F(u) = u3+9u2+4u                                                        (1)

 From (1) it follows that the stiffness of the bar is

k(u) = 3u^2+18u+4                                                           (2)

Explicit methods

Step 1
            u0 = 0.0, ⇒ k(u0) = 4, ∆F = 1, ∆u1 = ∆F/k(u0) = 1/4,                                                           

                                u1 = u0 + ∆u1 = 0+0.25=0.25

Step 2
        ⇒ k(u1) = 3(0.25)^2+18(0.25)+4=0.1875+4.5+4=8.6875,                                                     

   ∆F = 1, ∆u2 = ∆F/k(u1) = 1/8.6875 =0.11511,                                                                         

u2 = u1 + ∆u2 = 0.25+0.11511=0.36511

Step 3
        ⇒ k(u2) = 3(0.36)^2+18(0.36)+4 =0.3888+6.48+4 =10.8688,                                                   

∆F = 1, ∆u3 = ∆F/k(u2) = 1/10.8688=0.092                                                                       

     u3 = u2+∆u3= 0.36511+0.092=0.4571

Finally, from the three steps the following results are obtained.
Fext = ∆F + ∆F + ∆F = 3.0
fint  = u3+9u2+4u   = (0.4571)^3+9(0.4571)^2+4(0.4571)                                                                           

      =0.0955+1.880+1.828‬=3.804

It is evident from the results that external and internal forces are NOT in equilibrium because
Fext is not equal to fint. Table 1 summarizes the main results for the explicit analysis.

                                    Summary of explicit analysis results

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.
Basically, an incremental force is applied to advance the solution forward at the beginning
of a 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 must
be made to the displacement. This is accomplished by using Newton-Raphson iterations to
minimize the residual, R(u) = fint − Fext. Expanding the residual as a Taylor series about
the current displacement u
j gives

                Rj+1 = Rj(uj) + dR(uj)du δuj+1 + ... = Rj + k(uj)δuj+1 + ...                  . (3)

By neglecting higher order terms in the series, setting Rj+1 = 0 and solving for δuj+1 the
following correction is obtained

                    δuj+1 = −[k(uj)]^-1 x Rj                                                                      . (4)

Notice that the iteration variable is j and it may take several iterations for the norm of the residual, kRk, to be reduced below the chosen tolerance criteria (for the example below a tol-
erance of 10−2is chosen). The implicit analysis proceeds as follows (results are approximate,they are more precise if done in a computer to machine precision):

Step 1
          u0 = 0.0, ⇒ k(u0) = 4, ∆F = 1, ∆u1 = ∆F/k(u0) = 1/4,                                                             

u1 = u0 + ∆u1 = 0+0.25=0.25

Check the residual,

Fext = ∆F = 1, fint = u3+9u2+4u                                                                                                                       

                              =(0.25)^3+9(0.25)+4(0.25)                                                                                               

                             = 0.01565+0.5625+1                                                                                                 

     fint=1.578125                                                                                           

    R(0) = fint − Fext =1.578125-1=0.578125

R(0)=0.578125 , is greater than0.01,so Newton-Raphson iterations are necessary.

u1 = u(0)1

δu(1) = −[k(u(0)1)]^−1*R(0) = (-8.59)^-1*0.5781 =-0.067

u(1)1 = u(0)1 + δu(1) =0.25-0.06=0.184

Check the residual, Fext = 1,

                           fint =u3+9u2+4u =0.0062+0.3047+0.736=1.0469

        R(1) = fint − Fext =1.0469-1=0.0469

R(0)=0.0469 , is greater than0.01,so Newton-Raphson iterations are necessary.

u1 = u(0)1

δu(2) = δu(1) − [k(u(1)1)]^−1*R(1) =-0.06-(0.005)*0.04 =-0.0713

u(2)1 = u(0)1 + δu(2) = 0.25-0.0713 = 0.1787

R(2) = fint − Fext = 1.007-1=0.0079 is less than0.01,no further iterations are necessary.

Therefore, the final value is u1 = u(2)1 = 0.1787

simillarly same steps are followed by step2 and step 3

U2 = .2965

U3=0.3911

 The above figure shows the force value for exact, explicit and implicit value

 The above figure shows the force value for exact, explicit  value

Conclusion:

  From the graph, it is clearly shown that implicit scheme gives very small error compare to the explicit scheme. Implicit scheme gives more accurate result compare to the explicit scheme.

•   Explicit scheme is fast and it is always conditionally stable. But implicit method is time consuming and it depends on the time step and the mesh discretization.