sm8 55

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PROBLEM (8.55) Using Mohr’s circle, solve Prob. 8.15.
SOLUTION
For '''
30 : 200 35
oxxy
M
Pa MPa θσ τ ==− =
1
2
22 1
[(200 130) 35 ] 78.26 , ( 200 60) 130
2
avg
R
MPa MPa σ =−+= =−−=−
35
tan 200 130
α =−
26.5
o
α =6
4
33.4
o
β =
For 0:
θ
=
130 cos
xR
σ
β =− −
130 (78.26)cos(33.44 ) 195.3
o
x
σ =− − =−
sin (78.26)sin(33.44 ) 43.1
o
xy
R
MPa τβ =− =− =−
130 cos
yR
σ
β =− +
130 (78.26)cos(33.44 ) 64.7
o
M
Pa=− + =−
Sketch of results is as shown in solution of Prob. 8.15.
R
τ (MPa)
O
C
α
60o
x’
x
y’
y
β
(MPa)
σ
130
200
5
-3
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