shear stress failure

huangyhg

shear stress / failure
hey all,
sorry for the noob questions, its probably a simple one. but how do i go about working out the minimum thickness of a steel tube when i know the outer diameter and the torque??
i have found a shear stress in roarks for a hollow concentric circular section, but dont know if its right and where to go from there
from the specification for design of hss,
phixtn=0.9xfcrxc
c=hss torsional constant
for round hss,
fcr=larger of 1.23e/(sqrt(l/d)x(d/t)^5/4) or
0.6e/(d/t)^1.5 but greater than 0.6fy
tu<phixtn, tu is the factored torque
you can solve out for t in the above equations and with using c=pi(d-t)^2xt/2.
i think it's easier to pick a section and compare the torsion it gives you to the actual torsion than to solve for the thickness.
you may want to read through this:
you say hss, i was on about mild steel, does it matter
the provisions in that publication are for certain grades, see page 1. see if the steel you are using is listed.
also, i don't know what the application is but you may want to check for another code. for example if you are designing a part in a car.
the application is for a shear nut, the point of shear if basically like a thin wall tube, and so was going to analyse it like that.
so the equation you posted, is that still the one to use. could you explain it a little more, sorry :s
im trying to understand the equations you posted, but not having much look. how come they do not take into account the torque that the shear nut should fail at.
just to describe the situation, the component is a shear nut, the point at which it should shear resembles a thin wall tube, and so i presume it can be modelled as such. the nut is made from mild steel. the nut must shear at a minimum of 8nm and a maximum of 10nm. in order to correctly manufacture the component and its gauged proberly, i need to find out the min and max thickness that the wall of the tube at shear point can be.
hope this helps better
thanks for all your help
hello:
the formula that i am assuming that you found in roark's (taumax=(2tro)/(pi*(ro^4-ri^4)) is correct.  its derivation is found in most mechanics of materials textbooks.  that being the case, you should be able to determine the maximum shear stress for a given trial section.  
the question then becomes: what is the allowable maximum shear stress?  in structural engineering, we work with code prescribed stresses.  to further complicate matters, there are currently two steel design methods in use: one which utilizes service loads, and the other which utilizes ultimate loads.  i hesitate to offer any advice regarding the limit stress to use as i do not know whether you are dealing with service loads or ultimate loads.
for machine design, i have typically seen allowable stresses expressed as a fraction of the "ultimate strength" of the material.  for example, i have seen fu/5 typically used for tensile stresses.  i have not done a lot of machine design, so i am sorry that i can not offer you ony further advice on what to use as the limit stress.
i hope this helps.
hi navierstrokeseq
if you're designing a device to fail at a certain torque the
equations you need are in a previous thread which i answered a while back, take a look and if you need more help
just let me know.
thread 406-46941
regards desertfox
hi navierstrokes
the link doesn't seem to work but you will find the thread if you do a search.
ah got it this time i think
thanks desertfox,
in the equation you give in that thread, what is fp??, also how do i find out the yeilding shear stress of steel??
and how come the equation dont take in to account the thickness, which i need to find out??