Question

1. A solid steel bar of diameter d; = 25.0 mm is enclosed by a steel tube of outer diameter d3 = 37.5 mm and inner diameter d2 = 30.0 mm (see figure). Both bar and tube are held rigidly by a support at end A and joined securely to a rigid plate at end B. The composite bar has a length L = 600 mm and is twisted Tube by a torque T = 400 N-m acting on the end plate. The material has a shear modulus G= 80 GPa. Bar (a) Find the maximum shear stresses ti and T3 in the bar and tube, respectively. End plate (b) Find the stress in the inner face of the tube, T2, and sketch the shear stress distribution in the composite cross section. de -dz (c) Find the angle of rotation of the end plate, B. Answers: (a) Ti = 32.686 MPa T3 = 49.028 MPa (b) T2 = 39.223 MPa (sketch required) (c) 08 = 0.0196 rad = 1.1240 "to the right"

Answer 1

T Bar - End plate Т- 4000 L- 600мм , Зотм az:32 Смм d=25mm - Torque in Тъ- BY Tz + Torque intube Пр, - т (а.“): 5 (25х-1)* - 3-834 хог 4m Те, я (132-5x; 2)“. (бәх: 5* ) . 194 xo m“ ( 3-8 34 х -8 IP - да от Ту-т ( ) - 40° взчика П. 144 x 69 Scanned with ICS CamScanner

To= 100-29 83 N-m) Thule = T (A) - ( Hotel 1. IP2 11146X109 Ttute = 1 = 400 Ipit Ifa 1.146X10773-834xrö Trate=299.7217 N-m tu= To x die 100-2733 x 15X103 Ουουοοοοοοοοοοοοοοοιιιιιιιιιιι IPI 3.834X108 Ta max = 32.7 MPa] [1= 32.686 MPa Tz max = Tt (231). 299.7219 * 37.5 810-3 I pz 1.146X10-7 Zz max = 49.038 MPa . Scanned with CamScanner

by Iq Ca = tx / _ 299.7217 X 15X10 3 1.146X60-7 IP3 39.223 MPa 327 MPa = 39.223 MPa 49.0 38 MPa o Shaft in O= ToL = 3 IRIG Series TtL I 22 G 0 = 100-278 3 X 600X1ou? = 0.1962 radian 3.8348108 Os = 0.196 radion Scanned with CamScanner