Shear Stress / Failure
Shear Stress / Failure
(OP)
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
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






RE: Shear Stress / Failure
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:
http://www.aisc.org/ContentManagement/ContentDisplay.cf...
RE: Shear Stress / Failure
RE: Shear Stress / Failure
RE: Shear Stress / Failure
RE: Shear Stress / Failure
So the equation you posted, is that still the one to use. Could you explain it a little more, sorry :s
RE: Shear Stress / Failure
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
RE: Shear Stress / Failure
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.
RE: Shear Stress / Failure
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
RE: Shear Stress / Failure
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
Thread406-46941
RE: Shear Stress / Failure
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??
RE: Shear Stress / Failure
I believe the equations that Desertfox posted in his/her previous thread is for a solid shaft. that is why you do not see a thickness parameter.
RE: Shear Stress / Failure
RE: Shear Stress / Failure
RE: Shear Stress / Failure
RE: Shear Stress / Failure
RE: Shear Stress / Failure
RE: Shear Stress / Failure
RE: Shear Stress / Failure
The term Tfp means torsion to create a full plastic torque
across the shaft which is what you need to do to make it fail.
For a tube the only modification you need is to add the
inner radius of the tube, the formula looks like this
Tfp=(2*3.142/3)*(R^3-R1^3)* yielding shear stress
where R=outer radius and R1=inner radius
to get this device of yours to fail you probably have to do
so practical tests and not just rely on the mechanics, however to get you in the right ball park:-
okay so you know the torque, now this torque is going to have to generate the yielding shear stress throughout the wall of the tube therefore if we rearrange the formula above
we can say that:-
R^3-R1^3= 3*Tfp/(2*3.142*yielding shear stress)
further you know the outer diameter of the tube so subtract
R^3 from both sides and multiply both sides by -1
we get:-
R1^3= R^3-3*Tfp/(2*3.142*yielding shear stress)
find the cube root of the right hand side that will give you the inner radius of your tube.
Again this will be a ball park figure and I recommend you do some practical tests, again here look at the thread I gave you in the earlier posts.
regards desertfox
RE: Shear Stress / Failure
so Tfp is basically just the torque I want it to fail with
What value of yielding shear stress shall I use for mild steel
RE: Shear Stress / Failure
I would take the yieding shear stress to be 70% of the ultimate tensile stress for the mild steel.
regards desertfox
RE: Shear Stress / Failure
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
RE: Shear Stress / Failure
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
RE: Shear Stress / Failure
Τmax = (2*Tro)/(π*(ro4-ri4))
where π = pi
Please see FAQ731-376 for great suggestions on how to make the best use of Eng-Tips Fora.
RE: Shear Stress / Failure
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.