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Mechanics of materials question, angle of twist in shaft

  • #1
Hello, I'm studying machanics of materials.. I'm at the chapter that handles the angle of twist in shafts. There's one exercise that I get the most part of, but there's just a tiny part of it that I don't get!

here's the exercise: http://davidhansson.deviantart.com/art/Wp-20131015-001-407472489

and here's the solution from the book: http://davidhansson.deviantart.com/art/Wp-20131015-005-407472644

what I don't get, is what internal torque really means and how it become (1/2)*25x*x ,,
Where does the "(1/2)" comes from, and where does the extra x comes from..

My intuition is that it should be integrated like: 1/((pi/2)*75*10^9) ∫(15000/0.6)*x integration from 0 to 0.6 ,, which would be the twist in radians for the part AB.

I apologies if it's hard to understand

lot's of thanks/ David Hansson
 

Answers and Replies

  • #2
SteamKing
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It's tricky to explain, but I'll have a go.

The value of 15 kNm/m at point A is not the torque at A but the change in torque per unit length of the shaft from that point. Since the torque is zero at B, the slope of the torque distribution = 15 kNm/m divided by 0.6 m = 25 kNm /m^2. To calculate the value of the distributed torque at at distance x from point B, then

t(x) = 25 kNm/m^2 * x m = (25 x) kNm/m

The total torque applied would be the integral of the function t(x) from x = 0 to x = 0.6 m, so

T = (1/2)*(25 x^2) kNm

It's a weird way to express the torque distribution and I can understand your confusion.
 
  • #3
Hello, lots of thanks for the reply!

What you're saying sounds fully logic and was also my intuition. The weird thing though is that they make one additional integration to the (1/2)*(25 x^2) kNm ,, so we get (1/6)*(25 x^3) kNm (from 0 to 0.6).. and then multiply with the constants ofc,, this will give the final answer: 0.00895 rad that's written in the book. To me, this last integration feels totally unlogical, could it be that they actually have made a mistake in the book?

thanks/ David
 

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