Week - 2 - Explicit and Implicit Analysis
Objective:- Use the given equation and solve for implicit as well as explicit methods with tolerance of 0.001.
Procedure:-
Implicit method-
- This type of analysis is used in static analysis.
- In this method result for next step is calculated.
- This method considers equilibrium of the forces i.e. it will make sur ethe residual is close to zero untill repeated iterations are performed.
- For this analysis time does not matter.
- This method is used when there is small deformation with time.
- Accelertion vary linearly with time.
Explicit Method:-
- This type of analysis is used for dynamic analyis such as crash test,drop dest.
- In this method result for current step is calculated.
- This method does not consider equilibrium of the forces.
- In Explicit analysis time is important so called conditionally stable .
- This method is used in large deformation analysis such as car crash test.
Explicit Analysis:-
Here,three equal load increments (or steps) of ∆F = 1 unit.
The variables used are u for displacement, f for internal force,F for external force applied to the bar and k for stiffness. Incremental displacements or incremental externally applied forces are represented as ∆u or ∆F
respectively. Use is made of the relationship ∆F = k∆u. The analysis proceeds as follows
We have the given equation as `F(u)=u^3+9u^2+4u` .----->(1)
From 1 it follows that,
stifness can be given as,
k(u)=df/du.
therefore, k(u)=`3u^2+18u+4`---->(2)
STEP 1 :-
Our initial displacement point is 0 so, u=0
Now,`triangle F`=1 ---(steps)
So we have ∆F = k∆u.
rearranging the above equation we get ∆u= ∆F/k(u0)
substistuting the values, ∆u= 1/4=0.25 (incremental displacment value for next step)
Now,next displacement value u1=u0+ ∆u
i.e. u1=0+0.25=0.25
Similarly, For u2 & u3
Step 2 :-
u1=0.25
∆F=1
∆u= ∆F/k(u1)
k(u1)=3(0.25)^2+18(0.25)+4
k(u1)=8.6875
Therefore,
∆u=1/8.6875=0.1151
so, next displacement value u2=0.1151+∆u
u2=0.3651.
STEP 3:-
u2=0.3651
∆F=1
∆u= ∆F/k(u2)
k(u2)=3(0.3651)^2+18(0.3651)+4
k(u2)=10.9716
Therefore,
∆u=1/10.9716=0.0911
so, next displacement value u2=0.0911+∆u
u3=0.4562
We have done with 3 incremental steps, so further moving with to find whether both the quantities are in equilibrium or not.
F=∆F+∆F+∆F= 3.0
f=`u^3+9u^2+4u`
In above equation substitute the value of u3=0.4562.
f=`(0.4562)^3+9(0.4562)^2+4(0.4562)`
therefore, f=3.7928
We can conclue that F is not equal to f ie.internal force is not equal to external force.
| step | ∆F | ∆u | u | F | f | R=f-F |
| 1 | 1 | 0.25 | 0.25 | 1 | 0.5781 | 0.5781 |
| 2 | 1 | 0.1151 | 0.3651 | 2 | 2.7087 | 0.7087 |
| 3 | 1 | 0.0911 | 0.4562. | 3 | 3.7928 | 0.7928 |
Implicit Analysis :-
An implicit analysis is the same as the explicit analysis, except that at the end of each step
Newton-Raphson iterations are used to enforce equilibrium before moving to the next step.However, internal forces and external forces will not be in equilibrium unless the stiffness is linear for the given step. Hence, in order to achieve equilibrium, corrections mustbe made to the displacement. This is accomplished by using Newton-Raphson iterations to
minimize the residual, R(u) = fint − Fext. This quation is ittertaed till we get the value close to tolerance.
R^(j+1) = R^(j)(u^j) + dR(u^j)/du (δu^j+1) + ... = R^j + k(u^j)δu^j+1 + ...
now taking R^(j+1)=0 and solving only for δu^(j+1) we get,
`δu^(j+1) = −[k(u^j)]^(−1)R^j` Final Taylours equation.
Rewriting the equation,
δu^(j+1)=-R^j/[k(u^j)].
So starting with the first step:-
u0=0.0
∆F=1
∆u= ∆F/k(u0)
k(u0)=4
∆u= ∆F/k(u0)
∆u= 1/4=0.25
next displacement value u1=u0+ ∆u
u1=0+0.25=0.25
`k(u1)=`3(0.25)^2+18(0.25)+4`` =8.6875
Now to check for residual,
F(ext)=∆F=1,
f(int)=f(u1)= `u^3+9u^2+4u` =`0.25^3+9(0.25)^2+4(0.25)` =1.5781
since we have ext and internal forces lets find the value of residual
R(0)=f-F
R0=1.5781-1=0.5781
Since the value of residual> value of tolerance 0.001 therefore need further itherations using taylour series.
δu^(j+1)=-R^j/[k(u^j)]
`δu^(1)=-R^0/[k(u1)`
=-0.5781/8.6875= -0.066
Adding this step displacement value to u1 we get,
new u1=0.25+(-0.0660) =0.184.
Now, f=1.469 (substituting value of u1=0.184 in equation (1))
Now checking for residual with the new displacement value, u1=0.184
R(1)=f-F=1.0469=0.0469
since, R(1)>0.001
therefore need to iterrate further.
`δu^(2)=δu^(1)-R^1/[k(u^1)`
=-0.066-0.0469/8.6875=-0.0723
u1=0.25+(-0.0723)=0.1776
Now, checking the residual again
f=0.9998(substitute u=0.1776 in equation 1)
R(2)=f-F=0.9998-1=0.0002 <0.001
Hence tolerance is satistied.
SO, final value of displacement is 0.1776
STEP 2:-
u1=0.1776
k(u1)=7.72
F(ext)=∆F=1,
∆u2= ∆F/k(u1)=2/7.72=0.1295
u2=u1+ ∆u2=0.1776+0.2742=0.3071
k(u2)=9.810
Check for residual
F(ext)=F=2
fint=u2^3+9u2^2+4u2
fint=(0.3071)3+9(0.3071)^2+4(0.3071)=2.106
R(0)=fint-Fext=2.106-2=0.106 >0.001
iterations required.
Now calulation for u2(0)
`δu^(1)(0)=-R^0/[k(u2)`
=`-0.106/9.810=-0.011` `
corrected u2=`0.3071+(-0.011)=0.2961`
checking R(1)=f-F=0.005..hence tolerence reached.
therefore final value of displacement is 0.2961.
Now, similarly performed for step 3
we get u=0.3910
| step | ∆F | u | f | R=f-F |
| 1 | 1 | 0.1776 | 0.999 | 0.0002 |
| 2 | 1 | 0.2961 | 1.999 | 0.005 |
| 3 | 1 | 0.3910 | 3.000 | 0.00074 |
Conclusion:-
The hand calculation time by implicit method takes more than explicit method.