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 10, Problem 10.4.9P

The continuous frame ABC has a pin support at A, roller supports at B and C and a rigid corner connection at B (see figure). Members AB and BC each have flexural rigidity EI. A moment Müacts counterclockwise at A. Note: Disregard axial deformations in member AB and consider only the effects of bending.

  1. Find all reactions of the frame.

  • Find joint rotations 6 at A, B, and C.
  • Find the required new length of member AB in terms of L, so that θ A

  • in part (b) is doubled in size.
  •   Chapter 10, Problem 10.4.9P, The continuous frame ABC has a pin support at A, roller supports at B and C and a rigid corner

    (a)

    Expert Solution
    Check Mark
    To determine

    All the reactions of the frame.

    Answer to Problem 10.4.9P

    The reactions are : RA=RB=4M0/3L, HA=RC=2M0/3L.

    Explanation of Solution

    Given Information:

    The following figure is given with all relevant information,

    Mechanics of Materials (MindTap Course List), Chapter 10, Problem 10.4.9P , additional homework tip  1

    Calculation:

    Consider the following free body diagram,

    Mechanics of Materials (MindTap Course List), Chapter 10, Problem 10.4.9P , additional homework tip  2

    Take equilibrium of horizontal forces as,

    FH=0HARC=0                       ....(1)

    Take equilibrium of vertical forces as,

    FV=0RA+RB=0                       ....(2)

    Take equilibrium of moments about B as,

    MB=0RAL+M0+RCL/2=0                       ....(3)

    The bending moment at distance x from A along AB in part AB is given by,

    MAB=RAxM0

    Use second order deflection differential equation,

    d2vdx2=MEI=M ABEId2vdx2=1EI(RAxM0)                        

    Integrate above equation to get rotation as,

    dvdx=θAB=1EI(RAx2/2M0x+C1)                          

    Integrate above equation to get rotation as,

    v=δAB=1EI(RAx3/6M0x2/2+C1x+C2)            

    The bending moment at distance x from B along BC in part BC is given by,

    MBC=RCL/2RCx

    Use second order deflection differential equation,

      d2vdx2=MEI=M BCEId2vdx2=1EI(RCL/2RCx)                        

    Integrate above equation to get rotation as,

    dvdx=θBC=1EI(RCLx/2RCx2/2+C3)                          

    Integrate above equation to get rotation as,

    v=δBC=1EI(RCLx2/4RCx3/6+C3x+C4)            

    The constraint equations are,

    δAB(0)=0                          ...(4)δAB(L)=0                          ...(5)δBC(0)=0                          ...(6)δBC(L/2)=0                       ..(7)θAB(L)=θBC(0)                       ..(8)

    Solve equations (1-8) to get integration constants and reactions.

    So the reactions are RA=RB=4M0/3L, HA=RC=2M0/3L.

    Conclusion:

    Therefore, the reactions are: RA=RB=4M0/3L, HA=RC=2M0/3L.

    (b)

    Expert Solution
    Check Mark
    To determine

    Rotations at joints A, B, and C.

    Answer to Problem 10.4.9P

    Rotations at joints A, B, and C are θA=5M0L18EI, θB=M0L18EI, θC=M0L36EI

    Explanation of Solution

    Given Information:

    The following figure is given with all relevant information,

    Mechanics of Materials (MindTap Course List), Chapter 10, Problem 10.4.9P , additional homework tip  3

    Calculation:

    Consider the following free body diagram,

    Mechanics of Materials (MindTap Course List), Chapter 10, Problem 10.4.9P , additional homework tip  4

    Take equilibrium of horizontal forces as,

    FH=0HARC=0                       ....(1)

    Take equilibrium of vertical forces as,

    FV=0RA+RB=0                       ....(2)

    Take equilibrium of moments about B as,

    MB=0RAL+M0+RCL/2=0                       ....(3)

    The bending moment at distance x from A along AB in part AB is given by,

    MAB=RAxM0

    Use second order deflection differential equation,

    d2vdx2=MEI=M ABEId2vdx2=1EI(RAxM0)                        

    Integrate above equation to get rotation as,

    dvdx=θAB=1EI(RAx2/2M0x+C1)                          

    Integrate above equation to get rotation as,

    v=δAB=1EI(RAx3/6M0x2/2+C1x+C2)            

    The bending moment at distance x from B along BC in part BC is given by,

    MBC=RCL/2RCx

    Use second order deflection differential equation,

      d2vdx2=MEI=M BCEId2vdx2=1EI(RCL/2RCx)                        

    Integrate above equation to get rotation as,

    dvdx=θBC=1EI(RCLx/2RCx2/2+C3)                          

    Integrate above equation to get rotation as,

    v=δBC=1EI(RCLx2/4RCx3/6+C3x+C4)            

    The constraint equations are,

    δAB(0)=0                          ...(4)δAB(L)=0                          ...(5)δBC(0)=0                          ...(6)δBC(L/2)=0                       ..(7)θAB(L)=θBC(0)                       ..(8)

    Solve equations (1-8) to get integration constants and reactions.

    So the reactions are RA=RB=4M0/3L, HA=RC=2M0/3L.

    Substitute the integration constants and reactions in expressions of rotations to get,

    θAB(0)=θA=5M0L18EI, θAB(L)=θB=M0L18EI, θBC(L/2)=θC=M0L36EI

    Conclusion:

    Therefore, rotations at joints A, B, and C are θA=5M0L18EI, θB=M0L18EI, θC=M0L36EI

    (c)

    Expert Solution
    Check Mark
    To determine

    Length of AB.

    Answer to Problem 10.4.9P

    Length of AB is LAB=2.088L.

    Explanation of Solution

    Given Information:

    The following figure is given with all relevant information,

    Mechanics of Materials (MindTap Course List), Chapter 10, Problem 10.4.9P , additional homework tip  5

    Also θA=5M0L9EI .

    Calculation:

    Consider the following free body diagram,

    Mechanics of Materials (MindTap Course List), Chapter 10, Problem 10.4.9P , additional homework tip  6

    Take equilibrium of horizontal forces as,

    FH=0HARC=0                       ....(1)

    Take equilibrium of vertical forces as,

    FV=0RA+RB=0                       ....(2)

    Take equilibrium of moments about B as,

    MB=0RAL+M0+RCL/2=0                       ....(3)

    The bending moment at distance x from A along AB in part AB is given by,

    MAB=RAxM0

    Use second order deflection differential equation,

    d2vdx2=MEI=M ABEId2vdx2=1EI(RAxM0)                        

    Integrate above equation to get rotation as,

    dvdx=θAB=1EI(RAx2/2M0x+C1)                          

    Integrate above equation to get rotation as,

    v=δAB=1EI(RAx3/6M0x2/2+C1x+C2)            

    The bending moment at distance x from B along BC in part BC is given by,

    MBC=RCL/2RCx

    Use second order deflection differential equation,

      d2vdx2=MEI=M BCEId2vdx2=1EI(RCL/2RCx)                        

    Integrate above equation to get rotation as,

    dvdx=θBC=1EI(RCLx/2RCx2/2+C3)                          

    Integrate above equation to get rotation as,

    v=δBC=1EI(RCLx2/4RCx3/6+C3x+C4)            

    The constraint equations are,

    δAB(0)=0                          ...(4)δAB(L)=0                          ...(5)δBC(0)=0                          ...(6)δBC(L/2)=0                       ..(7)θAB(L)=θBC(0)                       ..(8)

    Solve equations (1-8) to get integration constants and reactions.

    So the reactions are RA=RB=4M0/3L, HA=RC=2M0/3L.

    Substitute the integration constants and reactions in expressions of rotations to get,

    θAB(LAB)=θA=5M0L9EI

    Solve above equation to get LAB=2.088L .

    Conclusion:

    Therefore, length of AB is LAB=2.088L .

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    Chapter 10 Solutions

    Mechanics of Materials (MindTap Course List)

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