FEA 28: Distributed Loads with Isoparametric Elements
Summary
TLDRThis video tutorial explains how to calculate distributed load vectors for isoparametric quadrilateral elements in structural analysis. It covers breaking down distributed loads into body and traction forces, integrating over element volume and surface, and using the Jacobian matrix for coordinate transformation. An example demonstrates calculating nodal force vectors for a specific element with a traction load, detailing the integration process and resulting forces at nodes.
Takeaways
- đ **Isoparametric Quadrilateral Elements**: The video discusses how to calculate distributed load vectors for isoparametric quadrilateral elements.
- đ **Distributed Load Conversion**: It explains the process of converting a distributed load over space into a load applied to individual nodes.
- đ **Element Load Vector**: The concept of an element load vector is introduced to represent the load at each node.
- 𧩠**Load Vector Components**: The load vector is broken down into body forces and traction forces.
- đ **Volume and Surface Integrals**: The body force term is integrated over the volume, while the traction force term is integrated over the surface.
- đ **Element Thickness**: The thickness of the element, denoted as H, is considered in the calculations.
- đ **Global vs. Local Coordinates**: The importance of the Jacobian matrix and its determinant in converting between global (X, Y) and local (s, t) coordinates is highlighted.
- đ **Jacobian Determinant**: The determinant of the Jacobian is used to relate differential areas in global and local coordinate systems.
- đ **Body Force Vector Calculation**: The body force vector is calculated as a double integral over the surface area in the X and Y coordinate system.
- đ **Mapping Dependencies**: If body forces depend on X and Y, the mapping from s and t to X and Y is necessary for integral evaluation.
- đ **Surface Traction Considerations**: The video provides an example of calculating nodal force vectors for surface tractions acting on the edges of the element.
Q & A
What is the purpose of calculating distributed load vectors for isoparametric quadrilateral elements?
-The purpose is to convert a load distributed over space into a load that can be applied to individual nodes of the element, allowing for a more precise analysis of the element's behavior under load.
How is the distributed load vector broken down in the script?
-The distributed load vector is broken down into two parts: body forces and traction forces.
What is the significance of the Jacobian matrix in this context?
-The Jacobian matrix is significant because its determinant represents the ratio of the areas of the elements in global versus local (natural) coordinate systems, which is crucial for transforming shape functions from local to global coordinates.
Why is it necessary to integrate over the volume for body forces and over the surface for traction forces?
-Integration over the volume is necessary for body forces to account for the load distribution throughout the element's thickness, while integration over the surface is for traction forces to account for the load applied on the element's surface.
How does the thickness of the element (H) affect the calculation of the load vector?
-The thickness (H) is used to scale the distributed load to account for the element's actual volume when calculating the body force vector.
What is the role of shape functions in calculating the element load vector?
-Shape functions are used to map the distributed load from the global coordinate system to the local coordinate system and are integral in evaluating the load vector at each node.
Why is it important to evaluate the shape functions at specific natural coordinates like s=-1, T?
-Evaluating shape functions at specific natural coordinates allows for the accurate calculation of the load vector components at each node, which is essential for determining the nodal forces.
What does the script suggest about the body force term if it depends on x and y?
-If the body force term depends on x and y, the mapping for x and y over to s and t must be used to evaluate the integral, as the shape functions are defined in terms of s and t.
How is the traction force distributed along the edges of the element in the example provided?
-In the example, the traction force is distributed linearly along the left edge of the element, with a peak value at node 1 and decreasing to zero at node 4.
What is the final result of the nodal force vector calculation in the example?
-The final result shows that node 1 experiences an upward force of 83.3 pounds, and node 4 experiences an upward force of 41.7 pounds.
Why are there no forces acting horizontally in the example calculation?
-There are no forces acting horizontally because the traction is applied only along the vertical edge of the element, as specified in the script.
Outlines
đ Calculating Distributed Load Vectors for Isoparametric Quadrilateral Elements
This paragraph explains the process of calculating distributed load vectors for isoparametric quadrilateral elements in structural analysis. It starts by discussing the need to convert distributed loads over space into loads applied to individual nodes, which is achieved by developing an element load vector. The element is affected by distributed loads, which are then broken down into body forces and traction forces. Body forces are integrated over the volume of the element, while traction forces are integrated over the surface where they act. The concept of the Jacobian matrix is introduced, which helps in transforming the integration from natural coordinates (s and t) to global coordinates (x and y). The determinant of the Jacobian is key in this transformation, as it represents the ratio of differential areas in global to local coordinate systems. The paragraph concludes with an example of calculating the nodal force vector for a specific isoparametric element with a traction load applied to its left edge.
đ Detailed Calculation of Nodal Force Vector for a Specific Element
The second paragraph delves into a detailed example of calculating the nodal force vector for an isoparametric element with a traction load applied to its left edge. It begins by defining the traction force in terms of the natural coordinate system, where the peak value is given as 500 pounds per square inch. The element's thickness is also provided as 0.1 inch. The paragraph then describes the process of mapping from the global XY coordinate system to the natural st coordinate system. The traction force is expressed in terms of the variable T, and a linear relationship is proposed and verified for the force distribution along the edge. The integral for the force vector is then set up, incorporating the shape functions evaluated at the left edge and the force vector in the Y direction. The paragraph concludes with the calculation of the force terms at nodes 1 and 4, which are 83.3 pounds and 41.7 pounds upward, respectively. These forces represent the static equivalent of the distributed load.
Mindmap
Keywords
đĄDistributed Load Vectors
đĄIsoparametric Quadrilateral Elements
đĄElement Load Vector
đĄBody Forces
đĄTraction Forces
đĄJacobian Matrix
đĄShape Functions
đĄNatural Coordinate System
đĄGlobal Coordinate System
đĄIntegration
đĄNodal Force Vector
Highlights
Calculating distributed load vectors for isoparametric quadrilateral elements involves converting distributed loads into individual node loads.
Element load vector is developed to determine the effect of distributed loads on an element.
Distributed load is divided into body forces and traction forces.
Body force term is integrated over the volume of the element.
Traction force term is integrated over the surface where the traction is acting.
The element's thickness, referred to as 'H', is considered in the calculations.
Shape functions are defined in terms of natural coordinates 's' and 't'.
The Jacobian matrix's determinant is key for converting between global and local coordinate systems.
The ratio of differential areas in global and local coordinates is represented by the Jacobian determinant.
Body force vector is calculated as a double integral over the surface area in the X and Y coordinate system.
If body force is dependent on X and Y, mapping to natural coordinates is required.
Surface traction is considered for the top, bottom, left, and right edges of the element.
Linear relationship is proposed for the traction force along the edge.
Force vector is calculated as the product of the shape function matrix and the traction force vector.
The length of the edge in the X&Y coordinate system is needed for the calculations.
Shape functions associated with the right edge nodes are zero when evaluating the left edge.
The final integral results in nonzero terms corresponding to the Y forces at nodes 1 and 2.
Nodal forces are calculated as 83.3 pounds upward at node 1 and 41.7 pounds upward at node 4.
The calculated nodal forces represent the static equivalent of the distributed load.
Transcripts
this short video goes through how we
calculate distributed load vectors for
isoparametric quadrilateral elements
whenever we have a load that's
distributed over space then we need to
find some way of converting that into a
load that's applied to individual notes
the way that we do that is by developing
an element load vector so we look at the
element and we see how it's affected by
that distributed load and then we
calculate a load vector that gives us
the loads at each one of the notes so
we're looking for the element load
vector for a isoparametric quadrilateral
element we start out with that load
vector call it F distributed we're going
to break it into two pieces portion four
body forces and a portion for traction
forces and then for our body force term
we're going to integrate over the volume
of the element and for the traction or
surface term we're going to integrate
over the surface where the traction is
acting now remember the surface and the
volume these are referencing the fact
that we're looking at a 2d element but
it has some thickness out of the screen
at us so that thickness we're going to
call H so looking at the body
specifically our volume integral becomes
a double integral over the surface area
in the X and y coordinate system so
we've got our shape functions defined in
terms of s and T we've got our body
force vector defined in the X and the y
directions and then we've got DX dy so
we've got a bit of a problem here
obviously we can't integrate our shape
functions when they're defined in terms
of s and T not x and y this leads us to
a very useful property of the Jacobian
matrix the determinant of the Jacobian
is the ratio of the areas of the
elements in global versus local
coordinate or in our case natural
coordinate system and specifically
what's useful here is that this can
happen on a differential basis so DX dy
is the size of a differential area in
global coordinates dsdt is the size in
local coordinate system and the
determinant of the Jacobian is the ratio
of those two so taking advantage of that
property that we can then write the body
force vector for this element as a
double integral going from negative 1 to
1 in both cases because remember
where this is our simple two by two
element in natural coordinate system our
shape functions are there our body force
vector our Jacobian determinant and then
dsdt
so that's a nice integral that we can
actually solve the only potential
complication is if your body force vet
body force term they're dependent on x
and y and then you'd have to use your
mapping for x and y over to S&T in order
to evaluate this integral so for the
surface traction we can break this into
surface tractions that act on the top or
the bottom of the element in which case
t is equal to either plus or minus one
or those surface tractions that act on
the left or the right of element of the
element in which case s is equal to plus
or minus one
in either case instead of okay let's do
an example of this let's find the nodal
force vector for the isoparametric
element shown here with the load applied
on its left edge now the left edge by
definition is the one for edge and this
is the edge that is going to correspond
to s equal to negative 1 we're also
given here that the peak value of this
traction is 500 pounds per square inch
and the thickness of the element is 0.1
inch so remember what we have is a
mapping from this XY space over to our
natural coordinate system the s and t
space where again s and t vary from
negative 1 to 1
attraction again is along the s equals
minus 1 edge so we use this traction
expression where we're going to evaluate
the shape functions at negative 1 comma
T and then we need the length of the
edge in X&Y coordinate system so that's
the big L there we're also going to need
to define FS in terms of T as their T
variable let's see how we do that T Y so
the traction at its peak value is equal
to 500 that's when we're at node 1 which
is where T equals minus 1 T y is equal
to 0
note 4 when T is from when little T is
equal to plus 1 so we can write or
propose a linear relationship here and
we'll just check to make sure this works
at each end and if it does we know it's
correct because it's a linear
relationship so we've got 500 divided by
2 times the quantity 1 minus T when T is
equal to negative 1 then 1 minus T
becomes 2 the 2's cancel and we get 500
psi at node 1 which is what we expect
when T is equal to a positive 1 1 minus
1 is 0 so T y is equal to 0 so it does
work at both locations so this means
that our force vector is equal to 500
divided by 2 or 250 times 0 and 1 minus
T psi so in the x-direction we have 0
because as we look at this figure we see
there is no force acting horizontally so
it's only in the Y direction that gives
us our FS so our integral now looks like
H times the integral from minus 1 to 1
times the shape function matrix with the
shape functions each evaluated at
negative 1 comma T times that FS vector
so 250 times 0 and 1 minus T and then L
over 2 times DT and of course we need to
evaluate L in this case it's pretty
simple it's just 5 inches for the length
now this particular matrix because we're
evaluating this along the left edge it's
a bilinear quadrilateral element I know
that the shape functions associated with
the right edge so nodes 2 & 3 are going
to be 0 when we're evaluating things on
the left edge and I know that to be a
fact because they are linear in terms of
s and T but you could also just plug in
the values and you'd see everything
would cancel all right so let's bring
this home here's what our integral looks
like we're going to do the
multiplication and when we do that we
end up with just two nonzero terms we're
here I've also plugged in that H equals
zero point 1 inches just two nonzero
terms and those correspond to the Y for
at node 1 and the Y force at node 2
that's what we'd expect we wouldn't
expect any other components given the
loading shown here plugging in the value
of n1 at negative 1 comma T that's going
to be 1/2 times 1 minus T and when I put
that into just this piece of the
integral the force becomes 83.3 pounds
when I evaluate n4 at negative 1 comma T
that becomes 1/2 times 1 plus T putting
that into just the integral for that
piece just that term I find 41.7 pounds
so these are the two nonzero force terms
at node 1 I have to apply 83.3 pounds
upward and at node 4 I apply 41.7 pounds
upward and these two forces together
form the static equivalent of the
distributed load shown
5.0 / 5 (0 votes)