# Homework Help: Strength of materials

1. Mar 19, 2013

### MMCS

A mild steel solid shaft is to be replaced by a stainless steel hollow shaft of the same
outside diameter. Calculate the ratio of internal diameter to external diameter of the
hollow shaft for equal strength. Also find the percentage saving in weight if density of
mild steel is the same as stainless steel.Take maximum shear stress for Mild Steel as
34 MPa and that for Stainless Steel as 57 MPa

I assume that for the strength to be the same the max torsion should be the same so i use the max allowed shear stress in this equationrearranged for torsion and equate them both

(34MPa * (pi*R^4)/2)/R = (57MPa * (pi*(R-Ri)^4)/2)/R - Ri

Is this the correct step? how would i get a ratio from this?

2. Mar 19, 2013

### PhanthomJay

(R^4 - Ri^4) is not the same as (R - Ri)^4.
And in both cases, max shear stress occurs at the same distance fron the centroid of the circular cross section.

3. Mar 19, 2013

### MMCS

Ok so

(34MPa * (pi*R^4)/2)/R = (57MPa * (pi/2*R^4)-(pi/2*Ri^4))/R

to simplify and multiply out pi terms i get

(-23*pi/2*r^4)/r = (57*pi/2*ri^4)/r

-36.128*r^4 = 89.535ri^4

so as a ratio i get

36.128/89.535 = 0.404

but i have the answer to be 0.8

4. Mar 19, 2013

### SteamKing

Staff Emeritus
You've got negative signs appearing and disappearing at random.

You should review basic algebra.

5. Mar 20, 2013

### MMCS

That is irrelevant. -36.128/89.535 produces the same ratio as 36.128/89.535.

6. Mar 20, 2013

### SteamKing

Staff Emeritus
And yet you have not solved this problem.

7. Mar 20, 2013

### MMCS

Another irrelevant response. I do not know how to solve the problem otherwise i wouldnt have posted the question, however, Your input had nothing to do with getting closer to the solution.

8. Mar 20, 2013

### PhanthomJay

the plus and minus signs heretofore not withstanding, you have calculated that (ri^4/r^4) = (ri/r)^4 = 0.404. So to calculate (ri/r), you need to take the 4th root of that number.
which is correct.