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Mechanics of materials question, angle of twist in shaft
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[QUOTE="SteamKing, post: 4540049, member: 301881"] 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. [/QUOTE]
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Mechanics of materials question, angle of twist in shaft
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