Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN: 9781337093347
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
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Chapter 4, Problem 4.5.29P

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.

  1. 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.
  2. 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).

  Chapter 4, Problem 4.5.29P, A beam of length L is designed to support a uniform load of intensity q (see figure). If the

a.

Expert Solution
Check Mark
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.

Answer to Problem 4.5.29P

a=0.5858L

Explanation of Solution

Given:

The given figure

Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.29P , additional homework tip  1

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)

The diagram of the shear force and the bending moment at the midpoint is symmetry,

At x=L2 the mid-section has the maximum bending moment,

  Mmax=q2(L2)2+qL2(L2La2)

  Mmax=q2L2+qL2(a)=q2(L22aL)(7)

As the maximum bending moment of the beam has the least numerical value,

  |Mmax|=|Mmax|

  q8(L22aL)=q8(La)2L22aL=L2a2+2aLa24aL+2L2=0a=(4L)±(4L)24(1)(2L2)2(1)a=(4L)±2L22(1)((2)±2)L,Buta<La=(22)L0.5858L

With the equation (7),

  Mmax=q8(L22(0.5858L)L)(7)

  =q8L2(11.1716)0.02145qL2

From equation 1,

  VA=q2(L0.5858L)0.2071qL

From equation 2,

  MA=q8(L0.5858L)20.0214qL2

From equation 3,

  VA=q2(0.5858L)0.2929qL

From equation 4,

  MA=q8(L0.5858L)20.021qL2

From equation (5),

  VA=q2(0.5858L)0.2929qL

From equation 6,

  MB=q8(La)20.0214qL2

Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.29P , additional homework tip  2

The bending moment is 0 from the given diagram, so we have,

  0=q2(b)2+qL2(bLa2)(b2)+L(bL0.5858L2)0=(b2)+L(b0.2071L)(b2)bL+0.2071L2=0

  b=(L)±(L)24(1)(0.2071L2)2=1±0.414252b=0.2929L

Conclusion:

Thus the distance a between the support is determined as a=0.5858L.

b.

Expert Solution
Check Mark
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.

Answer to Problem 4.5.29P

  b=0.3333L

Explanation of Solution

Given:

The given figure

Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.29P , additional homework tip  3

The length of the beam L supports the load which is uniform having the intensity as q0=q at the mid- point.

Concept Used:

Hence the above figure is in symmetry so, the calculations is done by considering this.

Resultant force,

  RA=RB=12×q0L2=q0L4

Calculation: Now the maximum moment is occurs at the centre of the beam then take,

  ΣΜ=0

  RB*a2=12(L2)q0×L6q0L4×a2=q0L224a=L3=0.3333L

Conclusion:

Thus the distance a between the support is determined as a=0.3333L .

The shear force and bending moment diagram is as follows:

Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.29P , additional homework tip  4

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