202hw13_soln

202hw13_soln - 1 1 1.1 Two solid cylindrical rods AB and BC...

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Unformatted text preview: 1 1 1.1 Two solid cylindrical rods AB and BC are welded lugether at Band loade Problem - shown. Knowing that d, = 50 mm and d2 = 30 mm, ﬁnd average normal stress a midsection of (a) rod AB, (6) rod BC. T_ (03 R091 AB .— Llo +30 = 70 w : 7oxlDE’N ,9 300m A = g .1: “EGO? - l.963SK103m-n1= L‘IESS'HD -3, 741w“)3 T :‘Pa B 6;” - A l-Q‘SSHQJ 35.7w 0 Gig -.-: 35.7 MP4 250mm. 09 R991 BC P = 30m = 3DXI03N C A : £26921: {1;(30): =' 706.86 mml= 705-94H6‘ W 5; _ 3. = m : 42_4x(o‘ Pa ‘ ' A 706. 86x10“ 65,; 3‘ 42.4 HP“. l________________________._ Problem 1.6 lino ?\ B 1:7- 1200 X 1.6 A strain gage located at C on the surface of bone AB indicates that the average nomial stress in the bone is 3.80 MPa when the bone is subjected to two 1200—N forces as shown. Assuming the cross section of the bone at C to be annular and knowing that its outer diameter is 25 mm, determine the inner diameter of the bone’s cross section at C. (#30200) 2 —3 2- 0'2 - (25”0 )‘ W 222.7 Ho" ml | l 5’2: H.93x10-3M 002: mash-an Prob|em 1 '15 1.15 When the force P reached 8 kN, the wooden specimen shown failed in shear along the surface indicated by the dashed line. Determine the average shearing stress along that surface at the time of failure. Area bein shearer} I. 1 A: CIOWIMX lSmM :3 1350mm 1' {Sfoxto‘a m Force “P? 3 *lOE' N f = 7? ~ --;—‘3'-“"—°.I*-- 5.93401" Paz5.93 We. 4 Problem 1 .1 7 1.1 7 A load P is applied to a steel rod supported as shown by an aluminum plate into which a 0.6-in.—diameter hole has been drilled. Knowing that the shearing stress must not exceed 18 ksi in the steel rod and 10 ksi in the aluminum plate, determine the largest load P that may be applied to the rod. For glee)? A. = Walt = Tl" (O.G\CO.‘H 0.7540 in‘ ’1: E :- ,1“, = (0.75740th : lg-g kI‘FS A2 = Tl’alt —- -.r(z.e)(o.25) .~ 1.2546 'm‘ P: A21; = U.25é6)<lo‘) : 12, 57 kip; P —_— 12.57 tops «- Problem 1 .18 1.18 Two wooden planks, each 12 mm thick and 225 mm wide, arejoined by the dry mortisejoint shown. Knowing that the wood used shears off along its grain when the average shearing stress reaches 8 MPa, determine the magnitude P of the axial load 15 mm which will cause the joint to fail. 16 mm ’ 36y (targets mus‘l loe slnem/‘eol 0?? , 25!];m.-150n’1m when ‘Hie Joim‘l lads, anln 0? ‘erSr L’ 50mm: 25mm 225mm—p.P areas l'tkS c’iwtenSHJHS lé mm X l2 MM) : l i‘ls area being -‘ Amman: V12“ —192>t/o"m A+ ¥aixpure ‘l’lxe \$0M? F Carries; lb)! 9.6;ch 0'? QV‘QQS {4 F : TA : (exioémaz Ho's) : ISSQ N = [.534 w Since erwe 60%: siy lad/Ute areas P: 6F ’5 (é )(l.53é) T (122 [(N " ...
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