FEA 28: Distributed Loads with Isoparametric Elements

Summary

This 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

• 📐 **Isoparametric Quadrilateral Elements**: The video discusses how to calculate distributed load vectors for isoparametric quadrilateral elements.
• 🔄 **Distributed Load Conversion**: It explains the process of converting a distributed load over space into a load applied to individual nodes.
• 📈 **Element Load Vector**: The concept of an element load vector is introduced to represent the load at each node.
• 🧩 **Load Vector Components**: The load vector is broken down into body forces and traction forces.
• 📊 **Volume and Surface Integrals**: The body force term is integrated over the volume, while the traction force term is integrated over the surface.
• 📏 **Element Thickness**: The thickness of the element, denoted as H, is considered in the calculations.
• 🌐 **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.
• 🔄 **Jacobian Determinant**: The determinant of the Jacobian is used to relate differential areas in global and local coordinate systems.
• 📉 **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.
• 🔗 **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.
• 📖 **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

00:00

📐 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.

05:02

🔍 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.

Keywords

💡Distributed Load Vectors

Distributed Load Vectors are mathematical representations used to describe how a load is distributed across different points in a structure. In the context of the video, these vectors are essential for converting a distributed load over space into a load that can be applied to individual nodes of an isoparametric quadrilateral element. The process involves calculating how each node is affected by the distributed load, which is crucial for structural analysis and ensuring the accuracy of the element's response to the load.

💡Isoparametric Quadrilateral Elements

Isoparametric quadrilateral elements are a type of element used in finite element analysis, characterized by their ability to map a parent element in a simpler coordinate system (like a square) to a more complex shape in the actual model. The video discusses how to calculate distributed load vectors for these elements, which is vital for understanding how loads are distributed across the element's nodes and for ensuring the structural integrity of the model.

💡Element Load Vector

An Element Load Vector is a vector that describes the forces acting on an element in a structural model. The video explains the process of developing an element load vector by breaking down the distributed load into body forces and traction forces, which are then integrated over the element's volume and surface, respectively. This vector is crucial for determining the load at each node of the isoparametric quadrilateral element.

💡Body Forces

Body forces are forces that act uniformly over the volume of an element, such as gravitational forces. In the video, body forces are integrated over the volume of the element to contribute to the element load vector. Understanding how body forces are calculated is important for accurately modeling the behavior of structures under load.

💡Traction Forces

Traction forces are forces that act on the surface of an element, such as friction or pressure. The video describes how these forces are integrated over the surface where they act to contribute to the element load vector. Traction forces are critical in determining the surface effects on the element and how they influence the overall structural response.

💡Jacobian Matrix

The Jacobian Matrix is a mathematical tool used to transform coordinates between different systems. In the video, it is used to relate the differential areas in global coordinates to those in the local (natural) coordinate system. The determinant of the Jacobian is particularly useful as it provides the ratio of these areas, which is essential for converting shape functions defined in terms of natural coordinates to global coordinates.

💡Shape Functions

Shape functions are mathematical functions used in finite element analysis to describe the shape of an element and how it deforms. The video mentions that these functions are defined in terms of natural coordinates (s and t) and are used to evaluate the body force vector and traction forces. Shape functions are key to understanding how loads are distributed within the element.

💡Natural Coordinate System

The natural coordinate system, also known as the local coordinate system, is a simplified coordinate system used to define the shape and properties of elements in finite element analysis. In the video, the natural coordinate system is used to define shape functions and to simplify the integration process for calculating load vectors. This system is crucial for the isoparametric formulation discussed.

💡Global Coordinate System

The global coordinate system is the coordinate system used to describe the overall structure or model. In the video, the global coordinate system is used in conjunction with the natural coordinate system to calculate the element load vector. The transformation between these systems, facilitated by the Jacobian matrix, is essential for accurately applying loads to the element.

💡Integration

Integration is a mathematical process used to calculate areas, volumes, or other quantities by summing up infinitesimally small contributions. In the video, integration is used to calculate the distributed load over the volume and surface of the element. This process is fundamental to developing the element load vector and understanding how loads affect the element.

💡Nodal Force Vector

The Nodal Force Vector is a vector that describes the forces acting at the nodes of an element. The video provides an example of calculating the nodal force vector for an isoparametric element, which is essential for understanding how the distributed load is translated into forces at each node. This vector is crucial for connecting the element's response to the overall structural analysis.

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