Question

CE 3350 Summer 2019 EXAM 2 NAME Mohama Alhemian Closed book, closed notes, closed hon sheet only formulas are allowed; you collected with the exam. You may us lowed; you cannot write One 8 " by 11" formula sheet is allowed. On the formul Write definitions or sample problems. Formula sheet will b blator. You cannot use a cell phone. 1. (35 points) Given: A rod is made from two segments. torque of 700 N.m. The steel portion has a AB is steel and BC is brass. It is fixed at its ends and subjected to a ameter of 30 mm. G -39 GPa. G-75 GPa Find: (a) Determine the required diameter of (b) determine the absolute maximum shear stres er of the brass portion so the reactions at the walls will be the same; and Shear stress in the shaft. TA=TB = 7/2 = 350 Nom Gbr = 39 GPa TATTo T c Get = 75 Gla 700 N.m I t = 700 Nom Comp. Fixed B Fixed angle of twist. iz II (15mm) = 23.3 (15mm) = 23.56 m. ? - 350 EQ-(7001(1.6m) 17001 17001 (0.75)

Answer 1

Tx(42.7,3 × 350x (1000) r = 22,907344.85 N/m2 - 22.907 N/mm² Samlarly T (shear stoess in shaft & AB) * Bojs *350* (lovos - 66053314.46 T= 660533 olar? marinum struss occans in shaft AB' & il, 660533 N/am? & shear strus in shaft Be is, 22.967 N/mm² SunOC

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(วะ) รล. (งด) 10 O.IT Jab Tbc - 4.10.95 Jab abc = V4.1025 dab - 1.423 (dah) - 1493 (30) Tabe- 42.7mm) manimum shear stress in the shafts 04 05 Absolute Im WA 06 07 a. 700/=350 350 (0 = 42. Frm (fr Bd som (for AB 3. 14 xD3 * 350