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 9, Problem 9.5.18P

A beam ABCD consisting of a simple span BD and an overhang AB\s loaded by a force P acting at the end of the bracket CEF (see figure),

  1. Determine the deflection at the end of the over h a tig.

  • Under what conditions is this deflection upward? Under what conditions is it downward?
  •   Chapter 9, Problem 9.5.18P, A beam ABCD consisting of a simple span BD and an overhang AB\s loaded by a force P acting at the

    (a)

    Expert Solution
    Check Mark
    To determine

    Deflection at A.

    Answer to Problem 9.5.18P

    The deflection at A is δA=PL2(10L9a)/324EI (Positive upward).

    Explanation of Solution

    Given Information:

    The following figure is given along with relevant information,

      Mechanics of Materials (MindTap Course List), Chapter 9, Problem 9.5.18P , additional homework tip  1

    Calculation:

    Consider the following diagram,

      Mechanics of Materials (MindTap Course List), Chapter 9, Problem 9.5.18P , additional homework tip  2

    Transfer the load P from F to C and draw reaction forces as shown in following figure,

      Mechanics of Materials (MindTap Course List), Chapter 9, Problem 9.5.18P , additional homework tip  3

    Take equilibrium of forces in horizontal direction as,

      Fx=0Bx=0

    Take equilibrium of moments about D as,

      MD=0ByL+Pa+2LP/3=0By=Pa+2LP/3L

    Take equilibrium of forces in vertical direction as,

      Fy=0By+Dy=PDy=PBy=PPa+2LP/3L

    The bending moment at distance x from point A is given by,

      M={0                                                                  xL/2By(xL/2)                                                xL/2+L/3By(xL/2)P(xL/2L/3)Pa       xL/2+L/3+2L/3

    The deflection and bending moment is related by following differential equation

      d2vdx2=MEI          ....(1)

    Integrate differential equation (1) with respect to x by putting expression for M to get angle of rotations, as,

      θ=dvdx=1EI{C1                                                                xL/2By(xL/2)2/2+C2                                                xL/2+L/3By(xL/2)2/2P(xL/2L/3)2/2Pax+C3       xL/2+L/3+2L/3

    Integrate angle of rotation with respect to x get deflections as,

      δ=v=1EI{C1x+C4                                                                xL/2By(xL/2)3/6+C2x+C5                                                x5L/6By(xL/2)3/6P(x5L/6)3/6Pax2/2+C3x+C6       x3L/2

    The following conditions are used to evaluate integration constants,

      limx(L/2)θ=limx(L/2)+θ                  ....(1)limx(5L/6)θ=limx(5L/6)+θ                ....(2)limx(L/2)δ=limx(L/2)+δ                  ....(3)limx(5L/6)δ=limx(5L/6)+δ                ....(4)δ(L/2)=0                               ....(5)δ(3L/2)=0                             ....(6)

    Then substitute values of constants and x=0 in δ to get

      δA=PL2(10L9a)/324EI (Positive upward)

    Conclusion:

    Therefore the deflection at A is δA=PL2(10L9a)/324EI (Positive upward) .

    (b)

    Expert Solution
    Check Mark
    To determine

    Condition for upward and downward deflection at A .

    Answer to Problem 9.5.18P

    The deflection at A is upward when a/L<10/9and downward when a/L>10/9

    Explanation of Solution

    Given Information:

    The following figure is given along with relevant information,

      Mechanics of Materials (MindTap Course List), Chapter 9, Problem 9.5.18P , additional homework tip  4

    Calculation:

    Consider the following diagram,

      Mechanics of Materials (MindTap Course List), Chapter 9, Problem 9.5.18P , additional homework tip  5

    Transfer the load P from F to C and draw reaction forces as shown in following figure,

      Mechanics of Materials (MindTap Course List), Chapter 9, Problem 9.5.18P , additional homework tip  6

    Take equilibrium of forces in horizontal direction as,

      Fx=0Bx=0

    Take equilibrium of moments about D as,

      MD=0ByL+Pa+2LP/3=0By=Pa+2LP/3L

    Take equilibrium of forces in vertical direction as,

      Fy=0By+Dy=0Dy=By=Pa+2LP/3L

    The bending moment at distance x from point A is given by,

      M={0                                                                  xL/2By(xL/2)                                                xL/2+L/3By(xL/2)P(xL/2L/3)Pa       xL/2+L/3+2L/3

    The deflection and bending moment is related by following differential equation

      d2vdx2=MEI          ....(1)

    Integrate differential equation (1) with respect to x by putting expression for M to get angle of rotations, as,

      θ=dvdx=1EI{C1                                                                xL/2By(xL/2)2/2+C2                                                xL/2+L/3By(xL/2)2/2P(xL/2L/3)2/2Pax+C3       xL/2+L/3+2L/3

    Integrate angle of rotation with respect to x get deflections as,

      δ=v=1EI{C1x+C4                                                                xL/2By(xL/2)3/6+C2x+C5                                                x5L/6By(xL/2)3/6P(x5L/6)3/6Pax2/2+C3x+C6       x3L/2

    The following conditions are used to evaluate integration constants,

      limx(L/2)θ=limx(L/2)+θ                  ....(1)limx(5L/6)θ=limx(5L/6)+θ                ....(2)limx(L/2)δ=limx(L/2)+δ                  ....(3)limx(5L/6)δ=limx(5L/6)+δ                ....(4)δ(L/2)=0                               ....(5)δ(3L/2)=0                             ....(6)

    Then substitute values of constants and x=0 in δ to get

      δA=PL2(10L9a)/324EI (Positive upward)

    Now, the deflection at A is upward when a/L<10/9 and downward when a/L>10/9

    Conclusion:

    Therefore the deflection at A is upward when a/L<10/9 and downward when a/L>10/9

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    Chapter 9 Solutions

    Mechanics of Materials (MindTap Course List)

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