Finite Element Method (FEM)

Finite Element Method (FEM) 1
FINITE ELEMENT METHOD (FEM)
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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
k 1=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
Q6 ]=
[−1000
+1000 ]
= [ 64 x 105 Q 4 +16 x 105 Q 6
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 1q 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 Q6
=( 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 1 lepi
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 ]
B 1=
[ −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 formula2 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:
k 1=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