Problem 1: Total Lagrangian
edit
Given:
Consider the tapered two-node element shown in Figure 1. The displacement field in the element is linear.
Figure 1. Tapered two-node element.
The reference (initial) cross-sectional area is
A
0
=
(
1
−
ξ
)
A
01
+
ξ
A
02
.
{\displaystyle A_{0}=(1-\xi )~A_{01}+\xi ~A_{02}~.}
Assume that the nominal (engineering) stress is also linear in the element, i.e.,
P
=
(
1
−
ξ
)
P
1
+
ξ
P
2
.
{\displaystyle P=(1-\xi )~P_{1}+\xi ~P_{2}~.}
Using the total Lagrangian formulation, develop expressions for the internal nodal forces.
The displacement field is given by the linear Lagrange interpolation expressed in terms of the material coordinate.
u
(
X
,
t
)
=
1
l
0
[
X
2
−
X
X
−
X
1
]
[
u
1
(
t
)
u
2
(
t
)
]
{\displaystyle \mathbf {u} (X,t)={\frac {1}{l_{0}}}[X_{2}-X\quad X-X_{1}]{\begin{bmatrix}u_{1}(t)\\u_{2}(t)\end{bmatrix}}}
where
l
0
=
X
2
−
X
1
{\displaystyle l_{0}=X_{2}-X_{1}}
ε
(
X
,
t
)
=
u
,
X
=
1
l
0
[
−
1
1
]
[
u
1
(
t
)
u
2
(
t
)
]
{\displaystyle {\boldsymbol {\varepsilon }}(X,t)=u_{,X}={\frac {1}{l_{0}}}[-1\quad 1]{\begin{bmatrix}u_{1}(t)\\u_{2}(t)\end{bmatrix}}}
which defines the
B
0
{\displaystyle \mathbf {B} _{0}}
B
0
=
1
l
0
[
−
1
1
]
.
{\displaystyle \mathbf {B} _{0}={\frac {1}{l_{0}}}[-1\quad 1].}
The internal nodal forces are then given by the usual relations.
f
e
int
=
∫
X
1
X
2
B
0
T
(
A
0
P
)
d
X
=
1
l
0
∫
X
1
X
2
(
(
1
−
ξ
)
A
01
+
ξ
A
02
)
(
(
1
−
ξ
)
P
1
+
ξ
P
2
)
[
−
1
1
]
d
X
{\displaystyle \mathbf {f} _{e}^{\mbox{int}}=\int _{X_{1}}^{X_{2}}\mathbf {B} _{0}^{T}(A_{0}P)dX={\frac {1}{l_{0}}}\int _{X_{1}}^{X_{2}}\left((1-\xi )A_{01}+\xi A_{02}\right)\left((1-\xi )P_{1}+\xi P_{2}\right){\begin{bmatrix}-1\\1\end{bmatrix}}dX}
Integrating the above integral with
ξ
=
(
X
−
X
1
)
/
l
0
{\displaystyle \xi =(X-X_{1})/l_{0}}
f
e
int
=
1
6
(
2
A
02
P
2
+
A
02
P
1
+
A
01
P
2
+
2
A
01
P
1
)
[
−
1
1
]
{\displaystyle {\mathbf {f} _{e}^{\mbox{int}}={\frac {1}{6}}\left(2A_{02}P_{2}+A_{02}P_{1}+A_{01}P_{2}+2A_{01}P_{1}\right){\begin{bmatrix}-1\\1\end{bmatrix}}}}
What are the internal nodal forces if the reference area and the nominal stress are constant over the element?
f
e
int
=
A
0
P
[
−
1
1
]
{\displaystyle {\mathbf {f} _{e}^{\mbox{int}}=A_{0}P{\begin{bmatrix}-1\\1\end{bmatrix}}}}
Assume that the body force is constant. Develop expressions for the external nodal forces for that case.
The external body forces arising from the body force,
b
{\displaystyle b}
f
e
ext
=
∫
X
1
X
2
N
T
ρ
0
A
0
b
d
X
=
ρ
0
b
l
0
∫
X
1
X
2
A
0
[
X
2
−
X
X
−
X
1
]
d
X
{\displaystyle \mathbf {f} _{e}^{\mbox{ext}}=\int _{X_{1}}^{X_{2}}\mathbf {N} ^{T}\rho _{0}A_{0}bdX={\frac {\rho _{0}b}{l_{0}}}\int _{X_{1}}^{X_{2}}A_{0}{\begin{bmatrix}X_{2}-X\\X-X_{1}\end{bmatrix}}dX}
f
e
ext
=
ρ
0
b
6
[
x
1
2
A
02
+
2
A
01
x
1
2
−
2
x
1
A
02
x
2
−
4
A
01
x
1
x
2
+
2
A
01
x
2
2
+
A
02
x
2
2
A
01
x
1
2
+
2
x
1
2
A
02
−
4
x
1
A
02
x
2
−
2
A
01
x
1
x
2
+
A
01
x
2
2
+
2
A
02
x
2
2
]
{\displaystyle {\mathbf {f} _{e}^{\mbox{ext}}={\frac {\rho _{0}b}{6}}{\begin{bmatrix}x_{1}^{2}A_{02}+2A_{01}x_{1}^{2}-2x_{1}A_{02}x_{2}-4A_{01}x_{1}x_{2}+2A_{01}x_{2}^{2}+A_{02}x_{2}^{2}\\A_{01}x_{1}^{2}+2x_{1}^{2}A_{02}-4x_{1}A_{02}x_{2}-2A_{01}x_{1}x_{2}+A_{01}x_{2}^{2}+2A_{02}x_{2}^{2}\end{bmatrix}}}}
What are the external nodal forces if the reference area and the nominal stress are constant over the element?
f
e
ext
=
b
A
0
ρ
0
l
0
2
2
[
1
1
]
{\displaystyle {\mathbf {f} _{e}^{\mbox{ext}}={\frac {bA_{0}\rho _{0}l_{0}^{2}}{2}}{\begin{bmatrix}1\\1\end{bmatrix}}}}
Develop an expression for the consistent mass matrix for the element.
The element mass matrix is
M
e
=
∫
X
1
X
2
ρ
0
A
0
N
T
N
d
X
{\displaystyle \mathbf {M} _{e}=\int _{X_{1}}^{X_{2}}\rho _{0}A_{0}\mathbf {N} ^{T}\mathbf {N} dX}
M
e
=
ρ
0
l
0
12
[
3
A
01
+
A
02
A
01
+
A
02
A
01
+
A
02
A
01
+
3
A
02
]
{\displaystyle {\mathbf {M} _{e}={\frac {\rho _{0}l_{0}}{12}}{\begin{bmatrix}3A_{01}+A_{02}&A_{01}+A_{02}\\A_{01}+A_{02}&A_{01}+3A_{02}\end{bmatrix}}}}
Obtain the lumped (diagonal) mass matrix using the row-sum technique.
Lumped mass matrix is given by
M
i
i
=
∫
ξ
1
ξ
2
ρ
0
A
0
N
i
d
X
{\displaystyle M_{ii}=\int _{\xi _{1}}^{\xi _{2}}\rho _{0}A_{0}N_{i}dX}
M
e
=
ρ
0
l
0
6
[
2
A
01
+
A
02
0
0
A
01
+
2
A
02
]
{\displaystyle {\mathbf {M} _{e}={\frac {\rho _{0}l_{0}}{6}}{\begin{bmatrix}2A_{01}+A_{02}&0\\0&A_{01}+2A_{02}\end{bmatrix}}}}
Find the natural frequencies of a single element with consistent mass by solving the eigenvalue problem
K
u
=
ω
2
M
u
{\displaystyle \mathbf {K} ~\mathbf {u} =\omega ^{2}~\mathbf {M} ~\mathbf {u} }
with
K
=
E
(
A
01
+
A
02
)
2
l
0
[
1
−
1
−
1
1
]
{\displaystyle \mathbf {K} ={\cfrac {E(A_{01}+A_{02})}{2l_{0}}}{\begin{bmatrix}1&-1\\-1&1\end{bmatrix}}}
where
E
{\displaystyle E}
l
0
{\displaystyle l_{0}}
The above equation can be rewrite as
(
K
−
ω
2
M
)
u
=
0
{\displaystyle (\mathbf {K} -\omega ^{2}\mathbf {M} )\mathbf {u} =0}
which only has a solution if
det
(
K
−
ω
2
M
)
=
0
{\displaystyle {\mbox{det}}(\mathbf {K} -\omega ^{2}\mathbf {M} )=0}
Solving the above determinant for
ω
{\displaystyle \omega }
ω
=
0
,
±
18
E
(
A
01
2
+
2
A
01
A
02
+
A
02
2
)
ρ
0
l
0
2
(
A
01
2
+
4
A
01
A
02
+
A
02
2
)
{\displaystyle {\omega =0,\pm {\sqrt {\frac {18E(A_{01}^{2}+2A_{01}A_{02}+A_{02}^{2})}{\rho _{0}l_{0}^{2}(A_{01}^{2}+4A_{01}A_{02}+A_{02}^{2})}}}}}
Problem 2: Updated Lagrangian
edit
Given:
Consider the tapered two-node element shown in Figure 1.
The current cross-sectional area is
A
=
(
1
−
ξ
)
A
1
+
ξ
A
2
.
{\displaystyle A=(1-\xi )~A_{1}+\xi ~A_{2}~.}
Assume that the Cauchy stress is also linear in the element, i.e.,
σ
=
(
1
−
ξ
)
σ
1
+
ξ
σ
2
.
{\displaystyle \sigma =(1-\xi )~\sigma _{1}+\xi ~\sigma _{2}~.}
Using the updated Lagrangian formulation, develop expressions for the internal nodal forces.
The velocity field is
v
(
X
,
t
)
=
1
l
0
[
X
2
−
X
X
−
X
1
]
[
v
1
(
t
)
v
2
(
t
)
]
{\displaystyle v(X,t)={\frac {1}{l_{0}}}\left[X_{2}-X\quad X-X_{1}\right]{\begin{bmatrix}v_{1}(t)\\v_{2}(t)\end{bmatrix}}}
In term of element coordinates, the velocity field is
v
(
ξ
,
t
)
=
1
l
0
[
1
−
ξ
ξ
]
[
v
1
(
t
)
v
2
(
t
)
]
ξ
=
X
−
X
1
l
0
{\displaystyle v(\xi ,t)={\frac {1}{l_{0}}}\left[1-\xi \quad \xi \right]{\begin{bmatrix}v_{1}(t)\\v_{2}(t)\end{bmatrix}}\qquad \xi ={\frac {X-X_{1}}{l_{0}}}}
The displacement is the time integrals of the velocity,
and since
ξ
{\displaystyle \xi }
u
(
ξ
,
t
)
=
N
(
ξ
)
u
e
(
t
)
{\displaystyle u(\xi ,t)=\mathbf {N} (\xi )\mathbf {u} _{e}(t)}
Therefore, since
x
=
X
+
u
{\displaystyle x=X+u}
x
(
ξ
,
t
)
=
N
(
ξ
)
x
e
(
t
)
=
[
1
−
ξ
ξ
]
[
x
1
(
t
)
x
2
(
t
)
]
ξ
,
ξ
=
x
2
−
x
1
=
l
{\displaystyle x(\xi ,t)=\mathbf {N} (\xi )\mathbf {x} _{e}(t)=\left[1-\xi \quad \xi \right]{\begin{bmatrix}x_{1}(t)\\x_{2}(t)\end{bmatrix}}\qquad \xi _{,\xi }=x_{2}-x_{1}=l}
where
l
{\displaystyle l}
ξ
{\displaystyle \xi }
ξ
=
x
−
x
1
x
2
−
x
1
=
x
−
x
1
l
,
l
=
x
2
−
x
1
,
ξ
,
x
=
1
l
{\displaystyle \xi ={\frac {x-x_{1}}{x_{2}-x_{1}}}={\frac {x-x_{1}}{l}},\quad l=x_{2}-x_{1},\quad \xi _{,x}={\frac {1}{l}}}
So
ξ
,
x
{\displaystyle \xi _{,x}}
x
,
ξ
{\displaystyle x_{,\xi }}
The
B
{\displaystyle \mathbf {B} }
B
=
N
,
x
=
N
,
ξ
ξ
,
x
=
1
l
[
−
1
1
]
{\displaystyle \mathbf {B} =\mathbf {N} _{,x}=\mathbf {N} _{,\xi }\xi _{,x}={\frac {1}{l}}[-1\quad 1]}
Using (146) in Handout 13, we have
f
e
int
=
∫
x
1
x
2
B
T
σ
A
d
x
=
∫
x
1
x
2
σ
A
l
[
−
1
1
]
d
x
{\displaystyle \mathbf {f} _{e}^{\mbox{int}}=\int _{x_{1}}^{x_{2}}\mathbf {B} ^{T}\sigma Adx=\int _{x_{1}}^{x_{2}}{\frac {\sigma A}{l}}{\begin{bmatrix}-1\\1\end{bmatrix}}dx}
Integrating the above equation to obtain
f
e
int
=
1
6
(
A
1
σ
2
+
2
(
A
2
σ
2
+
A
1
σ
1
)
+
A
2
σ
1
)
[
−
1
1
]
{\displaystyle {\mathbf {f} _{e}^{\mbox{int}}={\frac {1}{6}}\left(A_{1}\sigma _{2}+2(A_{2}\sigma _{2}+A_{1}\sigma _{1})+A_{2}\sigma _{1}\right){\begin{bmatrix}-1\\1\end{bmatrix}}}}
Assume that the body force is constant. Develop expressions for the external nodal forces for that case.
The external forces are given by
f
e
ext
=
ρ
b
6
[
2
x
2
A
1
+
x
2
A
2
−
2
x
1
A
1
−
x
1
A
2
2
x
2
A
2
+
x
2
A
1
−
2
x
1
A
2
−
x
1
A
1
]
{\displaystyle {\mathbf {f} _{e}^{\mbox{ext}}={\frac {\rho b}{6}}{\begin{bmatrix}2x_{2}A_{1}+x_{2}A_{2}-2x_{1}A_{1}-x_{1}A_{2}\\2x_{2}A_{2}+x_{2}A_{1}-2x_{1}A_{2}-x_{1}A_{1}\end{bmatrix}}}}
Problem 3: Modal Analysis
edit
![{\displaystyle \mathbf {u} (X,t)={\frac {1}{l_{0}}}[X_{2}-X\quad X-X_{1}]{\begin{bmatrix}u_{1}(t)\\u_{2}(t)\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/743aadc9c0b3f5a2603cee96f00e3ba2fcf68ad4)
![{\displaystyle {\boldsymbol {\varepsilon }}(X,t)=u_{,X}={\frac {1}{l_{0}}}[-1\quad 1]{\begin{bmatrix}u_{1}(t)\\u_{2}(t)\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd11485cb49ca08a5a0a3d58dfc0e9af76d41289)
![{\displaystyle \mathbf {B} _{0}={\frac {1}{l_{0}}}[-1\quad 1].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0db546525b3398f5a79cb747b39546bafca864cf)
![{\displaystyle v(X,t)={\frac {1}{l_{0}}}\left[X_{2}-X\quad X-X_{1}\right]{\begin{bmatrix}v_{1}(t)\\v_{2}(t)\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cacd53d734a24c7309aace1aa414976e31012fc6)
![{\displaystyle v(\xi ,t)={\frac {1}{l_{0}}}\left[1-\xi \quad \xi \right]{\begin{bmatrix}v_{1}(t)\\v_{2}(t)\end{bmatrix}}\qquad \xi ={\frac {X-X_{1}}{l_{0}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a66d1a56c0f1bcbf897ee2cf6c36a199088a1a8)
![{\displaystyle x(\xi ,t)=\mathbf {N} (\xi )\mathbf {x} _{e}(t)=\left[1-\xi \quad \xi \right]{\begin{bmatrix}x_{1}(t)\\x_{2}(t)\end{bmatrix}}\qquad \xi _{,\xi }=x_{2}-x_{1}=l}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a564acf494b0692b10a429c0544fd239b416bbc)
![{\displaystyle \mathbf {B} =\mathbf {N} _{,x}=\mathbf {N} _{,\xi }\xi _{,x}={\frac {1}{l}}[-1\quad 1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4702926e081737cff1b26a242008680fc3be6075)
