Finite Element Method (FEM)

This article discusses the Finite Element Method (FEM) and provides solutions to related questions. It covers topics such as slope calculation, vertical deflection, stress-displacement matrix, and principal stresses in a thick cylinder. The article also includes tables, equations, and matrices to aid in the calculations.

Finite Element Method (FEM) 1
FINITE ELEMENT METHOD (FEM)
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Course
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Finite Element Method (FEM) 2
Finite Element Method (FEM)
Question 1
Slope at 2 and 3
The three nodes form two elements that are considered in this question. The constraint of the
beam makes the following displacements to be zero: Q1 = Q2 = Q3 = Q5 = 0. However, Q4 and Q6
have to be determined. Since the sections and length of the two parts of the beam are equal, the
equation used to calculate the element matrices is:
k e= EI
l e3 [ 12 6≤¿−12 6≤¿ 6≤¿ 4 l e2 −6≤¿ 2l e2
−12 −6≤¿ 12 −6≤¿ 6≤¿ 2 l e2 −6≤¿ 4 l e2 ]
E = 200 GPa and I = 4 x 106 mm4
EI
l3 = ( 200 x 109 ) Pa∨N /m2 x ( 4 x 10−6 ) m4
13 m3 =8 x 105 N /m
Effective length, le = 1
k1 =k2=8 x 105
[ 12 6 −12 6
6 4 −6 2
−12 −6 12 −6
6 2 −6 4 ]
The global loads that are applied on the beam are calculated as follows:
F4 = −p l2
12 = 12 KNm x 12 m2
12 =−1,000 Nm
F6 = + p l2
12 = 12 KNm x 12 m2
12 =+1,000 Nm
Elimination method is used together with connectivity to give global stiffness as follows:

Finite Element Method (FEM) 3
K= [k 44
(1)+ k22
(2 ) k24
(2)
k42
(2) k 44
(2) ]=8 x 105
[8 2
2 4 ]
The matrix equation for the slopes at 2 and 3 are:
= 8 x 105
[8 2
2 4 ][Q 4
Q6 ]= [−1000
+1000 ]=
[64 x 105 16 x 105
16 x 105 32 x 105 ] [Q 4
Q 6 ]=[−1000
+1000 ]
= [ 64 x 105 16 x 105
16 x 105 32 x 105 ] [ Q 4
Q 6 ]= [ −1000
+1000 ]
= [64 x 105 Q 4+ 16 x 105 Q6
16 x 105 Q 4+ 32 x 105 Q 6 ]= [−1000
+ 1000 ]
Solving this simultaneously gives
[ Q 4
Q 6 ] = [−2.697 x 10−4
4.464 x 10−4 ]
Hence Q4 = -2.697 x 10-4 and Q6 = 4.464 x 10-4
Vertical deflection at midpoint of the uniformly distributed load
For the second element of the beam, q1 & q3 = 0, q2 = Q4, and q4 = Q6. Vertical deflection at the
midpoint of the second element is determined using ν = Hq at ξ = 0
ν=H 1 q 1+ ¿
2 H 2 q 2+ H 3 q 3+ ¿
2 H 4 q 6
Thus ν=0+ ¿
2 H 2Q 4 +0+ ¿
2 H 4 Q 6
=( 1
2 )( 1
4 ) (−2.679 x 10−4 ) + (1
2 )(−1
4 ) ( 4.464 x 10−4 )

Finite Element Method (FEM) 4
= (-3.348 x 10-5 m) + (-5.58 x 10-5 m) = 8.928 x 10-5 m = -0.08928 mm
Question 2
Elastic modulus, E = 200 GPa; Poisson’s ratio, ν = 0.3
The connectivity and coordinates tables for this question are as follows:
Coordinates
Node r z
1 40 10
2 40 0
3 60 0
4 60 10
Connectivity
Element 1 2 3
1 1 2 4
2 1 3 4
D= E (1−v)
(1+ v )(1−2 v )
[ 1 v
1−v 0 v
1−v
v
1−v 1 0 v
1−v
0 0 1−2 v
2(1−v ) 0
v
1−v
v
1−v 0 1 ]Substituting the values of E = 200 GPa and ν = 0.3 gives the value of D as
D=
[2.69 x 105 1.15 x 105 0 1.15 x 105
1.15 x 105 2.69 x 105 0 1.15 x 105
0 0 7.7 x 104 0
1.15 x 105 1.15 x 105 0 2.69 x 105 ]

Finite Element Method (FEM) 5
The elements are similar hence the det. J and Ae for each element is 200 mm2 and 100 mm2
respectively.
F1 and F3 are equal and are determined as follows:
F 1=F 3= 2 πr 1lepi
2 =2 π x 40 x 10 x 2
2 =2,514.3 N
The B matrices for element 1 and 2 are obtained as follows:
Element 1:
r =1
3 ( 40+ 40+60 ) =46.67 mm
B=
[ z 23
det J 0 z 31
det J 0 z 12
det J 0
0 r 32
det J 0 r 13
det J 0 r 21
det J
r 32
det J
z 23
det J
r 13
det J
z 31
det J
r 21
det J
z 12
det J
N 1
r 0 N 2
r 0 N 3
r 0 ]
B1=
[ −0.05 0 0 0 0.05 0
0 0.1 0 −0.1 0 0
0.1 −0.05 −0.1 0 0 0.05
0.0071 0 0.0071 0 0.0071 0 ]
Element 2:
r =1
3 ( 40+60+ 60 )=53.33 mm

Finite Element Method (FEM) 6
B 2=
[ −0.05 0 0.05 0 0 0
0 0 0 −0.1 0 0.1
0 −0.05 −0.1 0.05 0.1 0
0.00625 0 0.00625 0 0.00625 0 ]
To get stress-displacement matrix for each element, D is multiplied by B as follows:
For element 1:
D B 1=104
[ −1.26 1.15 0.082 −1.15 1.43 0
−0.49 2.69 0.082 −2.69 0.657 0.1
0.77 −0.385 −0.77 0 0 0.385
0.384 1.15 0.191 −1.15 0.766 0 ]
For element 2:
D B 2=104
[ −1.27 0 1.42 −1.15 0.072 1.15
−0.503 0 0.647 −2.69 0.072 2.69
0 −0.385 −0.77 0.385 0.77 0
0.407 0 0.743 −1.15 0.168 1.15 ]
Each element’s stiffness matrix is determined by using the formula 2 π r Ae BT D B. The
symmetric stiffness matrices of the two elements are as determined below, with the first row of
each matrix being the global degree of freedom (DOF).
Stiffness matrix for element 1:
k1 =1 x 107
[ 1 2 3 4 7 8
4.03 −2.58 −2.34 1.45 −1.932 1.13
8.45 1.37 −7.89 1.93 ¿ ¿
−0.24¿0.16 ¿−1.13¿7.89¿−1.93¿0¿ 2.25¿0¿0.565¿ ]
Stiffness matrix for element 2:
k 2=1 x 107
[ 3 4 5 6 7 8
2.05 0 −2.22 1.69 −0.085 −1.69
0.645 1.29 −0.645 −1.29 ¿ ¿
−3.46¿−2.42¿2.17¿9.66¿1.05¿−9.01¿2.62¿0.241

Finite Element Method (FEM) 7
Applying elimination method and assembling each element’s stiffness matrix to the DOF 1 and 3
gives
=1 x 107
[ 4.03 −2.34
−2.34 4.35 ][Q 1
Q3 ]= [2514
2514 ]; solving this equation simultaneously gives
Q1 = 1.4 x 10-4 mm, and Q3 = 1.33 x 10-4 mm
Question 3
From the dimensions given, the cylinder is a thick cylinder because Di
t < 20i . e .= 68
16 =4.15<20.
Therefore it is a thick cylinder subjected to internal pressure only. The principal stresses
distributed in the cylinder are hoop stress (σθ), radial stress (σr) and longitudinal stress (σL). Both
σθ and σr varies with radius but σL is constant/uniform across the cylinder’s thickness. These
stresses are calculated as follows:
Hook stress, σθ= PiR i2
( R o2−Ri2 ) ( r2 + R o2
r2 )
At inside diameter: σθ= 1 x 342
(502−342 ) ( 342 +502
342 )=0.86012 (3.1626 )=2.7202 MPa
At outside diameter, σθ= 1 x 342
(502−342 ) ( 502 +502
502 )=0.86012 ( 2 ) =1.7202 MPa
Radial stress, σr = PiR i2
( R o2−R i2 ) ( r2−R o2
r2 )
At inside diameter, σr = 1 x 342
(502−342) ( 342−502
342 )=0.86012 ( −1.16263 ) =−1 MPa

Finite Element Method (FEM) 8
At outside diameter, σr = 1 x 342
(502−342) ( 502−502
502 )=0.86012 ( 0 )=0 MPa
Longitudinal stress, σL= PiR i2
(R o2−R i2)
At inside diameter, σL= 1 x 342
(502−342)=0.86 Mpa
At outside diameter, σL= 1 x 342
(502−342)=0.86 Mpa
Maximum shear stress, τmax = PiR i2 R o2
r2 ( R o2−R i2)
At inside diameter, τmax = 1 x 342 x 502
342 (502−342) = 2890000
1553664 =1.86 MPa
At outside diameter, τmax = 1 x 342 x 502
502(502−342)= 2890000
3360000 =0.86 MPa
Change in diameter:
Change in inside diameter
δDi= 2 r
E ( σθ−vσr−vσl )= 2 x 34
70000 ¿
New inside diameter after deformation = 68mm + 2.68 = 70.68 mm
Change in outside diameter
δDi= 2 r
E ( σθ−vσr−vσl )= 2 x 50
70000 ¿
New inside diameter after deformation = 100mm + 2.089 = 102.089 mm

Finite Element Method (FEM) 9
Question 4
The cylinder in this question is thick because Di
t < 20i . e .= 68
16 =4.15<20
Change in diameter,δD= 2 r
E ( σθ−vσr −vσl ), where E = elastic modulus, r = any radius of the
cylinder, σθ = hoop stress, σr = axial stress, σl = longitudinal stress and ν = Poisson’s ratio.
At inside diameter where r = ri: σθ=Pi [ r o2 +r 12
r o2−r 12 ]; σr = -Pi; and σl= Pir i2
r o2−r 12
σθ=1.5 N /mm2
[ 502+342
502−342 ]=4.08 KN /m m2
σr = -1.5 N/mm2
σl= 1.5 N /m m2 x(342 )
502−342 =1.29 N / m m2
Hence δD= 2 x 34
200000 ( 4.08−0.3 (−1.5)−0.3(1.29) ) =0.00034(4.08+0.45−0.387)
= 0.00141 m
New inside diameter after deformation = 68mm + 1.41mm = 69.41 mm
At outside diameter where r = ro: σθ= [ 2 Pir 12
r o2−r 12 ]; σr = 0i; and σl= Pir i2
r o2−r 12
σθ= [ 2 x 1.5 x 342
502−342 ]=2.58 N / mm2
σl= 1.5 x 342
502−342 =1.29 N /mm2

Finite Element Method (FEM) 10
δD= 2 x 50
200000 ( 2.58−0−0.3(1.29) )=0.0005 ( 2.58−0.387 )=0.001097 m
δD= PiRi
E(R o2−R i2) [ ( 1−2 v ) R i2+ ( 1+ v ) R o2 ]
New outside diameter after deformation = 100mm + 1.0097mm = 101.01 mm
Question 5
The first step is to determine the Jacobian, J equation J=
[ ∂ x
∂ ξ
∂ y
∂ ξ
∂ z
∂ ξ
∂ x
∂ η
∂ y
∂η
∂ z
∂ η
∂ x
∂ ζ
∂ y
∂ζ
∂ z
∂ ζ ]= [ x 14 y 14 z 14
x 24 y 24 z 24
x 34 y 34 z 34 ]
using the nodal coordinates provided below
J= [0 1 1
0 0 1
1 0 1 ]. Det. J = 1
The equation A=J−1= 1
det J [ y 24 z 34− y 34 z 24 y 34 z 14− y 14 z 34 y 14 z 24− y 24 z 14
z 24 x 34−z 34 x 24 z 34 x 14−z 14 x 34 z 14 x 24− z 24 x 14
x 24 y 34−x 34 y 24 x 34 y 14−x 14 y 34 x 14 y 24− x 24 y 14 ] is
used to calculate the inverse of the Jacobian, A, which is found to be
A=
[0 −1 1
1 −1 0
0 1 0 ]
The value of B1 is
[0 0 0
0 1 0
0 0 0
0 0 1
0 0 0
1 0 0
]

Finite Element Method (FEM) 11
The matrix D of the stress-strain relation is
D=1 x 107
[4.038 1.731 1.731 0 0 0
1.731 4.038 1.731 0 0 0
1.731 1.731 4.038 0 0 0
0 0 0 1.154 0 0
0 0 0 0 1.154 0
0 0 0 0 0 1.154
]
Modifying the stiffness matrix gives
K=Ve B1
T D B1=1 x 106
[1.923 0 0
0 6.731 0
0 0 1.923 ], which is a diagonal matrix.
The force vector, F in this question is F = [0 0 -1000]T.
KQ = F is solved as follows:
106
[1.923 0 0
0 6.731 0
0 0 1.923 ] [ Q ]=[0 0 −1000]
Solving this gives the value of Q as Q= [ 0 0 −0.00052 ] T.
Question 6
The wall has a 3-element finite model with the connectivity matrices obtained from the formula
kT = ke
¿ [ 1 −1
−1 1 ]
Hence the connectivity matrices of the three elements are:
kT
(1)= 20
0.3 [ 1 −1
−1 1 ]

Finite Element Method (FEM) 12
kT
(2)= 30
0.15 [ 1 −1
−1 1 ]
kT
(3)= 50
0.15 [ 1 −1
−1 1 ]
The global K is obtained by finding the sum of the three stiffness matrices of the three elements,
as follows:
K=∑ kT =66.7
[ 1 −1 0 0
−1 4 −3 0
0 −3 8 −5
0 0 −5 5 ]
Conversion happens at node 1 and this is accommodated by adding the constant h = 25 W/m2°C
to the (1, 1) location of K to give:
K=66.7
[1.375 −1 0 0
−1 4 −3 0
0 −3 8 −5
0 0 −5 5 ]
There is no heat produced hence the heat rate vector R is only comprised of ht ∞, which is
present in the first row only as follows:
R= [ 25 x 800 , 0 , 0 , 0 ]T
The outer temperature of the wall, which acts as the boundary condition of the question, is To =
20°C. This boundary condition is applied using penalty method. In this case, C is selected on the
basis of:
C = max |Kij| x 104 = 66.7 x 8 x 104
C and CT4 are added to the K’s (4,4) location and R’s 4th row respectively to get:

Finite Element Method (FEM) 13
66.7
[1.375 −1 0 0
−1 4 −3 0
0 −3 8 −5
0 0 −5 80005 ][T 1
T 2
T 3
T 4 ]=
[ 25 x 800
0
0
10672 x 104 ]
Solving the above equation gives: T1 = 304.6 °C, T2 = 119.0 °C, T3 = 57.1 °C and T4 = 20.0 °C.
Question 7
The element matrices and connectivity of the bar are provided in the table below
BT = 1
det J [ y 23 y 31 y 12
x 32 x 13 x 21 ]
The matrix for each element is as follows:
Element 1:
BT
(1)= 1
0.06 [−0.15 0.15 0
0 −0.4 0.4 ]
Element 2:
BT
(2)= 1
0.12 [ −0.15 −0.15 0.3
0.4 −0.4 0 ]
Element 3:
BT
(3)= −1
0.06 [0.15 −0.15 0
0 −0.4 0.4 ]
Element → local 1 2 3
1 ↕ global 1 2 3
2 5 1 3
3 5 4 3

Finite Element Method (FEM) 14
KT is obtained using the formula KT = kAe BT
T BT . The kT matrices for each element are as
follows:
Element 1:
KT = kT
(1)= ( 1.5 ) ( 0.03 ) BT
(1)T BT
(1) =
[ 1 2 3
0.28125 −0.28125 0
−0.28125 2.28125 −2.0
0 −2.0 2.0 ]
Element 2:
KT = kT
(2)= ( 1.5 ) ( 0.03 ) BT
(2)T BT
(2) =
[ 5 1 3
1.14 −0.86 −0.28125
−0.86 1.14 −0.28125
−0.28125 −0.28125 0.5625 ]
Element 3:
KT = kT
(3)= ( 1.5 ) ( 0.03 ) BT
(3)T BT
(3 ) =
[ 5 4 3
0.28125 −0.28125 0
−0.28125 2.28125 −2.0
0 −2.0 2.0 ]
This is followed by developing matrices hT for elements that have convection ends. Both element
1 and element 3 have convection ends as 2-3 hence the formula used to determine hT is:
hT = hl 23
6 [0 0 0
0 2 1
0 1 2 ]
Using the above formula, the matrices for hT for the two elements are as follows:

Finite Element Method (FEM) 15
hT
(1)=
[1 2 3
0 0 0
0 2.5 1.25
0 1.25 2.5 ] and hT
(2)=
[5 4 3
0 0 0
0 2.5 1.25
0 1.25 2.5 ]
The next step is to assemble the global matrixK=∑ ( kr +hr ). This is done using elimination
method and taking into considering the boundary conditions that have been provided in the
question of T = 180°C at nodes 4 & 5. Applying this method means eliminating the rows and
columns but the R vector is modified using the 4th and 5th rows. This gives the matrix K as:
K=
[ 1 2 3
1.42125 −0.28125 −0.28125
−0.28125 4.78125 −0.75
−0.28125 −0.75 9.5625 ]
The element convection contributions are used to assemble the heat=rate vector R using the
formula
r ∞= hT ∞ l 2−3
2 [0 11]
Applying this formula gives the following for the two ends
r∞
(1)= ( 50 ) ( 25 ) (0.15)
2 [ 1 2 3
0 1 1 ]
r∞
(3)= ( 50 ) ( 25 ) (0.15)
2 [ 5 4 3
0 1 1 ]
Hence the global R is given as
R=93.75 [1 2 3
0 1 2 ]T
The temperature distribution in the bar is then determined using the formula KT = R as follows:

Finite Element Method (FEM) 16
KT =R=
[ 1 2 3
1.42125 −0.28125 −0.28125
−0.28125 4.78125 −0.75
−0.28125 −0.75 9.5625 ] [T 1
T 2
T 3 ]=93.75 [1 2 3
0 1 2 ]❑
Solving the above equation simultaneously gives:
T1 = 124.5 °C, T2 = 34.0 °C, and T3 = 45.4 °C.
Question 8
Wall thickness, t = 30cm = 0.3m
Thermal conductivity, K = 0.7W/m°C
Inner surface temperature = 40°C
Outer surface temperature = -10°C
Heat transfer coefficient, h = 50W/m2°C
A two-element model is used
Temperature distribution:
BT
(1)= 1
0.06 [−0.15 0.15 0
0 −0.4 0.4 ]
BT
(2)= 1
0.06 [−0.15 0.15 0
0 −0.4 0.4 ]
KT = kAe BT
T BT
KT = kT
(1)= ( 0.7 ) ( 0.03 ) BT
(1 )T BT
(1) =
[ 1 2 3
0.13125 −0.13125 0
−0.13125 1.0646 −0.9333
0 −0.9333 0.9333 ]

Finite Element Method (FEM) 17
KT = kT
(2)= ( 1.5 ) ( 0.03 ) BT
(2)T BT
(2) =
[ 5 1 3
0.532 −0.4013 −0.13125
−0.4013 0.532 −0.13125
−0.13125 −0.13125 0.2625 ]
hT is determined as follows:
hT = hl 23
6 [0 0 0
0 2 1
0 1 2 ]
Using the above formula, the matrices for hT for the two elements are as follows:
hT
(1)=
[1 2 3
0 0 0
0 1.1667 0.4667
0 0.4667 1.1667 ] and hT
(2)=
[5 4 3
0 0 0
0 1.1667 0.5833
0 0.5833 1.1667 ]
K=
[ 1 2 3
0.66325 −0.13125 −0.13125
−0.13125 2.23125 −0.35
−0.13125 −0.35 4.4625 ]
The assembled heat-rate vector is
r ∞= hT ∞ l 2−3
2 [0 11], which gives
r∞
(1)= ( 50 ) (−10 ) (0.07)
2 [ 1 2 3
0 1 1 ]
r∞
(3)= ( 50 ) (−10 ) (0.07)
2 [5 4 3
0 1 1 ]
The global R becomes

Finite Element Method (FEM) 18
R=−17.5 [1 2 3
0 1 2 ]T
The temperature distribution in the bar is then determined using the formula KT = R as follows:
K T =R=
[ 1 2 3
0.66325 −0.13125 −0.13125
−0.13125 2.23125 −0.35
−0.13125 −0.35 4.4625 ] [T 1
T 2
T 3 ]=−17.5 [1 2 3
0 1 2 ]❑
Solving the above equation simultaneously gives:
T1 = 27.8 °C, T2 = 12.4 °C, and T3 = -2.6 °C.
Heat flux:
Heat flux, q, through the wall is calculated as follows
q= Tsi−Tso
L
K
; Where Tsi = temperature of inner wall surface, Tso = temperature of outer wall
surface, L = wall thickness, and K = thermal conductivity of the wall.
q= 40 °C−(−10)° C
0.3 m
0.7 W /m° C
= 50° C
0.4286 m2 ° C /W =116.66 W /m2
Question 9
The problem in this question is solved using characteristic polynomial approach. The eigenvalue
problem is obtained from the mass and stiffness values of the corresponding DOF Q2 and Q3.
This gives the eigenvalue problem as:

Finite Element Method (FEM) 19
E
[ A 1
L1 + A 2
L2
− A 2
L 2
−A 2
L2
A 2
L 2 ] [ U 2
U 3 ]= λ ρ
6 [ A 1 L 1+ A 2 L 2 A 2 L2
A 2 L 2 2 A 2 L 2 ][ U 2
U 3 ]
Density of the bar is determined as follows: ρ= f
g = 0.283 lb /¿.³
32.2 x 12 =7.324 x 10−4 lb s2 /¿ .³
Substituting these values in the eigenvalue gives
30 x 106
[ 0.2 −0.1
−0.1 0.1 ] =1.22 x 10−4 λ [ 25 2.5
2.5 5 ][ U 2
U 3 ]
The characteristic equation for the problem is given as:
det [ (6 x 106−30.5 x 10−4 λ) (−3 x 106 −3.05 x 10−4 λ)
(−3 x 106−3.05 x 10−4 λ) (3 x 106 −6.1 x 10−4 λ) ]=0
Simplifying this equation gives
1.77 x 10-6λ2 – 1.465 x 104λ + 9 x 1012 = 0
Solving the above quadratic equation using the formula x=−b ± √b2−4 ac
2 a gives the eigenvalues
as:
λ1 = 6.68 x 108 and λ2 = 7.61 x 109
Calculation of the eigenvector for λ1 is done using the equation (K – λ1M)U1 = 0
This gives 106
[ 3.96 −3.204
−3.204 2.592 ][U 2
U 3 ]1 = 0
The determinant of the above matrix is zero hence the solution becomes
3.96U2 = 3.204U3

Finite Element Method (FEM) 20
Therefore U1
T =[U 2 ,1.236 U 2]
To normalize things, the following condition is applied: U1
T MU 1=1
Substituting U1 gives: U1
T =[14.53 , 17.96]
Following the same procedure, the eigenvector for λ2 is: U2
T =[11.57 ,−37.45]
Question 10
In this question, the shaft is of step discontinuity because its cross-section changes abruptly.
The bar is cylindrical hence their diameters are determined as follows:
A1 = 1,200mm2 = πr2 → r2 = 1200mm2/π = 381.972mm2 → r = 19.544mm = 0.01995m
A2 = 900mm2 = πr2 → r2 = 900mm2/π = 286.479mm2 → r = 16.926mm = 0.01693m
J 1= π x d 14
32 = π x 0.019954
32 =1.555 x 10−8 m4
J 2= π x d 14
32 = π x 0.016934
32 =8.065 x 10−9 m4
Assuming that modulus of rigidity of the bar is 0.8 x 1011 N/m.
Kt 1= GJ 1
l 1 = 0.8 x 1011 x 1.555 x 10−8
0.3 =4146.667 Nm/rad
Kt 2= GJ 2
l 2 = 0.8 x 1011 x 8.065 x 10−9
0.4 =1613 Nm /rad
Effective length of the bar with reference to the second shaft is calculated as follows:
¿= l 1
J 1 Je+ l 2
J 2 Je= 0.3
15.55 x 8.065+ 0.4
8.065 x 8.065=0.1556+0.4=0.5556 m

Finite Element Method (FEM) 21
The equivalent stiffness is calculated as follows:
Kte= GJe
¿ =0.8 x 1011 x 8.065 x 10−9
0.5556 =1161.27 Nm/ rad
Hence the natural frequency of the bar is calculated as follows:
Assume that Ipi = 0.015 kg/m2; Ip2 = 0.01 kg/m2
ωnf 2= √ ( Ip 1+Ip 2) Kle
Ip1 Ip 2 = √ ( 0.015+0.01 ) x 1161.27
0.015 x 0.01 =439.937 rad /sec
The position of the node depends on the characteristics of the bar and its stiffness. To determine
node position, relative displacement must be calculated first as follows:
lne1
lne2 = Ip 2
Ip 1 = 0.01
0.015 =0.667; → lne1 = 0.667lne2
But lne1 + lne2 = 0.5556m
Therefore
0.667lne2 + lne2 = 0.5556 → 1.667lne2 = 0.5556; lne2 = 0.3333 m and lne1 = 0.2223 m
The node shape of the bar is as shown below
0.01
0.015
0.2223m
0.3333m

Finite Element Method (FEM) 22