## Torsion in a beam

1. The problem statement, all variables and given/known data

The question is shown in the attached picture. I know how to do the first part fine,and i know how to do the second part but i keep getting the wrong answer (answers are on the back of the sheet)

2. Relevant equations

tau/r=T/J (tau = shear stress, r = radial distance from centre, T = Torque, J = Polar second moment of area)

J = (pi/2)*r^4 for a solid circular cross section

3. The attempt at a solution

T = 12k*0.75 = 9k Nm

J = (pi/2)*(37.5x10^-3)^4 = 3.106x10^-6 m^4

tau = (9x10^3*37.5x10^-3)/3.106x10^-6
tau = 108.7 MPa (answer given is tau = 45.8 MPa)

sigma = My/I

I = (pi/4)*r^4 = 1.553x10^-6

M = 18kNm
y = 37.5mm
sigma = 18x10^3*37.5x10^-3/1.553x10^-6
sigma = 434.6 MPa (answer given is sigma = 188 MPa)

both answers are wrong but if you use r and y = 15.8mm (but not when calculating second moments) the right answer comes out. Where does this 15.8mm come from or am I going wrong somewhere else?

Thanks for any help
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Recognitions:
Homework Help
 Quote by jaderberg J = (pi/2)*r^2 for a solid circular cross section
Well, first of all, isn't J given with $$J = \frac{d^4 \pi}{32} = \frac{r^4 \pi}{2}$$?

 Quote by radou Well, first of all, isn't J given with $$J = \frac{d^4 \pi}{32} = \frac{r^4 \pi}{2}$$?
oh yeah thats a typo

Recognitions:
Homework Help

## Torsion in a beam

Very interesting, I get the same answer for the shear stress due to the torsion moment. I'll have to consult my mechanics of materials book, I'll be back later.
 Recognitions: Homework Help Science Advisor jaderberg: Nice work. You did not do anything wrong. You got all answers correct. Both of the given answers from the back of the sheet are wrong.

 Quote by nvn jaderberg: Nice work. You did not do anything wrong. You got all answers correct. Both of the given answers from the back of the sheet are wrong.
cheers man, was beginning to wonder whether that was the case :p