Torsion in a beam

  • Thread starter jaderberg
  • Start date
  • #1
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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
 

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Answers and Replies

  • #2
radou
Homework Helper
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J = (pi/2)*r^2 for a solid circular cross section

Well, first of all, isn't J given with [tex]J = \frac{d^4 \pi}{32} = \frac{r^4 \pi}{2}[/tex]?
 
  • #3
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Well, first of all, isn't J given with [tex]J = \frac{d^4 \pi}{32} = \frac{r^4 \pi}{2}[/tex]?

oh yeah thats a typo
 
  • #4
radou
Homework Helper
3,115
6
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.
 
  • #5
nvn
Science Advisor
Homework Helper
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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.
 
  • #6
30
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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.

cheers man, was beginning to wonder whether that was the case :p
 

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