Lecture Notes: Introduction to Finite Element Method |
Chapter 2. Bar and Beam Elements |
Applying the result in (43) and carrying out the integration, we arrive at the same stiffness matrix as given in (38).
Combining the axial stiffness (bar element), we obtain the stiffness matrix of a general 2-D beam element,
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EA |
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12EI |
6EI |
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2EI |
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k = |
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EA |
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12EI |
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2EI |
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3-D Beam Element
The element stiffness matrix is formed in the local (2-D) coordinate system first and then transformed into the global (3- D) coordinate system to be assembled.
(Fig. 2.3-2. On Page 24 of Cook’s book)
© 1997-2002 Yijun Liu, University of Cincinnati |
57 |
Lecture Notes: Introduction to Finite Element Method |
Chapter 2. Bar and Beam Elements |
Example 2.5
Y |
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P |
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M |
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E,I |
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Given: The beam shown above is clamped at the two ends and acted upon by the force P and moment M in the midspan.
Find: The deflection and rotation at the center node and the reaction forces and moments at the two ends.
Solution: Element stiffness matrices are,
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v1 |
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v2 |
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12 |
6L |
−12 |
6L |
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k1 |
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EI |
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6L 4L2 |
−6L |
2L2 |
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L3 |
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−6L |
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−12 |
12 −6L |
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6L |
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2L |
−6L 4L |
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v2 |
θ2 |
v3 |
θ3 |
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12 |
6L |
−12 |
6L |
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EI |
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6L 4L2 |
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−6L |
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12 −6L |
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6L |
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2L |
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© 1997-2002 Yijun Liu, University of Cincinnati |
58 |
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
Global FE equation is,
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v1 |
θ1 |
v2 |
θ2 |
v3 |
θ3 |
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12 |
6L −12 6L |
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F1Y |
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6L |
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4L |
−6L 2L |
θ1 |
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EI −12 −6L 24 |
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−12 6L v2 |
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6L |
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2L |
8L |
−6L 2L |
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−12 −6L |
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12 −6L v3 |
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6L 2L2 |
−6L 4L2 |
θ3 |
M3 |
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Loads and constraints (BC’s) are, |
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F2Y = −P, |
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M2 = M , |
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v1 = v3 =θ1 =θ3 = 0 |
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Reduced FE equation, |
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EI |
24 |
0 v2 |
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0 8L2 |
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Solving this we obtain,
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24EI |
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θ2 |
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− PL23M
From global FE equation, we obtain the reaction forces and moments,
© 1997-2002 Yijun Liu, University of Cincinnati |
59 |
Lecture Notes: Introduction to Finite Element Method |
Chapter 2. Bar and Beam Elements |
F1Y |
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M1 |
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F |
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3Y |
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M3 |
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−12 EI −6L L3 −126L
6L
2L2 v2 = −6L θ2
2L2
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2P +3M / L |
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PL + M |
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2P −3M / L |
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− PL + M |
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Stresses in the beam at the two ends can be calculated using the formula,
σ = σx = − MyI
Note that the FE solution is exact according to the simple beam theory, since no distributed load is present between the nodes. Recall that,
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d 2v |
= M (x) |
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and |
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dMdx =V (V - shear force in the beam)
dVdx = q (q - distributed load on the beam)
Thus,
EI |
d 4v |
= q(x) |
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If q(x)=0, then exact solution for the deflection v is a cubic function of x, which is what described by our shape functions.
© 1997-2002 Yijun Liu, University of Cincinnati |
60 |