CAED Lec4 BarElement Sept2014

7/23/2019 CAED Lec4 BarElement Sept2014

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B A R E L E M E N T

COMPUTER AIDED

ENGINEERING



MOTIVATION

• Ductile material

• Behave like a “spring” in the elastic region.

• Hook’s Law, s=Ee

L L

EA F

L

L

E A

F

E

e s

Is the stiffness of the material L

EA

2

1

2

1

2

1

2

1

2

1

2

1

11

11

11

11

F

F

U

U

L

EA

F

F

U

U k

F

F

U

U

k k

k k

F U K

https://www.google.com/imgres?imgurl&imgrefurl=http://en.wikipedia.org/wiki/Ultimate_tensile_strength&h=0&w=0&sz=1&tbnid=0cuNkkcSuUqjAM&tbnh=225&tbnw=225&zoom=1&docid=5Ze7sZ8zy7sLrM&hl=en&ei=UwFSUrbYF86BrQe9jIHwCQ&ved=0CAEQsCU

https://www.google.com/imgres?imgurl&imgrefurl=http://en.wikipedia.org/wiki/Ultimate_tensile_strength&h=0&w=0&sz=1&tbnid=0cuNkkcSuUqjAM&tbnh=225&tbnw=225&zoom=1&docid=5Ze7sZ8zy7sLrM&hl=en&ei=UwFSUrbYF86BrQe9jIHwCQ&ved=0CAEQsCU



Applying equilibrium at nodes 1e and 2e of the

‘spring’ element e (using f = k Δ) :

e2

e1

e2

e1

F

F

u

u

k k

k k

ee

ee

}F{}u{]K [ eee

Element e1e 2e

u1e u2

e

F1e

F2e

k e

The stiffness Matrix

e2

e2

e1

e1

e2

e1

F)uu(k

F)uu(k

e

e

expressed in matrix form

Displacement

Matrix

Force

Matrix

Revisit: Single element



EXAMPLE 1:

A tapered square beam as shown in the figure

below is cantilevered at one end and is pulledby a force of 100 N at the other end. Calculatethe total elongation of the beam.

Nov-144

100 N

0 . 0 5 m 0.02 m

0.2 m

E = 200000 Pa

Square cross

section



EX 1: SOLUTION

• Use 2 uniform cross section elements.

Nov-145

100 N

E1 E2 100 N

Equation for side length,

For average side lengths ofthe elements,

hE1 = 0.0425m& hE2 = 0.0275m

x y 075.0025.0

0.05 m

0.15 m

mmm

mmm

0275.001375.0,15.0

0425.002125.0,05.0

075.0025.0

h y xat

h y xat

x y



E1 E2 100 N

100 N

k 1 k

2 Node 1 Node 2 Node 3

R 1

Two Elements Stiffness Matrix

100 N

0 . 0 5 m 0.02 m

0.2 m

E = 200000 Pa



K=

The resulted stiffness Matrix for the system is:



EX 1: SOLUTION

• Since this is a uniaxial 1-D problem, the elementscan be simplified as 2 springs in series.

Nov-148

100 N

k1 k2

Node 1 Node 2 Node 3

R1

with element stiffness,ki = E * Ai / L

k1 = [(200000*0.04252)/0.1] = 3612.5 N/mk2 = [(200000*0.02752)/0.1] = 1512.5 N/m



EX 1: SOLUTION

Nov-149

K= =

Global Stiffness Matrix**:

**next lecture



Nov-1410

Ex 1: Solution

Arranging the problem in matrix form (K*x = F) yield;

A known solution, u1 = 0. Thus the problem can be

reduced to (by eliminating first column & first row):

=

=



Nov-1411

Using the inverse matrix method, x = K * F ;

-1

Ex 1: Solution

= =

=

-1

0.027681661

0.093797363



Three springs system: ‘3 element’

δ

Force

L

A=Cross-

sectional Area

1 2 3 4

F – applied load

u 1 u 2 u 3 u 4

R 1 – unknown

reaction forcek

2 k 3

ele. 1 ele. 2 ele. 3

k 1



Use stiffness equations for each element,F = k (u i- ui-1)

to express the forces in terms of nodal

displacements and substitute into the

previous four equations

0)( 1211 uuk R

0)()( 232121 uuk uuk

0)()( 343232 uuk uuk

0)( 343 F uuk

node 1: (->)

node 2: (->)

node 3: (->)

node 4: (->)



Write these equations in matrix form

to obtain the global stiffness equations

F

R

u

uu

u

k k

k k k k k k k k

k k

00

00

00

00 1

4

3

2

1

33

3322

2211

11

FuK

1- D Three Elements



Systematic Assembly of the GlobalStiffness Matrix

Assembly of the stiffness matrix, [K] follows apattern based on element-node numberconnectivity shown in the table.

1 2 3

1 k 1 2 k 2 3 k 3 4

P

Element (e) Nodes 1e, 2e

1 1, 2

2 2, 3

3 3, 4

k 1

-k 1

-k 1

k 1+ k 2

-k 2

-k 2

-k 3

k 3-k 3

k 2+ k 3

K

1 2 3 4

2

1

3

4



Try a different element systemconfiguration and solve for the SystemStiffness Matrix

k 2

k 14

3

2

1

k 3

1

2

3

P

Systematic Assembly of theGlobal Stiffness Matrix



Solution to different configuration andsolve for the Stiffness Matrix

k 2

k 1

43

2

1

k 3

1

2

3

P

Element(e) Nodes 1e, 2e

1 1, 3

2 2, 3

3 3, 4

K

2

1

3

4

k 1

-k 1

-k 1

k 2

-k 2

-k 2

-k 3

k 3-k 3

k 1+k 2+ k 3

1 2 3 4

Systematic Assembly of the GlobalStiffness Matrix



HAVE A GREAT DAY! !

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CAED Lec4 BarElement Sept2014