HW2 - ,__ P I 2.] Two gage marks are placed exactly 250 mm...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ,__ P I 2.] Two gage marks are placed exactly 250 mm apart on a 12-mm—diam ' romem 2 1 aluminum rod. Knowing that, with an axial load of 6000 N acting on the rod‘ettfi‘: i} distance between the gage marks is 250.18 mm, determine the modulus of elastic-mr - of the aluminum used in the rod. ' g = AL = L — to r 250. IS - 150.00 = o. '18 mm ‘5 _ $5M!” . _ E : T ' agourim “ 0' 00°72 _... T .1 , _£ 3. A r if J2 r a," (in) = i13-0‘?7 hm" = [1103? #0 M P 600 o 6 2 if = = 53-05?“ ’0" Pa- 6“ 53 05:2 XJO‘ 1 a ' :—-—_~———'—-—-———-—=__¢|OP E=3.GP : E 8 0.00071 73 68 5‘ .a 7 7 CL «III- I _._....________.__......_.__...._. ! Problem 2.3 2-8 80411-10118 wire ofS-mm diameter is made ofa steel with Ea an ultlrnate tensfle strength of 400 MPa. If a factor of safet H 200 G determine (a) the largest allowable tension in the Wife 5 y of 3'2 [5 dc: elongation of the wire. ’ ( ) the Correspom.‘ CL 2' DO 6 -— _' _ 7. H 6“ L’ “a 9“ A'fidl—Ef51=rre.essw‘~ ' {9.7535 No"- R): GLA a [400xl0‘)(1?. gums“) r 725% N I a R) 725% PM" . ' 3.1 Re can 5: PL—W __ _ --3 AE (immSHFO-cjcgooxiog) :: 50,03“) 1m 5-2500an w :' N Pay”: 2.145 kw Problem 217 2.17 Two solid cylindrical rods are joined at 3 and loaded as shown. Rod AB is made of steel (E = 200 (We) and rod BC of brass (E = 105 GPa). Determine (a) the total deformation of the composite rod ABC, (2:) the deflection of point B. P = 30 MN Real AB: FAB = -*P -— -30KH03N 30mm Lm : 0.2mm as : 200x10“ 5P9. 40“” titJma = 14(30):: "£06,85th : 705-33xxd‘m‘ 30x 03} 0.150} 50mm 3”: M — _%* Em; A” ' (200 XIDQ}(706.3‘SX to") -53. 052x10“ w E”! BC: PM: Bow-10 = 70 Ma r 70x10! N H L54 - 9.300;“ EEC.— [05x {9* Pa Add = 3050)“ = 1.9e35xzo3mm‘= League“; 8 1. Far. LB: h (?O¥103](O-goo\ 6L _____ “ _' Eat/Ago (Ios‘x tO‘XtfléSS’tlo" ~ -101- 8551 Ha“ m (at) T041“? plepowma+tnont (la) 54.5-1- : 8AB+ SQ; : “IsLi-qytlfc'm 1' "GJSHctr-nm“ DCP}80+;091 o? Eofm+ B, 53: Sat 58: OJth mm ‘k"‘ Problem 2 20 2.20 ThejodABC is made of an aluminum for which E = 70 GPa. Knowin = 6 kN and Q = 42 kN, determine the deflection of (a) point A, (b) pointg, P ____ A I 1 Ti ’- -c 1. 1 Am = q 3% .- ‘q(0.020) -‘- 3H. M We: m 20-mmdiameler AM. : : 15":0‘060}1 T 2' KID-3 m2 0.4m ¢ P PM :- P = 6 V103 N r Pg; : P- Q = gm?— wuo‘r -39.!0“ N '_ w dj t Lara : 0.4M Lac: 0-5- M 1 8A : PflLm _, (leofiflo-q) Q a AME. (3I4.Jé>'!0")(?ovro“) _ :- IOfiJfi-S'W" ° W l a. g = P5CLB¢_ (rseuobflofii EC. A515 (2.32?‘f'yf§3r7o Hoe) : " filo-.ql'f7 VIC—5 :41 (an 3a = Sm + 55.1 r W. i3§*i6‘-¢io.747xzd‘m r Jamie" M = 0.01851 9mm? '4 3L3 -' 5:3; 1 - ?o-G¥ *{O-‘ M = "' MM 0" 0. OCH“? mm 1! '4 Problem 2.26 2.26 Each of the links AB and CD is made of steel (E = 29 x 10" psi) and . . l . . his; umform rectangular cross secuon of 4 x l m. Determme the largest load win. can be suspended from point E if the deflection of E is not to exceed [101 m = O Arenarfinki A:(—‘q)(r)= 0.251? = O LequH: L = 8 Eh. . _ HQL _ 0.5me _ , _ .c ———~—-————DeF°pm“+'°M‘ 5‘3" EA - L'MHO‘MOJS) ‘ I'Ebfiwo P 8:13 e L" L ‘ gz'gmfim = 2.7586540”- P E ' (Rfix:06\(Q253 DeHml-om. Pad 8. 53-. 5% 53: i.css:z»:o"m Pond C. St = 8m 5: = 2.75%esro'WL Geomehét eanaa. 5 c ” SB ‘ ‘= . g5 Pawn—1H (5';n.———»4 e = 3:1: Sc : Leg-fl no“ Tc; gnsgexmlp = O.HUI3SHD‘C P gs = SF + che = 9.75.86ufo'6P1-(153(ofl‘fI33xJO-‘P)= amaswd‘f’i UAflI-Je a"? 5.5: 5.9 7 0.01;“. Limknj mice a? ? 9.3793H0"P = 0.0: P: 104; ,9}; P: Loss ka5 4 ‘\ 2.27 Each of the links AB and CD is made of aluminum (E = 75 GPa) and has a Cross-sectional area of 125 mm}. Knowing that they support the rigid member BC, determine the deflection of point E. EB Fan USE mam her BC as q. 'F'PQQ. C 8 5)::03'N — (can) as + (o.qq)c$x:o’ )= o F”, = 3.4315323!" M 1—1:, = 1.5625 H03 u I 2MB = o '. (0.6.4) Fan “ (O-Q‘JKQ‘IOS ) .- 0 pp» ,Praks AB anal an, A= ii25'mm1 = 120104.." a 3 ' =W : 1310:: do“ m = 5e- (15%]qu +25‘x l0“) - L—L—"5‘25*03)(0’36) = £0.00x104m 3 SC. (75xlo‘x lu'xlo" ) _éu _ Sg—S: _ 72-00 “lb 549.0le 9 “ —'—“Ek — 'O-é‘h = 112.5 No" m! 51?: Sc+ PR6 = 60.00 no“ 4- (0.49 XML: no“) SE:O,IOq5‘MML -‘ gm”. = lo‘?..5' >40" in ...
View Full Document

Page1 / 5

HW2 - ,__ P I 2.] Two gage marks are placed exactly 250 mm...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online