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

angle of twist of the bar is 3.32 ×
`10^(-5)` degree

Step-by-step 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 =
`(\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 =
`(\pi (0.205^4 - 0.045^4)/(32)`

polar moment = 173.006 ×
`10^(-6)`
`m^(4)`

now we get here angle that is

angle =
`(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 =
`(774.17*0.944)/(173*10^(-6)*1.27^(11))`

angle = 3.32 ×
`10^(-5)` degree