Explicit and Implicit Analysis
Objective:
In this report, I will compare the results from explicit and implicit methods analyzing a simple bar in tension.
Suppose that the force in the bar is a nonlinear function of the displacement:
`f(u)=u^3+9u^2+4u`
It then follows that the stiffness of the bar is:
`(u)=df/du=3u^2+18u+4`
Explicit Analysis:
The analysis proceeds as follows:
Step 1:
`u_0=0.0, →k(u_0 )=4, ∆F=1, ∆u_1=∆F/k(u_0 ) =0.25, u_1=u_0+∆u_1=0.25`
Step 2:
`→k(u_1 )=3〖×0.25〗^2+18×0.25+4=8.6875, ∆F=1, ∆u_2=∆F/k(u_1 ) ≅0.1151, u_2=u_1+∆u_2=0.3651`
Step 3:
`→k(u_2 )=3〖×0.3651〗^2+18×0.3651+4=10.9717, ∆F=1, ∆u_3=∆F/k(u_2 ) ≅0.0911, u_3=u_2+∆u_3≅0.4562`
Finally, from the three steps the following results are obtained:
`F_ext=∆F+∆F+∆F=3.0`
`f_int=〖〖u_3〗^3+9〖u_3〗^2+4u〗_3=〖0.4562〗^3+9×〖0.4562〗^2+4×0.4562≅3.7928`
It is evident from the results that the external and internal forces are NOT in equilibrium because `F_ext≠f_int.`
The following table summarizes the main results for the explicit analysis:
Implicit Analysis:
The analysis proceeds as follows:
Step 1:
`u_0=0.0, →k(u_0 )=4, ∆F=1, ∆u_1=∆F/k(u_0 ) =0.25, u_1=u_0+∆u_1=0.25`
Check the residual, `F_ext=∆F=1, f_int=〖u_1〗^3+9〖u_1〗^2+4u_1=1.5781, 〖 R〗^((0) )=f_int-F_ext=0.5781`
`‖R^((0) ) ‖>〖10〗^(-2)→`Newton-Raphson iterations are necessary.
Calculate the correction to `u_1=〖u_1〗^((0) )`
`δu^((1) )=-[k(〖u_1〗^((0) ) )]^(-1) R^((0) )=-〖8.6875〗^(-1)×0.5781≅-0.0665.`
`→` The updated value of `〖u_1〗^((1) )=〖u_1〗^((0) )+〖δu〗^((1) )=0.25-0.0665=0.1835.`
Check the residual again, `F_ext=1, f_int=(〖u_1〗^((1) ) )^3+9(〖u_1〗^((1) ) )^2+4〖u_1〗^((1) )≅1.0432`
`〖 R〗^((1) )=f_int-F_ext=0.0432`
`‖R^((1) ) ‖>〖10〗^(-2)→` another Newton-Raphson iteration is necessary.
Calculate the new total correction to `u_1=〖u_1〗^((0) )`
`δu^((2) )=δu^((1) )-[k(〖u_1〗^((1) ) )]^(-1) R^((1) )=-0.0665-(〖7.404〗^(-1) )×0.0432≅-0.0723.`
The updated value of `〖u_1〗^((2) )=〖u_1〗^((0) )+〖δu〗^((2) )=0.25-0.0723=0.1777.`
Check the residual, `F_ext=1, f_int=(〖u_1〗^((2) ) )^3+9(〖u_1〗^((2) ) )^2+4〖u_1〗^((2) )≅1.0006`
`〖 R〗^((2) )=f_int-F_ext=0.0006`
`‖R^((2) ) ‖<〖10〗^(-2)→` no further iterations are necessary.
Therefore, the final value is `u_1=〖u_1〗^((2) )=0.1777.`
Step 2:
`k(u_1 )=3〖×0.1777〗^2+18×0.1777+4≅7.2933, ∆F=1, ∆u_2=∆F/k(u_1 ) ≅0.1371,`
`u_2=u_1+∆u_2=0.3148`
Check the residual, `F_ext=2×∆F=2, f_int=〖u_2〗^3+9〖u_2〗^2+4u_2≅2.1823`
`〖 R〗^((0) )=f_int-F_ext=0.1823`
`‖R^((0) ) ‖>〖10〗^(-2)→`Newton-Raphson iterations are necessary.
Calculate the correction to `u_2=〖u_2〗^((0) )`
`δu^((1) )=-[k(〖u_2〗^((0) ) )]^(-1) R^((0) )=-〖9.9637〗^(-1)×0.1823≅-0.0183.`
`→` The updated value of `〖u_2〗^((1) )=〖u_2〗^((0) )+〖δu〗^((1) )=0.3148-0.0183=0.2965.`
Check the residual again, `F_ext=2, f_int=〖〖u_2〗^((1) )〗^3+9〖〖u_2〗^((1) )〗^2+4〖u_2〗^((1) )≅2.0033`
`〖 R〗^((1) )=f_int-F_ext=0.0033`
`‖R^((1) ) ‖<〖10〗^(-2)→`no further iterations are necessary.
Therefore, the final value is `u_2=〖u_2〗^((1) )=0.2965.`
Step 3:
`k(u_2 )=3〖×0.2965〗^2+18×0.2965+4≅9.6007, ∆F=1, ∆u_3=∆F/k(u_2 ) ≅0.1042,`
`u_3=u_2+∆u_3=0.4007`
Check the residual, `F_ext=3×∆F=3, f_int=〖u_3〗^3+9〖u_3〗^2+4u_3≅3.1122`
`〖 R〗^((0) )=f_int-F_ext=0.1122`
`‖R^((0) ) ‖>〖10〗^(-2)→`Newton-Raphson iterations are necessary.
Calculate the correction to `u_3=〖u_3〗^((0) )`
`δu^((1) )=-[k(〖u_3〗^((0) ) )]^(-1) R^((0) )=-〖11.6943〗^(-1)×0.1122≅-0.0096.`
`→`The updated value of `〖u_3〗^((1) )=〖u_3〗^((0) )+〖δu〗^((1) )=0.4007-0.0096=0.3911.`
Check the residual again, `F_ext=3, f_int=〖〖u_3〗^((1) )〗^3+9〖〖u_3〗^((1) )〗^2+4〖u_3〗^((1) )≅3.0009`
`〖 R〗^((1) )=f_int-F_ext=0.0009`
`‖R^((1) ) ‖<〖10〗^(-2)→`no further iterations are necessary.
Therefore, the final value is `u_2=〖u_2〗^((1) )=0.3911.`
The results of the implicit analysis are summarized in the following table:
It is evident from the table that the external forces and internal forces are coming into equilibrium.
Plots of results:
We can see from the plots of comparison that the Implicit results are closest to the exact solutions. This is because the Implicit method uses Newton-Raphson iterations to enforce equilibrium in each step while the Explicit method does not take into account the equlibrium conditions.
Comparing with the results in the reference, using the new function of force vs displacement `f(u)=u^3+9u^2+4u` has helped the Implicit results reach equilibrium faster for each step.