Question

T= 2000 Nm

The following circular shaft of total length=1 meter, with a change of section diameter at distance a (see value in previous email), from end A, is subjected to a torque (moment) of value I (see previous email) which is applied at point C. The material is Aluminum 2014-T6 with G = 27 GPa . The shaft is attached to FULLY RIGID supports at ends A and D. Diameters and lengths of shaft are given in Figure, and radius of connection r between the two different sections is in previous email). Note: T, a , and r are in previous personalized email. It is required: (i) to find the torque reactions at rigid supports A an D. NOTE: there is a change of section, and of moments of inertia, then a compatibility equation MUST be used (30%) considering the change of section (Textbook Section 5.8) compute the MAXIMUM stress (30%) (ii) (iii) check if computed maximum stress value is within the elastic period of such material ( 10% . Circle "Yes or “NO”). Values of T, r and a are sent in separate personalized email. diameter 100 mm Diameter 200 mm 0.4m b=1m- Radius 0.6m meter.

a = 0.4 m

r = 15mm

Answer 1

1. Finding Torque reactions at A and D Th Ą 100 mm B 200mm cb d TAB LA 0.4m|k AR TCB = OBC J = Polar L = 0.6 m Bc compatibility equation TAB + TBC = T - 2000 N-M Also twisting angles, OAB But o= TL and TT xdat moment of GJ 32 Inertia TBe LBC G JAB -> TAB X 400 x 32 Твс х 600 х 32 IT X (100)4 ↑ x (20034 TAB 0.09375 TBC 400 X xat TAB LAB en JBC - => 4 600 x (100) (20034 => TBC 10.666 x TAB Now TAB + TBC 2000 Nom TAB + 10.666 TAB 2000 => TAB =) TAB 2000 1+1:438 - 11. 666 and TBC = 10.666 x 171: 438 = 11828.561 Nam change of 2. Maximum stress at the point of CYOSS -section (B). TK lo D 7

Theoretical stress concentration factor (K+) for a stepped shaft with a shoulder Butlet in Torsion For old and & 100 n. Kt - 200 -2 15 100 d 0.15 1.28 ( Using table] Now from torque equation, the theoretical shear stress is given by z = 16 T (To be calculated at the Td3 minimum section) 3 16 x 2000 X10 10. 186 N/mas x (10033 But the maximum shear stress due to the stress concentration is! Emane = 7 x Kt 10, 186 X 1:28 = 13,038 N mm > z= IT s 3 Yield tensile strength of Al 2014 - T6 material is Syt 345 MPa According to maximum shear stress theory of failure, the field shear stress is Sys - Syt 2 S 172.5 N / min 3 45 = => sys Since Tmax < sys 2 Yes, it is within the elastic limit.