.1 Introduction The numerical solution of non-linear problems is quite different from linear problems. The solution can not be obtained with a single step but by a superposition of linear solutions. Because the stiffness changes during the loading, a linear solution will in general not satisfy equilibrium and this means that it will not lie on the equilibrium path. In a non-linear analysis it is also not guaranteed that a correct solution can be obtained and in some cases different equilibrium paths are possible. The main problem here is that usually the loading is specified and the displacement computed (load control). In some cases the specified loading may be near or greater than the ultimate load that the structure can sustain. If the laod is near the ultimate load, finding the equilibrium state is difficult, if it is above the ultimate load the no equilibrium state exists. There are a number of solution methods that can be used, among them the incremental solution, the classical Newton-Raphson methods and the arc length method. Newton's method was first published in 1685 in A Treatise of Algebra both Historical and Practical by John Wallis. In 1690, Joseph Raphson published a simplified description in Analysis aequationum universalis. Raphson viewed Newton's method purely as an algebraic method and restricted its use to polynomials, but he describes the method in terms of the successive approximations instead of the more complicated sequence of polynomials used by Newton. Finally, in 1740, Thomas Simpson described Newton's method as an iterative method for solving general nonlinear equations. The solution methods will be explained on the examples given in the previous chapter, where analytical solutions were obtained via displacement control. -1
Incremental solution. Incremental solution In the incremental solution we divide the load into increments is also known as a load factor and apply a repeated solution of ΔP λp where λ 1 Δw K T ΔP The method is explained in the flow chart in Fig..1 for the geometrically non-linear problem in 1.3.3. (Problem 1) and the material non-linear problem in 1.5 (problem ). Compute initial stiffness Initialise displacement,load Equations: Problem 1 Problem K 0 ------- z L L -- w 0 0 P 0 0 K 0 ---------- L Divide Load into load steps P nλp DO i0,n Apply load increment ΔP λp Solve for displ. increment 1 Δw K i ΔP w Accumulate displacements i + 1 w i + Δw Accumulate Load P i + 1 P i + ΔP Compute N or σ ------- zw 1 + i + + --w 1 i+ 1 N i 1 L σ Ai, 1 + E w i + 1 ---------- L if σ Ai, + 1 σ YA σ A, i+ 1 σ YA i+ 1 Equilibrium check P i + 1 N z+ w i + 1 ------------------- + kw L i + 1 0 P i + 1 A ( σ A + σ B ) Update stiffness + + K 0 + ------- ( zw i+ 1 + w i+ 1 ) + ---------- + k L K i 1 L 3 N i 1 if σ i+ 1 σ YA + ------- L K i 1 Fig..1 Flow chart and equations for the incremental method The equilibrium check is optional. -
Incremental solution..1 Numerical solution of problem 1 Here we divide the total load of 100 N into 10 increments. Therefore the load factor λ is 0,1 and ΔP 0, 1 100 10N Assuming k1,35 for the moment the initial stiffness is K 0 ------- z L L -- + k 3, 35 The first value of w is computed by 1 10 w 1 Δw K 0 ΔP -----------, 98mm 335, The corresponding axial force is N 1 -------- ( zw L 1 + w 1 ) 560, 5N The tangent stiffness for the second load step is K 1 L 3 N 1 ------- z L L -- + ------- ( zw 1 + w 1 ) + ----- + k L K 1 35, 045, 0, 58N, mm PN ( ) Incremental solution 30 0 exact solution 10 wmm ( ) 10 0 30 Fig.. Incremental solution of the problem 1-3
Incremental solution We see that the stiffness is reduced due to the geometric non-linear effects. The next increment in displacement is computed by 1 Δw K 1 ΔP and the total displacement after the second step is The normal force in the bar is computed by 387mm, w w 1 + Δw 6, 85mm N 5( 75 + 37, 5) 1187, 5N If we substitute this into the equilibrium equation z+ w P N -------------- + kw L 17, 84 0N which is not equal to the applied load. This means that equilibrium conditions are not fulfilled and that the computed point does not lie on the equilibrium path. If we continue applying increments this drift from the equilibrium path will become more pronounced as show in Fig.... PN ( ) 10 5 wmm ( ) 1,5 5 37,5 50 Fig..3 Incremental solution of problem 1 (k0) The problem gets worse if k is zero. In this case the initial stiffness is K 0 ------- z L L -- N mm -4
Incremental solution The first increment of displacement is computed as w 1 1 10 Δw K 0 ΔP ----- 5mm The corresponding axial force is N 1 900N The tangent stiffness for the second load step is K 1 0, 7 0, 36 0, 9N mm and the second increment of displacement is computed as 1 10 Δw K 1 ΔP ----------- 09, 10, 8mm with the accumulated displacement w w 1 + Δw 15, 8mm The corresponding axial force is N 1 z L -- w 1 1 ----- -- w 1 ----- + L L 165N The tangent stiffnes for the third load step is K 1, 73 0, 866 0, 59 which is negative. We will discuss the implication of this in more detail later. We see that with this method no reasonable solution can be reached for this case... Problem : In this example with the two bars and E 10 4 N, A5mm, L1000mm, σ YA 10N/mm, σ YB 0N/mm we apply a load of 50 N in steps, therefore the load factor is 0,5 and ΔP 15N. The initial stiffness is K 0 ---------- 100N/mm L -5
Newton-Raphson method The first value of w is computed by 1 15 w 1 Δw K 0 ΔP -------- 15mm, 100 The stress in the bar A is σ A σ B E w 1 ----- 1, 5N/mm > 10N mm L which is greater than allowable for bar A. To satisfy the yield condition the stress in bar A has to be reduced to σ A 10N ( mm ). If we check the equilibrium we find that it is not fulfilled: 15 5( 10 + 1, 5) 1, 5 0 The iteration is graphically depicted in Fig..4. P(N) 15 150 100 1 w(mm) Fig..4 Incremental solution of problem It can be seen that the incremental solution procedure is not suitable. The main problem is that at the end of the increment the satisfaction of equilibrium conditions is not guaranteed..3 Newton-Raphson method In the classical Newton-Raphson method we introduce an additional step into the process that checks the equilibrium condition and restores equilibrium by applying the out-of-equilibrium force to the system. The method is explained in the flow chart in Fig..5 and on the problem 1 with k0. The increment of load is reduced to -7. -6
Newton-Raphson method DO i0, Number of iterations Equilibrium check δr i z + w i P i N ------------- i L yes converged? no Update stiffness K i 1 Compute increment in displacement δw i K i δr1 Accumulate displacements w i+ 1 w i + δw i Fig..5 Flow chart of Newton-Raphson solution Step 1: Compute the first increment of displacement and the corresponding axial force in the bar Step : Check equilibrium, compute the residual force with P 1 7 : Here the symbol δ is used for iteration increments. Step 3: Compute updated stiffness 1 7 w 1 K 0 ΔP -- 3, 5mm N 1 700N z + w 1 δr 1 P 1 N -------------- 7 + 6, 0 0, 98N 1 L K 1 0, 5 0, 8 1, N mm Step 4: Apply residual force to the system and compute correction to displacement δw 1 K 1 098, 1 δr1 ----------- 0, 816mm 1, -7
Newton-Raphson method Step 5: Update displacement w 35, 0816, 4, 316mm and compute the corresponding axial force in the bar N 788N Repeat step : Compute residual force δr 7 + 6, 5 0, 48N Repeat step 3: K 0, 63 0, 315 1, 055N mm Repeat Step 4: 1 048, δw K δr -------------- 0, 45mm 1, 055 Step 5: Update displacement w 3 4, 316 0, 45 4, 766mm and compute the corresponding axial force in the bar N 3 86N Repeat step : Compute residual force δr 3 7 + 6, 97 0, 03N The solution process is represented graphically in Fig..6 and should be repeated until a convergence criterion is satisfied. The term convergence means that the exact solution can be obtained after an infinite number of iterations and that the residual force decreases at each iteration. Convergence and the rate of convergence of the method depends on the problem and is not guaranteed. In general, if the load is near the ultimate load then convergence is slow and if it is above the ultimate load then no convergence can be achieved. We will present an example later. A measure of convergence, e, could be the ratio of the residual force at the ith iteration to the residual force at the first iteration, i.e. the criterion for convergence is δr i -------- < e R δr 1-8
Newton-Raphson method where e R is some specified value. 10 PN ( ) 7 ΔR ΔR 1 ΔR 3 w( mm) 4 8 Fig..6 Graphical representation of the Newton-Raphson method When the load is near to the ultimate load this criterion may not be appropriate. For the example above the ratio of the residual forces is 0.03 after iterations, but as can be seen in the graph, there is still a non-negligible increase in the displacement for the last step. Another criterion that can be applied is the ratio of the displacement increment for the ith iteration to the one for the first iteration. δw i --------- < e w δw 1 which for this example is 0.5 after iterations. After obtaining the solution for a load of -7N the load can be incremented further. Assuming ΔP 3N we first calculate the tangent stiffness K T 1, 03 0, 34 0, 63N mm and then the first estimate of the incremental displacement 1 3 Δw K T ΔP ----------- 4, 76mm 063, -9
Newton-Raphson method The first estimate of the total displacement is w w 1 + Δw 4, 76 4, 76 9, 5mm and the corresponding axial force in the bar is The residual force is: 1541N δr 1 10 + 9, 54 0, 46N The updated stiffness is K 1, 3 0, 61 0, 16N mm with this the increment of the displacement is computed as 1 046, δw K ΔR 1 -------------- 016,, 87mm and the next estimate of the total displacement is w 95, 87, 1, 39mm with the corresponding axial force in the bar 1864N and the updated stiffness is K 1, 49 0, 63 0, 1N mm We see that the stiffness becomes negative because we are on the downward leg of the equilibrium curve. As mentioned earlier, a negative stiffness is a paradox and in a problem with more than one degree of freedom would result in an indefinite (or negative definite) stiffness matrix where no meaningful inverse can be determined. It is obvious that this occurs here because the Newton-Raphson method attempts to follow the curve and a solution for a load level of -10N does not exist on the first part of the equilibrium curve. The next solution that exists for this load level is on the upward curve. The method proposed here would not be able to reach this equilibrium point and if one would continue, then the iterations will diverge or converge to a wrong solution. Indeed, this is typical of non-linear problems and adds another complexity to the simulation. In the next section we propose an alternative method that avoids the problem of a negative stiffness. -10
Modified Newton-Raphson.4 Modified Newton-Raphson The only change with respect to the classical Newton-Raphson is that instead of using an updated stiffness the stiffness is kept constant. We can either use the linear stiffness or the updated stiffness of the load step at the beginning of the iterations. We show how the previous example can be solved and we also see that the amount of numerical work is decreased for each iteration as a few steps (re-computation of the stiffness) may be left out. Translated to multiple degree of freedom systems this becomes more significant as a frequent inversion of the stiffness matrix is avoided. In the following we will recompute the example with the modified Newton-Raphson method using the linear stiffness. Here we apply the load of -10N in one step and seek an equilibrium solution, which can be successfully obtained. Compute the first estimate of displacement 1 10 w 1 K 0 ΔP ----- 5mm The corresponding axial force in the bar is The residual force is 900N δr 1 10 + 7,, 8N PN ( ) 10 5 wmm ( ) 1,5 5 37,5 50 Fig..7 Graphical representation of the modified Newton-Raphson method The first correction of the displacement is 1 8, δw 1 K 0 ΔR 1 -------- 1, 4mm -11
Arc length method and the total displacement after 1 iteration w 1 5 1, 4 6.4 As can be seen in Fig..7 this continues for a while until finally an equilibrium condition is reached. We observe that the number of iterations is large but also that unlike the previous example, the equilibrium condition will be reached..5 Arc length method As we have seen, the solution methods discussed so far have difficulties when the load is near to or larger than the ultimate load. In the latter case no solution exists for a given load level and therefore the load must be reduced. Let us re-write the equilibrium equation in a different way Rwλ (, ) λp+ P int 0 where λ is the load factor introduced earlier, P is the external applied load and P int is the internal load. The main essence of the arc length method is therefore that the load factor is no longer constant but becomes a variable. The main idea is to find the intersection of an arc with radius Δl (a specified arc length) with the equilibrium curve, which is obtained using Pythagora s theorem (see Fig..8): Δl Δw + ( ΔλψP) The parameter ψ is required because the load contribution depends on the scaling between load and displacement (i.e. the scale factor between the horizontal and vertical axes). This can be rewritten as a Δw + ( ΔλψP) Δl 0 Using a Taylor series expansion we can write for the equilibrium equation: δr( wλ, ) R R δw + δλ 0 w λ with R w R K T, λ P The Taylor expansion of the equation for a is δa Δwδw + Δλ( ψp) δλ 0-1
Arc length method After substitution of δr R n R o, δa a n a o where the subscripts n and o mean new and old value, we obtain a system of equations. R o a o K T P Δw, Δλ( ψp) δw δλ which can be solved for the change in displacement δλ. P δw and change in load factor δλ 1 R o ΔλP Δλ 1 P Δl Δw δw 1 w Fig..8 Explanation of the arc length method We start the iteration with applying an increment of condition we determine the residual and compute R o ΔλP. Using the equilibrium a 0 Δw + Δλ ψ P Δl With these values the first increment of displacement δw 1 δλ 1 can be obtained. and of the load factor This is repeated until the residual becomes sufficiently small. The process is graphically represented in Fig..8. -13
References The method just described was initially introduced by Riks in 1979. The problem with this approach is that the symmetric property of the stiffness matrix is destroyed and that it is sometimes difficult to determine a value for ψ. Therefore many variants of this method were introduced the most notable ones by Ramm (1981) and Al-Rasby (1991)..6 References Al-Rasby, S. N. (1991). Solution techniques in nonlinear structural analysis. Computers and Structures, 40(4):985-993. Ramm, E. (1981) Strategies for tracing the non-linear response near limit points. In: Nonlinear Finite Element Analysis in Structural Mechanics, Springer, New York. Riks, E. (1979). An incremental approach to the solution of snapping and buckling problems. International Journal of Solids and Structures, 15:59-551 Tjalling J. Ypma, Historical development of the Newton-Raphson method, SIAM Review 37 (4), 531 551, 1995..7 Exercises.7.1 Incremental solution of problem 1 Using MATLAB or similar complete the incremental analysis for k1,35 and k0 and plot the results. Use different increments of the load and comment on the results obtained..7. Incremental solution of problem Using MATLAB or similar to complete the incremental analysis of problem and plot the results. Use different increments of the load and comment on the results obtained..7.3 Newton-Raphson solution of problem 1 Complete the Newton-Raphson analysis for problem 1 and for k0; plot the results. Use different increments of the load and comment on the results obtained. -14
Exercises.7.4 Newton-Raphson solution of problem Complete the Newton-Raphson analysis for problem and plot the results. Use different increments of the load and comment on the results obtained..7.5 Modified Newton-Raphson solution of problem 1 Complete the modified Newton-Raphson analysis for problem 1 and for k0; plot the results. Use different increments of the load and comment on the results obtained..7.6 Modified Newton-Raphson solution of problem Complete the modified Newton-Raphson analysis for problem and plot the results. Use different increments of the load and comment on the results obtained..7.7 Arc length method of problem 1 Apply the arc length method to problem 1 with k0. -15