Homework_13_and_14 - Problem 102 10.2 Knowing that the...

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Unformatted text preview: Problem 102 10.2 Knowing that the Spring at A is of constant k and [but the bar AH '- ' detcnninc the critical load PC... I“ “W. Efflaro: FLcose — Px = o P kLzsinQ cosQ - PLsi-«D = 0 Using sir-lei Q 4.54 case 1:13 k‘L-‘Q — FAQ -' O (um—PUG = 0* POEM *‘ Pun l5 Propel-{Unmet} +3 I, Few Staff] “Gaga C'2JiaPJ I525“ -1 _ —.. C ‘- afgp)' FISMM Is: é;- (Is)?: 39‘7gfxfo MMH‘ ____ 76(yjo-9M., J:- -— I‘. "w my“ "0"; Cz=ii L In: #(c'}—c-’) (Pfih)h I11 Cy- C fil C sf. d -——____ -_-_- — r 4.. __ _ J _ x - 'f (Pan I: c‘“ “ I (c) * "ka _ f T - lfig") - I—re o , . . MM: 1-}; Mafia 515370 4 P r E. WEI; r15__7r‘(iosxro“Memento-q) cu» [6 L7. #6 “fir—m '-= 17.17 3:30 N PW - 17 I'? w «a Problem 10.22 10.22 Column AB carries a centric load P of magnitude 15 bps. Cables 3c and 30 are taut. and prevent motion of point B in the xz plane. Using Euler‘s formula ~ TIZEL _ _T_r EL Par ' L' 2 “EF ‘e Shuffle-f Uafige ‘pmf‘ L jaderrs. and a factor of safety of 2.2, and neglecting the maximum allowable length L. Use E = 29 >< 106 psi. lat/10:21: p .— 15 >610,5 9b. PC, —. (ES.}P= (magma - 33m? 4% W'- E 2 me tension in the cables, determine Id I“: Ham”. = |I_‘-l in" Buwu‘n 'In XZ- ppane. Le E 0.7L- _ FZEIJ H E P" ' {onw- * 0.71 _ l (gagioslflléll m . L“ 0.7 ‘/ 3:334:03 ‘ Wq‘z "‘- 3/‘_Z__:epane . L: = ((Z‘Nlo‘flilgd _ . —-—._____33x!03 .. 595.8 H1. Lr 4%.2 :n. 37. 4 4??“ «a _10 11.10 .Using E = 200 GPa. data-mine (0-) the strain energy ofthe mm m ‘ Problem 1 1 when P = 25 RN, {5) the corresponding sham-energy denstty in portions A}; of the rod. 20-mmdjameter - Am: atzoi‘ r sums m3 = swung-amt 16-min diameter Pier. 2 3.106327 20:.05 mm‘ = 201.0é'rio" m‘ 1’ p: zsxjos' N 2111 I _ EIL_ U ' 2 25A mummy (gifizoouo‘Xamc no“) (zsxro‘>‘(o.8> + (23(200210“ )(2ot.06n!0" ) (Q U = 5.953 + (3,213 = 12.18 N-M = 1248 I as (13) GM: 3. -_- fl: : 7c;_53x}o‘ Pa. Am 3.: this x to 1. 1. L.A : ELL: .(_79-53’”.°€)_ = (segue; : 15.83 Id/ms -- a 2i? [21(200 v: to“ } x 5 6M=-§: = 275% : l2‘i.2%xi0‘?a .5 - = 6“ _ (121.22.;30‘)‘ _ , s 2 8 5 __‘ u“ .27? _ wwwoouoe) _ 33x; 1'0 3 .6 kl/M Problem 11.11 11-" A 3041!. length of aluminum pipe of crass-sectional anaa 1.85 in2 is welded to a fixed supportA and to a rigid cap 8. The steel rod EF, of INS-in. diameter. is welded to cap 3. Knowing that the modulus of elasticity is 29 x 10" psi for the steel and 10.6 X It)6 psi for the aluminum, determine (a) the total strain energy of the system when P = 10 kips, (b) the corresponding strain—energy density of the pipe CD and in the rod EF. . = PZL t. (toktoyiTSo) _ . QT} Um, ZEA ~(—————._2movéimcm_8$) - 76.49 Wit; 3 2 F: Us: ELL - = 527.32. shdL. r 25A FRMZfim‘Meywsfi {a} To-iaf-i U = Um + UH .- U= ZéHMJL 4: . . __ ioooo _, _ — . _ 53' _ ~5405' Y" _ . .3 (in) C . 6- L85 4 590.5 P513 Li - 25 — m _. J'.37Q m—JL/m d F : _ 10000 = 3., 7 £1: ZZESS 2' _-_ ; _ ’. 3 _E__ 6’ owls 22635}: J u E —~————(Z)(2qxloq 2,33 "JL/n .4 Probtem 11.25 11.23 through 11.26 Taking into account only the effect of no determine the strain energy of me prismatic beam A3 for the loadin rm“ Slresfica. 3 Show. SJMMF§P€¢ Eco/v. 61311.4: 106.4;an +T7Fj30: PL+PB—‘ZP:O 933'Q-srp Ow’r lflov'i'l‘on AD 1 M 7 RAD-C = PX a - M 2 _ ‘P2 ‘1 z ‘ Pa 13 61 fl' P2031 DAD ' SO 15): oh ' 251 80 V at“; ' 2E1 3 JD _ GEI . . . e _ Pit: (L ~24} OUPV‘ Par+son DE_ - M — Pa. UDE - W” __ t = _ - _ a OUev‘ Boy-haw- Eg - SJ’WWDT’U U51? ' Um " 65]: 2 To+a_,9: U = UM} + UDE + Use, E 123: C3L~quj “3‘ probjem 1134 I 11.34 Rod AC is made of aluminum and is subjected to a torque T applied at C‘ Knowing that G = '33 GPa and that portion BC of the rod is hollow and has an inner diameter of 16 mm, delermirle the strain energy of the rod for a maximum shearing stress of 120 MPa. Ski-mm diameter C C°:%‘3=|2mm) Cg=%:8mm 1 J33 = 36:: 120233 32,572,:0‘M“: amaze}? wold A: (Cat CC.) = %Oz‘- 8') = 25,133er' raw" : 26. IBSM'DJt Mu : 3 _— J:..f.ar ‘ 29.:38xto‘quzono‘) : I 71” J“ T C ~ “(Hog 2.9.33 N m T: Lu: C2él.3*8)1('-iooh!o'”' 3 : “‘"— ’ = _ u? U” ZGJm {ZK'IBXIOWCSLSMHD'U 5 7 J - T‘an (2;:.33)‘(Sooxso-3) _ U“ ' 26:5; 7 (27(73xle“)(2c.133 moo-1) ' 8.95! J Tp+ml7= U= .U.G+U,¢ U_=M.7DJ a! _..__.-— .—.u Problem 1 1 33 11.38 The state of stress shown occurs in a machine componenth steel for which 0y = 65 ksi. Using the maximam'l-diSnorti(:)n-c.'11v::rg3,r criterion determine the factor of safety associaied with the yield strength when (a) a) = Ht»; ksi, (b) ay= — 16 ksi. (a) 6; = (SJ : lé. ks; (|8_5g+|o_$c)z+ (Hose—Ia): (16- 18.50‘ ‘ 2 f—iY 847.q7+ 703% + 6.55 - fix Pi, 2133 __. (b) 6; : 63. : 4c. ks; ([3'5‘¢+,Io_5{,}1+ (40.5w :4): 4, (4943-54. )1 = 26%)," 8q7,?7 + 29.59 + {£99.39 = ($5311 PFObiem 1 1 .59 11.58 and 11.59 Using I115 method ofwork and energy, determine the deflection an point D caused by the load P. 92MA=O: P¢+PEL10 Par—E3 Over Ecrh'aq DA: M: - PX 1" W 2‘ °‘ 1 1320.3 Um : So 251 “9“ = .26: [a x A“ " g5: W3 M = — L L a a L 1 Um 1. 5 M J". 7 P 5» S V25”, : a ZEI ZEIL‘ c. 6E1 13+»?- U= Um + Um — 021;”? .L _ _ $.14 Rflm L3 P59 U as P 5,, 39: $ :- 11.103 and H.104 Each member of me truss shown is m d Pmblem11‘104 and has a cross-sectional area of 500mm? Using E :3 zcflgfflfi determine the deflection indicated. ’ H.104 Horizontal deflection of joint B. Final Hie 19943111 6; earn-fl men he» es shouh_ AAA hu¢i20n+qP Q qJ+J-ol.v‘+ B- +- ZE=0z gag +§Fm—3,G—Q:o 38 Fan +1 Sam: éEB—égo—ma :o Safw'na sg‘MwE-hueeusly PM; .— 615 1- 0.6256) kn 'Jomt D F“ 2 +1.15 + O-62§Q kn Vega“ “213:0: E. F“, - - éFGD : 1.05 - 0 375:? (lo, N-m] 6.25 + 0.6250 AD [.05 + 0.375'61 ED - L75 4 0.6159 —2_FS75_ 6.0 o 0 L+4589 1 = J— F 9F L r" ———————__.__‘q 630'“) : qg_8>«ro“m 53= 0.0363 MIN-1“" "'1 ...
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Homework_13_and_14 - Problem 102 10.2 Knowing that the...

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