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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Textbook Question
Chapter 5, Problem 5.6.11P

A beam ABC with an overhang from B to C is constructed of a C 10 × 30 channel section with flanges facing upward (sec figure). The beam supports its own weight (30 lb/ft) plus a triangular load of maximum intensity g0 acting on the overhang. The allowable stresses in tension and compression arc IS ksi and 12 ksi, respectively.

  1. Determine the allowable triangular load intensity allow if tne distance L equals 4 ft.

  • What is the allowable triangular load intensity fallow ^tnc Dcam is rotated 180e about its longitudinal centroidal axis so that the flanges are downward?
  •   Chapter 5, Problem 5.6.11P, A beam ABC with an overhang from B to C is constructed of a C 10 × 30 channel section with flanges

    (a)

    Expert Solution
    Check Mark
    To determine

    The allowable triangular load intensity.

    Answer to Problem 5.6.11P

    The allowable triangular load intensity is 419.21lb/ft .

    Explanation of Solution

    Given information:

    The weight of the beam is 30lb/ft , the allowable stress in tension is 18kpsi , allowable stress in compression is 12ksi , the distance is 4ft .

    The following figure shows the free body diagram.

      Mechanics of Materials (MindTap Course List), Chapter 5, Problem 5.6.11P

    Figure-(1)

    Write the expression for the moment about point A .

      By×L=(q×2L)(L)+(q0×L2)(L+2L3)By=(q×2L)+(q0×L2)(53).....(I)

    Here, the vertical reaction on point B is By , the length of the beam is 2L , self weight of the beam per unit length is q ,and triangular load intensity is q0 .

    Write the expression for the equilibrium forces at point A and B .

      Ay+By=(q×2L)+(q0×L2).....(II)

    Here, the vertical reaction on point A is Ay .

    Write the expression for the maximum moment at point B .

      Mmax=Ay×L+(q×L)(L2).....(III)

    Here, the maximum moment is Mmax .

    Write the expression for the allowable bending moment based on tension.

      Mt=σt×Ic1.....(IV)

    Here, maximum tensile stress is σt , the moment of inertia is I , and the distance of neutral axis is c1 .

    Write the expression for the allowable bending moment based on compression.

      Mc=σc×Icc.....(V)

    Here, maximum compressible stress is σc , the moment of inertia is I , and the distance of neutral axis is c2 .

    Write the expression for the load intensity of triangular load considering tension.

      Mmax=L2(q03+q2) ….. (VI)

    Write the expression for the load intensity of triangular load considering compression.

      Mmax=L2(q03+q2) ….. (VII)

    Calculation:

    Substitute By=(q×2L)+(q0×L2)(53) for By in Equation (II).

      Ay+(q×2L)+(q0×L2)(53)=(q×2L)+(q0×L2)Ay=(q×2L)+(q0×L2)(53)(q×2L)+(q0×L2)Ay=(q0×L3)

    Substitute (q0×L3) for Ay in Equation (III).

      Mmax=(q0×L3)L+(q×L)L2=(( q 0 L)×( L 2 )3)+(q×L)L2=L2(q03+q2)

    Substitute 3.93in4 for I , 18kpsi for σt ,and 2.381in. for c1 in Equation (IV).

      Mt=(18kpsi)×(3.93 in4)(2.381in)=(18kpsi)( 1000psi 1kpsi)×(3.93 in4)(2.381in)=(18000psi)( 1 lb/ in 2 1psi)×(3.93 in4)(2.381in)=29710lbin.

    Substitute 3.93in4 for I , 12ksi for σc ,and 0.649in. for c2 in Equation (V).

      Mc=(12kpsi)×(3.93 in4)(0.649in)=(12kpsi)( 1000psi 1kpsi)×(3.93 in4)(0.649in)=(12000psi)( 1 lb/ in 2 1psi)×(3.93 in4)(0.649in)=29710lbin

    Substitute 29710lbin for Mmax , 4ft for L ,and 30lb/ft for q in Equation (VI).

      (29710lbin)=(4ft)2(q03+30lb/ft2)((29710lbin)×1ft12in)=(( 4ft)2( q 0 3+ 30 lb/ ft 2))q0=(( 2475.83lbft (4ft ) 2 30 lb/ ft 2)×3)q0=419.21lb/ft

    Substitute 72666lbin for Mmax , 4ft for L ,and 30lb/ft for q in Equation (VII).

      (72666lbin)=(4ft)2(q03+30lb/ft2)((72666lbin)×1ft12in)=(( 4ft)2( q 0 3+ 30 lb/ ft 2))q0=(( 6055.5lbft (4ft ) 2 30 lb/ ft 2)×3)q0=1090.41lb/ft

    Conclusion:

    The allowable triangular load intensity is 419.21lb/ft .

    (b)

    Expert Solution
    Check Mark
    To determine

    The allowable triangular load intensity when the beam is rotated 180° .

    Answer to Problem 5.6.11P

    The allowable triangular load intensity when the beam is rotated 180° is 264.48lb/ft .

    Explanation of Solution

    Write the expression for the allowable bending moment based on tension.

      Mt=σt×Ic1.....(VIII)

    Write the expression for the allowable bending moment based on compression.

      Mc=σc×Ic2.....(IX)

    Write the expression for the load intensity of triangular load considering tension.

      Mmax=L2(q03+q2) ….. (X)

    Write the expression for the load intensity of triangular load considering compression.

      Mmax=L2(q03+q2)….. (XI)

    Calculation:

    Substitute 3.93in4 for I , 18ksi for σt ,and 0.649in. for c1 in Equation (VIII).

      Mt=(18ksi×3.93 in40.649in)=(18ksi×3.93 in40.649in)( 103lb/ in 21ksi)=108988.45lbin

    Substitute 3.93in4 for I , 12ksi for σc ,and 2.381in for c2 in Equation (IX).

      Mc=(12ksi×3.93 in42.381in)=(12ksi×3.93 in42.381in)( 103lb/ in 21ksi)=19806.8lbin

    Substitute 108998.45lbin for Mmax , 4ft for L ,and 30lb/ft for q in Equation (VI).

      (108998.45lbin)=(4ft)2(q03+30lb/ft2)((108998.45lbin)×1ft12in)=(( 4ft)2( q 0 3+ 30 lb/ ft 2))q0=(( 9083.20lbft (4ft ) 2 30 lb/ ft 2)×3)q0=1658.10lb/ft

    Substitute 19806.8lbin for Mmax , 4ft for L ,and 30lb/ft for q in Equation (VII).

      (19806.8lbin)=(4ft)2(q03+30lb/ft2)((19806.8lbin)×1ft12in)=(( 4ft)2( q 0 3+ 30 lb/ ft 2))q0=(( 1650.57lbft (4ft ) 2 30 lb/ ft 2)×3)q0=264.48lb/ft

    Conclusion:

    The allowable triangular load intensity when the beam is rotated 180° is 264.48lb/ft .

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    Chapter 5 Solutions

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