Rectangular tubing torsional shear stress

[GhostRider117]

Salutations,

I’m taking a box bracket made out of a rectangular steel tube used to mount butterfly valves to actuators and am trying to find out if I can preemptively calculate the maximum allowable torque the bracket will withstand before it fails.

My approach thus far is to use the equation for Torsional Shear Stress:

Ƭau = Tc / J Ƭau = Material Yield Strength x .58 (Safety Factor)
T = Torque (what I’d be solving for)
c = radial distance from the center to the outside wall
J = Polar Moment of Inertia

My questions are:
1.) Is this an applicable formula to solve for maximum allowable torque?
2.) If so, how do I calculate the Polar Moment of Inertia for a hollow rectangular tube, or will the Moment of Inertia about the axis parallel with the torque suffice?


Roark’s gives the equation for the Moment of Inertia relative to both Ix and Iy to be: [(b(d)^3 - h(k)^3)] / 12
I’ve been told that the Polar Moment of Inertia will simply be these two values added together, but I have my suspicions about the legitimacy of that being accurate. I’ve come across an equation I found through a lecture regarding the polar moment of inertia for a rectangular tube, but I’ve not seen this before and would like to verify its accuracy.

J = [(2b)^2(h)^2(t1)(t2)] / [(b)(t1) + (h)(t2)]
**t1 = t2 in this case because the thickness of the tube doesn't change**

 

[rb1957]

no, for a colsed tube torsion stress = T/(2[A]*t) where [A] is the area enclosed by the mid-planes of the section ... (b-t)*(d-t), yes?

Quando Omni Flunkus Moritati

 

[desertfox]

This site polar moments for rectangular tube

http://www.roymech.co.uk/Useful_Tables/Sections/RHS_cf.html

 

[GregLocock]

You can't use J to represent the torsional constant for a rectangular section. In fact it only works accurately for circular sections or circular tubes.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[GhostRider117]

GregLocock, to all my recollection I thought the same thing but made the inquiry because I stumbled upon a lecture that gave an equation for it in a relevant problem.

That process was to calculate J with the equation [(2)(b^2)(h^2)(t1)(t2)] / [(b)(t1) + (h)(t2)] and from there use Ƭ = Tc / J

What's interesting is after running the numbers for this process and then doing the same with the formula given by rb1957 (which I believe is exactly what I should be using), the results are very close. I plan on running an FEA on the part to see if it will generate data comperable to I’m getting. Either way, I have to wonder if that lecture had a realistic and accurate application that I’ve yet to use.

If you care to satisfy any curiosity the values I’m referencing are:

Torque = 10000 inlbs
b = 1.78
h = 5.62
t1 = t2 = .25
c = 1.14
Ƭ = [30000 psi Y.S.] x .65 SF = 19500 => results from both processes were under this maximum

Bracket dims: 6.12” wide x 2.28” tall x .25” thick

Lecture Process end result was 1686, result with rb1957’s eqn was 1678 (where b = 6.12 - .25 and d = 2.28 - .25)

 

[rb1957]

"Tau = Material Yield Strength x .58 (Safety Factor)" ... if this means you're estimating the shear yield from the tension yield, then the 0.58 factor is NOT a safety factor.



Quando Omni Flunkus Moritati

 

[GhostRider117]

You're absolutely right, should've proof-reaf that more thoroughly!