Structural Analysis (10th Edition)
Structural Analysis (10th Edition)
10th Edition
ISBN: 9780134610672
Author: Russell C. Hibbeler
Publisher: PEARSON
Textbook Question
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Chapter 11, Problem 11.1P

Determine the moments at A,B, and C, then draw the moment diagram for the beam. The moment of inertia of each span is indicated in the figure. Assume the support at B is a roller and A and C are fixed. E = 29(103) ksi.

Chapter 11, Problem 11.1P, Determine the moments at A,B, and C, then draw the moment diagram for the beam. The moment of

Expert Solution & Answer
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To determine

The moments at A, B, and C, and to draw the moment diagram for the beam.

Answer to Problem 11.1P

The moment at A for member AB is, 38.4kft.

The moment at B for member BA is, 45.52kft.

The moment at B for member BC is, 45.52kft.

The moment at C for member CB is, 67.23kft.

The following figure shows the moment diagram of the beam.

  Structural Analysis (10th Edition), Chapter 11, Problem 11.1P , additional homework tip  1

Explanation of Solution

Calculation:

The following figure shows the beam diagram.

  Structural Analysis (10th Edition), Chapter 11, Problem 11.1P , additional homework tip  2

        Figure-(1)

Calculate the fixed end moments at every end as shown below.

Calculate fixed end moments of member AB.

   FEMAB=wl212

Here, given load is w and length is l.

Substitute 800lb/ft for w and 24ft for l.

   FEMAB=(800lb/ft)(24ft)212(1k1000lb)=460.812=38.4kft

Calculate fixed end moments of member BA.

   FEMBA=wl212

Substitute 800lb/ft for w and 24ft for l.

   FEMBA=(800lb/ft)(24ft)212(1k1000lb)=460.812=38.4kft

Calculate fixed end moments of member BC.

   FEMBC=wl8

Substitute 30k for w and 16ft for l.

   FEMBC=(30k)(16ft)8=4808kft=60kft

Calculate fixed end moments of member CB.

   FEMCB=wl8

Substitute 30k for w and 16ft for l.

   FEMBC=(30k)(16ft)8=4808kft=60kft

Calculate the member relative stiffness factors on either side of B.

Calculate relative stiffness of member BA.

   KBA=4EIBALBA(Forfarendfixed)

Here, KBA is relative stiffness of member BA, E is modulus of elasticity, I is the moment of inertia and L is length.

Substitute 29×103ksi for E, 24ft for L.

   KBA=4(29×103ksi)IBA24ft(1ft12in)=402.78IBA

Calculate relative stiffness of member BC.

   KBC=4EIBCLBC(Forfarendfixed)

Substitute 29×103ksi for E, 16ft for L.

   KBC=4(29×103ksi)IBA16ft(1ft12in)=604.16IBC

Calculate the distribution factors for different members.

Calculate distribution factors for AB.

   (DF)AB=0 (Since, far end is fixed)

Calculate distribution factors for BA.

   (DF)BA=KBAKBA+KBC .....(I)

Substitute 402.78IBA for KBA and 604.16IBC for KBC in Equation (I).

   (DF)BA=402.78IBA402.78IBA+604.16IBC .....(II)

Substitute 900in4 for IBA and 1200in4 for IBC in Equation (II).

   (DF)BA=402.78×900in4402.78×900in4+604.16×1200in4=362502362502+724992=0.33

Calculate distribution factors for BC.

   (DF)BC=KBCKBA+KBC .....(III).

Substitute 402.78IBA for KBA and 604.16IBC for KBC in equation (III).

   (DF)BC=604.16IBC402.78IBA+604.16IBC .....(IV)

Substitute 900in4 for IBA and 1200in4 for IBC in Equation (IV).

   (DF)BC=604.16×1200in4402.78×900in4+604.16×1200in4=724992362502+724992=0.67

Calculate distribution factors for CB.

   (DF)CB=0 (Since, far end is fixed)

Calculate the balance moment for each member as shown below.

Calculate balance moment foe BA.

   (Balancemoment)BA=(38.460)×(DF)BA=21.6×0.33=7.128kft

Calculate balance moment for BC.

   (Balancemoment)BC=(38.460)×(DF)BC=21.6×0.67=14.47kft

Draw the moment distribution table as shown below.

    JointsABC
    MemberABBABCCB
    DF 0 0.33 0.67 0
    FEM 38.4 38.4 60 60
    Balance- 7.128 14.47 -
    Carry over moment 3.564 0 0 7.236
    Balance 0 0 0 0
    ΣM 34.8kft 45.52kft 45.52kft 67.23kft

Consider the section AB as shown below.

  Structural Analysis (10th Edition), Chapter 11, Problem 11.1P , additional homework tip  3

        Figure-(2)

Write the Equation for sum of vertical forces.

   ΣFy=0Ay+By(800lb/ft×24ft)(1k1000lb)=0Ay+By=19.2k .....(V)

Here, vertical reaction at A and B are Ay and By.

Write the Equation for sum of moment about A.

   ΣMA=0(34.830.8×24×12+By×2445.52)kft=0By×24ft=241.09kftBy=10.04k

Substitute 10.04k for By in equation (V).

   Ay+10.04k=19.2kAy=9.16k

Calculate the distance at which shear force changing its sign for maximum moment.

Write the Equation for shear force at distance x from A.

   Vx=9.16k0.8x

Equate the value of shear force to zero for distance.

   9.16k0.8x=00.8x=9.16kx=11.45ft

Calculate the value of maximum moment at a distance of 11.45ft.

   M11.45=34.83+9.16×11.450.8×11.4522=17.611kft

Consider the section BC as shown below.

  Structural Analysis (10th Edition), Chapter 11, Problem 11.1P , additional homework tip  4

        Figure-(3)

Write the Equation for sum of vertical forces.

   ΣFy=0Cy+By30k=0Cy+By=30k .....(VI)

Write the Equation for sum of moment about B.

   ΣMB=0(45.5230×8+Cy×1667.23)kft=0Cy×16ft=261.71kftCy=16.35k

Here, vertical reaction at C is Cy.

Substitute 16.35k for Cy in equation (VI).

   16.35k+By=30kBy=13.65k

For member BC, moment will be maximum at the center.

Calculate maximum moment at the center of BC.

   M8=(45.52+13.65×8)kft=63.68kft

Conclusion:

The moment at A for member AB is, 38.4kft.

The moment at B for member BA is, 45.52kft.

The moment at B for member BC is, 45.52kft.

The moment at C for member CB is, 67.23kft.

The following diagram shows the moment diagram of the beam.

  Structural Analysis (10th Edition), Chapter 11, Problem 11.1P , additional homework tip  5

        Figure-(4)

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