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Homework Help: Ratio of shear stress to normal stress along principle place

  1. Nov 20, 2017 #1
    1. The problem statement, all variables and given/known data
    "In a component under multi-axial state of stress, the ratio of shear stress to normal stress along principle places is _____.

    A) 0.0 B) 0.5 C) 1.0 D) 1.5 E)2.0"

    2. Relevant equations
    σx' = (σxy)/2 + ((σxy)/2)*cos(2θ) + τxy*sin(2θ)

    σy' = (σxy)/2 - ((σxy)/2)*cos(2θ) - τxy*sin(2θ)

    τmax,in plane = √[((σxy)/2)2 + τxy2]

    σ1,2 = (σxy)/2) ± [((σxy)/2)2 + τxy2]

    τmax,absolute = σ1/2

    3. The attempt at a solution

    Ratio of shear stress to normal stress?? ->
    τmax,in planex' ->
    max,in plane = √[((σxy)/2)2 + τxy2]] / [σx' = (σxy)/2 + ((σxy)/2)*cos(2θ) + τxy*sin(2θ)]

    Impossible to simplify? Wrong ratio equations?

    I don't know what to do. Please help. I have heard from different sources that it is either 0 or 0.5 neither with explanations or work shown.

    Thank you
  2. jcsd
  3. Nov 20, 2017 #2
    What is the definition of a principal direction of stress?
  4. Nov 20, 2017 #3
    If I understand correctly the definition of a principal direction of stress is the stress vector at which the transformed normal stress vector is maximized.

    According to the book If I substitute the relations for ThetaP1 or ThetaP2, which are 90deg apart the shear stress will equal 0 and thus the ratio will equal zero. However I'm missing how it is that the shear stress would always equal 0 given the substitution
  5. Nov 20, 2017 #4
    My understanding is that the principal directions are those for which the shear component of the traction vector on a plane normal to the principal direction is equal to zero.
  6. Nov 20, 2017 #5
    If a plane is at some angle and the stresses are transformed such that one is normal and one is parallel with the plane then wouldn't the vector that is parallel be the shear component and thus not be 0 ? I'm definitively missing something here
  7. Nov 20, 2017 #6
    One equation you omitted from your original post is the shear stress in the transformed coordinate system. Please include that too. The principal directions are those for which this shear stress is zero. Let's determine the angles for which that is zero, and then see what we get for the normal stress components.
  8. Nov 20, 2017 #7
    Ah yea I used the wrong equation in my original post. However even with the other one I couldn't come up with any results.

    Here's the other one τx'y' = - ((σxy)/2)*sin(2θ) + τxy*cos(2θ)
  9. Nov 20, 2017 #8
    OK. So, from this equation, in terms theta, what is the equation for this shear stress being zero?

    The equation for ##\sigma_{x'}## can be rewritten as:
    $$\sigma_{x'}=\frac{(\sigma_x+\sigma_{y})}{2}+\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\ \left(\cos{\phi}\cos{2\theta}+\sin{\phi}\sin{2\theta}\right)$$where$$\cos{\phi}=\frac{\frac{\sigma_x-\sigma_y}{2}}{\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}}$$and $$\sin{\phi}=\frac{\tau_{xy}}{\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}}$$
    OK so far? For ##\tau_{x'y'}##, can you figure out how to express this in a similar way?
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