Week - 2 - Explicit and Implicit Analysis
Explicit Analysis
Clearly it is possible to plot a graph of force, f, versus displacement, u, by using equation (1) (see Figure 2). However, if the nonlinear force versus displacement relationship is
not known, but the stiffness is known, it is possible to construct the graph of force versus
displacement numerically. One method is to use an incremental explicit load control scheme.
This is illustrated next. For this example three equal load increments (or steps) of ∆F = 1
unit each are used to load the tension bar. The variables used are u for displacement, f for
internal force in the bar, F for external force applied to the bar and k for stiffness. Incremental displacements or incremental externally applied forces are represented as ∆u or ∆F,
respectively. Use is made of the relationship ∆F = k∆u. The analysis proceeds as follows
Step-1̲">Step−1
u0=0.0,⇒k(u0)=4,ΔF=1,Δu1=ΔFk(u0)=0.25,u1=u0+Δu1=0.25">u0=0.0,⇒k(u0)=4,ΔF=1,Δu1=ΔFk(u0)=0.25,u1=u0+Δu1=0.25
Fexternal=ΔF=1">Fexternal=ΔF=1
FInternal=u3+9u2+4u=1.5781">FInternal=u3+9u2+4u=1.5781 (∵u=u1)">(∵u=u1)
Step-2̲">Step−2
u1=0.25,⇒k(u1)=8.688,ΔF=1,Δu2=ΔFk(u1)=0.1151,u2=u1+Δu2=0.3651">u1=0.25,⇒k(u1)=8.688,ΔF=1,Δu2=ΔFk(u1)=0.1151,u2=u1+Δu2=0.3651
Fexternal=ΔF+ΔF=2">Fexternal=ΔF+ΔF=2
FInternal=u3+9u2+4u=2.708">FInternal=u3+9u2+4u=2.708 (∵u=u2)">(∵u=u2)
Step-3̲">Step−3
u2=0.3651,⇒k(u2)=10.9717,ΔF=1,Δu3=ΔFk(u2)=0.0911,u3=u2+Δu3=0.4562">u2=0.3651,⇒k(u2)=10.9717,ΔF=1,Δu3=ΔFk(u2)=0.0911,u3=u2+Δu3=0.4562
`F_(external)=DeltaF+DeltaF+DeltaF=3`
FInternal=u3+9u2+4u=3.7928">FInternal=u3+9u2+4u=3.7928 (∵u=u3)">(∵u=u3)
It is evident from the results that external and internal forces are NOT in equilibrium because
Fexternal≠FInternal">Fexternal≠FInternal
The following table summarizes the results for explicit analysis method.
| STEP i | Fexternal=ΔF+ΔF+ΔF=3">ΔFi | Fexternal=ΔF+ΔF=2">Δui | ui | (Fext)i | (Fint)i | R=(Fint)i-(Fext)i |
| 1 | 1 | 0.25 | 0.25 | 1 | 1.5781 | 0.5781 |
| 2 | 1 | 0.1151 | 0.3651 | 2 | 2.708 | 0.708 |
| 3 | 1 | 0.0911 | 0.4562 | 3 | 3.7928 | 0.7928 |
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 uj gives
Rj+1 = Rj(uj) + dR(uj)du δuj+1 + ... = Rj + k(uj)δuj+1 + ...
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)]−1Rj. (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 tolerance of 10−2
is chosen). The implicit analysis proceeds as follows (results are approximate,
they are more precise if done in a computer to machine precision):
Step-1̲">Step−1
u0=0.0,⇒k(u0)=4,ΔF=1,Δu1=ΔFk(u0)=0.25,u1=u0+Δu1=0.25">u0=0.0,⇒k(u0)=4,ΔF=1,Δu1=ΔFk(u0)=0.25,u1=u0+Δu1=0.25
Checking the residual,
Fexternal=ΔF=1">Fexternal=ΔF=1
FInternal=u13+9u12+4u1=1.578">FInternal=u31+9u21+4u1=1.578
R(0)=FInternal-Fexternal=0.578">R(0)=FInternal−Fexternal=0.578R(0)=FInternal-Fexternal=0.578
R(0)>10−2⇒">R(0)>10−2⇒Newton-Raphson iterations are necessary.
Calculate the correction to u1=u1(0)">u1=u(0)
δu(1)=−[k(u1(0))]−1⋅R(0)=-(8.6875)-1⋅0.578=-0.0665">δu(1)=−[k(u(0)1)]−1⋅R(0)=−(8.6875)−1⋅0.578=−0.0665
⇒">⇒⇒The updated value of u1(1)=u1(0)+δu(1)=0.25-0.0665=0.1835">u(1)1=u(0)1+δu(1)=0.25−0.0665=0.1835
Checking the residual again,
Fexternal=ΔF=1">Fexternal=ΔF=1
FInternal=(u1(1))3+9(u1(1))2+4u1(1)=1.0432">FInternal=(u(1)1)3+9(u(1)1)2+4u(1)1=1.0432
R(1)=FInternal-Fexternal=0.0432">R(1)=FInternal−Fexternal=0.0432
R(1)>10−2⇒">R(1)>10−2⇒Newton-Raphson iterations are necessary
Calculate the new total correction to u1=u1(0)">u1=u(0)
δu(2)=δu(1)−[k(u1(1))]−1⋅R(1)=-0.0665-(7.404)-1⋅0.0432=-0.0723">δu(2)=δu(1)−[k(u(1)1)]−1⋅R(1)=−0.0665−(7.404)−1⋅0.0432=−0.0723
The updated value of u1(2)=u1(0)+δu(2)=0.25-0.0723=0.1777">u(2)1=u(0)1+δu(2)=0.25−0.0723=0.1777
Checking the residual again,
Fexternal=ΔF=1">Fexternal=ΔF=1
FInternal=(u1(2))3+9(u1(2))2+4u1(2)=1.0006">FInternal=(u(2)1)3+9(u(2)1)2+4u(2)1=1.0006
R(2)=FInternal-Fexternal=0.0006">R(2)=FInternal−Fexternal=0.0006
R(2)<10−2⇒">R(2)<10−2⇒Newton-Raphson iterations are not necessary
Therefore, the final value is u1=u1(2)=0.1777">u1=u(2)1=0.1777
Step-2̲">Step−2
u1=0.1777,⇒k(u1)=7.2933,ΔF=1,Δu2=ΔFk(u1)=0.1371,u2=u1+Δu2=0.3148">u1=0.1777,⇒k(u1)=7.2933,ΔF=1,Δu2=ΔFk(u1)=0.1371,u2=u1+Δu2=0.3148
Checking the residual,
Fexternal=ΔF+ΔF=2">Fexternal=ΔF+ΔF=2
FInternal=u23+9u22+4u2=2.1822">FInternal=u32+9u22+4u2=2.1822
R(0)=FInternal-Fexternal=0.1822">R(0)=FInternal−Fexternal=0.1822
R(0)>10−2⇒">R(0)>10−2⇒Newton-Raphson iterations are necessary.
Calculate the correction to u1=u1(0)">u1=u(0)
δu(1)=−[k(u2(0))]−1⋅R(0)=-(9.9637)-1⋅0.1822=-0.0182">δu(1)=−[k(u(0)2)]−1⋅R(0)=−(9.9637)−1⋅0.1822=−0.0182
⇒">⇒⇒The updated value of u2(1)=u2(0)+δu(1)=0.3148-0.0182=0.2966">u(1)2=u(0)2+δu(1)=0.3148−0.0182=0.2966
Checking the residual again,
Fexternal=ΔF+ΔF=2">Fexternal=ΔF+ΔF=2
FInternal=(u2(1))3+9(u2(1))2+4u2(1)=2.0042">FInternal=(u(1)2)3+9(u(1)2)2+4u(1)2=2.0042
R(1)=FInternal-Fexternal=0.0042">R(1)=FInternal−Fexternal=0.0042
R(1)<10−2⇒">R(1)<10−2⇒Newton-Raphson iterations are not necessary
Therefore, the final value is u2=u2(1)=0.2966">u2=u(1)2=0.2966
Step-3̲">Step−3
u2=0.2966,⇒k(u2)=9.6027,ΔF=1,Δu3=ΔFk(u2)=0.1041,u3=u2+Δu3=0.4007">u2=0.2966,⇒k(u2)=9.6027,ΔF=1,Δu3=ΔFk(u2)=0.1041,u3=u2+Δu3=0.4007
Checking the residual,
Fexternal=ΔF+ΔF+ΔF=3">Fexternal=ΔF+ΔF+ΔF=3
FInternal=u33+9u32+4u3=3.1121">FInternal=u33+9u23+4u3=3.1121
R(0)=FInternal-Fexternal=0.1121">R(0)=FInternal−Fexternal=0.1121
R(0)>10−2⇒">R(0)>10−2⇒Newton-Raphson iterations are necessary.
Calculate the correction to u2=u2(0)">u2=u(0)
δu(2)=−[k(u3(0))]−1⋅R(0)=-(11.6942)-1⋅0.1121=-0.00958">δu(2)=−[k(u(0)3)]−1⋅R(0)=−(11.6942)−1⋅0.1121=−0.00958
⇒">⇒⇒The updated value of
u3(1)=u3(0)+δu(2)=0.4007-0.00958=0.3911">u(1)3=u(0)3+δu(2)=0.4007−0.00958=0.3911
Checking the residual again,
Fexternal=ΔF+ΔF+ΔF=3">Fexternal=ΔF+ΔF+ΔF=3
FInternal=(u3(1))3+9(u3(1))2+4u3(1)=3.0008">FInternal=(u(1)3)3+9(u(1)3)2+4u(1)3=3.0008
R(1)=FInternal-Fexternal=0.0008">R(1)=FInternal−Fexternal=0.0008
R(1)<10−2⇒">R(1)<10−2⇒Newton-Raphson iterations are not necessary
Therefore, the final value is u3=u3(2)=0.3911">u3=u(2)3=0.3911
The results of the implicit analysis are summarized in the following table.
| STEP i | Fexternal=ΔF+ΔF+ΔF=3">ΔFi | ui | (Fext)i | (Fint)i | R=(Fint)i-(Fext)i |
| 1 | 1 | 0.1777 | 1 | 1.0006 | 0.0006 |
| 2 | 1 | 0.2966 | 2 | 2.0042 | 0.0042 |
| 3 | 1 | 0.3911 | 3 | 3.0008 | 0.0008 |
It is evident from the table that the external forces and internal forces are coming into equilibrium.