Question

The composite shaft shown consists of a 0.2-in-thick brass jacket (G = 5.6 x 10 psi) bonded to a 1.2-in.-diameter steel core (Gsteel = 11.2 x 10 psi). Knowing that the shaft is subjected to 5 kip-in. torques, determine: 1. The torques in steel and brass. 2. The maximum shearing stress in the brass jacket. 3. The maximum shearing stress in the steel core. (Note: the angle of twist in steel equals to the angle of twist in brass) 6 ft T B Brass jacket T 1.2 in. Steel core 0.2 in.

Answer 1

Given : T: 5 Kip-in Applying the equilibrium condition 5 - Tst - Tbor = 0 Tst Tor: 5 o Angle of twist in steel = Angle of twist in bar & st & br Tst Lst Tor. Lor 2) Ist. Gst Ibr. Gbr o J = Polar moment of inertia Jst = ㅈ -4 where r= radius of steel core = (1.272) in = 0.6 in Jst = 1.2 0.204 in 2 4 4 Jor Rro rit) 2 where ro = outside radius of brass 0.6+0.2=0.8 in ri= inside radius of brass -0.bin

Joy = q (0.84-0.64) = 0.44 in! 2 Griven: Ist = lbr = 6ft = 72 in 6 Gbg = 5.6 x 10° psi -5.6x103 ksi Gst:11:2x10 psi = 1:32.10% ksi Putting the values in eq - ② Ist X 72 0.204 x 11. 2x10 3 Tor X 72 0.44 x 5.6x103 Tho= 1.07 8 Ist 3 1 from ea", we know Tst +T 68 = 5 Solving Eqhs ☺ and 6 Tst + 1.078 Tst - 5 = Tst = 2.41 kip-in - Tor = 5-Tst = 2.59 kip-in we get . Torque in Steel = 2.41 kip-in Torque in brass = 2.59 Kip-in Tmax = 3. Using To9ision fosimula T-C J T= unternal torque developed (= outer radius of the shaft Maximum Shear stress where

- (tbr) max Tor.ro Jor 2:59 x 0.8 = 4.71 ksi 0:44 ;. Maximum shearing stress in brass jucket = 4.71 Ksi 3: (est) max = Tstor Jst 2.41 X 0.6 = 7:09 ksi 0:204 stress in steel care = 7:09 Ksi Maximum Shearings