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 4.5.39P

A plane frame (see figure) consists of column AB and beam BC that carries a triangular distributed load (see figure part a). Support A is fixed, and there is a roller support at C. Beam BC has a shear release just right of joint B.

  1. Find the support reactions at A and C then plot axial-force (N), shear-force (V), and bending-moment (M) diagrams for both members. Label all critical N,K and M values and also the distance to points where any critical ordinates are zero.

  • Repeat part (a) if a parabolic lateral load acting to the right is now added on column AB (figure part b).
  • Chapter 4, Problem 4.5.39P, A plane frame (see figure) consists of column AB and beam BC that carries a triangular distributed

    a.

    Expert Solution
    Check Mark
    To determine

    The support reaction at point A and C and plot the shear, moment, and axial force diagram.

    Answer to Problem 4.5.39P

    Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.39P , additional homework tip  1

    Explanation of Solution

    Given: .

    The given figure.

    Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.39P , additional homework tip  2

    AB column and BC beam forms the plane frame that carries a load that is distributed in the triangular shape. At C there is roller support and support A is fixed. Below the B joint there is a moment release at column AB

    Concept Used: .

      .

      .

    Vertical force equilibrium is given as,

    FV=0.

    Ay+Cy=q0L2.

    Horizontal force equilibrium is given as,

      FH=0Ax=0.

    Calculation: .

    Vertical force equilibrium is given as,

    FV=0.

    Ay+Cyq0L2=0.

    Ay+Cy=q0L2.

    Horizontal force equilibrium is given as,

      FH=0Ax=0.

    At the top of the moment release the moment is given as,

    MB=0q0L2(2L/3)CyL=0q0L2(2L/3)=CyLCy=q0L3.

    In equation (1),Cy is substituted,

    Ay=q0L2q0L3q0L6.

    Below B there is moment release at point A,

    MA=0.

    At x the shear force is given and equated to 0,

      q0L6=(1/2)x(q0/L)xq0L6=q02Lx2x2=q0L6(2L/q0).

      =L23.

      0.57735L.

    At point x bending moment is maximum,

      Mx=q0L6(L/3)q02LL23(L/33).

      =q0L263q0L2183=3q0L2q0L2183.

      =2q0L2183.

      Mx=q0L2930.0641q0L2.

    Axial force critical valuesNmax=q0L6.

    Shear force critical valuesVmax=q0L3.

    Moment critical valuesMmax=0.06415q0L2.

    Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.39P , additional homework tip  3

    Conclusion: .

    Thus, the support reaction at point A and C and plot the shear, moment and axial force diagram.

    b.

    Expert Solution
    Check Mark
    To determine

    For the load that is parabolic, lateral acts to the right added to AB column and part (a) is repeated.

    Answer to Problem 4.5.39P

    Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.39P , additional homework tip  4

    Explanation of Solution

    Given: .

    The given figure:.

    Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.39P , additional homework tip  5

    AB column and BC beam forms the plane frame that carries a load that is distributed in the triangular shape. At C there is roller support and support A is fixed. Below the B joint, there is a moment release at column AB.

    Concept Used: .

      .

    Vertical force equilibrium is given as,

    FV=0.

    Ay+Cy=q0L2.

    Horizontal force equilibrium is given as,

      FH=0Ax=0.

    Calculation: .

    With the equilibrium force being vertical,

      Fy=0Ay+Cy=q0L2(2).

    Moment at point B above moment release,

      MB=0CyLq0L2(2L/3)=0CyL=q0L2(2L/3)Cy=q0L3.

    In (2) substituteCy.

      Ay=q0L2q0L3.

      =3q0L3q0L6=q0L6.

    With the equilibrium force being horizontal,

      FH=0.

    As left of x axis is negative,f(x)0.

      Ax=02L(1y/2L)q0dy.

      =q002L(1y/2L)dy.

      =q0(2L(1y/2L)3/23/2)02L=4q0L3((12L/2L)(10))=4q0L3(01).

      Ax=4q0L3.

    At A moment below the moment release,

      MA=02L(q0y1y/2L)dy+12q0L(2L/3)q0L3(L).

      =I1+q0L23q0L23.

      I1=02L(q0y1y/2L)dyu=y,dv=(1y/2L)1/2du=dy,v=(1y/2L)3/22L3/2.

      =4L3(1y/2L)3/2.

      d(uv)=uvvdu.

      =q0{(y(4L/3)(1y/2L)3/2)02L02L((4L/3)(1y/2L)3/2)dy}=q0(0+(4L/3)((1y/2L)5/25/2(1/2L))02L=q0(4L3(0(4L/5))=q0((4L/3)(4L/5)).

      MA=16q0L215.

    At x shear force is calculated and equated to 0,

      q0L6=(1/2)x(q0/L)xq0L6=q02Lx2.

      x2=q0L6(2L/q0).

      =L23x=0.57735L.

    At x bending moment is calculated

      Mx=q0L6(L/3)q02LL23(L/33).

      =q0L263q0L2183=3q0L2q0L2183=2q0L2183q0L293Mx=0.06415q0L2.

    For plane frame and critical values of N, V, M, the axial, shear and moment is given.

    Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.39P , additional homework tip  6

    Axial force critical valuesNmax=q0L6.

    Shear force critical values in beamVmax=q0L3.

    Shear force critical values in columnVmax=4q0L3.

    Moment critical values in beamMmax=0.06415q0L2.

    Moment critical values in columnMmax=16q0L215.

    Conclusion: .

    Thus, for the load that is parabolic lateral acts to the right added to AB column and part (a) is repeated.

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