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Mechanical Principles

  1. Nov 8, 2006 #1
    Hey guys im a first timer here. I need some help with a couple of questions im quite stumped.

    1. A bar in a 2-D loading system is 1 metre long and 20 millimeters square in section before loading. Determine the change in its "x" and "y" dimensions if E=200GPa and Poisson's Ratio is 0.3

    The force acting in the "x" direction is 60KN and in the "y" direction 40KN.

    ok, i know i have to find the stress in the x and y directions first. By dividing the Force by the Area of the bar. e.g. 60000 newtons/ mm2.

    i cant seem to work out what the area of the bar would be, i dnt understand "20 millimeters square in section" i need the area in mm2 also.

    Then i would work out the combined strain in the xx direction and yy direction.

    Finally to work out the change in length i multiply the strain by the original length.

    Any feed back would be great, thanks
     
    Last edited: Nov 8, 2006
  2. jcsd
  3. Nov 9, 2006 #2

    FredGarvin

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    You are directly given the area of the cross section. The area is [tex]A=20 mm^2[/tex] You will need to convert that to [tex]m^2[/tex]. So now you have the combined tensile stress with the bending stress. You also have Poisson's ratio for calculating the transverse strain.

    From there on out you seem to have a good grasp of the problem.
     
  4. Nov 10, 2006 #3
    Hi Fred thx for the reply :). I think i got a little messed up with my formula but hopefully its ok this time if you could take a look?

    To get the stress in the axial direction i divided the force (60,000N) by the area 20mm2 (0.0004m) which gave me 150,000,000 or 150x10 to the power of 6.

    Then to get the axial strain i divided the stress 150,000,000 by the Elastic modulus 200x10 to the power of 9 which gave me 0.00075.

    Then to get the change in the x direction i multiply the strain in the x direction by the orignal length, 0.00075 x 1 = 0.0075m or 0.75mm ?

    Then to get the lateral or transverse strain i multiply poisson's ratio by the axial strain. 0.3 x 0.0075 = 0.000225

    Finally to get the change in the y direction i multiply the y strain by the original length 0.000225 x 0.02 = -0.0045mm

    Does that sound right?
     
  5. Nov 10, 2006 #4

    radou

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    Watch out for the units! [tex]1 [mm^2] = 1 \cdot 10^{-6} [m^2][/tex].
     
  6. Nov 10, 2006 #5

    FredGarvin

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    Like was already mentioned, your units conversion is not correct on your area. The original area is [tex]A = 20 mm^2[/tex], that equates to [tex]A = 20x10^{-6} m^2 [/tex]
     
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