Compute the angle of twist in an aluminum tube that has an outside diameter of 8

Question

Compute the angle of twist in an aluminum tube that has an outside diameter of 80 mm and an inside diameter of 60 mm when subjected to a torque of 2250 N middot m. The tube is 1200 mm long. A steel rod with a length of 8.0 ft and a diameter of 0.625 in is used as a long wrench to unscrew a plug at the bottom of a pool of water. If it requires 40 lb middot ft of torque to loosen die plug, compute the angle of twist of the rod. For the rod described in Problem 10-25, what must the diameter be if only 2.0 deg of twist is desired when 40 lb middot ft of torque is applied? Compute the angle of twist of the free end relative to the fixed end of the steel bar shown in Figure P10-27.

Explanation / Answer

24).

=TL/(JG)

G for aluminium = 26 GPa

J = (r14-r24)/2 = (0.044-0.034)/2 = 1.3744*10-6 m4

therefore, =2250*1.2/(1.3744*10-6 *26*109) = 0.07556 radians or 4.330

25).

shear modulus of steel = 29,000,000 psi = 29*106*122 lb/ft2

=TL/(JG)

J = r4/2 = *(0.625/12)4/32 = 7.224*10-7 ft4

therefore, =40*8/(7.224*10-7*29000000*122) = 0.106 radians or 6.080

26).

=TL/(JG)

=20 = 2/180

therefore,

2/180 = 40*8/(J*29000000*122)

J = 2.1952*10-6 ft4

J = d4/32 =2.1952*10-6

therefore, d = 0.0688 ft or 0.825 in

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