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2.092/2.093 — Finite Element Analysis of Solids & Fluids I Fall ‘09

Lecture 6 - Finite Element Solution Process

Prof. K. J. Bathe MIT OpenCourseWare

In the last lecture, we used the principle of virtual displacements to obtain the following equations:

KU = R (1)

K = Σ K(m) ; K(m) = B(m)T C(m)B(m)dV (m)

m V (m)

R = RB + RS

RB = Σ RB (m) ; RB

(m) = H(m)T fB(m)dV (m)

m V (m)

f f

m S S i i(m) f S

RS = Σ R(m) ; R(m) = Σ HS

i(m)T fS

i(m)

dSi(m)

f

u(m) = H(m)U (2) ↓

ε(m) = B(m)U (3)

Note that the dimension of u(m) is in general not the same as the dimension of ε(m).

Example: Static Analysis

Reading assignment: Example 4.5

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Lecture 6 Finite Element Solution Process 2.092/2.093, Fall ‘09

Assume:

i. Plane sections remain plane

ii. Static analysis no vibrations/no transient response →

iii. One-dimensional problem; hence, only one degree of freedom per node

Elements 1 and 2 are compatible because they use the same U2. Next, use a linear interpolation function.

⎡

(1)(x) = x x ⎣u 1 − 100 100 0

H(1) ⎡

1 1ε(1)(x) = � − 100 ��100 0 � ⎣

B(1) ⎡ ⎤ 1 � 100 ⎢

− 100 ⎥� K = E 1 ⎢ 1 ⎥ 1 1 · · ⎣ 100 ⎦ −100 100

0 0 ⎡ ⎤ ⎡

= E ⎣⎢⎢

−1

1

−1

1

0

0

⎦⎥⎥ +

13E ⎣⎢⎢ 0

0

100 3 80· 0 0 0 0

⎤ ⎡ ⎤ U1 � � � � U1 ⎦ ; u(2)(x) = 0 x x ⎣ ⎦U2 1 − 80 80 U2

U3 U3 H(2) ⎤ ⎡ ⎤

U1 � � U1 ⎦ ε(2)(x) = 1 1 ⎣ ⎦U2 ; 0 80 80 U2 � −�� �U3 U3 B(2) ⎡ ⎤ � � 80 � x �2 ⎢

0 ⎥� � ⎢ 1 ⎥ 1 10 dx + E 0

1 + 40 ⎣ − 80 ⎦ 0 −80 80

dx 1 80 ⎤

0 0 ⎥1 −1 ⎦⎥

−1 1

13 2The “equivalent cross-sectional area” of element 2 is A = 3 cm . This equivalent area must lie between the areas of the end faces A = 1 and A = 9.

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Lecture 6 Finite Element Solution Process 2.092/2.093, Fall ‘09

⎤⎡

E K =

240

⎢⎢⎣

2.4 −2.4 0

−2.4 15.4 −13 ⎥⎥⎦

0 −13 13

We note:

• Diagonal terms must be positive. If the diagonal terms are zero or negative, then the system is unstable physically. A positive diagonal implies that the degree of freedom has stiffness at that node.

• K is symmetric.

• K is singular if rigid body motions are possible. To be able to solve the problem, all rigid body modes must be removed by adequately constraining the structure. i.e. K is reduced by applying boundary conditions to the nodes.

The K used to solve for U is, then, positive definite (det K > 0). This ensures that the elastic strain energy is positive and nonzero for any displacement field U . In the analysis, each element is in equilibrium under its nodal forces, and each node is in equilibrium when summing element forces and external loads.

Homework Problem 2

⎤⎡⎤⎡ ∂u εxx⎣ ⎦ = ⎣ ∂x ⎦

εzz ux

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Lecture 6 Finite Element Solution Process 2.092/2.093, Fall ‘09

εzz is frequently called the “hoop strain”, εθθ.

2π(u + x) − 2πx u εzz = =

2πx x⎡ ⎤ E 1 ν

C = ⎣ ⎦1 − ν2

ν 1

fB = ρω2R N/cm3 ; R = x

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