Finite Element Method AERSP 301 Finite Element Method Jose Palacios July 2008

AERSP 301Finite Element MethodFinite Element Method

Jose Palacios

July 2008

Today• HW 5 has been uploaded (Stationary Principle, MOS FOS)

– Short assignment– Due Thursday

• HW’s 4 due Tuesday– I will upload solutions Wednesday evening

• HW’s 3 & 4 due no later than Friday– Solutions will be uploaded Friday evening

• No Class on Thursday

• Friday during class, Exam Review

• Exam– Time– Equation Sheet

• Start Finite Element Method - Bars

Axial deformations in barsAxial deformations in bars

• We were able to write expressions for the strain energy, U, and external work, W, for a bar under axial loading.

• For the 1-DOF and 2-DOF spring problems, the next step was to apply the stationary principle.

– For the 1-DOF system, gave the equilibrium equation.

– Likewise, for the 2-DOF system,

gave the (system of two)

equilibrium equations.

As in HW #5, Problem 2


• For the bar, though being a continuum structure – it has an infinite number of degrees of freedom. – Displacement at every x will vary, u = u(x).

• Second issue – if you were applying the stationary principle, differentiating is a problem because the expressions for U and W have integrals.

• The finite element method represents continuum structures (with an infinite number of degrees of freedom) into a system with a finite number of degrees of freedom.– Represents structure in an approximate sense

– The structure is subdivided (“discretized”) into a number of segments (or finite elements)


• Variation of displacement (u) within each element is assumed to be very simple, such that U and W may be easily integrated across the element length

• U and W are related to displacements at a finite number of points which are the end-points of the element

• Solution of the problem is then reduced to the system of an algebraic system for a finite set of “end point” or NODAL values.


Entire Bar:• Displacements (u’s) must be

continuous along length of bar (no jumps)


• When the elements of the bar are reassembled:

• This GLOBAL system has 4 DOFs (q1…q4)

– Called GLOBAL degrees of freedom = Global nodal displacements

• Each element has 2 DOFs:

– Called LOCAL DOFs or Element DOFs or Element nodal displacements


• Every local DOF corresponds to a nodal displacement

• Looking closely at a single element:

u1, u2: local DOFs

x1, x2: element nodal coordinates

l = x2 – x1 = length of element


• Global x-coordinate system is difficult to use so we transform to a non-dimensional local coordinate system (s-coordinate) such that:

– At x = x1, s = 0

– At x = x2, s = 1

– Transformation expressed as

where (1-s) and (s) are mapping functions and dx = lds where l is the length scaling for mapping

ldsdx

lxxds

dx

21


• For individual element, express the displacements within the elements (in terms of values of nodes)

• For this, assume some displacement profile or shape within the element (in terms of element nodal displacements u1 and u2)

– An obvious choice is a linear profile wherein

– Such that at

– And at


• This is the expression for the axial displacement, u, within an element (at every point within the element) – in terms of local coordinates, s, and displacements at end points (nodes)

– u1, u2:

– Define N1(s) = 1 – s , and N2(s) = s displacement shape functions

• Shape functions are always in the local coordinate system

– Recall that

– The mapping functions are identical to the shape functions for this problem (this is not always the case)

where M1 and M2 are mapping functions

Axial deformations in bars – Axial deformations in bars – Element Stiffness MatrixElement Stiffness Matrix

• Recall , where – Total Potential

– U – Strain Energy

– W – External Work Potential

• Look at U (Strain Energy)

– For the bar in extension:

– Then discretize the bar into a number of elements:


• Since strain energy, U, is a scalar quantity, the integral

may be broken up– Strain energy may be calculated for each individual element and

summed to obtain the total strain energy of the entire structure:

• Consider a single element:

Also Recall:Mapping function

Assumed displacement within the element with shape fns, N1(s) & N2(s)


• Then

• This partial derivative can be written in two forms:

• Introducing into the strain energy for the ith element:

211 suusu

EA a function of s?

Are u1 and u2 functions of s?

How? Recall:ldsdx

lxxds

dx

21


• Look at

• Each term in the matrix may

be integrated individually.


• Individual terms are:

• This yields:

orELEMENT STIFFNESS MATRIX

What if an element moves through space without elongation (Rigid Body Translation)?

Can an element represent a constant strain condition?

Axial deformations in bars – Axial deformations in bars – Element Load VectorElement Load Vector

• Now look at the External Work Potential, W:

– Recall

– Consider the integral term::

• Discretize the bar using the finite element method

– Since Work is also a scalar and can be calculated for individual elements and added.


• Look at the contribution from a single element (putting integral in local coordinate system):


• Or in vector form:

• This is the contribution to the External Work Potential of the distributed force, f, over an element.

• What is the physical implication of representing:

Distributed load is represented as two individual forces (one at each node)


• Now consider a constant distributed force, f = c

– Half of total load, cl, is placed at each node. What about other loading conditions?

– Could integrate exactly across element length, if convenient.• Or, could use linear interpolation across the element

Try these.

L

xs

L

xsffsf

sfxf

xxs

xxs

sxxsx

2121

2

1

21

1

1

0

1

Axial deformations in bars Axial deformations in bars Global Load Vector and Global Stiffness Matrix Global Load Vector and Global Stiffness Matrix

• So far, have obtained expressions for strain energy, Ui, and external work potential, Wi, for any individual element.

• Now add the contributions from individual elements together to obtain the strain energy, U, and external work potential, W, for the entire structure.

• This process is called ASSEMBLY.

• Look at the External Work Potential, W, due to the distributed force, f.

Axial deformations in barsAxial deformations in barsGlobal Load Vector and Global Stiffness MatrixGlobal Load Vector and Global Stiffness Matrix

• The external work potential for element #1, W1, can be written as:

• Similarly, the external work potential for element #2, W2, can be written as:


• And, the external work potential for element #3, W3, can be written as:

• The total external work potential due to the distributed force, f, is:Element #1

Element #2

Element #3


• The external work potential due to the tip load, P, is W = q4P

• Similarly, there is a reaction force, R, acting at the root, x = 0 (because the bar is clamped at the end). The external work potential due to this reaction force, R, is W = q1R


• The total External Work Potential, W, is obtained by adding each load vector

• What if there were additional concentrated loads acting at some other locations on the bar?

Global Load Vector

(obtained from elemental load vectors and contributions of concentrated loads)


• Now, look at the strain energy, U, of the structure.

• The strain energy for element #1, U1, can be written as:


• Similarly, the strain energy for element #2, U2, can be written as:

• And, the strain energy for element #3, U3, can be written as:


• The total strain energy of the structure:

Global Stiffness Matrix

(obtained from 2x2 elemental stiffness matrices)

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Finite Element Method AERSP 301 Finite Element Method Jose Palacios July 2008