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normal force, must balance the applied point force. If the hemisphere has radius
a
,
the traction per unit area on it is given by relation (1.214), where the components
of the normal vector are ν
1
x
1
)/
a
, ν
2
x
2
)/
a
,andν
3
x
3
/
a
while the
stresses τ
ij
are the sum of those given by relations (1.346) through (1.351) and
those given by relations (1.353) through (1.358). The components of the tractions
per unit area are
=
(
x
1
−
=
(
x
2
−
=
3
A
x
1
x
1
x
3
a
4
2μ
B
x
1
x
1
−
−
F
ν
1
=−
−
x
3
)
,
(1.362)
a
2
(
a
+
3
A
x
2
−
x
2
x
3
a
4
2μ
B
x
2
−
x
2
a
2
(
a
F
ν
2
=−
−
x
3
)
,
(1.363)
+
⎝
⎠
−
3
x
3
A
1
a
2
μ
λ
+
μ
2μ
B
1
F
ν
3
=−
a
2
+
a
2
.
(1.364)
With (
a
,θ,φ) as the spherical polar co-ordinates on the surface of the hemisphere,
the components of the total traction on the hemisphere are
a
2
2π
0
π/2
F
ν
i
sinθ
d
θ
d
φ.
T
i
=
(1.365)
0
The Cartesian co-ordinates on the surface of the hemisphere, referred to its centre,
are
x
1
x
1
=
x
2
=
−
a
sinθcosφ,
x
2
−
a
sinθsinφ,and
x
3
=
a
cosθ. Carrying out the
integrations in (1.365), we find that
λ
+
2μ
B
2μ
λ
+
μ
T
1
=
T
2
=
0,
T
3
=−
2π
A
+
.
(1.366)
While the normal stresses, (1.348) and (1.355), vanish on the surface
x
3
=
0, the
shear stresses τ
13
and τ
23
do not vanish, but add, to become
μ
λ
+
μ
2μ
B
x
1
x
1
−
τ
13
=−
A
+
,
(1.367)
R
0
μ
λ
+
μ
2μ
B
x
2
x
2
−
τ
23
=−
A
+
.
(1.368)
R
0
The condition for the surface
x
3
=
0tobestressfreeisthen
μ
λ
+
μ
A
+
2μ
B
=
0.
(1.369)
The second condition, that the total traction on the surface of the hemisphere, given
by (1.366), balances the applied point force, becomes
λ
+
2μ
λ
+
μ
P
2π
.
A
+
2μ
B
=
(1.370)
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