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1. jal 1 n Lisia 2 6 Saiz j l Since there are many reactions R and just one force P in our example the easiest way to obtain strain energy variation is through the second of these two equations Starting from step 12 where the com puted deflection under force P was 6 3 33 mm we obtain step 12 3 33 mm 1 1x10x3 33x107 16 7 107 kNm step 11 9 3 05 mm 14 1x10x 3 33 3 05 x107 1 4x107 kNm L 16 7 1 4 x107 18 1x107 kNm o 1x10x1 2 17 1 75 x107 2 1x107 kNm step 6 1 75 mm L 1 22 54 2 1 x107 24 46x107 kNm The strain energy variation calculated in this way has been shown in a diagram in Fig 6 There are some interesting features to be observed in Fig 5 and 6 namely One can see that the elimination of the released DOF s takes place in a quite regular frontal man ner bearing in this respect some resemblances to other known elimination processes e g to the Gauss elimination DOF fixity which becomes definitely eliminated is here always the one which binds the big gest strain energy In more complex systems this will apply to certain groups of DOF s rather than the single DOF s which can make it less visible e We see that the total number of fixed DOF s does not always become smaller at every step Neither the total strain energy of these DOF s does always become smaller See e g the transition from step 7 to
2. Graphical DISCO is not a graphically orientated program Its graphical facilities serve only the purpose of a global control Also in this case the Graphical option will only prompt a view of the structure defor mations enlarged for a good survey The user can chose between e Graphical view of only the deformed structure model e Graphical view of in the background undeformed and in the foreground deformed model The deformed models are plotted using joint deformations only The members connecting those joints are drawn as straight lines You will therefore see polygonal lines instead of curves which is an obvi ous simplification for all structure types except the trusses The option Graphical gives only a screen presentation no hard prints can be obtained Exit It is possible to run the Output block as many times and in as many options as one wishes However with the exception of Graphical you should take care that the final output version is the last which has been processed anyhow the last which has been saved on the disk Do not run Output e g to check one thing still after you have completed the final version because DISCO will overwrite it then Press E for Exit after completing the output This will end the session with the following message Your data output file C DANCE LastDat txt Your solution file C DANCE LastSol txt Thank you Strike a key _ Pressing any key
3. uses a basic solution where the following limi tations and other assumptions have been applied e Structure model consists of straight one dimensional elements members Shells plates etc can not be modeled directly and must be simulated using members e All structure elements have default rigid or pinned nodes joints between each other depended on the type of the structure Trusses are assumed to have pinned joints all other structures are assumed to have rigid joints Modeling a hinge a slide joint etc in a structure of default rigid joints e g a frame can be done using simulation members e User can choose between the following 12 types of structures 1 Continuous beam 2 Discontinuous beam 3 Continuous plane truss 4 Discontinuous plane truss 5 Continuous grid 6 Discontinuous grid 7 Continuous plane frame 8 Discontinuous plane frame 9 Continuous space truss 10 Discontinuous space truss 11 Continuous space frame 12 Discontinuous space frame Loadings can be concentrated forces and or moments in joints or distributed over an entire member length All loads are stored in the same data files as the system data Joint loads in externally fixed directions are acceptable e Every joint can in principle be loaded by a pointed load force or moment in any direction but on the other hand to input such a load one must first define there a joint Also every member can carry a distributed load in any di
4. Dy m 1E3 My kNm Dz m 1E3 Joint 17 0 0000 13 0012 10 1731 31 3101 0 1260 Mz extr 1 3698 26 25221 13 7671 7 9211 0 2171 extr 0 5270 21 4035 11 5728 22 3119 0 1377 Dz extr 3 1404 3 9309 18 1855 222 919 0 0538 Joint 38 3 3800 2 5960 18 7718 26 4025 0 0560 Step 0 0 0000 13 0012 10 1731 ron 0 1260 Step 1 0 4225 20 0556 11 2967 24 0961 0 1372 Step 2 0 8450 24 5375 12 4080 16 8820 0 1342 Step 3 1 2675 26 4467 13 5041 9 6679 0 1213 Step 4 1 6900 25 7833 14 5839 2 4538 0 1026 Step 5 2125 22 5474 15 6477 4 7602 0 0825 Step 6 2 5350 16 7389 16 6977 11 9743 0 0652 Step 7 2 9575 8 3578 17 7373 19 1884 0 0550 Step 8 3 3800 2 5960 18 7718 26 4025 0 0560 Member 43 x m Mz kNm Dy m 1E3 My kNm Dz m 1E3 Joint 18 0 0000 23 7466 11 4167 2207 0 1271 Mz extr 2 1233 32 2229 17 2865 6 3565 0 0775 Dz extr 0 4711 27 0906 12 7582 22 7987 0 1365 Dz extr 3 0551 30 5906 19 7075 22 7987 0 0527 Joint 40 3 3800 29 2540 20 5295 28 5315 0 0570 0 stop 1 10 jump In the program the extreme bending moments have been computed using analytical approach which is in fact quite simple The computation of extreme deflections is however not simple from the program ming point of view DISCO uses a modern iteration method here called Illinois iteration 9 which is a modified very fast version of a classical regula falsi A discussion on this matter goes however be yond the su
5. E5 E E5 E E5 E E5 E E5 E E5 E E5 E E5 E E5 E E5 E E5 E E5 E mH k k KO O LU G ils OX O M OT KA y H 0 OT LU un m QO OT S NN HE M SN LO O G 0 OT co 4 Lo Ln O1 OT OT OT OT OT OT OT OT KA A A KA EY AA S S MKM KM KM KM KM KM NW PR Fe OO d 01 O W OT H NO o d OY MS Regarding the input of real values please observe the follovving e The Anglo Saxon notation should be used 1 e with decimal points and not decimal commas If you use a decimal point DISCO expects at least one digit behind it For the notations like 10 an error message will be returned e You can choose between the decimal e g 9 81 and exponential e g 0 981E1 notation by each individual data Both positive and negative exponents are acceptable As the output will be returned in a decimal notation it is advisable to choose such force and length units that the results will form very long numbers E g for a bridge it is better to input in the data kiloNewtons kN and meters m than in Newtons N and millimeters 34 Table 10 Format of data files for space frames Space frames files CSF TXT and DSF TXT General Example Space frame Figure
6. This can be solved by defining a number of conditional point wise supports in a line or over a surface Such models are in fact powerful tools in solving numerous structural contact prob lems see e g Fig 2 e and f what has been illustrated in a practical case in section D Contact problems have been studied broadly in recent years A linear incremental approach to such problems has e g been presented by Simunovic and Saigal in 5 Other approaches can e g be found in the works of Re faat and Meguid 6 or Wang and Nakamachi 7 6 3 PROPERTIES OF DISCONTINUOUS LINEAR MODELS Polygonally linear here called discontinuous models have some special properties which distinguish them from conventional smooth linear continuous models The most favored property is obviously that they allow for more adequate structural analyses Having seen the examples in Fig 2 and 3 it would not be wrong to say that most structures behave more or less polygonally The fact that they are usually subject to smooth modeling can be justified by limited precision requirements narrow ranges of load variation our convenience tradition etc a The second property is a warning Separately computed load cases 7 should as a rule not be combined The principle of superposition o t should be considered false in discontinuous analyses Observe what happens to the beam from the beginning of this manual Fig 1 b when sepa
7. internal discontinuities and some other problems is possible and will briefly be discussed later on Engineering practice proves that a great majority of discontinuities occur at transition points between negative and positive fixities of diverse degrees of freedom DOF s For clarity reasons this analysis is limited to such cases It is however a minor problem to program the levels of discontinuity as input data so that other than zero transition points can be defined Such modification requires no further changes to the presented algorithm In Fig 2 a number of polygonally linear problems have schematically been shown Cases a until f repre sent discontinuous fixities of reaction forces in various directions cases g and h are examples of discon tinuous moment fixities A short description follows below a Suspensions of pipelines tie rod fixities of expansion joints b Rails crane driveways with limited tension fixities c Cable stayed bridges bustle pipes of steel works blast furnaces d Cable stayed masts towers halls etc e Support rings e g of vertical pressure vessels free laid foundation grids L Concentrated loads on refractory lined oven shells traffic tunnels etc g Single sided moment fixities of columns and beams torsion fixity of a blade h Examples of discontinuous moment fixity in reinforced concrete Fig 2 Examples of discontinuous fixity problems 2 MODELING DISCONTINUOS FIXTTIES Note tha
8. ous and discontinuous data files not mentioning their names is that the continuous joint types Tci change into the discontinuous ones Tp as shown in section 10 3 The data in brackets are optional If you e g only want to enter the second of them you must obviously enter a zero for the first Table 6 Format of data files for beams and plane trusses Beams files CBE TXT and DBE TXT Plane trusses files CBE TXT and DBE TXT General Example General Example Beam Figure 11 Truss Figure 116 Profect ear ae 6 Project kNm Fu Lu n Fu Lun m free line 2 0 free line 2 0 0 0 0 T Xi Mz S S T Xi Y Fa Fr ae 0 0 100 To Xo Fe Mza eae 45 T X Y Fx Fre 11 lo 5 0 5 314 1 12 0 0 0 0 100 T Xi Fy Mz 1 16 10 T Xi Yi Fyi Fy 11 15 0 5 0 118 0 0 0 ONES 2E4 0 1 21 0 5 0 50 50 Tn Xn Fyn Mz 264 4 Th Xn Yn Fyn Fyn 4 24 0 0 0 free line 2E4 free line El qyx bi el 2 7 o EL 4x2 z EA 5 7 166 5 234 7 9 16 qy b EAy 2 4 166 4 6 1866 Ba OoXCocGckua aaa ae Elin ym bm EA 1 2 5E5 free line 2 3 2E5 on 3 4 2E5 li ja js 4 5 2E5 5 6 2E5 6 7 2E5 7 8 2E5 8 9 5E5 31 Table 7 Format of data files for grids Grids files CGR TXT and DGR TXT
9. tending to open the gate Due to the complex torsional rigidity of the gate it is impossible to deter mine the effective contact length directly as an input data This problem has been solved using a row of conditional elastic supports which can only bear compression see joints no 41 through 60 When ten sion is computed these joints become released to show the widths of a leakage gap between the two leaves The gap between the bottom members and the threshold is computed simultaneously Table 11 presents some input excerpts for a negative water head fall of 1 0 m which is a maximum for this lock operation By larger falls 1 0 to 3 0 m the navigation holds up the negatively faced gate goes open and the opposite positively faced gate bears the entire load Yet the negative fall of 1 0 m presents a more severe problem due to the leakage For demonstration reasons some more types of discontinuous fixities have been modeled The second one concerns the fixity of a rotation angle Az along the contact line In the upper part above water the buffer is in fact twice as wide as underneath in order to sustain frequent prestression When the whole gate deflects the compressed contact line will have a fixed rotation about the Z axis Since the E modulus of UHMPE is low 300 500 N mm this effect can practically be ignored but let s assume that we like to see it in the stiffer upper part only It can be done by fixing both sides of the rotati
10. 11f Project kN m 18 29 Fu Lunm free line 57 5 0 2 9 0 0 T Xi Y Zi Fxi Fy Fz Mx My Mazi 5 E 100 To X Y Z Fx Fr Fz Mx My Mz 25 0 0 5 8 0 0 sas 41 5 0 2 9 0 0 T Xi Yi Zi Fxi Fry Fz My Mz 41 5 0 2 3 0 0 57 1040 558 0 4 0 1 4 5 216 22 50 U 50 Tn Xn Yn Zn Fyn Fyn Fzn Myx My Mz 1 4 5 2 6 2 0 0 80 free line 1 4 0 2 3 4 0 b e EA Gla El Ela ay 41 S 7 150 0 0 120 b e Gle El El qx qv qz dr 00 46 40 1 4 0 2 3 4 0 b e EA El Ely ax ay dwl 1 4 0 0 0 4 0 0 0 0 120 L 4 02 2223 400 253 1 0 0 4 6 4 0 bin em EAxm Glin Elyn Elan dYm 1 0 0 Qu 5 5 60 free line ji j2 j3 ja iz 1E 1E 1E 1E 1E BE BE BE BE BE BE BE BE H G a HS S A S S LO G ARS O OT G CO 0 OT HB O KA 1 0 OT HS C KA 7 0 OT S re OY ND o co OO KA 10 OT S 9 N Ln LH O1 Ln Ln O1 Ln Ln N No N KA K K NS Ln OT OT Ln OT tri tri tri tri iri tri m tri iri tri tri iri tri iri tri iri iri iri iri iri iri ti ir WWWWWWWWW C LO 9 LO LO LO C WW WwW Ww No N N N oo OO 0 KA N NN NN ri tl iri iri tri ti iri irl tri ti iri irl ti ti iri irl ti ti iri irl ti 169 ir ri ri ri tri r L
11. 55 13300 000 100 000 1000 000 15 900 73 36 56 13300 000 100 000 1000 000 15 900 74 37 57 13300 000 100 000 1000 000 15 900 75 38 58 11500 000 100 000 1000 000 13 700 76 39 59 9500 000 100 000 1000 000 11 500 77 40 60 18000 000 100 000 1000 000 21 600 Joint loadings Joint Type FX kN FY kN FZ kN MX kNm Y kNm MZ kNm ib il 0 000 1000 000 0 000 0 000 0 000 0 000 15 1 0 000 0 000 220 000 0 000 0 000 0 000 Member loadings Member From To QX kN QY kN QZ kN 33 UES 10 590 31 830 0 000 34 15 20 5 280 15 870 0 000 35 20 27 5 280 15 870 0 000 36 6 16 22 950 69 000 0 000 37 16 32 m22 950 69 000 0 000 38 7 d k 4 180 1 2 5070 0 000 39 8 12 1 090 3 280 0 000 40 l 10 880 32 710 0 000 41 12 18 2 920 8 780 0 000 42 17 38 15 380 46 220 0 000 43 18 40 4 010 12 060 0 000 Members for extended output 42 43 DISCO needs 5 iteration steps to solve this sample problem The computation time on a 133 MHz Intel Pentium PC is about 120 sec This time was measured in the late 1990 s There has been much pro gress in microprocessor speeds since then therefore only a fraction of this time will be required today The performances of this range are typical for problems of medium until high complexity which may be considered the case here due to the 41 discontinuous fixities The solution is numerically stable there are e g no visible inaccuracies or traceable differences between the totals of
12. 9 0 0 75 3 0 9 0 20 E3 E3 E E3 E E3 E E3 E E3 E E5 E E5 E E5 E E5 E E5 E E5 E E5 E E5 E E5 E E5 E E5 E E5 E E5 E ae E F E E E4 E4 E4 E4 E4 E3 16 0 E3 18 0 E3 20 0 E3 E3 E3 E4 E4 ES 5 E5 1 No No KA KA M No No KM KM KM GO GO OO ONDAN M M C 01 Q dA CO OTD UB WN H S H 1 Ln re LA N Pe m S OT OT OT OT O O OY OY FU a m 0 COT OO O O OOOO C Co CD Co COK DED N m Ww La H Table 9 Format of data files for space trusses Space trusses files CST TXT and DST TXT General Project Fu Lunm free line T Xi Y Z Fxi Fy Fz T2 X Y 22 Fx Fn Fz Continued on the next page Example Space truss Figure lle kN m 15 39 8 0 20 20 0 0 0 40 8 0 0 2 0 0 0 0 0 120 1 0 0 0 0 5 0 50 1 4 0 2 0 0 0 0 0 80 1 4 0 220 050 1 4 0 0 0 5 0 0 30 33 Table 9 continued TRE T bi 0 1 8 0 0 0 5 0 0 30 Tn Xn Yn Za Fxn Fxa Fm 11 12 0 2 0 0 0 0 0 80 0020 b ei LA 8 16 0 2 0 0 0 0 90 40 b e2 g TEs 0 10 20 c oe 16 0 0 0 5 0 50 b T 5 E E5 E E5 E E5 E E5 E E5 E E5 E E5 E E5 E E5 E E5 E ae E e E F E a E E a E P E F E E E ae E E F E E E F E as E
13. As discussed in section 5 DISCO performs structural analyses for structure models of twelve different types Therefore you should first choose the type which suits your problem the best Keep in mind that the higher your structure type number will be the more complex and memory consuming computation it will require In extreme cases i e by very large space frame models the program may even run out of memory Special program architecture and the use of a band matrix optimization take care that this does not happen soon Exact limits can not be given but space frames up to about 150 nodes joints and 200 members should in general successfully be computed For other types of structures these limits usu ally exceed 800 The only programmed limitation is no more than 999 joints and 999 members There are six basic types of structures to be chosen from Fig 11 divided into two groups as follows 1 Continuous beam 2 Discontinuous beam 3 Continuous plane truss 4 Discontinuous plane truss 5 Continuous grid 6 Discontinuous grid 7 Continuous plane frame 8 Discontinuous plane frame 9 Continuous space truss 10 Discontinuous space truss 11 Continuous space frame 12 Discontinuous space frame Fig 11 Six basic types of structures examples 24 The structure geometry must be input in a global right handed Cartesian coordinate system For the types 1 8 a d in Fig 11 the position of the global coordinate axes is partly p
14. brings you back to the operation system You can now get the data file LastDat txt and the solution file LastSol txt from the hard disk using any word editor e g WordPad standard present by MS Windows and make hard prints of those files These prints are skipped in this manual for space reasons There is however an input data file cpf5 txt on the attached diskette You can run DISCO and process this file by yourself if you like to see the output 42 14 SAMPLE PROBLEM LEAKAGE OF A LOCK GATE The complex system of water management in the Netherlands faces the designers with still higher de mands One of them is the construction of locks with mitre gates which can bear water pressure from both sides the pointed so called positive side and the concave so called negative side The second load case is unfavorable as water tends then to open the gate instead of as in the first case keeping it closed Despite leakage problems the idea wins still more support since it reduces the number of neces sary mitre gate leaves The most recent project where this idea has been used is the check gates of the double lock aqueduct over a motorway Naviduct Enkhuizen This remarkable project gives a free navigation passage between two large lakes IJsselmeer and Markermeer which originate from the damming of the ancient Dutch internal see Zuiderzee in the early 1930 s The construction of this project was completed in 2003 The detailed design of
15. continuous approach both linear and nonlinear gives a tensile reaction on the third support which is obviously an error Moreover there is another quite principal argument in favor of discontinuous approach Nature behaves in fact non linear in a majority of problems For no other reason than our own convenience we often approximate it by using linear or piecewise linear ap proach The family of problems dealt with by DISCO represents a rather exceptional opposite case Here the nature itself behaves piecewise linear polygonal There is no need to approximate it it can be modeled the way it behaves In mathematical sense polygon angles are slope discontinuities There fore we will refer to them as discontinuities In engineering it is quite usual to refer in such a way to sudden sharp changes in structure properties without explicit mentioning the word slope see e g dis cussions on joints in shells of revolution by Roark 3 or by Bull 4 In this sense one can distinguish various types of discontinuities or discontinuous fixities e external e g discontinuous support conditions e internal e g ties or unfixed contacts between elements e mechanical caused by geometry or mechanical properties e physical caused by material properties e g plastic hinges fractural irreversible caused by breakage etc The DISCO algorithm presents a solution for structures with external discontinuities An extension for
16. data Joint il t 4 X 6 000 Y 0 200 0 000 FY 0 000 MZ 0 000 The same four options appear now in the top line Pressing F1 repeats the dialogue about this particular joint other options work as discussed above Press any other key to go to joint 2 then joint 3 etc When all joint data has been input and you have not taken a break by pressing F2 DISCO will go to the third part of the Input dialogue the member data You will now be asked to input in succession the member beginning and end joint the axial rigidity EAx the flexural rigidity El and the member distributed loads At the end of the first member input the screen will look like this Input 1 Repeat F2 Finish Escape Other Go on Cpf6 txt ember data Member il from 1 to 7 Elz 1 000 EAx 2000 000 QY 0 000 QX 0 000 37 The keys F1 F2 Esc and any other allow again for respectively repeating the member data input breaking the job erasing the entire new input and continuing the dialogue Press any other key to go to member 2 then member 3 etc When all member data has been input and you have not taken a break by pressing F2 DISCO will go to the fourth and last part of the Input dialogue the members for extended output You will be asked to enter the numbers of members for which you like to receive the locations and values of extreme deflections and or bending moments Entering those members proceeds in a dia logue sim
17. details on the behavior of some members All these parts must be separated from each other by a single free line The data file parts are 29 e General data project and or structure name force and length units total numbers of joints and members e Joint data per input line joint type joint global coordinates joint concentrated loads e Member data per input line beginning and end joint sectional rigidities member distributed loads e Members for extended output member numbers of for detailed output of extreme deflections and bending moments Observe that this data combines the data about structure geometry and stiffness with the data about structure loads The latter not only do not make a separate loading file but they are even given in the same input lines as the first This would be surprising in computer programs performing a conventional continuous analysis but is a logical and deliberate step in DISCO In discontinuous analyses super position of different loading cases is by definition an error because loadings co define the systems see discussion in section 3 Therefore no provisions should be made to encourage such a superposition 11 2 Input in a text file The data file can be prepared in two different manners e Using a text editor which can process the TXT files e Using the DISCO own interactive input dialogue The first way is faster and offers better review possibilities for a sk
18. display it and ask you to confirm it by prompting Y N behind You can change it then by pressing N or go to the next data by pressing Y For the plane frame from Fig 11d the screen will look as follows after the general data has been input Input 1 Repeat F2 Finish Esc Escape Other Go on Cpf6 txt General data Project Plane frame Figure lld Force units kN Length units m No of joints 18 No of members 22 Four options vvill appear in the top line makes the program go back to the entry Project and repeat this part of the dialogue e F2 allows you to save the completed part of the input and go back to the previous screen which will now show 3 processing options Delete Update and Append You can then press 0 to take a break or use Append to resume your work Esc will erase all new input If you are in the Input mode it will erase the new opened file If you are in the Append or Update mode it will only erase all additions or updates Any other key will continue the Input dialogue Press indeed any other key which will bring you to the second part the input of joint data DISCO will now ask the joint type joint coordinates and joint concentrated loads for all succeeding joints checking every entry and asking you to confirm it At the end of the first joint input the screen will look like this Input 1 Repeat F2 Finish Esc Escape Other Go on Cpf6 txt Joint
19. of joint fixed and free DOF s has a unique within the structure type joint type number This applies to continuous as well as discontinuous structures In the latter the number of combinations is much larger Coding joint types is one of the crucial points of the entire algorithm Note that each single DOF can be If continuous If discontinuous 1 free 1 free on positive and on negative side 2 fixed 2 free on positive fixed on negative side 3 fixed on positive free on negative side 4 fixed on positive and on negative side Since the number Np of joint DOF s varies from 2 for beams to 6 for space frames the number Nr of possible joint fixity combinations joint types will vary still stronger This has been shown in Table 1 Table 1 Numbers of joint types in different types of structures No Np of No N7 of joint types DOF s Continuous Discontinuous 5 s 2 6 nar Space truss E Space frame Let s assume that type no 1 represents a joint with all DOF s free i e no external fixities and type no Nr represents a joint with all DOF s fixed The procedure described below shows the way to determine any joint type from the range 1 Nr 0 The Pascal notation for ranges is used in this manual A notation a b means here a b or a till b in cluded The DISCO software has been developed by the author in Turbo Pascal of Borland International Inc 12 1 a ta
20. same time The truth is that the first block writes the second and the second one writes the third after erasing itself from your computer hard disk This makes it possible that the block two and three have the same names DANCE But still more important is the fact that the second processing block is in this way freed of all the tasks that can be separated from solving the simultaneous equation system Solving that system is the most memory con suming procedure in structural analysis programs As the program blocks erase and write themselves in every run they can also tailor themselves to the type of structure that is being processed This allows for still more efficiency in memory use However such a programming method requires that a new generated program is compiled during the actual com puting session Therefore a runtime compiler of Turbo Pascal 4 0 TPC EXE makes part of the pro gram package which has also been shown in Fig 15 Obviously you do not need to care about the prompt C DISCO if you add it to the PATH command in your AUTOEXEC BAT file 39 DISCO Enter DANCE Enter DANCE Enter COMPILING DANCE PAS FROM THE DESKTOP AND STARTING EXECUTION INPUT OR REVIEW OF COMPILING DANCE PAS DATA INCL PRINT FROM THE DESKTOP AND GRAPHICAL AND OTHER STARTING EXECUTION DATA 0 K SOLVING THE SYSTEM OF EQUATIONS IN MATRICES DEFINED BY DISCO EXE SAVING DATA FILE y BOUNDARIES COMPUTING MEMOR
21. the mitre gates was performed using the finite element analysis program DIANA 16 17 The contact and the leakage problems of the gates were investigated using DISCO Combining these two programs in one design proved to be successful in author s earlier hydrotechnical projects e g the storm surge barrier on the Hartel Canal in the harbor of Rotterdam 18 19 Due to the symmetry only one of leaf of the Naviduct mitre gate had to be modeled Since the global geometrical behavior was of prior interest not the local stresses the computer model used by DISCO was highly simplified Fig 16 It was a 3D frame model with the main body of the leaf in one plane Only the drive arm lever and the line support to the other leaf did not lay in that plane All elements were linear members all internal joints were rigid each with 6 DOF s WAP 8 41 Z vaa o Wun 43 AF n 44 MR 45 LAM 46 YA 47 48 Aare MA 50 WME 51 52 Nua 53 n 54 Win SS vv 66 Von 57 58 59 6 g0 8 1 80 6x 37 222 53 Fig 16 Structural analysis model for one of the tvvo gate leaves 43 When closed the gate drive cylinders prestress the gate with a force Fy x 1000 KN in order to limit the opening which appears along the contact line under negative hydraulic load This prestression results in a compression of the edge lining The resultant of the hydraulic load acts some meters lower
22. 0 s the hardware requirements are quite mild in relation to the current standards What the user needs is only running under any version of MS Windows or MS DOS e hard disk in drive C with about 1 MB memory space for the DISCO system files e graphical card on board enabling the emulation of one of the following cards CGA MCGA EGA VGA or Hercules diskette drive USB port or any other data storage device as long as it is configured to be The delivered software version can not address a data storage port other than It is also not tailored for running in a network system although it can be adapted to that by a skilled professional 9 PROGRAM INSTALLATION To install the DISCO software on your PC please do the following 1 Make a back up copy of your original DISCO diskette 2 Take a new diskette a USB memory key or any other data storage medium assigned to drive and copy all data file directories CBE through DSF into it Label it e g DISCO data For operation under MS Windows 3 Insert your DISCO diskette into a disk drive of your PC Get its directory on the screen 4 Use MS Windows Explorer to copy the entire directories names and contents DISCO and DANCE into drive C Attention Not into C Programs or any other directory on drive C 5 Click on CADISCO and get its directory on the screen 6 Link shortcut the DISCO EXE file to your MS Windows desktop Att
23. 3224 57 1027 5 4528 16 9015 0 1654 3 1816 1 1057 2 3851 58 1027 5 8557 18 0800 0 1630 3 1538 1 0695 2 4477 59 1027 6 1391 18 9171 0 1644 2391 1 0692 2 4804 60 1027 6 4224 19 7504 0 1638 3 1256 1 0688 25 12 Support reactions Joint Type RX kN RY kN RZ kN MX kNm My kNm MZ kNm 2 3841 305 7436 1058 9745 0 0000 0 0000 0 0000 0 0000 8 3969 199 3141 258 2148 220 0000 0 0000 0 0000 0 0000 41 1026 218 3036 0 0000 0 0000 0 0000 0 0000 1 72 T 42 1026 140 1272 0 0000 0 0000 0 0000 0 0000 1 7623 43 1026 41 1275 0 0000 0 0000 0 0000 0 0000 1 7359 44 1026 0 0000 0 0000 0 0000 0 0000 0 0000 1 7544 45 1027 0 0000 0 0000 0 0000 0 0000 0 0000 0 0000 46 1027 0 0000 0 0000 0 0000 0 0000 0 0000 0 0000 59 1027 0 0000 0 0000 0 0000 0 0000 0 0000 0 0000 60 1027 0 0000 0 0000 0 0000 0 0000 0 0000 0 0000 Observe that only the first 3 elastic supports foints 41 42 43 remain in contact vvith the gate other leaf These joints undergo no displacements in the X direction and bear compressive reactions varying from 218 to 41 KN In fact this distribution of compression has been used in dimensioning the UHMPE front post lining Below that part a gap begins to open reaching 6 4 mm twice due to the symmetry in the bottom joint no 60 The leakage gap along the threshold is a geometrical sum of displacements Dx and Dy and re
24. 6 20 Kollbrunner C F and Basler K Torsion in Structures An Engineering Approach Springer Verlag Berlin Heidelberg Nevv York 1969 pp 10 45 21 Dabrowski R Torsion of Hydrotechnical and Bridge Girderes of Closed Thin vvalled Sections in Polish Gdansk Technical University Gdansk 1955 pp 5 189 22 Srinivasan S Biggers S B Jr and Latour R A Jr Identifying global local interface boundaries us ing an Objective Search Method Int J Numer Meth Engng 39 805 828 1996 23 Paris F A Blazquez and J Canas Contact problems with nonconforming discretizations using boundary element method Comp Struct 57 829 839 1995 24 Ezawa Y and Okamoto N Development of contact stress analysis programs using the hybrid method of FEM and BEM Comp Struct 57 691 698 1995 25 Chia Ching Lin Lawton E C Caliendo J A and Anderson L R An iterative Finite Element Boundary Element algorithm Comp Struct 59 899 909 1996 48
25. 7 positive integer see section 10 3 XYZ global coordinates X Y and Z real values in Lu x y 2 local coordinates x y and z real values in Lu 1 length of member positive real value in Lu 10 The number of members is superfluous for beams In beams this number equals the number of joints minus one The member beginning and end joints are also superfluous there as the joints are numbered sequentially 30 Fyi Fyi Fz joint i concentrated force loads real values in Fu in global axes Mv Mz joint i concentrated moment loads real values in Fu Lu about global axes bi ej m beginning and end joint of member positive integer lt n section axial rigidity of member j positive real value in Fu section torsion rigidity of member positive real value in F n Lu El section bending rigidity of member positive real value in F u Lu about the local y axis 1 section bending rigidity of member positive real value in F u Lu about the local z axis xi drr qz member j distributed loads in global axes real values in Fu Lu With the exception of the data Project that occupies a whole input all other data have to be sorted out in lines and separated from each other by one or more blanks no commas Below Tables 6 10 are the data formats for the six types of structures from section 10 1 The only difference between continu
26. 81 0 000 fo kt oP Ultra High Molecular Polyethylene 20 The torsional rigidity is built up by diagonals between beam rear flanges In a plane model this can be simu lated e g using the approach presented by Kollbrunner 20 or Dabrowski 21 44 Member data Member From To EAx kN GIx kNm Ely kNm Elz kNm 1 1 3 10710000 000 2536 000 43200 000 738600 000 2 2 3 10550000 000 158000 000 21000000 000 222200 000 29 9 13 7346000 000 564 800 21000000 000 417900 000 30 10 14 7346000 000 56070 000 21000000 000 417900 000 31 13 19 7346000 000 564 800 21000000 000 417900 000 32 19 22 7346000 000 564 800 21000000 000 417900 000 33 I 7480000 000 157700 000 21000000 000 249100 000 34 20 7480000 000 111100 000 21000000 000 249100 000 58 21 41 25100 000 100 000 1000 000 101 100 59 22 42 25100 000 100 000 1000 000 101 100 60 23 43 25100 000 100 000 1000 000 101 100 61 24 44 25100 000 100 000 1000 000 101 100 62 25 45 13700 000 100 000 1000 000 16 400 63 26 46 13700 000 100 000 1000 000 16 400 64 27 47 13700 000 100 000 1000 000 16 400 65 28 48 13700 000 100 000 1000 000 16 400 66 29 49 13700 000 100 000 1000 000 16 400 67 30 50 13700 000 100 000 1000 000 16 400 68 31 51 13700 000 100 000 1000 000 16 400 69 32 52 13700 000 100 000 1000 000 16 400 70 33 53 13300 000 100 000 1000 000 15 900 dal 34 54 13300 000 100 000 1000 000 15 900 72 35
27. A LOCK GATE BIBLIOGHAPHY 11 14 14 15 17 19 22 23 23 24 26 27 29 29 36 39 41 43 47 22 S DELIVERY CONDITTONS HARDVVARE REQUIREMENTS DISCO has been developed by the author vvith no contribution of any third parties of persons The au thor does not intend to register this softvvare or to take any other steps to protect his rights and or dis tribute his product commercially As this software has been enclosed to the doctor s thesis submitted at the Civil and Environmental Engineering Department of the Gdansk University of Technology further called the University the University owns now its copy rights As such the University may take steps to protect these right and or impose any distribution or other restrictions according to its policy Although utmost care was taken to debug this software nor the author neither the University can be held responsible for any consequences of its applications In particular users are warned that unprofes sional modifications of the included Pascal and text files e g intended to adapt third party lay outs may cause the damage of the software The software is delivered in a set containing e his manual one 3 2 diskette named DISCO and containing e system files in directories DISCO and DANCE e data files in directories CBE DBE CPT DPT CGR DGR CPF DPF CST DST CSF DSF As the first software versions were developed in the late 198
28. Gdansk University of Technology Civil and Environmental Engineering Faculty DISCO OBLICZENTA KONSTRUKCJI O PODPORACH CHARAKTERU CIAGLEGO I NIECIAGLEGO DISCO AN ANALYSIS OF STRUCTURES WITH CONTINUOUS AND DISCONTINUOUS SUPPORT CONDITIONS DISCO BEREKENEN VAN CONSTRUCTIES MET STEUNPUNTEN VAN CONTINU EN DISCONTINU KARAKTER program manual enclosure to doctor s thesis Contact problems in lock gates and other hydraulic structures in view of investigations and field experiences Author Ryszard A Daniel M Sc Eng Address Ministry of Transport Public Works and Water management of the Netherlands Civil Engineering Department P O Box 59 NL 2700 AB Zoetermeer Supervisor Eugeniusz Dembicki Prof D Sc Eng Politechnika Gda ska Wydzia In ynierii L dowej i rodowiska ul G Narutowicza 11 12 80 952 Gda sk Poland Gda sk April 4 2005 CONTENTS PART A REFERENCE MANUAL DISCONTINUOUS FIXTTTES INTRODUCTION MODELING DISCONTINUOUS FIXITIES PROPERTIES OF DISCONTINUOUS LINEAR MODELS SOME THEORETICAL BACKGROUND CODING DISCONTINUOUS FIXITIES PROGRAMMING APPROACH 6 1 Program environment 6 2 Unconditional conversion 6 3 Conditional modification POSSIBLE EXTENSIONS OF THE ALGORITHM PART B APPLICATION MANUAL 8 9 10 11 12 13 14 DELIVERY CONDITIONS HARDWARE REQUIREMENTS PROGRAM INSTALLATION STRUCTURE MODELING 10 1 General assumptions 10 2 Glo
29. General Example Project Grid Figure 11c kN 16 24 Fu Lu n m C free line 8 00 T Xi Y Fz Mx My l X Yo Fz Mx My 8 0 9 settee 5 0 T Xi Yi Fz Myx My 5 3 5 6 0 50 5 9 Tn Xn Yn Mxn Mya 10 0 free line 10 3 b ei Gla Ely qa 5 e2 Elo qz 5 15 0 1 5 23 bi Gly Ely 1451 155 7 19 bm Glem 14 1 1 2 2E4 465 free line 2 3 2E4 4E5 a A a 4 3 4 2E4 4E5 fit jo fad 5 6 2 4 4E5 6 7 2H4 4E5 7 8 2E4 4E5 9 10 2E4 4E5 10 11 2E4 4E5 11 12 2E4 4E5 13 14 2E4 4E5 14 15 2E4 4E5 15 16 2E4 4E5 1 5 5E4 E6 2 6 5E4 E6 3 7 5E4 E6 4 8 5E4 1E6 5 9 5E4 1E6 6 10 5E4 1E6 20 7 11 5E4 1E6 8 12 5E4 1E6 9 13 5E4 1E6 10 14 5E4 1E6 11 15 5E4 1E6 12 16 5E4 1E6 Table 8 Format of data files for plane frames Plane frames files CPF TXT and DPF TXT General Example Plane frame Figure 11d Project kN m 18 22 Fu Lu n m free line Continued on the next page 32 Table 8 continued T Y Fx Fy Mz X Yo Fx Fy Mz T Xi Yi Fyi Fy Mz T Xn Y Fyn Fyn Mz free line b El qx ay l b2 e2 Ela qx qr bi Eli qx qy D EAxm ET qxm zm free line ji j2 j3 jad 4 6 0 0 2 4 4 0 0 2 2 27 L 02 4 2 0 0 2 4 4 0 0 2 4 6 0 0 2 6 0 0 0 4 0 0 0 250 0 0 2 0 0 0 1 4 0 0 0 5 6 0 0 0 4 0 3 0 4 0 3 0 e 6 0 0 0 50 0 6 0 0 0 0 3 0
30. H LH OT OT OT OT OT OT IS IS HS S HS S S OT OT OT OT O Ln KA KM KA oo 9 0 o oo 0 0 KM KM KM ti mi td Di mi Di mi 65 ml eee ni 29 29 29 69 69 eee E al HS RRR RR C C LO G WA ROB BORG 0 0 0 0 0 0 0 0 10 18 5H4 2E3 8E3 8E3 12 18 5 4 2E3 8E3 8E3 13 18 5H4 2E3 8E3 8E3 14 18 5E4 2E3 8H3 8H3 16 18 5H4 2E3 8E3 8E3 17 18 5H4 2E3 8E3 8E3 16 17 18 19 20 21 22 23 24 25 20 20 20 20 35 11 2 Input in a DISCO dialogue DISCO offers also an opportunity to input the data interactively in a dialogue vvith the user This op tion requires some more key strokes but it may still be convenient due to the built in data control sub routines vvhich vvill not let any incorrect value pass through The dialogue begins already on the pro gram opening screen which looks like this Gdansk Institute of Technology Faculty of Hydro amp Environmental Engineering DISCO ANALYSIS OF STRUCTURES WITH CONTINUOUS AND DISCONTINUOUS SUPPORT CONDITIONS R A Daniel Continuously supported Continuous beam Continuous plane truss Continuous grid Continuous plane frame Continuous space truss Continuous space frame Ho J U LU KA 1 Your choice 0 Exit DISCO computes the following types of structures Version 4 04 Discontinuo
31. Y ARRANGEMENTS BAND MATRIX OPTIMIZATION SAVING DIS MODIFYIN PLACEMENTS G FIXITIES READING DISPLACEMENTS SAVED BY OLD DANCE PAS COMPUTING REACTIONS AND MEMBER FORCES COMPILING DANCE PAS FROM THE DESKTOP AND STARTING EXECUTION 9 259 Q DC F 53 19 Es ERASING DANCE PAS OPENING NEW DANCE PAS SELF ERASING OPENING WITH COMPUTED AND TAILORING A NEW COMPILING DANCE PAS MEMORY ARRANGEMENTS DANCE PAS FROM THE DESKTOP AND STARTING EXECUTION ADDING FILE DANC1 TXT ADDING FILE DANC2 TXT N TO COMPLETE DANCE PAS TO COMPLETE DANCE PAS OTHER OPTION Fig 15 Three main program blocks of DISCO from left to right Input Processing and Output Let us go back to the last screen message Pressing a key brings you back to the operating system To receive and process the output you should now do the same as before i e e under MS Windows double click on the icon DANCE e under MS DOS type DANCE on the prompt C DISCO and press Enter The third block Output is activated now DISCO reads the joint displacements from the file Temp txt computes the reactions and member internal loads member forces and comes up with the message Output for Plane frame Figure lld Choose output option Complete Partial Selective Graphical Exit Your choice _ The output options are discussed in the following section 40 13 OUTPUT OF SOLUTION In order to save paper and spare the environment DISCO a
32. aches 20 8 mm in joint no 60 These values have been used in designing additional soft gaskets to prevent excessive leakage As foreseen the discontinuous fixities of the rotation angle Az along the contact line gave in its upper part some small reaction moments Mz and no rotations Below that part free positive rotations and no moments Mz were computed Discontinuous fixity of the Dz displacement in the pivot bearing resulted in no displacement and an upward reaction Rz That reaction is exactly equal to the own weight of the gate reduced by the buoyancy as input in joint 15 which is one of the signs that a numerically stable solution has been computed 46 The last part of the output covers the bending extremes for the members of the users particular interest Bending extremes are the extreme bending moments and deflections in the member local coordinate sys tem see section 10 2 In this case the user was particularly interested in two members no 42 and 43 Below are the bending extremes computed for those members Table 13 For the member 42 the entire lines of bending moments and deflections have also been computed This facility is only available in the Selective output mode in which DISCO will ask the user to specify the number of equal steps for such lines Entering 0 skips this facility for the member in question Table 13 Sample problem bending extremes Bending extremes Member 42 x m Mz kNm
33. active and bear reactions and which will be released there is no way to solve this discontinuous linear model directly A solution must involve an iteration process eliminating released fixities and converging in a model with a definite smooth fixity system At each step of such iteration an entire let s call it after Ralston 9 basic solution of the system must be computed A classical nonlinear approach to such problems as discussed e g by Bathe 1 new supports arising during structure deformation considers the externally applied loads thus also deformations stresses etc to be a function of time The resulting approach is an incremental step by step procedure using the solution for discrete time to compute the solution for discrete time f At Aside from the plea against nonlinear approach as such see Introduction there are several practical reasons why this strategy has been rejected here For the space reasons they are not discussed Some of them e g error accumula tion inconvenience of time functions will become clear further in this manual The proposed algorithm of solution can in the simplest terms be described as follows 1 Convert the structure model in such a way that all conditional single sided fixities become uncondi tional double sided Memorize conditional fixities in the original model 2 Solve the converted model for the entire load combination computing the vector of node displace
34. al coordinate system O Q as a local coordinate system Sny Fig 12 Right handed orthogonal coordinate system Global coordinate system is the system as allocated by the user within the assumptions discussed in section 10 1 As the name says that system shall be used for all input data and solution results that are globally orientated i e refer to the entire model rather than a particular member In particular the fol lowing data must be input in the global coordinate system joint fixities types joint coordinates e joint loads e member distributed loads The program will return the following solution results in the global coordinate system e joint displacements support reactions Local coordinate system is a system associated with a particular member of the structure Unlike the global system the position of the local system is defined by the program nor by the user Its origin lies always in the beginning of the member and the local x axis always coin cides with the member itself pointing at the end of it Fig 13 4z y X For beams the local system is further identical to the global one when moved parallel to the beginning of the member beginning 2 5 Fig 13 local coordinate system For plane trusses grids and plane frames the local system may also rotate about the z axis in order to let the x axis match the direction of the member The local y axis follows thi
35. bal and local coordinates 10 3 Coding discontinuous fixities INPUT OF DATA 11 1 Data format general 11 2 Input in a text file 11 3 Input in a DISCO dialogue PROGRAM OPERATION OUTPUT OF SOLUTION SAMPLE PROBLEM BIBLIOGHAPHY 11 14 14 15 17 19 22 23 23 24 24 26 27 29 29 30 36 39 41 43 47 1 DISCONTINUOUS FIXTTTES INTRODUCTION This manual presents a computer program DISCO for the linear analysis of structures that may con tain discontinuous fixities These are fixities showing different linear behavior in various load ranges e g tension free supports of foundation grids compression free stays of cable stayed bridges etc The load displacement diagrams of structures containing such fixities are polygonal lines rather than curves which distinguishes them from non linear structures although some authors consider this property as a form of non linearity The program presented here has successfully been used in many projects of the author s engineering practice Although polygonal or piecewise smooth behavior of structures is considered sometimes e g Bathe 1 to be a form of non linearity there are authors e g Szilard 2 who do not share this view It will not be adapted in this manual either Without entering into broad discussion on this matter a strict dis tinction will be made between both terms Let us consider a beam laid on some supports and loaded by its own weight q a varia
36. beginning of the program Nevertheless an effort has been done to minimize the computation One of the measures applied is the introduction of a logical variable Fix which provides exits from different loops as soon as their tasks are completed Ka Converting into continuous types means here converting into type coding of a continuous model using the stiff approach E g the type number of an entirely fixed i e in fact continuous joint of a discontinuous space frame changes from T 4096 into Te 64 see examples in section 5 15 Belovv are some other features of the flovv chart in Fig 8 The comments concerning programming approach apply largely to the next flow chart in this manual as well e In order to speed up the computing binary operations are used given here in the Turbo Pascal notation Two of them may require an explanation ishlj shifts the value of i by j bits to the left ishrj shifts the value of i by j bits to the right v succ U shr 1 no e The conversion is only activated for LN T joint types Tp gt 1 Logical because if int x 179 Tn 1 then Tc 1 All entirely free N Y joints usually a majority in structure Fix or LI gt Nrshru Inc r U z r models are in this way skipped what N fastens the procedure The integer variables i u v g h andj are counters Here are their ranges in case the bit level code presents some survey problems i 1 N u 1 2Nop Fix F
37. bject of the manual BIBLIOGTAPHY 1 Bathe K J Finite Element Procedures in Engineering Analysis Prentice Hall Inc Engelwood Cliffs New Jersey 1982 pp 301 314 2 Szilard R Finite Berechnungsmethoden der Strukturmechanik Band I Stabwerke W Ernst amp Sohn Berlin Miinchen 1982 pp 300 347 3 Roark R J Young W C Formulas for Stress and Strain McGraw Hill Book Co Singapore 1986 pp 498 501 4 Bull J W Finite Element Analysis of Thin Walled Structures Elsevier London New York 1988 pp 146 150 47 5 Simunovic S and Saigal S A linear programming formulation for incremental contact analysis Int J Numer Meth Engng 38 2703 2725 1995 6 Refaat M H and Meguid S A Updated Lagrangian formulation of contact problems using varia tional inequalities Int J Numer Meth Engng 40 2975 2993 1997 7 Wang S P and Nakamachi E The inside outside contact search algorithm for finite element analy sis Int J Numer Meth Engng 40 3665 3685 1997 8 Timoshenko S Strength of Materials Part I Elementary Theory and Problems Van Nostrand Reinhold Co Ltd New York Cincinnati Toronto 1978 pp 301 361 9 Ralston A and Rabinowitz Ph A First Course in Numerical Analysis 2 edition McGraw Hill Singapore 1986 pp 411 477 10 Daniel R A DISCO Analysis of discontinuous and continuous skeletal structures internal bro chure Veth engineering consul
38. ble for the type of structure under consideration with in the headline all DOF s as listed earlier in this section The tables for discontinuous structures should have double positive and negative columns under each DOF In the most complex case of a space frame the headlines of such tables should be similar to the ones shovvn in Table 2 Table 2 Examples of joint type determination in 3D frames Continuous space frame ZT Y Type no 4041 1345 4096 2 Assign to each column an integer value varying e for continuous models e for discontinuous models from 2277 down to 2 from 227 downto 2 as shown for space frames Np 6 in the table headlines above Table 2 3 List all the fixed joints supports of the considered model in the left column and check the cells representing joint fixities e g by The sums of values assigned to the checked cells increased by 1 represent the joint types It is not difficult to see resemblances to the binary system in this coding An advantage of such coding in computer programming is that it enables the use of very quick bit level operations for all the transitions from discontinuous to continuous fixities joint types and for all the modifications of joint types in the iteration process The details of this will be discussed in the next section Dha computer program this procedure can e g be realized by using overlay windows Checking fixed cells can
39. ble force P and a constant axial force T as shown in Fig 1 The beam supports are not fixed against vertical tension This simple model can in principle be analyzed in 4 different ways which are shown underneath in a matrix form see Fig 1 CONTINUOUS slope no discantinulties discontinulties A i 5 7 os tensile reaction error 5 L support release d slope a _ no discontinuities dl conninuidas 7 1 P P P 5 7 8 tensile reaction error LB support release Fig 1 Convention of non linearity and dis continuity assumed in this manual Observe that the division into continuous and discontinuous behavior runs across the one into linear and non linear one The criterion is here a presence or an absence of slope discontinuities in the behavior functions of the structure not a linear or non linear character of these functions Strictly speaking the terms continuous and discontinuous are not quite correct here in the mathematical sense as the functions 6 P remain continuous in all the four cases They are only not smooth in the right half of the matrix These terms are however correct with respect to the first derivates slopes of these functions and above all correct in the physical sense Loosing contact with beam supports introduces material discontinuities in the system In this physical sense we shall use these terms here In the considered example the
40. cements and reactions the algorithm would distinguish be tween the values below and above certain levels which should then be specified in the input data Also this possibility will not be used often in structural engineering but it can be helpful e g in simulations of plastic hinges supports on buoyancy tanks etc It can not be used for modeling fracture problems e g cracks as the algorithm handles only polygonal behavior where there is just one function value for each argument In fracture problems more values are possible for a single argument e Fracture discontinuities The above does not necessarily mean that no routines of the algorithm can be adapted in fractural dis continuity analyses Especially interesting for this purpose can be 1 The binary technique of coding joint types the number of types might be larger 2 The so called stiff approach see section 4 and the unconditional conversion 3 Conditional modification in an internal iteration within a load step In general it looks promising to use the discussed routines within the user defined load steps in fracture analyses As the algorithm does not contribute to error accumulation this will probably lead to fine tuning of load step results The error accumulation effect can in this way be limited to inaccuracies at transition points between the load steps e Non linear polygonal problems In non linear polygonal analyses see discussion on terminology at the b
41. conversion and conditional modification These blocks represent the essence of the algorithm and will be discussed further in this section e With the exception of discontinuous analysis this approach does not differ much from some early programs for skeletal structures e g STRESS 11 However it is a minor problem to adapt it to a more complex FEM environment The block unconditional conversion converts all discontinuous joint types Tp NT into continuous ones Tc N 10 As result the structure model becomes in fact Continuous The vector of discontinuous joint types Tp N remains in memory for the boundary tests at each step of the iteration These tests are performed in the block conditional modification The tested objects are vectors of joint reactions R Np and displacements D Np In general the procedure investigates whether reactions have only been computed on the fixed sides and displacements on the free sides of single sided sup ports Each time the answer is no a proper DOF gets released respectively fixed for the next iteration step 6 2 Unconditional conversion Unconditional conversion defines the initial model for the iteration All discontinuous joint types Tn are replaced by continuous types Tc in such a way that fixity on any side or of a considered DOF qualifies this DOF as fixed This procedure is shown in a flow chart in Fig 8 It does not make part of the iteration process and is only executed once at the
42. displacement and reaction show no numerical insta bility i e if one of the two can be considered 0 the appropriate DOF v fixity may undergo modification N e The algorithm checks first positive and then negative side of DOF v The switch is controlled by a sign switch w The check and the modi fication take place in one operation thanks to the use of a logical func tion Ord expr see the two main Tor Toi Ord Odd f shi Np Ord Dyy gt 4 gt 1000 Toi Tor Ordij gt Ny shr u 1 ahl No vy Ord 2 lt e lt 1000 e blocks middle in the flow chart The upper block is activated when there g gshit is a fixity on the positive side of 0 DOF v the lower block when the negative side of DOF v is fixed Nye Np TY C END Fig 9 Flow chart of DISCO conditional modification e Observe that the entire procedure is ruled by the output of current iteration steps The continuous joint type numbers 7c become increased respectively decreased without checking up if they do not already contain the fixity or the freedom of DOF v Such programming can only be successful if all possible output combinations are controlled including numerical instabilities This is indeed the strategy in DISCO e Both main operation blocks contain an exit option for numerical instabilities These are not the same form of instabilities as the one handled by the co
43. e irrelevant since they pass directly to reactions causing no strain effect in a system This is not necessarily true in discontinuous models The programs for discontinuous analyses must enable the input of those loads Obviously the importance of loadings in discontinuous linear models requires more care to their input One should e g be cautious in using load factors which is a common engineering practice nowadays Increasing a load factor e g for variable loads does not necessarily lead to a safer structure There is a chance that a number of nodes get released or fixed after such operation what may not be the intention The last property in this short overview concerns stability Discontinuous linear models require more consideration to stability problems providing in return a more reliable stability test Models which be come unstable e g due to single sided support releases cannot be computed A program should issue a warning in such a case On the other hand properly modeled discontinuous linear structures that are successfully computed are also certainly stable 4 SOME THEORETICAL BACKGROUND Let s consider a structure model meeting all Clapeyron s conditions 8 and supported by a number of point wise external fixities of forces and or moments Let s assume that some of those fixities or all of them are conditional discontinuous in the sense as discussed above Since it is not known which single sided fixities will be
44. eginning of this paper a strat egy opposite to the one mentioned above seems more promising Use the non linear routines within the algorithm iteration steps Such approach would possibly lead to a very accurate multi purpose struc tural analysis programming However the following two problems should be taken into consideration 1 The non linear procedures must then be highly accurate as well Their error should in principle not exceed the boundaries set by the e conditions see section 6 3 2 In case of large displacements some extra precautions may be necessary to ensure convergence Convergence problems are however not new in non linear analysis 21 CONTENTS PART A REFERENCE MANUAL DISCONTINUOUS FIXTTIES INTRODUCTION MODELING DISCONTINUOUS FIXITIES PROPERTIES OF DISCONTINUOUS LINEAR MODELS SOME THEORETICAL BACKGROUND CODING DISCONTINUOUS FIXITIES PROGRAMMING APPROACH 6 1 Program environment 6 2 Unconditional conversion 6 3 Conditional modification POSSIBLE EXTENSIONS OF THE ALGORITHM PART B APPLICATION MANUAL 8 9 10 11 12 13 14 DELIVERY CONDITIONS HARDWARE REQUIREMENTS PROGRAM INSTALLATION STRUCTURE MODELING 10 1 General assumptions 10 2 Global and local coordinates 10 3 Coding discontinuous fixities INPUT OF DATA 11 1 Data format general 11 2 Input in a text file 11 3 Input in a DISCO dialogue PROGRAM OPERATION OUTPUT OF SOLUTION SAMPLE PROBLEM LEAKAGE OF
45. ems This difference is visualized in Fig 10 Another source of high performance is the stability of solutions Note that the entire iteration process is ruled by logical algebra no numerical values are passed from one iteration step to the next one In con sequence there is no danger of error accumulation This advantage will not be found in diverse non linear programs which are often used to approximate discontinuous behavior 12 This comparison has not been studied further but a certain analogy can be drawn to the performance of itera tive Jacobi Gauss Seidler 14 15 and direct Gauss solutions of simultaneous equation systems The first ones perform better by large complex systems 19 Below is a brief discussion on extension and modification ideas vvhich can still be considered These ideas have not yet been tested in a computer program The dis cussion is therefore somewhat specula oe ls 1 4 tive Nevertheless it might be helpful to prospective programmers SIMULTANEOUS APPROACH Fig 10 Number of iterations in two strategies of the analysis Required number of iterations Complexity number of discontinultles e Extension for internal discontinuities Member begin and end joints can be given joint type numbers in the similar way as for the external dis continuities Then the two programming strategies can be considered 1 Expanding unconditional conversion and co
46. ention Not the DISCO files with other extensions e g PAS BAK 7 Link shortcut the DANCE BAT file MS DOS batch file to your MS Windows desktop Attention Not the DANCE files with other extensions e g EXE PAS BAK TXT 8 Get your desktop screen insert the DISCO data disk in drive AN click on DISCO and Voila Further in this manual we shall talk about data disk and drive A only However it refers also to e g USB data key and port A if this is the configuration of your computer 23 For operation under MS DOS 3 Insert your DISCO diskette into a disk drive of your PC Get the prompt 4 Copy the entire directories names and contents DISCO and DANCE into drive CA e g using the DOS Xcopy s command Attention Not into any other directory on drive C 5 Log into the DISCO directory e g by typing CD DISCO and pressing Enter 6 Insert the DISCO data disk in drive A type DISCO press lt Enter gt and Voila Your DISCO system is operational now and you can in principle start processing the example data files on your data disk and computing their solutions You can also input and compute your own data files It is advisable however to read the rest of this manual first In particular deleting the supplied data files or modifying them through the DISCO dialogue may result in a loss of valuable examples 10 STRUCTURE MODELING 10 1 General assumptions
47. f joint i is not effected by load in any sense It can appear e g when there is another sufficiently fixed joint between i and the load when the entire sys tem is not loaded in direction v or when the strains in this direction are in internal equilibrium in the vicinity of i Also in such case no fixity modification is performed 7 POSSIBLE EXTENSIONS OF THE ALGORITHM In the form presented in this manual the DISCO algorithm proved to give sufficient support in numer ous polygonally linear discontinuous problems in the recent 16 years of the author s engineering prac tice Nevertheless there are problems which can possibly be better approached in another way or which might require some extensions to the algorithm A good reason to consider such extensions is that the algorithm proves to be relatively high performing One of the sources of this performance use of the fast binary arithmetic has already been discussed Another one is a simultaneous approach strategy analyzing the whole set of system discontinuities at a time The practice shows that the number of iteration steps does not grow with the number of discon tinuities with exception of some trivial cases as the beam in section 4 but usually becomes stabilized at a certain level This is a very favorable feature An algorithm using a kind of successive approach i e solving the discontinuities successively would require more iteration steps for complex discontinu ous syst
48. for continuous structures or an iteration process described in chapters 4 and 6 for discontinuous structures There will be a cloud of numbers flying through your screen impossible to follow due to the high speed Do not pay attention to that These numbers represent the band matrix structure at different steps of the Gaussian elimination and the itera tion process They had been helpful during the programming when PC s worked much slower than now but they became incommunicative later The only reason to display them now is that they may still be helpful in case of program modifications or extensions in the future At the end of this computing DISCO finds the solution in terms of a displacement matrix of all joints and stores it in on drive in a temporary file Temp txt You should then receive the following message MODEL SOLVED Setting up for output Cpf6 txt DISCO ready to output Cpf6 txt Run DANCE Strike any key _ The user is asked to run DANCE again but do not be confused it is not the same DANCE as the one in section 11 2 In order to understand what happened you should take a look at the global program layout shown in Fig 15 DISCO consists actually of three main blocks which should be run in succes sion and which do not communicate with each other during operation These blocks are e Input block e Processing block e Output block Moreover these blocks do not even exist together at the
49. ified from free into fixed Since both cases involve testing of equalities to zero there may arise numerical accuracy problems The nature and the size of such problems depend on a number of factors e g e complexity of structure models e presence of so called ill conditioned areas in these models e precision of floating point variables and operations etc These problems are common in computer programming and do not need to be discussed here DISCO makes use of two boundary considered to be zero values which proved to produce satisfactory results in PC programming assuming no very disproportional force or length units are used in the input data These values are 107 and e 10 1 1S used in zero testing of both displacements and reactions The boundaries are Di lt g is considered to be Diy 0 Riv lt 1000 61 is considered to be Ri 22 is used to test numerical stability of the solution When the solution is numerically stable the prod ucts Div Ri must equal 0 If this is not the case the DOF v fixity of a particular joint 7 must not un dergo modification in order to preserve convergence The boundaries used in DISCO are Dive Riv 2 DOF v of joint i stable modification possible Diva Riv gt amp DOF vof joint i instable no subject to modification Fig 9 presents a flow chart of the conditional modification as programmed in DISCO Here are some additional comment
50. ik and v lt u shr 1 y 1 Np s 1 Nr binary ie 1 2 4 Inc 8 etc A 1 g 1 2 1 Fig 8 Flow chart of DISCO unconditional conversion The counters u and v match discontinuous fixities negative positive or both of joint DOF s with appropriate continuous ones In simple terms Each time u spots a fixity in a joint type number Tp i v increases the appropriate continuous type number Tc by 2097 or binary shi Vp v The counter g is actually a function of u g 2 or binary 8 1shl u 1 and represents the column values in tables of discontinuous joint type numbers see section 4 The counters h and j help u to find whether there is a fixity in these columns This is the case when N h 2215 or binary Tp shr 4 1 h 1 16 6 3 Conditional modification Conditional modification is the most essential procedure of the algorithm It modifies joint i fixities i e the continuous type numbers Tc to meet the discontinuous fixity conditions of that joint In accordance with the strategy presented in section 4 the modification takes place in the two following cases 1 When the basic solution produces a reaction Ri 0 on a free side of a single sided fixity in the DOF y 1 Np That DOF becomes then modified from fixed in into free 2 When the basic solution produces a displacement Dj 0 on a fixed side of a single sided fixity in the DOF v 1 Np That DOF becomes then mod
51. ilar to what has already been discussed and is terminated by entering 0 When this is suc cessfully completed the following screen should appear Input complete Cpf6 txt Your data file Cpf6 txt Available processing options 1 Input 2 Delete 3 Update 4 Append 5 Model 6 Output 7 Setup Your choice 0 Exit If there are less e g only the first four processing options at the end of your input it means that some incorrect data has been entered not detected by the DISCO verifying routines The program will usually help you localize it by issuing a message like Bad joint 15 or Bad member 20 Use e g Update to correct it The use of a txt file editor discussed in section 11 2 is less convenient because DISCO stores all real values in an exponential notation 11 digits long which is as not easy to survey as the decimal notation For this reason it is also better not to let it replace the data prepared using a text edi tor unless really necessary Having successfully completed the input you can now view the structure model using the option Model It gives a simple graphical presentation of the structure model meant only for screen control not for printing Using the option Output will produce another data file called LastDat txt decently arranged in tables and suitable for a hard print This data file will be stored in the directory CADANCE The option Setup computes some memory co
52. illed program user It may however be less convenient for a beginner As the only test of the data is then the program run itself it may cost a number of runs before the data file is debugged The second way is slower and offers less review pos sibilities but it makes it almost impossible to produce a file resulting in a runtime error That way will be discussed in the following section There are a number of text editors which can process TXT files All computers working under the MS Windows operating system are e g standard supplied with a WordPad editor This editor produces and processes the TXT files Such files are also produced by a range of old text editors running under the MS DOS operating system Also the compilers of popular high level computer languages contain edi tors of such files The author e g wrote his first data files in the editor of Borland s Turbo Pascal the same compiler that was used to develop and test the actual program Let us assign Project name of the project loading case etc string up to 56 characters Fu force units e g KN N T kG Lb string up to 2 characters Lu length units e g m cm mm ft in string up to 2 characters n number of joints positive integer lt 999 m number of members positive integer lt 999 i sequential joint number positive integer lt n J sequential member number positive integer lt m Ti type number of joint
53. input data for each run loading case is contained in one input file No division into e g system and loading files has been made According to the discussion in section 3 no procedures combining single loading cases into complex superposed cases have been programmed In the input data the loadings have the same status as the so called system data Vector Fi Np of joint i concentrated loads comes in fact right behind the vector of joint coordinates Vector Q INo of member j distributed loads follows the vector of member stiffnesses S Ns Incase of odd type number Ts see structure types earlier in this manual the structure is continuous and there is no need for iteration The first approach leads directly to the solution In case of even Ts the structure is discontinuous It is first converted into a continuous structure After the band ma trix optimization it undergoes an iteration process with joint type modifications at every step con verging in a solution that meets all discontinuous fixity conditions e The iteration process is memory and time saving Note that no operations on files are involved The vector of joint reactions R Np which rules the process along with joint displacements D Np be comes only computed for the joints of types Tc 1 and always to the same memory space e Compared to continuous programs the flow chart in Fig 7 contains only two really new blocks marked unconditional
54. iteration that solves the problem using only irreversible DOF modifications Naturally it should be proved that the procedure shown above is convergent From the engineering point of view however an instructive example is often more convincing than a strictly theoretical dis cussion A simple model enabling to observe the features of the algorithm in a transparent way is a weightless beam on a large number of rigid tensionless supports loaded by a single force in the middle of one span In Fig 5 a half of such a beam has been modeled due to the system symmetry 2 A third middle way is also thinkable if additional precautions are taken to ensure the convergence This might lead to further perfectioning of the algorithm It has not been investigated so far This example requires an exceptionally large number of iterations This and the problem triviality are chosen deliberately to picture some features of the procedure It must not be seen as a sign of its small efficiency 8 Strain energy of reactions Nm 10 0 KN a 12x20m R D T 4 4 A A A El 1000 kNm A A 5560 ee A nn aa k ml y5l gt gt x gt gt O O N OA bp WH O lt K a gt a L lt H p4 4 lt lt lt x gt x gt gt qm Step Step Step Step Step Step Step Step Step Step S
55. little sense to extend it in that way e Mutually related discontinuities In section 4 four fixity conditions of a discontinuous DOF are distinguished This covers most forth coming problems There are cases however where fixity of a single DOF depends on a fixity of another DOF rather than on a sign of displacement in the same DOF In the sample problem presented further in this manual it would probably be more convenient to relate the fixity of rotation angle Az to the fixity of displacement Dx Such relations can be realized e g by adding another joint type number this time for mutually related discontinuities to the current one and expanding the conditional modification For mutually related discontinuities the type numbers gt Tp can possibly be used Since such discontinuities are seldom a proper detection could be performed prior to entering the expanded routines 3 A quite deep analysis based however on a different incremental search algorithm has been presented in 22 a Special types of complex elements vvhich are in fact used novvadays to simulate contact problems are boundary elements 23 In particular the hybrid methods combining finite and boundary element approach 24 25 have been successful in this field 20 e Other than zero discontinuity levels As already mentioned it is a minor problem to install other than zero discontinuity levels Instead of or next to detecting positive and negative displa
56. llows you to view the solution first and to present only the essential parts of it in the final output Therefore a number of output options have been programmed Such additional facilities are possible thanks to the fact that the structure has already been computed and the large part of the computer memory has been freed The first three of the output four main options shown in the last screen one page back can be delivered as screen output or as both the screen and the disk output Here is that screen again after choosing e g the complete output Output for Plane frame Figure lld Choose output option Complete Partial Selective Graphical Exit Your choice C Screen Hard disk Screen Your choice _ Below is a short description of the available output options Complete The complete output consists of the following parts Program headlines and general data see section 11 1 Displacements and rotations of all nodes called joints Support reactions i e reactions in all externally fixed joints Member internal loads called member forces at the beginning and the end of each member Bending extremes extreme deflections and moments in all earlier specified members After choosing C for Complete the computer will output these data in decent tables easy to survey Due to the problems with screen control under MS Windows by other than MS software no data scroll routine has been program
57. loads and reactions The output excerpts interesting for this manual are presented in table 12 45 Table 12 Sample problem output excerpts Joint displacements Joint Type DX m 1e3 DY m 1e3 DZ m 163 AX le 3 AY le 3 AZ le 3 1 1 9 4280 14 3150 1 4698 0 9172 0 1801 7 7712 2 3841 0 0000 0 0000 0 2309 1 6380 0 1823 4 3305 3 1 0 0999 0 7688 0 2309 0 9172 0 1801 4 3305 7 1 0 0633 0 2379 0 0086 0 4589 0 1197 2 6146 8 3969 0 0000 0 0000 0 0000 0 4580 0 1189 3s 2261 9 1 1 0912 2 8855 0 1288 1 3718 0 3910 2 5717 41 1026 0 0000 2 5662 0 1144 3 1876 1 0380 0 0000 42 1026 0 0000 1 6082 0 1144 3 1876 1 0385 0 0000 43 1026 0 0000 0 3871 0 1164 3 2421 1 0375 0 0000 44 1026 0 2303 0 8526 0 1183 3 2739 1 0374 0 0000 45 1027 0 6247 2 1921 0 1204 3 2828 1 0383 he 8s 46 1027 1 0196 3 4440 0 1226 3 2690 1 0402 1 8612 47 1027 1 4154 4 6863 0 1249 3 2323 1 0431 1 9242 48 1027 1 8118 5 9244 0 1307 3 2659 1 0437 1 9538 49 1027 2 2086 7 1722 0 1364 3 2831 1 0443 1 9833 50 1027 2 6055 8 4234 0 1422 3 2841 1 0450 2 0128 51 1027 3 0027 9 6719 0 1479 3 2688 1 0457 2 0424 52 1027 3 4002 10 9115 021537 3 23412 1 0464 2 0719 53 1027 3 7960 12 17157 1593 S271 1 0904 2 1346 54 1027 4 2051 13 3188 0 1632 3 2316 1 1183 2 1972 55 1027 4 6216 14 5189 0 1656 3 2205 1 1302 2 2598 56 6 OZ b 0395 15 7138 0 1663 3 2038 1 1260 2
58. mbinations is possible Plane trusses can only bear pointed loads in joints These loads are force components F and Fy Truss members can have different sectional rigidity EA which is the only parameter determining the truss deformation c Grids Grids are 2D structures built of linear straight members with rigid internal joints and loaded perpen dicularly to the structure plane The DOF s of a grid joint are displacement D and rotation angles A and A Grids can be supported by any number of joint DOF fixities or their combinations The possible grid loads are joint force F joint moments M and M and member in DISCO equally distributed load qz Grid members have two sectional rigidities torsional GT and flexural ET d Plane frames Plane frames are 2D structures built of linear straight members with rigid internal joints and loaded in the structure plane The DOF s of a plane frame joint are displacements D and D and a rotation angle Also plane frames can be supported by any number of joint DOF fixities or their combinations The possible loads are joint forces F and joint moment M and member in DISCO equally distributed loads q and qy Plane frame members have two sectional rigidities axial FA and flexural EL e Space trusses Space trusses are 3D structures built of linear straight members with all joints also supports hinged The DOF s of a space truss joint are displacements D Dy and D Any
59. med If your output is long you can better use the H option first and examine it in the hard disk file There will still be an opportunity to shorten that file for the final presentation Partial This option gives you the opportunity to choose only those parts of the output that are of your particular interest The format and the size of the output parts are the same as in the option Complete but you can skip the parts of less significance The program will ask whether a part has to be output displayed in Screen mode or also put on the disk in Hard disk Screen mode before processing it It is especially useful for quick or limited analyses e g when only the system reactions are to be considered Selective The Selective output goes further than Partial allowing you to tailor the output exactly to the form in which want to present it The editing takes now place not only on the level of the output parts but also on the level of particular joints and members After printing each table headline the program will ask you to enter one by one the joint or member numbers of your interest Every entry is followed by 18 This term less common in Europe is largely used in America e g in the classical MIT programs 11 12 immediate output for that joint or member You may also enter them in your own succession or double the entries if you like This option is especially useful for comparing and sorting purposes
60. ments D and the vector of node external reactions R 3 Check if every R reaction occurs on the fixed side of a proper conditional DOF in the original model Each time it does not modify the converted model by releasing the particular DOF 4 Check if every D 0 displacement occurs on the free side of a proper conditional DOF in the origi nal model Each time it does not modify the model by fixing the particular DOF 5 If step 3 or 4 result in any modification go to step 2 Otherwise the system is solved The question concerning step 1 is why to convert the single sided DOF s into double sided In general two ways can be considered to convert a discontinuous model into a continuous one Let us call them Stiff approach replace all single sided fixities by double sided e Slack approach release all single sided fixities A disadvantage of the slack approach is that it may produce an unstable model during the iteration This endangers the convergence The stiff approach does not bear that risk Another possible question is why to bother checking the sides of displacements in step 4 One might suppose that since the stiff ap proach has been used it is the releasing of the DOF s that leads to the solution not the fixing This proves not to be sufficient A DOF that has once been released may require to be fixed again in one of the next iteration steps As far as this approach has been researched there exists no convergent
61. mination in all 12 types of structures that can be computed by DISCO along with some calculation examples Tables 5 Table 5 Joint type determination in 12 types of structures Continuous beam Type yx no Y 6 5 4 2 Y A Discontinuous beam Type Continuous grid Type no Type Type no Type Type 28 Discontinuous space truss 11 INPUT OF DATA 11 1 Data format general Type no 4033 641 1153 Your data must be submitted in a text file a file with extension TXT stored in drive on a diskette or other data storage medium in one of the following directories CBE CPT CGR CPF CST CSF for for for for for for Continuous beams Continuous plane trusses Continuous grids Continuous plane frames Continuous space trusses Continuous space frames DBE DPT DGR DPF DST DSF for for for for for for Discontinuous beams Discontinuous plane trusses Discontinuous grids Discontinuous plane frames Discontinuous space trusses Discontinuous space frames The data file names have the same names as the names of the directories followed by a sequential num ber from the range 1 99 E g the full address of the first discontinuous grid data file will always be A DGR DGR1 TXT The data file consists of three parts that must be submitted followed by one part that may be submitted in case you like additional
62. ndition Diva Riv lt g Basically two forms of numerical instability can be distinguished in the program e Inaccuracy problems Caused usually by too complex modeling and relatively low variable and or operation precision To recognize e g by unstable zero s in the output This form is primarily handled by the 1 conditions e Out of range problems Caused usually by so called ill conditioned features To recognize by the output of high real numbers usually a number of ranges higher than the input values This form is primarily handled by the 2 condition e The modification results in fixing or releasing the DOF v depending on the computed displacement Di and reaction R v If the use of the Ord function presents some survey inconveniences a simpler notation in Table 4 can be helpful 18 Table 4 Action of two main operation blocks in Fig 8 The upper block The lower block Riv lt 1000 gi Ri gt 1000 2 Ri gt 1000 gi Ri 1000 gi Positive side Releasing v Negative side Releasing v loaded correct Te decreases loaded correct l Te decreases No change by 2 7 i No change by 277 Fixing v Te in Numerical in Fixing v Ta Numerical in creases by stability exit increases by stability exit DADR No change i 22 No change There is one more special case covered by the g conditions It arises when both displacement D and reaction Ri are equal to i e when DOF v o
63. nditional modification in such a way that all internal and external fixities are handled at a time This leads to a single level iteration probably the fastest Additional convergence precautions may then be needed 2 Dividing the procedure Each step of external fixity iteration contains then an entire internal fixity iteration Such a double level iteration is probably slower but better convergent Naturally in internal fixity modifications the global joint displacements should be used as boundary val ues not zeros This presents some problems since these displacements may as well be effected by dis continuously connected members The iteration will probably require a deeper joint analysis then pre sented in this manual e Discontinuous edge or surface supports In order to simulate a linear or surface discontinuous support the user has to input a large number of pointed discontinuities This can obviously be avoided by defining special contact interfaces modules etc allowing to input entire contact edges or surfaces as single items Such procedures are known e g to generate complex finite element types and do not need to be discussed here This extension seems to be convenient for large FEM programs running on networks with powerful central units DISCO has been programmed for a small stand alone PC with a limited operation memory the used Turbo Pascal version can not address more than 64 kB therefore it made
64. nstants and performs a simple band matrix optimization especially useful for large models which might otherwise cause a memory overflow This optimization is an original DISCO routine Simply speaking it divides the band matrix into a number of dynamic sub matrices most of which are narrower than the band matrix width This saves the memory allowing to compute more complex models After performing Setup the problem is ready for actual computation which is announced in the following manner Computing memory constants Rock 6 Roll optimization Setting up for processing DISCO set up for Plane frame Figure 11d Joints 18 Members 22 matrix 1 from 1 to 18 band width 5 Run DANCE Strike any key _ 38 12 PROGRAM OPERATION The program operation starts by choosing the structure type and submitting the input data as discussed in chapter 11 VVhether the input has been prepared using a text file editor or the DISCO dialogue you still need to get a screen shovving the seven processing options see preceding page run the option Setup and get the screen telling you to run DANCE Pressing a key closes DISCO and brings you back to the operating system In order to perform the actual computing you can now do the following e under MS Windows double click on the icon DANCE e under MS DOS make sure you have the prompt C DISCO type DANCE and press Enter 5 DISCO vvill novv compute the problem in a single run
65. nts are given in Table 3 UNCONDITIONAL CONVERSION in ona resulting In Nol l gin 27 Table 3 Matrix formatting constants depending 11 LRA TAR on structure types Structures r 9 10 Fay boundaries tor gint O K JOINT TYPE MODIFICATION Tot Tot m di EE Pine tans G N PY 6 12 Space frames N any modifications gt OUTPUT DIN Np and the rest of the solution Fig 7 DISCO general flow chart CEND Peele Es of 2 3 2 1 121121 3 CONDITIONAL MODIFICATION Table 3 helps also to understand why a division into structure types has been used in DISCO Modern FEM programs offer a number of element types to be used in a model rather than conforming the model to one element type However in iterative algorithms where the entire basic solution must be computed a number of times it is preferable to use simple models In particular the low numbers of DOF s Np and sectional stiffnesses Ns in models simpler that 3D frames speed the computing considerably up It Band matrix optimization falls beside the scope of this manual DISCO uses an own simple optimization method A good introduction to more complex so called fractorization methods can e g be found in 9 14 also keeps the band matrix 14 1151 narrow limitting the memory consumption Further the following features should be observed in the flow chart in Fig 6 e The entire
66. number of DOF fixities sup ports or their combinations is possible Space trusses can only bear pointed loads in joints These loads are force components F Fy and F Truss members can have different sectional rigidity FA which is the only parameter determining the truss deformation f Plane frames Space frames are 3D structures built of linear straight members with rigid internal joints The DOF s of a space frame joint are displacements Dx Dy and D and rotation angles A Ay and A Space frames can be supported in any number of joint DOF fixities or their combinations The possible loads are joint forces F F and joint moments M M and M and member equally distributed loads qx qy and qy Space frame members have four sectional rigidities axial torsional GZ and two flexural EI and El 10 2 Global and local coordinates As mentioned in section 10 1 DISCO makes use of a right handed orthogonal Cartesian coordinate system This system including the positive sign convention is shown below Fig 12 In can be conven ient to memorize the positive rotation signs as clockwise when Z looking in the positive direction of proper axes Memorizing the mutual position of the system axes is essential Swapping two of Ae them will produce a left handed system which requires another interpretation than the one presented in this manual The pro gram uses the system from Fig 12 in two different manners o as a glob
67. on an gle Az in that part However just for demonstration single sided rotation fixities have been used The upper part first 4 supports is fixed against the positive Az rotation which is expected to occur The lower part remaining 16 supports is fixed against the negative Az rotation which is not expected in this case Finally there is a single sided fixity of a vertical displacement in a pivot bearing under the ro tation axis joint no 8 downwards fixed upwards free A double sided pinned support would be cor rect here as well since there is no doubt about the sign of the vertical support reaction However this is not always clear in more complex structures and or load cases Table 11 Sample problem input excerpts Joint data Joint Type X m Y m Z m DX DY DZ AX AY AZ 1 1 2 033 1 455 esau 4 2 ke az 6 q al 0 07 2 3841 0 000 0 000 7 100 X E o H H ic Spe x a 3 1 0 000 0 000 6 550 285555 5 T oh 0 000 0 000 04500 PE EP b dl 8 3969 0 000 0 000 0 000 4 B f f 9 1 1 101 0 366 6550 A AA Io a gt of a afi s 41 1026 6 657 2 181 6 850 5 44 42 1026 6 657 2 181 6 550 7 ae ae ae ae et ey ee ce 4 43 1026 6 657 2 181 6 170 Ed E E E 44 1026 2667 2 181 5 790 5 45 1027 6 657 2 181 5 410 ace CY CORY 46 1027 6 657 2 181 5 030 5 58 1027 2 181 0 530 59 1027 6 657 2 181 0 265 ny 4 af 60 1027 6 657 2 1
68. rately computed load cases are combined Fig 4 The superposed deflected line and reactions d strongly differ from the T correct ones a and are mutually inconsistent In consequence complete load combinations should each time be computed rather than computing single load cases and combining the results This property 15 important not only to the program users but also to the programmers A complete discussion on this subfect goes beyond the scope of this manual Nevertheless it s significance must de emphasized because superposition is one of the elementary d procedures in structural analyses Fig 4 Superposition results in an error Let us confine the discussion to the three following recommendations 1 The conventional programs for continuous structural analyses contain procedures to combine sin gle load cases in load combinations usually with user defined combination factors These pro cedures should not be programmed neither be used for discontinuous linear analyses 2 Most existing structural analysis programs make use of two separate input files called e g sys tem file and loading file In discontinuous analyses as already discussed loadings actually co define the systems For the sake of consistency they should better make part of one integral input file 3 Several existing programs do not allow to set concentrated loads on supports in the directions of fixity It is assumed that such loads ar
69. rection but is assumed to be equally distributed over the entire member The discussed software was originally developed in the 1980 s when memory consumption and computation time were of more significance than they are today For continuous models similar divisions have been used in some early structural analysis programs e g STRESS 11 ICES STRUNDL 12 Such approach leads to a quick memory saving basic solution 11 length To input an unequally distributed load or a load covering a part of a member length one must first divide the member into sectors by defining more joints e The result of the basic solution is a vector D of all joint displacements and a vector R of all joint re actions Both vectors are appropriate to the structure type and related to the global orthogonal XYZ axes The form of both vectors is the same The displacements D are in beams displacements Dy rotation angles Az in plane trusses displacements Dx Dy e in grids displacements Dz rotation angles Ax Ay in plane frames displacements Dx Dy rotation angles Az in space trusses displacements Dx Dy Dz in space frames displacements Dx Dy Dz rotation angles Ax Ay Az In the vector R of reactions the forces R come in place of displacements D and the moments M come in place of rotation angles A All indices remain the same e Joint external fixities supports are identified by joint types Every combination
70. redefined by assuming that the structure must lie in the global XY plane For beams types 1 and 2 a in Fig 11 the additional assumptions are that the global X axis coincides with the beam the loads act in the global XY plane and the joints and members are sequentially numbered in the positive direction of X For the types 9 12 e and f in Fig 11 any position of the global coordinate system can be chosen The difference between continuous and discontinuous structures has been discussed in section 1 of this 46 29 46 29 46 manual The terms beam plane truss grid plane frame space truss and space frame are widely known To avoid confusion however here is how DISCO sees these types of structures a Beams Beams are linear straight structures supported by any number of pointed supports fixing any degree of freedom DOF or a combination of DOF s The DOF s of a beam node joint are deflection D and ro tation angle A Beams can be loaded by pointed forces F and moments M as well as by member in DISCO equally distributed loads qy Beam members can have different sectional rigidity EZ which is the only parameter determining their flexural behavior b Plane trusses Plane trusses are 2D structures built of linear straight members with all joints also supports hinged The DOF s of a plane truss joint are displacements D and D Any number of DOF fixities supports or their co
71. s on this flow chart e In addition to the notation as discussed by flow charts in Fig 7 and 8 the use of a logical Pascal function Ord expr may require an explanation e Ifthe expression expr making the argument of Ord is true then Ord returns 1 e Ifthe expression expr is false then Ord returns 0 e Just as the unconditional conversion the conditional modification becomes only activated for joint types Tc gt 1 This speeds the computing considerably up e The integer variables u v g h andj are counters w is a sign switch The ranges of these variables are as follows u 1 2Np v 1 No g 1 N7 binary ie 1 2 4 8 etc A l g j 1 N7 2 w 1 1 With the exception of w the same notation is used here as in the flow chart of unconditional conversion 1D In exceptional cases these values may require to be tuned up For more output stability variable consid ered to be zeros can also be used e g belonging to the input data or resulting from the analysis of numerical input This has not been programmed but it may be considered in prosperous versions of DISCO 17 Fixity detection in the columns of CSTART joint type definition see examples in section 5 takes place in the same u 1 v 1 w 1 g 1 way as in unconditional conversion e 10 6 The initial model contains a fixity in a column u if N h F or binary Tp Nr shr Lu DS hk 1 e If the computed
72. s rotation and the local z axis remains parallel to the global Z axis In trusses also space trusses you may forget the local axes y and z since truss members can only bear loads in the x direction The members of plane frames can also bear shear in the y direction and bending moments about the z axis In space frames the local axes y and z are defined as follows e The y axis is parallel to the global XY plane In vertical mem bers it is directed the same as the global Y axis e The z axis lies in a vertical plane containing the x axis Its pro jection on the global Z axis is never negative Fig 14 Local system in the posts of a football goal 26 This definition applies when the global Z axis is vertical which is an advised choice A good example of it is the determination of local axes in the posts of a football goal see Fig 14 If the global Z axis is not vertical than vertical members should be read as members perpendicular to the global X Y plane and vertical plane should be read as plane parallel to the global Z axis 10 3 Coding discontinuous fixities As discussed in section 5 each joint of your structure has a joint type number that defines its external fixities It must explicitly be included into your input data The method to determine joint type numbers has globally been shown using the most complex case a space frame joint as an example Following are the table headlines for joint type deter
73. step 8 Yet this does not endanger the convergence 10 e The variation of the total strain energy of the fixed DOF s can globally be approximated by a con cave curve arc convex downward The energy losses in the first iteration steps are usually the big gest This favorable property will still be discussed e The approximation by a concave curve shows here an interesting irregularity which can for the present be called an energy wave This has not been studied any further It has however been ob served that this phenomenon becomes less visible in models on elastic instead of rigid supports This leads to some analogies with damping Finally let s observe that a nonlinear approach using time functions would be useless for this problem Time is irrelevant since any force P acting downwards gives basically the same solution P directed up wards gives instability For the same reason also the term polygonal approach may be controversial for this load case Yet since weightless beams are quite exceptional we shall drop that detail 5 CODING DISCONTINOUS FIXITIES Since every step of the iteration computes an entire basic solution of the system the method can be con siderably time consuming Therefore it is advisable to use a quick simple procedure for the basic solu tion even at the cost of diversity of modeling features The algorithm of the DISCO program has been developed for PC applications The program itself 10
74. t all discontinuous fixities can be modeled by defining a conditional DOF e g e force tension fixed compression free 0 or reverse e moment clockwise fixed anti clockwise free 0 or reverse This also applies to DOF s that are not simply fixed or free but that show different positive and nega tive fixity characteristics The solution is then an additional conditionally fixed element in the direction of the considered DOF This method can be illustrated by the three following examples see Fig 3 b 1 Fig 3 Modeling different fixity characteristics In the first example a conditional spring has been used to model the elastic tension fixity of a beam sup port In the second example tvvo conditional springs represent different fixity characteristics of a column base Note that the node coordinates of these springs can in fact be identical as long as no connection between them is defined The third example shows two ways of modeling discontinuous moment fixi ties In that example the two conditionally supported springs can optionally be fixed against displace ment or against rotation This technique applies also to 3D models where the number of conditional DOF s can be larger The limitation of the algorithm is however that only point wise fixities can be defined in this way A condi tional line or surface support becomes released over a certain distance or area what makes the solu tion more complex
75. tants Papendrecht NL 1989 11 Massachusetts Institute of Technology STRESS A User s Manual M LT Cambridge Mass 1965 12 Massachusetts Institute of Technology ICES STRUNDL II Engineering User s Manual M LT Cambridge Mass 1968 13 Gibshman M E Analysis theory of bridges of complex space systems in Russian Izdatjelstvo Transport Moskva 1973 pp 13 192 14 Stoer J and Burlisch R Introduction to Numerical Analysis corrected 2 printing Springer Verlag New York Heidelberg Berlin 1983 Polish edition Wstep do analizy numerycznej PWN VVarszavva 1987 15 Dahlquist G and Bj rck Numerical Methods Prentice Hall Inc Engelwood Cliffs N J 1974 Polish edition Metody numeryczne PWN Warszawa 1987 16 TNO Building and Construction Research DIANA Finite Element Analysis User s Manual Re lease 5 1 TNO Delft NL April 1993 17 Daniel R A and Gerrits E M W Design and analysis of a steel lock gate in Finite Elements in Engineering and Science Proceedings of the 2 International DIANA Conference Amsterdam NL 4 6 June 247 251 1997 18 Daniel R A and Leendertz J S Integrated design of the storm surge barrier in the Hartel Canal in Dutch Civiele Techniek 4 Gorinchem NL 9 14 1994 19 Daniel R A The Hartel Canal Barrier Shoving the options in Dutch Bouvven met Staal 130 Rotterdam NL 38 45 May June 199
76. tep Step Step 0 1 2 3 4 5 6 7 8 9 10 Fig 6 Strain energy of reactions in the 12 iteration steps Energy level 0 equals here in fact 16 7 Nm see calculations in the text 11 12 As known the solution of such a system is a free supported one span beam The way in which the al gorithm described above comes to this solution has been shown in steps 0 through 12 Note that each step results in eliminating sometimes also in adding of a number of supports from vvhich at least here always one is eliminated definitely That one support has each time been marked black in the drawing Note also that the supports guaranteeing the stability of the system here just one are not eliminated at any step These two features of the algorithm have been confirmed in a large number of tests on differ ent systems They are sufficient to make the procedure convergent We shall now follow the changes of strain energy bound by reactions in this iteration process It is convenient to do it in reverse order in which we expect this energy to grow Let s consider two neighboring iteration steps i and i 1 and sign by j the joint node numbers of the beam From the recip rocal theorem of Betti Maxwell 8 we have Yer R 3 P H HT By proper modification of this equation we can obtain a double formula work L ii required to bring the beam from the deformation state in step i back to the step i 1 1 n L a 2 y R 5
77. then take place by a mouse click 9 In the past the binary code was used much wider to program the analyses of complex structural problems An impressing example of such notation can e g be found in 13 This is obviously not the case here any more 13 6 PROGRAMMING APPROACH 6 1 Program environment In order to minimize the computation time it 1s important to choose an optimal program environment It is e g not advisable to incorporate time consuming operations on files opening reading writing etc in the iteration process On the other hand it is certainly advisable to use e g band matrix optimiza tion Fig 7 presents a flovv chart of a softvvare environment as programmed in DISCO incorporating the algorithm for discontinuous analysis Flovv charts of the actual algorithm vvill be shovvn later on START The notation used in this flovv chart may require an explanation INPUT N and M are total numbers of respectively BONE C N Ac P N joints and members M M 2 SI Ns QIM NJ Nc Ns and No are integer constants de Na v pending on structure type Ts and helping to format proper matrices They represent the characteristic numbers of respectively joint coordinates joint DOF s equal to possible concentrated loads member sectional stiff Input test memory arrangeme optimizing bani m b sl A l l nesses and member distributed loads These test etc Ready for computing consta
78. usly supported Discontinuous Discontinuous Discontinuous Discontinuous Discontinuous Discontinuous GA H NOONAN beam plane truss grid plane frame Space truss space frame Make sure that the DISCO data diskette is in drive A at this moment You may now enter the type of structure which you want to analyze Let us assume that it is a continuous plane frame Type 7 and press Enter The program response will be about as follows Available data files of this type 1 Cpf1 txt Cont plane frame 3 Cpf3 txt Small offshore rig 5 Cpf5 txt Frame Figure 11d Your choice 0 Exit EX1 2 Cpf2 txt Cont plane frame EX2 4 Cpf4 txt Hartelkering stijlen 6 Cpf6 txt Free If you want to process Delete Update Append Model Output or Setup for computing an existing data file you will enter its number in this case 1 5 Take care not to do it when you intend to Input a new file as it will overwrite the existing one We now discuss an entirely new input therefore type 6 and press Enter DISCO will then open a free file Cpf6 txt with only one processing option Input Available processing options 1 Input Your choice 0 Exit 36 Typing 1 and pressing Enter opens then the input dialogue in which DISCO asks the succeeding data about your structure and you enter those data from the keyboard If the entry is incorrect the program will not accept it otherwise it will

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