2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 19 - Slender structures
Prof. K.J. Bathe
Beam analysis,
t
L
MIT OpenCourseWare
� 1 (e.g.
t
L
=
1
1
100 , 1000 , · · · )
Reading:
Sec. 5.4,
6.5
(plane stress)
� L
�
0
2
J=
0 2t
1
h2 = (1 − r) (1 + s)
4
1
h3 = (1 − r) (1 − s)
4
(19.1)
(19.2)
(19.3)
Beam theory assumptions (Timoshenko beam theory):
v2∗ = v3∗ = v2
t
u3∗ = u2 + θ2
2
t
∗
u 2 = u2 − θ 2
2
⎡
B∗ = ⎢
⎢
⎢
⎢
⎣
(19.4)
(19.5)
(19.6)
u∗2
v2∗
u∗3
v3∗
− 41 (1 + s) L2
0
− 14 (1 − s) L2
0
0
1
4 (1
− r) 2t
1
4 (1
−
r) 2t
− 14 (1 + s) L2
− 14 (1
0
− 14 (1 − r) 2t
81
−
⎤
r) 2t
− 14 (1 − s) L2
⎥ etc
⎥
⎥
⎥
⎦
(19.7)
MIT 2.094
19. Slender structures
u2
⎡
Bbeam
⎢
= ⎢
⎢ ···
⎢
⎣
− L1
v2
θ2
t
2L s
0
0
�
0
�
0
�
0
− L1
− 21 (1 − r)
⎤
∂u
∂x
⎛
⎜
⎥ ⎜
⎥ ⎜
⎥∼⎜
⎥ ⎜
⎦ ⎜
⎝
⎞
0
∂v
�
∂y
�
∂u
∂y
+
∂v
∂x
⎟
⎟
⎟
⎟
⎟
⎟
⎠
(19.8)
1
(1 − r)v2
2
1
st
u(r) = (1 − r)u2 − (1 − r)θ2
2
4
v(r) =
(19.9)
(19.10)
at r = −1,
v(−1) = v2
(19.11)
st
u(−1) = − θ2 + u2
2
(19.12)
Kinematics is
u(r) =
1
(1 − r)u2
2
(19.13)
results into �xx
→ �xx =
∂u 2
1
· =−
∂r L
L
(19.14)
st
(1 − r)θ2
4
(19.15)
u(r, s) = −
results into �xx , γxy
→ �xx =
γxy =
v(r) =
st
2L
∂u 2
1
· = − (1 − r)
∂s t
2
1
(1 − r)v2
2
(19.16)
(19.17)
(19.18)
results into γxy
→ γxy = −
82
1
L
(19.19)
MIT 2.094
19. Slender structures
For a pure bending moment, we want
1
1
− v2 − (1 − r)θ2 = 0
L
2
(19.20)
for all r! ⇒ Impossible (except for v2 = θ2 = 0) ⇒ So, the element has a spurious shear strain!
Beam kinematics (Timoshenko, Reissner-Mindlin)
γ=
�
dw
−β
dx
1 3
I=
bt
12
(19.21)
�
(19.22)
Principle of virtual work
�
L
EI
0
dβ dβ
dx + AS G
dx dx
�
L
�
0
dw
−β
dx
��
�
� L
dw
− β dx =
pwdx
dx
0
As = kA = kbt
(19.23)
(19.24)
To calculate k
� �2
�
�
V
1
1
2
(τa ) dA =
dAs
A
2G
2G
s
A
AS
Reading:
p. 400
(19.25)
where τa is the actual shear stress:
� � �2
�
t
− y2
3 V
2
τa = ·
� t �2
2 A
(19.26)
2
and V is the shear force.
⇒k=
Reading:
Ex. 5.23
5
6
(19.27)
83
MIT 2.094
19. Slender structures
Now interpolate
w(r) = h1 w1 + h2 w2
(19.28)
β(r) = h1 θ1 + h2 θ2
(19.29)
Revisit the simple case:
1+r
w1
2
1+r
β=
θ1
2
w=
(19.30)
(19.31)
Shearing strain
γ=
1+r
w1
−
θ1
L
2
(19.32)
Shear strain is not zero all along the beam. But, at r = 0, we can have the shear strain = 0.
θ1
w1
−
can be zero
L
2
(19.33)
Namely,
θ1
w1
2
−
= 0 for θ1 = w1
L
L
2
(19.34)
84
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2.094 Finite Element Analysis of Solids and Fluids II
Spring 2011
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