Shear Stress / Failure

[NavierStokesEq]

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

 

[desertfox]

Hi again

I would take the yieding shear stress to be 70% of the ultimate tensile stress for the mild steel.

regards desertfox

 

[NavierStokesEq]

thanks for your help desert fox

im right in saying the yield strength for steel is 172400000 Pascals, correct??? Also, The Tfp value, do I just keep it as 9 Nm or does it need to be in another form of units

 

[desertfox]

Hi

A typical value for ultimate tensile strength of mild steel
I have here is 430000000Pa so the yielding shear stress
would be 430000000 * 0.7= 301000000Pa.
Bear in mind these are general values what you need is the
certificate from the steel supplier giving actual ultimate
tensile strength values.
Now you can use 9Nm thats okay however make sure you use
Newtons and metres for the rest of the equation.

regards desertfox

 

[Focht3]

Roark's formula in a more readable format:

&[ignore]Tau[/ignore];_max = (2*T_ro)/(&[ignore]pi[/ignore];*(r_o⁴-r_i⁴))

where &[ignore]pi[/ignore]; = pi



Please see FAQ731-376 for great suggestions on how to make the best use of Eng-Tips Fora.

 

[desertfox]

Hi Focht3

The formula you have given is valid for stresses within the elastic limit, NavierStrokes is looking to fail his device
and therefore needs a formula that give the full failure stress across the section.
In addition he wanted to calculate the wall thickness of his tube, so the formula for full failure stress I rearranged so he could find the internal radius of his required tube.

regards desertfox.