Maximum Torsion in a Round Bar
Maximum Torsion in a Round Bar
(OP)
Hi all,
I'm a materials guy so you'll have to forgive my lack of knowledge for mechanical engineering formulas. I have a 0.5" diameter round steel bar that is subjected to torsional loads, and I'm trying to determine the failure mechanism. I am trying to determine the maximum torque that this shaft can take before breaking. Imagine that the one end of the rod is fixed in a vice, and I'm applying torque with a 1 foot breaker bar, I want to calculate the max torque that the rod can withstand (I'm assuming this is fairly basic). The formula that I'm using is Tmax = (pi/16)*(USS)*(D^3) = (0.1963)(Tensile strength*0.75)*(0.5^3) = 2576.4 in-lbs = 214.7 ft-lbs. Is this right? We've been listing the max operational torque at 340 ft-lbs but maybe this is incorrect. Am I approaching this problem correctly?
Thanks for the help guys!
I'm a materials guy so you'll have to forgive my lack of knowledge for mechanical engineering formulas. I have a 0.5" diameter round steel bar that is subjected to torsional loads, and I'm trying to determine the failure mechanism. I am trying to determine the maximum torque that this shaft can take before breaking. Imagine that the one end of the rod is fixed in a vice, and I'm applying torque with a 1 foot breaker bar, I want to calculate the max torque that the rod can withstand (I'm assuming this is fairly basic). The formula that I'm using is Tmax = (pi/16)*(USS)*(D^3) = (0.1963)(Tensile strength*0.75)*(0.5^3) = 2576.4 in-lbs = 214.7 ft-lbs. Is this right? We've been listing the max operational torque at 340 ft-lbs but maybe this is incorrect. Am I approaching this problem correctly?
Thanks for the help guys!





RE: Maximum Torsion in a Round Bar
Tmax = maximum twisting moment (Nmm, in lb)
σmax = maximum shear stress (MPa, psi)
R = radius of shaft (mm, in)
Ip = "polar moment of inertia" of cross section (mm4, in4)
Also, you said that it is to torsional loads, plural. If fatigue is a consideration, the actual stress required to fail the shaft could be considerably lower.
Here is a good link
http://www.engineeringtoolbox.com/torsion-shafts-d...
RE: Maximum Torsion in a Round Bar
Ip/R = pi*D^3/16
so your expression is correct. this expression is for elastic stress due to torque, so there is an adjustment for plastic stress dist'n.
mostly however i'd use USS = 0.57*UTS or look it up for the specific material.
RE: Maximum Torsion in a Round Bar
Try this article
http://www-personal.engin.umd.umich.edu/~relittle/...
RE: Maximum Torsion in a Round Bar
(i've only mentioned this 'cause the OP asked for the maximum torque)
RE: Maximum Torsion in a Round Bar
RE: Maximum Torsion in a Round Bar
i'd've though your job was to define the requirements (65 ft.lbs ... gosh that sounds like a lot !) and leave it to the guy doing the work to accomplish them. if he wants to use a 500 ft.lbs wrench why should you care.
in fact by saying "you can use a 340 ft.lbs rated torque wrench", aren't you exposing yourself to suits ... "i used the tool they recommended and broke the shafts ..."
RE: Maximum Torsion in a Round Bar
RE: Maximum Torsion in a Round Bar
RE: Maximum Torsion in a Round Bar
i think your estimate of USS (= 0.75UTS) is high,i think it should be 0.57.
but then there's the plastic factor of 4/3 ...
0.57*4/3 = 0.76
so close it makes me think that you knew the two factors combined (and it wasn't a case of "numerical transposition").
RE: Maximum Torsion in a Round Bar
RE: Maximum Torsion in a Round Bar