Given:
The body
−
α
<
θ
<
α
{\displaystyle -\alpha <\theta <\alpha }
,
0
≤
r
<
a
{\displaystyle 0\leq r<a}
is supported at
r
=
a
{\displaystyle r=a}
and loaded only by a uniform antiplane shear traction
σ
θ
z
=
S
{\displaystyle \sigma _{\theta z}=S}
on the surface
θ
=
α
{\displaystyle \theta =\alpha }
, the other
surface being traction-free.
A body loaded in antiplane shear
Find:
Find the complete stress field in the body, using strong boundary conditions on
θ
=
±
α
{\displaystyle \theta =\pm \alpha }
and weak conditions on
r
=
a
{\displaystyle r=a}
.
[Hint: Since the traction
σ
θ
z
{\displaystyle \sigma _{\theta z}}
is uniform on the surface
θ
=
α
{\displaystyle \theta =\alpha }
, from the expression for antiplane stress we can see that the displacement varies with
r
1
=
r
{\displaystyle r^{1}=r}
. The most general solution for the equilibrium equation for this behavior is
u
(
r
,
θ
)
=
A
r
cos
θ
+
B
r
sin
θ
{\displaystyle u(r,\theta )=Ar\cos \theta +Br\sin \theta }
]
Step 1: Identify boundary conditions
at
r
=
0
;
u
r
=
0
,
u
θ
=
0
at
r
=
a
;
u
r
=
0
,
u
θ
=
0
,
u
z
=
0
at
θ
=
−
α
;
t
θ
=
0
,
t
r
=
0
,
t
z
=
0
at
θ
=
α
;
t
θ
=
0
,
t
r
=
0
,
t
z
=
S
{\displaystyle {\begin{aligned}{\text{at}}~r&=0~;~~u_{r}=0,u_{\theta }=0\\{\text{at}}~r&=a~;~~u_{r}=0,u_{\theta }=0,u_{z}=0\\{\text{at}}~\theta &=-\alpha ~;~~t_{\theta }=0,t_{r}=0,t_{z}=0\\{\text{at}}~\theta &=\alpha ~;~~t_{\theta }=0,t_{r}=0,t_{z}=S\end{aligned}}}
The traction boundary conditions in terms of components of the stress tensor are
at
θ
=
−
α
;
σ
θ
r
=
0
,
σ
θ
θ
=
0
,
σ
θ
z
=
0
at
θ
=
α
;
σ
θ
r
=
0
,
σ
θ
θ
=
0
,
σ
θ
z
=
S
{\displaystyle {\begin{aligned}{\text{at}}~\theta &=-\alpha ~;~~\sigma _{\theta r}=0,\sigma _{\theta \theta }=0,\sigma _{\theta z}=0\\{\text{at}}~\theta &=\alpha ~;~~\sigma _{\theta r}=0,\sigma _{\theta \theta }=0,\sigma _{\theta z}=S\end{aligned}}}
Step 2: Assume solution
Assume that the problem satisfies the conditions required for antiplane shear. If
σ
θ
z
{\displaystyle \sigma _{\theta z}}
is to be uniform along
θ
=
α
{\displaystyle \theta =\alpha }
, then
σ
θ
z
=
μ
r
∂
u
z
∂
θ
=
C
{\displaystyle \sigma _{\theta z}={\frac {\mu }{r}}{\frac {\partial u_{z}}{\partial \theta }}=C}
or,
∂
u
z
∂
θ
=
C
r
μ
{\displaystyle {\frac {\partial u_{z}}{\partial \theta }}={\frac {Cr}{\mu }}}
The general form of
u
z
{\displaystyle u_{z}}
that satisfies the above requirement is
u
z
(
r
,
θ
)
=
A
r
cos
θ
+
B
r
sin
θ
+
C
{\displaystyle u_{z}(r,\theta )=Ar\cos \theta +Br\sin \theta +C}
where
A
{\displaystyle A}
,
B
{\displaystyle B}
,
C
{\displaystyle C}
are constants.
Step 3: Compute stresses
The stresses are
σ
θ
z
=
μ
r
∂
u
z
∂
θ
=
μ
(
−
A
sin
θ
+
B
cos
θ
)
σ
r
z
=
μ
∂
u
z
∂
r
=
μ
(
A
cos
θ
+
B
sin
θ
)
{\displaystyle {\begin{aligned}\sigma _{\theta z}&={\frac {\mu }{r}}{\frac {\partial u_{z}}{\partial \theta }}=\mu \left(-A\sin \theta +B\cos \theta \right)\\\sigma _{rz}&=\mu {\frac {\partial u_{z}}{\partial r}}=\mu \left(A\cos \theta +B\sin \theta \right)\end{aligned}}}
Step 4: Check if traction BCs are satisfied
The antiplane strain assumption leads to the
σ
θ
θ
{\displaystyle \sigma _{\theta \theta }}
and
σ
r
θ
{\displaystyle \sigma _{r\theta }}
BCs being satisfied. From the boundary conditions on
σ
θ
z
{\displaystyle \sigma _{\theta z}}
, we have
0
=
μ
(
A
sin
α
+
B
cos
α
)
S
=
μ
(
−
A
sin
α
+
B
cos
α
)
{\displaystyle {\begin{aligned}0&=\mu \left(A\sin \alpha +B\cos \alpha \right)\\S&=\mu \left(-A\sin \alpha +B\cos \alpha \right)\end{aligned}}}
Solving,
A
=
−
S
2
μ
sin
α
;
B
=
S
2
μ
cos
α
{\displaystyle A=-{\frac {S}{2\mu \sin \alpha }}~;~~B={\frac {S}{2\mu \cos \alpha }}}
This gives us the stress field
σ
θ
z
=
S
2
(
sin
θ
sin
α
+
cos
θ
cos
α
)
;
σ
r
z
=
S
2
(
−
cos
θ
sin
α
+
sin
θ
cos
α
)
{\displaystyle \sigma _{\theta z}={\frac {S}{2}}\left({\frac {\sin \theta }{\sin \alpha }}+{\frac {\cos \theta }{\cos \alpha }}\right)~;~~\sigma _{rz}={\frac {S}{2}}\left(-{\frac {\cos \theta }{\sin \alpha }}+{\frac {\sin \theta }{\cos \alpha }}\right)}
Step 5: Compute displacements
The displacement field is
u
z
(
r
,
θ
)
=
S
r
2
μ
(
−
cos
θ
sin
α
+
sin
θ
cos
α
)
+
C
{\displaystyle u_{z}(r,\theta )={\frac {Sr}{2\mu }}\left(-{\frac {\cos \theta }{\sin \alpha }}+{\frac {\sin \theta }{\cos \alpha }}\right)+C}
where the constant
C
{\displaystyle C}
corresponds to a superposed rigid body displacement.
Step 6: Check if displacement BCs are satisfied
The displacement BCs on
u
r
{\displaystyle u_{r}}
and
u
θ
{\displaystyle u_{\theta }}
are automatically satisfied by the antiplane strain assumption. We will try to satisfy the boundary conditions on
u
z
{\displaystyle u_{z}}
in a weak sense, i.e, at
r
=
a
{\displaystyle r=a}
,
∫
−
α
α
u
z
(
a
,
θ
)
d
θ
=
0
.
{\displaystyle \int _{-\alpha }^{\alpha }u_{z}(a,\theta )d\theta =0~.}
This weak condition does not affect the stress field. Plugging in
u
z
{\displaystyle u_{z}}
,
0
=
∫
−
α
α
u
z
(
a
,
θ
)
d
θ
=
S
a
2
μ
∫
−
α
α
(
−
cos
θ
sin
α
+
sin
θ
cos
α
+
C
2
μ
S
a
)
d
θ
=
S
a
2
μ
∫
−
α
α
(
−
cos
θ
sin
α
+
sin
θ
cos
α
+
C
2
μ
S
a
)
d
θ
=
S
a
2
μ
[
(
−
sin
θ
sin
α
−
cos
θ
cos
α
+
C
θ
2
μ
S
a
)
]
−
α
α
=
S
a
2
μ
(
−
2
sin
α
sin
α
+
2
C
α
2
μ
S
a
)
=
−
S
a
μ
+
C
α
{\displaystyle {\begin{aligned}0&=\int _{-\alpha }^{\alpha }u_{z}(a,\theta )d\theta \\&={\frac {Sa}{2\mu }}\int _{-\alpha }^{\alpha }\left(-{\frac {\cos \theta }{\sin \alpha }}+{\frac {\sin \theta }{\cos \alpha }}+C{\frac {2\mu }{Sa}}\right)d\theta \\&={\frac {Sa}{2\mu }}\int _{-\alpha }^{\alpha }\left(-{\frac {\cos \theta }{\sin \alpha }}+{\frac {\sin \theta }{\cos \alpha }}+C{\frac {2\mu }{Sa}}\right)d\theta \\&={\frac {Sa}{2\mu }}\left[\left(-{\frac {\sin \theta }{\sin \alpha }}-{\frac {\cos \theta }{\cos \alpha }}+C\theta {\frac {2\mu }{Sa}}\right)\right]_{-\alpha }^{\alpha }\\&={\frac {Sa}{2\mu }}\left(-2{\frac {\sin \alpha }{\sin \alpha }}+2C\alpha {\frac {2\mu }{Sa}}\right)\\&=-{\frac {Sa}{\mu }}+C\alpha \end{aligned}}}
Therefore,
C
=
S
a
2
μ
α
{\displaystyle C={\frac {Sa}{2\mu \alpha }}}
The approximate displacement field is
u
z
(
r
,
θ
)
=
S
2
μ
(
−
r
cos
θ
sin
α
+
r
sin
θ
cos
α
+
a
1
α
)
{\displaystyle u_{z}(r,\theta )={\frac {S}{2\mu }}\left(-r{\frac {\cos \theta }{\sin \alpha }}+r{\frac {\sin \theta }{\cos \alpha }}+a{\frac {1}{\alpha }}\right)}