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homework2_f10_solution - ME471-Mechanical Design II...

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Unformatted text preview: ME471-Mechanical Design II Homework #2 Due Wednesday, September 22, 2010 Read Chapter 3 Sections 3-] to 3-14 Problem 1: Problem 3-40 from your book Problem 2: Problem 3-44 from your book (show shear & moment diagrams) Problem 3: For the cantilever beam shown below. find the maximum tensile stress (principal stress), maximum compressive stress (principal stress), and maximum shear stress. You must locate the point or points where they occur by drawing the shear and moment diagrams (and labeling all points of interest). Use the cross section given at the right of the beam. 6in 2in Problem 4: Problem 3-71 from your book Problem 5: Problem 3-74 from your book (shear in y—x plane and moment about 2 axis & shear in y—z plane and moment about x axis) Problem 6: Problem 3-77 from your book Problem 7: Problem 3-80 a & c from your book Problem 8: An 8 in by 4 in, solid, rectangular shafl is loaded with a 10,000 lb force in the center of the cross section in the horizontal direction and at the top of the beam as shown. Determine the principal stresses and maximum shear stress at the point where they are the highest. force application point 22% 10,000 lbf r -3 q 3 4/0 I- 1(zy.)”- 3068mm W - "I _3333 “all“ Moog/om“ :I373«3 Hahn M mm“... [mww flm'mw 3 Sbo\\./. Sp — 251 “all“ Fifi? KM: “T 1 m ’ §3\Z.S'lL-\N)(-L5—W\ a] ‘ 510 03549 : if" 254615' 3.0680 m” m“ 8° Zzié$=53qo {05' [5M6 As (—3 \I —m Sfmmfiomaxl 8 500m 2 Z ML ————5 5m 5w SID 00 \s maze consaeVM'WE ”<5 M €$7\/~«1€, M 9401' mbeé Maui/14?. c) (5 (“651' AccoME A) (A 5) Age stmpuev "m mama 7716 Glfi’t’ CHOUCE. 1800 lbf 300 lbffll = 30060) +£1800 = 6900 lbf 2 30 R2 = 30090) —fl1800 = 3900 lbf 2 30 R, R, , ). 5100 3900 . l um a=—=]3 m 300 l<— u _.{ 0 .i' . -1800 M3 = —1800(10)= —18 000 lbf-1n R1 .4 10in B 30in A. -3900 szzm = (1/2)3900(13) = 25 3501bf-in M 25350 (lbf-in) ' ‘ 0 }:A\ 7=W=15 in -18000 1 3 _ 4 lin I] =E(3)(1 )= 0.25 In —-l l-— 1 3 . 4 . I2 =E(1)(3 )= 2.25 In 3 in '25 in Applying the parallel-axis theorem, . 2— ——-— _ _ . I, =[0.25+3(1.5-0.5)2]+[2.25+3(2.5—1.5)2] :85 in“ “ni— l—Ty— 1‘5 m 3 in Atx=10 in, y=—1.5 in, 0' rwrmm psi " 8.5 Atx= 10 in, y=2.5 in, a, =—fl:gafl=5294 psi (a) ' Atx=27 in, y=—1.5 in, a, =—W=4474 psi Atx=27 in, y=2.5 in, a, =—333:LS(2'3=—7456 psi Max tension = 5294 psi Ans. Max compression = —7456 psi Ans. (b) The maximum shear stress due to Vis at B, at the neutral axis. V = 5100 lbf m" Q=7A'=1.25(2.5)(1)=3.125 in3 (I ) _Q _ 5100(3.125) “m V _ 1b _ 8.5(1) (c) There are three potentially critical locations for the maximum shear stress, all at x = 27 in: (i) at the top where the bending stress is maximum, (ii) at the neutral axis where = 1875 psi Ans. fa the transverse shear is maximum, or (iii) in the web just above the flange where bending stress and shear stress are in their largest combination. For (i): The maximum bending stress was previously found to be -—7456 psi, and the shear stress is zero. From Mohr’s circle, rm =M—“l=—7—425—6=3728 psi 2 For (ii): The bending stress is zero, and the transverse shear stress was found previously to be 1875 psi. Thus, rmx = 1875 psi. For (iii): The bending stress at y = ~ 0.5 in is a, = ———‘180205(‘0'5) = —1059 psi The transverse shear stress is Q = 7’ ’ = (1)(3)(1) = 3.0 in3 VQ _ 5100(3.0) r=———————=1800 psi 1b 8.5(1) From Mohr’s circle, 2 rm = (—1259) +18002 =1876 psi The critical location is at x = 27 in, at the top surface, where rmax = 3728 psi. Ans. ('8 f); Tm =0 Z‘Tlfifiau 7' 0 58600: 59,000 lb/Inl (T) PONT a GM“ = 0 farce ‘0 (MIN 7' O A @ 61 FROM PtGKT BUD qflu Z ML! 1‘ {300‘1’H‘93N LfS‘(3OQM-> +' ER} (832)“): O Q 9E},- 89.10 H [m aM"; : {36x94 “(S—N3 (as %5°(3°DM*) - [250* 315—“) (7001“) ’ 8313(5505' fl) 3 mm W N) / 24 [MM-m) W 'W.37 lZ—x Poms] m" 287.: mu, ’ >< M — z 1 w L/7-37 + 32 ‘3 —' 57.24 N-M 086w» = (‘9’;- = (5-7qu N‘M\(.Ol M) 1 \= Z [El-Cow)” 7 88 mPa ’“ _ T: C'rbesmu ‘ T 2 «45” 30"" T a. TR -300023’>+ 95073): O 15])" 1—; _e .: 3L87S— N'M 2235.0” = 3\,Q’7§ N-IW 3(olm) _ 01350 '-= 72-88 mPc- am 1: Zo'z’q MPC‘ 6’) Assomwc TM 15 NEZMGABLE JAWL : O (Fm) 0x = 7288 (“Pa 03:0 (C‘V : 2019 MR: 72.88 72.88 1 z — 02w 2 $W— 3w“ V1.7! TM: (/1-7/ MP4 Kn) Fe F“ = -6.135 Z=CX:J*C}5*CVL A. Zmo= (V06 Xng ((1; x25: 0 43:“? a»; (398; 3803‘34 92 at 39232: woe/T): -Ho7L€/Z- saw - 352,4LUL — 3070.42 Vac xZ= 698%)“va 63;» Cy?) = 6.1304 wag; “3070.“! —-é.I3Cz :0 m ALLoweo Tb QO‘VM‘E So mew @v/v. To W526 121 -/4/(9’I,é6 — 3SL6'7/ + é./3 (X = o = 287. z I}; HA 5) 2F : 03-3618 =0 C: = 39,28 15 Z F)< = ox + 2812 428 =0 OX 3' ’lql/fi/[b ZFa: OZ 'SDOC} 4808.6):0 DW:]‘ U) \‘H "I W‘x 2C? a ”zip (4.0L ‘8 Mp 36238689,): Home-7 ND ll é? C> MW= l/Ié7‘+’738f72 = [3814 lbw TMX = (808\b)(3v86m3 = 3880 How“ A . . €550 " “1%: W: 97,49 7 (35" 1[I_c;3uu)w 5102 = Tr‘ = 3|3S.D “it,” (1%”) : 11,065sz N2 a) 0X 5 (,L H = 9753 + 3gz= /0.// ks“; 03pm .—= //.07 erg /0.(I lf— ' 7 f [5955)Z+//.a71=5.oe :lzfl fig Zm 0%: (58% W. ‘15M3+(Zée7u)(375m) " A} (”57"“) '-'- 0 AF 2 384.3 N Z M 03 = 0500936175”) —A£ [5153»):(3 Ag: LHH-‘I N Z F‘s}: 2384.3—2447—sez,uo}=o 0%:8653 Id Z—FE" 474-‘7”/602) + D? >0 0% = ”55:6 N @ Q fih _ 4:: GLENN)" — mac ts; (c3 (Crime = 81% :0 (610) {mar 3% :6§?\:{)SN): $09 (as; PONY $ Ommto 0—3643 ‘ Q ”9503 Z92c63§$%:(i0\b iii—{€52 igg> : 33:7? 3’; mm) (W3 P 6g ___ (10.000 sw "20° ls.) 036-0: (“3.00.0 ('05 20° ‘3)(2‘lqn) ‘ q “3) J ‘ No.51 NH 3 5295167 P561.) 3540-2. = W A 170.57 mt! 1‘ "3204:” PSA (T) Ox=10b99+5295167+ 320 64: 5713.19 F5; (T) k 03: 5713.19 A (T) 0-1. :- O ,- S7(?,l7 LW: 7 :- 235560 Psi ”—W’# 51132 5723: ~__._____‘______________ Pour?“ B “ MK Taxsugspza: amt: mass ’09»: (T) 0.3910“; O ( 3:0) 7: : 3L9: :UDIOObcos 20°) ‘1) ‘1)(2 ' 62C? 44: m—LfiL—L: ‘/‘/0~‘{7 [09.4 07,; =_ Eff—9: Nagy +(t/qoxnl'll‘ 53.44 t «4330 0. = ‘I’q'Z/‘I pg ('7') (57': 390. 26 IPSR-(a ..- 7 A“ = 973,70 not. POINT C. 6M4; = /06~93 [54' (T) 02am" 5285.572: Bird—69’: $06.3! [p533 (c) a” :20 53(5th '5 O 03<= 5206-31406. 89= 5"”?7—‘7’3 t *4/7qz/3 ,£[c - WW commas: ' 0; fl: ) $772633 M: J2, ‘— 0 ffvt TW" 2749—7 fsi ...
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