2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 2 - Finite element formulation of solids and structures
Prof. K.J. Bathe
MIT OpenCourseWare
Assume that on t Su the displacements are zero (and t Su is constant). Need to satisfy at time t:
• Equilibrium of Cauchy stresses t τij with applied loads
�
�
t T
τ = t τ11 t τ22 t τ33 t τ12 t τ23 t τ31
(2.1)
(For i = 1, 2, 3)
t
τij,j + t fiB
t
=
t
τij nj
=
t Sf
fi
=
(e.g.
0 in t V (sum over j)
(2.2)
t Sf
fi on t Sf (sum over j)
t
τi1 t n1 + t τi2 t n2 + t τi3 t n3
(2.3)
)
(2.4)
S
And: t τ11 t n1 + t τ12 t n2 = t f1 f
• Compatibility The displacements t ui need to be continuous and zero on t Su .
• Stress-Strain law
t
� �
τij = function t uj
(2.5)
7
Reading:
Ch. 1, Sec.
6.1-6.2
MIT 2.094
2. Finite element formulation of solids and structures
Principle of Virtual Work∗
2.1
�
t
τij t eij d t V =
�
tV
tV
t B
fi
ui d t V +
�
tS
t Sf
fi
S
u i f d t Sf
(2.6)
f
where
t eij
�
�
with ui �
t
=
1
2
�
∂ui
∂uj
+ t
∂ t xj
∂ xi
�
(2.7)
= 0
(2.8)
Su
2.2
Example
Assume “plane sections remain plane”
Principle of Virtual Work
�
�
t
t
τ11 t e11 d V =
tV
tV
t B
f1
t
�
t
u1 d V +
tS
S
Pr u 1 f d t S f
(2.9)
f
Derivation of (2.9)
t
�t
∗ or
τ11,1 + t f1B = 0
�
τ11,1 + t f1B u1 = 0
by (2.2)
(2.10)
(2.11)
Principle of Virtual Displacements
8
MIT 2.094
Hence,
�
2. Finite element formulation of solids and structures
�t
tV
�
τ11,1 + t f1B u1 d t V = 0
� tS
τ11 u1 � t Sf −
� �� u�
t
�
t
tV
S
u1 f tτ11 tSf
t
(2.12)
�
u1,1 τ d V +
���� 11
tV
u1 t f1B
d t V = 0
(2.13)
te11
where tτ11 | tS = tPr .
f
Therefore we have
�
�
t
t
e
τ
d
V
=
t 11
11
tV
S
u1 t f1B d t V + u1 f t Pr t Sf
tV
(2.14)
From (2.12) to (2.14) we simply used mathematics. Hence, if (2.2) and (2.3) are satisfied, then (2.14)
must hold. If (2.14) holds, then also (2.2) and (2.3) hold!
Namely, from (2.14)
�
�
� tS
u1,1 tτ11 d tV = u1 tτ11 � tSf −
u
tV
tV
u1 tτ11,1 d tV =
�
tV
S
u1 tf1B d tV + u1 f
tPr tSf
(2.15)
or
�
u1
tV
�t
�
�
S �
τ11,1 + tf1B d tV + u1 f tPr − tτ11 tSf = 0
�
Now let u1 = x 1 −
x
tL
��
(2.16)
�
τ11,1 + tf1B , where tL = length of bar.
t
Hence we must have from (2.16)
t
τ11,1 + tf1B = 0
(2.17)
and then also
t
Pr = tτ11
(2.18)
9
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2.094 Finite Element Analysis of Solids and Fluids II
Spring 2011
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