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Principle Stresses

  1. Apr 4, 2009 #1
    imagine a solid metal shaft.

    If we apply a torque, there is shear stress in the material. and we get [tex] \tau_{xy} [/tex] from [tex] \frac {\tau}{r} = \frac {T}{J}[/tex]

    If we applied direct tension, a tensile force, we get a [tex] \sigma_x [/tex] value from [tex] \sigma_x = \frac {F}{A}[/tex]

    If we apply a bending moment, we get a bending stress of value [tex] \sigma_B [/tex] and bending moment, M = [tex] \frac {I \sigma_B }{y} [/tex]

    can anyone confirm the fact that only shear (tau) is created by torque, sigma(x) by direct tension and sigma(y) is created by bending moment in every case.

    i have a problem which has torque only and shows the principle stresses to be:

    [tex] \sigma_1 = \frac {\sigma_x + \sigma_y}{2} + \frac {1}{2} \sqrt{ (\sigma_x - \sigma_y)^2 + 4 {\tau_{xy}}^2 }[/tex]

    reduces to this [tex] \sigma_1 = \tau_{xy}[/tex] (and then minus for sigma(2) ) i.e. [tex] \sigma_2 = -\tau_{xy}[/tex]

    and when we work out the maximum shear stress from TRESCA, its zero ?
  2. jcsd
  3. Apr 5, 2009 #2
    oops, can someone kindly move this to engineering please
  4. Apr 5, 2009 #3


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    I'm not quite sure what you're asking here, but normal stress in one orientation is usually equivalent to shear stress in another orientation. The exception is hydrostatic stress (which does not involve shear stress), but that's not one of your examples. You can use Mohr's circle to visualize the transformation.

    You do know that these are normal stresses (all principal stresses are) with the magnitude of the shear stress that was applied in another orientation, right? Does this solve your problem?
  5. Apr 5, 2009 #4
    yes its a little helpful

    my lecturer solved a problem in the class, and this is what i was trying to get at >>

    there was a bending moment M, and a torque T. These produce a certain value of sigma(x) and tau(xy). now im just curious as to why sigma(y) was ignored, or 0, and what sort of force is necessary to make sigma(y) have a value.
  6. Apr 5, 2009 #5


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    Were you dealing with a rod/bar/beam (something long and thin)? The lateral stresses ([itex]\sigma_y[/itex] and [itex]\sigma_z[/itex] when the rod axis is in the x-direction) are relatively low for these geometries and are usually assumed to be zero. The reason is that the lateral length scales are small, so the material can easily contract or expand slightly to relieve any stresses that arise.

    If the rod were set on a rigid surface and a distributed load applied on the other side, either [itex]\sigma_y[/itex] or [itex]\sigma_z[/itex] would have a nonnegligible value depending on the coordinate system. Similarly, if you were modeling plate bending, the lateral stresses would be nonnegligible because more than a slight contraction or expansion would be needed to relieve these stresses.

    I simulated the transition from a beam to a plate in https://www.physicsforums.com/showthread.php?t=293287", which you might find interesting.
    Last edited by a moderator: Apr 24, 2017
  7. Apr 8, 2009 #6
    I think it's important to clarify that the existence of [tex] \sigma_y [/tex] depends on the type of load applied, not the geometry of the part.

    A beam in direct tension will only have [tex] \sigma_x [/tex] stress.
    A beam in torsion will only have [tex]\tau_{xy}[/tex] stress.
    A beam in direct shear will only have [tex]\tau_{xy}[/tex] stress.

    A beam in pure bending will only have [tex] \sigma_x [/tex] stress. (Two moments applied on each side - smiley face beams)
    A beam in (impure?) bending will only have both [tex] \sigma_x[/tex] and lateral stresses [tex] \sigma_y [/tex] and/or [tex] \sigma_z [/tex] stresses. Think of a cantilevered beam with a distributed load on top of it. In addition to the bending stresses, in real life there are contact stress at the lateral surface of the beam.

    Usually we ignore these contact stresses when discussing beams because they are usually very small compared to the bending, torsion, or tension stresses. This is the same reason you usually ignore direct shear forces. You can always verify these assumptions yourself by simply including them in your calculations.

    Finally, this discussion referenced stresses in the x y and z axes. Failure theories of ductile materials predict that failure occurs along the plane of maximum shear stress (Tresca). We rotate the axes to find the find the maximum shear stress plane. This operation is visualized using Mohr's circle.

    Good luck with your class
  8. Apr 12, 2009 #7
    thanks for that rybose, so regarding that 'impure' beinding. if a load has a distributed load, the would [tex] \sigma_y[/tex] be only over the load? i.e. the area which that load is applied.

    this question ill just latch on here rather than make new thread.

    there is a hollow shaft, thin cylinder which has an internal pressure P, a direct force F, and a torque, T. i realise axial stress due to pressure adds with direct stress from tensile force. but where does hoop stress come into it?

    normally, if we have a plane at angle theta, we can work out normal and shear stresses through 2 equations, and get [tex] \sigma_n[/tex] and [tex] \tau_x[/tex] or we could use mohr circle.

    in this case we dont have a plane and im unsure as how the hoop+axial, direct and shear all correspond to each other.

    a tutorial on principle stresses would be good, if anyone knows any good ones?

    thanks alot
  9. Apr 13, 2009 #8
    Attached is a picture of what I attempted to explain in my previous post.

    Yes, on the surface and underneath the load, the lateral stress is simply the load divided by the area.

    On the top surface with no load, the lateral stress is zero. On the bottom surface the lateral stress is zero (because there's no load). In more formal language, these are called "traction boundary conditions."

    Beneath the surface of the load, things get more complicated. But again in beam applications these are usually very small compared to the other loads.


    To visualize hoop stress, imagine cutting a cylinderical pressure vessel longitudinally. In order for the body to be in equilibrium the force from the material stress must equal the net upward force from the internal pressure.

    You can derive the thin-walled presure vessel equations by finding the net upward force on the semi-circle (hint: integrate). You might be able to find this in a reference.

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