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homework3_f10_solution

# homework3_f10_solution - ME471-Mechanical Design 11...

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Unformatted text preview: ME471-Mechanical Design 11 Homework #3 Due Wednesday, September 22, 2010 Problem 1: Three forces are applied to the steel post as shown. a) Find the principal stresses and maximum shear stress at point H. b) Find the principal stresses and maximum shear stress at point K. Problem 2: Two forces are applied to the small post as shown. Find the principal stresses and the maximum shear stress at point H. 1.75m Problem 3: A hollow 4”X 8” rectangular shaft with 0.5” thick walls is loaded with a 10,000 lb force on the upper front corner. Determine the principal stresses and maximum shear stress at the points A and B. Be sure to include shear stress due to torsion. force application point A\ 4 in s in \ EB f f 2 NW QE— —4 B X-section 24 in 10,000 lbf |._.. _.| Problem 4: A vertical 20 kN force is applied to the end of the bar as shown. Knowing that the cantilevered tube has a wall thickness of 6 mm, find the principal stresses and maximum shear stress at point a, b, and c. 20 kN \zo,ooo n (Saw A = m "' 1-967 “0% = Ié—W M (0 am .agéfzoooouymymzmz __ [53.000 )0 (.Im) .02n) Llano" mq 8.64 x10" m" K (T) )( )( ) a) __ (zo.ooo~ .OSn .O’Zm 8.54xlo'bm‘l 3" Z7378 “H.57lj— 13-894 (C) 53%: 2.313 mm (T3 f'\ ’ \JO. (50.000103635me ) L , -— = " m .O‘lm Foe“ 'rt 8.;oqx10v6mq ‘0‘ M) = 9.25"] mPa /'\ mane":I-§C =0 CT‘O) _ [Aé 7 6x - Omm+039¢o = a — 2.3: 5 ILL 332. (“pa (C) 03 = o (Ema = 9-239 ”7/04 a. a = 0.....v0‘u a); a; = 'fliSCﬂ—ii‘Ym-W‘ “-17.179’: Him == lessen ma; (c3 3;: 4/538 m (T) "Cw = 11.714 qu 01%: [been 111% (a) __ lzo.omu)(.05m) .06» *(SQOOONZIMMJM) 35‘“) ” 8.91 xzo m“ ammo-w (a) (c) 0;“ :: “by; + 34722 = 72389 mpa (c) C :V&__ ZOOOOMYJZm .03» .msm . “‘5 Ic’ Z.I(a¥lo"’m" .uzm) = 7.197 IN": ’6 “oesm = O Ox = lean + "76.38? = 623.0% Ma (a) 073 ‘0 €*%‘ 11.12“! (Mo 0 :0x+0~+i Kan-o, 2+” z‘_ 6130s!» “2 2 z) (‘2 ' i‘ (73%“ 34.11.7‘ Z = 76.528 ir 46,714 0', g 93.24: mPa (c) at: 0.186 ”We (ﬂ TM“: 76‘7/4 mPa 08949: .— 4801b)! 3 25-04 ’1 75w! +qu‘b)(\$1N)(,7S‘wj Ji—(Z ‘Im)(l- 0N)3(+ IL(Z-‘(0)[I-\$m)3 (c) (L) Tm“: = \JﬁiL (600mb C new» _=~\VL5~1W9__ ‘ t ‘12.,(I\$.)(zqm\’(l 8-04) ZSDO Pg“ ‘0 (Tm) 0;: [333+ 25”.] = Zéy‘ﬂs’psi (C) a c '42 2W4 iGwN/y (2329 My 3222:: 2828‘! N : ﬁ(84(qm)3- +z(7w)(3w)3= 26. 9/67 "J” 11‘ tOIYBP— tam)“: 841.9147 m“ am: (8~>(2~>(z~> -(7~>(/£~>(.7s~) = am” he EON—E WWW“: 354.27 PS“ (C) u w" 0’ : (/QMSIN ZO°Q(ZW~ >(ZN ) . 390m 26. 91‘? m" : é097 Z 109L (C> 0,360”, :- (0m (05 70° 2m 2m) __ \ 269/67 MI“ b /37é' I/ F?” CC> O—B—Q‘Oz : 0 (8,210) from? = 0 (61=0> Z- ~ T __ 00,00051N20°’L)(4IN) 1225(1).) “'Z‘tAn Z (0.5IN)(7-S.MJ>(3-5;N) : 62" I7 95* o'x ;- 3541,27 + 60912»r max/z £349-87 _ 834?,81 , z 07.; - Z : Eff???) + (ssz‘zu/mﬁy: 4207.34 5, = 83823 9st (C 03; 32H Psi (T\ Tm: 1207.3 psi 69 POINT B \ 0mm; = 33127 psi (C) 0260! = O C‘ézc’) 0:39” _-_- (I lb (cos zo«)(‘lm) [4,”) 84.?157 m‘l 275% = /01000 S‘INZo" 8.1 (N3 26.9167“)! )(\ m) = /770.57 Psi (c) I. : T‘g(.b\$mX~\"b3"/2k60wﬂ)<ogm)3 I : 2.008"? Xlo'é m " Tang 7 2/.337 mp0! J! r5 ' T:(Zo,oc)a~)(.lom) = ztoo uvm t = 6 mm = .006.“ I401 ’ (OWNCOWMF 0.004136, m” K. .. 2 ~ L WScou " (00 N (Y! W = 42-3” W“ K" (x; “21.33? + 423/! a 20.977 ”’62 o“ = 20.977 m/é (T) (j; : 203’” MA (c) [W ._. 20.974 mp4 UAW 5 — {1% MM 536“” ' I = ”Maggi 3:3." = 66.714 mpa TM = Egg". 1'. 204119” COS-n) -wom Coll?“ 2.0087q0'bm‘t) \O\ZIV\ Z [1.é9? ”(pa l 7mm 5 4'st qu’f 0-K '5 65:7’4 Mfa (T) Tuna =‘Il.m4+ (123” = 3am ﬂ/a G517! L ( J : Z i 2 *30'6'11 = 32857it/‘mo7 . C5, : 77.74% ”7,061 (—r} 0;— : [2.05 ("Pa (a) TM 5 Lin/.907 mp4 614sz O 0 5 {1:3} (20.0069 .lSm . ’n 834° * zoos-mugs“? = ‘74. 975’ mg (1') («Wee ‘ O Z‘msw :VZBH mPa _ 7%é7s'+ 4.675 2 an, ' Z — C77 1- L/ZBIIZ I 0;; 2: 37.3381: \$6430 ...
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homework3_f10_solution - ME471-Mechanical Design 11...

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