Homework 5 - 2.13 Rod ED is made of steel(E = 29 x 106 psi...

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Unformatted text preview: 2.13 Rod ED is made of steel (E = 29 x 106 psi) and is used to brace the axially compressed member ABC. The maximum force that can be developed in member ED is 0.02P. If the stress must not exceed 18 ksi and the maximum change in length of BD must not exceed 0.001 times the length of ABC, determine the l P = 13" kiPS smallest—diameter rod that can be used for member BD. Problem 2.13 F'BD = 0.02P = (0.09.)(Jso\= 2.61695 = 2.6XI0315. ConsiJem‘na s'l'ressi 6’: 13 ksi = [8*1’0: PM Re _ 6‘7: -- A=E€= 72% = 0.194% M COnEEJEJ‘Cn-z a$rmm+¢\om: $=(0.00H(}W-§3= O."+‘I iv“ 5 Fol—m. _ Feat-an (2.6ylo3)(54) _ 0.033g2in2. .. _-——-—-—————————————_, Larger area Saver-us. A = 0-!‘4W-N l‘n,‘ =3: 1 3 _ [L75 _ I HXOJQWHS _ - A Lfd ' d ’ Tr - o! - OJ'ilal \Yj,‘ Problem 2. 1 4 2.14 The 4‘-mm-diameter cable BC is made ofa steel with E = 200 GPa. Knowing that the maximum stress in the cable must not exceed 190 MPa and that the elongation of the cable must not exceed 6 mm, find the maximum load P that can be applied as shown. Lac = 1/ 91+ 4‘ = 7.2m m Use loaf A3 as a-pree loaaly. P - FB‘ QEMA = o 3.5? — (Glfijim Pack o 1: = 0.?509 FBC Ax Cans/Jena? aijouaéfl: S‘h‘ess: 6-: IqOXIOG Pa A7 A = %al1' = Egacm'r = [2.566 #10“ m‘ 6* z %‘ .1 FM: 5A = 090 x;0‘\(22.5;5xzo“) = 2.388MO‘ N Conga/aria? ajfiordwlaje epomjn+:‘an= a = g x/odm —6 q I g 7 FngL'I‘G 1‘: a 02-5%”: ilioo “to —-——-—x 6 xlofi) : 2.041. X103 U Swapper Vaduz. gave/As. lie—— 2.07/Xl03 N P .- 0.9593: F“ = (031509“;ow:x;o‘)=l.qzzwo’u Prmss IN a Problem 2 16 2.16 The specimen shown is made from a l-in.-diameter cylindrical steel rod with ' two 1.5-in.-outer-diameter sleeves bonded to the rod as shown. Knowing that E = 29 x 106 psi, determine (a) the load P so that the total deformation is 0.002 in., (b) 1%-in. diameter the corresponding deformation of the central portion BC. P ' A / l-in. diameter B Sum P =(RQX'O‘XO-0013C6flas)” = magnum: P: 91.53 Mn «:3 (H = "‘ T:13 if: ’ ————Z:fifi*g‘3’<a.zm) S=l.25"-l><lo’5{m_sn Problem 2.24 2.24 For the structure in Prob. 2.23, determine (a) the distance h so that th: deformations in members AB, BC, CD and AD are equal to 0.04 in., (b) tht corresponding tension in member AC. 2.23 Members AB and CD are 1;~in.-diameter steel rods, and members BC and AD 7 are Fin-diameter steel rods. When the turnbuckle is tightened, the diagonal member AC is put in tension. Knowing that E = 29 x 106 psi and h = 4 fl, determine the largest allowable tension in AC so that the deformations in members AB and CD do not exceed 0.04 in. (Q3 Sin-Hes? Use Join'l‘ B 0.5 m‘Free body. 8 F3: FV/DM simupar ‘l’n‘anjpesj Lo RB_1_F® F» F‘" F» F” h '5 'y, ‘ T ' E2 F3; 5 ' Ear“ i 3 N'An 9e Geome‘l'Ii For ea, um? daton 4+l‘0145) 1 b A S“ = 58‘ A FELAIZB : ‘— CA): ‘0' FAB = 7; ' A: Fer. Equajihnj expressions ‘Fov‘ FAB, problem 2 26 2.26 Each ofthe links AB and CD is made of steel (E = 29 x 106 psi) and has a uniform rectangular cross section of % x l in. Determine the largest load which can be suspended fi-om point E if the deflection of E is not to exceed 0.01 in. Staci-{cs ‘ FN‘ E, = O Awe“. 6F IQIwPI Aaflm = max = O LequH: " L = 8 in. . _ FEEL. __ (1.5?)(m _ _ .6 De-Fov‘wmi'ncms. SAB- EA -W - LBDSQVID P _ F L _ 2.5?)(23 : -6 SC”- ?F ' (muu‘MOJS) Ziggwlo P DC'FIe—e'ifons. Pofh‘i‘ B. 53': 5,45 Sg=|.6552>‘10-“P Q POM C. 5; = Sm 5: = 2.753910"? L Geomeiréfi plumage. 55 C E B ‘55 F—IDIni— [gin—>1 — -C -6 e: 53+3c: Lessnxlo P+ 7.758exlo P = oflulsguodp LB; ‘0 SE = 3., + LcEe = Q.7s.Séx[D'°P+(Is‘)(o.'-I‘H38>‘IO"P) = 7-37fi3x/6‘P\ Limf'i‘ihj VAIUC a? 5". SE 1' 0.0] in. Lmflma VaJue a??? 7.3'H3xzo"P = 0.0! P: [055 J]: P: LOéé hrs :1 ...
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Homework 5 - 2.13 Rod ED is made of steel(E = 29 x 106 psi...

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