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5 - Tl I ’A =6ftismade T-bT w Tarn-TT 1...

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Unformatted text preview: Tl - I. - ’A eompositeshaftoflengthL=6ftismade ' T-bT. w, Tarn:- . -: - . . .-TT.:-. ' ._ 1 1. -y.slnink-fitting a titanium-alloy sleeve.(element 1) over an :3.ch 2TH; .T- '.T -- T T. _: - -- r._ .- 3 T aluminum-alloy core (element 2), as shown in Fig. P4.6-7. G1". 5 X.—1.0’ ksi, G; = .4 x IO’ksi. (d.)1=225 In”: - _ - ’- The two-component shaft is fixed against rotation at end B. (d!) = d = 2 0' m :T‘ and a torque T4 = 20 kip -in. is applied at end A. (a) Deter- ‘ z ' ' Tr! mine the maximum shear stresses (1...). in the titanium -' sleeve and (1..)1' m the aluminum core. (b) Determine the ' - angle of rotation at A, ¢4.- ‘ fife/7,290 .’ (a) #742: lug am Mew .r/m'rcr. m =0: 77+ 7: = 72 m .5 . 5/6?» an / I‘d/fa: — fia/I'J’I‘ 561a Ufa/f '. Z . . =/— «71 ‘ rd '3 é 757/ }(z) L/Agxg 7g 44) 7751/74). ’ —/o/T) f’] da/zéfl 4994;» 3 64 = 74: 7.: 7(_ /_z_) 2’2 4 H, 4T 4 04 )‘z 'W ¢ = 0 O/I/ii’ 12/ m. 33 flame/r2 aé OéA/Mdfifl/l: ' T T TI ébT =¢z = 9’54 m 3 . FT é»: Ma, {/1 m ”.../M 7— 7% 7;; ’7 s 25%? 41/34», f 7477 7M: ,T/x; /7) '7? Manx f/I’trfcf.‘ TTTJ [Ml/z] /6 77(4). 7274: E 77 = /_0. 57/ KI}: m, __,_ ...-... . _ _ ... _-..:-;TLT-_-.:.:-.- .. T-_-- UCLA— : _____._ 1}: Tweak/xxx? Q 7:. 4 3 (771“ )z Fb/LJ : 4 6 i7? 7/ (5) flak/6‘04 d/IQJ ¢9. ¢¢=5é:74,77‘0./20¢f/é/ . ‘ . .. . -a.'fi.’Z.-— ”I: . .. El: Prob. 4.11-2. For the shaft in Fig. P4.11—2, (a) determine the value of the yield torque Ty. (b) Determine the angle of twist, ¢, for the following values of torque, T: Ty, 1.1 Ty, 1.2 Ty, and 1.3 Ty. Sketch the T versus 4t curve for 0 s T S 1.3Ty. (c) Determine the fully plastic torque, Tr, for this shaft. 60 . I 1:2 _T"1Yol5 ’lTflg gs'.)g5,0tn.23 _ , , TY— he ”.9 ” Wm If 015.4 klp' in. b) l; A; T . Bksu T: "’ [4(2)er5] G=Y$=OOOWI2€LW Y “i“ W ¢= eL= (CL)? fi/JE—JEF— MW GMT =TY3 = 5 443%“? “ [0 2r 1% k5! 'm' ‘I . ' 5 _ 12W. . .5 . i; . ¢(T‘- LIT—Y)" 55% " l ’Tl' l8 1 M = Q.gfilmz4 mi 3 . ‘.' ¢<T = [:2TY)= M— 2r” ‘8 l m = c C! flflia I W . 5 (5 om )3 to 1'5) 45 42km an ____ (NT = LEE)“ J—r— WrLi gum age. the ‘FolloWlhOJ pajc ‘rFor a 6kc£cln oV-the toque-kwist curve, a? 4. I l— 2 ( cbnt’a. )-- 5- 3.6%: a 7; 0.388888 (u d .J‘a: .fi‘i-‘E ELM ' Torque mm.) -1. } :«L'f- 311' 3"l L '.V ’ NJ? »; 1 n. . ‘ ’.‘ 0 0.02 0.04: _ 0.06 0.08 0.1 0.12 '.Twisto'ad)‘ I . C) iikgcfl): L 4T! = HHaHélbkigvin.) = '5 5 TP = [211,2 kip-in. /"W. I} I'.‘ '1 Prob. 4.11-8. The two ends of a solid circular shaft of diame» r , = in. and length L = 18 in. are rotated with respect to each other by exactly one revolution (Le., ¢ = 2a rad). The shaft is made of elastic, perfectly plastic material with 7 versus 7 curve as shown' in Fig. P4.11-8b. (:1) Determine the maximum shear strain for this lowing conditi_9_n_ _(b) Determine the radius of the elastic core for this loading_ con- dltlon. :L'y (rad) y, = 0.002 NJ 1-8 and P4.11-9 X)m CROW! gwmeianj 0F dJ—wmab‘on‘ mex = o. 1521 moi 95W 12 _K__SI G= XY = ooozm fluidlfl. atTarst 15w 12km mm @4575 ‘ (.103 ksifi/z InJ/Z “Wm =>6ln0e (1)?ch 4 4), there, 16 some Plats-be Uiddmfl' xmwfifilm =/e =Gr‘y __ TyL Fr ' C35: . I2 I In. =0, 105 n It. L'QQOElQfllLlD'. by = 53500—3) in. “:HV'TTFEW - Prob. 5.4-6. An bshaped frame ED is welded to the beam ABC in Fig. P5.4-6 (see Prob.'5.2-11). A downward vertical load of P = 5 kips is applied to the frame at D as shown. Let dimension a = 2 ft, and neglect the width of the connection at B. Plot the shear diagram and the moment diagram for beamABC. ,fi/k/fp/L' fill/'1‘ 1/1241») ,' 5/ étam 36; WW6 -—o: Ame) + Pa 4%) =0 ' ’97 '54 a a ._ IDA/bk] Jw/(ZMA ‘ .' (,[ral -P/24) - Pa =0 E 3 V0) 3 '9; x (J = ?P 6/f67fl16n/fi€(0<1(<6)) pp“ A M We; we: = ip 4 1) “x: I f 4 24. —" ' KM 9711);“ /)),/x)=,q,( 33::Px ) K For ffjmen/KC [[4<,((_{'4)/ , 3 . 277 0: K4): Al—p; "El” Pp“ War—35F) (EMA =0: Ma; : 47x +P¢ — P/x—Za') letl" 3/0 “ii/9K V fax [4. i; 435$ Prob. 5.4-31. Sketch the shear diagram and the moment diagram for the beam A5 in Prob. 5.4-11. fly? I baum A13 + 4 4 Wm (2km: ”(W 2(2M)b= -:o Agfl 2121M] 26%: 0 AB: 0 5 m A D B 9(2M).= oi 2kN(5mW' 4 kN'm 5 D} I” DU (4 MY‘M “II A kLm+m ——ak—Im—>k‘—lm—>l x r vm (m? m VCO‘) =13. (2)AVA=A.1'* V(o+‘)=’ o+os =_QéLN_ (3)1?oro4x<4m, aggipmio. I<IIAVI>= up =>VHm+)= 0. 5+1. 5— =L|sli 2!, _ _ Mi (sfior 4m<x4bm, x ’2qu _Lm. Imam) = 2+ Him“ = .0. X(m) (I)M(o)=_Q (simpl Suppor-Lcd beam) (211% o<2<42m x= V00 =_O_5.w x011) (5)M(2m")= 0+ EVMMX = lgN’m MAM;- ‘lkN =>M(2m+)= 1-4 =m {sfipr 2m4x<4m, 3L V00 =mm (HM (4m‘) 5 ‘5 *E VoOAx = m (meor4m 491410?" “35% V00, whlch IS positiw. 01ml lineaxhj Awash bfliVeMMHheflwem/l 6km”; (mam) * -2 + LVooon =.o. ‘Prob. 5.4-36. The distribution of the force between the bottom of a ski and the snow is approximated by the piecewise-linear distribution (force per unit length) depicted in Fig. P5.4-36. The skier’s boot exerts a downward force RC and a couple MC that result in the load distribution shown. (Neglect the fact that RC and MC are actually distn‘buted over the length of the boot and the bindings that attach the boot to the ski.) Sketch the shear diagram and the moment diagram for the ski; treating it as a straight beam AE. l2 in. 12 in. P5.4«36 93/4/40” .‘ lt/m ed flz, Vie/«tr 0/6 an o/Mc . flhzc fl: /éa// sly/412’» ff»; (/é/I'Cr [Alix/j flm 0° fa [’ ("10/ fl," (/00) /7 u/V/Jc are/M fa flu: :xfl/b/(K/y/ern‘mr y/D/ V/x/ /'/7 fl‘ft flue rc/rmcq/r a/flt 51/) [6mm [a ffi/I/Zrl'um ,‘ z Wm. #35 =0; 4% +(za/rn, Wag/M +£(5/m.l/!z/>..)(zl =0 IQ = /96 /6 4&7”); :0; Mr -{z/lfli.){6£/h.)fa¥r,i —i(r/1//a.)//zm.)(¢,z,l f z' (fl/m.) (/25. )(¢/;..) = 0 M4 = Z72 flak. 6r zw’n. < x < awn ,1 — r K L/L{1)=2x +Zi/X'Z4ll i f L_7£____, W» -Z(a5—x) “aha—xii lh. - r Z T/,(x}= 'fix #le —6/é f6: nac/ flat/5c f6/ [44%; 4nJ/Mpm :n/ fi‘éi/amf’ 20ml. £444 [tan/f) flax [J/QZHM (I) V/olm JV (2) fiv 0 “52413., $‘p/IJ‘Z IJ/IA, . u- V/xl [/1] ' [fJ‘Wz4)=04/pm.a = 0/4g=¢:/6 JJ 1 (5) (1) Fr 14/», <X<36/./a 12:,”1/1) (-51)£ [02 m 144; ,3 fa ddaf/z {frV/ié’) :46+ 1.me = 4£ “.4130 s/bz/K at 45 a) 2114 = '/76/6 => ‘V/rafjr mam/whim (7) ,6, 3M. < x5 43/3,) {I} 6.? Kfl'nf] J: (3) View) rVQa‘VLp/ucl‘ =44 +5'4=—1o/£ {fl/6r M1,}, 5‘5“]; 1'31? ”pk; 3/1/31, V310 l)’ (7) -74 ZHW/N‘f/C {10) Wk J=Vz4zl+ffmc1¢ :— *fdfwr—o Wm" ll) Mama JM {2) fax 05x Q41)”, .7; -V/x) = 2;: /A MR) [/é-inJ {iJ ht/MFMon/"VmA-a+1145)le =r7é Ill/n, / {41 5/24/44“ 2m.) :9 £ij Km ‘gxz—Jx P/zalll f6 (flflflic') = M/lf)£/1M41}Jg :rmm ‘ mm M, {6)AMC:-/"C==—> mm ”14- :72 ”44/144. /7) Fax 96/5. we Mm] .7? ”=70! Zap} ={-z§xl+zu -6/()/1 (g {5)M/451‘m0uhu r4011; = #4? ~ 744 = 400 Ill/n. (06/ m,szsam 4;” w; [luv/JIM ‘6 (/0) MMJ' m/re) r/fl V/FIJt = #00—{/¢o)/u)' 0 ...
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