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hw7-10sol - “W £7 Sowfiom 4—2 The copper shaft is...

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Unformatted text preview: “W £7 Sowfiom 4—2. The copper shaft is ‘subjected to the axial loads so in. 75 in. 60 in. shown. Determine the displacement of end A with respect , , ‘ - ' . 2k1p 1kip to end D. The diameters of each segment are d AB = 3 in., ' dBc = 2in., and day = 1 in. Take E,“ = 18(103) ksi. The normal forces developed. in segment AB, BC and CD are shown in the 6 . .: I FBDS of each segment in Fig. a, b and c respectively.‘ 0 kl The cross- -sectional area of segment AB, BC and CD are A A3 = — :(32) = 2. 2511- m2 , E: ABC- — —(22) = 11in andACD = —(1)= 0.2511-in. ‘ (0:) V Thus, P L PM, LAB PBC Lac Pchco . 5A”, = 2—‘-— — + + . A i'Ez . AAB ECu ABC ECu ACD E01 6.00 (50) 2.00 (75) —1.00 (60) =.(2 251T)[-18(103)]+ 1r [18(103)].(0251r),[18(103)] -= 0.766(10‘?) in. . Ans. The positiveksign indicates that end A moves away from D. I :2 a=~ a) Thaw 4 4 (ma $2M® J, m» GWEN (9 1 E‘r- E \3.\ (7t!) =\3 NO 0001:“ M T I V l: 1w T" A 0mm M f L21“ W ‘3 “I” ”W W? i “W Es ‘WL 3H3 W W i, W( T59 wwmwa433=Wumwwmmagfi;¢ R): \OW W‘W W @W%fi¢;wm—Sgfiw=¢ 3 8319.03, = Fla 8 may = . W I 2‘ 8:“ Chink)” 1&003 RY (\fblumyUO WWW/M1} \ 1 2,. : 15 v 37‘ 030m ““44 X‘ Y L071 0 W W 2 {\7 1 figng “if“ 115(1) " ”(.533 —’60 w ‘ ’5" \ 03W?) WWW/M : ‘ 0.000% 0‘ch M gm) 1 ‘ Offlb mm WNW VQ, 943x? W M AWN 9w “LA Yr My WW \8 SMU \bvu‘ guy \nu‘ s (WNWWWWWWW 4—18. The assembly consists of two A-36 steel rods and a rigid bar BD. Each rod has a diameter of 0.75 in. If a force of 10 kip is applied to the bar as shown, determine the vertical displacement of the load. Here, F El.- = 10 kip. Referring to the FBD shown in Fig. a, C+EMB = 0; Fa, (2) — 100.25) = 0 Fa, = 6.25 kip ‘ C+EMD = o; _ 10(0I75) — FAB(2) : 0 F33 = 3.75 kip The cross-sectional area of the rods is A = % (-0.752) = 014062511 i112. Since points A and C are fixed, 5,, = F‘BL" = fl— = 0007025 in. l A En 0.1406257T[29.0(103)] F L 6.25 3 12 5,, = CD CD = _______(_)(__)_,_ = 0.01756in l A Eu O._14062517[29.0(103)] From the geometry shown in Fig. b 85 = 0.007025 + 1522(001756 - 0.00725) = 0.01361 in. l Here, 10(1)'(12) 4 =0.009366in'-i . FEF LEI? = 0.1406257 [29.0(103)] 5371-: = A E“ Thus, (+1) 5; = as + sm— = 0.01361 + 0.009366 '= 0.0230111 l Ans. (AWL: 015’ E. _ V535“ 4.4%; Any «CUE—(5‘0 ; ‘22 1 Kay/5k ; \A‘\S \va (“9 _ Au; “(0.7775010 E} .\O\L Axially Loaded Member Elongation II . The figure below shows a tensile member with multiple cross-sectional areas. The member is sub- » jected to eight different forces applied at five diflerent locations.- The modulus of elasticity remains constant throughout the member (EA = E3 = Ec = ED). . Rank these situations, from greatest to least, on the basis of the elongation of each section. Kl , Greatest 1 C/ 2_A__ 3-3— 4— Or, the elongation is the same .for every section. . Least Please carefully explain your reasoning. _ 7 , y - . s 'P - 5 “ \ \ ’le'k ‘V L ’PL e- .) l : ~——-' I ’2: —w— A) ’ v? ‘ SA A¢EA <\ NEWE- I E A ' ’ v V [\3 3 : "" Av E3 (110) .E 4 E HOW sure are you of your ranking? (Circle one) V Basically Guessed 'Sure I _ Very Sure 1 2 _3 4 5 6 7 8 9 - 10 , 21 ...
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