Question

MEEN 230 NAME B. For questions 5 through 8, consider the aluminum rod ABC, shown (E-10.1 X 10 psi). It consists of two cylindrical portions AB and otbol BC with diameters as shown. The rod is fixed at C and free at A. yhose who have Sunt b ubble GOLO JOIN boso o dust to Dahid A doo bolella on omad 12 in. - 1.5 in. a 2 0 do lat 10A B ob boo bosto -2.25 in. 18 in. mob.OS MIZA V not ei oqib olamo 30x00.(d) 9000 L IC or lo son (b) OXO.SK 2. dio ob 0S 5. Calculate the cross-sectional area in portion AB in square inches. (a) 0.9231 (b) 1.167 (c) 1.767 osob) (d) none of the others 6. Calculate the cross-sectional area in portion BC in square inches. (a) 0.9231 (b) 1.167 (c) 1.767 SITE (d) none of the others (5) 7. Calculate the displacement of end A with respect to B in inches. (a) 0.01667 down niin to b (b) 0.01883 down to (c) 0.03138 down (d) none of the others (1) 8. Calculate the total displacement of end A in inches. OXOS () (a) 0.01667 down (b) 0.01883 down (c) 0.03138 down (d) none of the others

Answer 1

Answer - Csiver? E = 10.1X10° psc 15 Crosseescchional area for AB portion = x+?= I x (15)2 =1176625 So, option © a 1,767 inch? 2 (Aes) Be= I x (2125) 2 = 3.974 inch? So, option (@. 0 Drawing FBD of AB Siksi: 1000 pso} 125 kips - * x 25 kips (All AB (AI) AB = Pe AE 12" ! kis" = 25x 12 x1000 168" | 25 kip " 11767x10.1X100 = 0.0168 So, option @ 125 kips 18 FBD of BC K 2.25" (48) Bcc @ 18" H = (25x1000) X 18 3.974 X 10.1X106 V 106 125 kips 0.0112! eftione D. So, Ophion OD toteel displasennent = (49) mp +(88) # = 0.0168" + 0.0112' ophion D # = 0.0284