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Tensile & Shear Strain

  1. Nov 7, 2004 #1
    A 1260kg freight elevator is supported by a steel cable of diameter 34.9mm. It is loaded with a crowd of people collectively having a mass of 2850kg and it is descending.

    a) What is the strain on the cable while the elevator is descending at the constant speed of 2.30 m/s?

    b) What is the strain in the cable when it is brought to a stop in 0.600s?

    okay this doesn't seem to be very hard, but as i tried to do it, i just keep getting stuck.
    okay firstly: strain.
    i only know of 2 types of strain: tensile, and shear. this definitely isn't a shear strain problem since the force applied is not directed parallel to the surface.
    that leaves tensile strain. Tensile strain is (the change in length in response to applied stress) / (relaxed length) is it not? But if this is what i am supposed to use, shouldn't the problem have given me an initial legth of the cable? This leads me to believe that i am confused on the concept of strain. Is there a different type of strain that i am missing, or should there be another step?

    Thanks physics gurus!
  2. jcsd
  3. Nov 7, 2004 #2


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    The questions make more sense if you regard "strain" as a typo; I think it must be the stress they're after.
    Otherwise, they would have needed to give you a stress/strain-relationship+initial length.

    However, they just MIGHT be mean to you, and assume you find out the particular relationship in that case.
    In order to proceed then, you should derive a relationship between relative longitudinal strain, and relative radial strain.
    Hint: This can be done by assuming that the cable's volume must be constant!

    Rewrite the appropriate Hooke's law using the relative radial strain instead.
  4. Nov 7, 2004 #3


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    Likewise ... the problem sounds pretty basic in order to focus on strain. Thinking the problem axisymmetric makes it easy to work out in such a case no matter.
  5. Nov 7, 2004 #4
    relative radial strain?
    Hookes law= Y (Change in length in response to applied stress / relaxed length) where Y= Young's modules
    I must have the wrong Hookes law. This just works to take the same problem into a different equation? (i don't know the initial length of the cable.)

    It could be a typo.... but i don't want to trust my grade on a typo, y'know? any other suggestions?
  6. Nov 7, 2004 #5
    You do not need initial length for this problem.

    You know the diameter of the cable ...

    Compute Area , Acceleration,Force , Stress ... etc
    For constant speed there is no acceleration (or acceleration = 0 )

    Tensile Stress = (Axial Force) / (Cross- Sectional Area)
    Modulus of Elasticity = Stress / Strain ( this is hooke's law)

    For Steel the modulus of elasticity = 200 GPa (approx. or u can use 207 GPa)

    Then compute Strain

    For constant speed there is no acceleration (or acceleration = 0 )
    Last edited: Nov 7, 2004
  7. Nov 7, 2004 #6


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    If you assume that the volume of the cable is constant, with unkown length L, KNOWN radius R (you have this one!), then:
    We assume:
    Multiplying the brackets out, and neglecting products of small terms, yields:
  8. Nov 7, 2004 #7
    Okay, I did like you suggested, but nowhere in my work did i factor in the speed at which the elevator was moving?

    here's my work:
    (1260kg+2850kg)*9.81 = 40319.1 Newtons
    Cross-sectional area of the cable = 2piR = 2pi*17.45 = 109.64mm = .10964m
    Tensile Stress = (axial force) / (cross-sectional area) = 40319.1 N / .10964m = 367740.788 Pa

    Young's Modulus for steel = 20*10^10 = Stress/Strain
    20*10^10 = 367740.788 Pa / Strain
    Strain = 367740.788 Pa / 20*10^10
    Strain = 1.83*10^-6?????????

    That number MUST be wrong? did i make a mistake, or is the process incorrect?
    Again, where am i supposed to account for the speed at which the elevator is moving? (I.E. the second problem)
  9. Nov 7, 2004 #8
    cross sectional area = Pi * Radius^2
  10. Nov 7, 2004 #9

    Wow, this went way over my head! lol. Is this the derivation of the relationship you were talking about? If so, then (since it looks unfamiliar) i would assume this isn't wasn't what my instructor was looking for?
  11. Nov 7, 2004 #10


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    You bet that's not what he asked you for if you haven't seen it before.

    Stick with what the other poster is doing; the fact that you in a) seems to get a tiny strain is simply because steel hardly lengthens.
  12. Nov 7, 2004 #11
    and you should also know:

    Summation of Froces(using directions) = Mass * Acceleration

    and the v = u - at ( calculate the acceleration, v = 0 , t = 0.6 sec)
  13. Nov 7, 2004 #12

    oops! haha i knew that!
    Strain = 2.11*10^13

    That's more like it, but is it right?

    And what about the speed of the elevator? If this is the answer to part a, then how do i solve for b?
  14. Nov 7, 2004 #13
    part a)
    M = 1260 + 2850 = 4110
    F = M*g = 4110 * 9.81 = 40319.1 N
    A = pi * dia ^2 / 4 = 9.566E-4
    stress = F/A = 42.147E6 Pa
    Strain = stress / Y = 42.147E6 / 200E9 = 2.1E-4
  15. Nov 7, 2004 #14
    A= pi * dia ^2 / 4 = 9.566 ^ -4 ??
    shouldn't it be:
    A= pi * 34.9mm^2 / 4 = 956mm = .956m ?
    stress = F/A = 4.21 * 10^4 Pa ?
    Young's modulus = stress/strain
    strain = stress / young's
    strain = 4.21*10^4 / 20*10^10 = 2.105 * 10 ^ 13 ??

    or did i screw it up again?
  16. Nov 7, 2004 #15
    34.9mm^2 = (34.9mm)^2 = (0.0349 m) ^ 2 = 1.218E-3 (metre squared)
    1.218E-3 *Pi/4 = 9.566E-4 (metre squared)
  17. Nov 7, 2004 #16
    hey Tycho .. i joined here today .. u?
  18. Nov 7, 2004 #17
    I saw what you did in the last one. Area=pi * R^2. you used the diameter instead of the radius. Welcome to the physics forum! this is my second month on.

    do you know how to take the speed of the lift into effect?
  19. Nov 7, 2004 #18
    ooops on the diameter matter :)

    initially moving in a constant speed 'u ' the elevator stops
    that means 'v' = 0 ...
    noy from v = u - at ( a = acceleration)

    we get a = u/t

    Mass * a = Summation of forces (vector addition)
  20. Nov 7, 2004 #19
    no oooops ......

    Radius^2 = (Diameter)^2 / 4
    (R)^2 = (D/2)^2 = D^2 / 4
  21. Nov 7, 2004 #20

    you're going to have to put these two things together for me. how would i put this summation of forces into the equaition for strain?? therein is my biggest problem!
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