# File Tutorial6withAnswers.pdf - Tutorial 6 ' I \ v\‘4 W...

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Unformatted text preview: Tutorial 6 ' I \ v\‘4 W The offset 1ka has a width of W = .200 mm and a thickness of 40 m. If the allowable normal stress is (53110“, = 75 MPa, determine the maximum load P that can be appliedto the cables. Solution 8—21 M: Pf'ﬂJs-O) ' difﬂw P A = 0-2mm) = 0.008 m2 6 = iwmxoz)’ = 26.666700“) 6:“ 12 P MC 6: —— + --—- A I 0.150 P(O.l) 75 105 = -——-— + ( ) 0.0087 26.666700’6) P = 109 kN Ans Problem 8-24 The gondola and passengers have a weight of 7.5 kN and center of gravity at G. The suspender arm AB has a square cross—sectional area of 35 mm by 3.5 mm, and is pin connected at its ends A and E. Determine the largest tensile o? Solution 8—24 N55=0 7500 N 7500 N Segment AB: - P. 7500 0' - s“"—=W——*“-=6.12 IvIPa ( m)” .4 (35x55) . Segment CD: P 7300 on, =ﬂ=—:—- = 6.12 MPa .4 (32.)(35) _ .Mc # 3525(1o3x175) g6" 35 “Pa ”W=. r’. ; I 71—27(35)(35)3 (Um CD = at, + orb = 6.12 + 367.35 = 373.47 MP3 6-3 Problem 8-66 Determine the maximum ram force F that can be applied to the clamp at D if the allowable normal stress for the material is 03110“, = 180 MPa. W N=P 1 0-226 m i , 4 M=d-22-6 I I3. F P mg}; %at11m *— ﬁﬂmm 10mm x: .— 22?! ~- 0.04{0.01} + 0.06(0.01) A = _0.o4(o.01} + 0.86(0.01) = 0.001 m2 I = {gamxonf} + (0.04)(G.01)(0.026—0.005)2 1 + 13(0.01)(o.063) + 0.01(0.06)(0.040-—0.026)2 = 0.4773(10“) m‘ 02:31—52 A I 54:} 22A __ (omsxomxam) + O.04{Q05}{Q01} 8026 = . m Assume tension failure, P 0.226 P( 0.026) 180(10‘) = .. + - _ 0.001 0.477310%) P = 1352414 = 13.5kN Assume compression failure. —100(106) = 0.001 "- 0.4773004) P = 9076 N = 9.08kN (controls) p 0.226 Homo—0.025) Ans TRANSFORM ALL To WOOD, CALCULATE. (Lemarao‘x'ﬁ- Q (szoqurzmaowLwAr \LLlhi-ZKLYD_ ' _: ZonO‘Aq KLL.S+Z>:\BDK4‘3><%3 + \Llrurxzqoxxzq 'Jratsrwo"? '~: 6\ '25 4: ‘30 {(1:- G‘LB. “mm w 6-3 3 2; I‘m— "; quooﬂ +2qooquom—45 v2, Jr; Z[\BO>\<L¥%+ \goxqukﬁﬂ.%-%3> '2... 3 '2» ‘+ luwwo+ \urwmonMz-tzo) - ' \‘2... - 3’ ' . é ' 4' .' ' —_= .Ll-OS.C\%><\D "mm. ' ' , é M : 2.6 kM»m.—;zsxxo m-mm M‘fy ZSNOGVJM‘Z' .' 7 r M r ‘ 6"” = Wi— : ADSRQxWL’J w' “.035 /W ‘ A. \$435135 . 0742-51“) ._ . Gs : 'YN ail: 2 " 53 6 g _405R mo -_-;. gem WW? .: _ Em 7+0 _ A ‘ Trahsrgorm OLVL‘ to awwu‘ntum: u. ' 3 w % 3. : 3-2-5 Ty- .. - \Sffa 2m+.o%\$\0m” Maxlmm Wmanm Ska-es 0Lst (NC P‘I- Sn=\oo M/mmq'" '7 O ' L. ‘ a g __ ’ 1 w , IO M . M - 6A 34, /7, .. \DOngLLLL \$3.49 \57 - —— “,2? imam—mm W ——5 CHEUK amass. \«0 BRA\$\$r Maxim» OQUJNS at B ' g 6.1;. = 9111 =' W233: 2mm 1' 144.0%!xm'3 >l 60 ‘ \N/m . Mio‘ grew f- \ L.) 0 9W4“ ‘5'”: I Sirréss CO hJFTD\\$ M. Skim-.55.: h .3? \60 Z \\S7<M%Z\W "2.44.02:me3 \WCaOX’Zf-LLLDEXID LS “A21 '3 W—Uroxwos N-wm -" \-’Z_U.-~.(\Dé M~W~W‘ .d M Larﬂed- Pumiss‘MTQ behokn‘hﬂ .mmm is. \._’Z4 him-m ' 5.4; A wood beam is reinforced with ales] and bottom as shown. Determine [h stress devalnped in the wood and 51: jected to a banding momnnt of M = sires: dislributinn acting mm the cm 11 GPa, £1, = 200 an. straps at ils lop c maximum bending e] if {he beam ix sub— 5 kN - m, Sketch {ha 55 section. Take EV = I = §(3.63536)(0,34)3 —-1-1i(3.43636)(0.3)é = 4.17843(1o‘3)m“ Maximum stress in shad: (03)“! = "MCI = = 179 Mpa I 4.17848(10‘3) Maximum Stu-ass in wood: MC; 5005015) 0'» u=—---=*—-—'~——=0.l79MP A ( L“ I 4.17848(10‘3J a m i (0..) = nto‘WJm“ = 18.182(0.179) = 3.26 Mm .570 MP4 _ 34% HP. . 0'U9HP‘ Ans ._.-.. The member has a square cross section and is subjected to a resultant internal bending moment of 3 M m 850 N. m as shown. Determine the slress at each corner and sketch the stress distribution produced by M. M, v; ssocos4s' =- 601.04N-m Mz = 850sin45' = 601.04N-m 1 3 = 0.3255208(1o‘3)m‘ i, =5 = Emsxoan _‘M + M I. 1) 601.04(-0.125) 60LD4L—OJZS) = 0 Ans = h oszsszosues) + oszsszuscmr!) 601.04 (— 0.125) + 601.04(0.1'25) = 462kpa Ans = " 03255203304) 03255293003) 601.04 (0.125) + WILD-190.125) g _ 462 Ea Am ab = F 0.3255108L10“) 0.3255208(10”) 601.04(D.125) + 601.040.1125) :0 Ms "3 = " oszsszosuo-h 03255203004) The negaﬂva sign indicates compressive sutss. ...
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## This note was uploaded on 08/22/2011 for the course ENG 48331 taught by Professor Brown during the Three '11 term at University of Technology, Sydney.

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File Tutorial6withAnswers.pdf - Tutorial 6 ' I \ v\‘4 W...

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