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Collection of mechaincal design and machine elements problems Diego Zenari 182160 A. A. 2015 - 2016 Due to .tex importation problems, the character "=" is substitued by ":" in command listings

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Page 1: Collection of mechaincal design and machine elements problems · MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 1 1 Pratt truss bridge analysis Diego Zenari, 182160, M.Sc Mechatronics

Collection of mechaincal design and machineelements problems

Diego Zenari 182160

A. A. 2015 - 2016Due to .tex importation problems, the character "=" is substitued by ":" in

command listings

Page 2: Collection of mechaincal design and machine elements problems · MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 1 1 Pratt truss bridge analysis Diego Zenari, 182160, M.Sc Mechatronics

MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 1 1

Pratt truss bridge analysisDiego Zenari, 182160, M.Sc Mechatronics engineering

Abstract—This first problem introduces the concept of structure optimization: starting from the analysis of the "firstchoice" structure, some structure dimensions are varied in order to obtain the minimum nodal displacement.

Index Terms—displacement optimization, Truss, Beam189, non-lienar geometry.

F

1 INTRODUCTION

THE subject of the analysis is pratt trussbridge, developed in 3 dimensions. Two

series of concentrated forces, pointed down-wards, as resembling the weight of the structureor an eventual load, are applied on the bottomchords of the bridge. The aim of the finiteelement analysis is to estimate the axial stresses,reaction forces and the maximum nodal dis-placement of the structure. In advance a set ofbracings will be modified in length in order toreduce the overall structure deformation.

2 PROBLEM RESOLUTION

2.1 Model constructionThe structure is built by means of node gen-eration, after their positioning the linkage isprovided by the creation of 1-D elements, thusobtaining the reticular structure. The type ofelement used is LINK-180, provided with twoknots ’i’ and ’j’, every knot has 3 D.o.f. (Degreeof freedom) which are translation on axis x,y,z.As in this case, LINK-180 is used where onlythe axial force is transmitted. The parametersassigned to the elements are shown in the tablebelow: there are 3 different cross section areas,the first is applied to bottom chords, top chords,struts, the second to sway bracings, the third totop lateral bracings.

• Master course of mechanical design and machine elements, Pro-fessor Matteo Benedetti, academic year 2015/2016.See https://www.esse3.unitn.it/

Cross sectional area 1 1000 [mm2]Cross sectional area 2 600 [mm2]Cross sectional area 3 400 [mm2]Young modulus 210 [Gpa]Poisson’s ratio 0.3 [−]Loading force 20 [kNm ]Bridge width 3500 [mm]Bridge height 4000 [mm]link length (where possible) 4000 [mm]

Figure 1. Model structure

2.2 Model solution

Constraints are applied on the lateral nodes ofthe bottom chords. On the x=0 side, nodes areclamped, i.e. Ux, Uy, Uz are equal to zero. Onthe other side, where x = maximum, nodes arecarried, i.e. Uy, Uz are equal to zero.

The distributed force is elaborated in orderto obtain nodal concentrated force. The initialvalue in multiplied for the bottom chord length

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 1 2

and then divided for the number of knots.

20000[N/m] · 24[m] : 7[knots] = 68571[N ] (1)

Before launching the solution, it is impor-tant to notice that the non-linear geometry packageis activated. The reason lies in the fact that highdisplacements occur.

2.3 Post processing

Analysing the nodal solution, it is possible tosee that the maximum value is reached in thecentral part of the bridge. The maximum axialtension in reached in the central part of thebottom chords, on the other hand the maximumaxial compression is reached at the externalsway braces (e.g. element with knots 13-20).

Figure 2. Nodal displacement [mm]

Maximum displacement 70,97 [mm]Reaction force knot 1 -7x 240000y -10z [N ]Reaction force knot 7 0x 240000y -10z [N ]Reaction force knot 13 -7x 240000y -11z [N ]Reaction force knot 19 0x 240000y -11z [N ]Max axial compression -402,97 [MPa]Max axial tension 274,99 [MPa]

3 DISPLACEMENT MINIMIZATION

The six central nodes of the top chords(9,10,11,21,22,23) are modified by increasingtheir y coordinate. The process is guided bytwo parameters ∆h1,∆h2. The first is applied

to the two central nodes (10,22), the second tothe remaining.

By using a *DO command the process isre-executed various times, in order to test allthe possible conditions of ∆h1,∆h2 in a rangefrom 0 up to 6000mm. Every time the maximumdisplacement is saved in an appositely createdfile ’umax.txt’.

The first choice approach is based on thecentral node, which was seen before as themaximum displacement node. This logic leadsto a particular structure deformation whichgraphically resembles to the stationary wavesphenomenon, precisely to the 3rd harmonic.

Figure 3. Structure deformation, simlilar to 3rd harmonic

By extending the analysis to other knots,thus evaluating a combination of displace-ments, a new optimum point is reached. As aconsequence the particular deformation of thebottom chords disappears.

4 CONCLUSION

The minimization process leads to a new set ofresults:

The minimum displacement reaches theminimum of 38, 08mm with increment values:∆h1 = 3200,∆h2 = 3600.

A secondary goal is achieved: the axialstress is slightly reduced, in particular the ten-sion one. It is important to underline that the

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 1 3

Figure 4. Minimum elaboration (red dot)

Maximum displacement 38,08 [mm]∆h1 3200 [mm]∆h1 3600 [mm]Reaction force knot 1 -4x 240000y -8z [N ]Reaction force knot 7 0x 240000y -8z [N ]Reaction force knot 13 -4x 240000y -8z [N ]Reaction force knot 19 0x 240000y -8z [N ]Max axial compression -402,38 [MPa]Max axial tension 172,14 [MPa]

axial compression is quite high: further inves-tigations has to be focused on the bucklingphenomenon.

APPENDIX ALIST OF COMMANDS

data.txt

FINISH/CLEAR, START, NEW/FILNAM, diego_zenari_1.1/TITLE, pratt bridge 1.1

/cwd,’C:\Users\diego\Desktop\compitini\compito1’

!*************************!---BRDIGE MODIFICATION---!*************************! >>>>> MODEL PARAMETERS <<<

A1: 1000 !area mm^2A2: 600 !lunghezze mmA3: 400length: 4000width: 3500

! >>>>> PRE-PROCESSING <<<<</prep7!************************

Figure 5.

!---IMPOSTAZIONE CICLO---!************************!In case of loop cicle some othercommands have to be un-commented

!*do,dh1,2000,5000,200 !dh1,dh2 rappresentano!*do,dh2,2000,5000,200 !le variaz. in altezza

/prep7! ***** MAT-PROPERTIES *****Et,1,180 !element type, link 180Et,2,180Et,3,180

R,1,A1 !real constants: cross sectionR,2,A2R,3,A3

Mp,Ex,1,210000 !one type of material definitionMp,prxy,1,0.3

!_____________________!---NODE GENERATION---

N,1,0,0,0NGEN,7,1,1,,,length,0,0N,8,length,length,0NGEN,5,1,8,,,length,0,0N,13,0,0,widthNGEN,7,1,13,,,length,0,0N,20,length,length,widthNGEN,5,1,20,,,length,0,0

!first row modify!NMODIF,9,length*2,length+dh1,0!NMODIF,10,length*3,length+dh2,0!NMODIF,11,length*4,length+dh1,0

!second row modify!NMODIF,21,length*2,length+dh1,width!NMODIF,22,length*3,length+dh2,width!NMODIF,23,length*4,length+dh1,width

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 1 4

!_________________________!---ELEMENT CONNECTION---

type,1 !TOP & BOTTOM CHORDS, STRUTSreal,1mat,1

e,1,2egen,6,1,-1 !copies,increm,last creatione,2,8egen,5,1,-1e,8,9egen,4,1,-1e,13,14egen,6,1,-1e,20,21egen,4,1,-1e,14,20egen,5,1,-1e,13,1egen,7,1,-1e,20,8egen,5,1,-1

type,3 !TOP CHORDSreal,3mat,1

e,8,21egen,4,1,-1e,20,9egen,4,1,-1

type,2real,2mat,1

e,8,3 !INTERNAL SWAY BRACESe,9,4e,4,11e,5,12

e,20,15e,21,16e,16,23e,17,24

e,1,8 !external sway bracese,13,20e,12,7e,24,19

ALLSEL, ALL, ALLSAVEFINISH

!________________________! >>>>>> SOLUTION <<<<<<</solu! ******CONSTRAINTS******

D,1,all,0D,13,all,0

D,7,uy,0D,19,uy,0

D,7,uz,0D,19,uz,0

NSEL,S,node,,1,7,1F,all,fy,-68571

NSEL,S,node,,13,19,1F,all,fy,-68571

nlgeom,onsolcontrol,ontime,1pivcheck,off

ALLSEL, ALL, ALLSAVE

solveFINISH

!________________________! >>>>>> POSTPROC <<<<<<</post1

Prnsol,u,compEtable,axialstress,ls,1Pretab,axialstressEtable,force,smisc,1Pretab,axialforce

!*get,umax4,node,4,u,sum!*get,umax3,node,3,u,sum!*get,umax5,node,5,u,sum

!*cfopen,umax,txt,,append!*vwrite,dh1,dh2,umax3,umax4,umax5,!F5.0,F5.0,F20.10,F20.10,F20.10!*cfclos

finish!parsav,scalar,parametri,parm!/clear,start!parres,new,parametri,parm

!*enddo!*enddo

REFERENCES

[1] Slides and exercises of the course, Mecanichal design andmachine elements, M. Benedetti, University of Trento, 2015.

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 2 1

Hook load testDiego Zenari, 182160, M.Sc Mechatronics engineering

Abstract—In this second problem a hook is analysed: two types of section are introduced, the best one will bedetermined basing the analysis on maximum displacement and bending stress. Furthermore a short mesh refinementdiscussion will be examined.

Index Terms—section comparison, Beam189, internal actions.

F

1 INTRODUCTION

THE investigation manages to compare thedifferent behaviour of a hook by changing

its cross section. A concentrated force,as resem-bling an eventual load, is applied in centralpart at the bottom. The aim of the finite elementanalysis is to estimate the maximum nodaldisplacement, bending moments and stresses,shear forces.

2 PROBLEM RESOLUTION

2.1 Model constructionThe structure is built by means of keypointsgeneration, after their positioning the connec-tion is realized by the creation of line elements.The type of element used is BEAM-189, pro-vided with three knots: ’i’, ’j’ and ’k’. Everyknot has 6 D.o.f. (Degree of freedom) which aretranslation and rotations on axis x,y,z. Thusthe shape function either for translation androtation is characterised by 3 conditions, thatis: N(ξ) = aξ2 + bξ + c. A forth knot is used toprovide orientation for the section: it orientatesthe z-axis from the ’k’ knot to the orientationknot. The 6th keypoint was selected in this case:Figure 1 shows three examples of axis orienta-tion. The parameters assigned to the elementsare shown in the table below.

Both the sections are user-defined. In the firstcase it is built by keypoint generation, lineconcatenation and thus area generation. In the

• Master course of mechanical design and machine elements, Pro-fessor Matteo Benedetti, academic year 2015/2016.See https://www.esse3.unitn.it/

Hook circle center (0,-200,0) [mm]Hook circle radius 100 [mm]Circle development ( 5π4 ,−π

4 ) [rad]Fillets 30 [mm]Poisson’s ratio 0.3 [−]Young modulus 205 [Gpa]Loading force 20 [kN ]

Figure 1. Model keypoints and lines

Figure 2. Section 1: trapezium, Section 2: rail

second case three areas are generated: a rect-angle and two semi-circles. By subtracting thetwo last areas to the first one, the final shape isobtained. The trapezium section is easily andautomatically map meshed with quadrilateralelements. The rail section is freely mapped, dueto the geometric complexity, with quadrilateralelements.

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 2 2

2.2 Model solution

The structure is clamped at the top, i.e.Ux, Uy, Uz, Rotx, Roty, Rotz are equal to zero.

The concentrated force is applied in the low-est part of the section, in-line with the centerof the hook circle.

The element length, used to subdivide thelines, is fixed to 20 mm for displacement, 5 mmfor force diagrams.

2.3 Post processing

Analysing the nodal solution, it is possibleto see that the maximum value of nodal dis-placement is reached at the end of the hook.Obviously shear force and bending moment areequal in both the sections. Their value is purelyfixed by the line geometry: what changes isbending stress and shear stress, which dependon area and inertia of the cross section.

Trapezium sectionMaximum displacement 2.49 [mm]Maximum shear force z 14200 [N ]Minimum shear force z -20000 [N ]Maximum bending moment y 2 · 106 [N ·mm]Maximum bending stress 230.95[MPa]Rail sectionMaximum displacement 3.10 [mm]Maximum shear force z 14200 [N ]Minimum shear force z -20000 [N ]Maximum bending moment y 2 · 106 [N ·mm]Maximum bending stress 319.03[MPa]

Figure 3. Nodal displacement

The shear force y and bending moment z are notreported cause their values approaches zero.The trapezium section performs the best re-sults: the nodal displacement and the bendingstress is significantly lower than the rail section.

As noted before, shear and moment dia-grams are the same for both the sections, andthis is the reason why one plot for both sectionsis sufficient.

Figure 4. Shear z, bending moment y and bendingstress diagrams

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 2 3

The maximum bending moment and stressis reached at the shear inversion point (Figure4), located at the same height of the hook circlecenter.

3 MESH REFINEMENT

A mesh sensitivity investigation is carried out:the element length is varied in a range from 1to 50 mm. Using a *DO command the analy-sis is re-executed varying the element length.Almost all the results have a great stability:for example maximum displacement change isabout 0.5%, the bending stress varies about2%. There is only one convergent behaviour:the maximum negative bending moment. Byrefining the mesh, its value drives to zero. It islocalized near keypoint 8, at the end of the firststraight line starting from the clamped edge. Itis possible to see from Figure 6 that in element25, due to shape function approximation thebending moment undergoes an inversion. Asthe mesh is refined this behaviour tends todisappear. However, the moment estimationerror doesn’t lead to a significant error in bend-ing stress, the reason lies in the fact that it isnot comparable with the positive component,which is around 100 times greater.

4 CONCLUSION

The examination has shown that even if theelement length is enlarged, till reaching evidentgeometrical approximations, the result doesn’tsuffer significant changes or invalidations.

Figure 5. moment convergence

Figure 6. moment inversion

Also from the point of view of the cross sec-tion the variation of element dimension doesn’trealize a concrete change in result.

APPENDIX ALIST OF COMMANDS

data.txt

! ----------------------------------------! PROBLEM: HOOK! ----------------------------------------

Finish/CLEAR, START, NEW/FILNAM, hook_diegozenari/TITLE,hook_diegozenari/cwd,’C:\Users\diego\Desktop\compito2’

!---------MODEL PARAMETERS---------!Center of the circle mmCirclecenter : 200!Arc Radius mmRadius : 100!Fillet radius mmRfillet : 30!Elment length mmE_length : 20

!--------MAT. PROPE_young : 205000Poisson : 0.3

!--------FORCEF : 20000 !N -> MPa N / mm^2

!-----PROBLEM INITIZIALIZATION/PREP7ET,1,BEAM189

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 2 4

!section activation in dependece of the case!SECTYPE,1,BEAM,Mesh!SECREAD,sezione_hooke_1,sect,,mesh

SECTYPE,1,BEAM,MESHSECREAD,sezione_hooke_2,sect,,Mesh

MP,Ex,1,E_youngMP,prxy,1,Poisson

!-------MODEL CREATION!geometrical calculations!tangential keypointpi : acos-1x1 : sinpi/4*Radiusx2 : Circlecenter - x1x3 : x2 - x1x4 : Circlecenter + x1x5 : Circlecenter + Radius

K,1,0,0,0 !hook topK,2,0,-x3,0 !hook first curvatureK,3,-x1,-x2,0 !circle tangentK,4,0,-x5,0 !force pointK,5,x1,-x4,0 !hook endK,6,x1,-Circlecenter,0 !orientation pointK,7,-x1,-x4,0 !arc break, avoid section "torsion"L,1,2L,2,3

LARC,3,7,6,RadiusLARC,7,4,6,RadiusLARC,4,5,6,RadiusLFILLT,1,2,Rfillet

LSEL,ALLLATT,1,,1,,6,,1

LSEL,ALLLESIZE,ALL,E_lengthLMESH,ALL

!-------SOLUTION/SOLUTIONANTYPE,STATIC,NEW

!constraintsDK,1,ALL,0

!force applicationFK,4,FY,-F

SOLVEFinish

!-------POST-PROCESSING/POST1/ESHAPE,1

PLNSOL,S,EQV

!Internal actions

ETABLE, shearyI, SMISC, 6 !shear I

ETABLE, shearyj, SMISC, 19 !shear JETABLE, shearzI, SMISC, 5 !shear IETABLE, shearzj, SMISC, 18 !shear J

ETABLE, momzI, SMISC, 3 !moment z in IETABLE, momzJ, SMISC, 16 !moment z in JETABLE, momyI, SMISC, 2 !moment y in IETABLE, momyJ, SMISC, 15 !moment y in J

ETABLE, momstresI, SMISC, 34 !moment y in IETABLE, momstresJ, SMISC, 39 !moment y in J

PLLS,momstresi,momstresj

REFERENCES

[1] Slides and exercises of the course, Mecanichal design andmachine elements, M. Benedetti, University of Trento, 2015.

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 3 1

Shaft stress concentrationDiego Zenari, 182160, M.Sc Mechatronics engineering

Abstract—The aim of this third problem is to determine the stress concentration factor in a shaft generated by a sectionvariation with a specified fillet radii. Furthermore, the effect of a groove will be analysed.

Index Terms—stress concentration, plane182, groove, fillet.

F

1 INTRODUCTION

THE subject of the analysis is a shaft whichundergoes a section restriction. Geometric

specifications are in Figure 1.

Figure 1. Shaft geometry

The shaft is subjected to an axial load appliedat the tip (right side in figure 1). After deter-mining the stress concentration factor, a reliefgroove will be introduced, with a consequentre-analysis. The stress concentration factor isdefined as the maximum principal stress overthe external nominal pressure.

Kshape = max[ σiPest

]i = 1 (1)

• Master course of mechanical design and machine elements, Pro-fessor Matteo Benedetti, academic year 2015/2016.See https://www.esse3.unitn.it/

2 PROBLEM RESOLUTION

2.1 Model construction

Owing to the shaft axial and longitudinalsymmetry the model analysed is simply aplane representing half of the shaft section.The model is built by means of bottom upapproach: firstly the Keypoints are generated,then linked with lines, areas are defined on linesand in conclusion meshed. The type of elementused is PLANE-182, consisting in 4 knots planeelement where each knot is characterised by 2d.o.f: translation about x and y axis. As a result,shape functions are polynomials of first orderin two variables: N(β, γ) = (aβ+ bγ+ cβγ+d).

Figure 2. keypoints and lines, model subdivision

The mapped meshing method is applied.The model is sub-divided in several sub-areasand the mesh refinement of every "district"

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 3 2

is determined on the stress gradient obtainedfrom a first coarse mesh.

D / d 1.4r / d 0.05R (D − d)/2Poisson’s ratio 0.3 [−]Young modulus 205 [Gpa]External pressure 50 [Mpa]d 50 [mm]

2.2 Model solutionThe structure is clamped at the bottom, i.e.Ux, Uy, Uz, Rotx, Roty, Rotz are equal to zero.Owing to axisymmetry, the left side (wherex=0) is constrained with carriages orientedalong the x axis. The top line is subjected topressure oriented upward.

2.3 Post processingA mesh refinement sensitivity is carried out,the mesh is chosen evaluating the maxi-mum stress convergence, the iso-stress con-tours smoothness, elements number. As writtenbefore, in every test the mesh is locally refinedwhere the stress gradients are higher. That is,moving closer to the section restriction andfillet.

The 1st principal stress is the highest princi-pal stress produced.

Convergence analysis1st princ. stress max [MPa] elements number

131.36 5000131.27 7500131.09 12000130.94 18000130.8 28000

The first mesh (5000 elements) is chosen as’reference’ and it is built considering an in-creased mesh localized refinement, thus achiev-ing a greater element dimension gradient, themaximum stress achieved diverges only 0.45%from the most refined mesh (ANSYS studentlimits at 32000 elements). The stress concentra-tion factor is estimated around 2.6.

It is possible to validate this result comparingthe solution to the available literature. Figure 4shows a condition close to the subject of the

Figure 3. 1st principal stress distribution

analysis, in particular the section restriction issignificantly wider, and consequently inducts alower concentration.

Figure 4.

The value estimated by literature, around 2.3,is slightly lower to the numerical one.

3 GROOVE ANALYSIS

The introduction of the groove requires thegeneration of a new sub-area: this necessity isdue to the difficulty of creating a mapped meshwithout distorted elements. Furthermore, thenew sub-area introduced allows the handlingof a *DO process which translates by little stepsthe groove, enabling a minimum research. Thelength of x, that is the distance between the

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 3 3

section restriction and the first edge of thegroove, is varied from 2.3 up to 6.6 mm. Fig-ure 5. During the analysis distorted elementsare produced, but, owing to their position farfrom the maximum stress points, their presencedoesn’t affect the investigation.

The number of element is slightly increasedrespect to the previous model without groove.

The stress concentration factor significantlydecreases, and it is numerically estimatedaround 1.8 in dependence of the groove po-sition. This situation could be compared againwith Figure 4: imposing a fillet radius equal tothe groove radius then the shape factor standsbetween 1.6 - 1.7 As presented in the figure, the

Figure 5. 1st principal stress behaviour

maximum principal stress tends to decrease asthe groove moves closer to the fillet. Logically,the behaviour of the plate tends to literaturevalue of figure 4. Theoretically, the finite ele-ment process should be suspended only whenthe maximum stress passes from the fillet tothe groove. The minimum value achieved at 1.8mm is still located in the neighbourhood of thefillet, however, it differs of only 0.4 MPa fromthe stress sited in the vicinity of the groove.

4 CONCLUSION

The mesh chosen as reference aims to balancethe estimation accuracy and elements number.The stress concentration factor estimated iscompatible with the data available by litera-ture. As expected the introduction of the relief

Figure 6. 1st principal stress distribution

groove decreases the stress concentration. Theaim of the study was to find the minimumstress, without considering practical risks: thecurrent groove position, (or a further reductionof it) could be not reasonable: the thin layer be-tween fillet and relief groove could easily breakor deform under loading conditions differentfrom the simple tensile test, or in the presenceof shaft accessories.

APPENDIX AMESHES

Figure 7. fillet analysis mesh

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 3 4

Figure 8. groove analysis mesh

APPENDIX BLIST OF COMMANDS

data.txt

! ----------------------------------------! PROBLEM: shaft fillet! ----------------------------------------

Finish/CLEAR, START, NEW/FILNAM, fillet_diegozenari/TITLE,fillet_diegozenari/cwd,’C:\Users\diego\Desktop\compitini\compito3’

!---------MODEL PARAMETERS---------d : 50 !mmDi : 70 !mmRfillet : 2.5 !mmRgroove : 10 !mm

dista : Rgroove/6middle : 3.5*Di/8

!--------MAT. PROPE_young : 210000 !MPaPoisson : 0.3

!--------PRESSUREP : -50 !Mpa

ndiv1 : 15 !15ndiv2 : 25 !25ndiv3 : ndiv2+ndiv1ndiv4 : 30 !30ndiv5 : 10 !10

numbe : 100 ! ficticious length

*dim,smax,array,numbe!--------MODEL CONSTRUCTION-----------conta:1 !count parameter

*do,sp,0.6,4,0.2

/PREP7Et,1,PLANE182,3,,1 !plane element, stimpified

!strain formulation, axisymmMP,Ex,1,E_youngMP,prxy,1,Poisson

!key point construction, the object will!be subdivided in five areasK,1,0,0,0K,2,Di/2,0,0K,3,Di/2,d+2*Rgroove+sp,0K,4,middle-Rgroove,d+2*Rgroove+sp,0K,5,0,d-2*sp,0

K,6,Di/2,d+2*Rgroove+sp+Rgroove/6,0K,7,d/2,d+2*Rgroove+sp+Rgroove/6+Rfillet,0K,8,Di/2-7.5,d+2*Rgroove+sp+Rgroove/6,0K,9,middle-Rgroove,d+2*Rgroove+sp+Rgroove/6+Rfillet,0K,10,0,d+2*Rgroove+sp+Rgroove/6+Rfillet,0K,11,0,2*d+2*Rgroove+sp+Rgroove/6,0K,12,d/2,2*d+2*Rgroove+sp+Rgroove/6,0K,13,0,3.5*d/2,0K,14,d/2,3.5*d/2,0

!additive kp for groove areaK,20,Di/2,d-2*spK,21,Di/2,dK,22,Di/2,d+2*RgrooveK,23,middle-Rgroove,d-2*sp,0K,100,2*d,2*d,0

L,1,5L,1,2L,2,20L,4,3L,20,23L,23,5L,23,4L,20,21LARC,21,22,23,RgrooveL,22,3L,5,10L,4,9L,3,6L,9,7L,10,9L,10,13L,13,11L,11,12L,12,14L,14,7L,13,14L,6,8LARC,8,7,100,Rfillet

/pnum,line,1/pnum,kp,1

!area meshing containing filletal,14,23,22,4,12,13lsel,alllesize,14,,,0.6*ndiv1,0.4!10lesize,23,,,ndiv2!10lesize,22,,,ndiv2/2,0.3!15,0.3

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 3 5

lesize,13,,,0.6*ndiv1lesize,4,,,ndiv2!15,0.4lesize,12,,,ndiv2/2!15lsel,s,line,,4lsel,a,line,,12lccat,alllsel,s,line,,22,23,1lccat,allTYPE,1Mat,1mshape,0,2Dmshkey,1amesh,1

!groove arealsel,alllesize,5,,,ndiv2!10,3lesize,7,,,ndiv2!25lesize,10,,,ndiv1!30,0.3lesize,8,,,ndiv1!30,0.3lesize,9,,,3*ndiv2al,10,4,7,5,8,9lsel,s,line,,4,5,1lsel,a,line,,7lccat,allTYPE,1Mat,1amesh,2

!area beside filletlsel,alllesize,6,,,ndiv2,3!15lesize,15,,,ndiv2,0.2!10,0.3lesize,11,,,1.5*ndiv2,0.2al,6,11,15,7,12lsel,s,line,,7lsel,a,line,,12lccat,allTYPE,1Mat,1amesh,3

!bottom arealsel,alllesize,1,,,ndiv1!20,2lesize,3,,,ndiv1!20,0.5lesize,2,,,2*ndiv2,0.5!20,0.7AL,1,2,3,5,6LSEL,S,LINE,,5,6,1LCCAT,ALLTYPE,1Mat,1AMESH,4

!top arealsel,alllesize,20,,,ndiv4,0.2!20,0.7lesize,16,,,ndiv4,4!10lesize,21,,,ndiv2+0.6*ndiv1,0.3!10AL,15,20,21,14,16lsel,s,line,,14,15,1lccat,allTYPE,1Mat,1AMESH,5

!top arealsel,alllesize,18,,,ndiv2+0.6*ndiv1!20,0.7lesize,17,,,ndiv5!10lesize,19,,,ndiv5!10AL,17,18,19,21TYPE,1Mat,1AMESH,6

eplot

!-----SOLUTION-----/SOLUTION!constran bottom lineDL,2,,ALLnsel,s,loc,x,-0.01,0.01dsym,symm,xallsel,all!symmetry on central nodes!tensile pressure on top lineSFL,18,pres,Psolve

!-----POST-PROCESSING-----/post1PLNSOL,S,1

*get,smaxconta,PLNSOL,0,MAXconta : conta +1

finishparsav,all,parametri,parm/clear,STARTparres,new,parametri,parm

*enddo

*cfopen,prova,txt,,

*vwrite,smax1F20.10

finish

*cfclos

REFERENCES

[1] Slides and exercises of the course, Mecanichal design andmachine elements, M. Benedetti, University of Trento, 2015.

[2] Shigley Progetto e costruzione di macchine, R. Budynas, J.K. Nisbett, McGrawHill, 2014.

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 1 1

Pipe JunctionDiego Zenari, 182160, M.Sc Mechatronics engineering

Abstract—The fourth problem aims at determining the stress distribution in a "T" pipe junction, consisting of a main(and larger) pipe, and a secondary (and tider) pipe. The junction consists in a sharpened edge.

Index Terms—shell, shell-181, sharp edge junction.

F

1 INTRODUCTION

THE subject of the analysis is an hydraulicconnection subjected to an internal pres-

sure of 10 bar. The junction is characterised bya sharpened edge. Geometric specifications arein Figure 1.

Figure 1. Pipe junction

It is expected an high stress state due tothe fillet absence. Interest will be given tothe analysis of the pipe deformation, and theinvestigation aims to obtain meridional andcircumferential stress:

• Along the periphery of the junction• Far from the junction along the main tube• Far from the junction along the secondary

tube

• Master course of mechanical design and machine elements, Pro-fessor Matteo Benedetti, academic year 2015/2016.See https://www.esse3.unitn.it/

2 PROBLEM RESOLUTION

2.1 Model constructionThanks to the double symmetry of the compo-nent, the analysed model is simply 1/4 of theoriginal shape. The structure is built by meansof bottom up approach: firstly a generatingline is created, then rotated along the cylinderaxis,thus generating areas. Junction is built bymeans of area intersection: the unnecessaryareas are deleted, the remaining are meshed.The type of element used is SHELL-181, con-sisting in 4 knots shell element where each knotis characterised by 6 d.o.f: 3 translations + 3rotations. Axisymmetry keyoption is enabled.The mapped meshing method is applied. The

Figure 2. global view, areas plot

main tube is subdivided in four areas in orderto allow an easy mesh refinement in the neigh-bourhood of the junction.

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 1 2

Main pipe radius 200 [mm]Secondary pipe radius 100 [mm]Main pipe length 1000 [mm]Poisson’s ratio 0.3 [−]Young modulus 210 [Gpa]Internal pressure 1 [Mpa]

2.2 Model solutionAt the sections A− A and B −B the structureis constrained straight-ward, that is, carriagesare putted along the circumference, directedalong the cylinder axis: firstly as to resemblingthe tube continuity, secondly to provide theconditions for a static solution. Owing to ax-isymmetry all the remaining boundaries, gener-ated by the symmetry cutting, are subjected tosymmetry conditions (DSYMM command). Theinternal pressure is applied by selecting areas.Due to the 3-D extension of the model, it is fun-damental to check for element surface normalhomogeneity: it ensures a correct orientation ofthe pressure applied (/PSYMB command).

2.3 Post processingThe mesh is gradually refined, in particularnear the junction periphery. The evaluation ofthe mesh is based on the smoothness of the iso-stress contours.

Convergence analysisVon Mises stress max [MPa] elements number

96.10 10500104.18 22000105.32 26000106.33 30000

The student release of ANSYS allows a max-imum element count of 32000 thus limiting theinvestigation about the maximum stress, whichis obviously located very close to the junction.

The displayment of the results is precededby rsys,solu, which orients the results to theelement coordinate systems, allowing the se-lection of the hoop stress (s,x) and meridionalstress (s,y).

In order to display results the line sitedon the junction has been selected. Then othertwo knots, one on the main tube and theother on the secondary tube, are graphically se-lected: they represent the reference point for the

stresses on the tubes far from the junction, asto test the behaviour of the section in ordinaryconditions.

Figure 3.

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 1 3

Figure 4. meridional stress distribution

3 CONCLUSION

The sharpened junction is characterised bya great stress concentration. Probably, by in-creasing the elements number, it is possible toachieve higher stress values.

Both the tubes sections are deformed assum-ing an elliptical shape: in the main tube case,the greater axis is aligned with the secondarytube, while in the secondary tube case, thegreater axis is aligned with the main pipe. Inthe vicinity of the junction the deformation isquite complex.

Top shell results min maxMax Von Mises stress - - 106.32 [MPa]Max meridional stress -39.93 51.78 [MPa]

Max hoop stress -77.20 22.43 [MPa]Max displacement - - 0.38 [mm]

Table 1Maximum results are not achieved in the same point

Mid shell results min maxMax Von Mises stress - - 84.12 [MPa]Max meridional stress -88.22 3.92 [MPa]

Max hoop stress -26.27 10.15 [MPa]

Table 1 proves that the maximum values ofstress are not achieved strictly on the junctioncontour, however, as shown in figure 4 and5, the high-stress field is very close to thejunction.

Figure 5. hoop stress distribution

Figure 6. general deformation

APPENDIX ALIST OF COMMANDS

data.txt

!--------------------------------------------! cylindrical shell!--------------------------------------------

finish/clear,start,new/FILNAM, pipejoint_diegozenari/TITLE,pipejoint_diegozenari

/cwd,’C:\Users\diego\Desktop\compitini\compito4’

!-----------------------! parameters definition!-----------------------

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 1 4

*set,length1,1000

*set,rad1,100

*set,thick1,4

*set,length2,500

*set,rad2,50

*set,thick2,2

*set,ndiv1,70 !70

*set,ndiv2,60 !60

*set,ndiv3,30 !30

*set,space,8

*set,PRES,1

*set,pi,acos-1

!-----------------------! Elements definition!-----------------------

/PREP7!store data of bottom, mid!and top layers with keyopt 8

ET,1,SHELL181KEYOPT,1,8,2SECTYPE,1,shellSECDATA,thick1

ET,2,SHELL181KEYOPT,1,8,2SECTYPE,2,shellSECDATA,thick2

MP,EX,1,210000MP,PRXY,1,0.3

MP,EX,2,210000MP,PRXY,2,0.3

!-------------------! model construction!-------------------/pnum,kp,1/pnum,line,1/pnum,area,1

k,1,rad1-thick1/2k,2,rad1-thick1/2,length1-pi*rad1/2k,3,rad1-thick1/2,length1/2k,4,0,length1/2k,5,

l,1,2l,2,3arotat,1,2,,,,,4,5,90arotat,3,4,,,,,4,5,90

CLOCAL,11,1,+rad1+length2,length1/2,0,,90,-90

csys,11k,16,0,0,0k,17,0,0,length2+rad1

k,18,rad2-thick2/2,90,length2+rad1k,19,rad2-thick2/2,90,0l,19,18arotat,13,,,,,,16,17,-90

aptn,2,5

adele,6ldele,17ldele,18

adele,8ldele,23ldele,22ldele,16

nummrg,allnumcmp,all

!-----------------------! Meshing!-----------------------

!central area sizinglesize,3,,,ndiv1lesize,5,,,ndiv1lesize,12,,,2*ndiv1lesize,16,,,ndiv2,0.3lesize,15,,,ndiv2,0.3

lsel,s,line,,3,5,2lccat,alllsel,all

!side main arealesize,7,,,ndiv1lesize,10,,,ndiv3lesize,9,,,ndiv3

!main tube bottomlesize,4,,,ndiv1lesize,1,,,ndiv3lesize,2,,,ndiv3

!side bottomlesize,6,,,ndiv3lesize,8,,,ndiv3

!secondary pipelesize,12,,,2*ndiv1lesize,11,,,2*ndiv1lesize,14,,,2*ndiv2,3lesize,13,,,2*ndiv2,0.3

!element type, material propertiesasel,s,area,,1,3,1asel,a,area,,5aatt,1,,1

asel,s,area,,4aatt,1,,2

asel,allmshkey,1mshape,0,2D

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 1 5

amesh,all

!element reference frame plot/psymb,esys,1eplot

SAVEFINI!-------------------! solution!-------------------/SOLUcsys,0

!symmetry boundary conditionsnsel,s,loc,y,length1/2dsym,symm,y

nsel,s,loc,y,0d,all,uy,0

nsel,s,loc,x,length2+rad1d,all,ux,0

nsel,s,loc,z,0dsym,symm,z

!pressure application on elementsasel,allsfe,all,1,pressure,,PRES/PBC,ALL, ,1allsel,all

SAVESOLVEFINI

!-------------------! postprocessor!-------------------/post1plnsol,s,eqv

!midplane-bottom-top selectshell,midrsys,solu!element ref results orientedlsel,s,line,,12nsll,s,1

!do cycle along the pheripery!to get membrane stressescsys,11cm,nodi,node

*get,nnodi,node,,count

*dim,sx,array,nnodi

*dim,sy,array,nnodi

*dim,angolo,array,nnodi

*do,i,1,nnodi

*get,zmini,node,,mnloc,znsel,r,loc,z,zmini-0.001,zmini+0.001

*get,nmin,node,,num,max

*get,sxi,node,nmin,s,x

*get,syi,node,nmin,s,y

*get,angoloi,node,,mnloc,y

cmsel,s,nodinsel,u,node,,nmincm,nodi,node

*enddo

*cfopen,stress_junction,csv,

*vwrite,sx1,sy1,angolo1F20.10,F30.10,F40.10

*cfclos

!do cycle along the pheripery!to get membrane+bending stressesallsel,all/post1shell,botrsys,solufinish

lsel,s,line,,12nsll,s,1!do cycle along the pheripery!to get membrane stressescsys,11cm,nodi,node

*get,nnodi,node,,count

*dim,sx,array,nnodi

*dim,sy,array,nnodi

*dim,angolo,array,nnodi

*do,i,1,nnodi

*get,zmini,node,,mnloc,znsel,r,loc,z,zmini-0.001,zmini+0.001

*get,nmin,node,,num,max

*get,sxi,node,nmin,s,x

*get,syi,node,nmin,s,y

*get,angoloi,node,,mnloc,y

cmsel,s,nodinsel,u,node,,nmincm,nodi,node

*enddo

*cfopen,stressb_junction,csv,

*vwrite,sx1,sy1,angolo1F20.10,F30.10,F40.10

*cfclos

allsel,all/post1shell,midfinish!MAIN PIPE PERIPHERY!do cycle far from pheripery!node picked graphically under!the area division in order!to get same element orient.csys,0

*get,ipsilon,node,182,loc,y

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 1 6

nsel,s,loc,y,ipsilon-0.01,ipsilon+0.01cm,nodi2,node

*get,nnodi2,node,,count

*dim,sx_far,array,nnodi2

*dim,sy_far,array,nnodi2

*dim,angolo2,array,nnodi2

*do,i,1,nnodi2

*get,xmini,node,,mnloc,xnsel,r,loc,x,xmini-0.001,xmini+0.001

*get,nmin,node,,num,max

*get,sx_fari,node,nmin,s,x

*get,sy_fari,node,nmin,s,yangolo2i : xmini/rad1-thick1/2

cmsel,s,nodi2nsel,u,node,,nmincm,nodi2,node

*enddo

*cfopen,stress_far,csv,

*vwrite,sx_far1,sy_far1,angolo21F20.10,F30.10,F40.10

*cfclos

nsel,all!SECONDARY PIPE PERIPHERY!do cycle far from pheripery!node picked graphically under!the area division in order!to get same element orient.csys,11

*get,zeta,node,5817,loc,znsel,s,loc,z,zeta-5,zeta+1cm,nodi3,node

*get,nnodi3,node,,count

*dim,sx_far2,array,nnodi3

*dim,sy_far2,array,nnodi3

*dim,angolo3,array,nnodi3

*do,i,1,nnodi3

*get,ymini,node,,mnloc,ynsel,r,loc,y,ymini-0.001,ymini+0.001

*get,nmin,node,,num,max

*get,sx_far2i,node,nmin,s,x

*get,sy_far2i,node,nmin,s,yangolo3i : ymini

cmsel,s,nodi3nsel,u,node,,nmincm,nodi3,node

*enddo

*cfopen,stress_far2,csv,

*vwrite,sx_far21,sy_far21,angolo31F20.10,F20.10,F20.10

*cfclosSAVEFINI

REFERENCES

[1] Slides and exercises of the course, Mecanichal design andmachine elements, M. Benedetti, University of Trento, 2015.

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 5 1

Pipe junction sub-modellingDiego Zenari, 182160, M.Sc Mechatronics engineering

Abstract—The fifth problem involves the implementation of the pipe junction problem: a fillet is introduced at the junction.The objective is to achieve a more realistic stress state condition near the junction periphery

Index Terms—solid bricks, solid-186, filleted junction.

F

1 INTRODUCTION

THE subject of the analysis is an hydraulicconnection subjected to an internal pres-

sure of 10 bar. The junction is characterised bya filleted edge. Geometric specifications are inFigure 1.

Figure 1. Pipe junction

By resuming the analysis carried out in thepreceding work, a solid 3-D sub-model is built.It aims to get the maximum and general stressdistribution in the neighbourhood of the junc-tion; as a consequence the attention is focusedon the fillet behaviour, expecting a stress reduc-tion in comparison with the sharpened edgeanalysis.

• Master course of mechanical design and machine elements, Pro-fessor Matteo Benedetti, academic year 2015/2016.See https://www.dii.unitn.it/

2 PROBLEM RESOLUTION

2.1 Model construction

The solid model obviously requires the gen-eration of volumes: similarly to the precedinganalysis the volume is generated firstly bycreating the primitive lines, then rotating themalong the cylinder axis,thus generating areasand then forming volumes. The junction is builtby means of volume intersection: the unneces-sary objects are deleted. Sub-model borders:

• Main pipe: a co-axial cylinder cuts themain pipe sector in order to create anequidistant cutting boundary.

• Secondary pipe: the tube is simply short-ened, as being cut by an transversal plane.

Figure 2. global view, areas plot

The type of element used is SOLID-186, con-sisting in 20 knots solid element where each

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 5 2

knot is characterised by 3 d.o.f: 3 translations.The free meshing method is applied: in order toget reliable interpolation results it is required tohave almost two layers of elements distributedon the thickness 1. This condition is imposedby the composition of membrane stresses: mo-ments and forces.

Main pipe radius 200 [mm]Secondary pipe radius 100 [mm]Main pipe length 1000 [mm]Poisson’s ratio 0.3 [−]Young modulus 210 [Gpa]Internal pressure 1 [Mpa]Fillet radius 10 [mm]element size (suggested) 1 [mm]

2.2 Model solutionThe sub-modeling boundaries specified in sec-tion 1 are selected and saved (NWRITE com-mand). Later, when the analysis requires theresume of the complete model solution, ansyswill be able to automatically synchronize thetwo projects following the saved nodes byNWRITE. It is fundamental to check for rightorientation of the two projects, an eventual mis-match impedes the resolution of the problem.

The remaining boundaries, as in previousproject, are subjected to symmetry conditions(DSYMM). Internal pressure is applied by se-lecting internal areas.

2.3 Post processingThe results are affected by a stress "bordereffect" localized on the sub-model cutting bor-ders. It is generated by the application of theboundary conditions from the complete model:by moving the cutting boundaries and enrich-ing the number of elements of the sub-model,the effect changes its intensity. The problem iseasily tackled by selecting a sub-set of nodesfar enough from the affected borders, conse-quently the results plotted in GUI gain in clar-ity. A mesh sensitivity is carried out, bringingto the following results: The converge analysisbrought to a maximum Von Mises stress of

1. Thanks to the ansys teaching advanced release available atuniversity pc suites, the elements limit is brought to 256’000,allowing an enough refined mesh

Figure 3. Stress distribution

86.97 MPa even tough, the results are quiteunstable: by changing the sub-model sizes andkeeping an element number between 100’000 -200’000 the solution varies about 12%. In addi-tion, by extending the sub-model borders overdouble of secondary pipe radius, the results areno more in agreement with the preceding work(homework 4)1.

In order to store stress results and have anidea of the stress distribution, two line sited onthe fillet has been selected (line 31 and 34).

Figure 4. Stress distribution along fillet lines

1. Mechanical design and machine elements problem 4,Diego Zenari, 2015

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 5 3

3 CONCLUSION

The introduction of the fillet significantly re-duces the stress state of the junction, how-ever, as first analysis involving solid modelling,there is a strong increase in time computationcost respect to the preceding projects, enlargedalso by high elements number. This conse-quence stresses the importance to exploit sym-metries, which results in fundamental compu-tation time reduction

APPENDIX ALIST OF COMMANDS

data.txt

!--------------------------------------------! cylindrical shell!--------------------------------------------

finish/clear,start,new/FILNAM, pipejoint_diegozenari/TITLE,pipejoint_diegozenari

!-----------------------! parameters definition!-----------------------

*set,length1,1000

*set,rad1,100

*set,thick1,4

*set,length2,500

*set,rad2,50

*set,thick2,2

*set,ndiv1,70

*set,ndiv2,60

*set,ndiv3,30

*set,space,8

*set,PRES,1

*set,pi,acos-1

!-----------------------! Elements definition!-----------------------

/PREP7

ET,1,SHELL181KEYOPT,1,8,2SECTYPE,1,shellSECDATA,thick1

ET,2,SHELL181KEYOPT,1,8,2SECTYPE,2,shellSECDATA,thick2

MP,EX,1,210000MP,PRXY,1,0.3

MP,EX,2,210000MP,PRXY,2,0.3

!-------------------! model construction!-------------------/pnum,kp,1/pnum,line,1/pnum,area,1

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 5 4

k,1,rad1-thick1/2k,2,rad1-thick1/2,length1-pi*rad1/2k,3,rad1-thick1/2,length1/2k,4,0,length1/2k,5,

l,1,2l,2,3arotat,1,2,,,,,4,5,90arotat,3,4,,,,,4,5,90

CLOCAL,11,1,+rad1+length2,length1/2,0,,90,-90

csys,11k,16,0,0,0k,17,0,0,length2+rad1k,18,rad2-thick2/2,90,length2+rad1k,19,rad2-thick2/2,90,0l,19,18arotat,13,,,,,,16,17,-90

aptn,2,5

adele,6ldele,17ldele,18

adele,8ldele,23ldele,22ldele,16

nummrg,allnumcmp,all

!-----------------------! Meshing!-----------------------

!central area sizinglesize,3,,,ndiv1lesize,5,,,ndiv1lesize,12,,,2*ndiv1lesize,16,,,ndiv2,0.3lesize,15,,,ndiv2,0.3

lsel,s,line,,3,5,2lccat,alllsel,all!side main arealesize,7,,,ndiv1lesize,10,,,ndiv3lesize,9,,,ndiv3!main tube bottomlesize,4,,,ndiv1lesize,1,,,ndiv3lesize,2,,,ndiv3!side bottomlesize,6,,,ndiv3lesize,8,,,ndiv3!secondary pipelesize,12,,,2*ndiv1lesize,11,,,2*ndiv1lesize,14,,,2*ndiv2,3

lesize,13,,,2*ndiv2,0.3

!element type, material propertiesasel,s,area,,1,3,1asel,a,area,,5aatt,1,,1

asel,s,area,,4aatt,1,,2

asel,allmshkey,1mshape,0,2D

amesh,all

!element reference frame plot/psymb,esys,1eplot

SAVEFINI!-------------------! solution!-------------------/SOLUcsys,0!symmetry boundary conditionsnsel,s,loc,y,length1/2dsym,symm,y

nsel,s,loc,y,0d,all,uy,0

nsel,s,loc,x,length2+rad1d,all,ux,0

nsel,s,loc,z,0dsym,symm,z

!pressure application on elementsasel,allsfe,all,1,pressure,,PRES/PBC,ALL, ,1allsel,all

SAVESOLVEFINI!---------------------------------!---SUBMODELLING-PIPE-JUNCTION----!---------------------------------finish/CLEAR,START,NEW/FILNAM, pipesubmodel_diegozenari/TITLE,pipejoint_diegozenari

!-----------------------! parameters definition!-----------------------

*set,length1,1000

*set,rad1,100

*set,thick1,4

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 5 5

*set,length2,500

*set,rad2,50

*set,thick2,2

*set,ndiv1,25

*set,ndiv2,20

*set,space,8

*set,PRES,1

!----submodel-parameters-----

*set,ysub,length1/2-220

*set,ysub2,length2-30

*set,degree,70

*set,rfillet,10

*set,pi,acos-1

*set,distance,1.5

!-----------------------! Elements definition!-----------------------

/PREP7

ET,1,SOLID186 !element type 1

MP,EX,1,210000 !YoungâAZs modulusMP,PRXY,1,0.3 !PoissonâAZs ratio

!-------------------! model construction!-------------------/pnum,line,1/pnum,area,1/pnum,volu,1CLOCAL,11,1,+rad1+length2,length1/2,0,,90,-90csys,0

k,1,0k,2,,length1/2

k,3,rad1,ysubk,4,rad1,length1/2l,3,4

k,5,rad1-thick1,ysubk,6,rad1-thick1,length1/2l,5,6l,3,5l,4,6AL,1,2,3,4

AROTAT,1,2,3,4,,,1,2,-degreeAL,5,6,7,8

csys,11

k,11k,12,0,0,length2+rad1k,23,distance*rad2,90,0k,24,distance*rad2,90,length2+rad1l,23,24AROTAT,13,,,,,,11,12,-90APTN,7,2,4,3,1

ADELE,13,15,1ADELE,17,18ADELE,5,6ADELE,10

lsel,s,line,,1,3,1lsel,a,line,,5,16,1lsel,a,line,,23,24,1lsel,a,line,,27,32,1ldele,all

ksel,s,kp,,1,3,1ksel,a,,,5ksel,a,kp,,7,14,1ksel,a,kp,,19,24,1kdele,all

numcmp,allVA,1,2,3,4,5

csys,11k,11k,12,0,0,length2

k,13,rad2,90,ysub2,k,14,rad2,90,length2+rad1l,13,14

k,15,rad2-thick2,90,ysub2k,16,rad2-thick2,90,length2+rad1l,15,16l,13,15l,14,16AL,10,11,12,13AROTAT,10,11,12,13,,,11,12,-90AL,14,15,16,17VA,6,7,8,9,10,11

csys,0

vptn,1,2vdele,3vdele,5

asel,s,area,,1,4,1asel,a,area,,6,9,1asel,a,area,,11,15,1asel,a,area,,22,25,1adele,all

lsel,s,line,,1,3,1lsel,a,line,,5,6,1lsel,a,line,,10,11,1lsel,a,line,,13,16,1lsel,a,line,,19,20,1lsel,a,line,,22,23,1lsel,a,line,,25,26,1lsel,a,line,,38,39,1lsel,a,line,,41,42,1ldele,all

ksel,s,kp,,1,2,1ksel,a,kp,,8,9,1ksel,a,kp,,11,12,1

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 5 6

ksel,a,kp,,14,16,2kdele,all

allsel,allnumcmp,all

afillt,4,12,rfilletAL,41,26,44AL,23,42,45VA,12,18,4,25,23

NUMCMP,allVADD,1,2,3,4

VSEL,S,volu,,5vatt,1,,1

mshape,1,3Dmshkey,0

esize,2vmesh,5

!cutting contourasel,s,area,,1,2,1nsla,snwrite !write the nodes on the cutting planesallsel,allsavefini

/solu!symmetryasel,s,area,,7,10,3asel,a,area,,12,16,4da,all,symm

asel,s,area,,4,6,2asel,a,area,,11,15,4nsla,s,1dsym,symm,zallsel,allsavefini

/post1resume,pipejoint_diegozenari,dbfile,pipejoint_diegozenari,rstset,1,1csys,0!shell to solid submodelingcbdof,,,,,,,,,1,fini

/prep7resume,pipesubmodel_diegozenari,db/input,,cbdo/input,,cbdo,,:cb1

/solu

asel,s,area,,12,13asel,a,area,,5asel,a,area,,3asel,a,area,,9

nsla,s,1sf,all,PRES,PRESallsel,allsolve

/post1plnsol,s,eqv!plot results omitting borderscsys,11nsel,s,loc,z,length2-20,1000nsel,r,loc,x,1,1.3*rad2esln,splnsol,s,eqvallsel,all

!save results of line 34 - 31lsel,s,line,,34nsll,snsle,snsle,r,cornernplot

*get,nnodi,node,,count

*dim,s_eqv,array,nnodi

*dim,count_i,array,nnodieps : 0.01

cm,nodi,node

*do,i,1,nnodi

*get,ymax,node,,mxloc,ynsel,r,loc,y,zmax-eps,zmax+eps

*get,nmin,node,,num,max

*get,s_eqvi,node,nmin,s,eqvcount_ii : ymaxcmsel,s,nodinsel,u,node,,nmincm,nodi,node

*enddo

*cfopen,junction1,csv

*vwrite,count_i1,s_eqv1F30.10,F20.10

*cfclosfini

lsel,s,line,,31nsll,snsle,snsle,r,cornernplot

*get,nnodi,node,,count

*dim,s_eqv,array,nnodi

*dim,count_i,array,nnodieps : 0.01

cm,nodi,node

*do,i,1,nnodi

*get,ymax,node,,mxloc,ynsel,r,loc,z,ymax-eps,ymax+eps

*get,nmin,node,,num,max

*get,s_eqvi,node,nmin,s,eqvcount_ii : ymax

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 5 7

cmsel,s,nodinsel,u,node,,nmincm,nodi,node

*enddo

*cfopen,junction2,csv

*vwrite,count_i1,s_eqv1F30.10,F20.10

*cfclosfini

REFERENCES

[1] Slides and exercises of the course, Mecanichal design andmachine elements, M. Benedetti, University of Trento, 2015.

[2] Pipe junction, Mecanichal design and machine elements prob-lem 4, D. Zenari.

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 6 1

Deep metal formingDiego Zenari, 182160, M.Sc Mechatronics engineering

Abstract—The sixth problem analyses the deep forming of a metal blank, pushed by a round punch. The circular metalblank is clamped by a couple o planar dies. After a certain travel of the punch, the maximum force applied to the punch,the elastic spring-back will be determined and the necessary travel to achieve a circumferential strain energy of 5 % willbe calculated.

Index Terms—sub-steps, step-loading, plane-186, non-linear geometry, .

F

1 INTRODUCTION

THE subject of the analysis is a circular metalblank. The forming apparatus is composed

by a round punch travelling vertically. Geomet-ric specifications are in Figure 1.

Figure 1. Forming process

The study is focused on fur points:• The stress state on the upper and lower

surface of the blank after a certain punchtravel.

• The elastic spring-back after the punchremoval and the residual stresses.

• The maximum force applied to the punch(fundamental information for machine de-sign).

• Master course of mechanical design and machine elements, Pro-fessor Matteo Benedetti, academic year 2015/2016.See https://www.dii.unitn.it/

• The punch penetration necessary to pro-vide a circumferential strain energy accu-mulation in the blank of 5 %.

2 PROBLEM RESOLUTION2.1 Model construction

Figure 2. global view

The problem is considered axisymmetric,moreover a vertical axis of symmetry couldbe individuated. Firstly the blank area is gen-erated. The other components (punch,die) areconsidered as rigid, i.e. with an infinite Pascalmodulus: only the contact borders are createdby means of lines and meshed. Contact pairsare generated: punch-blank, upper die-blank,lower die-blank.

To create a contact pair, a set of real constantsis needed (in total three sets). TARGE-169 and

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 6 2

CONTA-172 elements are defined: the first aregenerated by simply meshing the "rigid" lines,the second are generated trough the ESURFcommand, over the already meshed blank sur-face. The element used for the blank is PLANE-183: the element size has to guarantee at least8 elements trough the thickness of the blank.

A pilot node is defined for every (total: 3)rigid target surfaces: it is used to impose move-ments to the entire object. In conclusion, beforelaunching the solution environment, targe andconta element normals has to be checked andeventually corrected: otherwise the contact pairwill not work adequately (ESURF„REVERSE).

2.2 Model solutionThe analysis loading history is divided intosub-steps:

• Firstly to the upper die pilot node is ap-plied a force of 10000kN , in order to holdthe blank. (1 sub-step)

• Secondly the punch pilot node is graduallymoved downwards, achieving a travel ofδ = 4 · t. (24 sub-steps)

• Lastly, the punch pilot node is graduallymoved upwards to the original position.(5 sub-steps)

for each sub-step the configuration is savedto the database and load history (OUTRESand LSWRITE command. Owing to the fact thatthere is condition of large deformation, thenon-linear geometry option is enabled (NL-GEOM,ON. The solution is launched by thecommand LSSOLVE, specifying the first andthe last sub-step.

2.3 Post processingAfter closing the dies the metal blank is sub-jected to a stress of 19 MPa approx. Then thepunch gradually pushes the blank into the cav-ity and at the end of the travel the maximumstress achieved is of 1306.37 MPa.

The maximum force applied to the punch,measured at the end of the travel, is approxi-matively 100kN .

The absolute hoop strain energy accumu-lated is 5.8%, located near the lower die uppercorner. To maintain the hoop stress below 5%,the maximum punch travel is limited to 3.7 · t.

Figure 3. Von Mises stress distribution

3 CONCLUSION

When dividing the loading history of a process,great attention has to be given to the number ofthe divisions: as in a mesh, where the element

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 6 3

Figure 4. Von Mises residual stress

Figure 5. displacement of the blank midplane

size is reduced over critic areas (high stressgradients, geometric complexities...), the num-ber of sub-steps has to be tightened when thesystem is undergoing significant changes, e.g.as punch moves forward. On the other hand, afew (or at least one) sub-steps is required overminor events (die holding, punch removal).

ResultsVon Mises stress max 1306.37 [MPa]

max punch force 100 [kN ]max hoop strain 5.7 % [−]

max displacement -6.3 [mm]

APPENDIX ALIST OF COMMANDS

data.txt

!--------------------------------------------! blank forming!--------------------------------------------

finish/clear,start,new/FILNAM, die_forming/TITLE, die_forming/cwd,’C:\Users\diego\Desktop\compitini\compito6’

!-----------------------! parameters definition!-----------------------

*set,w_punch,25

*set,r_punch,4

*set,gap,2

*set,thick,1.5

*set,r_die,3

*set,w_die,25

*set,E_p,4000

*set,sigmay,355

force:100000 ![N]

*set,pi,acos-1

*set,delta,4*thick

*set,l,50!-----------------------! Elements definition!-----------------------

/PREP7

ET,1,PLANE183KEYOPT,1,3,1MP,EX,1,205000MP,PRXY,1,0.3MP,MU,1,0.1

TB,BKIN,1,1TBTEMP,0TBDATA,1,sigmay,E_p

!-------------------! model construction!-------------------/pnum,kp,1/pnum,line,1/pnum,area,1

blc4,,,l,thickk,41,0,thickk,5,w_punch,thickk,6,w_punch,thick+w_punchl,41,5l,5,6lfillt,5,6,r_punchlplot

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 6 4

k,9,gap+w_punchk,10,gap+w_punch,-w_punchk,11,gap+2*w_punchl,9,10l,9,11lfillt,8,9,r_dieksel,s,kp,,11,13,2kgen,2,all,,,0,thickallsel,alll,14,15

REAL,1MAT,1esize,0.2mshape,0mshkey,1amesh,1

!-------------------! KONTAKT!-------------------

!contact top flange - screw headET,2,TARGE169 !4ET,3,CONTA172 !5

!Contact algorithm: Augmented LagrangianKEYOPT, 3, 2, 0

!Element level time incrementation control:!Automatic bisection of incrementKEYOPT, 3, 7, 1

!Include both initial geometrical penetration!or gap and offset, but with ramped effectsKEYOPT, 3, 9, 2

!Contact stiffness update: Each iteration!necessary to vary stiff o penetration!tolerance if not convergesKEYOPT, 3, 10, 2

/COM, CONTACT PAIR CREATION - START! Generate the target surface!punch-blankallsel,allLSEL,S,line,,5,7,1CM,_TARGET,LINETYPE,2REAL,1LMESH,ALLallsel,allKMESH,8CMDEL,_TARGET! Generate the contact surfaceTYPE,3REAL,1LSEL,s,LINE,,3NSLL,s,1ESURFALLSEL,all/COM, CONTACT PAIR CREATION - END

/COM, CONTACT PAIR CREATION - START! Generate the target surface! lower dieLSEL,S,LINE,,8,10,1CM,_TARGET,LINETYPE,2REAL,2LMESH,ALLallsel,allkmesh,12CMDEL,_TARGET! Generate the contact surfaceTYPE,3REAL,2LSEL,s,LINE,,1NSLL,s,1ESURFALLSEL,all/COM, CONTACT PAIR CREATION - END

/COM, CONTACT PAIR CREATION - START! Generate the target surface!upper dieLSEL,s,,,11CM,_TARGET,LINETYPE,2REAL,3LMESH,ALLallsel,allKMESH,14CMDEL,_TARGET! Generate the contact surfaceTYPE,3REAL,3LSEL,s,LINE,,3NSLL,s,1ESURFALLSEL,all/COM, CONTACT PAIR CREATION - END/psymb,esys,1esel,s,type,,2 !check normal all downesel,a,type,,3 !check normal all up

lsel,s,line,,9,11,1esllesurf,,reverseallsel,all

!-------------------! CONSTRAINTS!-------------------lsel,s,,,4nsll,s,1dsym,symm,x

allsel,allksel,s,kp,,8nslk,sd,all,ux,0d,all,rotz,0

allsel,allksel,s,kp,,14nslk,s

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 6 5

d,all,ux,0d,all,rotz,0

allsel,allksel,s,kp,,12nslk,s,1d,all,all,0

allsel,allnstep : 25nstep2 : 30/SOLU

TIME,1ksel,s,kp,,14nslk,sF,all,FY,-forceallsel,allOUTRES, all, lastLSWRITE,1

*do,ii,2,nstepTIME,iiksel,s,kp,,8nslk,sD,all,uy,-4*thick*ii/nstepallsel,allOUTRES, all, last!stores the results of the only last sub stepLSWRITE,ii

*enddo

*do,ii,nstep+1,nstep2TIME,iiksel,s,kp,,8nslk,sD,all,uy,4*thick*-1+ii-nstep/nstep2-nstep+1allsel,allOUTRES, all, lastLSWRITE,ii

*enddo

SOLCONTROL,ONEQSLV,PCG!Include geometric nonlinearitiesNLGEOM, ONLSSOLVE,1,nstep2,1

FINISH

/post1set,last !last load step/eshape,1plnsol,s,eqv

path,superior,2,,50ppath,1,,0,thickppath,2,,l,thickpdef,sforzo_mises_top,s,eqv,avgplpath,sforzo_mises_top

path,inferior,2,,50ppath,1,,0,0ppath,2,,l,0pdef,sforzo_mises_inf,s,eqv,avg

plpath,sforzo_mises_inf

path,displacement,2,,50ppath,1,,0,thick/2ppath,2,,l,thick/2pdef,displ_max,u,y,avgplpath,displ_max

REFERENCES

[1] Slides and exercises of the course, Mecanichal design andmachine elements, M. Benedetti, University of Trento, 2015.

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 7 1

Shaft stress concentration (2)Diego Zenari, 182160, M.Sc Mechatronics engineering

Abstract—The aim of this third problem is to determine the stress concentration factor in a shaft generated by a sectionvariation with a specified fillet radii. Furthermore, the effect of a groove will be analysed.

Index Terms—stress concentration, plane182, groove, fillet.

F

1 INTRODUCTION

THE subject of the analysis is a shaft whichundergoes a section restriction. Geometric

specifications are in Figure 1.

Figure 1. Shaft geometry

The analysis carried out in the third problemis implemented: in this case the shaft is sub-jected to bending and a torsion. Both momentsare applied at the tip (right side in figure1). After determining the stress concentrationfactor, either for torsion and bending, a reliefgroove will be introduced, with a consequentre-analysis. The stress concentration factor isdefined as the maximum principal stress overthe external nominal stress.

Kshape = max[σiσ0

]i = 1, 2, 3 (1)

• Master course of mechanical design and machine elements, Pro-fessor Matteo Benedetti, academic year 2015/2016.See https://www.esse3.unitn.it/

2 PROBLEM RESOLUTION

2.1 Model construction

Owing to the shaft axial and longitudinal sym-metry the model analysed is simply a planerepresenting half of the shaft section. Themodel construction is taken from the precedingproblem1. The unique difference consists onthe type of element: PLANE-25, used for axi-symmetric geometries subjected to non axi-symmetric loads; e.g. bending, shear, torsion, orharmonic loads, that is, loads varying with thecircumferential coordiante following an har-monic function.

Figure 2. keypoints and lines, model subdivision

1. Shaft stress concentration, Mechanical design and machineelements problem 3, Diego Zenari, 2015

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 7 2

D / d 1.4r / d 0.05R (D − d)/2Poisson’s ratio 0.3 [−]Young modulus 205 [Gpa]Bending moment 1 · 106 [Nmm]Torsional moment 1 · 106 [Nmm]

2.2 Model solutionOwing to the non axi-symmetric loading con-dition (which produces non axi-symmetric de-formations), the left side ,where x = 0, is nomore subjected to axi-symmetry conditions.

Every loading condition is defined and thensaved to the database (LSWRITE command)waiting for the LSSOLVE command as in themetal forming problem2.

In order to apply the equivalent bendingmoment pressure, the pressure gradient is cal-culated and assigned (SFGRAD command):

σgradient =σx=r − σx=0

r − 0[MPa/mm] (2)

whereσx =

Mflett

Ix· x [MPa] (3)

and

Ix =π ·R4

4[mm4] (4)

Then the linear varying pressure is applied tothe selected nodes (SF command).

Owing to the nature of element PLANE-25,it is not possible to apply distributed pressuredirected along the z − axis, consequently, thetorsion moment is divided in equivalent nodalloads:

2. Mechanical design and machine elements problem 6,Diego Zenari,2015

σgradient =σx=r − σx=0

r − 0[MPa/mm] (5)

where the shape function, generally corre-sponding to a plane equation, results in a linearrelation if we concern on the connecting linebetween every i and j knot:

[N e] =

{1− ζ

leζle

}with ζ ∈ [0, le] (6)

and

Ipolar =π · r4

2[mm4] (7)

le is obtained by dividing the shaft radius d/2by the number of elements at the tip.

Every knot, (except the first and the last),play as i and j knot of two adjacent elements,consequently the nodal load is composed bythe sum of two contributes. Nodal loads in ax-isymmetric elements must be input on the fullcircumference, that is, every nodal load must bemultiplied for the circumference length givenby its radial coordinate.

All of these calculations are made apartand then saved on .txt file, which couldbe read and saved in an array withTREAD command. In conclusion the forceis applied by F, all, fz,%array_name%, where%array_name% is the array previously savedby TREAD.

2.3 Post processing

Stress intensity factorsWithout groove

Kf 2.25Kt 1.25

Max eqv stress 182.6 [MPa]With groove

Kf 1.75Kt 1.06

Max eqv stress 148.4 [MPa]

Table 1Max Von Mises stress is obtained from combined loading

(bending+torsion)

The external nominal stresses are:

σ0 =Mf

Ix· d2= 81.5 [MPa] (8)

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 7 3

Figure 3. First principal stress distribution, subjected tobending moment.

Figure 4. First principal stress distribution, subjected tobending moment.

Figure 5. First principal stress distribution, subjected totorsion.

Figure 6. First principal stress distribution, subjected tobending moment.

Figure 7. First principal stress distribution, subjected tobending moment.

Figure 8. First principal stress distribution, subjected totorsion.

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 7 4

Figure 9.

τ0 =Mt

Ipolar· d2= 40.5 [MPa] (9)

It is possible to validate the results comparingthe solution to the available literature. Figure 9shows a condition very close to the subject ofthe analysis, in particular the section restrictionis significantly wider, and consequently inductsa lower concentration. Values estimated by lit-erature:

Stress intensity factorsKf 2Kt 1.65

3 CONCLUSION

The stress concentration factor estimated iscompatible with the data available by litera-ture only for kf . Unfortunately, the numericalsimulation significantly under-estimates the ktfactor. However, following the expectations, the

introduction of the relief groove decreases thestress concentration.

APPENDIX ALIST OF COMMANDS

data.txt

! ----------------------------------------------! PROBLEM: HARMONIC ELEMENTS! ----------------------------------------------

Finish/CLEAR, START, NEW/FILNAM, fillet_bend_diegozenari/TITLE,fillet_bend_diegozenari/cwd,’C:\Users\diego\Desktop\compitini\compito7’

!---------MODEL PARAMETERS---------d : 50 !mmDi : 70 !mmRfillet : 2.5 !mmRgroove : 10 !mm

ndiv1 : 20 !20ndiv2 : 30 !30ndiv3 : ndiv2+ndiv1ndiv4 : 30 !30ndiv5 : 10 !10

sp : 0.6dista : Rgroove/6middle : 3.5*Di/8dista2 : d+2*Rgroove+sp+Rgroove/6maxy : 2*d+2*Rgroove+sp+Rgroove/6

!--------MAT. PROPE_young : 210000 !MPaPoisson : 0.3

!--------BENDING and TORSION

Mf:1e6 !bending momentpi:acos-1!mom. of intertiaIxx:pi/4*d/2**4!max bending stresssigma_max:Mf/Ixx*d/2!through-thickness number of elementsn_el : ndiv2 + 0.6*ndiv1eps : 1e-3 !selection tolerance

!--------MODEL CONSTRUCTION-----------

/PREP7ET,1,PLANE25,,,,,,2!plane element, stimpified!strain formulation, axisymm

MP,Ex,1,E_youngMP,prxy,1,Poisson

!key point construction, the object will

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 7 5

!be subdivided in five areas/pnum,line,1/pnum,kp,1

K,1,0,0,0K,2,Di/2,0,0K,3,Di/2,d+2*Rgroove+sp,0K,4,middle-Rgroove,d+2*Rgroove+sp,0K,5,0,d-2*sp,0

K,6,Di/2,dista2,0K,7,d/2,dista2+Rfillet,0K,8,Di/2-7.5,dista2,0K,9,middle-Rgroove,dista2+Rfillet,0K,10,0,dista2+Rfillet,0K,11,0,maxy,0K,12,d/2,maxy,0K,13,0,3.5*d/2,0K,14,d/2,3.5*d/2,0

!additive kp for groove areaK,20,Di/2,d-2*spK,21,Di/2,dK,22,Di/2,d+2*RgrooveK,23,middle-Rgroove,d-2*sp,0K,100,2*d,2*d,0

L,1,5L,1,2L,2,20L,4,3L,20,23L,23,5L,23,4L,20,21LARC,21,22,23,RgrooveL,22,3L,5,10L,4,9L,3,6L,9,7L,10,9L,10,13L,13,11L,11,12L,12,14L,14,7L,13,14L,6,8LARC,8,7,100,Rfillet

/PNUM,LINE,1

!area meshing containing filletal,14,23,22,4,12,13lsel,alllesize,14,,,0.6*ndiv1,0.4!10lesize,23,,,ndiv2!10lesize,22,,,ndiv2/2,0.3!15,0.3lesize,13,,,0.6*ndiv1lesize,4,,,ndiv2!15,0.4lesize,12,,,ndiv2/2!15lsel,s,line,,4lsel,a,line,,12lccat,all

lsel,s,line,,22,23,1lccat,allTYPE,1Mat,1mshape,0,2Dmshkey,1amesh,1

!groove arealsel,alllesize,5,,,ndiv2!10,3lesize,7,,,ndiv2!25lesize,10,,,ndiv1!30,0.3lesize,8,,,ndiv1!30,0.3lesize,9,,,3*ndiv2al,10,4,7,5,8,9lsel,s,line,,4,5,1lsel,a,line,,7lccat,allTYPE,1Mat,1amesh,2

!area beside filletlsel,alllesize,6,,,ndiv2,3!15lesize,15,,,ndiv2,0.2!10,0.3lesize,11,,,1.5*ndiv2,0.2al,6,11,15,7,12lsel,s,line,,7lsel,a,line,,12lccat,allTYPE,1Mat,1amesh,3

!bottom arealsel,alllesize,1,,,ndiv1!20,2lesize,3,,,ndiv1!20,0.5lesize,2,,,2*ndiv2,0.5!20,0.7AL,1,2,3,5,6LSEL,S,LINE,,5,6,1LCCAT,ALLTYPE,1Mat,1AMESH,4

!top arealsel,alllesize,20,,,ndiv4,0.2!20,0.7lesize,16,,,ndiv4,4!10lesize,21,,,ndiv2+0.6*ndiv1,0.3!10AL,15,20,21,14,16lsel,s,line,,14,15,1lccat,allTYPE,1Mat,1AMESH,5

!top arealsel,alllesize,18,,,ndiv2+0.6*ndiv1!20,0.7lesize,17,,,ndiv5!10lesize,19,,,ndiv5!10

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MECHANICAL DESIGN AND MACHINE ELEMENTS PROBLEM 7 6

AL,17,18,19,21TYPE,1Mat,1AMESH,6

/soluantype,staticDl,2,,ux,0Dl,2,,uy,0allsel,all

!bending momentmode,1,1nsel,s,loc,y,maxy-eps,maxy+epssfgrad,pres,0,x,0,sigma_max/d/2sf,all,pres,0allsel,all

LSWRITE,1

!torqueLSCLEAR,allmode,0

Dl,2,,allallsel,all

*dim,force,table,n_el+1,,,X

*tread,force,forza_torsione,txtnsel,s,loc,y,maxy-eps,maxy+epsf,all,fz,%force%allsel,all

LSWRITE,2

lssolve,1,2

/post1set,1,1,,,,0plnsol,s,1 !y

set,2,1plnsol,s,1 !yz

!superposition of the two loading casesset,1,1,,,,0LCWRITE ,1set ,2 ,1LCOPER ,ADD ,1LCWRITE ,1plnsol,s,eqv

REFERENCES

[1] Slides and exercises of the course, Mecanichal design andmachine elements, M. Benedetti, University of Trento, 2015.

[2] Shigley Progetto e costruzione di macchine, R. Budynas, J.K. Nisbett, McGrawHill, 2014.

[3] Shaft stress concentration, Mechanical design and machineelements problem 3, Diego Zenari, 2015.