Week - 2 - Explicit and Implicit Analysis
Consider the force in the bar is a non linear function of displacement
f(U)=U^3+9U^2+4U----(1)
the above function follows the stiffness of the bar is
k(u)=df/du=3u^2+18u+4
Explicit Analysis:
it is possible to plot a graph of force(f) versus displacement(u) numerically by using equation1
One method is to use an incremental explicit load control scheme.
For this example three equal load increments (or steps) of ∆F = 1unit 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 and ∆F respectively. Use is made of the relationship ∆F = k∆u. The analysis proceeds as follow:
STEP1:
U0=0,deltaF=1
K(U0)=3(0)^2+18(0)+4=4,
delta F=k*deltau
now,delta(u1)=deltaF/k(u0)=1/4=0.25
u1=u0+u1=0+0.25=0.25
STEP2:
u1=0.25,deltaF=1
K(0.25)=8.6875
deltau2=1/8.6875=0.115
u2=u1+deltau2=0.25+0.115=0.365
STEP3:
step3 is complted similar to the step1 and step2 the final value is u3=0.456
F(ext)=1+1+1=3
f(int)1=(u1)^3+9(u1)^2+4(u1)=0.25^3+9*0.25^2+4*0.25=1.57
similarly from 2 and 3values aref(int)2=2.69,f(int)3=3.78
it is evident from the results that external and internal forces are not in equilibrium because
F(ext)not equal to f(int)
summery of explicit analysis results
| step(i) | deltaF(i) | deltau(i) | u(i) | F(ext)i | f(int)i | fint-Fext=R |
| 1 | 1 | 0.25 | 0.25 | 1 | 1.57 | 0.57 |
| 2 | 1 | 0.115 | 0.365 | 2 | 2.69 | 0.69 |
| 3 | 1 | 0.091 | 0.456 | 3 | 3.78 | 0.78 |
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
R^j+1=R^j(u^j)+dR(u^j)/du*deltau^j+1.....=R^j+k(u^j)deltau^j+1+....
By neglecting higher order terms in the series, setting Rj+1 = 0 and solving for δuj+1 the
following correction is obtained
deltau^j+1=-[K(u^j)]^-1*R^j.
Notice that the iteration variable is j and it may take several iterations for the norm of the residual, ||R||, to be reduced below the chosen tolerance criteria (for the example below a tolerance of 10^−2is chosen). The implicit analysis proceeds as follows
STEP1:
U0=0,K(U0)=4,deltaF=1
deltau1=1/4=0.25
u1=u0+deltau1=0.25
check the residual
f(int)1=1.57,deltaF=1
R^0=fint-Fext=1.57-1=0.57
R^0>10^-2 newton-raphson iterations Re necessary.
calculate the corrections tou1=u1^(0)
deltaU(1)=-[k(u1)^0]^-1*R^0=[-3*0.25^2+18*0.25+4]^1*0.57=-0.0656
Updated the value of u1^(1)=u1^(0)+deltau(1)=0.25-0.0656=0.1844
check the residuals again,F(ext)=1
f(int)=u1^3+9*u1^2+4*u1=1.0499
R(1)=fint-Fext=1.0499-1=0.0499
calculate the new total corrections tou1=u1(0)
deltau(2)=delta(u)1-[k(u1)]^-1*R1=-0.07454
The updated value of u1(2)=u1(0)deltu(2)=0.25-0.07454=0.17546
check the residuals,fint=0.98431,deltaF=1
R(2)=0.98431-1=0.01569
STEP2:
K(u1)=7.25063
deltau2=0.13791
u2=u1+deltau2=0.17546+0.13791=0.31337
check the residuals,f(int)u2^3+9u2^2+4u2=2.1655
R(0)=2.1655-1=1.1655
calculate the iteration to u2=u2(0)
deltau(1)=-[ku2(0)]^-1*R(0)=-0.11733
the updated value of u2(1)=u2(0)+deltau(1)=0.3133-0.11733=0.196
check the residual
R(1)=fint-Fext=1.137-1=0.1372
R(1)>0.01 another newton raphson iteration is necessary.
calculate the new total correction tou2=u2(0)
deltau(2)=deltau(1)-[ku2(1)]^-1*R(1)=-0.1352
the updated value of u2(2)=u2(0)+deltau(2)=0.313-0.1352=0.1781
check the residual R(2)=1.0034-1=0.00344
R(2)<0.01no further iterations are necessary.
the final value is u2=u2(2)=1.00344
STEP3:
step3 is completed similar to step1 and step2
the final value is u3=1.935
the results of implicit analysis are summarized.
| step(i) | deltaF(i) | u(i) | F(ext)i | f(int)i | f(int)-F(ext)=R |
| 1 | 1 |
0.17546 |
1 | 0.98387 | 0.0161 |
| 2 | 1 | 1.00344 | 2 | 14.0759 | 12.0759 |
| 3 | 1 | 1.9354 | 3 | 48.702 | 45.7025 |
PLOT RESULTS:
A plot of the results is shown in graph It is evident from the plot that the explicit analysis drifts from the exact solution.However, even with more increments the explicit analysis still drifts from the exact solution. The drift from the exact solution illustrates the lack of equilibrium between the internal and external forces.To correct for this problem an implicit analysis is required. The results for the implicit analysis are
shown in the graph. The Newton-Raphson iterations correct the incremental steps so that they land on the exact solution according to the specified tolerance. Excellent agreement is achieved with just 3 increments and in each step only two Newton-Raphson iterations are required.
CONCLUSION:
A nonlinear analysis is illustrated using an incremental explicit and implicit procedure.The results are plotted for comparison. It is evident from the results that the explicit analysis drifts from the true solution.To overcome this problem an implicit analysis is used,which includes Newton-Raphson iterations to enforce equilibrium between internal and external forces. The techniques demonstrated are for a load control scheme.