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Shear modulus and shear strain relationship?

  1. Mar 27, 2013 #1
    If a shear force is acting on a rectangular beam, how do I use the value of the shear modulus to calculate if the force applied will fracture the beam? Below I've submitted the data involved.
    The shear force applied is 1000 N and the cross section area is 28 mm.
    So the shear force applied is 1000/28 N/mm^2 = 35.7 N/mm^2.

    As far as I'm concerned the shear modulus tells me how much force I can apply per mm^2 before it fractures. Am I right?

    Material: PVC
    E-modulus: 1300 N/mm^2
    Poisson's ratio: 0.35
    Applied shear force: 100 N
    Resisting area: 28 mm^2

    Shear modulus: E = 2G(1+v) --> G = E/(2(1+v)) = 1300/(2(1+0.35) = 481 N/mm^2
     
  2. jcsd
  3. Mar 27, 2013 #2

    AlephZero

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    It's not clear if you have given all the data that was in the question, or just what you thought was relevant.

    If you apply a shear force to a cantilever beam, you probably get other stresses as well as the shear stress, and the other stresses are often much bigger.
     
  4. Mar 27, 2013 #3
    But you see, you can split a force into two vectors Fx and Fy. You can do the same thing with a momentum. But my point is that you only take into consideration the force Fy which works in parallel to the cross section area. But by only looking at the given shear modulus, it is supposed to indicate what the maximum shear force is allowed, but no??

    I just figured out that the technical term that I'm looking for is called: "shear yield strength or shear yield stress. It can be represented by τy ≤ (σ1 - σ2) / 2 is the allowed shear stress before fracture.
    So I guess I need to derive the σ1 and σ2 somehow from the shear modulus.
     
  5. Mar 27, 2013 #4
    No, the shear modulus does not tell you about the shear resistance.
    Same as Young modulus does not tell about tensile resistance.
    They just determine how much will be the strain (shear or tensile) at a given stress.
    What you need is some information about the maximum shear strain (or stress) of the material.
     
  6. Mar 27, 2013 #5
    Update:
    The relationship between shear yield point and uniaxial yield point based on the Von Mises criterion:

    τy = ((1+μ) / sqrt(3))/sqrt(3) * σy

    μ is the property of the material that determines the effect of pressure on the yield point.
    As a material parameter, μ is the change in yield point with respect to a change in pressure.

    You're asking me to get the information about the maximum shear stress.
    Well that's what I'm asking you. You're making me sound like George Bush right now.
    But joke aside. σy is the uniaxial yield. Which by my definition is the stress at which plastic deformation takes place. Do you agree? In a stress test, the plotted diagram does show this point, and it seems to be close to the e-module measured in N/mm^2 or MPa.

    So if you don't have any other suggestion. It appears that my remaining problem is to define μ, which I have no clue how to achieve. What kind of test (and machine) should I perform in the lab?

    I have calculated that at the point on the diagram where it starts to become plastically deformed, a force of 844 N was exerted. The area was 40 mm^2. So the uniaxial yield point must then be 844N/40mm^2 = 21.1 N/mm^2 or 21.1 MPa.
     
    Last edited: Mar 27, 2013
  7. Mar 27, 2013 #6
    I am not asking you anything.
    I simply replied to your question below:
    "As far as I'm concerned the shear modulus tells me how much force I can apply per mm^2 before it fractures. Am I right?"

    I don't have any information about the maximum shear stress or the plasticity limit of your material. Sorry.
     
  8. Mar 28, 2013 #7
    I have already given you that data in my previous message. The uniaxial yield point is the maximum plasticity limit for the material, which is 21.1 N/mm^2.
     
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