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Unformatted text preview: 85. The openended 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 crosssectional 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 loo1500 Oa=—=—————=666.67si _,
A (1.5)(1.5) p "5 gleh
 Ii
PADm 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
lOOImndiameter supporting post. Show the results on a
volume element located at each of these points. [M
35.. 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 = 353’2 = mum—W J I: §(o.05)‘(o.1) $ t. x 14.8 MP: Ans 97. Solve Prob. 9—2 using the stresstransformation
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
inplane 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” ‘ﬁggJi (ﬁgL)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.91todumnhelhepdncipalphneofa. “0;: a.. a + EggLeos 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... ‘/(—JE)’+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 0,40 ,
m= 2 WW" 2"’)2+rx,2 = —l96.243+01 (496330), “46.47? a, = 1.37 MPa Ans ﬂﬂillﬁfa
0'; = l98 MP8 Ans ._... ...
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 Winter '07
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