Maximum Torsion in a Round Bar 1

[metalman8357]

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!

 

[satchmo]

Tmax = σmax Ip / R

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_947.html

 

[rb1957]

Ip = pi*D^4/32
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.

 

[desertfox]

Hi

Try this article

http://www-personal.engin.umd.umich.edu/~relittle/me265/design_analysis/chapter12.pdf

 

[rb1957]

looks like the plastic factor is 4/3 (pg 12.15)
(i've only mentioned this 'cause the OP asked for the maximum torque)

 

[metalman8357]

Thanks for the help everyone. I'm interested in finding out the maximum torque that this shaft can take before fracture. You guys seem to be in agreement that my equation was correct, so is it fair to assume that the max torque for a 0.5" shaft of steel with tensile strength of 140 ksi is only 214.7 ft-lbs? Even with the plastic factor that rb1957 mentioned, this brings the total to 286ft-lbs. We specify a max installation torque of no more than 65 ft-lbs, but we say that you can use an impact torque wrench rated up to 340ft-lbs. If someone were to use an impact wrench at it's maximum capacity, by my calculation it would be possible to fracture these parts (which is what was happening). Thoughts?

 

[rb1957]

if someone's dumb enough not to set the torque wrench properly, they deserve what they get ...

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 ..."

 

[satchmo]

Agree with rb, if you've given the max installation torque, why would you tell them what tool to use?

 

[metalman8357]

I agree with both. My calculated breaking torque of 214 ft-lbs was not far off from the actual breaking torque of 217 ft-lbs (I went out in the lab and just tested it myself). We need to change our literature. We specify a max installation torque of 65 ft-lbs, but then we say that you are permitted to use an impact wrench rated at 340 ft-lbs. Clearly this might confuse some people, since I know most pneumatic impact wrenches do not allow you to accurately set the torque value. Thanks everyone for the help.

 

[rb1957]

without meaning to be -ve (?), i think your answer was close 'cause of a "comedy of errors".

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").

 

[metalman8357]

HAH! Life is funny sometimes.

 

[Tmoose]

I'nm guessing this 1/2 inch shaft has a much larger hex formed on it for the wrench. it would be hard to grip the 1/2 inch diameter in a vice and develop the full strength of the 1/2 inch diameter?