notes - fem

8/11/2019 Notes - FEM

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1

Finite element is an approximate numerical solution technique in which continuous system

are discritised into many small and simple pieces called FINITE ELEMENT. For each

element it is necessary to make an assumption as to how the primary variable (such as

Displacement, temperature etc) distributed in terms of geometric position. This

assumption is the basis for the development of FINITE ELEMENT PROCEDURE.

AdvantageCan readily handle very complex geometry

Can handle a wide variety of engineering problems

o

Solid mechanics

o Dynamics

o

Heat problems

o

Fluids

o Electrostatic problems

Can handle complex restraints

Can handle complex loading

o Nodal load (point loads)

o Element load (pressure, thermal, inertial forces)

o

Time or frequency dependent loading

Disadvantage of FEM

o

A general Closed-form solution, which would permit one to examine system

response to changes in various parameters is not produced

o

The FEM obtains only approximate solutions

o

The FEM has inherent errors

o Mistakes by users can be fatel

Power of FE method is its versatility. Structure analysed may have

Arbitrary shape

Arbitrary supports

Arbitrary loads

SHORT History

Year

1943 The mathematician Courant described a piecewise solution for torsion

problem.

His work was not noticed by engineers and the procedure was impractical

at the time due to lack of digital computer.

1950 Aircraft industry started application

1960 The name finite element was coined by Clough

By 1963 The mathematical validity of FE method was recognized and the methodwas expanded from its structural beginnings to include heat transfer,

ground water flow, magnetic fields and other areas.

1965 A paper on heat conduction and seepage flow using FEA.

1967 First text book by O.C.Zienkiewicz and Chung

1970s Most general purpose software package originated in the 1970s. Abaqus,

Adina, Ansys, etc.)

By late 1980s The s/w was available on microcomputers, complete with color graphics

and pre- and post processes.

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By mid

1990s

Roughly 40,000 paper and books bout the FE method and applications

had been published.

FEA steps

1. Discretize the continuum

2.

Select interpolation function3. Find the element properties

4. Assemble the element properties to obtain the system equations

5.

Impose the boundary conditions

6.

Solve the system equations

7. Make additional computations if desired

Element type

Why so many no of elements?

It may because; of the intuition of the human being to search for the best solution and so

has tried with lots of option.

Node

Nodes usually lies on the element boundaries where adjacent elements are connected. In

addition to boundary nodes, an element may also have a few interior nodes. The nodal

values of the field variable and the interpolation functions are the elements completely

define the behavior of the field variable within the element.

Based on dimension

1-dimension elementExample:

(a)

Straight bar loaded axially; Bar element resist only axial load

(b) Straight beam loaded laterally; beam element can resist axial. Lateral and

twisting loads.

Deformation in a single plane, ex. In x-direction only, the dimension in one plane is

exceeding in other two directions.

It includes a straight bar loaded axially, a straight beam that can be loaded axially, laterally

as well as with twisting load, a bar that conducts heat or electricity, and so on.

In structural terminology, bar can resist only axial load. Beam in its most general sense,

can resist axial, lateral and twisting loads.

2-dimension

Deformation in both x and y direction.

3-dimension

Deformation in all x, y and z directions.

Linear element: Distribution of primary variable for an ex. Displacement between nodal

points are linear.

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Ex. Linear bar element, linear triangular element, linear tetrahedron, linear brick element.

It should be noted that: all the three dimensional element is included in this type of

element.

One-Dimensional elements and computational procedure

Based on order of interpolation between nodes.

Linear bar element

xu a bx

For x = 0,0

u a

For x = l, lu a bl ; or 0lu u

b

l

2 11 1 2 1 1 2 2

1 x

u u x xu x u u u N u N u

l l l

Where, N 1 and N 2 are called the shape function.

Shape function (Interpolation/Displacement function)

Shape function describes how the primary variable (such as displacement, temperature etc)

is distributed over an element in terms of geometric position. This function is estimated

from the assumed polynomial (linear, quadratic, cubic, quartic etc) for element type. It is

written for each nodes of a finite element and usually denoted by N .

Properties of shape function:(1)

1 for

0 for

i

i

N n i

N n i

For ith

node, the shape function at node i is Ni. The value of Ni at other nodes (viz.

1,2,3,…i-1, i+1….n) is 0. where, n is the total number of nodes in the finite element

and i is the node under consideration.

(2)1

1n

i

i

N

Class work : Find shape function for bar element of length l.

2

xu a bx cx

For x = 0, 0u a

For 2l x , 2

2 2 2u l a b l c l

1 2

1u 2

u 1F

2F

1 31u

3u

2

2u

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For l x , 2

u l a b l c l

1a u ; 1 2 3

13 4b u u u

l ; 1 2 32

22c u u u

l

2 2 2

1 2 32 2 2

3 2 4 4 21

x

x x x x x xu u u u

l l l l l l

1 1 2 2 3 3 xu N u N u N u

Where, N 1 and N 2 are called the shape function.

Cubic element2 3

xu a bx cx dx

1a u ; 1 2 3 4

133 44 27 6

6b u u u u

l ; 1 2 3 42

1 9518 36 9

6 3c u u u u

l

;

1 2 3 43

19 17 27 9

2d u u u u

l

2 3 2 3 2 3 2 3

1 2 3 42 3 2 3 2 3 2 3

11 9 45 27 9 27 9 91 9 9 18

2 2 2 2 2 2 2 2 x

x x x x x x x x x x x xu u u u u

l l l l l l l l l l l l

1 1 2 2 3 3 4 4 xu N u N u N u N u

Quartic element2 3 4

xu a bx cx dx ex

Strain, Stress and Stiffness Matrix for linear bar element

1

2

1 x

u x xu

ul l

Strain:

1

2

1 1 x x

uu

u x l l

Stress:

x x E

F x defines the force in the bar element at a distance of x unit from the node 1.

1 1

2 2

1 1 1 1 x x

u u EAF EA k

u ul

F 1 and F 2 are the nodal force at node 1 and 2.

Figure 1: Free body diagram, which shows element with nodes, nodal displacement, and

nodal forces

1 2

1u 2

u 1F

2F xF

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Figure shows the direction of Fx towards right. They are opposite to each other. If we

break the bar element into two pieces then we can see that the left side is pushing the right

side and in effect the right side is resisting the push (Newton’s third law). As, one opposes

the other.

For static equilibrium of above figure

1 10 x xF F F F

2 20 x xF F F F

1 1

1

2 2

1 1 1 1u u

F k k u u

1 1

2

2 2

1 1 1 1u u

F k k u u

1 1

2 2

1 1

1 1

F u

k F u

F k u

So, Stiffness matrix for a linear bar element isk k

k k

in element or local coordinate

system. u is the column matrix (vector) of nodal displacement, and F is the column

matrix (vector) of element nodal forces. Bar and a spring have the same behavior under

axial load and are represented by same stiffness matrix.

Stiffness matrix: stiffness matrix originated from structural analysis. The term is used to

to describe the matrix relation between force and displacement. The term is now usedregardless of the application. The matrix relation between temperature and heat flux is also

called stiffness matrix.

Finite element terminology defines two stiffness matrices. The local stiffness

matrix corresponds to an individual element. The global stiffness matrix is the assemblage

of all local stiffness matrix matrices and defines the stiffness of the entire system.

A symmetric matrix has off-diagonal terms such as ij jik k . Symmetry of the stiffness

matrix is indicative of the fact that the body is linearly elastic and each displacement is

related to the other by the same physical phenomenon. For example, if a force F (positive,

tensile) is applied at node 2 with node 1 held fixed, the relative displacement of the two

nodes is the same as if the force is applied symmetrically (negative, tensile) at node 1 withnode 2 fixed.

A linear elastic spring is a mechanical device capable of supporting axial loading only and

constraint such that, over a reasonable operating range (meaning extension or compression

beyond undeformed length), the elongation or contraction of proportionality between

deformation and load is referred to as the spring constant, spring rate, or spring stiffness,

generally denoted by k, and has units of force per unit length.

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Stiffness Matrix of Quadratic Bar Element

The quadratic bar element is a one-dimensional finite element were the local and global

coordinates coincide. It is characterized by quadratic shape functions. The quadratic bar

element has modulus of elasticity E, cross-sectional area A, and length L. Each quadratic

bar element has three nodes as shown in Figure. The third node is at the middle of the

element. In this case the element stiffness matrix is given by

7 1 8

1 7 83

8 8 16

EAk

L

It is clear that the quadratic bar element has three degree of freedom – one at each node.

Consequently for a structure with n nodes, the global stiffness matrix K will be of size n x

n (since we have one degree of freedom at each node). The order of the nodes for this

element is very important – the first node is the one at the left end, the second is the one at

the right end, and the third node is the one in the middle of the element.

Example:

Consider a structure built of two uniform elastic bars attached end to end.If a cantilever beam is discritized by two linear bar element. Where F 1, F 2, F 3 are the nodal

forces. It should be noted that both the element share the common nodal point, which is

acted upon by the force F 2.

Draw free body diagram

1 1 1

2 2 2

1 1 1 1

1 1 1 1

F u u AE k F u ul

F k u

k = Element stiffness matrix

u = Displacement vector

F3 F2

L1, A1, E1 L2, A2, E2

A B

F1

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Assemble for global matrix

For element 1

1 1 11 11

2 2 21

1 1 1 1

1 1 1 1

F u u A E k

F u ul

For element 2

2 2 22 22

3 3 32

1 1 1 1

1 1 1 1

F u u A E k

F u ul

Assemble

1 1

2 1 2

3 3

1 1 0

1 1 0

0 0 0

F u

F k u

F u

1 1

2 2 2

3 3

0 0 00 1 1

0 1 1

F uF k u

F u

Add both the matrix element by element, which form the global stiffness matrix

1 1 1 1

2 1 1 2 2 2

3 2 2 3

0

0

F k k u

F k k k k u

F k k u

Consider a structure built of three elastic bars attached end to end.

1 1

1 1 2 2

2 2 3 3

3 3

0 0

0

0

0 0

k k

k k k k

k k k k

k k

Assumption of global matrix

1. It has to be symmetric about major diagonal

2. All the diagonal position should be positive

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Problem 1: For the spring system with arbitrarily numbered nodes and elements, as

shown below, find the global stiffness matrix.

Fig :

Answer:

4 4

4 1 2 4 2 1

2 2 3 3

1 1

3 3

0 0 0

0K= 0 0

0 0 0

0 0 0

k k

k k k k k k k k k k

k k

k k

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Problem 2

Find the stiffness matrix of 3 bar element in the form of an equilateral triangle. The

stiffness of each bar element is same.

Answer:

5 √ 3 4 0 1 √ 33 0 0 √ 3 35 √ 3 1 √ 33 √ 3 3sym. 2 06

2D INCLINED BAR ELEMENT

For general, elements can assume any orientation in space. Let, a bar element is inclined at

an angle with the global Cartesian co-ordinate system (x-y) as shown in figure below.

In local co-ordinate system (x’-y’), the element lies along the x’ axis. The stiffness matrix

of the bar element in local co-ordinate system is already derived in previous section.

To find the stiffness matrix in global co-ordinate system, Let , are displacements atnode 1 and node 2 in local co-ordinate system and the corresponding displacements in

global co-ordinate system at node 1 and 2 are 1 , v1 and 2 , v2 respectively.

We can write: 1 cos v1 sin 0 1 sin v1 cos

x

y

v

v

L

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0 c o s s i n sin cos 1v1

Similarly, for node 2

0 c o s s i n

s i n c os 2

v2

Combining the above relation

00 c o s s i n 0 0 s i n c os 0 00 0 c o s s i n 0 0 sin cos 1v12v2

(1)

is the rotational transformation matrix.

Here, we are taking the component of global co-ordinate in local co-ordinate. So the

confusion comes as the bar element is defined to take load axial why vertical movement.

This is because the lateral displacement at both the nodes does not contribute to the stretchof the bar, within the linear theory.

Similarly, the forces acting in local co-ordinate system is related to global co-ordinate

system by the rotational matrix as: (2)

The force and displacement in local co-ordinate system can be written as: (3)

Putting the values of forces and displacement from equation (2) and (3) in (1):

(4)

Hence the above equation gives the relation between the force and displacement in global

co-ordinate system. Where, T

K R k R is the stiffness matrix of the element in

global axes system.

Augmenting k as

1 0 1 0

1 1 0 0 0 0

1 1 1 0 1 0

0 0 0 0

k k k

in the relation given below:

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0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 00 0 0 0

0 0 0 0

0 0 0 0

0 0 1 0 1 0 0 0

0 0 0 0 0 0 0 0

0 0 1 0 1 0 0

0 0 0 0 0 0

T c s c s

s c s cK k

c s c s

s c s c

c s c ss c s c

K k c s c s

s c s c

c s c s

s c s cK k

c s

s c

2 2

2 2

2

2

0

0 0

Sym.

c s

s c

c cs c cs

s cs s EAK L c cs

s

The global coordinate system is that system in which the behavior of a complete

structure is to be described. By complete structure is meant the assembly of many finite

elements (at this point, several springs) for which we desire to compute response to

loading condition.

Inclined bar in 2D

Fig :

1.83 m

300 600

600

1

2

3

4

Fx

Fy

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Note: For all the elements the inclination of the element should be measured either

clockwise or anti-clockwise. Here, for calculation anti-clockwise direction is taken.

2 0 0 0 2 0 0 0

0 0 2 0 0 0 2 0

13 2 0 0 0 2 0 0 0

13

0 0 2 0 0 0 2 0

2 0

23

23

cos 30 sin 30 cos30 cos 30 sin 30 cos30

sin 30 cos30 sin 30 sin 30 cos30 sin 30

cos 30 sin 30 cos30 cos 30 sin 30 cos30

sin 30 cos30 sin 30 sin 30 cos30 sin 30

cos 120 si

EAK

L

EAK

L

0 0 2 0 0 0

0 0 2 0 0 0 2 0

2 0 0 0 2 0 0 0

0 0 2 0 0 0 2 0

2

34

34

n120 cos120 cos 120 sin120 cos120sin120 cos120 sin 120 sin120 cos120 sin 120

cos 120 sin120 cos120 cos 120 sin120 cos120

sin120 cos120 sin 120 sin120 cos120 sin 120

cos 1

EAK

L

0 0 0 2 0 0 0

0 0 2 0 0 0 2 0

2 0 0 0 2 0 0 0

0 0 2 0 0 0 2 0

50 sin150 cos150 cos 150 sin150 cos150


cos 150 sin150 cos150 cos 150 sin150 cos150


0 0

0 0 0 0

0 0 0 0

1 3sin 30 ;cos30

2 2

3 1sin120 cos30 ;cos120 sin 30

2 2

1 3sin150 cos 60 ;cos150 sin 60

2 2

13 34 23

1.831.83;

3 L L L

1.83 m

300

600

600

1

2

3

4

Fx

Fy

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13

13

3 3 3 3

4 4 4 43 3 3 3

3 1 3 1

3 1 3 14 4 4 4

4 1.833 3 3 3 3 3 3 3

4 4 4 4 3 1 3 1

3 1 3 1

4 4 4 4

EA EAK

L

23

23

1 3 1 3

4 4 4 43 3 3 3

3 3 3 3

3 3 3 3 3 34 4 4 4

4 1.831 3 1 3 3 3 3 3

4 4 4 4 3 3 3 3 3 3

3 3 3 3

4 4 4 4

EA EAK

L

34

34

3 3 3 3

4 4 4 43 3 3 3

3 1 3 1

3 1 3 14 4 4 4

4 1.833 3 3 3 3 3 3 3

4 4 4 43 1 3 1

3 1 3 1

4 4 4 4

EA EAK

L

7.32

3 √ 3 0 0 3 √ 3 0 01 0 0 √ 3 1 0 0√ 3 3 √ 3 3 0 03√ 3 3 3√ 3 0 06 √ 3 3 3 √ 3

2 3√ 3 √ 3 1. 3 √ 31

Element coordinate or local coordinate system is defined for a particular element and it

varies from element to element with in a complete structure.

Problem 1: Find the stresses in the two bar assembly which is loaded with force P, and

constrained at the two ends, as shown in the figure.

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Fig :

Solution: ;

Problem 2: Determine the support reaction forces at the two ends of the bar shown above,

give that following.4 4 2 2

6.0 10 N, 2.0 10 N/mm , A 250 mm , L 150 mm, 1.2 mmP E

Fig :

Solution: 5.0 10N; 1.0 10N

Problem 3: A simple plane truss is made of two identical bars (with E, A, and L), and

loaded as shown in figure. Find the displacement of node 2 and stress in each bar.

Fig :

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Solution: v ; √ ; √

Problem 4: For the plane truss shown below, 1000 N, 1 m; 210 GPa;

6 10 . For elements 1 and 2,

6√ 2 10 for element 3. Determine

the displacements and reaction forces.

Solution: 0.011910.003968 m;

5005000.0500500 kN

Problem 5: (P.209, C S Krishnamoorthy)

Consider a two-dimensional truss structure shown in the figure below. The geometry and

loading are symmetrical about the center line. Assume the area of cross section of all the

members is the same. E = 2 x 104

kN/cm2

. Find the force in the vertical member.

Answer: 3.720 kN (tension)

50kN

100kN

0.5m

1.0m

50kN

100kN

x

y

1

2

3

4

5

4

3

1

2

300

60

0

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3-Noded Triangular Element

Pascal Triangle

It is an useful aid for determining the combination of terms which should be used to write

displacement function in the form of polynomial.

Let’s assume that the displacement field in the body can be defined with a continuous

polynomial function of x and y having p degree as

p

n

p

n

p p

p

n

p

n

p p

yb xyb y xb xbb y xv

ya xya y xa xaa y xu

1

1

1

210

1

1

1

210

),(

),(

where ai and bi are constants. In order to obtain the constants ai and bi the physical body

has to be discretized with triangular or quadrilateral elements in 2D and tetrahedron or

brick-type elements in 3D. For triangular elements, a complete polynomial can be defined

in Cartesian coordinates using all terms of a Pascal triangle as shown in Figure **. Based

on the degree of polynomial, number of nodes is defined as

2

)2)(1(

p pn

Pascal Triangle

Degree of

polynomial, p

Number of

nodes / terms,n

Name

1 0 1 Constantx y 1 3 Linear2

x xy 2 y 2 6 Quadratic

3 x

2 x y

2 xy

3 y 3 10 Cubic

4 x

3 x y

2 2 x y

3 xy

4 y 4 15 Quartic

5 x

4 x y

3 2x y

2 3 x y

4 xy

5 y 5 21 Quintic

6 x 5 x y

4 2 x y 3 3 x y

2 4 x y 5 xy

5 y 6 28 Hexadic

7 x 6 x y

5 2 x y 4 3 x y

3 4 x y 2 5 x y

6 xy 6 y 7 36 Septic

Figure ***. Pascal triangle for estimating polynomial of triangular element

Thus if the degree of the polynomial is 1, there are three nodes in the vertices of thetriangle and it is termed as linear triangle. For degree 2 polynomial or quadratic triangle,

node at each vertex and at the mid point of each side will be needed. A triangular element

of degree 1, 2 and 3 are called linear, quadratic and cubic triangular element respectively

as shown in Figure ***. Quadratic and higher order triangular element can be used with

curved sides. For cubic and higher order triangular elements, internal node(s) will be

present in the element. Figure 3.5 shows a discretized tunnel boundary with 6-noded

triangular elements. In general, smaller elements are formed near the boundary or curved

surfaces and relatively bigger elements are used to model surface away from the boundary.

Linear triangle Quadratic triangle Cubic triangle

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One of the widely used formulations for finite element analysis is based on the assumption

for the variation of displacement in the element and such models are called displacement

models or displacement formulation.

Displacement Models

Assume displacement function to be linear

, x y a bx cy For displacement in x direction the function can be written as

,u x y a bx cy

For nodal displacement at 1, 2, and 3

i i i

j j j

k k k

u a bx cy

u a bx cy

u a bx cy

In matrix form

1

1

1

i i i

j j j

k k k

u x y a

u x y b

u x y c

The unknown a, b, and c cab be found by taking the inverse of above relation1

1

1

1

i i i

j j j

k k k

a x y u

b x y u

c x y u

i

k

( xi, yi)

( x j, y j)

( xk , yk )

x

y

ui

u j

uk

vk

v j

vi

o

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1 1 1 2 1 3

2 1 2 2 2 3

3 1 3 2 3 3

1 11 1 1

1 1

1 11 1 1

1 1

1 11 1 11 1

T

j j j j

k k k k

i i i i

k k k k

i i i i

i j j j j

j

k

x y y x

x y y x

x y y y

x y y y

x y y xa u

x y y xb u

c u

1 1

1 1

1 1

1 1

1 1

1 1

1

1

1

T

j j j j

k k k k

i i i i

k k k k

i i i i

i j j j j

j

i i

k

j j

k k

x y y x

x y y x

x y y y

x y y y

x y y x

a u x y y xb u

x yc u

x y

x y

2

T

j k k j j k k j

k i i k k i i k

i

i j j i i j j i

j

k

x y x y y y x x

x y x y y y x xa u

x y x y y y x xb u

Ac u

2

j k k j k i i k i j j i

j k k i i j

i

k j i k j i

j

k

x y x y x y x y x y x y

y y y y y ya u

x x x x x xb u

Ac u

1

2

i j k i

i j k j

i j k k

a a a a u

b b b b u

c c c c u

Where,

i j k k j

i j k

i k j

a x y x y

b y y

c x x

Similarly, the other coefficients are obtained by a cyclic

permutation of subscript in order of i, j, and k i j

k

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For example

j k i i k

j k i

j i k

a x y x y

b y y

c x x

and

m i j j i

m i j

m j i

a x y x y

b y y

c x x

Where, A is the area of the triangle

11

1 or 22 21

i i

j j

k k

x y

A x y A x y

1

2

1

2

1

2

i i j j k k

i i j j k k

i i j j k k

a a u a u a u

b bu b u b u

c c u c u c u

1

,

2

i i j j k k i i j j k k i i j j k k u x y a u a u a u b u b u b u x c u c u c u y

1

,2

i i i i j j k j k k k k u x y a b x c y u a b x c y u a b x c y u

Similarly,

1

,2

i i i i j j k j k k k k v x y a b x c y v a b x c y v a b x c y v

, 1

, 2

i i i i j j k j k k k k

i i i i j j k j k k k k

a b x c y u a b x c y u a b x c y uu x y

v x y a b x c y v a b x c y v a b x c y v

0 0 0, 1

0 0 0, 2

i

i

i i i j j k k k k j

i i i j j k k k k j

k

k

uv

a b x c y a b x c y a b x c yu x y u

a b x c y a b x c y a b x c yv x y v

u

v

,

I I I,

e e

i j k

u x y N N N a N a

v x y

[N] is shape function for node i, j, and k.

1 0

I0 1

and

1

2i i i i N a b x c y

01

I02

i i i

i

i i i

a b x c y N

a b x c y

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0

0

x

y

xy

u

x x

uv

v y y

u v

y x y x

e e

L u

L N a B a

00 0

1 0 10 0 0

1 0 2

i

i i i

ii i i i i i

i i

i i i i i i

N a b x c y

x x x

N B L I N N a b x c y

y y y

N N

a b x c y a b x c y y x y x y x

01

02

i

i i

i i

b

B c

c b

similarly,

01

02

j

j j

j j

b

B c

c b

and

01

02

k

k k

k k

b

B c

c b

so,

0 0 01

0 0 02

i j k

i j k i j k

i i j j k k

b b b

c c c B B B B

c b c b c b

Here, it can be mentioned that the B-matrix is independent of co-ordinate axis and hence

strain is not dependent on the co-ordinate system. Hence the strain is constant throughout

the element so called constant strain element (CST) element. Some times it is also called

strain-displacement matrix as this matrix relates strain with displacement.

Problem 1: For a linear equilateral triangle of side a, show that if one side is parallel to

any of the axis x or y, the strain-displacement matrix will be the only function

of a.

Let us suppose the coordinates of the nodes are 11 , y x ; 22 , y x ; 33 , y x , as the

22 , y x 11 , y x

33 , y x y

x

1 2

3

8/11/2019 Notes - FEM


21

As it is given that the any of the side of isosceles triangle is parallel to any of the axis x or

y .

Use the equation 3.41a, 3.41b, and 3.41c to determine the [B] for CST element. Here

from figure it is clear that: a x x 12 ; 213 a x x ; 21 y y ; A y y2

313 ,

using these relations the [B] is given below.

013131

101010

000303

4

02

3

22

3

2

02

02

0

0002

30

2

3

2

1 a

aaa

aa

aaa

aa

B

Area of isosceles triangle of side a, 2

4

3a

013131

101010

000303

3

1

a B

It can be written as:

a f B

1

Hence the [B] is function of a.

Problem 2: For a linear triangular element shown in figure ***, obtain the matrix B and

also determine the strain vector, .

ANSWER: will depend on the order of nodes you are taking for answering.

Use the equation to determine the [B].

x

y

(1,1)

4,2

(3,4)

1

2

3 q T= { 0 0 0.008 -0.01 0 - 0.005 }

P

8/11/2019 Notes - FEM


0.2857 0.0000 0.4286 0.0000 0.1429 0.0000

0.0000 0.1429 0.0000 0.2857 0.0000 0.4286

0.1429 0.2857 0.4286 0.4286 0.4286 0.1429

B

Given,

1

1

2

2

3

3

0.000

0.000

0.008

0.010

0.000

0.005

u

v

uq

v

u

v

0.000

0.0000.2857 0.0000 0.4286 0.0000 0.1429 0.0000

0.008

0.0000 0.1429 0.0000 0.2857 0.0000 0.4286 0.0100.1429 0.2857 0.4286 0.4286 0.4286 0.1429

0.000

0.005

xx

yy

xy

ε

Bq

0.0034

0.0007

0.0059

xx

yy

xy

ε

notes - fem

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