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 5, Problem 5.10.2P

-1 through 5.10-6 A wide-flange beam (see figure) is subjected to a shear force V. Using the dimensions of the cross section, calculate the moment of inertia and then determine the following quantities:

  1. The maximum shear stress tinixin the web.

  • The minimum shear stress rmin in the web.
  • The average shear stress raver (obtained by dividing the shear force by the area of the web) and the ratio i^/t^
  • The shear force carried in the web and the ratio K b/K.
  • Note: Disregard the fillets at the junctions of the web and flanges and determine all quantities, including the moment of inertia, by considering the cross section to consist of three rectangles.

    5.10-2 Dimensions of cross section: b = 180 mm, v = 12 mm, h = 420 mm,

    i = 380 mm, and V = 125 kN.

      Chapter 5, Problem 5.10.2P, -1 through 5.10-6 A wide-flange beam (see figure) is subjected to a shear force V. Using the

    (a).

    Expert Solution
    Check Mark
    To determine

    To Find:

    The moment of inertia and the maximum shear stress in the web.

    Answer to Problem 5.10.2P

    The maximum shear stress in the web is = τmax=4.9 N/mm2 and moment of inertia is I=198.93×107 mm4

    Explanation of Solution

    Given Information:

    Shear Force V=125 kN

    Dimensions:

      b=180 mm.t=12 mm.h=420 mm.h1=380 mm.

    Concept Used:

    Following formula will be used:

    Maximum shear stress, τ=VQIb .

    Moment of inertia of rectangle, I=bh312 .

    Calculation:

    Area of upper and lower flanges/rectangles:

    A1=b(h2h12)=A3

    Mechanics of Materials (MindTap Course List), Chapter 5, Problem 5.10.2P , additional homework tip  1

    Second rectangle is efcb :

      A2=t(h12y1)

    In which y1 is the distance from neutral axis to line ef.

    The first moment of areas of A1 and A2 about the neutral axis.

      Q=A1(h12+h/2h1/22)+A2(y1+h1/2y12)

    By putting the values of A1 and A2, we get:

    Q=b8(h2h12)+t8(h124y12)So,τ=VQIt=VIt[b8(h2h12)+t8(h124y12)]             ...(1)

    The moment of inertia for the I section is given by following formula:

      I=bh312bh1312+th1312 about the neutral z axis.I=180×420312180×380312+12×380312I=198.93×107 mm4

    The maximum value of shear stress will be at neutral axis when y1=0

      τmax=VQIt=VIt[b8(h2h12)+t8(h124y12)]             ...(1)τmax=125198.93×107×12[1808(42023802)+128(380202)]τmax=4.9 N/mm2

    Conclusion:

    The maximum shear stress in the web is τmax=4.9 N/mm2 and moment of inertia is I=198.93×107 mm4 .

    (b).

    Expert Solution
    Check Mark
    To determine

    To Find:

    The minimum shear stress in web.

    Answer to Problem 5.10.2P

    The minimum shear stress in the web is τmin=3.770 N/mm2 .

    Explanation of Solution

    Given Information:

    Shear Force V=125 kN

    Dimensions,

      b=180 mm.t=12 mm.h=420 mm.h1=380 mm.

    Concept Used:

    Following formula will be used:

    Minimum shear stress, τ=VQIb .

    Calculation:

    Area of upper and lower flanges:

    A1=b(h2h12)=A3

    Mechanics of Materials (MindTap Course List), Chapter 5, Problem 5.10.2P , additional homework tip  2

    Second rectangle is efcb :

      A2=t(h12y1)

    In which y1 is the distance from neutral axis to line ef .

    The first moment of areas of A1 and A2 about the neutral axis.

      Q=A1(h12+h/2h1/22)+A2(y1+h1/2y12)

    By putting the values of A1 and A2, we get:

    Q=b8(h2h12)+t8(h124y12)So,τ=VQIt=VIt[b8(h2h12)+t8(h124y12)]             ...(1)

    The moment of inertia for the I section is given by following formula:

      I=bh312bh1312+th1312 about the neutral z axis.I=180×420312180×380312+12×380312I=198.93×107 mm4

    The minimum value of shear stress will be at cb when y1=h12

      τmin=VQIt=VIt[b8(h2h12)+t8(h124y12)]             ...(1)τmin=125198.93×107×12[1808(42023802)+128(38024×( 380 2)2)]τmin=3.770 N/mm2

    Conclusion:

    The minimum shear stress in the web is τmin=3.770 N/mm2 .

    (c).

    Expert Solution
    Check Mark
    To determine

    To Find:

    The average shear stress in the web.

    Answer to Problem 5.10.2P

    The average shear stress in the web is τaverage=27.41 N/mm2 and Ratio=τmaxτaverage=0.179 .

    Explanation of Solution

    Given Information:

    Shear Force V=125 kN

    Dimensions:

      b=180 mm.t=12 mm.h=420 mm.h1=380 mm.

    Concept Used:

    Following formula will be used:

    Average shear stress, τaverage=Vt×h1 .

    Calculation:

    The average shear stress in the web is:

      τaverage=Vt×h1τaverage=125×10312×380τaverage=27.41 N/mm2Ratio=τmaxτaverage=4.927.41=0.179

    Conclusion:

    The average shear stress in the web is τaverage=27.41 N/mm2 and Ratio=τmaxτaverage=0.179 .

    (d).

    Expert Solution
    Check Mark
    To determine

    To Find:

    Shear force Vweb carried by web.

    Answer to Problem 5.10.2P

    Shear force in the web is Vweb=20626.4 N and Ratio=VwebV=0.165 .

    Explanation of Solution

    Given Information:

    Shear Force V=125 kN

    Dimensions,

      b=180 mm.t=12 mm.h=420 mm.h1=380 mm.

    Concept Used:

    Following formula will be used

    Shear stress in the web, Vweb=th13(2τmax+τmin) .

    Calculation:

    Shear stress in the web:

      Vweb=th13(2τmax+τmin)Vweb=12×3803(2×4.9+3.77)Vweb=20626.4 NRatio=VwebV=20626.4125000=0.165 .

    Conclusion:

    Shear force in the web is Vweb=20626.4 N and Ratio=VwebV=0.165 .

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