Mechanical Engineering: Cantilever Beam Finite Element Analysis

This project focuses on the finite element analysis (FEA) of a cantilever beam. The solution begins by defining the problem parameters, including the coordinates of the nodes, and calculates the relevant dimensions. The solution then determines the displacement at each node and utilizes the CST element to create a finite element model, calculating the stiffness matrix [K] and the force vector [F]. The analysis evaluates the surface traction along the beam, using shape functions N2 and N3, and performs integration to determine the final solution. The project involves detailed calculations to determine the displacement and stiffness characteristics of the cantilever beam, providing a comprehensive understanding of its behavior under load.

Solution
a) q = 198
(x1, y1) = (1.198, 1.198) mm
(x2, y2) = (5.198, 1.198) mm
(x3, y3) = (5.198, 3.198) mm
(x4, y4) = (1.198, 3.198) mm
2b = 5.198 – 1.198 = 4 mm
b = 2 mm
2h = 3.198 – 1.198 = 2 mm
h = 1 mm
Displacements
(u1, v1) = (0.001, 0,0005) mm
(u2, v2) = (0.0015, 0,0006) mm

(u3, v3) = (0.0012, 0,0008) mm
(u4, v4) = (0.0025, 0,001) mm
[ B ] ¿ 1
4 [−(h− y) 0 (h− y )
0 −(b−x ) 0
0 ( h+ y ) 0
−(b+x ) 0 ( b+ x)
−(h+ y) 0
0 (b−x )
−(b−x) −(h− y) −( b+x)(h− y ) ( b+x) (h+ y)(b−x ) −( h+ y ) ]
¿ 1
4 [ − ( 1−1.198 ) 0 ( 1−1.198 ) 0 ( 1+3.198 ) 0− (1−3.198 ) 0
0− ( 2−1.198 ) 0− ( 2+5.198 ) 0 ( 2+5.198 ) 0 ( 2−1.198 )
− ( 2−0.001 )− ( 1−0.0005 )− ( 2.0015 ) ( 1−0.0006 ) ( 2+0.0012 ) ( 1+0.0008 ) ( 2−0.0025 )−1.001 ]
[B] = 1
4 [ 0.198 0 −0.198
0 −0.802 0
−0.802 0.198 −7.198
0 4.198 0
−7.198 0 7.198
−0.198 7.198 4.198
2.198 0
0 0.802
0.802 2.198 ]
b)
Coordinate point (3.198, 2.198)
Displacement
u ( x , y )= 1
4 ¿
¿ 1
4 ¿ mm
v ( x , y ) = 1
4 ¿
¿ 1
4 ¿ mm

Solution
{fs } = ∫
−1
1
[ Ns ]
T
{ T } h L
2 dt
Length
l2−3= √ ( x2−x3 ) 2+ ( y3− y2 )
2
= √ ( 8−5 )2+ ( 4−0 )2
= 5 mm
Shape function N2 and N3 must be used as we evaluate the surface traction alongside 2-3 (as s = 1)
{fs } = ∫
−1
1
[ Ns ]
T
{ T } h L
2 dt
= ∫
−1
1
[ N2 0 N3
0 N 2 0
0
N3 ] T
{ PS
PT } h∗L
2 dt

Evaluating along, s = 1
N2 = (1+ s)(1−t )
4 = s−t−st +1
4
N3 = (1+s)(1+t)
4 = s +t ∓ t+ 1
4
{ T } = { ps
pt }= [ 20+ 0. q
0 ]
Substituting L and {T} and h = t= 2.0 + 0.q
{fs } = ∫
−1
1
[ Ns ]
T
{ T } h L
2 dt
¿∫
−1
1
[N 2 0
0 N2
N 3 0
0 N3
] {20+0.q
0 }(2.0+ 0. q)∗5
2 dt
Evaluating along, s = 1
Simplifying
{fs } = 2.5(2.0+ 0. q)∫
−1
1
[(20+ 0.q )N2
0
(20+ 0.q) N3
0 ]dt
= 2.5(2.0+ 0. q)(20+0. q)∫
−1
1
[ N2
0
N3
0 ]dt
= 2.5(2.0+ 0. q)(20+0. q)∫
−1
1
[ s−t−st+ 1
4
0
s+t +st +1
4
0
]dt
Evaluating along, s= 1

Substituting for s = 1 into the integrands and performing integration
= 2.5(2.0+ 0. q)(20+0. q)∫
−1
1
[ 2−2 t
4
0
2 t+2
4
0
]dt
{fs} = 2.5(2.0+ 0. q)(20+0. q)
[0.5 t−t2
4
0
0.5 t+ t2
4
0
]−1
1
¿ 2.5 ( 2.0+0. q ) ( 20+0. q )
[0.50−0.25
0
0.50+0.25
0 ]−2.5 ( 2.0+0. q ) ( 20+0.q )
[−0.50−0.25
0
−0.50+0.25
0 ]
=
(2.5 ( 2.0+0. q ) ( 20+0.q )
[0.25
0
0.75
0 ] )−
(2.5 ( 2.0+0. q ) ( 20+0. q )
[−0.75
0
−0.25
0 ] )
But q = 198
¿ 2.5 ( 2.0+0. q ) ( 20+0. q )= 2(20 + 0.198) + 0.198(20 + 0.198)
= 40.396 + 3.999204
= 44.395204
{fs} = 44.395204
[ 1
0
1
0 ] N

Nodal displacement at each node for CST element to create finite element model of a cantilever beam

D = E
1−v2
[1 v 0
v 1 0
0 0 1−v
2 ]
= 1
1−0.32 [ 1 0.3 0
0.3 1 0
0 0 0.35 ]
= 2.308 [ 1 0.3 0
0.3 1 0
0 0 0.35 ]
Element 1
Area of element A1 = 1
2 |det [ J ]|
= 1
2∗2.00198∗0.3=0.300297 m2
Stiffness matrix is given
[K](1) = t1A1[B]T[D][B]1
[F] = [K][Q]
σ =¿ [D][B](2)[d](2)
Check excel for all calculations