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})}}}}}
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}}}}
Given:
Consider the axially loaded bar in problem VM 59 of the ANSYS Verification manual. Assume that the bar is made of Tungsten carbide.
The input file for ANSYS is as shown in VM 59 except the following material properties are used:
E
=
100
{\displaystyle E=100}
ρ
=
0.567
/
386
{\displaystyle \rho =0.567/386}
⋅
{\displaystyle \cdot }
2
/
{\displaystyle ^{2}/}
4
{\displaystyle ^{4}}
Find the fundamental natural frequency of the bar.
f
1
=
36.963
Hz
{\displaystyle {f_{1}=36.963{\mbox{ Hz}}}}
Find the first three modal frequencies for a load of 40,000 lbf.
f
1
=
33.174
Hz
f
2
=
144.107
Hz
f
3
=
328.443
Hz
{\displaystyle {f_{1}=33.174{\mbox{ Hz}}\quad f_{2}=144.107{\mbox{ Hz}}\quad f_{3}=328.443{\mbox{ Hz}}}}
Given:
Consider the stretched circular membrane in problem VM 55 of the ANSYS
Verification manual. Assume that the membrane is made of OFHC (Oxygen-free
High Conductivity) copper.
The input file for ANSYS is as shown in VM 55 except the following material properties are used:
E
=
17
{\displaystyle E=17}
ρ
=
0.322
/
386
{\displaystyle \rho =0.322/386}
⋅
{\displaystyle \cdot }
2
/
{\displaystyle ^{2}/}
4
{\displaystyle ^{4}}
Find the fundamental natural frequency of the bar.
f
1
=
1.509
Hz
{\displaystyle {f_{1}=1.509{\mbox{ Hz}}}}
Find the first five modal frequencies for a load of 10,000 lbf.
f
1
=
88.329
Hz
f
2
=
203.114
Hz
f
3
=
320.174
Hz
f
4
=
441.655
Hz
f
5
=
571.350
Hz
{\displaystyle {f_{1}=88.329{\mbox{ Hz}}\quad f_{2}=203.114{\mbox{ Hz}}\quad f_{3}=320.174{\mbox{ Hz}}\quad f_{4}=441.655{\mbox{ Hz}}\quad f_{5}=571.350{\mbox{ Hz}}}}
![{\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)
