Recent content by sssooo

I understand! Thanks for the explanation and all the help!

Indeed. 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...

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...

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

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?

Oh, cool. v.m: √(σφ2+σztot2-σφ⋅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...

Thanks for your replies. Well, 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...

Homework Statement 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...