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Torque and Shear stress this time

by chrisking2021
Tags: shear, stress, time, torque
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Mar24-07, 11:39 AM
P: 7
I'm I'm OK at electrical theory but mechanical just doesn't compute.

I have the following question if anyone can help.

Determine the external diameter of a tube needed to transmit a torque of 30KNm if it has an external diameter twice that of its internal diameter. The sheer stress is not to exceed 80 MPa.

Any help would again be gratefully received.

Also could anyone tell me how to put formula into this thread as i don't really know how.
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Mar24-07, 11:50 AM
P: 7
J = (pi/32)*(D^4-d^4)

T/J = Sheer Stress/r

T = Torque
r = Radius of external diameter
D = External diameter
d = Internal diameter

These are some formula ive tried working through but i need J or D to work it through.
Mar24-07, 12:48 PM
Cyrus's Avatar
P: 4,777
You have everything you need. J is a function of D only. D and D/2.

Nov17-11, 01:28 PM
P: 1
Torque and Shear stress this time

As Cyrus mentioned. All the info was given.
To finish off the question, for those like me found the thread.
The formula should look like:

D= external diameter in meters

30*10^3 [Nm] = 80*10^6 [Pa=N/m^2] / (D/2) * (pi/32)*(D^4-(D/2)^4)

When solved

Thank you for the comments. I used this thread to help study for the FE exam.

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