Hollow shaft outer diameter for 100Nm torque with 1/8 thickness ratio

[bigbang42]

Homework Statement

the statement:
You are taked to design a 2m long shaft which is to be manufactured either as a solid circular bar or a circular tube. the shaft is required to transmit a torque of 100Nm without exceeding the alloqable strength of material.

Find the required outer diameter of the tube if the thickness of the tube is specified as one-eigth of the outer diameter?

shear stress = 320Mpa

Homework Equations

T/J = $\tau$ /r

J= [pi(D^4 - d^4)] / 32

The Attempt at a Solution

thickness is 1/8
1/8 + 1/8 = 1/4

J= pi( D^4 - 0.75^4 D^4) / 32

work it out in sections
0.75^4 = 81/256 >>> 1 - 81/256 = 175/256 >>> 175/256 x pi = 2.14 >> 2.14 / 32 = 0.067D^4

then using the formula T/J = t/r

1000/0.067D^4 = 320x10^6 / D/2

get D on one side i get

1000 / 2 x 320 x 10^6 = 0.067D^4 / D

worked all that out then cube root it to get 0.028 which i then x1000 to get 28mm


I feel I'm on the right path ( i hope) but the whole 1/8 of the diameter has confused me and I'm wondering whether I've done that right. finding the "J" has to be my main concern i feel I've got the method right for finding D later on. If anyone can help my go through it and check everything is correct or if its wrong, advice on how to correct it would be fantastic.
thanks.

 

[PhanthomJay]

Looks like you got it all correct, good step by step work. I assume the torque is 1000 Nm and not 100 Nm as originally stated.