CHAP 4 FEA For Elastoplastic Problems
CHAP 4FEA for Elastoplastic ProblemsNam-Ho Kim Introduction Elastic material: a strain energy is differentiated bystrain to obtain stress– History-independent, potential exists, reversible, no permanentdeformation Elatoplastic material:– Permanent deformation for a force larger than elastic limit– No one-to-one relationship between stress and strain– Constitutive relation is given in terms of the rates of stress andstrain (Hypo-elasticity)– Stress can only be calculated by integrating the stress rate overthe past load history (History-dependent) Important to separate elastic and plastic strain– Only elastic strain generates stress
Table of Contents 4.2. 1D Elastoplasticity 4.3. Multi-dimensional Elastoplasticity 4.4. Finite Rotation with Objective Integration 4.5. Finite Deformation Elastoplasticity withHyperelastcity 4.6. Mathematical Formulation from Finite Elasticity 4.7. MATLAB Code for Elastoplastic Material Model 4.8. Elastoplasticity Analysis Using Commercial Programs 4.9. Summary 4.10. Exercises 4.21D Elastoplasticity
Goals Understand difference between elasticity and plasticity Learn basic elastoplastic model Learn different hardening models Understand different moduli used in 1D elastoplasticity Learn how to calculate plastic strain when total strainincrement is given Learn state determination for elastoplastic material Plasticity Elasticity – A material deforms under stress, but thenreturns to its original shape when the stress is removed Plasticity - deformation of a material undergoing nonreversible changes of shape in response to applied forces– Plasticity in metals is usually a consequence of dislocations– Rough nonlinearity Found in most metals, and in general is a good descriptionfor a large class of materials Perfect plasticity – a property of materials to undergoirreversible deformation without any increase in stressesor loads Hardening - need increasingly higher stresses to result infurther plastic deformation
Behavior of a Ductile MaterialTermsProportional limitExplanationThe greatest stress for which the stress is still proportional tothe strainElastic limitThe greatest stress without resulting in any permanent strain onrelease of stressYoung’s ModulusYield stressStrain hardeningUltimate stressNeckingSlope of the linear portion of the stress-strain curveThe stress required to produce 0.2% plastic strainA region where more stress is required to deform the materialThe maximum stress the material can resistCross section of the specimen reduces during deformationVUltimatestressFractureYield nghardeningH Elastoplasticity Most metals have both elastic and plastic properties– Initially, the material shows elastic behavior– After yielding, the material becomes plastic– By removing loading, the material becomes elastic again We will assume small (infinitesimal) deformation case– Elastic and plastic strain can be additively decomposed byHH e Hp– Strain energy density exists in terms of elastic strainU01 E( H )2e2– Stress is related to the elastic strain, not the plastic strain The plastic strain will be considered as an internalvariable, which evolves according to plastic deformation
1D Elastoplasticity Idealized elastoplastic stress-strain behavior– Initial elastic behavior with slope E (elastic modulus) until yieldstress Y (line o–a)– After yielding, the plastic phase with slope Et (tangent modulus)(line a–b).ʍ– Upon removing load, elastic unloadingďwith slope E (line b-c)Ğ Ăƚ– Loading in the opposite direction,the material will eventually yieldin that direction (point c)– Work hardening – more force isrequired to continuously deformin the plastic region (line a-b or c-d)ŽɸĐĚWork Hardening ModelsʍĞ Kinematic hardening– Elastic range remains constant– Center of the elastic region movesparallel to the work hardening line– bc de 2oa– Use the center of elastic domainas an evolution variableďĂŽɸĐĚʍ Isotropic hardening– Elastic range (yield stress) increasesproportional to plastic strain– The yield stress for the reversed loadingis equal to the previous yield stress– Use plastic strain as an evolutionĚvariableĞďĂŚŽĐŚɸ No difference in proportional loading (line o-a-b)
Elastoplastic Analysis Additive decomposition– Only elastic strain contributes to stress (but we don’t know howmuch of the total strain corresponds to the elastic strain)– Let’s consider an increment of strain: 'H– Elastic strain increases stress by 'V'He 'HpE'He– Elastic strain disappears upon removing loads or changing directionʍȴʍůĂƐƚŝĐ ƐůŽƉĞ͕ ʍ ƚƌĂŝŶ ŚĂƌĚĞŶŝŶŐ ƐůŽƉĞ͕ ƚ/ŶŝƚŝĂů �ŽĂĚŝŶŐɸzɸȴɸɸ Elastoplastic Analysis cont. Additive decomposition (continue)– Plastic strain remains constant during unloading– The effect of load-history is stored in the plastic strain– The yield stress is determined by the magnitude of plastic strain– Decomposing elastic and plastic part of strain is an importantpart of elastoplastic analysis For given stress V, strain cannot be determined.– Complete history is required (path- or history-dependent)– History is stored in evolution variable (plastic strain)ʍŽɸ
Plastic Modulus Strain increment 'H'He 'HpE'He'V'Hpʍ Stress increment 'V Plastic modulus Hȴʍ ƚƌĂŝŶ ŚĂƌĚĞŶŝŶŐ ƐůŽƉĞ͕ ƚʍz Relation between moduli'VE'H e'VEt'V 'V EHHH'HpEEtE EtȴɸĞEt 'H 1EtEtEHE HɸzȴɸƉɸȴɸ1 1 E HEtHE ·§E 1 E H ¹ 'He'Hp Both kinematic and isotropic hardenings have the sameplastic modulus Analysis Procedure Analysis is performed with a given incremental strain– N-R iteration will provide 'u º 'HH– But, we don’t know 'HHe or 'HHp When the material is in the initial elastic range, regularelastic analysis procedure can be used When the material is in the plastic range, we have todetermine incremental plastic strain'H'V 'HpE'He 'Hp§H·'Hp 1 E¹ 'HpH'HpEʍ 'Hpȴʍʍz'H1 H/EOnly when the material is on the plastic curve!!ȴɸĞɸzȴɸƉȴɸɸ
1D Finite Element Formulation Load increment– applied load is divided by N increments: [t1, t2, , tN]– analysis procedure has been completed up to load increment tn– a new solution at tn 1 is sought using the Newton-Raphson method– iteration k has been finished and the current iteration is k 1 Displacement increments– From last increment tn:'dkn 1 k– From previous iteration:Gdkn 1 k 13 d nddd n 1dkX X [ [ / u1 ½ ¾ u2 ¿3 1D FE Formulation cont. Interpolation'u(x) 'u ½[N1 N2 ] 1 ¾ 'u2 ¿d'udx'Hª 1« L N 'd1 º 'u1 ½ ¾L »¼ 'u2 ¿B 'd Weak form (1 element)GuGHN GdB GduN dHB dd u1 ½ ¾ u2 ¿– Internal force external forceLd T ³ BT n 1 Vk 1Adx0d T n 1F, d R2
1D FE Formulation cont. Stress-strain relationship (Incremental)n 1Vk 1 n 1Vk wVGHwHn 1V k D ep GH– Elastoplastic tangent modulusDep E if elastic Et if plastic Linearization of weak formLd T ª ³ BTDepBAdx º Gd« 0»¼d T n 1F d T ³ BT n 1 VkAdxTangent stiffnessResidualL0 1D FE Formulation cont. Tangent StiffnessADep ª 1 1 ºL « 1 1 »¼kT Residualn 1 kRLF ³ Bn 1T n 1 k0V Adx State Determination: n 1 Vk n 1F1 n 1 VkA ½ n 1¾n 1 kFA V ¿ 2f( n V, n Hp , 'Hk , !)Will talk about next slides Incremental Finite Element Equation– N-R iteration until the residual vanisheskT Gdkn 1 kR
Isotropic Hardening Model Yield strength gradually increases proportional to theplastic strain– Yield strength is always positive for both tension or compressionTotal plastic strainVnYV0Y HHnpInitial yield stress– Plastic strain is always positive and continuously accumulated evenin cycling loadingsV YnV Y ʍV YnV Y ɸƉɸɸĞʍɸɸƉV YnV Yn State Determination (Isotropic Hardening) How to determine stress– Given: strain increment ( H) and all variables in load step n(E,H, V0Y , Vn , Hnp )1. Computer current yield stress (pont d)VnYV0Y HHnpʍʍƚƌ2. Elastic predictor (point c)'VtrE'HV trVn 'V trTrial yield function (c – e)ftrV tr Vnyftr(1 R)E'HȴɸƉȴʍƚƌZȴʍƚƌ3. Check yield statusĐʍŶнϭĚV YnV Y ʍŶĂĨȴɸR: Fraction of 'Vtr to the yield stressĞȴɸĞƉс ;ϭʹZͿȴɸďɸ
State Determination (Isotropic Hardening) cont. If ftr d 0 , material is elasticVn 1VtrEither initial elastic region or unloading If ftr ! 0 , material is plastic (yielding)Either transition from elastic to plastic or continuous yielding– Stress update (return to the yield surface)Vn 1ʍVtr sgn(Vtr )E'Hp/ŶŝƚŝĂů ůŽĂĚŝŶŐ– Update plastic strainHnp 1Hnp 'HphŶůŽĂĚŝŶŐZĞůŽĂĚŝŶŐPlastic strain increment is unknown'Hɸ'He 'HpFor a given strain increment, how much is elastic and plastic? State Determination (Isotropic Hardening) cont. Plastic consistency condition– to determine plastic strain incrementfn 1Vn 1 Vny 10– Stress must be on the yield surface after plastic deformation V tr sgn( V tr )E'Hp ( Vny H'Hp ) V tr Vny (E H) 'Hp'Hp'HpVtr VnyE H(1 R)R00ʍʍƚƌĐȴɸƉtrfE HE'HE H1 ȴʍZȴʍʍŶнϭV YnV Y ʍŶĂĚĨftr'Vtr%Note: 'Hp is always positive!!ȴɸĞȴɸĞƉс ;ϭʹZͿȴɸďɸ
State Determination (Isotropic Hardening) cont. Update stressVn 1Vn 1Vtr sgn(Vtr )E'Hpʍʍƚƌ(1 R)E2 sgn(Vtr )'HE HVtrȴɸƉȴʍElastic trialPlastic compensation(return mapping)ĐZȴʍʍŶнϭV YnV Y ʍŶĂĚ AlgorithmĨĞȴɸĞƉс ;ϭʹZͿȴɸďȴɸ1) Elastic trialɸ2) Plastic return mapping– No iteration is required in linear hardening models Algorithmic Tangent Stiffness Continuum tangent modulus– The slope of stress-strain curveDep Algorithmic tangent modulus E if elastic Et if plastic– Differentiation of the state determination algorithmDalgw'Hpw'HDalgw'Vw'Hw'H pw tr V sgn( tr V)Ew'Hw'H1 w trfE H w'Hsgn( tr V)EE H E if elastic Et if plastic Dalg Dep for 1D plasticity!!– We will show that they are different for multi-dimension
Algorithm for Isotropic Hardening Given: 'H, E, H, V0Y , Vn , Hnp1. Trial state VtrVn E'HVnYV0Y HHnpVtr VnYftr1. If ftr d 0 (elastic)n 1Remain elastic: V–Hnp 1V tr ,Hnp; exit2. If ftr ! 0 (plastic)ftrE Hb. Update stress and plastic strain (store them for next increment)Calculate plastic strain: 'Hpa.Vn 1Hnp 1Vtr sgn(Vtr )E'HpHnp 'Hp Ex) Elastoplastic Bar (Isotropic Hardening) E 200GPa, H 25GPa, 0Vy 250MPa nV 150MPa, nHp 0.0001, 'H 0.002 Yield stress: n V Y0V Y H n Hp252.5MPa– Material is elastic at tn Trial stress: ' tr VtrVE 'Hn400MPaV ' tr V550MPaNow material is plastic Plastic consistency conditiontrf1.322 u 10 3'HpE H State updaten 1n 1VtrHpnV sgn( tr V)E'HpHp 'Hp285.6MPa1.422 u 10 3
Kinematic Hardening Model Yield strength remains constant, but the center of elasticregion moves parallel to the hardening curve Effective stress is defined using the shifted stressKV D Use the center of elastic domain as an evolution variableDn 1Dn sgn(K)H'HpVnV Y ʍVnV Y ʍV Y DnɸƉ V Y D n ϮɸƉɸɸĞBack stressɸ State Determination (Kinematic Hardening) Given: Material properties and state at increment n:( 'H,E,H, V0Y , Vn , Dn , Hnp ) Elastic predictorVtrVn E'H,D trDn ,KtrVtr D trʍʍƚƌ Check yield statusTrial yield functionftrKtr V 0yʍŶнϭĚVtrEither initial elastic region or unloadingĞKtr If ftr d 0, material is elasticVn 1ĐʍŶɲŶнϭɲŶďĂĨŐȴɸɸ If ftr ! 0, material is plastic (yielding)Either transition from elastic to plastic or continuous yielding
State Determination (Kinematic Hardening) cont. Updating formulas for stress, back stress & plastic strainVn 1Vtr sgn(Ktr )E'HpDn 1D tr sgn(Ktr )H'HpHnp 1Hnp 'Hp Plastic consistency condition– To determine unknown plastic strain increment– Stress must be on the yield surface during plastic loadingfn 1Kn 1 V 0y0 V tr sgn( Ktr )E'Hp D tr sgn( Ktr )H'Hp V 0y V tr D tr V 0y (E H) 'HpKtr Vny'HpftrE HE H00%Note: the same formula withisotropic hardening model!! Algorithm for Kinematic Hardening Given: 'H, E,H, V0Y , Vn , Dn , Hnptr1. Trial state VVn E'HD trDnKtrVtr D trftrKtr V0Y2. If ftr d 0 (elastic)–n 1Remain elastic: VV tr ,Dn 1Dn ,Hnp 1Hnp ; exit3. If ftr ! 0 (plastic)ftra. Calculate plastic strain: 'HpE Hb. Update stress and plastic strain (store them for next increment)Vn 1Vtr sgn(Ktr )E'HpHnp 1Dn 1Dn sgn(Ktr )H'HpHnp 'Hp
Ex) Elastoplastic Bar (Kinematic Hardening) E 200GPa, H 25GPa, 0Vy 200MPa nV 150MPa, nD 50MPa, 'H í0.002 SincenKnV nD100 0 VY , elastic state at tn Trial stress:' tr VtrDE 'HnD Since trf 400MPa,tr50MPa,tr Plastic strain State updatetrKVtrnV V ' tr VtrD 250MPa 300MPaK 0 VY ! 0, material yields in compression'HptrfE H0.444 u 10 3n 1VtrV sgn( tr K)E'Hp 161.1MPan 1DtrD sgn( tr K)H'Hp38.9MPa Ex) Elastoplastic Bar (Kinematic Hardening)
Combined Hardening Model Baushinger effect– conditions where the yield strength of a metal decreases when thedirection of strain is changed– Common for most polycrystalline metals– Related to the dislocation structure in the cold worked metal. Asdeformation occurs, the dislocations will accumulate at barriersand produce dislocation pile-ups and tangles. Numerical modeling of Baushinger effect– Modeled as a combined kinematic and isotropic hardeningVnY 1VnY (1 E)H'HpDn 1Dn sgn(K)EH'Hp0dEd1E 0: isotropic hardeningE 1: kinematic hardening Combined Hardening Model cont. Trial stateVtrVn E'HD trDnKtrVtr D trftrKtr VnY Stress updateVn 1Vtr sgn(Ktr )E'HpDn 1D tr sgn(Ktr )EH'HpVnY 1Vny (1 E)H'Hp Show that the plastic increment is the same'HpftrE H
MATLAB Program combHard1D%% 1D Linear combined isotropic/kinematic hardening model%function [stress, alpha, ep] combHard1D(mp, deps, stressN, alphaN, epN)% Inputs:% mp [E, beta, H, Y0];% deps strain increment% stressN stress at load step N% alphaN back stress at load step N% epN plastic strain at load step N%E mp(1); beta mp(2); H mp(3); Y0 mp(4);%material propertiesftol Y0*1E-6;%tolerance for yieldstresstr stressN E*deps;%trial stressetatr stresstr - alphaN;%trial shifted stressfyld abs(etatr) - (Y0 (1-beta)*H*epN);%trial yield functionif fyld ftol%yield teststress stresstr; alpha alphaN; ep epN;%trial states are finalreturn;elsedep fyld/(E H);%plastic strain incrementendstress stresstr - sign(etatr)*E*dep;%updated stressalpha alphaN sign(etatr)*beta*H*dep;%updated back stressep epN dep;%updated plastic strainreturn; Ex) Two bars in parallel Bar 1: A 0.75, E 10000, Et 1000, 0VY 5, kinematic Bar 2: A 1.25, E 5000, Et 500, 0VY 7.5, isotropic MATLAB program%% Example 4.5 Two elastoplastic bars in parallel%DU %E1 10000; Et1 1000; sYield1 5;5LJLGE2 5000; Et2 500; sYield2 7.5;mp1 [E1, 1, E1*Et1/(E1-Et1), sYield1];%DU mp2 [E2, 0, E2*Et2/(E2-Et2), sYield2];nS1 0; nA1 0; nep1 0;nS2 0; nA2 0; nep2 0;A1 0.75; L1 100;A2 1.25; L2 100;tol 1.0E-5; u 0; P 15; iter 0;Res P - nS1*A1 - nS2*A2;Dep1 E1; Dep2 E2;conv Res 2/(1 P ual');fprintf('\n %3d %7.4f %7.3f %7.3f %7.3f %7.3f %8.6f %8.6f %10.3e',.iter,u,nS1,nS2,nA1,nA2,nep1,nep2,Res);
Ex) Two bars in parallel cont.while conv tol && iter 20delu Res / (Dep1*A1/L1 Dep2*A2/L2);u u delu;delE delu / L1;[Snew1, Anew1, epnew1] combHard1D(mp1,delE,nS1,nA1,nep1);[Snew2, Anew2, epnew2] combHard1D(mp2,delE,nS2,nA2,nep2);Res P - Snew1*A1 - Snew2*A2;conv Res 2/(1 P 2);iter iter 1;Dep1 E1; if epnew1 nep1; Dep1 Et1; endDep2 E2; if epnew2 nep2; Dep2 Et2; endnS1 Snew1; nA1 Anew1; nep1 epnew1;nS2 Snew2; nA2 Anew2; nep2 epnew2;fprintf('\n %3d %7.4f %7.3f %7.3f %7.3f %7.3f %8.6f %8.6f ,WHUDWLRQXV V HS HS 5HVLGXDO ( ( ( ( Summary Plastic deformation depends on load-history and itsinformation is stored in plastic strain Stress only depends on elastic strain Isotropic hardening increases the elastic domain, whilekinematic hardening maintains the size of elastic domainbut moves the center of it Major issue in elastoplastic analysis is to decompose thestrain into elastic and plastic parts Algorithmic tangent stiffness is consistent with the statedetermination algorithm State determination is composed of (a) elastic trial and (b)plastic return mapping
1D Elastoplastic Analysis Using ABAQUS Material Card*MATERIAL,NAME ,.0029Plastic strainYield stress 1D Elastoplastic Analysis Using ABAQUS*HEADINGUniaxialPlasticity*NODE,NSET .,1.6,1.,0.,1.7,1.,1.,1.8,0.,1.,1.*ELEMENT,TYPE C3D8,ELSET ALLE1,1,2,3,4,5,6,7,8*SOLID SECTION,ELSET ALLE,MATERIAL ALLE*MATERIAL,NAME *STEP,INC ,3,,.0048,3,,.004*EL PRINT,FREQ 1S,E,EP,*NODE PRINTU,RF*END STEP
Stress Curve 4.3Multi-DimensionalElastoplastic Analysis
Goals Understand failure criteria, equivalent stress, andeffective strain Understand how 1D tension test data can be used fordetermining failure of 3D stress state Understand deviatoric stress and strain Understand the concept of elastic domain and yieldsurface Understand hardening models Understand evolution of plastic variables along with thatof the yield surface Multi-Dimensional Elastoplasticity How can we generalize 1D stress state (V11) to 3D state (6components)?– Need scalar measures of stress and strain to compare with 1D test– Equivalent stress & effective strain– Key ingredients: yield criteria, hardening model, stress-strainrelation We will assume small (infinitesimal) strains Rate independent elastoplasticity- independent of strainrate Von Mises yield criterion with associated hardening modelis the most popular
Failure Criteria Material yields due to relative sliding in lattice structures Sliding preserves volume º plastic deformation is relatedto shear or deviatoric part Tresca (1864, max. shear stress)– Material fails when max. shear stress reaches that of tension test– Tension test: yield at V1 VY, V2 V3 0WmaxV1 V3d WY2– Yielding occurs when Wmax ĂĨĞ ƌĞŐŝŽŶVϮVzVY2&ĂŝůƵƌĞ ƌĞŐŝŽŶʹVzWYVzVϭʹVz Failure Criteria cont. Distortion Energy Theory (von Mises)– Material fails when distortion energy reaches that of tension testUd d Ud (tension test)– We need preliminaries before deriving Ud Volumetric stress and mean strainVm1 tr( V )31 (V113 V22 V33 )Hm1 tr( H )31H3 v1 (H3 11 H22 H33 ) Deviatoric stress and strainsV Vm1Idev : VIijkl(Gik G jl GilG jk ) / 2eH Hm 1Idev : HIdevI 31 11
Failure Criteria cont. Example: Linear elastic materialV[O11 2PI] : H { D : HVO(3Hm )1 2P( e Hm 1)Vm(3O 2P)Hm 1 2PeN volumetric deviatoric(3O 2P)Hm2PesBulk modulus Distortion energy densityUUd1V2:H1s21 (V 1 m2:e1 s4PKs) : ( H m 1 e )3V H2 m m 21 s : e3O 2P3:s Failure Criteria cont.ª2 00 º«3»V « 0 31 0 »«0 0 1 » 3¼ 1D CaseV11UdV1DVm1 s4P1V3s1 2 V24P 3:s1 V26P Material yields whenUd1 s4P:s1 V26P YUd1D Let’s define an equivalent stressVe3s2:s Then, material yields whenVeVYvon Mises stress Stress can increase from zero to VY, but cannotincrease beyond that
Equivalent Stress and Effective Strain Equivalent stress is the scalar measure of 3D stress statethat can be compared with 1D stress from tension test Effective strain is the scalar measure of 3D strain statethat makes conjugate with equivalent stressUd1s2Ud1 s4Pee1 V3P e:e1V e2 e e1 V26P e:s3s213PEffective strain1V e2 e e:s13P3 2Pe2: 2Pe2e3:e Equivalent Stress and Effective Strain cont. 1D Case cont.H11eeeHH22H33 QHª2 0 0 º(1 Q)H «»010 »3 « « 0 0 1 ¼»2e3:e2(1 Q)H3ɸHm1 2QH32e:e§ (1 Q)H ·6 3 ¹Effective strain for 1D tensionɸ
Von Mises Criterion Material yields when Ve VYVe3s2:s1s2J23J2:s2nd invariant of sJ21ª (Vx V y )2 (V y Vz )2 ( Vz Vx )2 º W2xy W2yz W2zx¼6 J21
Yield strength remains constant, but the center of elastic region moves parallel to the hardening curve Effective stress is defined using the shifted stress Use the center of elastic domain as an evolution variable n1 n sgn( )H p (( ) Back stress)( ˇ n Y ˇ ˇ (n Y ˇ n Y "ˇ (n Y State Determination (Kinematic Hardening)
Preparing the living compartment (inside)P.4 Chap. III CArrYInG PASSEnGErS P.5 Chap. IV LoADS AnD LoADED wEIGHt P.6 Chap. V wInDowS AnD SKYLIGHtS P.8 Chap. VI DrIVInG P. 10 Chap. VII tYrES P.11 Chap. VIII SIGnALInG, lights P.12 Chap. IX PArKInG mAnEuVErS P. 14 Chap. X tHE CAmPEr DrIVEr'S CHArtEr P. 16 Chap. XI oPErAtIon oF APPLIAnCES P.17
MINISTRY Of FINANCE 01 O i I Companies ( Chap 75: 04) 02 Other Companies ( Chap 75: 02) 03 Individuals (Chap 75: 01) 04 Withholding Tax ( Chap 75: 01) 05 Insurance Surrender Tax ( Chap 75: 01) 07 Business Levy ( Chap 75: 02) 09 Hea I th Surcharge ( Chap 75: 05) Total Estimates 2022 Revised Estimates 2021 .
Bruksanvisning för bilstereo . Bruksanvisning for bilstereo . Instrukcja obsługi samochodowego odtwarzacza stereo . Operating Instructions for Car Stereo . 610-104 . SV . Bruksanvisning i original
10 tips och tricks för att lyckas med ert sap-projekt 20 SAPSANYTT 2/2015 De flesta projektledare känner säkert till Cobb’s paradox. Martin Cobb verkade som CIO för sekretariatet för Treasury Board of Canada 1995 då han ställde frågan
service i Norge och Finland drivs inom ramen för ett enskilt företag (NRK. 1 och Yleisradio), fin ns det i Sverige tre: Ett för tv (Sveriges Television , SVT ), ett för radio (Sveriges Radio , SR ) och ett för utbildnings program (Sveriges Utbildningsradio, UR, vilket till följd av sin begränsade storlek inte återfinns bland de 25 största
Hotell För hotell anges de tre klasserna A/B, C och D. Det betyder att den "normala" standarden C är acceptabel men att motiven för en högre standard är starka. Ljudklass C motsvarar de tidigare normkraven för hotell, ljudklass A/B motsvarar kraven för moderna hotell med hög standard och ljudklass D kan användas vid
LÄS NOGGRANT FÖLJANDE VILLKOR FÖR APPLE DEVELOPER PROGRAM LICENCE . Apple Developer Program License Agreement Syfte Du vill använda Apple-mjukvara (enligt definitionen nedan) för att utveckla en eller flera Applikationer (enligt definitionen nedan) för Apple-märkta produkter. . Applikationer som utvecklas för iOS-produkter, Apple .
MEDICARE COP 1. Home Health Patient Bill of Rights D HHII.1 484.10 2. Intake Process D HHII.4 484.18 3. Admission Criteria and Process D HHII.4a 484.18 . CHAP MEDICARE . CHAP MEDICARE . Visiting Nurse & Hospice Care CHAP 4. Visiting Nurse & Hospice Care CHAP .
