Week - 2 - Explicit and Implicit Analysis
Objective: To solve the above given equation by using explicit and implicit analysis method
Lets try to solve the above equation by using above methods
F(u)=u3+9u2+4u --1
By derivating the above equation we get the below equation for stiffness
K(u)=3u2+18u+4 --2
For Explicit
Step 1:
- The initial displacemnt is zero when force is applied on the body so u=0
- Update the value of u=0 in the above equation 2
K(0)=3(0)^2+18(0)+4
K(0)=4
The icremental load is Δf=1
As we are solving static problem the governing equation is K(X)=F(t) by applying the value of K and f we can find incremental time step
Δu= Δf/k
Δu=1/4
Δu=0.25
now u1=u0+ Δu
u1=0+0.25
u1=0.25
Internal force Fint=u13+9u12+4u1
Fint=(0.25)3+9(0.25)2+4(0.25)
Fint=1.578
Residual= ΔF+Fint
=-1+1.578
ΔF =0.578
Step 2:
- now u1=0.25 is the input in the time step 2 and we have a second load increment Δf=1
Now in this we have to update the stiffness matrix.
K(u1)=3u2+18u+4
K(0.25)=3(0.25)2+18(0.25)+4
K(0.25)=8.687
now use the governing equation to find the incremental Disp
Δu=Δf/k
=1/8.687
Δu=0.115
the final displacement is u2=u1+Δu
u2=0.25+0.115
u2=0.365
Fint=u23+9u22+4u2
Fint=(0.365)3+9(0.365)2+4(0.365)
Fint=2.707
Residual=ΔF+Fint
=-2+2.707
=0.707
Step 3
Now u2=0.365 is the input in step 3 and we have a second load increment Δf=1
Now in this we have updated stiffness matrix
K(u2)=3u22+18u2+4
K(0.365)=3(0.365)2+18(0.365)+4
K(0.365)=10.96
No use the governing equation to find the incremental displacement
Δu=Δf/k
=1/10.96
Δu=0.0912
The final diplacement
u3=u2+Δu
u3=0.365+0.0912
u3=0.4562
The internal force Fint=u33+9u32+4u3
Fint=(0.4562)3+9(0.4562)2+4(0.4562)
Fint=3.792
Residual=-ΔF+Fint
=-3+3.792
=0.792
For Implicit
- The initial displacemnt is zero when force is applied on the body so u=0
- Update the value of u=0 in the above equation 2
K(0)=3(0)^2+18(0)+4
K(0)=4
The icremental load is Δf=1
As we are solving static problem the governing equation is K(X)=F(t) by applying the value of K and f we can find incremental time step
Δu= Δf/k
Δu=1/4
Δu=0.25
now u1=u0+ Δu
u1=0+0.25
u1=0.25
Fint =u13+9u12+4u1
=(0.25)3+9(0.25)2+4(0.25)
=1.578
Residual =-ΔF+Fint
=-1+1.578
=0.578
Newton raphsons iterations are neccesary
K(u1^0)=3u2+18u+4
=3(0.25)+18(0.25)+4
=8.68
|R(0)|>10-3
δu1=-[K(u1^0)]R(0)
=-8.68^-1(0.578)
δu1=-0.0665
u1^1=u1+δu1
=0.25-0.0665
u1(1)=0.1835
Fint=u13+9u1^2+4u1
=(0.1835)3+9(0.1835)2+4(0.1835)
=1.0071>0.001
Residual=-ΔF+Fint
=-1+1.0071
=0.0071<0.001
Same above steps to be followed for u2=0.2966,u3=0.3911
Conclusion:Hence solved the above static loading equation using Implicit and explicit dynamic method and came to a conclusion that no need to carry out iterations for explixit dynamics