Solutions_for_HW5

# Solutions_for_HW5 - Chapter 5 5-1 Sy = 350 MPa S MSS...

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Unformatted text preview: Chapter 5 5-1 Sy = 350 MPa. S MSS: 01—0-3=Sy/n 2) n= y (‘71—‘73) DE 0" =(0'121 —0'A0'B +03"; )1/2 = (a: —0'x0'y +0"; + 1;)1/2 O’ (a) MSS: 01 = 100 MPa,0'2 = 100 MPa,0'3 = 0 n = 350 =3.5 Ans. 100—0 DE: 0" = (1002 —100(100)+1002)“2 = 100 MPa, n = % = 3.5 Ans. (b) MSS: 0'1 = 100 MPa,02 = 50 MPa,c73 = 0 n: 350 =35 Ans. 100—0 DE: 0' = (1002 —100(50) +502)“2 = 86.6 MPa, n = 5396 = 4.04 Ans. 2 (c) 0A, GB =¥t (13—0) (—75)2 =140, —40 MPa 01:140, 0'2 = 0, 03 = ~40 MPa MSS: n = i = 1.94 Ans. 140—(—40) / DE: 0" =[1oo2 +3(—752)]1 2 =164 MPa, n=§§9=z13 Ans. ‘ 164 2 (d) GAJB = ‘50; 75 i (‘50; 75] +(—-50)2 =—11.0,—114.0 MPa 0'1: 0, 0'2 =——11.0, 0'3 =—114.0 MPa MSS: n=i=3ﬂ7 Ans. 0—(—114.0) ‘ A DE: 0" = [(--50)2 — (—50)(—75) + (—75)2 +3(—50)2]“2 = 109.0 MPa n = 350 =3.21 Ans. 109.0 2 (e) (34,03 =100;20: [1002—20) +(—20)2 =104.7, 15.3 MPa Chapter 5-Rev B, Page 1/55 5-3 01 =104.7, 0—2 =15.3, 0‘3 =0 MPa MSS: n=~i=334 Ans. 104.7—0 1/2 DE: o—'=[1002 —100(20)+202 +3(—20)2] =98.0 MPa n=ﬂ=357 Ans. 98.0 sy = 350 MPa. S 01—03: Sy/n 3 n=__y__ (01"03) . S S DE: o"=(0‘:—0'Ao-B +03%)“2 =__Y_ :> n:__y' n o- 350 ' (a) MSS: 01:100 MPa, 03 =0 :> n = =3.5 Ans. 100—0 DE: n=~——2————3—50————2—U2=3.5 ANS. [100 —(100)(100)+100 ] (b) MSS: 0'1 =100 , 0'3 =—100 MPa => n=———%-50————=1.75 Ans. 100—(—-100) DE: n= 350 1/2 =2.02 Ans. [1002 — (100)(—100) + (-100)? 350 (c) MSS: 01:100 MPa, 03 =0 ::> n= =35 Ans. 100—0 DE: n PAW—7524.04 Ans. [1002 — (100)(50) + 502] 350 (d) MSS: 01:100, 03 =—50 MPa :> n=——-—=2.33 Ans. IOU—(~50) DE: n =——=—ﬂ———W=2.65 Ans. [1002 —(100)(—50)+ (~50) ] 350 (e) MSS: 0'1 =0, (73 =—100MPa => n=——~——=3.5 Ans. 0—(—100) ‘ DE: n = —-~——-—-———§:q———~ = 4.04 Ans. [(—50)2 —(—50)(—100)+(—100)2]U2 From Table A—20, sy = 37.5 kpsi Chapter 5-Rev B, Page 2/55 5-11 Sy =295 MPa, 0A,0'B = ~80+30i [—80—30 2 2 ) +(—10)2 = 30.9,—80.9 MPa 5 (a) n— y = 295 =2.95 Ans. (02 WAGE +01%)1/2 [30.92 —3O.9(—80.9)+(480.9)2]w (b) Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. ‘73 (MPa) f’ Ix.) \0 L11 — OB ~£= 2.93 Ans. n _—-—— _. GA 0.87" 1" = 100 NIPa 5—12 Syt = 60 kpsi, Syc = 75 kpsi. Eq. (5-26) for yield is —1 n=(_a__§i] Syt Syc (a)o*—25ksi (7-0 :> n-[£—£ 4—240 Ans 1 p ’ 3 6O 75 ‘ ~ ' 15' —15 ‘1 ' (b) 01:15, U3 =—15 kpsi :> n=[—6-6——7—5-) =2.22 Ans. Chapter 5-Rev B, Page 11/55 2 , (c) 0A, 03:?i +(—10)2 =24.1, —4.1kpsi, —1 01:24.1, 0'2 =0, 0'3 =—4.1kpsi => n=[ﬂ—:i'~1 =2.19 Ans. 60 75 2 (d) 0A,ch = ~122+15i [422—15) +(—9)2 =17.7,—14.7 kpsi —1 01:17.7, 02 =0, 0'3 =—14.7 kpsi :> =2.04 Ans. . 60 75 2 (e) 0A,0'B=_242—24i (123233] +(—15)2 =I—9, —39 kpsi —1 01:0, 02=—9, 03 =—39 kpsi => n=(—O——ﬂ) =1.92 Ans. 60 75 5-13 Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. 0'5 (kpsi) (a) (7A =25, 0'3 =15 kpsi " 1" : 20 kpsi , (a) n=2€=ﬂ=239 Ans. GA 1.46" . " f 60 (b) O'A =15, 0'3 =—15 kp51 I ' F(c) n=0—D=236—=2.23 Ans. 0C 1.06" (c) l 2 (c) O'A, aB=—2—Oi- +(—10)2 =24.1, ~4.1kpsi- 2 2 n=2€=ﬂ=219 Ans. 0E 1.22" (d) 2 . aA,aB=”122+15i [422—15) +(—9)2 =17.7,—14.7 kpsi Chapter 5-Rev B, Page 12/55 ...
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Solutions_for_HW5 - Chapter 5 5-1 Sy = 350 MPa S MSS...

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