Civil Engineering - Strength of Materials - Discussion

The strength of shaft means max. torque or max power transmitted capacity.
Here asking their strength equal means weight ratio of the shaft.

So weight depends on the size of the shaft.
We taking maximum torque equation of finding dia of the shaft.
Consider d for outer dia of hollow shaft and nd for inner dia of the shaft.
Then formula = (π) * [ d^4 - (nd)^4] * shear stress /16/d^4.

Assume dia of solid shaft D.

Then we equating torque of solid shaft = torque of the hollow shaft
we get D = { [d^4 - (nd)^4 ]/ d^4 }
The weight depends on the area of the shaft section so,
Hollow shaft cross-section area / solid shaft cross-section area.

By substituting the value of solid shaft dia in area formula we get the final answer.

Assume External diameter = d then internal diameter = nd.

As both the cases have equal strength so equate both strength which is polar modulus
case 1 polar modulus = pi()*(d^4-nd^4)/32/d/2
case 2 polar modulus = pi()*(D^3)/16.

Equating both find D in terms of n pi()*(d^4-nd^4)/32/d/2 = pi()*(D^3)/16
D = d((1-n^4)^1/3.

Now weight is directly proportional to Area.
area in case 1 = pi()*(d^2-nd^2)/4,
Area in case 2 = PI()*(d((1-n^4)^1/3)^2/4,
dividing both we get the ratio as (1-n^2)/(1-n^4)^2/3.

Now, d = nD, D = d/n.

Ratio of the weight = ratio of C/A = (d^2 * (1/n^2-1))/D^2 ...1.
Find d^/D^2 as follows:

((d/n)^3 - d^3)/D^3 = Th/Ts = 1
(d/D)^2 = (1-n^2)/(1-n^4)^(2/3).

Substitute in eq 1 you will get the answer.

I am going to prepare with this help after second carrying after two years so please explain with notations and clearly.

Given, d/D = n and stress ratio D^4 - d^4/D^4 = Th/Ts.
Now, d = nD, D = d/n putin above and get 1 - n^2/(1 - n^4)^2/3.

I can't understand. Can anyone explain clearly?

What is polar modulus?

And what is p=, I=?

Anyone give explanation for the answer.

Please expain clearly.

Thank you @Krishna.