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# lesson25 - I— V~"\Jll Pr0b|em 2 2 2.2 A 60-m-long...

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Unformatted text preview: I—\ V~"\Jll Pr0b|em 2 2 2.2 A 60-m-long steel wire is subjected to 6 kN tensile force. Knowing that E = 200 GPa and that the length of the rod increases by 48 mm, determine (a) the smallest diameter that may be selected for the wire, (b) the corresponding normal stress 13:91:03»), 3:43>‘IO—3m) E:zooxzo"Pa 5: SEE— , _ EL. _ (gxlo3){go7 _ —c z A" E ‘ (200%109)L48¥lo‘3 _ 37'5Xl0 m (al A : 72,10“ dtﬁ r/(H‘L—Lr-SE‘ ”Io-G) : BFHYIO" m gm.“ 4 ‘5 (b3 6 = -:— = ﬁrs/i9“? = 150 xzo‘ Pq teqo MPCLA 2.14 The aluminum rod ABC (E = 10.1 X 10" psi), which consists of two cylindrical portions AB and BC, is to be replaced with a cylindrical steel rod DE (E = 29 X 10" psi) of the same overall length. Determine the minimum required diameter d of the steel rod if its vertical deformation is not to exceed the deformation of the aluminum rod under the same load and if the allowable stress in the steel rod is not to exceed 24 ksi. DeFormaii‘I‘ow 0‘? Q/PQWU‘WUM V‘oé PLA: PL“ _ 19(ng + Lee) 1- _— __ AnE ARE ‘5 AM3 A“ Z : zsxm3 (42.— "5 > 1: 0,0313% ih SA= 10.! xno‘ \$0.5? +' 1!;(2-25)‘ Sleep T009 g 1 0.03I376 tin _ (23 x 109(1)”) - L . 2 PL. .ht 5 ‘ T “ A ES ‘ (zqwoonoaIsm) 0.72317. _ _ P 28x103 _ .L — ~' A - B: — ZLinO3 — 1.14.47 m A = 14647:».1 2.15 The specimen shown has been cut from a 5—mm-thick sheet of vinyl (E = 3.10 GPa) and is subjected to a 1.5—kN tensile load. Determine (a) the total deformation of the specimen, (b) the deformation of its central portion BC. Problem 2.15 Dimensions in mm A B 10 c D 5 : e 7 ~ A8 pu:1,5kx ,; _ i .- '-: 11:) l\.\ PL“; _ LSXIOingLLQfL__., EA", (3.:ou10°)(25x/o‘3)(5xzu“5 = 15434 >c/o‘é w _ PLBL _ (LSXIOslcﬁ‘Oxhp‘g) \$6; — EA“, - (SJDK qu lOOxld\$3<5x (0‘3) '1 1+83.%7>1IO-‘ m 19ch g“? : EACD (a) E+aﬂ 09€¥ofmq7iiot43 5 : §A8+ gu'i' \$50 '3 7q3-52>‘/D_L W) 0.79“} mm 4 :- Sng 7 (59-8%‘XIO—Gm (bl Oxier BC. gee = 483.87xl0" 03434 mm d 2.21 For the steel truss (E = 29 X 10" psi) and loading shown, determine the deformations of members AB and AD, knowing that their cross-sectional areas are 4.0 in2 and 2.8 inz, respectively. Problem 2.21 30 kips S+¢l€c\$ 1 Remal‘lons al A aw, C . 2.5 hips T MQMLHBV ED {3 q zero ‘FOWC{ me--Ler_ Lna: ./|3"+ 32 : 15.79% H = 123-172 in- LAD: :3 l—‘t .— /5é m. Use join‘l A as “Free L'OaL/ PM; H‘ZB = O 1 25 + {55—22173 RB =0 REL-~47. 70! Lap. . l3 Aﬁ‘ao ifF—y—‘OI Rolm FMS: O PM): 40.52; kt'rs 2 anLn-a _ (~47.70/>tl03)(183.l72‘) _ _ > £0,075 ‘I, 4 l? AA: (20' " l06)( ”20) K 3 H E05: - (HO_Q2§>el03)(I5é‘) : , — 0.0780 ' _ an EA“) (247 Ho°)(2.3) m ...
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