Question

Problem 2 (25 pts): A 3 in. diameter shaft of length L = 8 ft is fixed at end A. End C is free to rotate through an angle of 2.5º, but is rigidly prohibited from rotating more than 2.5º. Find torque To (applied at B), in kip-in, that will twist the crossbar at C into contact with the stoppers. Take G = 12,000 ksi. [Formula: Torsional Angle of Twist, 0 = (TL) (JG); Torsional Shear Stress, T = (Tr)/J; J = (T)(C)/2] Crossbar 4.5 ft 3.5 ft Stopper D=3" B Stopper Cross-sectional end view

Answer 1

we know ,

$\theta$ = TL / GIp where T torque applied , G is shear modulus , $\theta$ is rotation at C

here ,$\theta$ is permited to 2.5 degree and torque is required to applied on B means only AB is subjected to torque and will lead to the rotation of C point as the rotation of B point .

please take care of units of length and formula of Ip while putting in the formula i have mentioned in the answer sheet

ΤΟ We know, 4:57 35 А B ? TO TO 0 3" A B с SLAB = 45X12" 6 G - 12,000 KSC * Max tarole which can applied Matation 2.5 $ 2.5 X n المحور الحر to purisseble radan 180 we know = TIL G Pp CO Rotation torque lengt) * torque is applied for AB only. 2.5 X TX 4.5812 (ºp 3,493 180 32 12000 x7x34 32 A Hence T = 77k-inch