TAGGED: ansys-apdl, apdl, reinf264, reinforcement
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July 29, 2024 at 10:39 ammuhammadaziz.sarwarSubscriber
I am working on 11 layer rubber pipe with 4 layers of steel wire arranged at particular helix angle in Mechanical APDL2024.I want to model these wires as reinforcements with other layers as rubber material.I am thinking that smeared reinforcing section would be good,as only one type of steel wire is used and at same distance from each other.I am attaching the 90 sector model of pipe with one end closed.Please recommend which type of reinforcing and options ccould be used to model this pipe. I want to apply reinforcement only on V1 .I am available for more information.I tried different options on reinforcing section .But model is giving same deformation with or without reinforcement.I have also studied REINF263,REINF264 and REINF265.But i have not succeded to apply to my model.Please share more information for applying reinforcement at particular angle for each layer.
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July 31, 2024 at 6:50 amAshish KhemkaForum Moderator
Hello,
Please see if the following link helps: 14.1. Reinforcing Workflow (ansys.com)
Regards,
Ashish Khemka
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August 2, 2024 at 11:41 ammuhammadaziz.sarwarSubscriber
Thanks for your information.Unfortunately problem is still there beacuse of contact and particular angle requirements and continuity of the mesh..Could you please expalin any other option ?
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August 2, 2024 at 7:11 pmwrbulatAnsys Employee
Volume 1 appears to be subdivided (sliced) in such a way that there is a fairly sharp angle between the plane of the cut and the inner cylindrical surface. For what reason is the the geometry subdivided in this way? You mentioned contact. Is there contact between one of the surfaces on the right end of the reinforced pipe, with the pipe subjected to an axial load? Did you end up meshing with tetrahedra? If the subdivision was made to clearly define contact surface, I think you can do away with it... the pressure distribution from the contacting body (if you model it explicity) should be fairly well represented.
There are multiple options for defining smeared reinforcing. I believe that in the option you are using, the location and orientation of the fibers is defined with respect to the order of nodes in each element:
If you have a tetrahedral mesh, with random node number order within each element and multiple layers of elements through the thickness of the pipe wall, you might end up with a corresponspondingly random reinforcing locations and orientations.
I can offer you the following (copy the lines below into a text file and read the text file into MAPDL with the /INPUT command). Try two different values of parameter r_reinf (0.001 & 0.0005). The calculated results should differ substantially. I have the luxury of using very simple geometry (not sliced as yours is). Even with this "unfair advantage", I had to resort to a "trick" to align my SOLID185 elements (initially define them to be layered so I could use EORIENT to renumber the elements in such a way as to get consistent REINF265 orientation in the cylinder wall). I added comments to explain the numerous other settings I used in the setup.
Another option would be to create MESH200 elements and assign the reinf section properties to them.
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I hope this helps!
Bill
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fini/cleÂ/sys,del file*.pngÂ/vie,1,1,1,1/vup,1,zÂÂC********************************************************C*** PARAMETERSC********************************************************pi=acos(-1)Âr1=0.010 ! CYLINDER INNER RADIUSr2=0.020 ! CYLINDER OUTER RADIUSl=0.01 ! CLINDER LENGTHÂr_reinf=0.001 ! REINFORCING FIBER RADIUS (COMPARE 0.001 & 0.0005)a=pi*r_reinf**2 ! CORRESPONDING REINF CROSS SECTION AREAfacenum=3 ! FACE NUMBER FOR "ELEF" REINF SECTION PARAMETERs=0.0025 ! REINF SPACINGthta=30 ! MAGNITUDE OF REINF FIBER ORIENTATION ANGLEÂE_base=0.05e9 ! BASE ELEMENT ELASTIC MODULUS (~RUBBER)nu_base=0.20 ! BASE ELEMENET POISSON'SÂE_reinf=2e11 ! REINF ELASTIC MODULUSnu_reinf=0.20 ! REINF POISSON'SÂu_top=0.001 ! MAGNITUDE OF AXIAL DISPLACEMENT IMPOSED ON CYLINDER IN TRIAL SOLVEf_top=2e3 ! MAGNITUDE OF AXIAL FORCE IMPOSED ON CYLINDER IN TRIAL SOLVEÂÂ/title,R_fiber=%r_reinf%, S_fiber=%s%, E_fiber=%E_reinf%ÂC********************************************************C*** GEOMETRY & BASE ELEMENT MESHC********************************************************wpcs,-1,0 ! CYLINDRICAL ELEMENT COORDINATE SYSTEMcswp,11,1Â/prep7 ! GEOMETRY (FOUR 90 DEG SECTORS)cyli,r1,r2,0,l,0,90cyli,r1,r2,0,l,90,180cyli,r1,r2,0,l,180,270cyli,r1,r2,0,l,270,360numm,kp ! MERGE COINCIDENT KEYPOINTSvatt,1,1,1,11 ! ASSIGN ATTRIBUTE NUMBERS (MAT, REAL, TYPE, ESYS)Âet,1,185,,,1 ! LAYERED SOLID185 ELEMENT TYPEmp,ex,1,E_base ! ELASTIC MODULUSmp,nuxy,1,nu_base ! POISSON'Svmes,all ! MESH VOLUMESÂÂC********************************************************C*** REORIENT BASE ELEMENTSC********************************************************eorient,lysl,posz ! RENUMBER SO THAT FACE 1 IS PARALLEL +Z ESYS AXISÂÂC********************************************************C*** REINFORCING ELEMENT PROPERTIESC********************************************************mp,ex,2,E_reinf ! ELASTIC MODULUSmp,nuxy,2,nu_reinf ! POISSON'SÂsect,2,reinf,smear ! REINF SECTION FOR INNER RADIUSsecd,2,A,s,,thta,elef,facenum,0.5seccontrol,0,1,3 ! REMOVE BASE MATERIAL IN DOMAIN OCCUPIED BY REINFORCING, INCLUDE TRANSVERSE SHEAR & BENDINGÂsect,3,reinf,smear ! REINF SECTION FOR OUTER RADIUSsecd,2,A,s,,-thta,elef,facenum,0.5seccontrol,0,1,3 ! REMOVE BASE MATERIAL IN DOMAIN OCCUPIED BY REINFORCING, INCLUDE TRANSVERSE SHEAR & BENDINGÂÂC********************************************************C*** CREATE REINFORCING ELEMENTSC********************************************************csys,11 ! DETERMINE INNER/OUTER RADII*get,rmin,node,,mnloc,x*get,rmax,node,,mxloc,xÂnsel,s,loc,x,rmin ! REINFORCE BASE ELEMENTS ON INNER RADIUSeslnesel,r,type,,1nslesecn,2ereinfÂnsel,s,loc,x,rmax ! REINFORCE BASE ELEMENTS ON OUTER RADIUSeslnesel,r,type,,1nslesecn,3ereinfÂÂC********************************************************C*** MAKE BASE ELEMENT HOMOGENEOUS, PLOT ELEMENTSC********************************************************et,1,185 ! CHANGE BASE ELEMENT TYPE FROM LAYERED TO HOMOGENEOUS OPTIONÂesel,s,type,,1 ! MAKE BASE ELEMENTS TRANSLUCENT/trlcy,elem,0.9alls/psy,layr,-1 ! DISPLAY LAYERS (REINFORCING)/esh,1/dev,vect,0 ! RASTER FILL DISPLAYeplo ! PLOT ELEMENTS/sho,png $eplo $/sho,close $/wait,2Âesel,s,type,,2 ! SELECT REINFORCING ELEMENTS/dev,vect,1 ! WIRE FRAME DISPLAYeplo ! PLOT ELEMENTS/sho,png,,1 $eplo $/sho,close $/wait,2Â/dev,vect,0 ! REVERT TO RASTER FILLesel,s,type,,1 ! MAKE BASE ELEMENTS OPAQUE/trlcy,elem,0allsÂfiniÂÂC********************************************************C*** TRIAL SOLVEC********************************************************/soluÂnlgeom,on ! MAY BE NECESSARY TO INVOKE TENSION-ONLY REINF BEHAVIORÂautots,offoutr,all,allÂnsel,s,loc,z ! FIX BASEd,all,allÂnsel,s,loc,z,lcp,1,uz,allnd_top=ndnext(0)ÂallsÂ!d,nd_top,uz,u_top ! TENSION ON TOP SURFACEf,nd_top,fz,f_top ! TENSION ON TOP SURFACEnsub,5solvÂ!d,nd_top,uz,-u_top ! COMPRESSION ON TOP SURFACEf,nd_top,fz,-f_top ! COMPRESSION ON TOP SURFACEnsub,10solvÂfiniÂÂC********************************************************C*** POST PROCESSINGC********************************************************Â/post26!rfor,2,nd_top,f,znsol,2,nd_top,u,z/axl,y,Axial Displacementplva,2/sho,png $plva,2 $/sho,close $/wait,2ÂÂ/post1set,1Âesel,s,ename,,265ples,s,x/sho,png $ples,s,x $/sho,close $/wait,2Âesel,s,ename,,185rsys,1plns,s,z/sho,png $plns,s,z $/sho,close $/wait,2Â -
August 8, 2024 at 10:29 ammuhammadaziz.sarwarSubscriber
Thank you so much. I have applied your model to check .But problem for me is angle of reinforcement should be 54.04 to 54.103. and area of cross section should be 0.02. and i am not sure about the contact.I have total 11 layers of steel and rubber (among 4 of them were steel reinforcements at particular angle and i am not sure about distance)I want to see as well reinforcement like steel wire at helix.
I believe this approach is really good.It can save simulation cost and can be easily edited fior future use.when i am trying to change the angle and area of cross section it shows different style of reinforcements.
Could you please guide me is this reinforcing method can be applied to hydraulic hoses as shared above.If yes could you please guide me regarding reinforcing sections and data.For meshing any meshing can be used around 6mm thickness of pipe but because of different material involved we just need contact and correct result for deformation.Thanks again.
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