Question

The rod shown below is composed of two pieces: the diameter of piece AC is 3 in, and the diameter of piece CD is 2 in. Both pieces are made of brass: Young's modulus is E = 15x10 psi and the shear modulus is G = 5.6x10 psi. The applied external axial forces are Pc = 50 kips and Po = 30 kips. The applied extemal torques are Tg = 25 kip in and To= 10 kip in. (Hint: for parts a-d, you can decompose this into two less complex problems: 1) the axial forces (parts a and b) and 2) the torques (parts c and d)) Fixed a) Determine the deflection, &, of the entire rod. Specify longer or shorter. b) Determine the normal stresses, in pieces AB, BC and CD. c) Determine the twist, of the entire rod. d) Determine the shear stresses, in pieces AB, BC and CD. c) Using the normal stress, o, and shear stress, for piece CD, plot a Mohr's circle. (Hint: the figures below show the directions of the stresses at point E, which is on the surface of piece CD). 1) From Mohr's circle, determine the maximum normal stress, the minimum normal stress, ... and the maximum shear stress, Car Label these stresses on the circle. Cea o Ans: 8= 9.620x10in (longer-shorter). O = Onc = 2.829 ksi (T). Op = 9.549 ksi (C). o=1.306, 0 2.829 ksi, c = 1.886 ksi, ED=6.366 ksi, = +3.183 ksi, G -12.733 ksi, 7.958 ksi,

Answer 1

Page 0 Solution: Given that © 2014. ® 10 in © 2uin 24 Радок, Pc =50kips TB = 25 kipin Toc lokipin. F= 15x10 psi; b=5.6 X10 psi (a) calculation of s'e daca 3in, deo-rin. 30 kips > 50 Ribo. F.B.A ZOKID -30kips - 24 in. - pokip. 20kips & -Din 2okine soin - 207n. Rokips. sign Convention Tension (+) comprension() 's'=dArtodo, adeo = PABLAB & BBC ABC & Peoles AAB FAR ABLEBI Aco Eco

278=+20x105x20 + 20x108x10 – 30x10²x18824 Page - I x 3? x 15X105 4x33x15X10" & A 2+2 15305 = +3.773x103 in + 4.886x103 n., ~ 15,279 410 in. = -9.620x10"} in. (Contrusion) from F.B.A. - DARBA = 20 X = 2.829 Ksi (Tennion) - AAR 3² BBC = Psc = 20%D0-21829 ksi ( Temmon) 6 2.8 2 x 3² PCP 5-30 = 9.549 ksi(comprension) ACD Ex22

ge (3) To = 25 kipit 10 kapin 24in. -2014 a lotul 10 kibin 2uin PO KLI 10 kipin 15 kilink 10) = 15 kipin. Sign convention - 2oin- Runs, P-BA DC CB - Teoles & bro Jes = 10 x103 x24_ 5.681 06 xnx2" BA Teg los at TBA IBA the TR GIBA IBA + 1OXIDXTO – 15x1ồx20 5.6*06** x3 5.6 X10 Xa 13" 32 32

page- = +265 +24,284x10°+ 3.246X1636.739 X = 22.792X10° radrens ligos Don = 1.306° Thors, ($=13060 (1) we know that ea Tip TAB= TABOAB = 154 2 = 2.829 Kai JAB IX39 Toc TBC NBC = 10*1.5) 21.886 ksi JBC Zx3" T = TEDxos = 10x/2/2) = 6.366 ksi JO 32 X2" 32 Elies in CD -) C = OCD = -9,549 ksi 8 Takes - 6.366 ksi

Page ① Now, coordinates of Point = (-9.54, -6.366) coordinate of Point ☺ = (0, 6,366) Drawing Mohrs Circle using points ♡ and on 40, 6.366) Smin sty 6 Omary (-4.7750) (-9.540, -6.366) Centre of Mohr's cunde = -9.549+0 -6.366 7636 (1) Radius of Mohris Circle = {(2.775-0) 1631 J44.725-)+(1.366-0) ? R = 7.958 units using mohris arde - Omera = -4.775 +R=-4.775+7.958 =+3.183 6.E^4.775-R=-4.775-7.9582 Toman - 3.183,67 min & -12.733 kai tam temp. at Ksi -4.775 .958 Termion -12.733KS: c