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Homework Help: Statics Question (Using Modulus of Rigidity)

  1. Jan 17, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the displacement (mm) in the horizontal direction of point A due to the force, P. P=100kN w1=19mm w2=15mm

    2. Relevant equations

    [itex]\tau[/itex] = G * [itex]\gamma[/itex]
    [itex]\tau[/itex] = Shear stress = P / A
    [itex]\gamma[/itex] = Shear strain = (pi / 2) - [itex]\alpha[/itex]

    3. The attempt at a solution

    I haven't attempted to work out a solution here yet, but I do have a question regarding the separate G values that are given.

    Can I just look at the top layer, the layer where P is acting, and use that G value to determine [itex]\delta[/itex]? Or do I need to do something with the other G value as well?

    If I were to try something, I would find tau by doing 100[kN] / (100[mm] * 2[mm]). So tau would be equal to 1[kN]/2[mm2] = 0.5[GPa]. Next I would find gamma by dividing tau by G (100[GPa]) giving me [itex]\gamma[/itex] = .005rad. I can use trig to define gamme as [itex]\gamma[/itex]=sin-1([itex]\delta[/itex]/40). Setting this equal to .005 I would get [itex]\delta[/itex]= .20[mm].

    Even if I do have to do something with both of the G values, I feel like my method is correct. Any help is appreciated, thanks in advance.

    Attached Files:

    Last edited: Jan 17, 2012
  2. jcsd
  3. Jan 17, 2012 #2
    Hi papasmurf
    Last edited: Jan 17, 2012
  4. Jan 18, 2012 #3
    If delta at A is relative to the fixed base, then all the shear displacements of the various layers must be taken into account.
  5. Jan 18, 2012 #4
    Am I correct in assuming the shear force will be the same at all the various layers?
  6. Jan 18, 2012 #5
    Hi Dr.PSMokashi
  7. Jan 18, 2012 #6
    I'm getting closer to the correct answer. First I set V/A, where V is the internal shear force and A is the area of the cross section where the shear force is acting, equal to G*[itex]\gamma[/itex], where G is the modulus of rigidity and gamma is the shear strain.
    I rewrote gamma as pi/2 - θ, where θ=cos-1([itex]\delta[/itex]/h), h is the height of the "layer", and put it all together so that my equation looks like this:

    V/A = G * ( pi/2 - cos-1([itex]\delta[/itex]/h) )

    Solving for [itex]\delta[/itex] I come up with
    [itex]\delta[/itex] = h * cos( (pi/2) - V/AG)

    I used this formula for each "layer" and added up all of the deltas.

    However after plugging my numbers in and making sure of correct units, I still am off by fractions of a millimeter.
    Last edited: Jan 18, 2012
  8. Jan 18, 2012 #7
    Also, should the h value be the height of the layer only or should it go from the base to the top of the layer I am looking at? For example if I am looking at the first layer where G=0.1MPa, would my h be simply w2 [mm] or would it be w2+2 [mm]?
  9. Jan 18, 2012 #8
    I keep getting an answer that is off by fractions of a millimeter. I can not figure out what I am doing/not doing that keeps giving me a wrong answer.
  10. Jan 21, 2012 #9
    How do you know that "an answer that is off by fractions of a millimeter" is "a wrong answer"?
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