2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 20 - Beams, plates, and shells
Prof. K.J. Bathe
MIT OpenCourseWare
Timoshenko beam theory
The fiber moves up and rotates and its length does not change.
Principle of virtual displacement
�
EI
L
�
� � �T
β
β � dx + (Ak)G
0
0
L
(Linear Analysis)
�
dw
−β
dx
�T �
�
� L
dw
− β dx =
wT pdx
dx
0
(20.1)
Two-node element:
Three-node element:
For a q-node element,
�T
�
û = w1 · · · wq θ1 · · · θq
w = Hw û
β = Hβ û
�
�
Hw = h1 · · · hq 0 · · · 0
�
�
Hβ = 0 · · · 0 h1 · · · hq
dx
J=
dr
(20.2)
(20.3)
(20.4)
(20.5)
(20.6)
(20.7)
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MIT 2.094
20. Beams, plates, and shells
dw
= J −1 Hw,r û
dx
� �� �
(20.8)
dβ
ˆ
= J −1 Hβ,r u
dx � �� �
(20.9)
Bw
Bβ
Hence we obtain
� � 1
�
EI
BβT Bβ det(J )dr + (Ak)G
−1
1
�
T
(Bw − Hβ ) (Bw − Hβ ) det(J )dr û
−1
�
1
=
HwT p det(J )dr
(20.10)
−1
Kû = R
(20.11)
K is a result of the term inside the bracket in (20.10) and R is a result of the right hand side.
For the 2-node element,
w1 = θ1 = 0
(20.12)
w2 , θ2 = ?
(20.13)
γ=
1+r
w2
−
θ2
L
2
(20.14)
We cannot make γ equal to zero for every r (page 404, textbook). Because of this, we need to use
about 200 elements to get an error of 10%. (Not good!)
Recall almost or fully incompressible analysis: Principle of virtual displacements:
�
�
�T � �
� C � dV +
�v (κ�v )dV = R
V
(20.15)
V
u/p formulation
�
�
T
�� C � �� dV −
�v pdV = R
V
V
�
�p
�
p
+ �v dV = 0
κ
V
(20.16)
(20.17)
But now we needed to select wisely the interpolations of u and p. We needed to satisfy the inf-sup
condition
�
qh � · vh dVol
≥β>0
(20.18)
inf sup Vol
����
����
�qh ��vh �
qh ∈Qh vh ∈Vh
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MIT 2.094
20. Beams, plates, and shells
4/1 element:
We can show mathematically that this element does not satisfy inf-sup condition. But, we can also
show it by giving an example of this element which violates the inf-sup condition.
v1 = �,
above
�
v2 = 0 ⇒ � · vh for both elements is positive and the same. Now, if I choose pressures as
qh �vh dVol = 0,
hence (20.18) is not satisfied!
(20.19)
Vol
9/3 element
satisfies inf-sup
9/4-c
satisfies inf-sup
Getting back to beams
� L
�
�
EI
β βdx + (AkG)
0
�
0
L
�
�
dw
− β γ AS dx = R
dx
(20.20)
L
�
�
γ AS γ − γ AS dx = 0
(20.21)
dw
− β,
dx
(20.22)
0
where
γ=
from displacement interpolation
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MIT 2.094
20. Beams, plates, and shells
γ AS = Assumed shear strain interpolation
(20.23)
2-node element, constant shear assumption. From (20.21),
�
L
�
0
�
� L
dw
�dx =
�dx
−β �
γ AS
γ AS �
γ AS
dx
0
�
+1
�
⇒−
−1
⇒ γ AS =
1+r
θ2
2
�
·
Reading:
Sec. 4.5.7
(20.24)
L
dr + w2 = γ AS · L
2
(20.25)
w2 − L2 θ2
L
(20.26)
γ AS (shear strain) is equal to the displacement-based shear strain at the middle of the beam.
Use γ AS in (20.20) to obtain a powerful element. For “our problem”,
γ AS = 0
�
⇒EI
0
L
θ2
2
(20.27)
�
�
β β � dx = M β �x=L
(20.28)
hence
L
w2 =
�� �
�
2
1
· L θ2 = M
⇒ EI
L
⇒ θ2 =
ML
,
EI
w2 =
(20.29)
M L2
2EI
(20.30)
(exact solutions)
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MIT 2.094
20. Beams, plates, and shells
Plates
Reading:
Fig. 5.25,
p. 421
⎧
⎪
⎨ w = w(x, y) is the transverse displacement of the mid-surface
v = −zβy (x, y)
⎪
⎩
u = −zβx (x, y)
(20.31)
For any particle in the plate with coordinates (x, y, z), the expressions in (20.31) hold!
We use
w=
q
�
hi wi
(20.32)
i=1
βx = −
βy = +
q
�
i=1
q
�
hi θyi
(20.33)
hi θxi
(20.34)
i=1
where q equals the number of nodes. Then the element locks in the same way as the displacement-based
beam element.
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2.094 Finite Element Analysis of Solids and Fluids II
Spring 2011
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