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 5.10.3P

-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 V^tV.
  • Noie: 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.

    5.10-3 Wide-flange shape, W 8 x 28 (see Table F-L Appendix F); V = 10 kChapter 5, Problem 5.10.3P, -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 5.10.3P

    The maximum shear stress is τmax=5.795 ksi and moment of inertia is I=333.422 in4 .

    Explanation of Solution

    Given Information:

    Shear Force V=10 k

    Dimensions for beam W8×28 ,

      b=6.535 in.t=0.285 in.h=8.06 in.h1=7.13 in.

    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,

    A1=b(h2h12)=A3

    Total area of cross section:

      A=A1+A3+t×h1A=2b(h2h12)+t×h1A=b(hh1)+t×h1A=6.535(8.067.13)+0.285×7.13=8.11 in2

    Mechanics of Materials (MindTap Course List), Chapter 5, Problem 5.10.3P , 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=6.535×8.063126.535×7.13312+0.285×7.13312I=141.771 in4

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

      τmax=VQIt=VIt[b8(h2h12)+t8(h124y12)]             ...(1)τmax=10141.771×0.285[6.5358(8.0627.132)+0.2858(7.13202)]τmax=3.304 ksi

    Conclusion:

    The maximum shear stress in the web is τmax=3.304 ksi and moment of inertia is I=141.771 in4 .

    (b).

    Expert Solution
    Check Mark
    To determine

    To Find:

    The minimum shear stress in web.

    Answer to Problem 5.10.3P

    The minimum shear stress in the web is τmin=2.86ksi .

    Explanation of Solution

    Given Information:

    Shear Force V=10 k

    Dimensions for beam W8×28 ,

      b=6.535 in.t=0.285 in.h=8.06 in.h1=7.13 in.

    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 5.10.3P , 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=6.535×8.063126.535×7.13312+0.285×7.13312I=141.771 in4

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

      τmin=VQIt=VIt[b8(h2h12)+t8(h124y12)]             ...(1)τmin=10141.771×0.285[6.5358(8.0627.132)+0.2858(7.1327.132)]τmin=2.86ksi

    Conclusion:

    The minimum shear stress in the web is τmin=2.86ksi .

    (c).

    Expert Solution
    Check Mark
    To determine

    To Find:

    The average shear stress in web.

    Answer to Problem 5.10.3P

    The average shear stress in the web is τaverage=4.92 ksi and Ratio=τmaxτaverage=0.67155 .

    Explanation of Solution

    Given Information:

    Shear Force V=10 k

    Dimensions for beam W8×28 ,

      b=6.535 in.t=0.285 in.h=8.06 in.h1=7.13 in.

    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=100.285×7.13τaverage=4.92 ksiRatio=τmaxτaverage=3.3044.92=0.67155

    Conclusion:

    The average shear stress in the web is τaverage=4.92 ksi and Ratio=τmaxτaverage=0.67155 .

    (d).

    Expert Solution
    Check Mark
    To determine

    To Find:

    Shear force Vweb carried by web.

    Answer to Problem 5.10.3P

    The shear force in the web is Vweb=6.413k and Ratio=VwebV=0.641 .

    Explanation of Solution

    Given Information:

    Shear Force V=10 k

    Dimensions for beam W8×28 ,

      b=6.535 in.t=0.285 in.h=8.06 in.h1=7.13 in.

    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=0.285×7.133(2×3.304+2.86)Vweb=6.413kRatio=VwebV=6.41310=0.641 .

    Conclusion:

    The shear force in the web is = Vweb=6.413k and Ratio=VwebV=0.641 .

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