Solutions 8

# Solutions 8 - 8-5 The open-ended pipe has a wall thickness...

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Unformatted text preview: 8-5. The open-ended pipe has a wall thickness of 2 mm and an internal diameter of 40 mm. Calculate the pressure that ice exerted on the interior wall of the pipe to cause it to burst in the manner shown. The maximum stress that the material can support at freezing temperatures is om, = 360 MPa. Show the stress acting on a small element of material just before the pipe fails. Hoop Stress for Cylindrical Pipe : Since ; = 329 = ID. 0; 3360 MP4. then thin wdl analysis can be used. Applying Eq.8 - l v r '0 a! = allov- =£:- o; 360( 10‘) = T3322) p 8 36.0 MPa Ans Longitudinal Stress for Cylindrical Pipe : Since the pipe IS open at both ends. then 02:0 *8—24. The gondola and passengers have a weight of 1500 lb and center of gravity at G. The suspende r arm A12” has a square cross-sectional area of 1.5 in. by 1.5 in.. and is pin connected at its ends A and E. Determine the largest tensile stress developed in regions AB and DC of the arm. SegmentAB: PA, 1500 . a..." = — = --——-—- = 667 s: Ans ( m A (1.5)(1.5) p Segment CD: " Pco 1500 loo-1500 Oa=—=—————-=666.67si _, A (1.5)(1.5) p "5 gleh - Ii PAD-m Maﬂg] I‘ m = if: = [875(12)(0.75) = “mpg I ,12-(1.5)(1.53) "Mtg (0....)CD = a, + a, = 666.67 + 40000 = 40666.67 psi - 40.7 ksi Ans ,5 8—49. The sign is subjected to the uniform wind loading. Determine the stress components at points A and B on the lOO-Imn-diameter supporting post. Show the results on a volume element located at each of these points. [M 3-5.. 3 "" PoinA: 3“). m I {(0.05)‘ 15.219(10‘)=15.3Mh A... mm. J 1(0'05) 1:35"? Point 8: ﬂ 0- =0 Au . .05 1 r- = 35-3’2 = mum—W J I: §(o.05)‘(o.1) \$ t. x 14.8 MP: Ans 9-7. Solve Prob. 9—2 using the stress-transformation equations developed in Sec. 9.2. a,=5ksi a,=3ksi r,,=8ksi 9:13o° 0 + “'0 0,. = .13."; + 2’—2—Lcos 20+r,,sin 29 = §+—3+-5—:§cos 260°-+8 sin 260° =-4.os ksi Ans 2 2 The negative sign indicates 0,. is a compressive stress. 1.}, = --a—':-£Lsin 20+ n, cos 20 2 = —(§—;—3-)sin 260° + 800s 260° = -0.404 ksi Ans The negative sign indicm Q,- is in the - y' direction. 9—13. The state of stress at a point is shown on the element. Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the element in each case. ()0 M Pa 30 MPa 45 MPH a. usurp: a, "soup. «.3, =30MP. 3) 0” ‘ﬁ-g-gJ-i (ﬁg-L)2+ttyz t 1 + 30 = 2 ( 2 ) ( Y a. = 53.0 MP. An: 0'; = ~68.0 MP1 Au Orientation of principal was: In 20, .-.-. _E£L__. g A... .051“ (a. — 6, )I2 (45 - (-60))I2 a, a 14.81. -7s.|3 UscEq.9-1todumnhelhepdncipalphneofa. “0;: a.. a + Egg-Leos 29+ t.,sin29. where o s 14.8? - 4S +(-60) + 45- (-60) 2 2 00: 29.14“ + 30 sin 29.74’ I 33.0 MP. Therefore 9,. I14.9° Am and 0,; I-‘ISJ' Ans b) .- 45-(-60) r... -‘/(-—J-E)’+r.’ :- (—-—-)3+30' I60.5MPa Au 5 “"" 2 ’ 2 , your: a". = a, +0: . 43+(-60) 2 2 Orientation of maximum in - plane shear amass: I ~7.50 MP: All: mza, . M . W .t ..,75 t., 30 0, I -30.l' Au and 9, I 59.9‘ An: By mm. in and: to We equilibrium don; AB. t... ha to act in the dictation shown. *9—40. Solve Prob. 9—39 for point B located on the web at the top of the bottom flange. _!_ —- ‘ 12mm 10 mm I 250 mm B 1 = l(o.2)(o.247)3 - —‘—(o.19)(o.25)3 = 95.451233(10'°) m‘ — i 12 mm 12 12 H Q, = (0.131)(o.012)(o.2) = 0314400”) 200 mm a, z 12 = -w = -196,” We. I 95.45 123300“) -3 t. 3 Q = 50(103)(o.3144)(10 ) 316.47 MP3 I: 95.451233(10—6)(0.01) a, = -l96.43 MPa a, = o 1,, = -l6.47 MPa 0,4-0 ,- m= 2 WW" 2"’)2+rx,2 = —l96.243+01 (49633-0), “46.47? a, = 1.37 MPa Ans ﬂﬂillﬁfa 0'; = -l98 MP8 Ans ._... ...
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