Week - 2 - Explicit and Implicit Analysis
Implicit method:
- Implicit method computes only stiffness matrix(F=ku) and displacement is obtained by inverting the stiffness matrixand multiplyinf to residual force.
- This method is unconditionally stable.
- computationally expensive and requires large amount of memory.
- assumes acceleration varies linearly.
explicit method:
- Explicit method uses mass matrix to compute then acceleration is calculated
- this method is conditionally stable i,e it follows courant condition (timestep< wave propogation speed in element)
- computationally fast and requires less memory
- assumes constant acceleration.
Aim :To solve F(u)=u^3+9u^2+4u in both explicit and implicit method with tolerance 10^-2
explaination: given equation F(u)=u^3+9u^2+4u ..(1)eq
stiffness relation K(u)=df/du=3u^2+18u+4 ..(2)eq
where u-displacement k-stiffness F-force `delta`u-increment in displacement `delta`F- increment in force fext=`delta`F -external force Fint-internal force
explicit method:
step 1:`delta`F=1 ,Uo=0,0 k(u0)=3*0+18*0+4 =4 by substituting in eq2
`delta`u1=`delta`F/k(u0)=1/4 or 0.25 u1=u0+`delta`u1=0+0.25
Fint=o.25^3+(9*0.25^2)+4*0.25 =1.5781 ...sub in eq(1)
residual force R=Fint -Fext=1.5781-1=0.5781
step2: simplifying as above we get u2=0.36511 ...{calculation shown in spread sheet}
step3:simplifying as above we get u3=0.456 ...{calculation shown in spread sheet}
summary of explicit analysis:
| step | delta Fi | delta Ui | Ui | Fext(i) | Fint(i) | R=Fint=Fext | |
| 1 | 1 | 0.25 | 0.25 | 1 | 1.5781 | 0.5781 | |
| 2 | 1 | 0.2206 |
|
2 | 3.9809 | 1.9809 | |
| 3 | 1 | 0.2548 | 0.6199 | 3 | 6.1771 | 3.1771 |
from the above table it is evident that internal fore is not in equilibrium with external force.
Implicit Analysis method: This method follows the inital steps as in explicit method but the difference is that after the completion of step using newton raphson iteration is solved to obtain equilibrium. correction is made to to displacement this is accomplished by using newton raphson iteration to minimise residual forces to a given tolerance value R(u)=Fint-Fext
expanding using taylor series we get Rj+1=Rj(uj)+dR(uj)/du*`del`uj+1+...
neglecting higher order and sub Rj+1=0 and solving we get
`del`uj+1=-[k(uj)^-1*Rj] ...eq3 where j is iteration variable and R to be reduced below tolerance value
Step1: U0=0,0 K(u0)=4, `delta`u1=0.25 u1=0.25
check for residual Fint=u3+9u^2+4u=1.5781
R0=1.57813-1=0.57813 |R0|>10^-2 Newton raphson iteration is necessary
calculate the correction u1=u1^(0)
`del`u^1=-(ku1^0)^-1 R^0
=-0.14453
updated value of u^1=u1^0+`del`u^1=-0.14453+0.25=0.10546
check the residual again Fint=0.52316 by sub in eq1
R=0.52316-1=-0.4768 second iteration is necessary
`del`u2=`del`u1-[k(u1)]^-1 *R1 =-0.14453-[5.93]^-1*-0.47684 =-0.0641
updated u1^2=ui^0+`del`u2=0.25+(-0.0641)=0.185
calculate residual R=1.0603-1=0.06
since R is greator than tolerance third iteration is necessary
by solving as above we get u1^3=0.17781
and residual R=i.00133-1=0.00133 |R|<10^-2 therefore no further iteration is necessary
Step2: calculating as above we get u2^2=0.2962 and R=0.0004 [calculations shown in spread sheet}
Step3: calculating as above we get u2^2=0.3910 and R=3.908e^-5 [calculations shown in spread sheet}
conclusion: from above calculation method it is evident that implicit method is more accurate but requires high computational time.
refer spreadsheet for calculation