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Homework Help: Cross section and torque- please help

  1. Dec 21, 2009 #1
    A 3.5 m long steel member with a W310 x 143 cross- section is subjected to 4.5KNm torque. Knowing that G=77GPa, determine (a) the maximum shearing stress along the line a-a, (b) the maximum shearing stress along the line b-b, (c) the angle of twist

    http://lh4.ggpht.com/_H4Iz7SmBrbk/Sysfg6WJmLI/AAAAAAAACBI/SsE4jsJMYAE/22.jpg [Broken]

    i know that

    [tex]\tau[/tex]=[tex]\frac{M*t}{Jeq}[/tex]

    while Jeq=[tex]\frac{1}{3}[/tex][tex]\sum[/tex]hibi3

    Jeq=1/3*((0.309*0.02293) +(0.309*0.02293) + (0.2872*0.0143))

    Jeq=2.7365e-6

    [tex]\tau[/tex]=[tex]\frac{4500*t}{2.7365e-6}[/tex]

    a)

    [tex]\tau[/tex]a-a=[tex]\frac{4500*0.0229}{2.7365e-6}[/tex]

    =3.7657e7
    =37.657MPa

    b)

    [tex]\tau[/tex]b-b=[tex]\frac{4500*0.014}{2.7365e-6}[/tex]

    = 2.3022e7
    =23.02MPa

    (according to my book the correct answers are meant to be 39.7MPa and 24.2MPa)


    for the angle

    [tex]\phi[/tex]=M*L/(G*Jeq*[tex]\eta[/tex])

    [tex]\eta[/tex]=1.29)

    [tex]\phi[/tex]=[tex]\frac{4500*3.5}{77e9*2.7365e-6*1.29}[/tex]

    =0.0579rad
    = 3.3199 degrees,

    but again the correct answer is meant to be 4.72 degrees,

    what am i doing wrong??
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 22, 2009 #2
    i think i need to SOMEHOW find the torque seperately on the flange and web and them somehow solve the problem,, please help someone
     
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