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Solid Mechanics of Materials

  1. Oct 15, 2014 #1
    1. The problem statement, all variables and given/known data

    A long circular steel tube having a mean diameter of 254 mm and 3.2 mm wall thickness is subjected to an internal pressure of 4.83 MPa. The ends of the tube are closed. The yield stress of the steel is 227 MPa. Find the additional axial tensile load F which is needed to cause yielding of the tube, based on the von Mises criterion and Tresca criterion, respectively.

    2. Relevant equations ...

    v.M = sqrt(sigma^2 + 3*tau^2) - sigma_y0 = 0 where sigma is the stress and tau is the shear stress. sigmay_0 is the yield stress.

    2. The attempt at a solution
    Well, actually I am not entirely sure how to start this. Anyhow I try I get an answer of the load about ten times higher than the answer (v.M which should be 395 kN).
    So I guess I am entirely wrong and would probably just need a hint on how I should procede with the problem.
    My solid mechanics certainly need refreshed.

    I attempt that
    So I calculate tau by multiplying the pressure with the diameter and dividing by 2 times thickness. then I solve the equation.

    So I guess my actual question is to refresh my solid mechanics for this type of question (the problem is in a constitutive modelling course...)

    PS; Sorry about lack of latex...
  2. jcsd
  3. Oct 15, 2014 #2


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    Instead of trying to work this problem on the fly, set it up in a formal fashion and calculate the stresses acting in the wall of the tube. The internal pressure is going to cause a hoop stress and a radial stress, and what about the axial stress due to the tube being closed at the ends?

    It's not clear if you have used the correct formulas for stress so far. I wouldn't worry too much about the solid mechanics; you need to make sure the basic strength of materials questions are answered properly first.
  4. Oct 15, 2014 #3
    SteamKing is right. Before you start trying to apply the yield criteria, you first need to establish the 3 principal stresses? Do you know how to do that? If so, what are they for the base case they gave?

  5. Oct 21, 2014 #4
    Thanks for your replies.
    I figure the only stresses affecting the tube due to the internal pressure would be the hoop stress and the axial stress (z-dir).
    No shearing would be applied since if so the tube would rotate...

    So there would be two principal stresses, sigma_fi and sigma_z, as

    sigma = [sigma_fi,0,0;
    0, sigma_z + sigma_f, 0;

    where sigma_fi would be (hoop stress) (P*D)/(2*t)
    P - internal pressure
    D - internal diameter
    t - thickness of wall

    and sigma_z would be (axial stress) (P*D)/(4*t) = sigma_fi/2

    I get these values to be:
    sigma_fi = 186.7 MPa
    sigma_z = 93.4

    The V.M yield criterion would then give: v.M: sqrt(sigma_fi^2 + sigma_z_tot^2 -sigma_fi*sigma_z_tot)
    where sigma_z_tot = sigma_z + sigma_f
    , sigma_f being the sought stress.

    I get sigma_z_tot = 252 Mpa (or 2nd root -60 MPa) and then sigma_f = 155 MPa.

    The answer should be 395 kN so I am still lost...

    Thank you.
  6. Oct 21, 2014 #5
    Your principal stresses look correct. I don't think you applied the von Mises yield criterion correctly. Please write out the criterion using the excellent equation editors that Physics Forums has provided you so I can tell exactly what you doing. I really can't follow your equations well the way you are presenting them.

  7. Oct 21, 2014 #6
    Oh, cool.

    v.m: √(σφ2ztot2φ⋅σztot) - σy = 0

    Couldn't make a large square root sign by some reason... what's in the paranthesis is square-rooted....

    Anyways, this expression is for plane stress (σ1, σ2) which would apply in this case well since we only have 2 principal stresses.

    So I am solving this 2nd grade equation for σztot.
  8. Oct 21, 2014 #7
    You got the stress right (I think), but you forgot to multiply by the cross sectional area to get the force.

  9. Oct 21, 2014 #8
    If I do multiply the cross sectional area (the internal right)... Then I get σf⋅πr2 = 155 ⋅ π⋅123.82 = 7.46⋅106 N....

    Or am I thinking wrong here?
  10. Oct 21, 2014 #9
    You used the wrong cross sectional area. Ask yourself, what is the metal cross sectional area that the axial stress acts on? Also, watch the units.

  11. Oct 21, 2014 #10
    Ehrm... I would suppose they would cause stress on the "lids" of the tube....

    Units; I am calculating in MPa, shouldn't I be calculating in mm then?

    Sorry I feel quite stupid, but I can't follow that last hint. :L
  12. Oct 21, 2014 #11
    No problem. You have done very well so far.

    In a previous post, you determined the axial stress under pure pressure loading, and, in deriving this equation, you used the correct cross sectional area. Do you remember what cross section you used to get that result?

  13. Oct 21, 2014 #12
    Thanks for the kind words.

    Well, I used a textbook definition - σ = (P⋅D)/(4⋅t) which I would assume would be an approximation due to the very long tube. D being the internal diameter...
  14. Oct 21, 2014 #13
    So you didn't derive this equation yourself. This equation was derived from:
    ##\pi r^2P=(2\pi r)t \sigma_z##
    Does this latter relationship make sense to you?

  15. Oct 21, 2014 #14

    Ok (interesting! never saw that one), so what it says there is that F = A⋅P, the pressure on the lid, equals

    the circumference of the lid times the thickness....
    So basically the cross sectional area I am seaking and where the stress is working on is the wall?

    Well it gives me the correct result.


    Follow-up question if I may;
    Does this say that the pressure works on the lid but the stresses develop in the wall?
  16. Oct 21, 2014 #15
    The stresses always develop in the wall (or, more generally, in the material). In this problem, the pressure is a distributed external load applied to the body. The stress tensor describes the internal state of stress within the body necessary to hold each arbitrary parcel of material within the body in equilibrium. If you were to slice the pipe perpendicular to its axis, you would have to apply a loading on the newly exposed circumference of the pipe section equal to the axial stress to continue to hold the section in place (assuming that the pressure were still pushing at its end).

  17. Oct 21, 2014 #16
    I understand!
    Thanks for the explanation and all the help!
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