

At a distance from x to C, the section between CA is considered,

Fy=0Vqx=0V=qxMx=0M+q(x)(x2)=0M=qx22

When x=0,

VC=0,MC=0,

When x=La2 ,

VA=q2(La)(1)MA=q2(La2)2MA=q8(La)2(2)

At a distance from x to C, the section between AB is considered,

Fy=0qL2Vqx=0V=qx+qL2Mx=0M+q(x)(x2)qL2(xLa2)=0M=qx22+qL2(xLa2)

When x=La2 ,

VA=q(La2)+qL2VA=qa2(3)MA=q2(La2)2qL2(0)MA=q8(La)2(4)

When, x=LLa2 ,

=L+a2

VB=q(L+a2)+qL2VB=qa2(5)MB=q2(L+a2)2+qL2(L+a2La2)MB=q8(L+a)2+qL2(a)MB=q8(L2+a2+2aL4aL)MB=q8(L2+a22aL)MB=q8(La)2(6</

b.

To determine

The distance ‘a’ between the support by having the small numerical value of maximum bending moment and to draw the diagrams of shear force and bending moment.

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