ME6603 - FINITE ELEMENT ANALYSIS FORMULA BOOK

FINITE ELEMENT ANALYSIS FORMULA BOOK BY ASHOK KUMAR .R (AP / MECH) 1
ME6603 – FINITE ELEMENT ANALYSIS
FORMULA BOOK

DEPARTMENT OF MECHANICAL ENGINEERING ME6603 – FEA
WEIGHTED RESIDUAL METHODS:
General Trial function:
𝑢 = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥2
+ 𝑎3 𝑥3
Substitute the trial function in differential equation to get the residue function.
Point Collocation Method:
𝑅( 𝑥) = 0
R(x) Residue function
Here the point of x is between these limits. If the limit is 0 to 1
For two unknowns any two point between these limits. For example
x value is
1
4
x value is
3
4
For three unknowns any two point between these limits. For example
x value is
1
4
x value is
1
2
x value is
3
4
Sub Domain Method:
∫ 𝑅( 𝑥) . 𝑑𝑥 = 0
Here the limit should be taken into two intervals. If the limit is 0 to 1
The limit varies between 0 – 0.5 ∫ 𝑅( 𝑥)
0.5
0
𝑑𝑥 = 0
The limit varies between 0.5 – 1 ∫ 𝑅( 𝑥)
1
0.5
𝑑𝑥 = 0
Least Square Method:
∫ 𝑅( 𝑥) .
𝑑𝑅
𝑑𝑎𝑖
𝑑𝑥 = 0 𝑖 = 0, 1, 2, 3 … … ….
Differentiate the R(x) with respect to a1, a2
Galerkin Method:
∫ 𝑅( 𝑥) . 𝜙 ( 𝑥). 𝑑𝑥 = 0
UNIT I – INTRODUCTION

𝜙 ( 𝑥) Weighting function (Function associated with unknown
trial function)
𝜙 ( 𝑥) = 𝜙0 + 𝑎2 𝜙1 + 𝑎3 𝜙2
RAYLEIGH – RITZ METHOD:
For Beam the Fourier series equation
𝑦 = ∑ 𝑎 sin
𝑛𝜋𝑥
𝑙
∞
𝑛=1,3,5
𝑦 = 𝑎1 sin
𝜋𝑥
𝑙
+ 𝑎2 sin
3𝜋𝑥
𝑙
For Cantilever Bar (Axial Loading)
𝑢 = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥2
+ 𝑎3 𝑥3
+. . . . . +𝑎 𝑛 𝑥 𝑛
Where a1 a2 & a3 are Ritz parameters
The Total Potential Energy of the Beam:
𝜋 = 𝑈 − 𝐻
U Strain Energy
H Work Done by External Force
Strain Energy for Transverse Loading:
𝑈 =
𝐸𝐼
2
∫ (
𝑑2
𝑦
𝑑𝑥2
)
2𝑙
0
𝑑𝑥
Strain Energy for Axial Loading:
𝑈 =
𝐸𝐴
2
∫ (
𝑑𝑢
𝑑𝑥
)
2𝑙
0
𝑑𝑥
Work Done by External Force:
𝐻 = ∫ 𝜔 𝑦 𝑑𝑥
𝑙
0
For SSB with UDL throughout its length
𝐻 = 𝑊 𝑦 𝑚𝑎𝑥 For SSB with Point Load at its mid – point of length

𝐻 = ∫ 𝜔 𝑦 𝑑𝑥
𝑙
0
+ 𝑊 𝑦 𝑚𝑎𝑥 For SSB with UDL throughout its length
and Point Load at its mid – point of length
𝐻 = ∫ 𝑃 𝑑𝑥
𝑙
0
= ∫ 𝜌 𝐴 𝑢 𝑑𝑥
𝑙
0
For Cantilever Bar with axial loading
Bending Moment:
𝑀 = 𝐸𝐼
𝑑2
𝑦
𝑑𝑥2
E Young’s modulus
I Moment of Inertia
Exact Solution:
SSB Max Deflection with UDL throughout its length
𝑦 𝑚𝑎𝑥 =
5
384
𝑤𝑙4
𝐸𝐼
SSB Max Bending Moment with UDL throughout its length
𝑀𝑐𝑒𝑛𝑡𝑟𝑒 =
𝑤𝑙2
8
SSB Max Deflection with Point Load at its mid – point of length
𝑦 𝑚𝑎𝑥 =
𝑊𝑙3
48𝐸𝐼
SSB Max Bending Moment with Point Load at its mid – point of length
𝑊𝑙
4
SSB Max Deflection with UDL throughout its length and Point Load at its mid
– point of length
𝑦 𝑚𝑎𝑥 =
5
384
𝑤𝑙4
𝐸𝐼
+
𝑊𝑙3
48𝐸𝐼
SSB Max Bending Moment with UDL throughout its length and Point Load at
its mid – point of length
𝑤𝑙2
8
+
𝑊𝑙
4

GENERAL TRIGONOMETRIC FUNCTIONS USED IN RITZ METHOD:
𝑠𝑖𝑛2
𝑥 =
1 − 𝑐𝑜𝑠2𝑥
2
𝑐𝑜𝑠2
𝑥 =
1 + 𝑐𝑜𝑠2𝑥
2
sin 𝐴 sin 𝐵 =
cos( 𝐴 − 𝐵) + cos(𝐴 + 𝐵)
2
SPRINGS:
The Total Potential Energy of the Beam:
𝜋 = 𝑈 − 𝐻
U Strain Energy
H Work Done by External Force
Strain Energy:
𝑈 =
1
2
𝑘 𝛿2
k Stiffness of Spring
δ Deflection
Work done by External Force:
𝐻 = 𝐹𝑢
F Force acting on the spring
u Displacement

Displacement Function:
For 1D linear bar element
𝑢 = 𝑁1 𝑢1 + 𝑁2 𝑢2
𝑁1 = 1 −
𝑥
𝑙
𝑁2 =
𝑥
𝑙
For 1D quadratic element
𝑢 = 𝑁1 𝑢1 + 𝑁2 𝑢2 + 𝑁3 𝑢3
𝑁1 = 1 −
3𝑥
𝑙
+
2𝑥2
𝑙2
𝑁2 = −
𝑥
𝑙
+
2𝑥2
𝑙2
𝑁3 =
4𝑥
𝑙
−
4𝑥2
𝑙2
UNIT II – ONE DIMENSIONAL PROBLEMS

N1 Shape Function at node 1
u1 Displacement at node 1
Stiffness Matrix:
[ 𝑘] =
𝐴𝐸
𝑙
[
1 −1
−1 1
]
[ 𝑘] =
𝐴𝐸
3𝑙
[
7 1 −8
1 7 −8
−8 −8 16
]
A Area of the element - mm2
E Young’s Modulus of the element - N/mm2
l Length of the element - mm
General Force Equation:
{ 𝐹} = [ 𝑘] { 𝑢}
{𝐹} Force vector (Global)
[𝑘] Global stiffness matrix
{𝑢} Displacement matrix
Reaction Force:
{ 𝑅} = [ 𝑘] { 𝑢} − { 𝐹}
{𝑅} Reaction force
If The Body Is Subjected To Self – Weight:
{ 𝐹} =
𝜌𝐴𝑙
2
{
1
1
}

{ 𝐹} = 𝜌𝐴𝑙
{
1
6
1
6
2
3}
ρ Unit weight density of the element - N/mm3
Stress On The Element:
𝜎 = 𝐸 .
𝑑𝑢
𝑑𝑥
If stress on element one should found then the formula will be
𝜎 = 𝐸 .
𝑢2 − 𝑢1
𝑙1
𝑙1 Length of the element one - mm
𝑢1 Displacement at nodal point 1 - mm
𝑢2 Displacement at nodal point 2 - mm
FOR TAPER PLATE:

For rectangular cross section
Area = Width * Thickness
At any point of x
𝐴 𝑥 = 𝐴1 − (𝐴1 − 𝐴3)
𝑥
𝑙
Area at node 1:
𝐴1 = 𝑊1 ∗ 𝑡1
Area at node 2(Applicable only for mid – point):
𝐴2 = 𝑊2 ∗ 𝑡2
𝐴2 = (
𝑊1 + 𝑊3
2
) ∗ 𝑡2
Area at node 3:
𝐴3 = 𝑊3 ∗ 𝑡3
Average area of element 1:
𝐴
=
𝐴1+ 𝐴2
2
Average area of element 2:
𝐴
=
𝐴2 + 𝐴3
2
For circular cross section
𝐴𝑟𝑒𝑎 =
𝜋
4
𝑑2
TEMPERATURE EFFECTS ON STRUCTURAL PROBLEM:
Stiffness Matrix:
[ 𝑘] =
𝐴𝐸
𝑙
[
1 −1
−1 1
]

Thermal Load:
{ 𝐹} = 𝐸𝐴𝛼∆𝑇 {
−1
1
}
α Coefficient of thermal expansion - /˚C
∆T Temperature difference
Thermal Stress:
{ 𝜎} = 𝐸
𝑑𝑢
𝑑𝑥
− 𝐸𝛼∆𝑇
𝐸𝛼∆𝑇 Thermal Strain
For element 1
{ 𝜎} = 𝐸1
𝑢2 − 𝑢1
𝑙1
− 𝐸1 𝛼1∆𝑇
Reaction Force:
{ 𝑅} = [ 𝑘] { 𝑢} − { 𝐹}
{𝑅} Reaction force
[𝑘] Global stiffness matrix
SPRINGS:
Stiffness Matrix:
[ 𝑘] = 𝑘 [
1 −1
−1 1
]
k Stiffness of the spring - N/mm
HEAT TRANSFER:
ONE DIMENSIONAL HEAT TRANSFER ON WALL:
Global Stiffness Matrix When Wall Subjected To Conduction:
[ 𝐾] { 𝑇} = { 𝐹}

[𝐾] Global stiffness matrix
{𝑇} Temperature matrix
[ 𝐾] =
𝐴𝑘
𝑙
[
1 −1
−1 1
]
𝐴𝑘
𝑙
[
1 −1
−1 1
] {
𝑇1
𝑇2
} = {
𝐹1
𝐹2
}
A Area of the wall - m2
k Thermal conductivity of wall - W/mK
l Length of the wall - m
Global Stiffness Matrix When Wall Subjected To Conduction &
Convection:
[ 𝐾𝑐] =
𝐴𝑘
𝑙
[
1 −1
−1 1
]
[ 𝐾ℎ] = ℎ𝐴 [
0 0
0 1
]
[ 𝐾] = [ 𝐾𝑐] + [ 𝐾ℎ]
[ 𝐾] { 𝑇} = { 𝐹}
[
𝐴𝑘
𝑙
[
1 −1
−1 1
] + ℎ𝐴 [
0 0
0 1
]] {
𝑇1
𝑇2
} = ℎ𝑇∞ 𝐴 {
0
1
}
A Area of the wall - m2
k Thermal conductivity of wall - W/mK
l Length of the wall - m
h Heat transfer coefficient - W/m2
K
𝑇∞ Atmospheric air temperature - K

ONE DIMENSIONAL HEAT TRANSFER ON FIN:
Global Stiffness Matrix When Fin Subjected To Conduction &
Convection:
[ 𝐾𝑐] =
𝐴𝑘
𝑙
[
1 −1
−1 1
]
[ 𝐾ℎ] =
ℎ𝑃𝑙
6
[
2 1
1 2
]
[ 𝐾] = [ 𝐾𝑐] + [ 𝐾ℎ]
[ 𝐾] { 𝑇} = { 𝐹}
[
𝐴𝑘
𝑙
[
1 −1
−1 1
] +
ℎ𝑃𝑙
6
[
2 1
1 2
]] {
𝑇1
𝑇2
} =
𝑄𝐴𝑙 + 𝑃ℎ𝑇∞ 𝑙
2
{
1
1
}
𝐴 = 𝐿𝑒𝑛𝑔𝑡ℎ ∗ 𝑇ℎ𝑖𝑐𝑘𝑒𝑛𝑠𝑠 = 𝑙 ∗ 𝑡
𝑃 = 2 ∗ 𝑙 (𝐴𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒𝑙𝑦)
A Area of the fin - m2
P Perimeter of the fin - m
k Thermal conductivity of fin - W/mK
l Length of the fin - m
K
Q Heat Generation - W

TRUSSES:
Stiffness Matrix:
[ 𝑘] =
𝐸𝑒 𝐴 𝑒
𝑙 𝑒
[
𝑙2
𝑙𝑚 −𝑙2
−𝑙𝑚
𝑙𝑚 𝑚2
−𝑙𝑚 −𝑚2
−𝑙2
−𝑙𝑚 𝑙2
𝑙𝑚
−𝑙𝑚 −𝑚2
𝑙𝑚 𝑚2
]
Ae Area of the element - mm2
Ee Young’s Modulus of the element - N/mm2
le Length of the element - mm
Length of element 1
𝑙 = √( 𝑥2 − 𝑥1)2 + ( 𝑦2 − 𝑦1)2
𝑙 =
𝑥2 − 𝑥1
𝑙
𝑚 =
𝑦2 − 𝑦1
𝑙
Stress:
Stress of element 1
𝜎 =
𝐸
𝑙
[−𝑙1 −𝑚1 𝑙1 𝑚1] {
𝑢1
𝑢2
𝑢3
𝑢4
}
u matrix varies according the element

BEAMS:
Displacement Function:
𝑢 = 𝑁1 𝑢1 + 𝑁2 𝑢2 + 𝑁3 𝑢3 + 𝑁4 𝑢4
𝑁1 =
1
𝐿3
(2𝑥3
− 3𝑥2
𝐿 + 𝐿3)
𝑁2 =
1
𝐿3
( 𝑥3
𝐿 − 2𝑥2
𝐿2
+ 𝑥𝐿3)
𝑁3 =
1
𝐿3
(− 2𝑥3
+ 3𝑥2
𝐿)
𝑁4 =
1
𝐿3
( 𝑥3
𝐿 − 𝑥2
𝐿2)
Stiffness Matrix:
[ 𝑘] =
𝐸𝐼
𝐿3
[
12 6𝐿 −12 6𝐿
6𝐿 4𝐿2
−6𝐿 2𝐿2
−12 −6𝐿 12 −6𝐿
6𝐿 2𝐿2
−6𝐿 4𝐿2
]
I Moment of Inertia - mm4
E Young’s Modulus - N/mm2
L Length of the beam - mm

Nodal Forces and Bending Moments:
S.No
LOADING
CASE
F1 m1 F2 m2
1)
−𝑊
2
−𝑊𝐿
8
−𝑊
2
𝑊𝐿
8
2)
−𝑊𝑏2
(𝐿 + 2𝑎)
𝐿3
−𝑊𝑎𝑏2
𝐿2
−𝑊𝑎2
(𝐿 + 2𝑏)
𝐿3
𝑊𝑎2
𝑏
𝐿2
3) −𝑊 −𝛼(1 − 𝛼) 𝑊𝐿 −𝑊 𝛼(1 − 𝛼) 𝑊𝐿
4)
−𝑤𝐿
2
−𝑤𝐿2
12
−𝑤𝐿
2
𝑤𝐿2
12

S.No
LOADING
CASE
F1 m1 F2 m2
5)
−7𝑤𝐿
20
−𝑤𝐿2
20
−3𝑤𝐿
20
𝑤𝐿2
30
6)
−𝑤𝐿
4
−5𝑤𝐿2
96
−𝑤𝐿
4
5𝑤𝐿2
96
7)
−13𝑤𝐿
32
−11𝑤𝐿2
192
−3𝑤𝐿
32
5𝑤𝐿2
192
LONGITUDINAL VIBRATION OF BAR:
General Finite Element Equation:
{[ 𝐾] − [ 𝑚] 𝜔2} { 𝑢} = { 𝐹}
[𝐾] Stiffness matrix
[𝑚] Mass matrix
{𝐹} External Force

𝜔 Natural Frequency - rad / sec
Stiffness Matrix:
[ 𝐾] =
𝐴𝐸
𝑙
[
1 −1
−1 1
]
Lumped Mass Matrix:
NOTE: Unless mention in the problem to use lumped mass matrix we should not
use this. If nothing is mentioned use consistent mass matrix only
[ 𝑚] =
𝜌𝐴𝑙
2
[
1 0
0 1
]
Consistent Mass Matrix:
NOTE: If nothing is mentioned use consistent mass matrix only
[ 𝑚] =
𝜌𝐴𝑙
6
[
2 1
1 2
]
TRANSVERSE VIBRATION OF BEAMS:
{[ 𝐾] − [ 𝑚] 𝜔2} { 𝑢} = { 𝐹}
[𝑚] Mass matrix
{𝐹} External Force

Stiffness Matrix:
[ 𝑘] =
𝐸𝐼
𝐿3
[
12 6𝐿 −12 6𝐿
6𝐿 4𝐿2
−6𝐿 2𝐿2
−12 −6𝐿 12 −6𝐿
6𝐿 2𝐿2
−6𝐿 4𝐿2
]
I Moment of Inertia - mm4
L Length of the beam - mm
Lumped Mass Matrix:
NOTE: Unless mention in the problem to use lumped mass matrix we should not
use this. If nothing is mentioned use consistent mass matrix only
[ 𝑚] =
𝜌𝐴𝑙
2
[
1 0 0 0
0 0 0 0
0 0 1 0
0 0 0 0
]
Consistent Mass Matrix:
NOTE: If nothing is mentioned use consistent mass matrix only
[ 𝑚] =
𝜌𝐴𝑙
420
[
156 22𝑙 54 −13𝑙
22𝑙 4𝑙2
13𝑙 −3𝑙2
54 13𝑙 156 −22𝑙
−13𝑙 −3𝑙2
−22𝑙 4𝑙2
]
EIGEN VALUES & EIGEN VECTORS:
{[ 𝐾] − [ 𝑚] 𝜆} { 𝑢} = 0
[𝑚] Mass matrix

𝜆 Eigen values
Determinant Method:
|[ 𝐾] − [ 𝑚] 𝜆| = 0
[𝑚] Mass matrix
𝜆 Eigen values
Natural Frequency of Bar Element:
𝜆 = 𝜔2
𝑓 =
1
2𝜋
𝜔
𝑓 Frequency of bar - Hz
[𝑚] Mass matrix
𝜆 Eigen values

CONSTANT STRAIN TRIANGULAR ELEMENT:
Shape Function:
𝑁1 =
𝑝1 + 𝑞1 𝑥 + 𝑟1 𝑦
2𝐴
𝑁2 =
𝑝2 + 𝑞2 𝑥 + 𝑟2 𝑦
2𝐴
𝑁3 =
𝑝3 + 𝑞3 𝑥 + 𝑟3 𝑦
2𝐴
𝑝1 = 𝑥2 𝑦3 − 𝑥3 𝑦2
𝑝2 = 𝑥3 𝑦1 − 𝑥1 𝑦3
𝑝3 = 𝑥1 𝑦2 − 𝑥2 𝑦1
𝑞1 = 𝑦2 − 𝑦3
𝑞2 = 𝑦3 − 𝑦1
𝑞3 = 𝑦1 − 𝑦2
𝑟1 = 𝑥3 − 𝑥2
𝑟2 = 𝑥1 − 𝑥3
𝑟3 = 𝑥2 − 𝑥1
Area:
𝐴 =
1
2
|
1 𝑥1 𝑦1
1 𝑥2 𝑦2
1 𝑥3 𝑦3
|
Shape Function:
𝑁1 + 𝑁2 + 𝑁3 = 1
When all the shape function is given
𝑥 = 𝑁1 𝑥1 + 𝑁2 𝑥2 + 𝑁3 𝑥3
UNIT III – TWO DIMENSIONAL SCALAR VARIABLE
PROBLEMS

𝑦 = 𝑁1 𝑦1 + 𝑁2 𝑦2 + 𝑁3 𝑦3
When two or one shape function is given
𝑥 = ( 𝑥1 − 𝑥3) 𝑁1 + ( 𝑥2 − 𝑥3) 𝑁2 + 𝑥3
𝑦 = ( 𝑦1 − 𝑦3) 𝑁1 + ( 𝑦2 − 𝑦3) 𝑁2 + 𝑦3
Shape Function Equation:
When Pressure is given
𝑃(𝑥, 𝑦) = 𝑃1 𝑁1 + 𝑃2 𝑁2 + 𝑃3 𝑁3
When Temperature is given
𝑇(𝑥, 𝑦) = 𝑇1 𝑁1 + 𝑇2 𝑁2 + 𝑇3 𝑁3
When Displacement is given
𝑢(𝑥, 𝑦) = 𝑢1 𝑁1 + 𝑢2 𝑁2 + 𝑢3 𝑁3
To Find The Coordinate Points:
To find x coordinates when pressure is given
𝑥 − 𝑥1
𝑥2 − 𝑥1
=
𝑃 − 𝑃1
𝑃2 − 𝑃1
To find y coordinates when pressure is given
𝑦 − 𝑦1
𝑦2 − 𝑦1
=
𝑃 − 𝑃1
𝑃2 − 𝑃1
To find x coordinates when temperature is given
𝑥 − 𝑥1
𝑥2 − 𝑥1
=
𝑇 − 𝑇1
𝑇2 − 𝑇1
To find y coordinates when temperature is given
𝑦 − 𝑦1
𝑦2 − 𝑦1
=
𝑇 − 𝑇1
𝑇2 − 𝑇1
RECTANGULAR ELEMENT:

Shape Function:
𝑁1 = ( 1 −
𝑥
1
−
𝑦
ℎ
+
𝑥𝑦
𝑙ℎ
)
𝑁2 =
𝑥
𝑙
−
𝑥𝑦
𝑙ℎ
𝑁3 =
𝑥𝑦
𝑙ℎ
𝑁4 =
𝑦
ℎ
−
𝑥𝑦
𝑙ℎ
When Pressure is given
𝑃( 𝑥, 𝑦) = 𝑃1 𝑁1 + 𝑃2 𝑁2 + 𝑃3 𝑁3 + 𝑃4 𝑁4
When Temperature is given
𝑇( 𝑥, 𝑦) = 𝑇1 𝑁1 + 𝑇2 𝑁2 + 𝑇3 𝑁3 + 𝑇4 𝑁4
When Displacement is given
𝑢( 𝑥, 𝑦) = 𝑢1 𝑁1 + 𝑢2 𝑁2 + 𝑢3 𝑁3 + 𝑢4 𝑁4
To Find The Coordinate Points:
To find x coordinates when pressure is given
𝑥 − 𝑥1
𝑥2 − 𝑥1
=
𝑃 − 𝑃1
𝑃2 − 𝑃1
To find y coordinates when pressure is given
𝑦 − 𝑦1
𝑦2 − 𝑦1
=
𝑃 − 𝑃1
𝑃2 − 𝑃1
To find x coordinates when temperature is given
𝑥 − 𝑥1
𝑥2 − 𝑥1
=
𝑇 − 𝑇1
𝑇2 − 𝑇1
To find y coordinates when temperature is given
𝑦 − 𝑦1
𝑦2 − 𝑦1
=
𝑇 − 𝑇1
𝑇2 − 𝑇1

TWO DIMENSIONAL HEAT TRANSFER:
Stiffness Matrix When Subjected To Conduction:
[ 𝐾𝑐] = [ 𝐵] 𝑇 [ 𝐷] [ 𝐵] 𝐴 𝑡
[𝐵] Strain Displacement matrix
[𝐷] Material Property matrix
t Thickness of the triangular element - mm
𝐴 Area of the triangular element - mm2
Strain Displacement Matrix:
[ 𝐵] =
1
2𝐴
[
𝑞1 𝑞2 𝑞3
𝑟1 𝑟2 𝑟3
]
𝑞1 = 𝑦2 − 𝑦3
𝑞2 = 𝑦3 − 𝑦1
𝑞3 = 𝑦1 − 𝑦2
𝑟1 = 𝑥3 − 𝑥2
𝑟2 = 𝑥1 − 𝑥3
𝑟3 = 𝑥2 − 𝑥1
Area:
𝐴 =
1
2
|
1 𝑥1 𝑦1
1 𝑥2 𝑦2
1 𝑥3 𝑦3
|
Material Property Matrix:
[ 𝐷] = [
𝑘 𝑥𝑥 0
0 𝑘 𝑦𝑦
]
kxx & kyy Thermal conductivity - W/mK

Stiffness Matrix When Subjected To Conduction:
[𝐾𝑐] =
𝑘
4𝐴
[
𝑏1
2
+ 𝑐1
2
𝑏1 𝑏2 + 𝑐1 𝑐2 𝑏1 𝑏3 + 𝑐1 𝑐3
𝑏1 𝑏2 + 𝑐1 𝑐2 𝑏2
2
+ 𝑐2
2
𝑏2 𝑏3 + 𝑐2 𝑐3
𝑏1 𝑏3 + 𝑐1 𝑐3 𝑏2 𝑏3 + 𝑐2 𝑐3 𝑏3
2
+ 𝑐3
2
]
𝑏1 = 𝑦2 − 𝑦3
𝑏2 = 𝑦3 − 𝑦1
𝑏3 = 𝑦1 − 𝑦2
𝑐1 = 𝑥3 − 𝑥2
𝑐2 = 𝑥1 − 𝑥3
𝑐3 = 𝑥2 − 𝑥1
k Thermal conductivity - W/mK
Stiffness Matrix When Subjected To Convection:
If side 1-2 is subjected to convection
[ 𝐾ℎ] =
ℎ1−2 𝑙1−2 𝑡
6
[
2 1 0
1 2 0
0 0 0
]
[ 𝐾ℎ] =
ℎ2−3 𝑙2−3 𝑡
6
[
0 0 0
0 2 1
0 1 2
]
𝐾ℎ =
ℎ1−3 𝑙1−3 𝑡
6
[
2 0 1
0 0 0
1 0 2
]
Stiffness Matrix When Subjected To Conduction & Convection:
[ 𝐾] = [ 𝐾𝑐] + [ 𝐾ℎ]

[ 𝐾] = [ 𝐵] 𝑇 [ 𝐷] [ 𝐵] 𝐴 𝑡 +
ℎ1−2 𝑙1−2 𝑡
6
[
2 1 0
1 2 0
0 0 0
]
[ 𝐾] = [ 𝐾𝑐] + [ 𝐾ℎ]
[ 𝐾] = [ 𝐵] 𝑇 [ 𝐷] [ 𝐵] 𝐴 𝑡 +
ℎ2−3 𝑙2−3 𝑡
6
[
0 0 0
0 2 1
0 1 2
]
[ 𝐾] = [ 𝐾𝑐] + [ 𝐾ℎ]
[ 𝐾] = [ 𝐵] 𝑇 [ 𝐷] [ 𝐵] 𝐴 𝑡 +
ℎ1−3 𝑙1−3 𝑡
6
[
2 0 1
0 0 0
1 0 2
]
Where
𝑙1−2 = √(𝑥2 − 𝑥1)2 + (𝑦2 − 𝑦1)2
𝑙2−3 = √(𝑥3 − 𝑥2)2 + (𝑦3 − 𝑦2)2
𝑙1−3 = √(𝑥3 − 𝑥1)2 + (𝑦3 − 𝑦1)2
[𝐷] Material Property matrix
l Length of the side subjected to convection - m
K
Force Vector For Convection:
𝐹 =
ℎ1−2 𝑇∞ 𝑙1−2 𝑡
2
[
1
1
0
]
𝐹 =
ℎ2−3 𝑇∞ 𝑙2−3 𝑡
2
[
0
1
1
]

𝐹 =
ℎ1−3 𝑇∞ 𝑙1−3 𝑡
2
[
1
0
1
]
l Length of the side subjected to convection - m
K
Force Vector If Heat Source Is Given:
𝐹 =
𝑄𝑉
3
[
1
1
1
]
𝑉 = 𝐴 ∗ 𝑡
Q Heat Source - W/m3
V Volume of triangular element - m3
A Area of triangular element - m2
Force Vector If Heat Flux Is Given:
If side 1-2 is subjected to heat flux
𝐹 =
𝑞𝑙1−2 𝑡
2
[
1
1
0
]
If side 2- 3 is subjected to heat flux
𝐹 =
𝑞𝑙2−3 𝑡
2
[
0
1
1
]
If side 1-3 is subjected to heat flux
𝐹 =
𝑞𝑙1−3 𝑡
2
[
1
0
1
]

TORSION:
Shape Function:
𝑁1 =
𝑝1 + 𝑞1 𝑥 + 𝑟1 𝑦
2𝐴
𝑁2 =
𝑝2 + 𝑞2 𝑥 + 𝑟2 𝑦
2𝐴
𝑁3 =
𝑝3 + 𝑞3 𝑥 + 𝑟3 𝑦
2𝐴
𝑝1 = 𝑥2 𝑦3 − 𝑥3 𝑦2
𝑝2 = 𝑥3 𝑦1 − 𝑥1 𝑦3
𝑝3 = 𝑥1 𝑦2 − 𝑥2 𝑦1
𝑞1 = 𝑦2 − 𝑦3
𝑞2 = 𝑦3 − 𝑦1
𝑞3 = 𝑦1 − 𝑦2
𝑟1 = 𝑥3 − 𝑥2
𝑟2 = 𝑥1 − 𝑥3
𝑟3 = 𝑥2 − 𝑥1
𝜑(𝑥, 𝑦) = 𝜑1 𝑁1 + 𝜑2 𝑁2 + 𝜑3 𝑁3
[ 𝐵] =
1
2𝐴
[
𝑞1 𝑞2 𝑞3
𝑟1 𝑟2 𝑟3
]
𝑞1 = 𝑦2 − 𝑦3
𝑞2 = 𝑦3 − 𝑦1
𝑞3 = 𝑦1 − 𝑦2
𝑟1 = 𝑥3 − 𝑥2

𝑟2 = 𝑥1 − 𝑥3
𝑟3 = 𝑥2 − 𝑥1
Area:
𝐴 =
1
2
|
1 𝑥1 𝑦1
1 𝑥2 𝑦2
1 𝑥3 𝑦3
|
Stiffness Matrix When Subjected To Torsion:
[ 𝐾] = [ 𝐵] 𝑇[ 𝐵] 𝐴
We know that
[ 𝐾] [ 𝜑] = [ 𝐹]
[𝜑] Shear Function
[𝐹] Force Vector
Force Vector in Torsion:
𝐹𝑇 =
2𝐺𝜃𝐴
3
[
1
1
1
]
𝜃 Angle of Twist - ◦
G Modulus of Rigidity - N / mm2
Total Torque:
𝑇 =
2𝐴
3
( 𝜑1 + 𝜑2 + 𝜑3)
𝜑 Shear Function
For an example if an element is taken as shown in Fig

Then
Total Torque T = (T
+ T
) * 4 (Multiply by 4 since four quadrant)
Shear Stress:
𝜏 𝑥𝑧 =
𝜕𝜑
𝜕𝑦
𝜏 𝑦𝑧 = −
𝜕𝜑
𝜕𝑥
Where
𝜑 = 𝜑1 𝑁1 + 𝜑2 𝑁2 + 𝜑3 𝑁3

CONSTANT STRAIN TRIANGULAR ELEMENT:
Shape Function:
𝑁1 =
𝑝1 + 𝑞1 𝑥 + 𝑟1 𝑦
2𝐴
𝑁2 =
𝑝2 + 𝑞2 𝑥 + 𝑟2 𝑦
2𝐴
𝑁3 =
𝑝3 + 𝑞3 𝑥 + 𝑟3 𝑦
2𝐴
𝑝1 = 𝑥2 𝑦3 − 𝑥3 𝑦2
𝑝2 = 𝑥3 𝑦1 − 𝑥1 𝑦3
𝑝3 = 𝑥1 𝑦2 − 𝑥2 𝑦1
𝑞1 = 𝑦2 − 𝑦3
𝑞2 = 𝑦3 − 𝑦1
𝑞3 = 𝑦1 − 𝑦2
𝑟1 = 𝑥3 − 𝑥2
𝑟2 = 𝑥1 − 𝑥3
𝑟3 = 𝑥2 − 𝑥1
Area:
𝐴 =
1
2
|
1 𝑥1 𝑦1
1 𝑥2 𝑦2
1 𝑥3 𝑦3
|
Shape Function:
𝑁1 + 𝑁2 + 𝑁3 = 1
UNIT IV – TWO DIMENSIONAL VECTOR VARIABLE
PROBLEMS

When all the shape function is given
𝑥 = 𝑁1 𝑥1 + 𝑁2 𝑥2 + 𝑁3 𝑥3
𝑦 = 𝑁1 𝑦1 + 𝑁2 𝑦2 + 𝑁3 𝑦3
When two or one shape function is given
𝑥 = ( 𝑥1 − 𝑥3) 𝑁1 + ( 𝑥2 − 𝑥3) 𝑁2 + 𝑥3
𝑦 = ( 𝑦1 − 𝑦3) 𝑁1 + ( 𝑦2 − 𝑦3) 𝑁2 + 𝑦3
[ 𝐵] =
1
2𝐴
[
𝑞1 0 𝑞2 0 𝑞3 0
0 𝑟1 0 𝑟2 0 𝑟3
𝑟1 𝑞1 𝑟2 𝑞2 𝑟3 𝑞3
]
A Area of the triangular element - mm2
𝑞1 = 𝑦2 − 𝑦3
𝑞2 = 𝑦3 − 𝑦1
𝑞3 = 𝑦1 − 𝑦2
𝑟1 = 𝑥3 − 𝑥2
𝑟2 = 𝑥1 − 𝑥3
𝑟3 = 𝑥2 − 𝑥1
Stress Strain Relationship Matrix:
For Plane Stress condition:
[ 𝐷] =
𝐸
1 − 𝛾2
[
1 𝛾 0
𝛾 1 0
0 0
1 − 𝛾
2
]
For Plane Strain condition:
[ 𝐷] =
𝐸
(1 + 𝛾)(1 − 2𝛾)
[
(1 − 𝛾) 𝛾 0
𝛾 (1 − 𝛾) 0
0 0
(1 − 2𝛾)
2
]
γ Poisson’s ratio

Stiffness Matrix:
[ 𝐾] = [ 𝐵] 𝑇 [ 𝐷] [ 𝐵] 𝐴 𝑡
[𝐷] Stress – Strain Relationship matrix
Force Vector:
When self – weight is considered or density of material is given
𝐹 =
𝜌𝐴𝑡
3
[
1
1
1
]
𝜌 Density - Kg/m3
𝐴 Area - m2
𝑡 Thickness - m
To Convert Traction Force Into Nodal Force:
𝐹 =
1
2
𝑇 𝐴
𝐹 =
1
2
𝑇 (𝑏 ∗ 𝑡)
𝐹 Nodal force - N
𝑇 Surface traction - N / mm2
t Thickness of the element - mm

b Breadth of the element - mm
To Convert Pressure Into Nodal Force:
If side j - k is subjected to pressure in x direction (node 2 & 3)
{ 𝐹} =
{
0
0
𝑃𝐿𝑡
2
0
𝑃𝐿𝑡
2
0 }
If side j - k is subjected to pressure in y direction (node 2 & 3)
{ 𝐹} =
{
0
0
0
𝑃𝐿𝑡
2
0
𝑃𝐿𝑡
2 }
If side i - j is subjected to pressure in x direction (node 1 & 2)

{ 𝐹} =
{
𝑃𝐿𝑡
2
0
𝑃𝐿𝑡
2
0
0
0 }
If side i - j is subjected to pressure in y direction (node 1 & 2)
{ 𝐹} =
{
0
𝑃𝐿𝑡
2
0
𝑃𝐿𝑡
2
0
0 }
If side i - k is subjected to pressure in x direction (node 1 & 3)
{ 𝐹} =
{
𝑃𝐿𝑡
2
0
0
0
𝑃𝐿𝑡
2
0 }
If side i - k is subjected to pressure in y direction (node 1 & 3)
{ 𝐹} =
{
0
𝑃𝐿𝑡
2
0
0
0
𝑃𝐿𝑡
2 }
𝐹 Force vector - N
𝑃 Pressure - N / mm2

L Length of the element - mm
Element Stress:
{ 𝜎} = [ 𝐷] [ 𝐵] { 𝑢}
{
𝜎𝑥
𝜎 𝑦
𝜏 𝑥𝑦
} = [ 𝐷] [ 𝐵]
{
𝑢1
𝑣1
𝑢2
𝑣2
𝑢3
𝑣3}
𝜎𝑥 Normal stress in x direction - N/mm2
𝜎 𝑦 Normal stress in y direction - N/mm2
𝜏 𝑥𝑦 Shear Stress - N/mm2
u & v Nodal displacement - mm
Maximum Normal Stress:
𝜎 𝑚𝑎𝑥 = 𝜎1 =
𝜎𝑥 + 𝜎 𝑦
2
+ √(
𝜎𝑥 − 𝜎 𝑦
2
)
2
+ 𝜏2
𝑥𝑦
Minimum Normal Stress:
𝜎 𝑚𝑎𝑥 = 𝜎1 =
𝜎𝑥 + 𝜎 𝑦
2
− √(
2
)
2
+ 𝜏2
𝑥𝑦
Principle Angle:
tan 2𝜃 𝑝 =
2𝜏 𝑥𝑦
Element Strain:
{ 𝑒} = [ 𝐵] { 𝑢}

{
𝑒 𝑥
𝑒 𝑦
𝛾𝑥𝑦
} = [ 𝐵]
{
𝑢1
𝑣1
𝑢2
𝑣2
𝑢3
𝑣3}
𝜎𝑥 Normal strain in x direction - N/mm2
𝜎 𝑦 Normal strain in y direction - N/mm2
𝜏 𝑥𝑦 Shear Stress - N/mm2
u & v Nodal displacement - mm
TEMPERATURE EFFECTS:
Initial Strain:
{ 𝑒0} = {
𝛼∆𝑇
𝛼∆𝑇
0
}
For Plane Strain condition:
{ 𝑒0} = (1 + 𝛾) {
𝛼∆𝑇
𝛼∆𝑇
0
}
∆T Temperature difference ˚C
Element Temperature Force:
{𝐹} = [ 𝐵] 𝑇 [ 𝐷] { 𝑒0} 𝐴 𝑡
{𝑒0} Initial strain

AXIS SYMMETRIC ELEMENTS:
Shape Function:
𝑁1 =
𝛼1 + 𝛽1 𝑥 + 𝛾1 𝑦
2𝐴
𝑁2 =
𝛼2 + 𝛽2 𝑥 + 𝛾2 𝑦
2𝐴
𝑁3 =
𝛼3 + 𝛽3 𝑥 + 𝛾3 𝑦
2𝐴
𝑟 =
𝑟1 + 𝑟2 + 𝑟3
3
𝑧 =
𝑧1 + 𝑧2 + 𝑧3
3
𝛼1 = 𝑟2 𝑧3 − 𝑟3 𝑧2
𝛼2 = 𝑟3 𝑧1 − 𝑟1 𝑧3
𝛼3 = 𝑟1 𝑧2 − 𝑟2 𝑧1
𝛽1 = 𝑧2 − 𝑧3
𝛽2 = 𝑧3 − 𝑧1
𝛽3 = 𝑧1 − 𝑧2
𝛾1 = 𝑟3 − 𝑟2
𝛾2 = 𝑟1 − 𝑟3
𝛾3 = 𝑟2 − 𝑟1
Area:
𝐴 =
1
2
|
1 𝑟1 𝑧1
1 𝑟2 𝑧2
1 𝑟3 𝑧3
|

[ 𝐵] =
1
2𝐴
[
𝛽1 0 𝛽2 0 𝛽3 0
𝛼1
𝑟
+ 𝛽1 +
𝛾1 𝑧
𝑟
0
𝛼2
𝑟
+ 𝛽2 +
𝛾2 𝑧
𝑟
0
𝛼3
𝑟
+ 𝛽3 +
𝛾3 𝑧
𝑟
0
0 𝛾1 0 𝛾2 0 𝛾3
𝛾1 𝛽1 𝛾2 𝛽2 𝛾3 𝛽3]
A Area of the triangular element - mm2
𝛼1 = 𝑟2 𝑧3 − 𝑟3 𝑧2
𝛼2 = 𝑟3 𝑧1 − 𝑟1 𝑧3
𝛼3 = 𝑟1 𝑧2 − 𝑟2 𝑧1
𝛽1 = 𝑧2 − 𝑧3
𝛽2 = 𝑧3 − 𝑧1
𝛽3 = 𝑧1 − 𝑧2
𝛾1 = 𝑟3 − 𝑟2
𝛾2 = 𝑟1 − 𝑟3
𝛾3 = 𝑟2 − 𝑟1
[ 𝐷] =
𝐸
(1 + 𝛾)(1 − 2𝛾)
[
1 − 𝛾 𝛾 𝛾 0
𝛾 1 − 𝛾 𝛾 0
𝛾 𝛾 1 − 𝛾 0
0 0 0
1 − 2𝛾
2 ]
Stiffness Matrix:
[ 𝐾] = 2𝜋𝑟𝐴 [ 𝐵] 𝑇 [ 𝐷] [ 𝐵]
𝑟 =
𝑟1 + 𝑟2 + 𝑟3
3

r Radius of the triangular element - mm
Element Stress:
{ 𝜎} = [ 𝐷] [ 𝐵] { 𝑢}
{
𝜎𝑟
𝜎 𝜃
𝜎𝑧
𝜏 𝑟𝑧
} = [ 𝐷] [ 𝐵]
{
𝑢1
𝑤1
𝑢2
𝑤2
𝑢3
𝑤3}
𝜎𝑟 Radial stress - N/mm2
𝜎 𝜃 Circumferential stress - N/mm2
𝜎𝑧 Longitudinal stress - N/mm2
𝜏 𝑟𝑧 Shear Stress - N/mm2
u & w Nodal displacement - mm
TEMPERATURE EFFECTS:
Thermal Force Vector:
[ 𝐹] 𝑡 = 2𝜋𝑟𝐴 [ 𝐵] 𝑇 [ 𝐷] { 𝑒0}
𝑟 =
𝑟1 + 𝑟2 + 𝑟3
3
{ 𝑒0} Strain matrix
r Radius of the triangular element - mm
Strain:

{ 𝑒0} = {
𝛼∆𝑇
𝛼∆𝑇
0
𝛼∆𝑇
}
∆T Temperature difference ˚C

ISOPARAMETRIC ELEMENT:
FOUR NODED QUADRILATERAL:
Shape Function:
𝑁1 =
1
4
(1 − 𝜀) (1 − 𝜂)
𝑁2 =
1
4
(1 + 𝜀) (1 − 𝜂)
𝑁3 =
1
4
(1 + 𝜀) (1 + 𝜂)
𝑁4 =
1
4
(1 − 𝜀) (1 + 𝜂)
𝑥 = 𝑁1 𝑥1 + 𝑁2 𝑥2 + 𝑁3 𝑥3 + 𝑁4 𝑥4
𝑦 = 𝑁1 𝑦1 + 𝑁2 𝑦2 + 𝑁3 𝑦3 + 𝑁4 𝑦4
Jacobian Matrix:
[ 𝐽] =
[
𝜕𝑥
𝜕𝜀
𝜕𝑦
𝜕𝜀
𝜕𝑥
𝜕𝜂
𝜕𝑦
𝜕𝜂]
[ 𝐽] = [
𝐽11 𝐽12
𝐽21 𝐽22
]
UNIT V – ISOPARAMETRIC FORMULATION

𝐽11 =
1
4
[− (1 − 𝜂) 𝑥1 + (1 − 𝜂) 𝑥2 + (1 + 𝜂) 𝑥3 − (1 + 𝜂) 𝑥4 ]
𝐽12 =
1
4
[− (1 − 𝜂) 𝑦1 + (1 − 𝜂) 𝑦2 + (1 + 𝜂) 𝑦3 − (1 + 𝜂) 𝑦4 ]
𝐽21 =
1
4
[− (1 − 𝜀) 𝑥1 − (1 + 𝜀) 𝑥2 + (1 + 𝜀) 𝑥3 + (1 − 𝜀) 𝑥4 ]
𝐽21 =
1
4
[− (1 − 𝜀) 𝑦1 − (1 + 𝜀) 𝑦2 + (1 + 𝜀) 𝑦3 + (1 − 𝜀) 𝑦4 ]
[ 𝐵] =
1
| 𝐽|
[
𝐽22 −𝐽12 0 0
0 0 −𝐽21 𝐽11
−𝐽21 𝐽11 𝐽22 −𝐽12
] ∗
1
4
[
−(1 − 𝜂) 0 (1 − 𝜂) 0 (1 + 𝜂) 0 −(1 + 𝜂) 0
−(1 − 𝜀) 0 −(1 + 𝜀) 0 (1 + 𝜀) 0 (1 − 𝜀) 0
0 −(1 − 𝜂) 0 (1 − 𝜂) 0 (1 + 𝜂) 0 −(1 + 𝜂)
0 −(1 − 𝜀) 0 −(1 + 𝜀) 0 (1 + 𝜀) 0 (1 − 𝜀) ]

[ 𝐷] =
𝐸
1 − 𝛾2
[
1 𝛾 0
𝛾 1 0
0 0
1 − 𝛾
2
]
[ 𝐷] =
1
(1 + 𝛾)(1 − 2𝛾)
[
(1 − 𝛾) 𝛾 0
𝛾 (1 − 𝛾) 0
0 0
(1 − 2𝛾)
2
]
Stiffness Matrix:
For Isoparametric Quadrilateral Element:
[ 𝐾] = 𝑡 ∬[ 𝐵] 𝑇 [ 𝐷] [ 𝐵] 𝜕𝑥 𝜕𝑦
For Natural Co – Ordinates
[𝐾] = 𝑡 ∫ ∫ [ 𝐵] 𝑇 [ 𝐷] [ 𝐵] ∗ | 𝐽| ∗ 𝜕𝜀 ∗ 𝜕𝜂
1
−1
1
−1
|𝐽| Determinant of the Jacobian matrix
ε,η Natural Co – Ordinates
SIX NODED QUADRILATERAL:
Shape Function:
𝑁1 = 2 (𝐿1 −
1
2
) 𝐿1

𝑁2 = 2 (𝐿2 −
1
2
) 𝐿2
𝑁3 = 2 (𝐿3 −
1
2
) 𝐿3
𝑁4 = 4 𝐿1 𝐿2
𝑁5 = 4 𝐿2 𝐿3
𝑁6 = 4 𝐿1 𝐿3
EIGHT NODED QUADRILATERAL:
Shape Function:
𝑁1 = −
1
4
(1 − 𝜀) (1 − 𝜂) (1 + 𝜀 + 𝜂 )
𝑁2 = −
1
4
(1 + 𝜀) (1 − 𝜂) (1 − 𝜀 + 𝜂 )
𝑁3 =
1
4
(1 + 𝜀) (1 + 𝜂) (1 − 𝜀 − 𝜂 )
𝑁4 =
1
4
(1 − 𝜀) (1 + 𝜂)(1 + 𝜀 − 𝜂 )
𝑁5 =
1
2
(1 − 𝜀2) (1 − 𝜂)
𝑁6 =
1
2
(1 + 𝜀) (1 − 𝜂2)
𝑁7 =
1
2
(1 − 𝜀2) (1 + 𝜂)
𝑁8 =
1
2
(1 − 𝜀) (1 − 𝜂2)

NINE NODED QUADRILATERAL:
Shape Function:
𝑁1 =
1
4
( 𝜀 − 𝜀2) ( 𝜂 − 𝜂2)
𝑁2 = −
1
2
(1 − 𝜀2) ( 𝜂 − 𝜂2)
𝑁3 = −
1
4
( 𝜀 + 𝜀2) ( 𝜂 − 𝜂2)
𝑁4 = −
1
2
( 𝜀 − 𝜀2) (1 − 𝜂2)
𝑁5 = (1 − 𝜀2) (1 − 𝜂2)
𝑁6 =
1
2
( 𝜀 + 𝜀2) (1 − 𝜂2)
𝑁7 = −
1
4
( 𝜀 − 𝜀2) ( 𝜂 + 𝜂2)
𝑁8 =
1
2
(1 − 𝜀2) ( 𝜂 + 𝜂2)
𝑁9 = −
1
4
( 𝜀 + 𝜀2) ( 𝜂 + 𝜂2)
GAUSSIAN QUADRATURE:
∫ 𝑓( 𝑥) 𝑑𝑥 = ∑ 𝑤𝑖 𝑓(𝑥𝑖)
𝑛
𝑖=1
1
−1
∫ 𝑓( 𝑥) 𝑑𝑥 = 𝑤1 𝑓(𝑥1)
1
−1
+ 𝑤2 𝑓( 𝑥2) + 𝑤2 𝑓( 𝑥2) + 𝑤3 𝑓( 𝑥3) + ⋯ + 𝑤 𝑛 𝑓(𝑥 𝑛)
𝑤𝑖 Weight function
𝑓(𝑥𝑖) Values of the function at pre – determined sampling points

Number
of points
n
Location
xi
Weights
wi
1 𝑥1 = 0.000 𝑤1 = 2.000
2
𝑥1, 𝑥2 = ± √
1
3
= ±0.577350269189
𝑤1, 𝑤2 = 1.000000
3
𝑥1, 𝑥3 = ± √
3
5
= ±0.774596669241
𝑥2 = 0.0000
𝑤1, 𝑤3 =
5
9
= 0.55555 …
𝑤2 =
8
9
= 0.88888 …
4
𝑥1, 𝑥4 = ±0.8611363116
𝑥2, 𝑥3 = ±0.3399810436
𝑤1, 𝑤4 = 0.3478548451
𝑤2, 𝑤3 = 0.6527451549

ME6603 - FINITE ELEMENT ANALYSIS FORMULA BOOK

More Related Content

What's hot (20)

Similar to ME6603 - FINITE ELEMENT ANALYSIS FORMULA BOOK (20)

More from ASHOK KUMAR RAJENDRAN (20)

Recently uploaded (20)

ME6603 - FINITE ELEMENT ANALYSIS FORMULA BOOK