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Adam Chapman
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 Posted: Sat May 12, 2007 10:01 pm Post subject: finite element set up for beam |
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Hi,
I think this question should be pretty simple.
I have a rectangular 2D cantilever beam with an external point load in
the downward direction at the free (right-hand) top corner of the
beam (the beam is horizontal). The beam cannot delfec in the x-
direction because it is constrained with rollers against a vertical
wall.
This beam is broken into 2 triangular elements (split by a diagonal
line which joins the bottom left corner to the top right. There is a
node at each of the four corners of the beam, and the beam is fixed to
the wall at its left-hand end.
Im just setting the equations up to anallyse the beam, and i am
wondering if the vertiacl forces acting on the two left-hand nodes
should both be equal to the applied force in the opposite direction,
or both equal to half the applied force?
also would I be right in thinking that a horizontal force would need
to act at the bottom left node to balance moments? - this would mean
the bottom right element is not in horizontal equilibrium.
Thanks for any help,
Adam |
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Adam Chapman
Guest
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| Posted: Sat May 12, 2007 10:15 pm Post subject: Re: finite element set up for beam |
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30N
60N
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\
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3cm |
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\|/
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----- |
W
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---------- |
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------------ |
A
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------------ | 1 cm
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L /|\ 3
---------------------
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-----------
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L |
-----------
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-------
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|_-_________\__________________________________________________|
/
180N |
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Adam Chapman
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| Posted: Sat May 12, 2007 10:16 pm Post subject: Re: finite element set up for beam |
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| that was my attempt at drawing it- whoops! |
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Tom S.
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| Posted: Mon May 14, 2007 10:18 pm Post subject: Re: finite element set up for beam |
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"Adam Chapman" <adam.chapman@student.manchester.ac.uk> wrote:
| Quote: |
I have a rectangular 2D cantilever beam with an external point load in
the downward direction at the free (right-hand) top corner of the
beam (the beam is horizontal). The beam cannot delfec in the x-
direction because it is constrained with rollers against a vertical
wall.
This beam is broken into 2 triangular elements (split by a diagonal
line which joins the bottom left corner to the top right. There is a
node at each of the four corners of the beam, and the beam is fixed to
the wall at its left-hand end.
|
First off, these sounds like very odd elements...you normally want all the
element dimensions to be approximately equal or your assumed basis functions
may be too far from reality. These elements sound like they are extremely
skewed.
| Quote: |
Im just setting the equations up to anallyse the beam, and i am
wondering if the vertiacl forces acting on the two left-hand nodes
should both be equal to the applied force in the opposite direction,
or both equal to half the applied force?
|
Why are you setting these forces at all? You should be able to set the
displacements (zero) and get the reactions as a solution to the problem, not
as an input.
| Quote: |
also would I be right in thinking that a horizontal force would need
to act at the bottom left node to balance moments? - this would mean
the bottom right element is not in horizontal equilibrium.
|
You do need a horizontal reaction force to balance moments...that's the
realization of the compressive and tensile forces in the top and bottom of
the beam. The bottom element should be in eqilibrium because it will have a
reaction force at the bottom left node and a balancing force at the top
right node (transmitted throught the top left element).
Tom. |
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BobK207
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| Posted: Thu May 17, 2007 9:54 am Post subject: Re: finite element set up for beam |
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On May 12, 10:01 am, Adam Chapman
<adam.chap...@student.manchester.ac.uk> wrote:
| Quote: |
Hi,
I think this question should be pretty simple.
I have a rectangular 2D cantilever beam with an external point load in
the downward direction at the free (right-hand) top corner of the
beam (the beam is horizontal). The beam cannot delfec in the x-
direction because it is constrained with rollers against a vertical
wall.
This beam is broken into 2 triangular elements (split by a diagonal
line which joins the bottom left corner to the top right. There is a
node at each of the four corners of the beam, and the beam is fixed to
the wall at its left-hand end.
Im just setting the equations up to anallyse the beam, and i am
wondering if the vertiacl forces acting on the two left-hand nodes
should both be equal to the applied force in the opposite direction,
or both equal to half the applied force?
also would I be right in thinking that a horizontal force would need
to act at the bottom left node to balance moments? - this would mean
the bottom right element is not in horizontal equilibrium.
Thanks for any help,
Adam
|
Adam-
Call me crazy but shouldn't beams be modeled with beam elements?
Just apply loads & specify boundary conditions....the analysis will
take care of the rest.
cheers
Bob |
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Adam Chapman
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| Posted: Fri May 18, 2007 1:50 am Post subject: Re: finite element set up for beam |
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Thanks to both of you who replied.
I have since used the equation KU=F where k is the nxn stifness
matrix, U is the nx1 displacement vector and F is the nx1 force
vector.
Basically, what I dont understand is why in the question (in which
this funny element set-up is defined), the stiffness matrices are
given for both the elements.
I think that the displacements the bottom right-hand corner can be
found from analysing the bottom right elelemnt alone, using the row of
the equation KU=F as simultaneous equations.
Is the stiffness matrix of the top left-hand element needed at all? I
only need to find the displacenment of te bottom right-hand corner of
the beam |
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Jeff Finlayson
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| Posted: Fri May 18, 2007 9:08 am Post subject: Re: finite element set up for beam |
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Adam Chapman wrote:
| Quote: |
Thanks to both of you who replied.
I have since used the equation KU=F where k is the nxn stifness
matrix, U is the nx1 displacement vector and F is the nx1 force
vector.
Basically, what I dont understand is why in the question (in which
this funny element set-up is defined), the stiffness matrices are
given for both the elements.
I think that the displacements the bottom right-hand corner can be
found from analysing the bottom right elelemnt alone, using the row of
the equation KU=F as simultaneous equations.
Is the stiffness matrix of the top left-hand element needed at all? I
only need to find the displacenment of te bottom right-hand corner of
the beam
|
Yes. K is for the assembly of elements. Then apply the constraints
and solve for the unknown Us. |
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Tom S.
Guest
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| Posted: Fri May 18, 2007 9:48 pm Post subject: Re: finite element set up for beam |
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"Adam Chapman" <adam.chapman@student.manchester.ac.uk> wrote:
| Quote: |
I have since used the equation KU=F where k is the nxn stifness
matrix, U is the nx1 displacement vector and F is the nx1 force
vector.
Basically, what I dont understand is why in the question (in which
this funny element set-up is defined), the stiffness matrices are
given for both the elements.
|
Because you need both. A single element won't capture the true response of
the beam.
| Quote: |
I think that the displacements the bottom right-hand corner can be
found from analysing the bottom right elelemnt alone, using the row of
the equation KU=F as simultaneous equations.
|
They can't, because the global K matrix will have contributions from both
elements at the top right and bottom left nodes. If you leave out the upper
element, the K matrix will be wrong.
| Quote: |
Is the stiffness matrix of the top left-hand element needed at all?
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Yes. It influences the stiffness at the nodes it has in common with the
bottom right element.
| Quote: |
I only need to find the displacenment of te bottom right-hand corner of
the beam.
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Right, but the displacement of the bottom-right node (bottom right corner of
the lower element) is a function of applied load and the displacement of the
other nodes of that element. The left node is fixed by the support, but the
upper right node is free to move. This node is also attached to the top
left element, so the displacement of the bottom right corner is a function
of the displacement of both elements.
Tom. |
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