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Torsion in a beam |
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| Jan3-09, 09:27 AM | #1 |
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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 |
| Jan3-09, 10:12 AM | #2 |
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| Jan3-09, 10:14 AM | #3 |
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| Jan3-09, 10:23 AM | #4 |
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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.
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| Jan3-09, 06:10 PM | #5 |
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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.
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| Jan3-09, 06:15 PM | #6 |
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