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Unformatted text preview: PROBLEM 7.31 7.31 Solve Probs. 7.7 and 7.11, using Mohr’s circle. 7.5 through 7.8 For the given state of stress, determine (a) the principal planes, (b) 7.9 through 7.12 For the given state of stress, determine (a) the orientation of the
planes of maximum inplane shearing stress, (b) the maximum in—plane shearing mess,
(c) the corresponding normal stress. SOLUTION
5,.=GOMPQ S,:—4DMP« 6mm: €;_Z'6'3 = “‘50 “PO. Poin'is
>< 461,423,) = ( 6mm, ~35HPQ Y‘(ﬁ‘m 11.3): (wMPa, 35 MPa)
C: (Gave) 0 ) ‘ (“SDHPAJ O) = 36.4 MP4
5036.’I = —86.‘f MP4
.S'0+36.'+ ' 3.G MP4
93+ 45° = 7.97“
9» +45" ’ ”17.77"
12 = 3G}! MP6;
6‘... = —50 MW» PROBLEM 7.34 7.34 SOIVC Prob. 7.10, usmg MOhl' S cn‘cle_ 7.9 through 7.12 plan“ of maximum inplanc shearing stress,
(6) the mp0!) ' g normal stmss. SOLUTION P0in+$ X: (6x0 —'1:3)=(.2 165316;)
Y: (5:,J 7,19) = (ID lag3kg)
c: (6% o ) : (651:5?) o) 0.75 ' ’ circle.
PROBLEM 7.39 7.39 Solve Prob. 7.17, usmg Mohr s 7.17 and 7.18 The grain ofa wooden member forms an angle of15° with the vertical.
For the state of stress shown, determine (a) the in—plan'e shearing stress parallel to the
grain, (b)thenormal su'essperpendiailartothegrain. SOLUTION 6x:—3Mpa. G“):.8 M‘Pq 221:0
sm=ﬂzL§t = —2.4 MPa. 29 = 30° a: 0.6 MPa. 12: 0.6 MPa. (a) 2163’ ': “EXI $13130 1‘  R sin 30‘ 7 0.6 Siva 30. ? O.3 MP‘L l (m 6:. : G... — Ex'mso‘ = ~2fl  0.4 cosso" . — 2.92 MPa. l 7.40 Solve Prob. 7.18, using Mohr’s circle. 7.17 and 7.18 The grain ofa wooden member forms an angle of15° with the vertical.
For the state of stress shown, determine (a) the inplane shearing stress parallel to the PROBLEM 7.40 400 psn  . .
gﬁrgg/éﬁ 8min, (b) the normal stress perpendicular to the gram.
{ [IXV‘ZW SOLUTION
”if, :53 = = o ’r = 00 s‘
xiii 55%;? 5* 5: *1 4 P .
‘— 15° 6'“,e :. 5x + 61 = O 2 (a) 73.]. 2 R :05 30" r 400 cos 30" = 346 psi, 4 0)) 6" t 6;", " Rsin 30‘ 7 —‘+00 Sivx30a = 200 P$i 4 PROBLEM 7.54 7.54 through 7.57 Determine the principal planes and the principal Suﬁsm for the
state of plane stras multing ﬁ'om the superposmon of the two stata of sums shown. 25 “Pa SOLUTION Molnw‘s cichc 'Fol‘ 2nd shes: sick 6x:,20+20 as 40'
= 30 MPG. 6:, = 20—20025 ‘0'
: l0 MP4 Tl: : 20 $;ﬂ 6°. 5 \7.32 MFA
Resup‘l‘qd 5+ asses 20 al
5,=35+3o= 55qu [:_.,°
5 r 25 + Io = 35 M90.
7:) = O +1132 = l7.32 HPa. tn (MP0)  (62%;) = %(é$+55) = 50 MP4 _ (2 X1232 ) C5 _35 = 1.1547 9‘: 29.6 95:14.6 «a ’= 22.9/ M94.
625 “+1? 72.9: MP“ 61,: GM — 12 , 27.061 HPaL. ...
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 Fall '09
 MOORTHY

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