Analytical solution of stress and displacement in axisymmetric inhomogeneous half space under normal concentrated surface loading

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1 mme.modares.ac.ir * aataee@ut.ac.ir * : : : Analyticalsolutionofstressanddisplacementinaxisymmetric inhomogeneoushalfspaceundernormalconcentratedsurfaceloading MohammadBayat AliAsgharAtai 2* -DepartmentofMechanicalEngineering,IslamicAzadUniversityKarajBranch,Karaj,Iran 2-DepartmentofMechanicalEngineering,UniversityofTehran,Tehran,Iran *P.O.B Tehran,Iran,aataee@ut.ac.ir ARTICLEINFORMATION ABSTRACT OriginalResearchPaper Received2February25 Accepted22March25 AvailableOnline8April25 Keywords: Halfspace axisymmetric Inhomogeneous concentratedloading analyticalsolution Inthispaper,ananalyticalsolutionforstressanddisplacementinaninhomogeneoushalfspace undertheactionofconcentratednormalsurfaceloadingisinvestigated.theyoung smodulusis consideredtovarywiththesphericalradiusinpowerlawformofordern,whilethepoisson s ratioistakentobeconstant.theproblemissolvedanalyticallyusinganelasticityapproachand consideringsemi-inversemethodinwhich,basedonequilibriumequationsonthesurfaceofan arbitrary hemisphere in the half-space and centered at the point of application of load, some stresscomponentsareassumedtobeproportionalto/r 2 Itisthenshownthatthisassumption isvalidand all stresscomponentsin this axisymmetric problemare proportionalto/r 2 while displacementsareproportionalto/r (n+) andtheirvariationwithazimuthalcoordinateisin theformofspecialfunctioncalledhyper-geometricfunction.illustrativeexamplesarepresented whichshowvariationsofstressesanddisplacements inbothand directions.itisseen that theinhomogeneityparameterhassignificanteffectonbothofthesefieldvariables. [5]. [3] [2][] [7]. [8].. [9] []. -.. Pleasecitethisarticleusing: : M. Bayat, A. A. Atai, Analytical solution of stress and displacement in axisymmetric inhomogeneous half space under normal concentrated surface loading, Modares MechanicalEngineering Vol.5,No.6,pp.5-57,25(InPersian)

2 z= ( (,,) (2) cos() sin() + = = 2 sin() (3).(3) (2) (3).. (5) (4) = () (4) = () (5) () () (5) (4). cos() (). sin() (5) (4) (7) (6) (6) cot() = (7) + + cot() + 3 = (7) (6) (5) (4) (8) () + (cot() tan())() 2 + tan() tan() = (8) (8) (9). () tan() = () = () () () (8) () () + (cot() tan())() (2() + ()tan()) = () () () ().(2) () (9) [2].. [3] [4].. [5] [6]. [7]. [8].. [9] [2]. [2]. [22]. [23]. () -2.().() = (). (24).( 25 8) ) z

3 = (() + cot() () ( + ) + ()) (2) (7) (23).(24) = ( + ) ( ()( + ) tan() ()( + 2) + ()( ( + )) tan()) (24) (25) (2) (6) ()(( + ) )tan () + ()(( + ) ) tan() + ()( +)cot() ()( +)cot() ()( +)( ()( +)( + ())()( + + ())()( + )tan()= )tan()=.(26) () ()(25) () = C cos() + (( + )(()cot() + ()))/( + + ) (26) (5) (4) (3) (25) (26).(27) ()(2) C = 3 ((sin () ( + )/( + + )cos ())((+ )/( + + )cos()sin()()) 3 2 (22) (27) (26) (8) C (27) (28) ( + )( ( + + ) cos(2))csc ()() +2(+ )cot() () +2(+ ) () + 2 sin()( (3 + 2) + ( ) )C = () = = (28) ()(28).(32-29) () = /( + )(3sin()hypg([, ], [ /2], cos ())( + )C I ( +2)) + cos() sin() hypg([, ], [3 /2], cos ()) + /( + )(3( +) sin() C (I hypg([, ], [ /2], cos ()) + I cos()hypg([, ], [3 /2], cos ()))( + 2) ) (29) I = ( (hypg([, ], [3/2], 3 cos ())cos())/((sin())((( + )hypg([, ], [3/2], cos ()) /3hypg([, ], [5/2], cos ())(( 6 + ) = cot()() + C ( + cot ()) = C (3) (2). = cot()() (6) (3) (2) C = (3).(4) = () (4) () (). () () () () (8-5) = (5) = (6) ( + ) = (7) sin() (sin() + cos() ) = (8) 2 ( + ) = ( + ) = ( ( + )) = ( + ) (22-9) (9) (2) (2) = + (22) (9) (5).(23)

4 -3 =.25. /5 () /5....( 7 25) ) [25] n= ( E (n+)r (n+) /P n= n= n=.2 n=.4 n=.6 n=.8 n= n=.5 Ref ( ) 2 + 6)cos ())hypg([, ], [/2], cos ()) + (( 2 + ) + 2)cos ()hypg([, ], [3 /2], cos ())hypg([, ], [3 /2], cos ())))) I = ( (hypg([, ], [3/2], cos ())cos()) 3 /((sin())((( + )hypg([, ], [3 /2], cos ()) /3hypg([, ], [5 /2], cos ())(( 6+) + 6)cos ())hypg([ ], [ /2], cos ()) + (( 2+) + 2)cos ()hypg([, ], [3 /2], cos ())hypg([, ], [3 /2], cos ())))) I = ( (hypg([, ], [/2], cos ()))/(((( + )hypg([, ], [ /2], cos ()) 3hypg([, ], [3 /2], cos ())(( 2+) + 2)cos ())hypg([, ], [3 /2], cos ()) + (( 6+) + 6)cos ()hypg([, ], [5 /2], cos ())hypg([, ], [ /2], cos ()))sin())) (3) (3) (32),,,,,,,,,,,, hypg(,, ). 26 ).. ()(26) () (27) = cos() sin().(34) (33) = 2 ( + + cos(2) 2 sin(2)) (33) (34)

5 E (n+)r (n+) /P n= n= n=.2 n=.4 4 n=.6 n=.8 n= n= n=.5 Ref.25 E (n+)r (n+) /P n= n=.2 n=.4 n=.6 n=.8 n= n=.5 Ref.25 R 2 /P () n= n= -.2 n=.2 n= n= n=.8 n=.8 n= n= -.5 n=.5 Ref () R 2 /P R 2 /P ( ) 3.2 n=. n=.2 n=.4 n=.6 -. n=.8 n= -.2 n= Ref ( ) 4.4 n= n=.35 n=.2 n=.4.3 n=.6.25 n=.8 n=.2 n=.5.5 Ref () 7 / /2 ()

6 /P(/m 2 ) Ref (m) n= n=.2 n=.4 n=.6 n=.8 n= n=.2.. E (n+)/p(/m) 35 n= 3 n=.2 25 n=.4 n=.6 2 n=.8 5 n= n=.2 Ref.25 5 E (n+)/p(/m) (m) R 2 n= n= n=.2 n=.4 n=.6 n=.8 n= n=.2 Ref.25 E (n+)/p(/m) (m) (m) n= n=.2 n=.4 n=.6 n=.8 n= n=.2 Ref.25-4 n= n= n=.2 9 /P(/m 2 ) (m) n=.4 n=.4 n=.6 n=.8 n= n= n=.2 Ref.25 /P(/m 2 ) n= n=.2 n=.4 n=.6 n=.8 n= 3-4 n= (m) Ref

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