Week - 2 - Explicit and Implicit Analysis
LS Dyna is usually used to solve transient dynamic problems.Transient dynamic problems are those which have structural deformation due to time depended load. There are two methods to solve a transient dynamic problem.Implicit and Explicit method.
Implicit method: In general implicit methods are used to slove static problems where we care about only the end result, not on what happens in between.eg. When applied force at one end of a cantilever beam, the change is slow and static.
Explicit menthod:In this method there is rapid force and acceleration and within short amount of time state of the system changes, hence it is a dynamic problem Eg: a crash of a vehicle is an example where changes in a structur is due to rapid force and is in short amount of time.
In this challenge we are aiming to understand what all kinds of problems or situations are solved in a implicit method and explicit method.
We can solve an implicit problem using mathematical equation.
Consider du/dt=F(u)
Explicit Method:
Equation can be written as u(n+1)=u(n)+ΔtF(u(n))
Where n is the time & (n+1) Is the time step after n
Equation to be solved: Md²u/dt²+ku=F(t), where d²u/dt² = acceleration
At t=0, if we have the value of u &v, we can compute the values of M&K
Once we can compute the values of M&k, we can use them to find acceleration from which velocity and displacement can be calculated at Δt.
Relationship between acceleration, velocity and displacement is
V=u+at
S=ut+1/2at²
where: v= velocity, u=displacement, a= acceleration, t=time
a=m^-1(F(t)-Ku)
v(Δt )= v(0)+aΔt to find velocity at Δt
u(Δt)=v(0)Δt+1/2 aΔt² to find displacement at Δt
In implicit method:
At t=Δt we can compute M&K values and use them to find acceleration &velocity
a=m^-1(F(Δt)-Ku)
and velocity and displacement can be calculated.
PROBLEM TO SOLVE:
F(u)=u³+9u²+4u
Solve by Explicit and implicit method using tolerance 0.001
k(u)=dF/du=3u²+18u+4
Explicit analysis :
Step 1: u₀=0 k(u₀)=4 ΔF=1 Δu₁=ΔF/k(u₀)=1/4
Therefore u1=u0+Δu1=0+1/4=0.25
Step 2: u1=0.2 ΔF=1
K(u1)=3(0.25)²+18(0.25)+4=8.6875
Δu2=ΔF/k(u1)=1/8.6875=0.1151
Therefor u2=u1+Δu2=0.25+0.1151=0.3651
Step 3: u2=0.3651
K(u2)=3(0.365)²+18(0.365)+4=10.969
K(u2)=10.969
ΔF=1 Δu3=ΔF1/k(u2)=1/10.96=0.091
U3=u2+Δu3=0.3651+0.091=0.456
U3=0.456
Finaly, from the three steps , the following results were obtained.
ΔFext=ΔF+ΔF+ΔF=3
Fint=u₃³+9u₃²+4u₃
=0.456³+9(0.456)²+4(0.456)=3.790
Fext≠fint
Step I ΔFi Δui ui (Fext)I (fint)I R=Fext-fint
1 1 0.25 0.25 1 1.578 0.578
2 1 0.115 0.365 2 2.707 0.707
3 1 0.091 0.456 3 3.790 0.790
Implicit Analysis:
F(u)=u³+9u²+4u
K(u)=3u²+18u+4
Consider u0=0.0 ΔF=1 Δu1=ΔF/k(u0)=1/4
U1=u0+Δu1=0+1/4=0.25
Check the residual:
Fext=ΔF=1
Fint=u₁³+9u₁²+4u₁=0.25³+9(0.25)²+4(0.25)=1.578
R=fint-Fext=1.578-1=0.578
R>10^-2 Newton Raphson iteration are necessary
K(u1)=3(0.25)²+18(0.25)+4
Calculate the correction to u1=u1⁰
ꝺu₁=-(k(u))^-1R⁰=-(8.68)^-1 X 0.578=-0.066
The updatedvalue of u1=u1⁰+ꝺu1=0.25+(-0.066)=0.184
Check the residual Again
Fext=1 Fint=u1³+9u1²+4u1=0.184³+9(0.184)²+4(0.184)=1.046
R1=1-1.046=0.046
0.046>0.001
Calculate the new total correction
U1=u1(0)
(k(u1))=K(0.184)=3(0.184)²+18(0.184)+4=7.41
ꝺu2=ꝺu1-(K9u10)^-1R1=-0.066-(7.41)^-1X 0.046=-0.0722
Therfor the updated value of u2=u1+ꝺu2=0.25+(-0.0722)=0.1778
Check the residual
Fext=1
Fint=u1³+9u1²+4u1=0.1778³+9(0.1778)²+4(0.1778)=1.0013
R=Fext-fint=1-1.0013=0.0013
R<=0.001. No further interation required
U1=0.1778
Step2
K(u1)=3u²+18u+4=3(0.1778)²+18(0.1778)+4 =7.29
ΔF=1 Δu2=ΔF/k(u1)=1/7.295=0.137
U2=u1+Δu2=0.1778+0.137=0.314…
Hence check the residual and follow the same procedure to find the correct u2.
But implicit method requires more calculations and hence it is time consuming and difficult.
Therefor Explicit method is the best way to solve a transient dynamic problem