Civil Engineering - Strength of Materials - Discussion

Anyone give explanation for the answer.

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.

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 can't understand. Can anyone explain clearly?

Please expain clearly.

Explain it please.

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.

Thanks @Krishna.

Thanks @Krishna.

What is polar modulus?

And what is p=, I=?