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Chap01Student - LECTURE NOTE EGM 5533 APPLIED ELASTICITY...

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Unformatted text preview: LECTURE NOTE EGM 5533 APPLIED ELASTICITY AND ADVANCED MECHANICS OF SOLIDS By Nam-Ho Kim Spring 2006 CH1. W 11W 1W... Fluid (3:83 bada‘ Med/IMICS fiat—has {Degrmme bala- Sane! ‘ Rigid bad’é} MlC{D¢RPmalole (finch 1W a. ‘6an b~ KU‘E'HCS c. cons+i+wfiue velafiém (man‘Qn‘afl pmpefifi) 3W 4. Elemevrfana. v.2 651ng fleohamfcs 0]? fimhao E’— a) Elma/um ‘ one dimenst‘anafl Wme‘d‘ 'fintph’fyma basic assumptm ~ 6:) plane M-fidfin ~ érmplest s+hess drs-tribdwn b) Adwmced P firm-00A- 6MP: \ :1 ’29 ' i Bax/few flame}. Loa W (’96 9m A be M r» C Ba Y‘) I K ‘ C 1‘0 53 “5? 61" Ian r em aim 5 plan e ll b, Kine—He mafia sis / erfiw’iaé 4::3—4P @dWibu-ted 0M4 4 F<——-—! I—->[’ A 23:03 F=P E::] C. Mechanicafl Properka~ I~D Hooke’S Law 2. Beam a. Kinema-l-ic amaflasfs I) There au‘sts a nefimfl planeCN~A-) ("0 31W) 2) Cross—semen perpendfcwa» {-0 MA. W5 plane (We! perpendicflar '(‘0 MA, & ZEc=oi ZH=03 C. Hormn‘afl propeH-a‘ : <ch = E80: 3. C rrwlar $146494: 2—4 a: Kinemafic MabsfS ~ Pia/me cmss—serl'ron m-l-a+es M a rigid plmae linear tn Y‘ b Kih€+-IC M53 £7: MIA 2'9 Palm" mowed a? met-ha ‘3 C. ConS‘H'hxl'We relaiwn : 29.29 = Gr Yes A T=£Gng‘PclA = GLY‘SEiiV‘AA‘ '3 Egggr’zd’q 4: gummava‘fi 1‘5. Dispflacamw <————> @mefimm’ Applied \o‘ods 6-» 1mm fines waves; } Hooke’s ‘OWO For cfmlah cross—Saran Bar/m DB-fiecticn Using. MaclaurmS Egg; Bi ~Represevrfvrfion a? a Con-Hnuous 'Fn ‘30:) log a Power sem‘es «P x, whose coePficieuis are. aprermfl m +ems 0P +he higher derivatives (3‘? <1» 0:“ or=o. am) = \a’a) = 5%)?- -‘ Baumcb/ua Condt‘fians am. given 4:! 2:0 , such #0:! and-ms, yaw—a: , 3”ro)=:2a;, ‘6”’(o)=la_3,-w dam) :: " [390m a= am 3 \1w 5 <5: “\4-9 (7 §+H .. .. 4.413,, 3 4. 2&9 ( <90= mkmwn af 73 Elia. +710 51131 Determine. 9‘, 43mm addi-h‘omfl WW9”! 3mm _. 4&4,“ 1:2. 4 3W~9~1~72fi4fia% =0 600 = (901 ~%:13+ZELI¢W§ +213€Ié'b "a—gamtfi 0H9 Del-ermine 90 9mm (who mdi'h‘bn. ~ Alfiwafive 0111;1th : am Pmm 50:) am! infigmfo. 313“) = 2’61). 13333“) = P<x~a5’+ g<u~b>°—~ga~c>°——r1(x~a$" E13” = P<x-a>°+ 3<x-b>' ~ga—c>'—-H(x~d>"+ C. E13” = p(x—a>’+ §<x~b>£ §<X-C>z-H (1-d>°+ax +02 513’ {Ea-aft» got—bf“ §(X~C>3“M<X~d>‘+§1’2+éz+ Cs m 2 1.2111111411111915 <11>4~ 11111111 5111 211111114 "APPIH‘ Web/161 cmdrfi‘ons = __, V, 3 Zn 4 I 3 gm ‘9“ Z'EEX + 2.45:1 + 730%15~ 2%: (at->00 lay) ': o .. Va 2 i 3 g! a Q 251% + (BI X + 3.451 754—- 3%(1‘1>2 Bomda/‘Ua WW5 ”W = 0 = V°1~-§‘-’J2"+ Eff + @1414.) -.~. 0 I—g...(1(g°+&0LQQ-' (3J1+1(?e-za ).Q1+ 1(&+?flu1~ 1(&+&)‘Qfll)=o _ _. vs 3 a 52’ w —- o — m n JWH 721 .-o«3 "3- W. iWsmse ° T97: Sid/l +8512 “ S‘H’ESS ——Tme shes: (Hf—“g": .32 ~Tme sfmm 43:1» amen MCVEVHQWL' 5P, #8 191x514} dmges (95, db: ‘ihfi‘ni'fesimd Mcrament 019 31% rewlfina ‘B’Gm dP z. Hmrafi Pro perfies [i ‘- Engineering STress~sfan ’ ’ * ,7 ~ * r ,, ~ ~ ~ 7 'Qqa. ‘. elaan‘c limit 12‘, ' 03L I pmportimafi la‘mit VP!- 3 07%.. «if; : permnenfi strum Stress b" [M Pa] W) I Strain: . ' 05‘ 3 ulfimorfe WSFIe swam o E IMGdulus 0-? 2&3}:be ° Pencewf eloreafion = Efi'XIwZ (duCHIF-rg) aHOdH’Llj O-P RES‘UOME (51min anew ((914938) 2-51? (enema Permfi‘x/olume) c. Madam: <4) Eughness =0}? firm?» Bragg (49,5743 a} gamma. o Appmxim-Hon a? We 51% (19mm volcme Comb") 1" AtL 49mm ag=lm+e) L5; 1'4- WM -FM Can be deapmed usa‘na fires, am, di‘Spfl, . load, number 010 bad caches, eta. ~ Radar 0‘? 21%;; SF? = I.o~3.0. ~ Design ”3%. A) 4 .1823. F6?» -— SF Tefljat 07/) 11% (cad. . — Design w‘n‘h Rela‘abilu‘tg. >,./ SWISfiQzQ mr‘famre 2K6?) é ¢Rn (J‘P 'aAC‘ J‘ A”. K ‘ ¢ 31311th VamoWe 0 {+18 0 0165 _ 1. Excessive elaS‘h‘C deaect‘bn P af) 1% . 9(er AQ‘QSY‘VM'HGH «Bucklfna ~AmPli+ude 0P vibration a. Matefiafl Yn‘eldn‘na‘ Cfnelash‘c) 'Pevwxanw de%r~ma~fion emu! b3 du‘slom—eian m +he @W [90th -9h‘w?n hardening tan-ivmalWewG 0P clu‘Siocq-I-rans 3. 17me . ‘ Err-Hie madam? «Fafiaue 4. Bu cklfng. HWfiii §olve Prnblems 110/ Z. MW 1.26. A steel bar and an aluminum bar are joined end to end and fixed between two rigid walls as shown in Figure P126. The Cross-sectional area of the steel bar is As and that of the alumi- num bar is Aa. Initially, the two bars are stress free. Derive gen— oral expressions for the deflection of point A, the stress in the steel bar, and the stress in the aluminum bar for the following conditions: a. A load P is applied at point A. b. The left wall is displaced an amount 5 to the right. FIGURE P116 Bi-metallic rod. 1.27. In South African gold mines, cables are used to lower Worker cages down mine shafts. Ordinarily, the cables are made of steel. To save weight, an engineer decides to use cables made of aluminum. A design requirement is that the stress in the cable resulting from self-weight must not exceed one—tenth of the ulti- inate strength 0'u of the cable. A steel cable has a mass den- sity p: 7. 92 Mg/m3 and 0'u =103O MPa. For an aluminum‘ 'cable, p= 2. 77 Mg/rn3 and cu = 570 MPa. a. Determine the lengths of two cables, one of steel and the other of aluminum, for which the stress resulting from the self— weight of each cable equals one-tenth of the ultimate strength of the material. Assume that the cross—sectional area A of a cable is constant over the length of the cable. b. Assuming that A is constant, determine the elongation of each cable when the maximum stress in the cable is 0.1001,. The steel cable has a modulus of elasticity E = 193 GPa and for the aluminum cable E = 72 GPa. c. The cables are used to lower a cage to a mine depth of 1 km. Each cable has a cross section with diameter D = 75 mm. Determine the maximum allowable weight of the cage (includ— ing workers and equipment), if the stress in a cable is not to exceed 0.2001,. ...
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