Manual FE method - Triangular

Name: Abdulaziz Aldawish
ID : 1874641

1. Background:.................................................................................................................4
2. Manual FE method – Triangular....................................................................................5
2.1 The Stiffness Matrix [K1] for element (1):.......................................................................6
2.2 The Stiffness Matrix [K2] for element 2..........................................................................9
2.3 The Stiffness Matrix [K3] for element 3........................................................................12
2.4 The Stiffness Matrix [K4] for element 4........................................................................15
2.6 The solution of the displacements...........................................................................................20
2.7 Calculation of the element stresses:........................................................................................22
2.8 Strength Assessment:..............................................................................................................29
3. Software Triangular Solutions:....................................................................................31
3.1 (4) Elements:...........................................................................................................................31
3.1.1. MATLAB.................................................................................................................................................31
3.1.2 LUSAS:..................................................................................................................................32
3.2 (16) Elements:.........................................................................................................................34
3.2.1. MATLAB :...............................................................................................................................................34
3.2.2. LUSAS:....................................................................................................................................................36
3.3 (32) Elements:.........................................................................................................................38
3.3.1. LUSAS:................................................................................................................................................38
3.4 (64) Elements:.........................................................................................................................39
3.4.1. LUSAS:....................................................................................................................................................39
3.5 (288) Elements:.......................................................................................................................40
3.5.1. LUSAS:....................................................................................................................................................40
4. Manual FE Method Quadrilateral...................................................................................41
The total stiffness matrix for the whole system [K] total...............................................................48
The Stresses at the Gauss Points..........................................................................................52
(a) Strain and stress in element 1...............................................................................................52
(b) Strain and stress in element 2...............................................................................................54
5.Quadrlateral software Method:.......................................................................................57
5.1 2 rectangular elements:...........................................................................................................57
5.1.1 MATLAB..................................................................................................................................................57
5.1.2. LUSAS:....................................................................................................................................................58
5.2 4 Elements Rectangular...........................................................................................................59
5.2.1. MATLAB.................................................................................................................................................59
5.2.2. LUSAS:....................................................................................................................................................61
5.3 64 rectangular Elements:.........................................................................................................62
5.3.1. MATLAB.................................................................................................................................................62
5.3.2. LUSAS:....................................................................................................................................................64
6. Quick Analytical check....................................................................................................68
7.Discussion and recommendations:...................................................................................69
8. Exam Question:..............................................................................................................71

References:.........................................................................................................................71
5- Appendix:.......................................................................................................................71
5.1 MATLAB Codes :......................................................................................................................71
5.1.1 Rectangular MATLAB codes :..................................................................................................................71
5.1.2 Triangular Matlab Codes:.......................................................................................................................79
5.2 MATLAB Results:.....................................................................................................................87
Results of 64 rectangular element using MATLAB.........................................................................................88
Results of 288 rectangular elements using MATLAB.....................................................................................95
Result of 16 triangular elements:..................................................................................................................127

1. Background:
A uniform steel plate with a dimension are shown in the figure 1. With
thickness of 0.1 m. To analyze the plate, we consider it as a 2D plane stress
problem. The Young’s modulus of the material is 100 GPa and the Poisson
ratio is 0.3. The applied load is P  2106 N. Find the stress distribution in
the plane of the ‘deep beam’. If the allowable stress of the steel material is
   100 MPa , make the strength assessment.
Figure 1: Plate of steel

2. Manual FE method – Triangular
First of all, Triangular constant strain elements will be used to solve this question. The plate is
divided into four elements as it is shown in figure 2. The procedure of solution is listed below.
Figure 2: triangular elements
 Find the stiffness matrices of the elements 1-4 which are noted as K1, K2 , K3 and
K4.
 KtB DB, where t is the plate thickness,  is the area of the element,
and B is the strain matrix in  Bu. The elasticity matrix D is the same for all
elements. The stiffness matrices for the elements are
K1tB  DB ,  K2 tB  DB 
K3tB  DB,  K4 tB  DB 
 Assemble the four matrices, which will each be of order [66], into a
single Global Matrix. This will be of order [1212] because there are
6 nodes in total each with two displacement components or two force
components.
 Condense the Global Matrix by eliminating the rows and columns
corresponding to the fixed displacement components and the
unknown force components. This gives a [99] matrix.

 The condensed matrix is inverted to give the nine unknown
displacements.
 Those nodal displacements, together with zero displacements of the
fixed nodes, are used to calculate the element stresses using  
DB ufor the four elements.
2.1 The Stiffness Matrix [K1] for element (1):
Element 1 are shown in the figure 3. In this case, assume the vertical force P is assigned at
node 5 to element 1. The node numbering sequence is conducted in an anticlockwise manner.
Figure 3: Element 1
Table 1 represents all the data for the element. The units for the force and the length are ‘N’
and ‘m’ respectively.
Table 1: All the data for element 1
Node x y u v Fu Fv
i(1) 0 0 0 0 Fu1 Fv1
J(5) 1 1 u5 v5 0 -2x106
K(4) 0 1 u4 v4 0 0
The element 1 area is △ = 0.5 m2
The elasticity matrix [K1] is expressed as follows, in units of N/m2,
[ D ] = E
1−v2
[ 1 v 0
v 1 0
0 0 1−v
2 ]
1

[ D ] = 100 x 106
1−0.32
[ 1 0.3 0
0.3 1 0
0 0 1−0.3
2 ]
[ D ] =1.099 x 1011
[ 1 0.3 0
0.3 1 0
0 0 0.35 ]
[ D ]= [109890000000 32970000000 0
32970000000 109890000000 0
0 0 38946000000 ]
The expression for B matrix [B] can be produced by the following function:
[ B1 ] = 1
2 △ [ b1 0 b2 0 b3 0
0 a1 0 a2 0 a3
a1 b1 a2 b2 a3 b3 ]
Where, △ is the area of the triangular element and can be obtained using:
△ = 1 x 1
2 =0.5 m2
a1=(xk – xj)= -1 a2=(xi-xk)= 0 a3=(xj-xi)= 1
b1=(yj -yk)= 0 b2=(yk-yi)= 1 b3=(yi-yj)= -1
Substituting the nodal co-ordinates of element 1 in
[ B1 ]= 1
2∗0.5 [ 0 0 1 0 −1 0
0 −1 0 0 0 1
−1 0 0 1 1 −1 ]
[ B1 ]T=
[0 0 −1
0 −1 0
1 0 0
0 0 1
−1 0 1
0 1 −1
]Therefore
K1t B1T D B1 

¿ 0.1 x 0.5 x 1.099 x 1011
2∗0.5
[0 0 −1
0 −1 0
1 0 0
0 0 1
−1 0 1
0 1 −1 ] [ 1 0.3 0
0.3 1 0
0 0 0.35 ][ 0 0 1
0 −1 0
−1 0 0
0 −1 0
0 0 1
1 1 −1 ]
=5.495 x 109
[0.35 0 0
0 1 −0.3
0 −0.3 1
−0.35 −0.35 0.35
0 0.3 −1
0 −1 0.3
−0.35 0 0
−0.35 0.3 −1
0.35 −1 0.3
0.35 0.35 −0.35
0.35 1.35 −0.65
−0.35 −0.65 1.35 ]0.35 0 0 -0.35 -
0.35
0.35
0 1 -
0.3
0 0.3 -1
=5.495 x 109 0 -0.3 1 0 -1 0.3
-
0.35
0 0 0.35 0.35 -0.35
-
0.35
0.3 -1 0.35 1.35 -0.65
0.35 -1 0.3 -0.35 -
0.65
1.35
Therefore, the stiffness matrix is symmetrical. The relation between the nodal displacements
and nodal forces is given by the matrix equation K1{u} = {F}. Substituting the values for
element 1 and get
[ 0.35 0 0
0 1 −0.3
0 −0.3 1
−0.35 −0.35 0.35
0 0.3 −1
0 −1 0.3
−0.35 0 0
−0.35 0.3 −1
0.35 −1 0.3
0.35 0.35 −0.35
0.35 1.35 −0.65
−0.35 −0.65 1.35
] {
u1
v1
u5
v5
u4
v4
}
¿ 1
5.495 x 109
{ Fu1
Fv 1
Fu5
Fv 5
Fu 4
F v 4
}= 1
5.495 x 109
{ Fu 1
Fv1
0
−2 x 106
0
0
}