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Trusses

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Element Formulation by Virtual Work

Use virtual work to derive element stiffness matrix based on assumed displacements– Principle of virtual work states that if a general

structure that is in equilibrium with its applied forces deforms due to a set of small compatible virtual displacements, the virtual work done is equal to its virtual strain energy of internal stresses.

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At element level, Ue = We

– Ue = virtual strain energy of internal stresses

– We = virtual work of external forces acting through virtual displacements

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We now assume a simple displacement function to define the displacement of every material point in the element.

Usually use low order polynomials Here

u = a1 + a2x

– u is axial displacement

– a1, a2 are constants to be determined

– x is local coordinate along member

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The constants are found by imposing the known nodal displacements ui, uj at nodes i and j

ui = a1 + a2xi

uj = a1 + a2xj

ui, uj are nodal displacements

xi, xj are nodal coordinates

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letting xi = 0, xj = L, we get

– a1 = ui

– a2 = (uj-ui)/L

We can write

– [N] = matrix of element shape functions or interpolation functions

– {d} = nodal displacements

ux

L

x

L

u

ud

i

j

1 [ ]{ }N

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1 2

1

2

i

i

1 2

N N N

xN 1 ,

Lx

NL

Pr operties

N 1 at node i and zero at all other nodes

N 1

i.e. at any point in the element N N 1

N1=1

Variation of N1

Variation of N2

N2=1

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Strain is given by

where [B] is a matrix relating strain to nodal displacement (matrix of derivatives of shape function)

du d[N]{d} [B]{d}

dx dx1

B 1 1L

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Now

= E(= E[B]{d}-E – Stress and strain are constant in a member

Define internal virtual strain energy for a set of virtual displacements {d} to be

Te VU ( ) dV

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= virtual strain = stress level at equilibrium– dV = volume

Virtual work of nodal forces is

We = {d}T{f}

Then, virtual work is given by

T T

Vdv d f

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Substituting and rearranging gives

Canceling {d}T gives [k]{d}={F} where

T ToV

T T T T ToV V

[B]{ d} E[B]{d} E dV { d} {f}

{ d} [B] E[B]{d}dV { d} [B] E dv { d} {f}

[ ] [ ] [ ]k B B T

VE dV

o

1F f EA

1

o T For thermal problem

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for a truss we get

this formulation method also applies to 2-d and 3-d elements

[ ]k

EA

L

1 1

1 1

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Procedure for Direct Stiffness Method (Displacement Method)

1. Discretize into finite elements, Identify nodes, elements and number them in order.

2. Develop element stiffness matrices [Ke] for all the elements.

3. Assemble element stiffness matrices to get the global stiffness matrix ([KG] = [Ke]). The size of of global stiffness matrix = total d.o.f of the structure including at boundary nodes. Assembly is done by matching element displacement with global displacements. Also develop appropriate force vector (by adding element force vectors) such that equation of the type [KG] {u}={F} is obtained.

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4. Apply kinematic boundary conditions. Without applying boundary conditions, [KG] will be singular. (minimum number of boundary conditions required is to arrest ‘Rigid Body’ displacements).

5. Solve for unknown displacements {u} ( {u}= [KG] –1{F}).

6. Once displacements are determined find

(a) reactions by picking up appropriate rows from the equation {F}=[KG] {u}, (b) Find element forces {f}=[Ke] {ue}, (c) Element stresses given by {e}= [D][B]{ue}.

Procedure for Direct Stiffness Method

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1 2 3

F1, u1F2, u2 F3, u3

Boundary Conditionsu1=0, u2=0

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G

2 -2 AE

K -2 2+1 -1L

-1 1

2A, L, E A, L, E

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{

Reactions

1

3

0AE PL 2P

F 2 -2 0 1L 3AE 3

0

0AE PL P

F 0 -1 1 1L 3AE 3

0

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Element Forces

1 1

2 2

21

32

Element 1

f u 1 -12AE

f -1 1 uL

1 -1 0 2p / 32AE PL

-1 1 1 2p / 3L 3AE

Element 2

uf 1 -1AE

uf -1 1L

1 -1AE =

-1 L

1 p / 3PL

1 0 p / 33AE

A, L, E

f1f2

2A, L, E

2P

3

2P

3

A, L, EP

3

P

3

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1 2

Element 1 Element 2

1 1 1 1AE AEK K

1 1 1 1L L

u1 u2 u2 u3

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1 11 1

2 2

4 22 1

2 22 2

3 3

43 2

Stress in element 1

u u1 1E EB E

u uL L

u u 1.5 0 = E 2.0 10 200N / mm

L 150Stress in element 2

u u1 1E EB E

u uL L

u u 1.2 1.5 = E 2.0 10

L 150

240N / mm

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Direct Element Formulation

truss element acts like 1-d spring– l >> transverse dimensions– pinned connection to other members (only

axial loading).– usually constant cross section and modulus of

elasticity

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» A = cross section area

» E = modulus of elasticity

» L = length

kAE

L

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Assume displacements are much smaller than overall geometry– vertical displacements of horizontal member

produce no vertical force Stiffness matrix is written in local element

coordinates aligned along element axis want stiffness matrix for arbitrary

orientation

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rotate coordinate systems using rotation matrix [R]

displacement components in global coordinates are related to displacement components in local coordinates by {d’}=[R]{d}– {d} = displacement in global coordinates– {d’} = displacement in local element

coordinates

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L

AE

x’

y’

i ju’i u’i

v’i v’j

p’i

P’j

q’i

q’j

u’i=1p’i = k=AE / L p’j = k = AE / L

q’i = 0 q’j = 0

st

k

01 column

k

0

u’i=1p’i = k=AE / L

q’i = 0q’j = 0

rd

k

03 column

k

0

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v’j =1

q’i = 0

q’j = 0

p’j = 0

p’i = 0

th

0

04 column

0

0

nd

0

02 column

0

0

v’i =1

q’i = 0q’j = 0

p’j = 0

p’i = 0

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start with member on x axis, element equations are

or {k’}{d’}={f’} Note that y equations are all zero

i i

i i

j j

j j

u ' p 'k 0 k 0

v ' q '0 0 0 0

u ' p 'k 0 k 0

v ' q '0 0 0 0

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pi uix

x’y’

y

pj uj

vj

qi

vi

qj

p’ i

u’ i

u’ j

p’ j

q’ i

v’ i

q’ j

v’ j

i i i i i i

i i i i i i

at node i

u ' u cos( ) v sin( ) p ' p cos( ) q sin( )

v ' u sin( ) v cos( ) q ' q sin( ) q cos( )

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i i

i i

u ' ucos sin

v ' sin cos v

At node i

i i

i i

p ' pcos sin

q ' sin cos q

A similar matrix can be obtained at node j

i i

i i

j j

j

u ' ucos sin 0 0v ' vsin cos 0 0u ' u 0 0 cos sin

0 0 sin cos v '

jv

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Matrix [R] is:

cos sin

sin cos

cos sin

sin cos

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

c s

s c

c s

s c

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Similarly , force components are related by {f’} = [R]{f}

Local force displacement relation is [k’]{d’} = {f’}

global force displacement relation is [k][R]{d} = [R]{f}

using fact that [R]-1 = [R]T, we get [R]T[k][R]{d} = {f}

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then [k] = stiffness matrix in global coordinates is [R]T[k’][R]

2 2

2 2

2 2

2 2

c cs c cs

cs s cs s[k] k

c cs c cs

cs s cs s

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Structure equation is [k] {D} = {F}

– [k] = structure stiffness matrix

– {D} = nodal displacement vector

– {F} = applied load vector

' 'i j i i i

i

i

i

i

j

j

DB{u u } note u ' u cos( ) v sin( )

u ={c s}

v

u

vc s 0 0-1 1 E

u0 0 c sL L

v

i

i

j

j

u

vEc -s c s

uL

v

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2

2

E le m e n t i- n o d eC o o r d in a t e

x y

C o o r d in a te

x y

L e n g t hj- n o d e Cs

1 1 0 0 2 L c o s 4 5 L s in 4 5 c o s 4 5 s in 4 5L

3 0 2 L s in 4 5 2 L c o s 4 5 L s in 4 5L c o s 4 5 -sin 4 5

[ ' ]k k

c cs c cs

cs s cs s

c cs c cs

cs s cs s

2 2

2 2

2 2

2 2

j i j i

2 2

j i j i

x x y yC , m

L L

L x x y y

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u3

v3

u’3v’3

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Finite Element Model

usually use existing codes to solve problems

user responsible for – creating the model– executing the program– interpreting the results

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arrangement of nodes and elements is known as the mesh

plan to make the mesh model the structure as accurately as possible

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for a truss– each member is modeled as 1 truss element– truss members or elements are connected at

nodes– node connections behave like pin joints– truss element behaves in exact agreement with

assumptions– no need to divide a member into more than 1

element

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– such subdivision will cause execution to fail» due to zero stiffness against lateral force at the node

connection where 2 members are in axial alignment

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there is geometric symmetry– often possible to reduce the size of problem by

using symmetry– need loading symmetry as well

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Fig. 3-5 and 3-6 show symmetric loads and the reduced model– need to impose extra conditions along the line

of symmetry» displacement constraints: nodes along the line of

symmetry must always move along that line

» changed loads: the load at the line of symmetry is split in two

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Computer input assistance

a preprocessor is used to assist user input required inputs are

– data to locate nodes in space– definition of elements by node numbers – type of analysis to be done– material properties– displacement conditions– applied loads

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interactive preprocessors are preferable – you can see each node as it is created– elements are displayed as they are created– symbols are given for displacement and load

conditions– usually allow mesh generation by replication or

interpolation of an existing mesh– allow inserting nodes along lines– allow entering a grid by minimum and maximum

positions plus a grid spacing

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truss element consists of 2 node numbers that connect to form element

other information for truss is – modulus of elasticity– cross sectional area

data can form a material table assign element data by reference to the

table

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boundary or displacement conditions are set by selecting a node and setting its displacement

do not over constrain a structure by prescribing zero displacements where there is no physical support

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loading conditions are set by selecting nodes and specifying force or moment components

check model carefully at this point

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Analysis Step

mostly transparent to user small truss models have enough accuracy

and performance for an accurate solution a large model has a large number of

elements and nodes

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numerical solution may not be accurate if there are full matrices

get better accuracy if the nonzero terms are close to the diagonal– reduces the number of operations and round off

error (banded matrix)

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in FE model, element or node numbering can affect bandwidth– good numbering pattern can minimize

bandwidth– different methods based on node or element

numbering– to minimize, plan numbering pattern so nodes

that connect through an element have their equations assembled close together

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In Fig. 3-7, node numbers are considered, X’s show nonzero terms

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In Fig. 3-8, node numbers are considered

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many programs have bandwidth or wavefront minimizers available

most programs will keep original numbering for display but use the minimized number scheme

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numerical algorithms, numerical range of the computer affect solution

relative stiffness of members can influence results– problems when members of high and low

stiffness connect– can exceed precision of computer– physical situation is usually undesirable

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Approximation error for truss is zero Most common error messages (errors)

come from– incorrect definition of elements– incorrect application of displacement boundary

conditions

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– may get non-positive definite structure stiffness matrix from not enough boundary conditions to prevent rigid body motion

» two elements connect in-line zero lateral stiffness

» truss structure not kinematically stable (linkage)

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next look at stress components– in continua, stress components are related to

averaged quantities at the nodes– trusses have a stress in each member (not easy

to plot) truss model is exact so it does not usually

need refinement

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Output Processing and Evaluation

Get numerical results with input data followed by all nodal displacements and element stresses

first graphic to look at is the deformed shape of the structure– nodal displacements are exaggerated to show

structure deformation– check to ensure model behaves as expected

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linear elastic analysis, failure is by– overstressing– buckling (have to find members with

significant compression and use Euler's buckling equation)

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Final Remarks

few situations where a truss element is the right element for modeling behavior