Question

Determine the angle of twist of the bar. the bar ia subjected to a torque of t= 571 lb-ft and is of a material of g = 18,431, 356 lb/ in^2. express the results in degrees. the dimensions of the bar are: d1 = 8.1 in d2 = 1.8 in L = 3.1 ft

Answer 1

Answer:

angle of twist of the bar is 3.32 × $10^{-5}$ degree

Explanation:

given data

torque t = 571 lb-ft = 774.172 Nm

g = 18,431, 356 lb/ in² = 1.27 × $10^{11}$ Pa

d1 = 8.1 in = 0.205 m

d2 = 1.8 in = 0.045 m

L = 3.1 ft = 0.944 m

to find out

angle of twist of the bar

solution

first we get here polar moment that is express as

polar moment = $\frac{\pi (do^4 - d1^4}{32}$     ...........1

here do is outer diameter that is 0.205 m and d1 is inner diameter i.e 0.045 m

put here value we get

polar moment = $\frac{\pi (0.205^4 - 0.045^4}{32}$

polar moment = 173.006 × $10^{-6}$ $m^{4}$

now we get here angle that is

angle = $\frac{TL}{IG}$   ..............2

here T is torque and L is length and J is polar moment and G is share modulus

put these value in equation 2 we get

angle = $\frac{774.17*0.944}{173*10^{-6}*1.27^{11}}$

angle = 3.32 × $10^{-5}$ degree