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in the differential equation of equilibrium [Eq. (1.144a)] to obtain
2
2
∂
y
2
+
∂
ω
ω
2
z
2
=
0or
∇
ω
=
0
(1.151)
∂
∂
2
y
z
. This is often written in terms of the displacement
u
[Eq. (1.141)]
with
∇
=
∂
+
∂
2
u
2
u
∂
y
2
+
∂
2
u
z
2
=
∇
=
0or
0
(1.152)
∂
∂
This type of differential relationship is called
Laplace's equation
. A solution to Laplace's
equation is called a
harmonic function
. Thus, the warping function is a harmonic function.
If the expression for stresses obtained from Eqs. (1.142) and (1.143) are substituted in Eq.
(1.145), it is apparent that the boundary condition [Eq. (1.145)] in terms of the warping
function
ω
can be written as
∂ω
∂
y
a
y
+
∂ω
z
a
z
+
za
y
−
ya
z
=
0
(1.153)
∂
The expression for the torsional constant
J
in terms of the warping function
ω
is [Eq.
(1.150)]
M
t
G
1
G
J
=
φ
=
(τ
xz
y
−
τ
xy
z
)
dy dz
φ
A
y
y
z
z
dy dz
−
∂ω
∂
+
∂ω
∂
=
+
z
y
A
z
∂ω
∂
z
2
dy dz
y
∂ω
∂
y
2
=
y
−
z
+
+
(1.154)
A
where Eqs. (1.142) and (1.143) have been introduced.
EXAMPLE 1.6 Constant Warping Function
Suppose the warping function
ω
is constant, that is
ω(
y, z
)
=
c
where
c
is a constant.
2
In this case,
∇
ω
=
0 is satisfied by
ω
=
c
. From Eqs. (1.142) and (1.143),
φ
z
,
φ
+
∂ω
∂
+
∂ω
∂
τ
xy
=−
G
τ
xz
=−
G
−
y
(1)
y
z
ω
=
Substitute
c
into (1) and use the surface condition of Eq. (1.145) to find
dy
ds
y
dz
ds
z
−
−
=
0
(2)
Then
y
2
z
2
d
ds
+
=
0
(3)
2
or
y
2
z
2
r
2
+
=
(4)
is constant on the boundary. Since (4) is the equation of a circle of radius
r
, it follows that
ω
=
c
represents the warping function for the torsion of a bar of circular cross-section.
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