Torque and Shear stress this time

AI Thread Summary
To determine the external diameter of a tube that transmits a torque of 30 kNm without exceeding a shear stress of 80 MPa, the relationship between the external diameter (D) and internal diameter (d) is crucial, with D being twice d. The relevant formulas include J = (π/32)(D^4 - d^4) and T/J = shear stress/r. By substituting the values into the equations, it is calculated that the external diameter D is approximately 0.1268 meters. This calculation assists in understanding mechanical principles and is beneficial for exam preparation. The discussion also highlights the importance of proper formula application in solving engineering problems.
chrisking2021
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I'm :confused: 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.:smile:
 
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Useful formula

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 I've tried working through but i need J or D to work it through.
 
You have everything you need. J is a function of D only. D and D/2.
 
Chrisking2021,
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:

T=τ/r*(J)
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
D=.126781154m

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