Finite Elements Spring System Analysis

Third lecture, fall 2015

Illustration of direct method with rigid bodies connected by springs

For a single spring, oriented along the x-direction, obeying the simple linear spring law (f = kd, where d is spring elongation and k is spring rate, and f is force)

f1 is force felt at node 1 (that is, local node number 1), and f2 is force felt at (local) node number 2

u1 is displacement in the positive x-direction of node 1, and u2 is displacement of node 2

d is calculated as the difference between u2 and u1.

2

1

2

1

11

11

u

uk

f

f

Some definitions:

Node: a point in space, possessing one or more degrees of freedom;

Degree of freedom (DOF): the independent response of a node to external stimulus.

For structures, the response is displacement. For thermal problems the response is either temperature or temperature change. For other problems, there are other DOF.

They are theoretically changeable independent of each other.

Element: a piece of the solution domain (i.e., the field) bounded by nodes and possessing an interpolation function (more later.)

Illustration of the direct method: system of rigid bodies connected by springs

The example problem:

Element nodal information

Element Local I j

stiffness

1 1 2 k1

2 2 3 k2

3 2 4 k3

4 2 5 k4

5 4 5 k5

6 4 6 k6

Balance of forces at each node

Assembly into a global equation set For each element, write the element stiffness equation, f = kq.

Recall that the local node numbers 1 and 2 correspond to global node numbers that may differ. For example, element number 4 connects nodes 2 and 5, so we have, for that element only,

Note that rows of the stiffness matrix correspond to force components, and columns multiply displacements. This is a useful concept for assembly.

5

2

44

44

)4(

5

2

u

u

kk

kk

f

f

Assembly, continued

We next establish a n-by-n matrix of all-zero values to serve as our GLOBAL stiffness matrix, K. Into this matrix we add in stiffnesses of the individual elements. For example, with element 4, only, we have

6

5

4

3

2

1

44

44

4

5

4

2

000000

0000

000000

000000

0000

000000

0

0

0

0

u

u

u

u

u

u

kk

kk

f

f

As in your text, note that the parentheses in the superscripts have been removed for ease of expression. The forces shown above are on element 4, not raised to the fourth power.

Assembly, continued

Other elements also share nodes, so they go into (in some cases) the same location as terms that have already been placed into the global K matrix. In this case, we simply add terms. For elements 1 and 4, only:

6

5

4

3

2

1

44

4411

11

4

5

4

2

1

2

1

1

000000

0000

000000

000000

000)(

0000

0

0

0

u

u

u

u

u

u

kk

kkkk

kk

f

ff

f

Assembly, completed

After all elements have been included, we get

6

5

4

3

2

1

66

5454

656533

22

43243211

11

6

6

5

5

4

5

6

4

5

4

3

4

2

3

4

2

3

2

2

2

1

2

1

1

0000

0)(00

)(00

0000

0)(

0000

u

u

u

u

u

u

kk

kkkk

kkkkkk

kk

kkkkkkkk

kk

f

ff

fff

f

ffff

f

The above represents the GLOBAL stiffness equation, written in short form as KSQS = FS. The subscript S stands for system, and indicates a fully formed problem (which needs to be reduced prior to solution).

Forces; Boundary Conditions and Reactions The summed forces at the nodes which appear in the global

equation are easily dealt with: they represent the sum of internal forces on the elements, and must exactly balance the externally applied forces. These sums can thus be replaced by the net external force acting at the node, when that force is known.

The external force must always be known at every node, except in one instance: if the node is constrained to have some specific DOF value at the end of the motion, then there is an unknown force at the node.

At all unconstrained nodes, the force is known or zero, but displacement is unknown. At all constrained nodes, the displacement is known, but the net external force is not.

Forces; Boundary Conditions and Reactions, continued Forces that must exist in order to ensure proper final DOF

values are called reactions. They do not appear in our system equations, except as unknown values.

Unconstrained nodes have zero-valued reactions.

To account for this, we proceed as follows: Substitute Ri for the force term for any constrained node;

Substitute the known external force for the force term at any unconstrained node;

Separate the global equation into two parts: Set aside any line that corresponds to a reaction force;

Multiply known values of displacement by the appropriate terms of the stiffness matrix, and remove to the other side of the stiffness equation.

Result is a REDUCED, GLOBAL stiffness equation, (KQ = F) which is solvable so long as enough boundary conditions have been asserted.

Forces; Boundary Conditions and Reactions, Numerical example

Consider the case where all Ks are 100. and external forces are 3 = 5 = 1000

By applying boundary conditions, 1 = 6 = 0,the system of equations become

Forces; Boundary Conditions and Reactions, continued

The reduced system of equations

Which can be solved for 2, 3, 4, 5

The reaction at each wall can be obtained from the first and sixth equations (removed from the reduces system of equations), once 2, 4 are known:

Solving the reduced set of equations

The answer is:

The reaction forces must balance the applied loads. For this problem, we asserted a 1000 N force; The reactions must thus add to -1000 N, to balance the applied load.

Bars:

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The bar element, in 1-D space

For a bar of constant cross-sectional area and constant modulus of elasticity, we have the equation from Strength of Materials

where s is uniaxial stress, E is the modulus of elasticity and e is the uniaxial strain

We also have the equation for strain as (using coordinate x to define location and u to define motion in the x-direction)

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es E

dx

duE

dx

du se

The 1-D bar, continued

Again from Strength of Materials, we have the following approximation:

And if the force on the bar is evenly distributed over a cross-section, then we further have

Putting all of the pieces together, then, we get

18

L

uu

L

L

dx

du )( 12

Af s

)( 12 uuL

EAf

The 1-D bar, concluded

As with the spring, we note that

Sum of the nodal forces must be zero, hence the two end forces are equal, opposite

The equation found so far is for tensile response (u2 > u1)

Hence, we arrive to the equation for the bar element, in 1-D space,

which is exactly analogous to the FE equation set for a spring, only substituting the extensional stiffness (EA/L) for the spring rate k. 19

j

i

j

i

u

u

L

EA

f

f

11

11

1-D Steady-State Heat Conduction

By Fourier Law, 1-D conduction is defined by

=

=

(2 1)

Heat Flux is positive if heat flows INTO the bar, so the equation defines q1; No heat build-up in the bar, so q2 = -q1

Thus

2

1

2

1

11

11

T

T

L

kA

q

q

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Direct Method

For each of the cases shown, we started with a known, simplified solution to the differential equation:

=

=

21

21

= = 2 1

We simply asserted the known solution into the DE and manipulated to get a linear algebraic set

IN ESSENCE, what we have done is insert a chosen solution into the DE, and the linear algebraic equivalent materializes as a result of our choice.

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Notes:

See the amazing similarity between the three equation sets:

Except for the leading constants (k, EA/L or kA/L) and the terms used for displacement (true displacement or temperature), the equation sets are identical.

Each has transformed a first-order DE into a linear algebraic equation set.

Each element has two nodes, each node has one DOF, so each element has two DOF (two nodes x one DOF/node) and each has a (2x2) equation set. THIS IS NOT COINCIDENTAL. Element equation sets will always be (nxn) where there are n DOF in the element.

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Assumptions built into the equations: Each bar has constant modulus E and constant cross-sectional

area A (though bars can differ from each other.) Similarly all conduction elements have constant k, A and all springs have constant k.

Use multiple connected bars, each with constant modulus and area to emulate either material variations or cross-sectional area changes; similar with springs, conduction bars

A tapered bar, for example, is modeled as a set of bars that have areas that step to approximate the taper

All loading is done at the nodal locations 1 and 2.

Ensure this by choosing to put a node at every loading or support point

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Sets of Elements: Assembly

A single element is typically not very useful.

A structure or system is usually made up of multiple elements.

Real problems are solved by using many elements, connected at the nodes.

The process of putting together multiple elements to form an equation set for the connected whole is called ASSEMBLY.

In the assembly phase the elemental equation sets f = kq yield up the GLOBAL equation set F = KQ

24

Simple Compound Bar (1)

25

Simple Compound Bar (2)

Putting in values for E1, E2, etc., in the stiffness matrix completes the assembly, for this simple bar problem. We will input values in the force side and the displacement vectors later.

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