Lecture Notes: Introduction to Finite Element Method |
Chapter 2. Bar and Beam Elements |
Equivalent Nodal Loads of Distributed Transverse Load
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x L |
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qL/2 |
qL/2 |
qL2/12 |
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qL2/12 |
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This can be verified by considering the work done by the distributed load q.
q
L L
qL qL/2
qL2/12
L L
© 1997-2002 Yijun Liu, University of Cincinnati |
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Lecture Notes: Introduction to Finite Element Method |
Chapter 2. Bar and Beam Elements |
Example 2.6
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p |
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E,I |
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Given: A cantilever beam with distributed lateral load p as shown above.
Find: The deflection and rotation at the right end, the reaction force and moment at the left end.
Solution: The work-equivalent nodal loads are shown below,
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E,I |
2 |
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where
f = pL / 2, |
m = pL2 / 12 |
Applying the FE equation, we have
© 1997-2002 Yijun Liu, University of Cincinnati |
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Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
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12 |
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6L −12 6L v1 |
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F1Y |
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EI |
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6L |
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4L |
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−6L |
2L |
θ1 = |
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−12 |
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−6L 12 −6L v2 |
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F2Y |
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6L |
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−6L |
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2L |
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4L |
θ |
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Load and constraints (BC’s) are, |
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F2Y = − f , |
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M2 = m |
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v1 =θ1 = 0 |
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Reduced equation is, |
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EI |
12 |
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−6L v2 |
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− f |
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L3 |
−6L |
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4L2 θ |
= m |
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Solving this, we obtain, |
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v2 |
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−2L2 f |
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3Lm |
− pL4 |
/ 8EI |
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−3Lf |
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(A) |
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θ2 |
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6EI |
6m |
− pL3 |
/ 6EI |
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These nodal values are the same as the exact solution. Note that the deflection v(x) (for 0 < x< 0) in the beam by the FEM is, however, different from that by the exact solution. The exact solution by the simple beam theory is a 4th order polynomial of x, while the FE solution of v is only a 3rd order polynomial of x.
If the equivalent moment m is ignored, we have,
v2 |
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−2L2 |
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− pL4 |
/ 6EI |
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−3Lf |
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θ2 |
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6EI |
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− pL3 |
/ 4EI |
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The errors in (B) will decrease if more elements are used. The
© 1997-2002 Yijun Liu, University of Cincinnati |
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Lecture Notes: Introduction to Finite Element Method |
Chapter 2. Bar and Beam Elements |
equivalent moment m is often ignored in the FEM applications. The FE solutions still converge as more elements are applied.
From the FE equation, we can calculate the reaction force and moment as,
F1Y |
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L3 −12 |
6L v2 |
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pL / 2 |
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M |
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−6L 2L2 θ |
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= 5pL2 / 12 |
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EI |
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where the result in (A) is used. This force vector gives the total effective nodal forces which include the equivalent nodal forces for the distributed lateral load p given by,
− pL / 2− pL2 / 12
The correct reaction forces can be obtained as follows,
F |
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pL |
/ 2 |
− pL / 2 |
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pL |
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1Y |
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= |
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− |
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= |
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M1 |
5pL2 |
/ 12 |
− pL2 / 12 |
pL2 |
/ 2 |
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Check the results!
© 1997-2002 Yijun Liu, University of Cincinnati |
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