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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
sin150 cos150 sin 150 sin150 cos150 sin 150
cos 150 sin150 cos150 cos 150 sin150 cos150
sin150 cos150 sin 150 sin150 cos150 sin 150
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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20
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
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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
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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