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

 

[haynewp]

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:

http://www.aisc.org/ContentManagement/ContentDisplay.cfm?ContentID=4203

 

[NavierStokesEq]

you say Hss, I was on about mild steel, does it matter

 

[haynewp]

The provisions in that publication are for certain grades, see page 1. See if the steel you are using is listed.

 

[haynewp]

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.

 

[NavierStokesEq]

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

 

[NavierStokesEq]

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

 

[jec67]

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 &quot;ultimate strength&quot; 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.

 

[desertfox]

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

 

[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
thread406-46941

 

[NavierStokesEq]

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??

 

[jec67]

The shear yield stress for mild steel is estimated as between 1/2 and 5/8 of the tension yield stress (according to AISC - American Institute of Steel Construction). In their formulas, AISC uses Fytension/(sqrt3) as the value of the shear yield stress.

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.

 

[NavierStokesEq]

Got ya, is there a similar equation with thicknesses

 

[jec67]

In Roark (I'm using the sixth edition) section 9.5, there is a discussion that relates a tube tipy item to a solid shaft. I have not used the forula before, but I am thinking you could use Desertfox's formula coupled with Roark to determine your failure torque. The section in Roark is titled &quot; Ultimate Strength of Bars in Torsion&quot;. For an thin-walled section, the modulus of rupture for the section is related to the modulus of rupture for a solid shaft.

 

[NavierStokesEq]

I have the sixth edition but cant find the section you mention, my section 9 is bending of curved beams

 

[jec67]

It's toward the end of the chapter that covers torsion (page 386 in the 6th ed). Let me know if you cannot find it.

 

[NavierStokesEq]

Got it, thanks, it dont make much sense to me know, goes on about soap films!!! which equation should I use

 

[NavierStokesEq]

do I use the equation on 398, if so, how do I find the modulus of rupture for a solid bar, this dont take in to account of the torque i want to apply though???

 

[desertfox]

Hi NavierStrokesEq

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

 

[NavierStokesEq]

thank you

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