FINITE ELEMENT
ANALYSIS
Dr. Richa Agrawal
MODULE 3 - ONE DIMENSIONAL PROBLEMS
Finite Element Method (FEM)
1 1 2 2 3 3 4 4 5
P
L
• Divide the domain in no. of small parts – ‘elements’
• Solve the Governing Differential Equation (GDE) over the
general element and obtain the equation in matrix form –
‘Element Matrix Equation’ (EME)
• Assemble all the equations – Global Matrix Equation
(GME)
• Impose the global boundary conditions and solve to get
the solution
Terms in FEM
1 1 2 2 3 3 4 4 5
P
• Discretization- Process of dividing the domain in no. of
elements
• Element: Any part within the domain – specific length,
geometric and material properties remain consistent.
• Order of element – linear, quadratic, cubic or any polynomial
• Nodes – End point of the element
• Internal nodes – increase with order of element
• Tribology – Process of numbering the nodes
Terms in FEM
1 2 3 4
P
• Global Coordinates – Coordinates with respect to starting point
of domain
• Local Coordinates – Coordinates within the general element for
EME with respect to the starting point of element
• Global Boundary Conditions – Pertain to specific points in
given domain (EBC or NBC)
• Local Boundary Conditions – Holds true at any point of domain
(NBC)
FEM
Preprocessor Solver Postprocessor
• Geometry (Area, • EME • Validation
length, MOI) • GME • Representation
• Material ( E, • GBC
etc.) • Determine
• Discretization unknown PV &
• Applying LBC SV
Rod Subjected to Axial Load
1 1 2 2 3
P
• GDE in Local Coordinate
𝑑 𝑑𝑢
𝐸𝐴 =0 (𝑑𝑥 = 𝑑𝑥)
𝑑𝑥 𝑑𝑥
• Boundary Conditions
𝑑𝑢 𝑑𝑢
𝐸𝐴 = −𝑃𝑒 1 𝐸𝐴 = 𝑃𝑒 2
𝑑𝑥 𝑥=0
𝑑𝑥 𝑥=ℎ𝑒
Rod Subjected to Axial Load
𝑑 𝑑𝑢
𝐸𝐴 =0 (0 < 𝑥 < ℎ𝑒 )
𝑑𝑥 𝑑𝑥
𝑑𝑢 𝑑𝑢
𝐸𝐴 = −𝑃𝑒 1 𝐸𝐴 = 𝑃𝑒 2
𝑑𝑥 𝑥=0
𝑑𝑥 𝑥=ℎ𝑒
ℎ𝑒
𝑤𝑖 𝑅 𝑑𝑥 = 0
0
ℎ𝑒
𝑑 𝑑𝑢
𝑤𝑖 𝐸𝐴 𝑑𝑥 = 0
0 𝑑𝑥 𝑑𝑥
𝑑𝑢 ℎ𝑒 ℎ𝑒 𝑑𝑤𝑖 𝑑𝑢
𝑤𝑖 𝐸𝐴 − 0 𝑑𝑥
𝐸𝐴 𝑑𝑥 = 0
𝑑𝑥 0 𝑑𝑥
𝑑𝑢 𝑑𝑢 ℎ𝑒 𝑑𝑤𝑖 𝑑𝑢
𝑤𝑖 ℎ𝑒 𝐸𝐴 - 𝑤𝑖 0 𝐸𝐴 − 0 𝑑𝑥
𝐸𝐴 𝑑𝑥 = 0
𝑑𝑥 ℎ𝑒 𝑑𝑥 0 𝑑𝑥
Rod Subjected to Axial Load
𝑑 𝑑𝑢 𝑑𝑢 𝑑𝑢
𝐸𝐴 = 0 (0 < 𝑥 < ℎ𝑒 ) 𝐸𝐴 = −𝑃𝑒 1 , 𝐸𝐴 = 𝑃𝑒 2
𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑥=0
𝑑𝑥 𝑥=ℎ𝑒
𝑢 = 𝐶1 ∅1 + 𝐶2 ∅2
𝑥 𝑥
𝑢 = 𝐶1 1 − + 𝐶2
ℎ𝑒 ℎ𝑒
𝑥 𝑥
∅1 = 1 − ℎ an d ∅2 = ℎ𝑒
𝑒
𝑥
i) 𝑤1 = 1 − ℎ
𝑒
𝑑𝑢 𝑑𝑢 ℎ𝑒 𝑑𝑤𝑖 𝑑𝑢
𝑤𝑖 ℎ𝑒 𝐸𝐴 - 𝑤𝑖 0 𝐸𝐴 − 0 𝐸𝐴 𝑑𝑥 =0
𝑑𝑥 ℎ𝑒 𝑑𝑥 0 𝑑𝑥 𝑑𝑥
𝑒 ℎ𝑒 1 1 1
0-(−𝑃 1 )-EA 0 − ℎ −𝐶1 ℎ + 𝐶2 ℎ 𝑑𝑥=0
𝑒 𝑒 𝑒
𝑃𝑒 1 1 1
𝐸𝐴
= 𝐶1 - 𝐶2 ……..(1)
ℎ𝑒 ℎ𝑒
Rod Subjected to Axial Load
𝑑 𝑑𝑢 𝑑𝑢 𝑑𝑢
𝐸𝐴 = 0 (0 < 𝑥 < ℎ𝑒 ) 𝐸𝐴 = −𝑃𝑒 1 , 𝐸𝐴 = 𝑃𝑒 2
𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑥=0
𝑑𝑥 𝑥=ℎ𝑒
𝑥 𝑥
𝑢 = 𝐶1 1− + 𝐶2
ℎ𝑒 ℎ𝑒
𝑥
i) 𝑤2 =
ℎ𝑒
𝑑𝑢 𝑑𝑢 ℎ𝑒 𝑑𝑤𝑖 𝑑𝑢
𝑤𝑖 ℎ𝑒 𝐸𝐴 - 𝑤𝑖 0 𝐸𝐴 − 0 𝑑𝑥
𝐸𝐴 𝑑𝑥 = 0
𝑑𝑥 ℎ𝑒 𝑑𝑥 0 𝑑𝑥
ℎ𝑒 1 1 1
𝑃𝑒 2 - 0-EA 0 ℎ𝑒
−𝐶1 + 𝐶2 𝑑𝑥 =0
ℎ𝑒 ℎ𝑒
𝑃𝑒 2 1 1
𝐸𝐴
= −𝐶1
ℎ𝑒
+𝐶2
ℎ𝑒
…….(2)
Rod Subjected to Axial Load
𝑑 𝑑𝑢 𝑑𝑢 𝑑𝑢
𝐸𝐴 = 0 (0 < 𝑥 < ℎ𝑒 ) 𝐸𝐴 = −𝑃𝑒 1 , 𝐸𝐴 = 𝑃𝑒 2
𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑥=0
𝑑𝑥 𝑥=ℎ𝑒
𝑃𝑒 1 1 1
𝐸𝐴
= 𝐶1
ℎ𝑒
- 𝐶2 ……..(1)
ℎ𝑒
𝑃𝑒 2 1 1
𝐸𝐴
= −𝐶1
ℎ𝑒
+𝐶2 …….(2)
ℎ𝑒
𝐸𝐴
(𝐶1 − 𝐶2 ) = 𝑃𝑒 1 ……..(1)
ℎ𝑒
𝐸𝐴
(−𝐶1 + 𝐶2 ) = 𝑃𝑒 2 …….(2)
ℎ𝑒
𝐸𝐴 1 −1 𝐶1 𝑃1 𝑒 𝐸𝐴 1 −1 𝑢1 𝑃1 𝑒
= → 𝑢2 = 𝑃2 𝑒
ℎ𝑒 −1 1 𝐶2 𝑃2 𝑒 ℎ𝑒 −1 1
Rod Subjected to Axial Load
2 2 3 4 4
𝐸𝐴 1 −1 𝑢1 𝑃1 𝑒 1 1 3 5
ℎ𝑒 −1 1 𝑢2 = 𝑃2 𝑒 P
𝑈 = 𝑈1 ∅1 + 𝑈2 ∅2 L
𝑥 𝑥
𝑈 = 𝑈1 1− + 𝑈2
ℎ𝑒 ℎ𝑒
𝑑𝑢 1 1
= 𝜀 = 𝑈1 − + 𝑈2
𝑑𝑥 ℎ𝑒 ℎ𝑒
1 𝑈1
𝜀= −1 1 𝑈
𝐿 2
𝐸 𝑈
σ = −1 1 𝑈1
𝐿 2
Problem 1
40 mm 15 mm 30 mm 15 KN
E = 200 x 103 N/mm2
200 mm 500 mm 200 mm
𝐸𝐴 1 −1 𝑢1 𝑃1 𝑒
ℎ𝑒 −1 1 𝑢2 = 𝑃2 𝑒
Problem 2 E = 5 x 1010 Pa
E = 10 x 1010 Pa
A = 1 cm2 3000 N A = 2 cm2
10 cm 10 cm 10 cm
𝐸𝐴 1 −1 𝑢1 𝑃1
𝐿 −1 1 𝑢2 = 𝑃2
𝐸1 𝐴1 1 −1 105 ×100 1 −1 5 1 −1
𝐾1 = = =10 = 𝐾2
𝐿1 −1 1 100 −1 1 −1 1
𝐸3 𝐴3 1 −1 54 ×200 1 −1 1 −1
𝐾3 = = =105
𝐿3 −1 1 100 −1 1 −1 1
Problem 2
1 2 2 3 3 4
5 1 −1 1 1 −1
5 2 5 1 −1 3
𝐾1 = 10 𝐾2 = 10 𝐾3 = 10
−1 1 2 −1 1 3 −1 1 4
1 2 3 4
1 −1 0 0 1
−1 2 −1 0 2
𝐾= 105
0 −1 2 −1 3
0 0 −1 1 4
1 −1 0 0 𝑈1 𝑃1
−1 2 −1 0 𝑈2 𝑃2
105 =
0 −1 2 −1 𝑈3 𝑃3
0 0 −1 1 𝑈4 𝑃4
2 −1 𝑈2 𝑃
105 = 2
−1 2 𝑈3 𝑃3
2 −1 𝑈2 3000
105 =
−1 2 𝑈3 0
𝑈2 = 0.02𝑚𝑚 𝑈3 = 0.01 𝑚𝑚
𝑅1 = 𝐾 𝑈 − P
𝑈1 0
𝑈2 0.02
1 −1 0 0 − 𝑃 = 105 1 −1 0 0 = −2000 N
𝑈3 0.01
𝑈4 0
0
0.02
𝑅4 = 105 0 0 −1 1 = −1000 N
0.01
0
1 𝑈
𝜀= −1 1 𝑈1 , σ = 𝜀 × 𝐸
𝐿 2
−4
𝜀1 = 2 × 10 𝜎1 = 20𝑀𝑃𝑎
𝜀2 = −1 × 10−4 𝜎2 = −10 𝑀𝑃𝑎
𝜀3 = −1 × 10−4 𝜎3 = −5𝑀𝑃𝑎
Problem 3
55 kN
E = 2 x 105 MPa
A = 110 mm2 A = 220 mm2
1.2 m 2.4 m
1.2 mm
𝐸𝐴 1 −1 𝑢1 𝑃1
𝐿 −1 1 𝑢2 = 𝑃2
𝐸1 𝐴1 1 −1 105 ×100 1 −1 5 1 −1
𝐾1 = = =10
𝐿1 −1 1 100 −1 1 −1 1
𝐸2 𝐴2 1 −1 54 ×200 1 −1 1 −1
𝐾2 = = =105
𝐿2 −1 1 100 −1 1 −1 1
Problem 3
55 kN
E = 2 x 105 MPa
A = 110 mm2 A = 220 mm2
1.2 m 2.4 m
1.2 mm
𝐸𝐴 1 −1 𝑢1 𝑃1
𝐿 −1 1 𝑢2 = 𝑃2
𝐸1 𝐴1 1 −1 1 −1
𝐾1 = = =105
𝐿1 −1 1 −1 1
𝐸2 𝐴2 1 −1 1 −1
𝐾2 = = =105
𝐿2 −1 1 −1 1
Problem 3
1 2 3
2
5 0.183 −0.183 1 50.183 −0.183 2
𝐾1 = 10 𝐾2 = 10
−0.183 −0.1831 2
2
−0.183 0.183 3
1 3
0.183 −0.183 0 1
𝐾 = 105 −0.183 0.366 −0.186 2
0 −0.186 0.186 3
0.183 −0.183 0 𝑈1 𝑃1
105 −0.183 0.366 −0.186 𝑈2 = 𝑃2
0 −0.186 0.186 𝑈3 𝑃3
Problem 4
P = 20 KN
Rigid Plate
4 −4
𝐾1 = 104
−4 4
Al.
500 mm
5.2 −5.2
Steel 𝐾2 = 104
Steel −5.2 5.2
500 mm Brass 6.5 −6.5
𝐾3 = 104
−6.5 6.5
Properties Steel Aluminum Brass
C/S Area (mm2) 200 370 370
E (N/mm2) 2 x 105 7 x 104 8.8 x 104
1 1 1
Problem 4 2
1 2 4
3 3 3
4 + 5.2 + 4 −5.2 −4 − 4 𝑈1 𝑃1
104 −5.2 5.2 + 6.5 −6.5 𝑈2 = 𝑃2
−4 − 4 −6.5 4 + 6.5 + 4 𝑈3 𝑃3
13.2 −5.2 −8 𝑈1 𝑃1
104 −5.2 11.7 −6.5 𝑈2 = 𝑃2
−8 −6.5 14.5 𝑈3 𝑃3
Problem 5 – Taper Bar
𝐷1 − 𝐷2
𝐷𝑥 = 𝐷2 + (L − x)
𝐿
Dx
Problem 5 – Taper Bar
20 mm2
80 mm2
500 N
60 mm
𝐷1 − 𝐷2
𝐷𝑥 = 𝐷2 + (L − x)
𝐿
Problem 6
E = 200 GPa
1.5 m
150 mm2
60 kN
240 mm2
3m
1.5 mm
Vertical Taper Bar E = 200 Gpa
Thickness = 20 mm
= 7800 kg/m3
180 mm
160 mm
500 mm 500 mm
140 mm
200 mm 50 kN 200 mm 50 kN
110mm
80 mm
𝐷1 − 𝐷2
𝐷𝑥 = 𝐷2 + (L − x)
𝐿
Vertical Taper Bar E = 200 Gpa
Thickness = 20 mm
= 7800 kg/m3
21.3 −21.3
𝐾1 = 105
−21.3 21.3
160 mm
500 mm 22 −22
𝐾2 = 105
−22 22
21.3 −21.3 0 𝑈1 𝑃1
105 −21.3 43.3 −22 𝑈2 = 𝑃2
200 mm 50 kN 0 −22 22 𝑈3 𝑃3
110 mm
𝑊1 = 𝐴1 𝐿1 𝜌 𝑔 = 160 × 20 × 300 × 7800 × 10−9 × 9.81 = 𝑁
𝑊2 = 𝐴2 𝐿2 𝜌 𝑔 = 110 × 20 × 200 × 7800 × 10−9 × 9.81 = 𝑁
Vertical Taper Bar E = 200 Gpa
Thickness = 20 mm
= 7800 kg/m3
21.3 −21.3 0 𝑈1 𝑃1
105 −21.3 43.3 −22 𝑈2 = 𝑃2
150 mm
0 −22 22 𝑈3 𝑃3
500 mm
21.3 −21.3 0 0 36.7
105 −21.3 43.3 −22 𝑈2 = 50053.5
0 −22 22 𝑈3 16.8
200 mm 50 kN
100 mm 𝑈2 = 0.023 = 𝑈3
