Week - 2 - Explicit and Implicit Analysis
Aim:
To solve the given equation using Explicit and Implicit methods as per prescribed tolerance value.
Procedure:
Explicit Method:
Given f(u) = u3+9u2+4u
This implies that Force f is non linear function of displacement u.
K = df/du = 3u2+18u+4
Here by using incremental explicit load scheme, three equal increment loads of ΔF = 1 is used to load the tension bar. The variables used are u for displacement, F for external load applied for the bar, f for internal force in the bar and k for stiffness. Incremental displacements are represented by Δu and incremental force by ΔF.
Now by using the relation ΔF = k *Δu. Analysis will move forward as follows.
Step 1
u 0(0) = 0
So k = 0+0+4 = 4
ΔF = 1
Hence Δu1 = ΔF/k = 1/4 = 0.25
u1 = u 0 + Δu1
= 0+0.25
= 0.25.
so u1 = 0.25
Step 2
k (u1) = 3 * 0.252+ 18*0.25 + 4
= 8.68
ΔF = 1
Δu2 = ΔF/k = 1/8.68 = 0.11
u2 = u1 + Δu2
= 0.25+0.11 = 0.36
so u2 = 0.36
Step 3
k (u2) = 3 * 0.362+ 18*0.36 + 4
= 10.86
ΔF = 1
Δu3 = ΔF/k = 1/10.86 = 0.092
u3 = u2 + Δu3
= 0.36+ 0.092
= 0.45
so u3 = 0.45.
|
Step (i) |
ΔFi |
ui |
(Fext)i |
(fint)i |
R = fint - Fext |
|
1 |
1 |
0.25 |
1 |
1.578 |
0.578 |
|
2 |
1 |
0.36 |
2 |
2.653 |
0.653 |
|
3 |
1 |
0.45 |
3 |
3.713 |
0.713 |
Implicit Method:
Here the Residue is expanded as Taylor series about current displacement uj.
Rj+1 = Rj(uj) + (dR(uj)/du) * δuj+1 +.... = Rj + k(uj) * δuj+1+..
Neglecting higher order terms and setting Rj+1 =0 and solving for δuj+1, The following relation is obtained.
δuj+1 = -[ k(uj)]-1 Rj.
Step 1
u 0(0) = 0
So k = 0+0+4 = 4
ΔF = 1
Hence δu1 = ΔF/k = 1/4 = 0.25
u1 = u 0 + δu1
= 0+0.25
= 0.25.
(Fext)1 = 1
(fint)1 = 1.5781
R(0) = (fint)1 - (Fext)1
= 0.5781 > 10-2
Hence Newton Raphson Iteration should be performed
δu1 = -[ k(u0)]-1 R0
= -0.0665
u1(1) = u 1(0) + δu1
= 0.25 – 0.0665
= 0.1835
(Fext)1 = 1
(fint)1 = 1.04
R(1) = (fint) - (Fext)
= 0.04 > 10-2
Hence Newton Raphson Iteration should be performed
δu2 = δu1 - [ k(u0)]-1 R1
= -0.0719
u1(1) = u 1(0) + δu2
= 0.25 – 0.0719 = 0.1781
(Fext)1 = 1
(fint)1 = 1.0035
R(1) = (fint) - (Fext)
= 0.0035 < 10-2
No further Iterations are needed since the residue value is below the tolerance.
Hence u1 = 0.1781
Step 2
k (u1) = 3* (0.1781)2 + 18* 0.1781 + 4
= 7.3009
ΔF = 1
Hence δu2 = ΔF/k = 1/7.3009 = 0.1369
u2 = u1 + δu2
= 0.3150
(fint)2 = 2.1842
(Fext)2 = 2
R(2) = (fint) - (Fext)
= 0.1842 > 10-2
Newton Raphson Iteration should be performed
δu1 = -[ k(u0)]-1 R0
= -0.01847
u2(1) = u 2(0) + δu1
= 0.2965
(Fext)1 = 2
(fint)1 = 2.0032
R(1) = (fint) - (Fext)
= 0.0032 < 10-2
No further Iterations are needed since the residue value is below the tolerance.
Hence u2 = 0.2965
Step 3
k (u2) = 3* (0.2965)2 + 18* 0.2965 + 4
= 9.6007
ΔF = 1
Hence δu3 = ΔF/k = 1/9.6007 = 0.1041
u3 = u2 + δu3
= 0.4006
(fint)2 = 3.1110
(Fext)2 = 3
R(2) = (fint) - (Fext)
= 0.1110 > 10-2
Newton Raphson Iteration should be performed
δu2 = -[ k(u0)]-1 R0
= -0.0094
u3(1) = u 3(0) + δu2
= 0.4006 – 0.0094 = 0.3912
(Fext)1 = 3
(fint)1 = 3.0020
R(1) = (fint) - (Fext)
= 0.0020 < 10-2
No further Iterations are needed since the residue value is below the tolerance.
Hence u3 = 0.3912
|
Step (i) |
ΔFi |
ui |
(Fext)i |
(fint)i |
R = fint - Fext |
|
1 |
1 |
0.1781 |
1 |
1.0035 |
0.0035 |
|
2 |
1 |
0.2965 |
2 |
2.0032 |
0.0032 |
|
3 |
1 |
0.3912 |
3 |
3.0020 |
0.0020 |
Conclusion:
Residue is less in implicit analysis as compared to explicit analysis. Hence Implicit analysis should be used to solve this equation.
Learning Outcomes:
- Explit method to solve Equation.
- Implicit method to solve Equation.