For each increment of load, data is accepted by INCLOD to control the upper limit to the number of time steps, the output frequency, the size of load increment and the convergence tolerance limit. These quantities are specifically input as:

NSTEP	Maximum permissible number of time steps. This is a safety measure to cover situations where steady state conditions are not achieved. After performing NSTEP time steps the program will then stop.
NOUTP	This parameter controls the frequency of output of results:0—Print the results on convergence to steady state conditions only, for each load increment.1—Print the results after the first time step and at steady state, for each load increment.2—Print the results for each time step for each load increment.
FACTO	This quantity controls the magnitude of any load increment. The applied loading is accepted by subroutine DATA and stored in array RLOAD. The size of any load increment is then RLOAD factored by FACTO. Therefore if FACTO is input for the first three increments as respectively 0·3, 0·3 and 0·1, the total loading applied to the structure during the third increment is 0·7 times the loading input in subroutine DATA.
TOLER	This item of data controls the tolerance permitted on the steady state convergence process, and has been described in Section 4.9.

Subject to the replacement of NITER by NSTEP, the form of this subroutine for the present application is identical to that provided in Section 3.7. # 4.11 The main, master or controlling segment This master segment controls the calling, in order, of the other sub-routines. This program segment also controls the time-stepping process and also the incrementing of the applied loads, where appropriate. The following channel numbers are employed by the program: 5 (card reader), 6 (line printer), 1 (scratch file).

MASTER UNVISC	UVIS	1
C**********	UVIS	2
C	UVIS	3
C *** PROGRAM FOR THE 1-D SOLUTION OF NONLINEAR PROBLEMS	UVIS	4
C	UVIS	5
C**********	UVIS	6
COMMON/UNIM1/NPOIN,NELEM,NBOUN,NLOAD,NPROP,NNODE,IINCS,ISTEP,	UVIS	7
KRESL,NCHEK,TOLER,NALGO,NSVAB,NDOFN,NINCS,NEVAB,	UVIS	8
NSTEP,NOUTP,FACTO,TAUFT,DTINT,FTIME,FIRST,PVALU,	UVIS	9
DTIME,TTIME	UVIS	10
COMMON/UNIM2/PROPS(5,5),COORD(26),LNODS(25,2),IFPRE(52),	UVIS	11
FIXED(52),TLOAD(25,4),RLOAD(25,4),ELOAD(25,4),	UVIS	12
MATNO(25),STRES(25,2),PLAST(25),XDISP(52),	UVIS	13
TDISP(26,2),TREAC(26,2),ASTIF(52,52),ASLOD(52),	UVIS	14
REACT(52),FRESV(1352),PEFIX(52),ESTIF(4,4),VIVEL(25)	UVIS	15

TTIME=0.0	UVIS 16
CALL DATA	UVIS 17
CALL INITIAL	UVIS 18
CALL STUNVP	UVIS 19
DO 30 IINCS=1,NINCS	UVIS 20
CALL INCLOD	UVIS 21
DTIME=0.0	UVIS 22
DO 10 ISTEP=1,NSTEP	UVIS 23
TTIME=TTIME+DTIME	UVIS 24
CALL NONAL	UVIS 25
CALL ASSEMB	UVIS 26
IF(KRESL.EQ.1) CALL GREDUC	UVIS 27
IF(KRESL.EQ.2) CALL RESOLV	UVIS 28
CALL BAKSUB	UVIS 29
CALL INCVP	UVIS 30
CALL CONVP	UVIS 31
IF(NCHEK.EQ.0) GO TO 20	UVIS 32
IF(ISTEP.EQ.1.AND.NOUTP.EQ.1) CALL RESULT	UVIS 33
IF(NOUTP.EQ.2) CALL RESULT	UVIS 34
10 CONTINUE	UVIS 35
WRITE(6,900)	UVIS 36
900 FORMAT(1H0,5X,'STEADY STATE NOT ACHIEVED')	UVIS 37
STOP	UVIS 38
20 CALL RESULT	UVIS 39
30 CONTINUE	UVIS 40
STOP	UVIS 41
END	UVIS 42

UVIS 16 Initialise the total time to zero. UVIS 17 Call the subroutine which reads the input data as described in Section 3.2. UVIS 18 Call Subroutine INITIAL which: (i) Initialises to zero the viscoplastic strain vector and the stress vector. (ii) Initialises the array, ELOAD, which will contain the pseudo loads to be applied during each time step. (iii) Initialises the vector of applied loads. (iv) Initialises the vector of total displacements and total reactions. UVIS 19 Call the subroutine which evaluates the stiffness matrix for each element. UVIS 20 Enter the DO LOOP over the number of load increments. UVIS 21 Call Subroutine INCLOD which: (i) Reads and writes the input data required for each load increment as described previously in Section 4.10. (ii) Adds the current increment of load into the pseudo load vector, ELOAD, and into the total applied load vector, TLOAD. UVIS 23 Begin the time-stepping process. UVIS 24 Calculate the total time elapsed (note that the first time step corresponds to the elastic solution). UVIS 25 Call the subroutine which sets the parameter KRESL controlling equation resolution facility. UVIS 26-29 Call the subroutines which assemble the element stiffnesses and solve for the unknown displacements and reactions. UVIS 30 Call the subroutine which evaluates quantities at the end of the time step and evaluates the loads for the next time step. UVIS 31 Check whether or not steady state conditions have been achieved. UVIS 32 If so, terminate the time-stepping process for the current load increment. UVIS 33–34 Output the results at a frequency controlled by parameter, NOUTP. UVIS 35 End of time-stepping loop. UVIS 36–38 If steady state conditions have not been achieved when the upper time-step limit has been reached, write a message and terminate the execution. UVIS 40 End of load increment loop. # 4.12 Numerical examples The first example considered is the viscoplastic deformation of a single element under constant applied loading. The element is of length 10 units and the applied load is 15 units. The material properties assumed are included in Fig. 4.5, where it is noted that the strain hardening parameter is taken to be zero. The finite element prediction is seen to be in excellent agreement with the theoretical result (4.17) for this problem. The problem was then reanalysed for a strain-hardening material with $H' = 5000$ . The finite element results are compared with the theoretical expression (4.16) in Fig. 4.6 for three different values of the time-stepping parameter, $\tau$ , defined in Section 4.4. For a value of $\tau = 0.01$ excellent agreement is obtained, but as the time-step length is increased ( $\tau = 0.05$ and $\tau = 0.1$ ) comparison with the theoretical solution deteriorates. In particular, an increase in the time-step length progressively overestimates the viscoplastic strain increment, which is a characteristic of the Euler method of time stepping. It is noted that the time-step length is not so critical in the perfectly viscoplastic case of Fig. 4.5 since the exact viscoplastic strain increment is in fact linear for this case. For the material properties assumed, the theoretical value of the limiting time step is given from (4.36) to be 1·0. It is seen from Figs. 4.5 and 4.6 that the time-step lengths employed in solution are well within this critical value. However, Fig. 4.6 shows that to achieve an accurate result even smaller time-step lengths must be taken. Thus although the theoretical value of the limiting time-step length guarantees numerical stability of the solution process it may not always lead to an accurate solution. The second example considered illustrates the redistribution of stress with time which generally takes place in viscoplastic problems. Figure 4.7 shows two members in parallel which are subjected to an end load P which 

line

| Time | End displacement, φ₂ | |------|---------------------| | 0.0 | 0.015 | | 0.1 | 0.020 | | 0.2 | 0.025 | | 0.3 | 0.030 | | 0.4 | 0.035 | | 0.5 | 0.040 | | 0.6 | 0.045 | | 0.7 | 0.050 | | 0.8 | 0.055 | | 0.9 | 0.060 | | 1.0 | 0.065 |

Fig. 4.5 End displacement with time for a single viscoplastic element under constant applied load—No strain hardening. 

line

| Time | End displacement (Theoretical) | Finite element τ=0.1 Toler=0.1% | Finite element τ=0.05 | Finite element τ=0.01 | |------|----------------------------------|----------------------------------|------------------------|------------------------| | 0.0 | 0.016 | 0.015 | 0.016 | 0.016 | | 0.1 | 0.018 | 0.018 | 0.019 | 0.019 | | 0.2 | 0.020 | 0.020 | 0.021 | 0.021 | | 0.3 | 0.022 | 0.022 | 0.023 | 0.023 | | 0.4 | 0.023 | 0.023 | 0.024 | 0.024 | | 0.5 | 0.024 | 0.024 | 0.024 | 0.024 | | 0.6 | 0.024 | 0.024 | 0.024 | 0.024 | | 0.7 | 0.024 | 0.024 | 0.024 | 0.024 | | 0.8 | 0.024 | 0.024 | 0.024 | 0.024 | | 0.9 | 0.024 | 0.024 | 0.024 | 0.024 | | 1.0 | 0.024 | 0.024 | 0.024 | 0.024 |

Fig. 4.6 End displacement with time for a single viscoplastic element under constant applied load showing finite element results for different time-step lengths—Linear strain hardening. 

line

| Time | End displacement, φ₃ | |-------|----------------------| | 0.00 | 0.0011 | | 0.05 | 0.0012 | | 0.10 | 0.0012 | | 0.15 | 0.0014 | | 0.20 | 0.0015 | | 0.25 | 0.0015 | | 0.30 | 0.0016 | | 0.35 | 0.0017 | | 0.40 | 0.0017 | | 0.45 | 0.0018 | | 0.50 | 0.0019 | | 0.55 | 0.0020 |

Fig. 4.7 End displacement with time for an elasto-viscoplastic parallel bar model subjected to an incrementally applied end load showing the attainment of steady state conditions. is incrementally applied. The material properties for each element are included in Fig. 4.7 with the only difference between the two members being the initial yield stress of the materials. The load is applied in four increments and steady state conditions are allowed to develop for each increment before application of further load. The end displacement with time is shown in Fig. 4.7. Steady state conditions are achieved for the first three load increments but not for the fourth since both elements, which behave perfectly plastically, have become yielded at this stage. # 4.13 Problems 4.1 Develop the relationship between the applied stress, $\sigma$ , and the total strain, $\epsilon$ , for the rheological model shown in Fig. 4.8. Plot the strain response with time when the model is subjected to a constant applied stress, $\sigma_{A}$ . 4.2 Repeat Problem 4.1 for the rheological model shown in Fig. 4.9. In this case the friction slider becomes active for $\sigma \geqslant Y$ where, for a linear strain hardening material, $Y = \sigma_{Y} + H' \epsilon_{vp}$ . 

text_image

E₁ E₂ γ σ σ ε

Fig. 4.8 Problem 4.1. 

text_image

E1 E2 Y γ1 γ2 σ εa εw εvp ε

Fig. 4.9 Problem 4.2. · 4.3 Use the unidimensional computer code developed in this chapter to determine the stress relaxation with time when the Maxwell model shown in Fig. 4.10 is subjected to a constant displacement condition. The critical time-step length for this model can be shown to be $\Delta t = 2 / \gamma E$ . Solve the problem for several time-step lengths up to the critical value, thereby showing that numerical divergence occurs as soon as the limiting value is reached. For computation let $E = 100$ , $\gamma = 0.01$ and $\phi_p = 0.1$ . 

text_image

σ E γ φp

Fig. 4.10 Problem 4.3. 4.4 Modify the computer code developed in this chapter to allow solution of the material model of Problem 4.1. 4.5 In Section 4.9, Subroutine CONVP, monitoring convergence to steady state conditions, was based on a global criterion. Modify this subroutine so that convergence is based upon the condition $$ \frac {\left| \Delta \epsilon_ {v p} {} ^ {n} \right|}{\left| \Delta \epsilon_ {v p} {} ^ {1} \right|} \times 1 0 0 \leqslant \text { TOLER }, \tag {4.42} $$ for each individual element. 4.6 Develop the elastic stiffness matrix, $K^{(e)}$ , for a two-node finite element in the form of a sphere and which is to be subjected to spherically symmetrical radial loading only. Assume a linear variation between nodes and note the following relationships $$ \epsilon_ {r} = \frac {\partial u}{\partial r} = \frac {1}{E} [ \sigma_ {r} - \nu (\sigma_ {\theta} + \sigma_ {\phi}) ]; \quad \sigma_ {\theta} = \sigma_ {\phi}; $$ $$ \epsilon_ {\theta} = \epsilon_ {\phi} = \frac {u}{r} = \frac {1}{E} [ (1 - \nu) \sigma_ {\theta} - \nu \sigma_ {r} ], \tag {4.43} $$ in which u is the radial displacement and $\epsilon_{r}$ , $\epsilon_{\theta}$ , $\epsilon_{\phi}$ and $\sigma_{r}$ , $\sigma_{\theta}$ , $\sigma_{\phi}$ are respectively the strain and stress components. Also express the stress components in terms of the nodal displacements. 4.7 Use the stiffness matrix evaluated in Problem 4.6 to modify the one-dimensional viscoplastic program UNVIS to allow solution of spherically symmetrical problems. Assume a Tresca yield criterion which implies commencement of yielding when $\sigma_{r}-\sigma_{\theta}=\sigma_{Y}$ . 4.8 Employ the program developed in Problem 4.7 to determine the variation of the elasto-viscoplastic stress distribution with time in a sphere which is instantaneously loaded by an internal pressure of 500 N/mm $^{2}$ . The internal and external radii of the sphere are 10 cm and 25 cm respectively, the elastic modulus $E = 2 \times 10^{5} \, N/mm^{2}$ , Poisson's ratio $\nu = 0 \cdot 3$ , the uniaxial yield stress $\sigma_{Y} = 300 \, N/mm^{2}$ , hardening parameter, $H' = 0$ and take the fluidity parameter $\gamma = 0 \cdot 001$ . Compare your steady state solution with the theoretical elasto-plastic results of Ref. 2. # 4.14 References 1. CORMEAU, I., Numerical stability in quasistatic elasto-visco-plasticity, Int. J. Num. Meth. Engng., 9, 109–127 (1975). 2. HILL, R., The Mathematical Theory of Plasticity, Oxford University Press, 1950.