# Torsion in a beam

## Homework Statement

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)

## Homework 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

## 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

#### Attachments

• DSC00352.jpg
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Homework Helper
J = (pi/2)*r^2 for a solid circular cross section

Well, first of all, isn't J given with $$J = \frac{d^4 \pi}{32} = \frac{r^4 \pi}{2}$$?

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

Homework Helper
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.

nvn