Given:
The differential equations governing the bending of straight beams are
d
N
x
x
d
x
+
f
(
x
)
=
0
d
V
d
x
+
q
(
x
)
=
0
d
M
x
x
d
x
−
V
+
N
x
x
d
w
0
d
x
=
0
{\displaystyle {\begin{aligned}{\cfrac {dN_{xx}}{dx}}+f(x)&=0\\{\cfrac {dV}{dx}}+q(x)&=0\\{\cfrac {dM_{xx}}{dx}}-V+N_{xx}{\cfrac {dw_{0}}{dx}}&=0\end{aligned}}}
Show that the weak forms of these equations can be written as
∫
x
a
x
b
d
v
1
d
x
N
x
x
d
x
=
∫
x
a
x
b
v
1
f
d
x
+
v
1
N
x
x
|
x
a
x
b
∫
x
a
x
b
[
d
v
2
d
x
(
d
w
0
d
x
N
x
x
)
−
d
2
v
2
d
x
2
M
x
x
]
d
x
=
∫
x
a
x
b
v
2
q
d
x
+
v
2
(
d
w
0
d
x
N
x
x
+
d
M
x
x
d
x
)
|
x
a
x
b
−
d
v
2
d
x
M
x
x
|
x
a
x
b
{\displaystyle {\begin{aligned}\int _{x_{a}}^{x_{b}}{\cfrac {dv_{1}}{dx}}N_{xx}~dx&=\int _{x_{a}}^{x_{b}}v_{1}f~dx+\left.v_{1}N_{xx}\right|_{x_{a}}^{x_{b}}\\\int _{x_{a}}^{x_{b}}\left[{\cfrac {dv_{2}}{dx}}\left({\cfrac {dw_{0}}{dx}}N_{xx}\right)-{\cfrac {d^{2}v_{2}}{dx^{2}}}M_{xx}\right]~dx&=\int _{x_{a}}^{x_{b}}v_{2}q~dx+\left.v_{2}\left({\cfrac {dw_{0}}{dx}}N_{xx}+{\cfrac {dM_{xx}}{dx}}\right)\right|_{x_{a}}^{x_{b}}-\left.{\cfrac {dv_{2}}{dx}}M_{xx}\right|_{x_{a}}^{x_{b}}\end{aligned}}}
First we get rid of the shear force term by combining the second and third equations to get
d
N
x
x
d
x
+
f
(
x
)
=
0
(1)
d
2
M
x
x
d
x
2
+
q
(
x
)
+
d
d
x
(
N
x
x
d
w
0
d
x
)
=
0
(2)
{\displaystyle {\begin{aligned}{\cfrac {dN_{xx}}{dx}}+f(x)&=0{\text{(1)}}\qquad \\{\cfrac {d^{2}M_{xx}}{dx^{2}}}+q(x)+{\cfrac {d}{dx}}\left(N_{xx}{\cfrac {dw_{0}}{dx}}\right)&=0{\text{(2)}}\qquad \end{aligned}}}
Let
v
1
{\displaystyle v_{1}}
be the weighting function for equation (1) and let
v
2
{\displaystyle v_{2}}
be the weighting function for equation (2).
Then the weak forms of the two equations are
∫
x
a
x
b
v
1
(
d
N
x
x
d
x
+
f
(
x
)
)
d
x
=
0
(3)
∫
x
a
x
b
v
2
[
d
2
M
x
x
d
x
2
+
q
(
x
)
+
d
d
x
(
N
x
x
d
w
0
d
x
)
]
d
x
=
0
(4)
{\displaystyle {\begin{aligned}\int _{x_{a}}^{x_{b}}v_{1}\left({\cfrac {dN_{xx}}{dx}}+f(x)\right)~dx&=0{\text{(3)}}\qquad \\\int _{x_{a}}^{x_{b}}v_{2}\left[{\cfrac {d^{2}M_{xx}}{dx^{2}}}+q(x)+{\cfrac {d}{dx}}\left(N_{xx}{\cfrac {dw_{0}}{dx}}\right)\right]~dx&=0{\text{(4)}}\qquad \end{aligned}}}
To get the symmetric weak forms, we integrate by parts (even though the symmetry is not obvious at this stage) to get
(
v
1
N
x
x
)
|
x
a
x
b
−
∫
x
a
x
b
d
v
1
d
x
N
x
x
d
x
+
∫
x
a
x
b
v
1
f
(
x
)
d
x
=
0
(5)
(
v
2
d
M
x
x
d
x
)
|
x
a
x
b
−
∫
x
a
x
b
d
v
2
d
x
d
M
x
x
d
x
d
x
+
∫
x
a
x
b
v
2
q
(
x
)
d
x
+
(
v
2
N
x
x
d
w
0
d
x
)
|
x
a
x
b
−
∫
x
a
x
b
d
v
2
d
x
(
N
x
x
d
w
0
d
x
)
d
x
=
0
(6)
{\displaystyle {\begin{aligned}\left.(v_{1}~N_{xx})\right|_{x_{a}}^{x_{b}}&-\int _{x_{a}}^{x_{b}}{\cfrac {dv_{1}}{dx}}N_{xx}~dx+\int _{x_{a}}^{x_{b}}v_{1}~f(x)~dx=0{\text{(5)}}\qquad \\\left.\left(v_{2}~{\cfrac {dM_{xx}}{dx}}\right)\right|_{x_{a}}^{x_{b}}&-\int _{x_{a}}^{x_{b}}{\cfrac {dv_{2}}{dx}}{\cfrac {dM_{xx}}{dx}}~dx+\int _{x_{a}}^{x_{b}}v_{2}~q(x)~dx+\\&\left.\left(v_{2}~N_{xx}{\cfrac {dw_{0}}{dx}}\right)\right|_{x_{a}}^{x_{b}}-\int _{x_{a}}^{x_{b}}{\cfrac {dv_{2}}{dx}}\left(N_{xx}{\cfrac {dw_{0}}{dx}}\right)~dx=0{\text{(6)}}\qquad \end{aligned}}}
Integrate equation (6) again by parts, and get
(
v
2
d
M
x
x
d
x
)
|
x
a
x
b
−
(
d
v
2
d
x
M
x
x
)
|
x
a
x
b
+
∫
x
a
x
b
d
2
v
2
d
x
2
M
x
x
d
x
+
∫
x
a
x
b
v
2
q
(
x
)
d
x
+
(
v
2
N
x
x
d
w
0
d
x
)
|
x
a
x
b
−
∫
x
a
x
b
d
v
2
d
x
(
N
x
x
d
w
0
d
x
)
d
x
=
0
(7)
{\displaystyle {\begin{aligned}\left.\left(v_{2}~{\cfrac {dM_{xx}}{dx}}\right)\right|_{x_{a}}^{x_{b}}&-\left.\left({\cfrac {dv_{2}}{dx}}~M_{xx}\right)\right|_{x_{a}}^{x_{b}}+\int _{x_{a}}^{x_{b}}{\cfrac {d^{2}v_{2}}{dx^{2}}}M_{xx}~dx+\int _{x_{a}}^{x_{b}}v_{2}~q(x)~dx+\\&\left.\left(v_{2}~N_{xx}{\cfrac {dw_{0}}{dx}}\right)\right|_{x_{a}}^{x_{b}}-\int _{x_{a}}^{x_{b}}{\cfrac {dv_{2}}{dx}}\left(N_{xx}{\cfrac {dw_{0}}{dx}}\right)~dx=0{\text{(7)}}\qquad \end{aligned}}}
Collect terms and rearrange equations (5) and (7) to get
∫
x
a
x
b
d
v
1
d
x
N
x
x
d
x
=
∫
x
a
x
b
v
1
f
(
x
)
d
x
+
(
v
1
N
x
x
)
|
x
a
x
b
(8)
∫
x
a
x
b
[
d
v
2
d
x
(
N
x
x
d
w
0
d
x
)
−
d
2
v
2
d
x
2
M
x
x
]
d
x
=
∫
x
a
x
b
v
2
q
(
x
)
d
x
+
[
(
v
2
d
M
x
x
d
x
)
−
(
d
v
2
d
x
M
x
x
)
+
(
v
2
N
x
x
d
w
0
d
x
)
]
x
a
x
b
(9)
{\displaystyle {\begin{aligned}\int _{x_{a}}^{x_{b}}{\cfrac {dv_{1}}{dx}}N_{xx}~dx&=\int _{x_{a}}^{x_{b}}v_{1}~f(x)~dx+\left.(v_{1}~N_{xx})\right|_{x_{a}}^{x_{b}}{\text{(8)}}\qquad \\\int _{x_{a}}^{x_{b}}\left[{\cfrac {dv_{2}}{dx}}\left(N_{xx}{\cfrac {dw_{0}}{dx}}\right)-{\cfrac {d^{2}v_{2}}{dx^{2}}}M_{xx}\right]~dx&=\int _{x_{a}}^{x_{b}}v_{2}~q(x)~dx+\\&\left[\left(v_{2}~{\cfrac {dM_{xx}}{dx}}\right)-\left({\cfrac {dv_{2}}{dx}}~M_{xx}\right)+\left(v_{2}~N_{xx}{\cfrac {dw_{0}}{dx}}\right)\right]_{x_{a}}^{x_{b}}{\text{(9)}}\qquad \end{aligned}}}
Rewriting the equations, we get
∫
x
a
x
b
d
v
1
d
x
N
x
x
d
x
=
∫
x
a
x
b
v
1
f
d
x
+
v
1
N
x
x
|
x
a
x
b
∫
x
a
x
b
[
d
v
2
d
x
(
N
x
x
d
w
0
d
x
)
−
d
2
v
2
d
x
2
M
x
x
]
d
x
=
∫
x
a
x
b
v
2
q
d
x
+
[
v
2
(
d
M
x
x
d
x
+
N
x
x
d
w
0
d
x
)
]
x
a
x
b
−
[
d
v
2
d
x
M
x
x
]
x
a
x
b
(10)
{\displaystyle {\begin{aligned}\int _{x_{a}}^{x_{b}}{\cfrac {dv_{1}}{dx}}N_{xx}~dx&=\int _{x_{a}}^{x_{b}}v_{1}f~dx+\left.v_{1}N_{xx}\right|_{x_{a}}^{x_{b}}\\\int _{x_{a}}^{x_{b}}\left[{\cfrac {dv_{2}}{dx}}\left(N_{xx}{\cfrac {dw_{0}}{dx}}\right)-{\cfrac {d^{2}v_{2}}{dx^{2}}}M_{xx}\right]~dx&=\int _{x_{a}}^{x_{b}}v_{2}q~dx+\left[v_{2}\left({\cfrac {dM_{xx}}{dx}}+N_{xx}{\cfrac {dw_{0}}{dx}}\right)\right]_{x_{a}}^{x_{b}}-\left[{\cfrac {dv_{2}}{dx}}~M_{xx}\right]_{x_{a}}^{x_{b}}\end{aligned}}{\text{(10)}}\qquad }
Hence shown.
The von Karman strains are related to the displacements by
ε
x
x
=
ε
x
x
0
+
z
ε
x
x
1
ε
x
x
0
=
d
u
0
d
x
+
1
2
(
d
w
0
d
x
)
2
ε
x
x
1
=
−
d
2
w
0
d
x
2
{\displaystyle {\begin{aligned}\varepsilon _{xx}&=\varepsilon _{xx}^{0}+z\varepsilon _{xx}^{1}\\\varepsilon _{xx}^{0}&={\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\\\varepsilon _{xx}^{1}&=-{\cfrac {d^{2}w_{0}}{dx^{2}}}\end{aligned}}}
The stress and moment resultants are defined as
N
x
x
=
∫
A
σ
x
x
d
A
M
x
x
=
∫
A
z
σ
x
x
d
A
{\displaystyle {\begin{aligned}N_{xx}&=\int _{A}\sigma _{xx}~dA\\M_{xx}&=\int _{A}z\sigma _{xx}~dA\end{aligned}}}
For a linear elastic material, the stiffnesses of the beam in extension and bending are defined as
A
x
x
=
∫
A
E
d
A
←
extensional stiffness
B
x
x
=
∫
A
z
E
d
A
←
extensional-bending stiffness
D
x
x
=
∫
A
z
2
E
d
A
←
bending stiffness
{\displaystyle {\begin{aligned}A_{xx}&=\int _{A}E~dA\qquad \leftarrow \qquad {\text{extensional stiffness}}\\B_{xx}&=\int _{A}zE~dA\qquad \leftarrow \qquad {\text{extensional-bending stiffness}}\\D_{xx}&=\int _{A}z^{2}E~dA\qquad \leftarrow \qquad {\text{bending stiffness}}\end{aligned}}}
where
E
{\displaystyle E}
is the Young's modulus of the material.
Derive expressions for
N
x
x
{\displaystyle N_{xx}}
and
M
x
x
{\displaystyle M_{xx}}
in terms of the displacements
u
0
{\displaystyle u_{0}}
and
w
0
{\displaystyle w_{0}}
and the extensional and bending stiffnesses of the beam assuming a linear elastic material.
The stress-strain relations for an isotropic linear elastic material are
[
σ
x
x
σ
y
y
σ
z
z
σ
y
z
σ
z
x
σ
x
y
]
=
E
(
1
+
ν
)
(
1
−
2
ν
)
[
1
−
ν
ν
ν
0
0
0
ν
1
−
ν
ν
0
0
0
ν
ν
1
−
ν
0
0
0
0
0
0
(
1
−
2
ν
)
0
0
0
0
0
0
(
1
−
2
ν
)
0
0
0
0
0
0
(
1
−
2
ν
)
]
[
ε
x
x
ε
y
y
ε
z
z
ε
y
z
ε
z
x
ε
x
y
]
{\displaystyle {\begin{bmatrix}\sigma _{xx}\\\sigma _{yy}\\\sigma _{zz}\\\sigma _{yz}\\\sigma _{zx}\\\sigma _{xy}\end{bmatrix}}={\cfrac {E}{(1+\nu )(1-2\nu )}}{\begin{bmatrix}1-\nu &\nu &\nu &0&0&0\\\nu &1-\nu &\nu &0&0&0\\\nu &\nu &1-\nu &0&0&0\\0&0&0&(1-2\nu )&0&0\\0&0&0&0&(1-2\nu )&0\\0&0&0&0&0&(1-2\nu )\end{bmatrix}}{\begin{bmatrix}\varepsilon _{xx}\\\varepsilon _{yy}\\\varepsilon _{zz}\\\varepsilon _{yz}\\\varepsilon _{zx}\\\varepsilon _{xy}\end{bmatrix}}}
Since all strains other than
ε
x
x
{\displaystyle \varepsilon _{xx}}
are zero, the above equations reduce to
σ
x
x
=
E
(
1
−
ν
)
(
1
+
ν
)
(
1
−
2
ν
)
ε
x
x
;
σ
y
y
=
E
ν
(
1
+
ν
)
(
1
−
2
ν
)
ε
x
x
;
σ
z
z
=
E
ν
(
1
+
ν
)
(
1
−
2
ν
)
ε
x
x
.
{\displaystyle \sigma _{xx}={\cfrac {E(1-\nu )}{(1+\nu )(1-2\nu )}}\varepsilon _{xx};\qquad \sigma _{yy}={\cfrac {E\nu }{(1+\nu )(1-2\nu )}}\varepsilon _{xx};\qquad \sigma _{zz}={\cfrac {E\nu }{(1+\nu )(1-2\nu )}}\varepsilon _{xx}~.}
If we ignore the stresses
σ
y
y
{\displaystyle \sigma _{yy}}
and
σ
z
z
{\displaystyle \sigma _{zz}}
, the only allowable value of
ν
{\displaystyle \nu }
is zero. Then the stress-strain relations become
σ
x
x
=
E
ε
x
x
.
{\displaystyle \sigma _{xx}=E\varepsilon _{xx}~.}
Plugging this relation into the stress and moment of stress resultant equations, we get
N
x
x
=
∫
A
E
ε
x
x
d
A
M
x
x
=
∫
A
z
E
ε
x
x
d
A
{\displaystyle {\begin{aligned}N_{xx}&=\int _{A}E\varepsilon _{xx}~dA\\M_{xx}&=\int _{A}zE\varepsilon _{xx}~dA\end{aligned}}}
Plugging in the relations for the strain we get
N
x
x
=
∫
A
E
(
ε
x
x
0
+
z
ε
x
x
1
)
d
A
=
∫
A
E
ε
x
x
0
d
A
+
∫
A
z
E
ε
x
x
1
d
A
M
x
x
=
∫
A
z
E
(
ε
x
x
0
+
z
ε
x
x
1
)
d
A
=
∫
A
z
E
ε
x
x
0
d
A
+
∫
A
z
2
E
ε
x
x
1
d
A
.
{\displaystyle {\begin{aligned}N_{xx}&=\int _{A}E(\varepsilon _{xx}^{0}+z\varepsilon _{xx}^{1})~dA=\int _{A}E\varepsilon _{xx}^{0}~dA+\int _{A}zE\varepsilon _{xx}^{1}~dA\\M_{xx}&=\int _{A}zE(\varepsilon _{xx}^{0}+z\varepsilon _{xx}^{1})~dA=\int _{A}zE\varepsilon _{xx}^{0}~dA+\int _{A}z^{2}E\varepsilon _{xx}^{1}~dA~.\end{aligned}}}
Since both
ε
x
x
0
{\displaystyle \varepsilon _{xx}^{0}}
and
ε
x
x
1
{\displaystyle \varepsilon _{xx}^{1}}
are independent of
y
{\displaystyle y}
and
z
{\displaystyle z}
, we can take these quantities outside the integrals and get
N
x
x
=
ε
x
x
0
∫
A
E
d
A
+
ε
x
x
1
∫
A
z
E
d
A
M
x
x
=
ε
x
x
0
∫
A
z
E
d
A
+
ε
x
x
1
∫
A
z
2
E
d
A
.
{\displaystyle {\begin{aligned}N_{xx}&=\varepsilon _{xx}^{0}\int _{A}E~dA+\varepsilon _{xx}^{1}\int _{A}zE~dA\\M_{xx}&=\varepsilon _{xx}^{0}\int _{A}zE~dA+\varepsilon _{xx}^{1}\int _{A}z^{2}E~dA~.\end{aligned}}}
Using the definitions of the extensional, extensional-bending, and bending stiffness, we can then write
N
x
x
=
ε
x
x
0
A
x
x
+
ε
x
x
1
B
x
x
M
x
x
=
ε
x
x
0
B
x
x
+
ε
x
x
1
D
x
x
.
{\displaystyle {\begin{aligned}N_{xx}&=\varepsilon _{xx}^{0}A_{xx}+\varepsilon _{xx}^{1}B_{xx}\\M_{xx}&=\varepsilon _{xx}^{0}B_{xx}+\varepsilon _{xx}^{1}D_{xx}~.\end{aligned}}}
To write these relations in terms of
u
0
{\displaystyle u_{0}}
and
w
0
{\displaystyle w_{0}}
we substitute the expressions for the von Karman strains to get
N
x
x
=
[
d
u
0
d
x
+
1
2
(
d
w
0
d
x
)
2
]
A
x
x
−
d
2
w
0
d
x
2
B
x
x
M
x
x
=
[
d
u
0
d
x
+
1
2
(
d
w
0
d
x
)
2
]
B
x
x
−
d
2
w
0
d
x
2
D
x
x
.
{\displaystyle {\begin{aligned}N_{xx}&=\left[{\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]A_{xx}-{\cfrac {d^{2}w_{0}}{dx^{2}}}B_{xx}\\M_{xx}&=\left[{\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]B_{xx}-{\cfrac {d^{2}w_{0}}{dx^{2}}}D_{xx}~.\end{aligned}}}
These are the expressions of the resultants in terms of the displacements.
Express the weak forms in terms of the displacements and the extensional and bending stiffnesses.
The weak form equations are
∫
x
a
x
b
d
v
1
d
x
N
x
x
d
x
=
∫
x
a
x
b
v
1
f
d
x
+
v
1
N
x
x
|
x
a
x
b
(11)
∫
x
a
x
b
[
d
v
2
d
x
(
N
x
x
d
w
0
d
x
)
−
d
2
v
2
d
x
2
M
x
x
]
d
x
=
∫
x
a
x
b
v
2
q
d
x
+
[
v
2
(
d
M
x
x
d
x
+
N
x
x
d
w
0
d
x
)
]
x
a
x
b
−
[
d
v
2
d
x
M
x
x
]
x
a
x
b
(12)
{\displaystyle {\begin{aligned}\int _{x_{a}}^{x_{b}}{\cfrac {dv_{1}}{dx}}N_{xx}~dx&=\int _{x_{a}}^{x_{b}}v_{1}f~dx+\left.v_{1}N_{xx}\right|_{x_{a}}^{x_{b}}{\text{(11)}}\qquad \\\int _{x_{a}}^{x_{b}}\left[{\cfrac {dv_{2}}{dx}}\left(N_{xx}{\cfrac {dw_{0}}{dx}}\right)-{\cfrac {d^{2}v_{2}}{dx^{2}}}M_{xx}\right]~dx&=\int _{x_{a}}^{x_{b}}v_{2}q~dx+\left[v_{2}\left({\cfrac {dM_{xx}}{dx}}+N_{xx}{\cfrac {dw_{0}}{dx}}\right)\right]_{x_{a}}^{x_{b}}-\left[{\cfrac {dv_{2}}{dx}}~M_{xx}\right]_{x_{a}}^{x_{b}}{\text{(12)}}\qquad \end{aligned}}}
At this stage we make two more assumptions:
The elastic modulus is constant throughout the cross-section.
The
x
{\displaystyle x}
-axis passes through the centroid of the cross-section.
From the first assumption, we have
B
x
x
=
E
∫
A
z
d
A
.
{\displaystyle B_{xx}=E\int _{A}z~dA~.}
From the second assumption, we get
∫
A
z
d
A
=
0
⟹
B
x
x
=
0
.
{\displaystyle \int _{A}z~dA=0\qquad \implies \qquad B_{xx}=0~.}
Then the relations for
N
x
x
{\displaystyle N_{xx}}
and
M
x
x
{\displaystyle M_{xx}}
reduce to
N
x
x
=
[
d
u
0
d
x
+
1
2
(
d
w
0
d
x
)
2
]
A
x
x
(13)
M
x
x
=
−
d
2
w
0
d
x
2
D
x
x
(14)
.
{\displaystyle {\begin{aligned}N_{xx}&=\left[{\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]A_{xx}{\text{(13)}}\qquad \\M_{xx}&=-{\cfrac {d^{2}w_{0}}{dx^{2}}}D_{xx}{\text{(14)}}\qquad ~.\end{aligned}}}
Let us first consider equation (11). Plugging in the expression for
N
x
x
{\displaystyle N_{xx}}
we get
∫
x
a
x
b
d
v
1
d
x
[
d
u
0
d
x
+
1
2
(
d
w
0
d
x
)
2
]
A
x
x
d
x
=
∫
x
a
x
b
v
1
f
d
x
+
v
1
N
x
x
|
x
a
x
b
{\displaystyle \int _{x_{a}}^{x_{b}}{\cfrac {dv_{1}}{dx}}\left[{\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]A_{xx}~dx=\int _{x_{a}}^{x_{b}}v_{1}f~dx+\left.v_{1}N_{xx}\right|_{x_{a}}^{x_{b}}}
We can also write the above in terms of virtual displacements by defining
δ
u
0
:=
v
1
{\displaystyle \delta u_{0}:=v_{1}}
,
Q
1
:=
−
N
x
x
(
x
a
)
{\displaystyle Q_{1}:=-N_{xx}(x_{a})}
, and
Q
4
:=
N
x
x
(
x
b
)
{\displaystyle Q_{4}:=N_{xx}(x_{b})}
. Then we get
∫
x
a
x
b
d
(
δ
u
0
)
d
x
[
d
u
0
d
x
+
1
2
(
d
w
0
d
x
)
2
]
A
x
x
d
x
=
∫
x
a
x
b
(
δ
u
0
)
f
d
x
+
δ
u
0
(
x
a
)
Q
1
+
δ
u
0
(
x
b
)
Q
4
.
{\displaystyle \int _{x_{a}}^{x_{b}}{\cfrac {d(\delta u_{0})}{dx}}\left[{\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]A_{xx}~dx=\int _{x_{a}}^{x_{b}}(\delta u_{0})f~dx+\delta u_{0}(x_{a})Q_{1}+\delta u_{0}(x_{b})Q_{4}~.}
Next, we do the same for equation (12). Plugging in the expressions for
N
x
x
{\displaystyle N_{xx}}
and
M
x
x
{\displaystyle M_{xx}}
, we get
∫
x
a
x
b
{
d
v
2
d
x
(
[
d
u
0
d
x
+
1
2
(
d
w
0
d
x
)
2
]
A
x
x
d
w
0
d
x
)
+
d
2
v
2
d
x
2
(
d
2
w
0
d
x
2
)
D
x
x
}
d
x
=
∫
x
a
x
b
v
2
q
d
x
+
[
v
2
(
d
M
x
x
d
x
+
N
x
x
d
w
0
d
x
)
]
x
a
x
b
−
[
d
v
2
d
x
M
x
x
]
x
a
x
b
{\displaystyle {\begin{aligned}\int _{x_{a}}^{x_{b}}\left\{{\cfrac {dv_{2}}{dx}}\left(\left[{\cfrac {du_{0}}{dx}}+{\cfrac {1}{2}}~\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]A_{xx}{\cfrac {dw_{0}}{dx}}\right)\right.&+\left.{\cfrac {d^{2}v_{2}}{dx^{2}}}\left({\cfrac {d^{2}w_{0}}{dx^{2}}}\right)D_{xx}\right\}~dx=\int _{x_{a}}^{x_{b}}v_{2}q~dx+\\&\left[v_{2}\left({\cfrac {dM_{xx}}{dx}}+N_{xx}{\cfrac {dw_{0}}{dx}}\right)\right]_{x_{a}}^{x_{b}}-\left[{\cfrac {dv_{2}}{dx}}~M_{xx}\right]_{x_{a}}^{x_{b}}\end{aligned}}}
We can write the above in terms of the virtual displacements and the generalized forces by defining
δ
w
0
:=
v
2
δ
θ
:=
d
v
2
d
x
Q
2
:=
−
[
d
M
x
x
d
x
+
N
x
x
d
w
0
d
x
]
x
a
Q
5
:=
[
d
M
x
x
d
x
+
N
x
x
d
w
0
d
x
]
x
b
Q
3
:=
−
M
x
x
(
x
a
)
Q
6
:=
M
x
x
(
x
b
)
{\displaystyle {\begin{aligned}\delta w_{0}&:=v_{2}\\\delta \theta &:={\cfrac {dv_{2}}{dx}}\\Q_{2}&:=-\left[{\cfrac {dM_{xx}}{dx}}+N_{xx}{\cfrac {dw_{0}}{dx}}\right]_{x_{a}}\\Q_{5}&:=\left[{\cfrac {dM_{xx}}{dx}}+N_{xx}{\cfrac {dw_{0}}{dx}}\right]_{x_{b}}\\Q_{3}&:=-M_{xx}(x_{a})\\Q_{6}&:=M_{xx}(x_{b})\end{aligned}}}
to get
∫
x
a
x
b
{
d
(
δ
w
0
)
d
x
[
d
u
0
d
x
+
1
2
(
d
w
0
d
x
)
2
]
d
w
0
d
x
A
x
x
+
d
2
(
δ
w
0
)
d
x
2
(
d
2
w
0
d
x
2
)
D
x
x
}
d
x
=
∫
x
a
x
b
(
δ
w
0
)
q
d
x
+
δ
w
0
(
x
a
)
Q
2
+
δ
w
0
(
x
b
)
Q
5
+
δ
θ
(
x
a
)
Q
3
+
δ
θ
(
x
b
)
Q
6
.
{\displaystyle {\begin{aligned}\int _{x_{a}}^{x_{b}}\left\{{\cfrac {d(\delta w_{0})}{dx}}\right.&\left[{\cfrac {du_{0}}{dx}}+{\cfrac {1}{2}}~\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]{\cfrac {dw_{0}}{dx}}A_{xx}+\left.{\cfrac {d^{2}(\delta w_{0})}{dx^{2}}}\left({\cfrac {d^{2}w_{0}}{dx^{2}}}\right)D_{xx}\right\}~dx=\\&\int _{x_{a}}^{x_{b}}(\delta w_{0})q~dx+\delta w_{0}(x_{a})Q_{2}+\delta w_{0}(x_{b})Q_{5}+\delta \theta (x_{a})Q_{3}+\delta \theta (x_{b})Q_{6}~.\end{aligned}}}
Assume that the approximate solutions for the axial displacement and the transverse deflection over a two noded element are given by
u
0
(
x
)
=
u
1
ψ
1
(
x
)
+
u
2
ψ
2
(
x
)
(15)
w
0
(
x
)
=
w
1
ϕ
1
(
x
)
+
θ
1
ϕ
2
(
x
)
+
w
2
ϕ
3
(
x
)
+
θ
2
ϕ
4
(
x
)
(16)
{\displaystyle {\begin{aligned}u_{0}(x)&=u_{1}\psi _{1}(x)+u_{2}\psi _{2}(x){\text{(15)}}\qquad \\w_{0}(x)&=w_{1}\phi _{1}(x)+\theta _{1}\phi _{2}(x)+w_{2}\phi _{3}(x)+\theta _{2}\phi _{4}(x){\text{(16)}}\qquad \end{aligned}}}
where
θ
=
−
(
d
w
0
/
d
x
)
{\displaystyle \theta =-(dw_{0}/dx)}
.
Compute the element stiffness matrix for the element.
The weak forms of the governing equations are
∫
x
a
x
b
d
(
δ
u
0
)
d
x
[
d
u
0
d
x
+
1
2
(
d
w
0
d
x
)
2
]
A
x
x
d
x
=
∫
x
a
x
b
(
δ
u
0
)
f
d
x
+
δ
u
0
(
x
a
)
Q
1
+
δ
u
0
(
x
b
)
Q
4
(17)
∫
x
a
x
b
{
d
(
δ
w
0
)
d
x
[
d
u
0
d
x
+
1
2
(
d
w
0
d
x
)
2
]
d
w
0
d
x
A
x
x
+
d
2
(
δ
w
0
)
d
x
2
(
d
2
w
0
d
x
2
)
D
x
x
}
d
x
=
∫
x
a
x
b
(
δ
w
0
)
q
d
x
+
δ
w
0
(
x
a
)
Q
2
+
δ
w
0
(
x
b
)
Q
5
+
δ
θ
(
x
a
)
Q
3
+
δ
θ
(
x
b
)
Q
6
.
(18)
{\displaystyle {\begin{aligned}\int _{x_{a}}^{x_{b}}{\cfrac {d(\delta u_{0})}{dx}}&\left[{\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]A_{xx}~dx=\int _{x_{a}}^{x_{b}}(\delta u_{0})f~dx+\delta u_{0}(x_{a})Q_{1}+\delta u_{0}(x_{b})Q_{4}{\text{(17)}}\qquad \\\int _{x_{a}}^{x_{b}}\left\{{\cfrac {d(\delta w_{0})}{dx}}\right.&\left[{\cfrac {du_{0}}{dx}}+{\cfrac {1}{2}}~\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]{\cfrac {dw_{0}}{dx}}A_{xx}+\left.{\cfrac {d^{2}(\delta w_{0})}{dx^{2}}}\left({\cfrac {d^{2}w_{0}}{dx^{2}}}\right)D_{xx}\right\}~dx=\\&\int _{x_{a}}^{x_{b}}(\delta w_{0})q~dx+\delta w_{0}(x_{a})Q_{2}+\delta w_{0}(x_{b})Q_{5}+\delta \theta (x_{a})Q_{3}+\delta \theta (x_{b})Q_{6}~.{\text{(18)}}\qquad \end{aligned}}}
Let us first write the approximate solutions as
u
0
(
x
)
=
∑
j
=
1
2
u
j
ψ
j
(19)
w
0
(
x
)
=
∑
j
=
1
4
d
j
ϕ
j
(20)
{\displaystyle {\begin{aligned}u_{0}(x)&=\sum _{j=1}^{2}u_{j}\psi _{j}{\text{(19)}}\qquad \\w_{0}(x)&=\sum _{j=1}^{4}d_{j}\phi _{j}{\text{(20)}}\qquad \end{aligned}}}
where
d
j
{\displaystyle d_{j}}
are generalized displacements and
{
d
1
,
d
2
,
d
3
,
d
4
}
≡
{
w
1
,
θ
1
,
w
2
,
θ
2
}
.
{\displaystyle \{d_{1},d_{2},d_{3},d_{4}\}\equiv \{w_{1},\theta _{1},w_{2},\theta _{2}\}~.}
To formulate the finite element system of equations, we substitute the expressions from
u
0
{\displaystyle u_{0}}
and
w
0
{\displaystyle w_{0}}
from equations (19) and (20) into the weak form, and substitute the shape functions
ψ
i
{\displaystyle \psi _{i}}
for
δ
u
0
{\displaystyle \delta u_{0}}
,
ϕ
i
{\displaystyle \phi _{i}}
for
δ
w
0
{\displaystyle \delta w_{0}}
.
For the first equation (17) we get
∫
x
a
x
b
d
ψ
i
d
x
[
(
∑
j
=
1
2
u
j
d
ψ
j
d
x
)
+
1
2
d
w
0
d
x
(
∑
j
=
1
4
d
j
d
ϕ
j
d
x
)
]
A
x
x
d
x
=
∫
x
a
x
b
ψ
i
f
d
x
+
ψ
i
(
x
a
)
Q
1
+
ψ
i
(
x
b
)
Q
4
.
{\displaystyle \int _{x_{a}}^{x_{b}}{\cfrac {d\psi _{i}}{dx}}\left[\left(\sum _{j=1}^{2}u_{j}{\cfrac {d\psi _{j}}{dx}}\right)+{\frac {1}{2}}{\cfrac {dw_{0}}{dx}}\left(\sum _{j=1}^{4}d_{j}{\cfrac {d\phi _{j}}{dx}}\right)\right]A_{xx}~dx=\int _{x_{a}}^{x_{b}}\psi _{i}f~dx+\psi _{i}(x_{a})Q_{1}+\psi _{i}(x_{b})Q_{4}~.}
After reorganizing, we have
∑
j
=
1
2
[
∫
x
a
x
b
A
x
x
d
ψ
i
d
x
d
ψ
j
d
x
d
x
]
u
j
+
∑
j
=
1
4
[
1
2
∫
x
a
x
b
(
A
x
x
d
w
0
d
x
)
d
ψ
i
d
x
d
ϕ
j
d
x
d
x
]
d
j
=
∫
x
a
x
b
ψ
i
f
d
x
+
ψ
i
(
x
a
)
Q
1
+
ψ
i
(
x
b
)
Q
4
.
{\displaystyle {\begin{aligned}\sum _{j=1}^{2}\left[\int _{x_{a}}^{x_{b}}A_{xx}{\cfrac {d\psi _{i}}{dx}}{\cfrac {d\psi _{j}}{dx}}~dx\right]u_{j}&+\sum _{j=1}^{4}\left[{\frac {1}{2}}\int _{x_{a}}^{x_{b}}\left(A_{xx}{\cfrac {dw_{0}}{dx}}\right){\cfrac {d\psi _{i}}{dx}}{\cfrac {d\phi _{j}}{dx}}~dx\right]d_{j}=\\&\int _{x_{a}}^{x_{b}}\psi _{i}f~dx+\psi _{i}(x_{a})Q_{1}+\psi _{i}(x_{b})Q_{4}~.\end{aligned}}}
We can write the above as
∑
j
=
1
2
K
i
j
11
u
j
+
∑
j
=
1
4
K
i
j
12
d
j
=
F
i
1
{\displaystyle \sum _{j=1}^{2}K_{ij}^{11}u_{j}+\sum _{j=1}^{4}K_{ij}^{12}d_{j}=F_{i}^{1}}
where
i
=
1
,
2
{\displaystyle i=1,2}
and
K
i
j
11
=
∫
x
a
x
b
A
x
x
d
ψ
i
d
x
d
ψ
j
d
x
d
x
K
i
j
12
=
1
2
∫
x
a
x
b
(
A
x
x
d
w
0
d
x
)
d
ψ
i
d
x
d
ϕ
j
d
x
d
x
F
i
1
=
∫
x
a
x
b
ψ
i
f
d
x
+
ψ
i
(
x
a
)
Q
1
+
ψ
i
(
x
b
)
Q
4
.
{\displaystyle {\begin{aligned}K_{ij}^{11}&=\int _{x_{a}}^{x_{b}}A_{xx}{\cfrac {d\psi _{i}}{dx}}{\cfrac {d\psi _{j}}{dx}}~dx\\K_{ij}^{12}&={\frac {1}{2}}\int _{x_{a}}^{x_{b}}\left(A_{xx}{\cfrac {dw_{0}}{dx}}\right){\cfrac {d\psi _{i}}{dx}}{\cfrac {d\phi _{j}}{dx}}~dx\\F_{i}^{1}&=\int _{x_{a}}^{x_{b}}\psi _{i}f~dx+\psi _{i}(x_{a})Q_{1}+\psi _{i}(x_{b})Q_{4}~.\end{aligned}}}
For the second equation (18) we get
∫
x
a
x
b
{
d
ϕ
i
d
x
[
(
∑
j
=
1
2
u
j
d
ψ
j
d
x
)
+
1
2
d
w
0
d
x
(
∑
j
=
1
4
d
j
d
ϕ
j
d
x
)
]
d
w
0
d
x
A
x
x
+
d
2
ϕ
i
d
x
2
(
∑
j
=
1
4
d
j
d
2
ϕ
j
d
x
2
)
D
x
x
}
d
x
=
∫
x
a
x
b
ϕ
i
q
d
x
+
ϕ
i
(
x
a
)
Q
2
+
ϕ
i
(
x
b
)
Q
5
+
d
ϕ
i
d
x
(
x
a
)
Q
3
+
d
ϕ
i
d
x
(
x
b
)
Q
6
{\displaystyle {\begin{aligned}\int _{x_{a}}^{x_{b}}\left\{{\cfrac {d\phi _{i}}{dx}}\right.&\left[\left(\sum _{j=1}^{2}u_{j}{\cfrac {d\psi _{j}}{dx}}\right)+{\frac {1}{2}}{\cfrac {dw_{0}}{dx}}\left(\sum _{j=1}^{4}d_{j}{\cfrac {d\phi _{j}}{dx}}\right)\right]{\cfrac {dw_{0}}{dx}}A_{xx}+\left.{\cfrac {d^{2}\phi _{i}}{dx^{2}}}\left(\sum _{j=1}^{4}d_{j}{\cfrac {d^{2}\phi _{j}}{dx^{2}}}\right)D_{xx}\right\}~dx=\\&\int _{x_{a}}^{x_{b}}\phi _{i}q~dx+\phi _{i}(x_{a})Q_{2}+\phi _{i}(x_{b})Q_{5}+{\cfrac {d\phi _{i}}{dx}}(x_{a})Q_{3}+{\cfrac {d\phi _{i}}{dx}}(x_{b})Q_{6}\end{aligned}}}
After rearranging we get
∑
j
=
1
2
[
∫
x
a
x
b
(
A
x
x
d
w
0
d
x
)
d
ϕ
i
d
x
d
ψ
j
d
x
d
x
]
u
j
+
∑
j
=
1
4
{
1
2
∫
x
a
x
b
[
A
x
x
(
d
w
0
d
x
)
2
]
d
ϕ
i
d
x
d
ϕ
j
d
x
d
x
}
d
j
+
∑
j
=
1
4
(
∫
x
a
x
b
D
x
x
d
2
ϕ
i
d
x
2
d
2
ϕ
j
d
x
2
d
x
)
d
j
=
∫
x
a
x
b
ϕ
i
q
d
x
+
ϕ
i
(
x
a
)
Q
2
+
ϕ
i
(
x
b
)
Q
5
+
d
ϕ
i
d
x
(
x
a
)
Q
3
+
d
ϕ
i
d
x
(
x
b
)
Q
6
{\displaystyle {\begin{aligned}\sum _{j=1}^{2}\left[\int _{x_{a}}^{x_{b}}\left(A_{xx}{\cfrac {dw_{0}}{dx}}\right){\cfrac {d\phi _{i}}{dx}}{\cfrac {d\psi _{j}}{dx}}~dx\right]u_{j}&+\sum _{j=1}^{4}\left\{{\frac {1}{2}}\int _{x_{a}}^{x_{b}}\left[A_{xx}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]{\cfrac {d\phi _{i}}{dx}}{\cfrac {d\phi _{j}}{dx}}~dx\right\}d_{j}\\&+\sum _{j=1}^{4}\left(\int _{x_{a}}^{x_{b}}D_{xx}{\cfrac {d^{2}\phi _{i}}{dx^{2}}}{\cfrac {d^{2}\phi _{j}}{dx^{2}}}~dx\right)d_{j}=\\&\int _{x_{a}}^{x_{b}}\phi _{i}q~dx+\phi _{i}(x_{a})Q_{2}+\phi _{i}(x_{b})Q_{5}+{\cfrac {d\phi _{i}}{dx}}(x_{a})Q_{3}+{\cfrac {d\phi _{i}}{dx}}(x_{b})Q_{6}\end{aligned}}}
We can write the above as
∑
j
=
1
2
K
i
j
21
u
j
+
∑
j
=
1
4
K
i
j
22
d
j
=
F
i
2
{\displaystyle \sum _{j=1}^{2}K_{ij}^{21}u_{j}+\sum _{j=1}^{4}K_{ij}^{22}d_{j}=F_{i}^{2}}
where
i
=
1
,
2
,
3
,
4
{\displaystyle i=1,2,3,4}
and
K
i
j
21
=
∫
x
a
x
b
(
A
x
x
d
w
0
d
x
)
d
ϕ
i
d
x
d
ψ
j
d
x
d
x
K
i
j
22
=
∫
x
a
x
b
{
1
2
[
A
x
x
(
d
w
0
d
x
)
2
]
d
ϕ
i
d
x
d
ϕ
j
d
x
+
D
x
x
d
2
ϕ
i
d
x
2
d
2
ϕ
j
d
x
2
}
d
x
F
i
2
=
∫
x
a
x
b
ϕ
i
q
d
x
+
ϕ
i
(
x
a
)
Q
2
+
ϕ
i
(
x
b
)
Q
5
+
d
ϕ
i
d
x
(
x
a
)
Q
3
+
d
ϕ
i
d
x
(
x
b
)
Q
6
.
{\displaystyle {\begin{aligned}K_{ij}^{21}&=\int _{x_{a}}^{x_{b}}\left(A_{xx}{\cfrac {dw_{0}}{dx}}\right){\cfrac {d\phi _{i}}{dx}}{\cfrac {d\psi _{j}}{dx}}~dx\\K_{ij}^{22}&=\int _{x_{a}}^{x_{b}}\left\{{\frac {1}{2}}\left[A_{xx}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]{\cfrac {d\phi _{i}}{dx}}{\cfrac {d\phi _{j}}{dx}}+D_{xx}{\cfrac {d^{2}\phi _{i}}{dx^{2}}}{\cfrac {d^{2}\phi _{j}}{dx^{2}}}\right\}~dx\\F_{i}^{2}&=\int _{x_{a}}^{x_{b}}\phi _{i}q~dx+\phi _{i}(x_{a})Q_{2}+\phi _{i}(x_{b})Q_{5}+{\cfrac {d\phi _{i}}{dx}}(x_{a})Q_{3}+{\cfrac {d\phi _{i}}{dx}}(x_{b})Q_{6}~.\end{aligned}}}
In matrix form, we can write
K
=
[
K
11
11
K
12
11
⋮
K
11
12
K
12
12
K
13
12
K
14
12
K
21
11
K
22
11
⋮
K
21
12
K
22
12
K
23
12
K
24
12
…
K
11
21
K
12
21
⋮
K
11
22
K
12
22
K
13
12
K
14
12
K
21
21
K
22
21
⋮
K
21
22
K
22
22
K
23
12
K
24
12
K
31
21
K
32
21
⋮
K
31
22
K
32
22
K
33
22
K
34
22
K
41
21
K
42
21
⋮
K
41
22
K
42
22
K
43
22
K
44
22
]
{\displaystyle \mathbf {K} ={\begin{bmatrix}K_{11}^{11}&K_{12}^{11}&\vdots &K_{11}^{12}&K_{12}^{12}&K_{13}^{12}&K_{14}^{12}\\K_{21}^{11}&K_{22}^{11}&\vdots &K_{21}^{12}&K_{22}^{12}&K_{23}^{12}&K_{24}^{12}\\&&\dots &&&&\\K_{11}^{21}&K_{12}^{21}&\vdots &K_{11}^{22}&K_{12}^{22}&K_{13}^{12}&K_{14}^{12}\\K_{21}^{21}&K_{22}^{21}&\vdots &K_{21}^{22}&K_{22}^{22}&K_{23}^{12}&K_{24}^{12}\\K_{31}^{21}&K_{32}^{21}&\vdots &K_{31}^{22}&K_{32}^{22}&K_{33}^{22}&K_{34}^{22}\\K_{41}^{21}&K_{42}^{21}&\vdots &K_{41}^{22}&K_{42}^{22}&K_{43}^{22}&K_{44}^{22}\\\end{bmatrix}}}
or
K
=
[
K
11
⋮
K
12
⋮
K
21
⋮
K
22
]
.
{\displaystyle \mathbf {K} ={\begin{bmatrix}\mathbf {K} ^{11}&\vdots &\mathbf {K} ^{12}\\&\vdots &\\\mathbf {K} ^{21}&\vdots &\mathbf {K} ^{22}\end{bmatrix}}~.}
The finite element system of equations can then be written as
(21)
[
K
11
11
K
12
11
⋮
K
11
12
K
12
12
K
13
12
K
14
12
K
21
11
K
22
11
⋮
K
21
12
K
22
12
K
23
12
K
24
12
…
K
11
21
K
12
21
⋮
K
11
22
K
12
22
K
13
12
K
14
12
K
21
21
K
22
21
⋮
K
21
22
K
22
22
K
23
12
K
24
12
K
31
21
K
32
21
⋮
K
31
22
K
32
22
K
33
22
K
34
22
K
41
21
K
42
21
⋮
K
41
22
K
42
22
K
43
22
K
44
22
]
[
u
1
u
2
d
1
d
2
d
3
d
4
]
=
[
F
1
1
F
2
1
F
1
2
F
2
2
F
3
2
F
4
2
]
{\displaystyle {\text{(21)}}\qquad {\begin{bmatrix}K_{11}^{11}&K_{12}^{11}&\vdots &K_{11}^{12}&K_{12}^{12}&K_{13}^{12}&K_{14}^{12}\\K_{21}^{11}&K_{22}^{11}&\vdots &K_{21}^{12}&K_{22}^{12}&K_{23}^{12}&K_{24}^{12}\\&&\dots &&&&\\K_{11}^{21}&K_{12}^{21}&\vdots &K_{11}^{22}&K_{12}^{22}&K_{13}^{12}&K_{14}^{12}\\K_{21}^{21}&K_{22}^{21}&\vdots &K_{21}^{22}&K_{22}^{22}&K_{23}^{12}&K_{24}^{12}\\K_{31}^{21}&K_{32}^{21}&\vdots &K_{31}^{22}&K_{32}^{22}&K_{33}^{22}&K_{34}^{22}\\K_{41}^{21}&K_{42}^{21}&\vdots &K_{41}^{22}&K_{42}^{22}&K_{43}^{22}&K_{44}^{22}\\\end{bmatrix}}{\begin{bmatrix}u_{1}\\u_{2}\\\\d_{1}\\d_{2}\\d_{3}\\d_{4}\end{bmatrix}}={\begin{bmatrix}F_{1}^{1}\\F_{2}^{1}\\\\F_{1}^{2}\\F_{2}^{2}\\F_{3}^{2}\\F_{4}^{2}\end{bmatrix}}}
or
[
K
11
⋮
K
12
⋮
K
21
⋮
K
22
]
[
u
d
]
=
[
F
1
F
2
]
.
{\displaystyle {\begin{bmatrix}\mathbf {K} ^{11}&\vdots &\mathbf {K} ^{12}\\&\vdots &\\\mathbf {K} ^{21}&\vdots &\mathbf {K} ^{22}\end{bmatrix}}{\begin{bmatrix}\mathbf {u} \\\\\mathbf {d} \end{bmatrix}}={\begin{bmatrix}\mathbf {F} ^{1}\\\\\mathbf {F} ^{2}\end{bmatrix}}~.}
Show the alternate procedure by which the element stiffness matrix can be made symmetric.
The stiffness matrix is unsymmetric because
K
12
{\displaystyle \mathbf {K} ^{12}}
contains a factor of
1
/
2
{\displaystyle 1/2}
while
K
21
{\displaystyle \mathbf {K} ^{21}}
does not. The expressions of these terms are
K
i
j
12
=
1
2
∫
x
a
x
b
(
A
x
x
d
w
0
d
x
)
d
ψ
i
d
x
d
ϕ
j
d
x
d
x
,
i
=
1
,
2
,
and
j
=
1
,
2
,
3
,
4
K
i
j
21
=
∫
x
a
x
b
(
A
x
x
d
w
0
d
x
)
d
ϕ
i
d
x
d
ψ
j
d
x
d
x
,
i
=
1
,
2
,
3
,
4
,
and
j
=
1
,
2
.
{\displaystyle {\begin{aligned}K_{ij}^{12}&={\frac {1}{2}}\int _{x_{a}}^{x_{b}}\left(A_{xx}{\cfrac {dw_{0}}{dx}}\right){\cfrac {d\psi _{i}}{dx}}{\cfrac {d\phi _{j}}{dx}}~dx,\qquad i=1,2,~{\text{and}}~j=1,2,3,4\\K_{ij}^{21}&=\int _{x_{a}}^{x_{b}}\left(A_{xx}{\cfrac {dw_{0}}{dx}}\right){\cfrac {d\phi _{i}}{dx}}{\cfrac {d\psi _{j}}{dx}}~dx,\qquad i=1,2,3,4,~{\text{and}}~j=1,2~.\end{aligned}}}
To get a symmetric stiffness matrix, we write equation (18) as
∫
x
a
x
b
{
d
(
δ
w
0
)
d
x
[
1
2
d
u
0
d
x
+
1
2
(
d
w
0
d
x
)
2
+
1
2
d
u
0
d
x
]
d
w
0
d
x
A
x
x
+
d
2
(
δ
w
0
)
d
x
2
(
d
2
w
0
d
x
2
)
D
x
x
}
d
x
=
∫
x
a
x
b
(
δ
w
0
)
q
d
x
+
δ
w
0
(
x
a
)
Q
2
+
δ
w
0
(
x
b
)
Q
5
+
δ
θ
(
x
a
)
Q
3
+
δ
θ
(
x
b
)
Q
6
.
{\displaystyle {\begin{aligned}\int _{x_{a}}^{x_{b}}\left\{{\cfrac {d(\delta w_{0})}{dx}}\right.&\left[{\frac {1}{2}}{\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}+{{\frac {1}{2}}{\cfrac {du_{0}}{dx}}}\right]{\cfrac {dw_{0}}{dx}}A_{xx}+\left.{\cfrac {d^{2}(\delta w_{0})}{dx^{2}}}\left({\cfrac {d^{2}w_{0}}{dx^{2}}}\right)D_{xx}\right\}~dx=\\&\int _{x_{a}}^{x_{b}}(\delta w_{0})q~dx+\delta w_{0}(x_{a})Q_{2}+\delta w_{0}(x_{b})Q_{5}+\delta \theta (x_{a})Q_{3}+\delta \theta (x_{b})Q_{6}~.\end{aligned}}}
The quantity is green is assumed to be known from a previous iteration and adds to the
K
22
{\displaystyle \mathbf {K} ^{22}}
terms.
Repeating the procedure used in the previous question
∫
x
a
x
b
{
d
ϕ
i
d
x
[
1
2
(
∑
j
=
1
2
u
j
d
ψ
j
d
x
)
+
1
2
d
w
0
d
x
(
∑
j
=
1
4
d
j
d
ϕ
j
d
x
)
]
d
w
0
d
x
A
x
x
+
1
2
d
ϕ
i
d
x
d
u
0
d
x
(
∑
j
=
1
4
d
j
d
ϕ
j
d
x
)
A
x
x
+
d
2
ϕ
i
d
x
2
(
∑
j
=
1
4
d
j
d
2
ϕ
j
d
x
2
)
D
x
x
}
d
x
=
∫
x
a
x
b
ϕ
i
q
d
x
+
ϕ
i
(
x
a
)
Q
2
+
ϕ
i
(
x
b
)
Q
5
+
d
ϕ
i
d
x
(
x
a
)
Q
3
+
d
ϕ
i
d
x
(
x
b
)
Q
6
{\displaystyle {\begin{aligned}\int _{x_{a}}^{x_{b}}\left\{{\cfrac {d\phi _{i}}{dx}}\right.&\left[{\frac {1}{2}}\left(\sum _{j=1}^{2}u_{j}{\cfrac {d\psi _{j}}{dx}}\right)+{\frac {1}{2}}{\cfrac {dw_{0}}{dx}}\left(\sum _{j=1}^{4}d_{j}{\cfrac {d\phi _{j}}{dx}}\right)\right]{\cfrac {dw_{0}}{dx}}A_{xx}+{{\frac {1}{2}}{\cfrac {d\phi _{i}}{dx}}{\cfrac {du_{0}}{dx}}\left(\sum _{j=1}^{4}d_{j}{\cfrac {d\phi _{j}}{dx}}\right)A_{xx}+}\\&\left.{\cfrac {d^{2}\phi _{i}}{dx^{2}}}\left(\sum _{j=1}^{4}d_{j}{\cfrac {d^{2}\phi _{j}}{dx^{2}}}\right)D_{xx}\right\}~dx=\\&\int _{x_{a}}^{x_{b}}\phi _{i}q~dx+\phi _{i}(x_{a})Q_{2}+\phi _{i}(x_{b})Q_{5}+{\cfrac {d\phi _{i}}{dx}}(x_{a})Q_{3}+{\cfrac {d\phi _{i}}{dx}}(x_{b})Q_{6}\end{aligned}}}
After rearranging we get
∑
j
=
1
2
[
1
2
∫
x
a
x
b
(
A
x
x
d
w
0
d
x
)
d
ϕ
i
d
x
d
ψ
j
d
x
d
x
]
u
j
+
∑
j
=
1
4
{
1
2
∫
x
a
x
b
[
A
x
x
(
d
w
0
d
x
)
2
]
d
ϕ
i
d
x
d
ϕ
j
d
x
d
x
}
d
j
+
∑
j
=
1
4
[
1
2
∫
x
a
x
b
A
x
x
d
u
0
d
x
d
ϕ
i
d
x
d
ϕ
j
d
x
d
x
]
d
j
+
∑
j
=
1
4
(
∫
x
a
x
b
D
x
x
d
2
ϕ
i
d
x
2
d
2
ϕ
j
d
x
2
d
x
)
d
j
=
∫
x
a
x
b
ϕ
i
q
d
x
+
ϕ
i
(
x
a
)
Q
2
+
ϕ
i
(
x
b
)
Q
5
+
d
ϕ
i
d
x
(
x
a
)
Q
3
+
d
ϕ
i
d
x
(
x
b
)
Q
6
{\displaystyle {\begin{aligned}\sum _{j=1}^{2}&\left[{\frac {1}{2}}\int _{x_{a}}^{x_{b}}\left(A_{xx}{\cfrac {dw_{0}}{dx}}\right){\cfrac {d\phi _{i}}{dx}}{\cfrac {d\psi _{j}}{dx}}~dx\right]u_{j}+\sum _{j=1}^{4}\left\{{\frac {1}{2}}\int _{x_{a}}^{x_{b}}\left[A_{xx}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]{\cfrac {d\phi _{i}}{dx}}{\cfrac {d\phi _{j}}{dx}}~dx\right\}d_{j}+\\&{\sum _{j=1}^{4}\left[{\frac {1}{2}}\int _{x_{a}}^{x_{b}}A_{xx}{\cfrac {du_{0}}{dx}}{\cfrac {d\phi _{i}}{dx}}{\cfrac {d\phi _{j}}{dx}}~dx\right]d_{j}}+\sum _{j=1}^{4}\left(\int _{x_{a}}^{x_{b}}D_{xx}{\cfrac {d^{2}\phi _{i}}{dx^{2}}}{\cfrac {d^{2}\phi _{j}}{dx^{2}}}~dx\right)d_{j}=\\&\int _{x_{a}}^{x_{b}}\phi _{i}q~dx+\phi _{i}(x_{a})Q_{2}+\phi _{i}(x_{b})Q_{5}+{\cfrac {d\phi _{i}}{dx}}(x_{a})Q_{3}+{\cfrac {d\phi _{i}}{dx}}(x_{b})Q_{6}\end{aligned}}}
We can write the above as
∑
j
=
1
2
K
i
j
21
u
j
+
∑
j
=
1
4
K
i
j
22
d
j
=
F
i
2
{\displaystyle \sum _{j=1}^{2}K_{ij}^{21}u_{j}+\sum _{j=1}^{4}K_{ij}^{22}d_{j}=F_{i}^{2}}
where
i
=
1
,
2
,
3
,
4
{\displaystyle i=1,2,3,4}
and
K
i
j
21
=
1
2
∫
x
a
x
b
(
A
x
x
d
w
0
d
x
)
d
ϕ
i
d
x
d
ψ
j
d
x
d
x
K
i
j
22
=
∫
x
a
x
b
{
1
2
A
x
x
[
d
u
0
d
x
+
(
d
w
0
d
x
)
2
]
d
ϕ
i
d
x
d
ϕ
j
d
x
+
D
x
x
d
2
ϕ
i
d
x
2
d
2
ϕ
j
d
x
2
}
d
x
F
i
2
=
∫
x
a
x
b
ϕ
i
q
d
x
+
ϕ
i
(
x
a
)
Q
2
+
ϕ
i
(
x
b
)
Q
5
+
d
ϕ
i
d
x
(
x
a
)
Q
3
+
d
ϕ
i
d
x
(
x
b
)
Q
6
.
{\displaystyle {\begin{aligned}K_{ij}^{21}&={\frac {1}{2}}\int _{x_{a}}^{x_{b}}\left(A_{xx}{\cfrac {dw_{0}}{dx}}\right){\cfrac {d\phi _{i}}{dx}}{\cfrac {d\psi _{j}}{dx}}~dx\\K_{ij}^{22}&=\int _{x_{a}}^{x_{b}}\left\{{\frac {1}{2}}A_{xx}\left[{\cfrac {du_{0}}{dx}}+\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]{\cfrac {d\phi _{i}}{dx}}{\cfrac {d\phi _{j}}{dx}}+D_{xx}{\cfrac {d^{2}\phi _{i}}{dx^{2}}}{\cfrac {d^{2}\phi _{j}}{dx^{2}}}\right\}~dx\\F_{i}^{2}&=\int _{x_{a}}^{x_{b}}\phi _{i}q~dx+\phi _{i}(x_{a})Q_{2}+\phi _{i}(x_{b})Q_{5}+{\cfrac {d\phi _{i}}{dx}}(x_{a})Q_{3}+{\cfrac {d\phi _{i}}{dx}}(x_{b})Q_{6}~.\end{aligned}}}
This gives us a symmetric stiffness matrix.
Derive the element tangent stiffness matrix for the element.
Equation (21) can be written as
K
(
U
)
U
=
F
{\displaystyle \mathbf {K} (\mathbf {U} )\mathbf {U} =\mathbf {F} }
where
U
1
=
u
1
,
U
2
=
u
2
,
U
3
=
d
1
,
U
4
=
d
2
,
U
5
=
d
3
,
U
6
=
d
4
F
1
=
F
1
1
,
F
2
=
F
2
1
,
F
3
=
F
1
2
,
F
4
=
F
2
2
,
F
5
=
F
3
2
,
F
6
=
F
4
2
{\displaystyle {\begin{aligned}U_{1}&=u_{1},~U_{2}=u_{2},~U_{3}=d_{1},~U_{4}=d_{2},~U_{5}=d_{3},~U_{6}=d_{4}\\F_{1}&=F_{1}^{1},~F_{2}=F_{2}^{1},~F_{3}=F_{1}^{2},~F_{4}=F_{2}^{2},~F_{5}=F_{3}^{2},~F_{6}=F_{4}^{2}\end{aligned}}}
The residual is
R
=
K
U
−
F
.
{\displaystyle \mathbf {R} =\mathbf {K} \mathbf {U} -\mathbf {F} ~.}
For Newton iterations, we use the algorithm
U
r
+
1
=
U
r
−
(
T
r
)
−
1
R
r
{\displaystyle \mathbf {U} ^{r+1}=\mathbf {U} ^{r}-(\mathbf {T} ^{r})^{-1}\mathbf {R} ^{r}}
where the tangent stiffness matrix is given by
T
r
=
∂
R
r
∂
U
.
{\displaystyle \mathbf {T} ^{r}={\frac {\partial \mathbf {R} ^{r}}{\partial \mathbf {U} }}~.}
The coefficients of the tangent stiffness matrix are given by
T
i
j
=
∂
R
i
∂
U
j
,
i
=
1
…
6
,
j
=
1
…
6
.
{\displaystyle T_{ij}={\frac {\partial R_{i}}{\partial U_{j}}},\qquad i=1\dots 6,j=1\dots 6~.}
Recall that the finite element system of equations can be written as
∑
p
=
1
2
K
m
p
11
u
p
+
∑
q
=
1
4
K
m
q
12
d
q
=
F
m
1
,
m
=
1
,
2
∑
p
=
1
2
K
n
p
21
u
p
+
∑
q
=
1
4
K
n
q
22
d
q
=
F
n
2
,
n
=
1
,
2
,
3
,
4
{\displaystyle {\begin{aligned}\sum _{p=1}^{2}K_{mp}^{11}u_{p}&+\sum _{q=1}^{4}K_{mq}^{12}d_{q}=F_{m}^{1},\qquad m=1,2\\\sum _{p=1}^{2}K_{np}^{21}u_{p}&+\sum _{q=1}^{4}K_{nq}^{22}d_{q}=F_{n}^{2},\qquad n=1,2,3,4\end{aligned}}}
where the subscripts have been changed to avoid confusion.
Therefore, the residuals are
R
m
1
=
∑
p
=
1
2
K
m
p
11
u
p
+
∑
q
=
1
4
K
m
q
12
d
q
−
F
m
1
,
m
=
1
,
2
R
n
2
=
∑
p
=
1
2
K
n
p
21
u
p
+
∑
q
=
1
4
K
n
q
22
d
q
−
F
n
2
,
n
=
1
,
2
,
3
,
4
.
{\displaystyle {\begin{aligned}R_{m}^{1}&=\sum _{p=1}^{2}K_{mp}^{11}u_{p}+\sum _{q=1}^{4}K_{mq}^{12}d_{q}-F_{m}^{1},\qquad m=1,2\\R_{n}^{2}&=\sum _{p=1}^{2}K_{np}^{21}u_{p}+\sum _{q=1}^{4}K_{nq}^{22}d_{q}-F_{n}^{2},\qquad n=1,2,3,4~.\end{aligned}}}
The derivatives of the residuals with respect to the generalized displacements are
T
m
k
11
=
∂
R
m
1
∂
u
k
=
∑
p
=
1
2
∂
∂
u
k
(
K
m
p
11
u
p
)
+
∑
q
=
1
4
∂
∂
u
k
(
K
m
q
12
d
q
)
,
m
=
1
,
2
;
k
=
1
,
2
T
m
l
12
=
∂
R
m
1
∂
d
l
=
∑
p
=
1
2
∂
∂
d
l
(
K
m
p
11
u
p
)
+
∑
q
=
1
4
∂
∂
d
l
(
K
m
q
12
d
q
)
,
m
=
1
,
2
;
l
=
1
,
2
,
3
,
4
T
n
k
21
=
∂
R
n
2
∂
u
k
=
∑
p
=
1
2
∂
∂
u
k
(
K
n
p
21
u
p
)
+
∑
q
=
1
4
∂
∂
u
k
(
K
n
q
22
d
q
)
,
n
=
1
,
2
,
3
,
4
;
k
=
1
,
2
T
n
l
22
=
∂
R
n
2
∂
d
l
=
∑
p
=
1
2
∂
∂
d
l
(
K
n
p
21
u
p
)
+
∑
q
=
1
4
∂
∂
d
l
(
K
n
q
22
d
q
)
,
n
=
1
,
2
,
3
,
4
;
l
=
1
,
2
,
3
,
4
.
{\displaystyle {\begin{aligned}T_{mk}^{11}={\frac {\partial R_{m}^{1}}{\partial u_{k}}}&=\sum _{p=1}^{2}{\frac {\partial }{\partial u_{k}}}(K_{mp}^{11}u_{p})+\sum _{q=1}^{4}{\frac {\partial }{\partial u_{k}}}(K_{mq}^{12}d_{q}),&&\qquad m=1,2;~~k=1,2\\T_{ml}^{12}={\frac {\partial R_{m}^{1}}{\partial d_{l}}}&=\sum _{p=1}^{2}{\frac {\partial }{\partial d_{l}}}(K_{mp}^{11}u_{p})+\sum _{q=1}^{4}{\frac {\partial }{\partial d_{l}}}(K_{mq}^{12}d_{q}),&&\qquad m=1,2;~~l=1,2,3,4\\T_{nk}^{21}={\frac {\partial R_{n}^{2}}{\partial u_{k}}}&=\sum _{p=1}^{2}{\frac {\partial }{\partial u_{k}}}(K_{np}^{21}u_{p})+\sum _{q=1}^{4}{\frac {\partial }{\partial u_{k}}}(K_{nq}^{22}d_{q}),&&\qquad n=1,2,3,4;~~k=1,2\\T_{nl}^{22}={\frac {\partial R_{n}^{2}}{\partial d_{l}}}&=\sum _{p=1}^{2}{\frac {\partial }{\partial d_{l}}}(K_{np}^{21}u_{p})+\sum _{q=1}^{4}{\frac {\partial }{\partial d_{l}}}(K_{nq}^{22}d_{q}),&&\qquad n=1,2,3,4;~~l=1,2,3,4~.\end{aligned}}}
Differentiating, we get
T
m
k
11
=
∑
p
=
1
2
(
u
p
∂
K
m
p
11
∂
u
k
+
K
m
p
11
∂
u
p
∂
u
k
)
+
∑
q
=
1
4
(
d
q
∂
K
m
q
12
∂
u
k
+
K
m
q
12
∂
d
q
∂
u
k
)
,
m
=
1
,
2
;
k
=
1
,
2
T
m
l
12
=
∑
p
=
1
2
(
u
p
∂
K
m
p
11
∂
d
l
+
K
m
p
11
∂
u
p
∂
d
l
)
+
∑
q
=
1
4
(
d
q
∂
K
m
q
12
∂
d
l
+
K
m
q
12
∂
d
q
∂
d
l
)
,
m
=
1
,
2
;
l
=
1
,
2
,
3
,
4
T
n
k
21
=
∑
p
=
1
2
(
u
p
∂
K
n
p
21
∂
u
k
+
K
n
p
21
∂
u
p
∂
u
k
)
+
∑
q
=
1
4
(
d
q
∂
K
n
q
22
∂
u
k
+
K
n
q
22
∂
d
q
∂
u
k
)
,
n
=
1
,
2
,
3
,
4
;
k
=
1
,
2
T
n
l
22
=
∑
p
=
1
2
(
u
p
∂
K
n
p
21
∂
d
l
+
K
n
p
21
∂
u
p
∂
d
l
)
+
∑
q
=
1
4
(
d
q
∂
K
n
q
22
∂
d
l
+
K
n
q
22
∂
d
q
∂
d
l
)
,
n
=
1
,
2
,
3
,
4
;
l
=
1
,
2
,
3
,
4
.
{\displaystyle {\begin{aligned}T_{mk}^{11}&=\sum _{p=1}^{2}\left(u_{p}{\frac {\partial K_{mp}^{11}}{\partial u_{k}}}+K_{mp}^{11}{\frac {\partial u_{p}}{\partial u_{k}}}\right)+\sum _{q=1}^{4}\left(d_{q}{\frac {\partial K_{mq}^{12}}{\partial u_{k}}}+K_{mq}^{12}{\frac {\partial d_{q}}{\partial u_{k}}}\right),&&\qquad m=1,2;~~k=1,2\\T_{ml}^{12}&=\sum _{p=1}^{2}\left(u_{p}{\frac {\partial K_{mp}^{11}}{\partial d_{l}}}+K_{mp}^{11}{\frac {\partial u_{p}}{\partial d_{l}}}\right)+\sum _{q=1}^{4}\left(d_{q}{\frac {\partial K_{mq}^{12}}{\partial d_{l}}}+K_{mq}^{12}{\frac {\partial d_{q}}{\partial d_{l}}}\right),&&\qquad m=1,2;~~l=1,2,3,4\\T_{nk}^{21}&=\sum _{p=1}^{2}\left(u_{p}{\frac {\partial K_{np}^{21}}{\partial u_{k}}}+K_{np}^{21}{\frac {\partial u_{p}}{\partial u_{k}}}\right)+\sum _{q=1}^{4}\left(d_{q}{\frac {\partial K_{nq}^{22}}{\partial u_{k}}}+K_{nq}^{22}{\frac {\partial d_{q}}{\partial u_{k}}}\right),&&\qquad n=1,2,3,4;~~k=1,2\\T_{nl}^{22}&=\sum _{p=1}^{2}\left(u_{p}{\frac {\partial K_{np}^{21}}{\partial d_{l}}}+K_{np}^{21}{\frac {\partial u_{p}}{\partial d_{l}}}\right)+\sum _{q=1}^{4}\left(d_{q}{\frac {\partial K_{nq}^{22}}{\partial d_{l}}}+K_{nq}^{22}{\frac {\partial d_{q}}{\partial d_{l}}}\right),&&\qquad n=1,2,3,4;l=1,2,3,4~.\end{aligned}}}
These equations can therefore be written as
T
m
k
11
=
K
m
k
11
+
∑
p
=
1
2
u
p
∂
K
m
p
11
∂
u
k
+
∑
q
=
1
4
d
q
∂
K
m
q
12
∂
u
k
,
m
=
1
,
2
;
k
=
1
,
2
T
m
l
12
=
K
m
l
12
+
∑
p
=
1
2
u
p
∂
K
m
p
11
∂
d
l
+
∑
q
=
1
4
d
q
∂
K
m
q
12
∂
d
l
,
m
=
1
,
2
;
l
=
1
,
2
,
3
,
4
T
n
k
21
=
K
n
k
21
+
∑
p
=
1
2
u
p
∂
K
n
p
21
∂
u
k
+
∑
q
=
1
4
d
q
∂
K
n
q
22
∂
u
k
,
n
=
1
,
2
,
3
,
4
;
k
=
1
,
2
T
n
l
22
=
K
n
l
22
+
∑
p
=
1
2
u
p
∂
K
n
p
21
∂
d
l
+
∑
q
=
1
4
d
q
∂
K
n
q
22
∂
d
l
,
n
=
1
,
2
,
3
,
4
;
l
=
1
,
2
,
3
,
4
.
{\displaystyle {\begin{aligned}T_{mk}^{11}&=K_{mk}^{11}+\sum _{p=1}^{2}u_{p}{\frac {\partial K_{mp}^{11}}{\partial u_{k}}}+\sum _{q=1}^{4}d_{q}{\frac {\partial K_{mq}^{12}}{\partial u_{k}}},&&\qquad m=1,2;~~k=1,2\\T_{ml}^{12}&=K_{ml}^{12}+\sum _{p=1}^{2}u_{p}{\frac {\partial K_{mp}^{11}}{\partial d_{l}}}+\sum _{q=1}^{4}d_{q}{\frac {\partial K_{mq}^{12}}{\partial d_{l}}},&&\qquad m=1,2;~~l=1,2,3,4\\T_{nk}^{21}&=K_{nk}^{21}+\sum _{p=1}^{2}u_{p}{\frac {\partial K_{np}^{21}}{\partial u_{k}}}+\sum _{q=1}^{4}d_{q}{\frac {\partial K_{nq}^{22}}{\partial u_{k}}},&&\qquad n=1,2,3,4;~~k=1,2\\T_{nl}^{22}&=K_{nl}^{22}+\sum _{p=1}^{2}u_{p}{\frac {\partial K_{np}^{21}}{\partial d_{l}}}+\sum _{q=1}^{4}d_{q}{\frac {\partial K_{nq}^{22}}{\partial d_{l}}},&&\qquad n=1,2,3,4;l=1,2,3,4~.\end{aligned}}}
Now, the coefficients of
K
11
{\displaystyle \mathbf {K} ^{11}}
,
K
12
{\displaystyle \mathbf {K} ^{12}}
, and
K
21
{\displaystyle \mathbf {K} ^{21}}
of the symmetric stiffness matrix are independent of
u
1
{\displaystyle u_{1}}
and
u
2
{\displaystyle u_{2}}
. Also, the terms of
K
11
{\displaystyle \mathbf {K} ^{11}}
are independent of the all the generalized displacements. Therefore, the above equations reduce to
T
m
k
11
=
K
m
k
11
,
m
=
1
,
2
;
k
=
1
,
2
T
m
l
12
=
K
m
l
12
+
∑
q
=
1
4
d
q
∂
K
m
q
12
∂
d
l
,
m
=
1
,
2
;
l
=
1
,
2
,
3
,
4
T
n
k
21
=
K
n
k
21
+
∑
q
=
1
4
d
q
∂
K
n
q
22
∂
u
k
,
n
=
1
,
2
,
3
,
4
;
k
=
1
,
2
T
n
l
22
=
K
n
l
22
+
∑
p
=
1
2
u
p
∂
K
n
p
21
∂
d
l
+
∑
q
=
1
4
d
q
∂
K
n
q
22
∂
d
l
,
n
=
1
,
2
,
3
,
4
;
l
=
1
,
2
,
3
,
4
.
{\displaystyle {\begin{aligned}T_{mk}^{11}&=K_{mk}^{11},&&\qquad m=1,2;~~k=1,2\\T_{ml}^{12}&=K_{ml}^{12}+\sum _{q=1}^{4}d_{q}{\frac {\partial K_{mq}^{12}}{\partial d_{l}}},&&\qquad m=1,2;~~l=1,2,3,4\\T_{nk}^{21}&=K_{nk}^{21}+\sum _{q=1}^{4}d_{q}{\frac {\partial K_{nq}^{22}}{\partial u_{k}}},&&\qquad n=1,2,3,4;~~k=1,2\\T_{nl}^{22}&=K_{nl}^{22}+\sum _{p=1}^{2}u_{p}{\frac {\partial K_{np}^{21}}{\partial d_{l}}}+\sum _{q=1}^{4}d_{q}{\frac {\partial K_{nq}^{22}}{\partial d_{l}}},&&\qquad n=1,2,3,4;l=1,2,3,4~.\end{aligned}}}
Consider the coefficients of
T
12
{\displaystyle \mathbf {T} ^{12}}
:
T
m
l
12
=
K
m
l
12
+
∑
q
=
1
4
d
q
∂
K
m
q
12
∂
d
l
,
m
=
1
,
2
;
l
=
1
,
2
,
3
,
4
.
{\displaystyle T_{ml}^{12}=K_{ml}^{12}+\sum _{q=1}^{4}d_{q}{\frac {\partial K_{mq}^{12}}{\partial d_{l}}},\qquad m=1,2;~~l=1,2,3,4~.}
From our previous derivation, we have
K
m
q
12
=
1
2
∫
x
a
x
b
(
A
x
x
d
w
0
d
x
)
d
ψ
m
d
x
d
ϕ
q
d
x
d
x
.
{\displaystyle K_{mq}^{12}={\frac {1}{2}}\int _{x_{a}}^{x_{b}}\left(A_{xx}{\cfrac {dw_{0}}{dx}}\right){\cfrac {d\psi _{m}}{dx}}{\cfrac {d\phi _{q}}{dx}}~dx~.}
Therefore,
∂
K
m
q
12
∂
d
l
=
1
2
∫
x
a
x
b
[
A
x
x
∂
∂
d
l
(
d
w
0
d
x
)
]
d
ψ
m
d
x
d
ϕ
q
d
x
d
x
=
1
2
∫
x
a
x
b
[
A
x
x
(
∑
T
=
1
4
∂
d
T
∂
d
l
d
ϕ
T
d
x
)
]
d
ψ
m
d
x
d
ϕ
q
d
x
d
x
=
1
2
∫
x
a
x
b
[
A
x
x
d
ϕ
l
d
x
]
d
ψ
m
d
x
d
ϕ
q
d
x
d
x
{\displaystyle {\begin{aligned}{\frac {\partial K_{mq}^{12}}{\partial d_{l}}}&={\frac {1}{2}}\int _{x_{a}}^{x_{b}}\left[A_{xx}{\frac {\partial }{\partial d_{l}}}\left({\cfrac {dw_{0}}{dx}}\right)\right]{\cfrac {d\psi _{m}}{dx}}{\cfrac {d\phi _{q}}{dx}}~dx\\&={\frac {1}{2}}\int _{x_{a}}^{x_{b}}\left[A_{xx}\left(\sum _{T=1}^{4}{\frac {\partial d_{T}}{\partial d_{l}}}{\cfrac {d\phi _{T}}{dx}}\right)\right]{\cfrac {d\psi _{m}}{dx}}{\cfrac {d\phi _{q}}{dx}}~dx\\&={\frac {1}{2}}\int _{x_{a}}^{x_{b}}\left[A_{xx}{\cfrac {d\phi _{l}}{dx}}\right]{\cfrac {d\psi _{m}}{dx}}{\cfrac {d\phi _{q}}{dx}}~dx\end{aligned}}}
The tangent stiffness matrix coefficients are therefore
T
m
l
12
=
K
m
l
12
+
∑
q
=
1
4
d
q
{
1
2
∫
x
a
x
b
[
A
x
x
d
ϕ
l
d
x
]
d
ψ
m
d
x
d
ϕ
q
d
x
d
x
}
m
=
1
,
2
;
l
=
1
,
2
,
3
,
4
.
=
K
m
l
12
+
1
2
∫
x
a
x
b
A
x
x
d
ϕ
l
d
x
d
ψ
m
d
x
(
∑
q
=
1
4
d
q
d
ϕ
q
d
x
)
d
x
=
K
m
l
12
+
1
2
∫
x
a
x
b
A
x
x
d
ϕ
l
d
x
d
ψ
m
d
x
d
w
0
d
x
d
x
=
2
K
m
l
12
.
{\displaystyle {\begin{aligned}{T_{ml}^{12}}&=K_{ml}^{12}+\sum _{q=1}^{4}d_{q}\left\{{\frac {1}{2}}\int _{x_{a}}^{x_{b}}\left[A_{xx}{\cfrac {d\phi _{l}}{dx}}\right]{\cfrac {d\psi _{m}}{dx}}{\cfrac {d\phi _{q}}{dx}}~dx\right\}\qquad m=1,2;~~l=1,2,3,4~.\\&=K_{ml}^{12}+{\frac {1}{2}}\int _{x_{a}}^{x_{b}}A_{xx}{\cfrac {d\phi _{l}}{dx}}{\cfrac {d\psi _{m}}{dx}}\left(\sum _{q=1}^{4}d_{q}{\cfrac {d\phi _{q}}{dx}}\right)~dx\\&=K_{ml}^{12}+{\frac {1}{2}}\int _{x_{a}}^{x_{b}}A_{xx}{\cfrac {d\phi _{l}}{dx}}{\cfrac {d\psi _{m}}{dx}}{\cfrac {dw_{0}}{dx}}~dx\\&={2K_{ml}^{12}}~.\end{aligned}}}
Next, {consider the coefficients of
T
21
{\displaystyle \mathbf {T} ^{21}}
:
T
n
k
21
=
K
n
k
21
+
∑
q
=
1
4
d
q
∂
K
n
q
22
∂
u
k
.
{\displaystyle T_{nk}^{21}=K_{nk}^{21}+\sum _{q=1}^{4}d_{q}{\frac {\partial K_{nq}^{22}}{\partial u_{k}}}~.}
The coefficients of
K
22
{\displaystyle \mathbf {K} ^{22}}
are
K
n
q
22
=
∫
x
a
x
b
{
1
2
A
x
x
[
d
u
0
d
x
+
(
d
w
0
d
x
)
2
]
d
ϕ
n
d
x
d
ϕ
q
d
x
+
D
x
x
d
2
ϕ
n
d
x
2
d
2
ϕ
q
d
x
2
}
d
x
.
{\displaystyle K_{nq}^{22}=\int _{x_{a}}^{x_{b}}\left\{{\frac {1}{2}}A_{xx}\left[{\cfrac {du_{0}}{dx}}+\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]{\cfrac {d\phi _{n}}{dx}}{\cfrac {d\phi _{q}}{dx}}+D_{xx}{\cfrac {d^{2}\phi _{n}}{dx^{2}}}{\cfrac {d^{2}\phi _{q}}{dx^{2}}}\right\}~dx~.}
Therefore, the derivatives are
∂
K
n
q
22
∂
u
k
=
∫
x
a
x
b
1
2
A
x
x
[
∑
T
=
1
2
∂
u
T
∂
u
k
d
ψ
T
d
x
+
2
d
w
0
d
x
(
∑
T
=
1
4
∂
d
T
∂
u
k
d
ϕ
T
d
x
)
]
d
ϕ
n
d
x
d
ϕ
q
d
x
d
x
=
1
2
∫
x
a
x
b
A
x
x
d
ψ
k
d
x
d
ϕ
n
d
x
d
ϕ
q
d
x
d
x
{\displaystyle {\begin{aligned}{\frac {\partial K_{nq}^{22}}{\partial u_{k}}}&=\int _{x_{a}}^{x_{b}}{\frac {1}{2}}A_{xx}\left[\sum _{T=1}^{2}{\frac {\partial u_{T}}{\partial u_{k}}}{\cfrac {d\psi _{T}}{dx}}+2{\cfrac {dw_{0}}{dx}}\left(\sum _{T=1}^{4}{\frac {\partial d_{T}}{\partial u_{k}}}{\cfrac {d\phi _{T}}{dx}}\right)\right]{\cfrac {d\phi _{n}}{dx}}{\cfrac {d\phi _{q}}{dx}}~dx\\&={\frac {1}{2}}\int _{x_{a}}^{x_{b}}A_{xx}{\cfrac {d\psi _{k}}{dx}}{\cfrac {d\phi _{n}}{dx}}{\cfrac {d\phi _{q}}{dx}}~dx\\\end{aligned}}}
Therefore the coefficients of
T
21
{\displaystyle \mathbf {T} ^{21}}
are
T
n
k
21
=
K
n
k
21
+
∑
q
=
1
4
d
q
(
1
2
∫
x
a
x
b
A
x
x
d
ψ
k
d
x
d
ϕ
n
d
x
d
ϕ
q
d
x
d
x
)
=
K
n
k
21
+
1
2
∫
x
a
x
b
A
x
x
d
ψ
k
d
x
d
ϕ
n
d
x
(
∑
q
=
1
4
d
q
d
ϕ
q
d
x
)
d
x
=
K
n
k
21
+
1
2
∫
x
a
x
b
A
x
x
d
ψ
k
d
x
d
ϕ
n
d
x
d
w
0
d
x
d
x
=
2
K
n
k
21
.
{\displaystyle {\begin{aligned}{T_{nk}^{21}}&=K_{nk}^{21}+\sum _{q=1}^{4}d_{q}\left({\frac {1}{2}}\int _{x_{a}}^{x_{b}}A_{xx}{\cfrac {d\psi _{k}}{dx}}{\cfrac {d\phi _{n}}{dx}}{\cfrac {d\phi _{q}}{dx}}~dx\right)\\&=K_{nk}^{21}+{\frac {1}{2}}\int _{x_{a}}^{x_{b}}A_{xx}{\cfrac {d\psi _{k}}{dx}}{\cfrac {d\phi _{n}}{dx}}\left(\sum _{q=1}^{4}d_{q}{\cfrac {d\phi _{q}}{dx}}\right)~dx\\&=K_{nk}^{21}+{\frac {1}{2}}\int _{x_{a}}^{x_{b}}A_{xx}{\cfrac {d\psi _{k}}{dx}}{\cfrac {d\phi _{n}}{dx}}{\cfrac {dw_{0}}{dx}}~dx\\&={2K_{nk}^{21}}~.\end{aligned}}}
Finally, for the
T
22
{\displaystyle \mathbf {T} ^{22}}
coefficients, we start with
T
n
l
22
=
K
n
l
22
+
∑
p
=
1
2
u
p
∂
K
n
p
21
∂
d
l
+
∑
q
=
1
4
d
q
∂
K
n
q
22
∂
d
l
,
n
=
1
,
2
,
3
,
4
;
l
=
1
,
2
,
3
,
4
{\displaystyle T_{nl}^{22}=K_{nl}^{22}+\sum _{p=1}^{2}u_{p}{\frac {\partial K_{np}^{21}}{\partial d_{l}}}+\sum _{q=1}^{4}d_{q}{\frac {\partial K_{nq}^{22}}{\partial d_{l}}},\qquad n=1,2,3,4;l=1,2,3,4}
and plug in the derivatives of the stiffness matrix coefficients
K
n
p
21
=
1
2
∫
x
a
x
b
(
A
x
x
d
w
0
d
x
)
d
ϕ
n
d
x
d
ψ
p
d
x
d
x
K
n
q
22
=
∫
x
a
x
b
{
1
2
A
x
x
[
d
u
0