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BAR ELEMENT

Formulation of the finite element characteristics of an elastic bar

on the following assumptions:

1. The bar is geometrically straight.

2. The material obeys Hookes law.

3. The bar supports axial loading only;4. the bar element can be used in modelling both two - and thre

structures.

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Consider a tapered steel plate of uniform thickness t=25 mm.

E=2x105N/mm2,weight density ,=0.82x10-4N/mm3.In addition

weight the plate is subjected to a point load of 100 N at its mid

Calculate the following by modelling the plate with two finite e

1. Global force vector

2. Global stiffness matrix

3. Displacements in each element4. Stresses in each element

5. Reaction forces at the support

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A rod is subjected to an axial load P=600 KN is applied as showthe domain into two element. Determine

1. Displacements in each element

2. Stresses in each element

3. Reaction forces at the support.

Take A=250mm2 , E=2x105N/mm2

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AE

PLL

mmL 8.1

mmu 5.12

2

2

2

1

/400/2000mmNmmN

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An axial load of 4*105is appli ed at 30

c to the rod as shown in f ig.

The temperatur e is then raised to 60

c. Calculate the following

Displacements

Stresses

Reaction forces

c

mmNE

mmA

c

mmNE

mmA

iumaluFor

/10*12

2/10*2

1500

steelFor

/10*23

2/10*7.0

1000

min

61

5

1

2

1

61

5

1

2

1

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1

1)( TAEF

)( TAEdx

duE

In case temperature isgiven

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The strain displacement matrix is given by

Now we know that for bar element

uBe

llB

11

eE

From stress- strain relationship

UenergyStrain

The strain energy e

22

1 AL

F

F

If self weight is considforce is

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Consider a 4 bar truss as shown in fig.

1. Determine the element stiffness matrix

2. Assemble the structure stiffness matrix K for the entire truss

3.Solve for the nodal displacements

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Size of the global stiffness matrix = no of nodes X DOF per node

= 4 X 2 = 8

Coordinates for node 1 =(0,0)

Coordinates for node 2 =(1000,0)

Coordinates for node 3 =(1000,750)

Coordinates for node 4 =(0,750)

ConsiderElement 1

1. Find length of element 1

2. Find c and s values

Similarly for element 2,3 and 4

2122

121 )( yyxxl

nodes 1 2

x 0 1000

y 0 0

1

12

1l

xxc

1

121

l

yys

elements 1 2

nodes 1,2 2,3

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The general stiffness matrix for truss element is given by

By using the above formula find stiffness matrix for

Element 1 = K1

Element 2= K2

Element 3= K3

Element 4= K4

22

22

22

22

SCSSCS

CSCCSC

SCSSCS

CSCCSC

K

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FuK Thus the assembled stiffness matrix for all 3 elements will be of form

We know that

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By solving the above matrix we can find the nodal displacements

u3, u5,u6

6

5

3

6

5

3

666563

565553

363533

FF

F

uu

u

AAAAAA

AAA