sm8 3 5

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PROBLEMS (8.3 through 8.5) The states of stress at three points in a loaded body are represented
in Figs. P8.3 through P8.5. Determine for each point the normal and shearing stresses acting on the
indicated inclined plane. As was done in the derivations given in Sec. 8.2, use an approach based upon
the equilibrium equations applied to the wedge shaped element shown in each figure.
SOLUTION (8.3)
y
x
’
x
60 A
Δ
sin15o
y
’
80 AΔsin15o
'xA
σ
Δ
''xy A
τ
Δ
60 A
Δ
cos15o15o15o
50 AΔcos15o
oo
xx AAF 15sin8015cos50:0 22 '' Δ−+Δ=
∑
σ
0)15cos15sin60(2 =Δ− oo
A
MPa
x3.113036.565.46 '
−
=
+
+−=
σ
oo
yxy AAF 15cos15sin50:0 ''' Δ−Δ=
∑
τ
ooo AA 15cos6015cos15sin80 2
Δ−Δ−
015sin60 2=Δ+ o
A
MPa
yx 5.8402.498.55205.12 ''
=
−
++=
τ
SOLUTION (8.4)
o
xx AAF 30cos25:0 2 '' Δ−Δ=
∑
σ
0)30cos30sin10(2 =Δ+ oo
A
ksi
x09.1066.875.18 '
=
−=
σ
x
’
10 A
Δ
sin30o
y
’
'xA
σ
Δ
''xy A
τ
Δ
10 A
Δ
cos30o
30o
25 A
Δ
cos30o
Continued on next slide
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'''
0: 25 cos30 sin30
oo
yxy
FAA τ =Δ+Δ
∑
030sin1030cos10 22 =Δ−Δ+ oo AA
'' 10.83 7.5 2.5 15.83
xy ksi
τ
=− − + =−
SOLUTION (8.5)
70 AΔsin22.5o
x
’
40 A
Δ
cos22.5o
y
’
'xA
σ
Δ
''xy A
τ
Δ
10 A
Δ
cos22.5o
22.5o
10 AΔsin22.5o
o
xx AAF 5.22cos40:0 2 '' Δ+Δ=
∑
σ
0)5.22cos5.22sin10(25.22sin70 2=Δ−Δ− ooo AA
MPa
x8.1607.725.1014.34 '
−
=
+
+−=
σ
'''
0: 40 sin22.5 cos22.5
oo
yxy
FAA τ =−Δ−Δ
∑
−Δ
2
70 sin22.5 cos22.5 10 cos 22.5
oo o
AA−Δ 05.22sin10 2=Δ+ o A
MPa
yx 4646.154.875.2414.14 ''
=
−
+
+=
τ
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or
instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States
Copyright Act without the permission of the copyright owner is unlawful.