Nghiên cứu phương pháp tính toán tấm bê tông xi măng mặt đường có xét ảnh hưởng của biến dạng trượt ngang

Nội dung tài liệu: Nghiên cứu phương pháp tính toán tấm bê tông xi măng mặt đường có xét ảnh hưởng của biến dạng trượt ngang

1. MINISTRY OF EDUCATION AND TRAINING THE UNIVERSITY OF COMMUNICATION AND TRANSPORT NGUYEN ANH TUAN RESEARCH FOR METHOD ANALYSING CONCRETE PAVEMENT AND CONSIDERING THE EFFECTS OF TRANSVERSE SHEARING DEFORMATION Specialization: Construction of Highways and City Streets Code: 62.58.30.01 DISSERTATION SUMMARY HA NOI-2013
2. Completion in: THE UNIVERSITY OF COMMUNICATION AND TRANSPORT Scientific Guider: 1. As-Prof. Dr La Van Cham 2. Prof. Dsc Ha Huy Cuong Critic 1: Prof. Dsc Nguyen Van Lien Critic 2: Prof. Dr Vu Dinh Phung Critic 3: Prof. Dr Nguyen Xuan Dao Dissertation is protected in The Council in the University of Communication and Transport In time .hour .on date month .year 2013. Can find out about this dissertation in: - University Library - National Library
3. ISSUE The current sheet method, based on Kirchhoff plate theory, without considering the effects of transverse shear strain induced by shear forces, while not directly allow satisfying 3 the boundary conditions on the edge of plate. Solve plate on elastic foundation, according to Kirchhoff plate theory, can not be determined accurately on the boundary and internal plate corner, is not the ground state of stress. Thesis research allows us to identify the state of stress and deformation of the plate and of the foundation simultaneously, directly satisfy the boundary conditions on the edges 3 plates. CHAPTER 1 ELASTIC FOUNDATION MODELS AND PLATE ANALYSIS 1.1. Elastic foundation and Structure-Soil Interaction. - “Plate on elastic foundation” is solved two basic problems: Plate and Elastic foundation. - Elastic foundation models have done. Winker foundation and half-space foundation are used to the most popular. 1.2. Plate theory of G.R.Kirchhoff . 1.2.1. Basic assumptions. * Surface of the average plate in not distortion. * The section is plane and perpendicular with repect to the average section of the plate. * Separate layers are not prevention together. Based on this assumptions, they only study average section where has vertical displacement w x, y and internal forces impaction. 1.2.2. Balance equations and boundary conditions. 1.2.2.1. Balance equation between external forces and deflection: D w q (1.24) 1.2.2.2. Boundary conditions of rectangular plate: a/ Edge plate is associated restraint:
4. Deflection and Rotation are zero. b/ Edge plate is associated joint : Deflection and bending moment are zero. c/ Edge plate is free: Bending moment and converted shear are zero. To do this work, Kirchhoff’s theory is not considering effect of shears, and considered equivalent to the slip modul of materials G . So, the problem is much sipler, and it is true when what is a thin plate, and when the impact load is not on edge plate. 1.2.3. Effects of shear deformation. 1.2.3.1. Basic content of E.Reissner’s plate theory 2  D w q h2  q (1.40) 10 1  Eq. (1.40) is satisfied for 4 boundary conditions on the two edges of the plate similarly Kirchhoff’s theory. To find QQxy, functions, E.Reissner chosed any stress function  , what is satisfy conditions : 10  0 (1.41) h2 E.Reissner’s theory used to deflection function w and rotation function xy, of average section of the plate to solve problems while have considering effects of shear deformation. Anh so, it is satisfy three boundation conditions but is not two boundation conditions like Kirchhoff’s plate theory. However,  stress function is not general. Do this work, will have difficulty to satisfy boundary conditions and is approximately for stress function  when use numerical method. By side, condition M xy 0 on the free edge is not mentioned. 1.2.3.2. Plate theory has based on Timoshenko beam theory.
5. E.Reissner’s theory and Timoshenko beam theory are only different a one constant. 1.3. Basic on Gauss’s principle extreme to construct balance equations of the plate. Functional of forced Z: 2 2 2 2 Z = D x  x(0) 2  xy  xy (0)  y  y (0) dV k w w (0) dV Min VV (1.58) Plate has size a b, associated, is: a b2w  2 w a b  2 w  2 w a b  2 w  2 w a b Z 2 D  dxdy 2 D  dxdy 4 D  dxdy 2 q  wdxdy 0 2 2 2 2 0 0x  x 0 0  y  y 0 0  x  y  x  y 0 0 (1.60) Integrals eq. (1.60), will find balance functions and boundary conditions of the retangular plate. 1.4. Conclude chapter 1 Basic on Gauss’s principle extreme to construct balance equations and boundary conditions of the plate. E.Reissner’s theory and Timoshenko beam theory are only different the constant and Post-Graduate will use Gauss’s principle extreme and Timoshenko beam theory to construct and solve “Plate on the elastic foundation” in myself thesis. CHAPTER 2 PLATE ON ELASTIC FOUNDATION THEORY HAS CONSIDERING INFLUENCE OF SHEAR DEFORMATION 2.2. Plate on Winkler foundation. - Solve the plate on Winkler foundation problem followwing Kirchhoff’s theory, only have to solve equation (2.10) to definite deflection function w( x, y ) , and then to definite internal forces in the plate: 4w  4 w  4 w (2.10) D 4 2 2 2 4 kw q x  x  y  y
6. - Solve the plate on Winkler foundation problem while have considering effects of shear deformation have to solve eq. (2.58), (2.59) and (2.60) to definite functions , and : Qx x, y Qy x, y 2 h  (2.58) D w kw q  Qxy  Q 51  xy  2233 QQQx 4 hh  x  x  2 23 x 1  10  x  y 10  x (2.59) 4w  4 w h 2  2 q  2 w2  2 q  2 w D 4 2 2 q kw 2 k 2 2 k 2 x  x  y 10  x  x 1   y  y 2233 QQQy 4 hh  y  y  2 23 y 1  10  y  x 10  y 4w  4 w h 2  2 q  2 w2  2 q  2 w D 4 2 2 q kw 2 k 2 2 k 2 y  x  y 10  y  y 1   x  x (2.60) Boundary conditions of rectangular plate: On x = 0 and on x = a edge, and not impaction of load on that: a/ Edge plate is associated joint : w 0  22 w w D Q Qy x (2.52) MDx 00 22  x  y Gh  x  y  2 Q  w Qx y M xy 00 x  y2 Gh  y  x QQxy 0 ; 0  b/ Edge plate is associated restraint : w 0  w Q 0 (2.53) xxx Gh w( x, y ) 2w Q Q  x y M xy 00 x  y2 Gh  y  x QQxy 0 ; 0  c/ Edge plate is free :
7. 22 w w D Qx Qy MDx 00 22  x  y Gh  x  y (2.54) 2  w Qx Qy M xy 00  x  y2 Gh  y  x QQ 0 ; 0 xy  On y = o and y = b edge, and not impaction of load on that: like over . 2.3. Plate on elastic half-space foundation theory while have considering effects of shear deformation. 2.3.1. Situation. Assuming, investigation of the soil under impaction of P straight load, as Fig 2.3. Based on R.D.Mindlin’s solution, we have defined: o o o o o o x,,,,,  y  z  xy  xz  zy and displacements to axis XYZ,, . P Elastic half-space o X o o o x y z o o o Fig 2.3. Sitiation xy xz yz 1 for the plate on elastic half-space z Following Elastic Theory, has: uvwd  x 2GGG o  o  ;  y 2 o  o  ;  z 2 o  o  ; x  y  z (2.61)  u  v wwdd  v  u xy GGG o ;;  yz o  xz o y  x  y  z  x  z  Ignore soil weight: ooo  yx  x zx 0 x  y  z o o o (2.62) xy   y   yz 0 x  y  z  o  o  o xz yz z 0 x  y  z 
8. And now, a V is soil volume, as Fig 2.4. That is impacted P straight load, as (1) volume, be showing dashed line. The aim is be to definited x,,,,,  y  z  xy  xz  zy of V volume under impaction of P by the through that’s knowns components based on R.D.Mindlin’s or J. Boussinesq’s solutions. P Elastic half-space o X Fig 2.4. Using comparisons x y z xy xz yz based on Gauss’s principle 2 extremme for solve plate on elastic half-space z o o o o o o x,,,,,  y  z  xy  xz  zy In this: u,, v wd are displacements to axis XYZ,, . Based on Gauss’s principle extreme, by through comparison, has functional Zd : u  vw  u  v Z  o dV  o dV  od dV  o dV dxx yy zz xyxy VVVVx  y  z  y  x vuww  oo  dddV    dV 0 ,(2.63) yz yz zx zx VV z  y  x  z Dissociation (2.63) : ou o  u o  u  x  x   xy  xy   zx  zz  dV 0 x  y  z V (2.64) v  v  v  o    o    o  dV 0 xy xy y y yz yz  V x  y  z w  w  w  o d   o  d   o  d dV 0 zx zx yz yz z z V x  y  z  With u,,  v  wd are small and independence, has:
9. ooo yx     yx    x zx x zx ()a x  y  z  x  y  z o o o xxy,,,,,  y    z y  xy    zy xz  zy  xy   y   zy (2.66) ()b  x  y  z  x  y  z o o o xz yz zz xz yz ()c x  y  z  x  y  z  Knowledge from eq (2.66), the what we have defined by through under impaction of P external force. This means, could substitute P external load impacte calculating soil o o o o o o volume by x,,,,,  y  z  xy  xz  zy what have defined o o o o o o 2.3x.2.,,,,,  yPlate  z on xy elastic  xz  zy half-space theory and when considering influence of shear deformation. Binding condition: wdt x, y , z 0 w x , y w (2.67) P Calculating foundation Elastic half-space o Plate x y z xy xz yz o o o Known x y z o o o xy xz yz z Plate Have to calculate x y z a xy xz yz x o b y Fig 2.5. Place on elastic half-space model, have considering effects of shear deformation, based on Gauss’s princiole extreme
10. When z 0 and in scope  ab of the plate: 2M 22MM     o  o  o x 2 xy y zz yz xz yz xz (2.71) 22 x  x  y  y  z  y  x  z  y  x M M      oo  o   x xy Q x yx xz x yx xz (2.72) x x  y  x  y  z  x  y  z MM        o   o   o y xy Q y xy zy y xy zy (2.73) y y  x  y  x  z  y  x  z o o o y   xy   zy   y   xy   zy  y  x  z  y  x  z (2.74)  ooo xyx   zx   x yx   zx x  y  z  x  y  z  When z 0 and out side of the plate, has (2.66). Substitution (2.16) into (2.71), has:  o o o h2  D wzz yzxz yz xz   Q Q (2.75) xy z  y  x  z  y  x51   x  y Q Q If G , then  x 0; y 0 , has: xyGh Gh  o o o Dw zz yz xz yz xz (2.76) z  y  x  z  y  x Equation (2.76) is balance equation of plate on elastic half-space when ignoring effects of shear deformation. 2.4. Conclude chapter 2 Solve elastic Plate-Foundation based on Gauss’s principle extreme and while onsidering effects of shear deformation not only define status of stress-deformation of plate but also to could define status of stress-deformation of the elastic foundation. Satisfying three boundary conditions on edge plate by the way considering effects of shear deformation.
11. Only have to solve equation (2.10) to define deflection function w x, y of plate for calculating plate on Winkler foundation. Q x, y Q x, y Solve equations (2.58), (2.59)x and (2.60) couldy define deflection function w x, y and shear function and shear function while considering effects of shear deformation. Solve plate on elastic half-space based on Gauss’s could completion about bending plate theory. CHAPTER 3 SOLVE PLATE ON ELASTIC FOUNDATION, [5] In this, Post-Graduate use 4 nodes rectangular element has 16 degrees of freedoms, called is BFS-16, interpolation about displacement and shear function of the plate by Hermite function. Subgrade element is cubic finite-8 nodes and based on Mindlin’s solution. The feature of Winkler foundation model is coefficient (k).And the feature elastic half-space model is elastic modulus (Eo) and coefficient poisson (o). 3.1. Establish programe by FEM. 3.1.1. Displacement element w . 1 x = 1 2 x Fig 3.2. Rectangular element – 4 = 1 nodes y 3 4 y Interpolation functions Ni :
12.  wNfxfy1 ,()(),()() 1 1 1 y 1 Nfxfyy 9 1 3 wNfxfy2 ,()(),()() 2 2 1 y 2 Nfxfyy 10 2 3 (3.19) wNfxfy3 ,()(),()() 3 1 2 y 3 Nfxfyy 11 1 4 wNfxfy4 ,()(),()() 4 2 2 y 4 Nfxfyy 12 2 4  x1 ,()(),()()Nfxfyx 5 3 1 xy 1 N 13 fxfyxy 3 3 x2 ,()(N 6 f 4 x f 1 y),()() x xy2 N 14 f 4 x f 3 y x y x3 ,()(),()()Nfxfyx 7 3 2 xy 3 N 15 fxfyxy 3 4 ,()(),()()Nfxfyx N fxfyxy x4 8 4 2 xy 4 16 4 4  Called Nw is vector having 16 interpolation function , then: e (3.20) wew  N w  3.1.2. Shear element Qx .  NQQQQ f()()()() x f12 y N f x f y xx1411 (3.25) N f()()()() x f y N f x f y QQQQxx2212 25 NQQQQ f()()()() x f12 y N f x f y xx3633 With: fx( ) 1 3x 2x2  Q1 fx( ) 4x 4x2 Q2  fx( ) x 2x2 Q3  x = 1 1 2 3 x Q1 Q2 Q3 Fig 3.4. Shear element Qx = 1 y Q Q Q of 6 nodes plate element 4 5 6 Ni 4 5 6 y Called N e is cector having 6 interpolation functions (3.25): Qx e NNNNNNNQQQQQQQ (3.26) x x1 x 2 x 3 x 4 x 5 x 6 e and called Qx  is vector having QQQQQQ1,,,,, 2 3 4 5 6 : T QQQQQQQe (3.27) x x1 x 2 x 3 x 4 x 5 x 6
13. then: QNQe e e (3.28) x Qx x  3.1.3. Shear element Qy . 1 x = 1 4 x Q1 Q4 Q Q 2 2 5 5 = 1 = Fig 3.5. Shear element Q y y Q3 Q6 3 6 of 6 nodes plate element y Like shear element : QNQe e e (3.33) y Qx y  3.1.4. Stiffness matrix of plate element: Displacement element w has 16 unknowns, displacement element Qx has 6 unknowns and displacement element Qy has 6 unknowns .Ttotal is 28 unknowns. Called U is vector having 28 unknowns: T (3.34) U w Qxy Q Expension vector becoming what have 28 ingredients: Nw  N  (3.35) N Nw , z e ro 1,12 Deflection function in any point of the plate element, eq. (3.20), rewrite: w e w  N U (3.36) e Qx N in eq.(3.2), becoming vector NQ : Similar so, vector Qx x NQQ ze ro 1,16, N ,e z ro 1,6 (3.37) xx Vector N e in eq.(3.33), becoming vector N : Qy Qy N ze ro 1,22 , N (3.38) QQyy Finally: - Shear function Qx in any point of plate element:
14. e QNUx Qx  (3.39) - Shear function Qy in any point of plate element: QNUe  (3.40) yQ y From (3.36), (3.39) and (3.40) could easy define relations: shear deformation; rotation; bending deformation and curly deformation. Element plate like plate and that has two-way link, so, extreme condition (2.29), becoming: Z 11      Mxx My 20 M xy Qee Q y dxdy x y xy x y (3.48) ui00  u i  u i  u i  u i  u i i 1,2, 28 Have done this work with i = 1,2, ,28 , could have stiffness matrix of plate element Ae 28 28 . Have connect stiffness matrixs of element, with external force function and boundary conditions and continuous conditions, could establish general stiffness matrix. 3.1.5. Element of subgrade reaction R based on Winkler model: Subgrade reaction in any point of element: e R kw k N U (3.50) 3.1.6. Subgrade element following elastic half-space foundation model: a/ (-1,-1,-1) 8 7 (1,-1,-1) b/ 5 6 o x = 1 1 (-1,1,-1) (1,1,-1) x o x y = 1 (-1,-1,1) 3 2 (1,-1,1) 1 (-1,1,1) (1,1,1) 4 1 y z y z Fig 3.8. a/ Subgrade element -8 nodes and b/ Plate element BFS-16
15. Subgrade Plate-4 free edge 8 nodes element BFS-16 element Load x P y x y Subgrade z 8 nodes element z Fig 3.9. Plate on elastic half-space foundation model, based on FEM Deflection following x,, y z of any point in element: 8  u x,, y z  Nii u i 1 (3.51) 8 v x,, y z  Nii v  i 1 8 wd x,, y z  N i w di i 1  General function to define deflection of any point: 1  Ni 1 xx o 1 yy o 1 zz o (3.52) 8  i 1,2 8  In this, xo,, y o z o are coordenates of node; - are coordenates of calculating point. Composition of u has 8, and composition of v has 8 , and composition wd has 8, too. Total, it has 24 unknown. So, K is vector having 24 unknown this: T K  u v wd  (3.53) Deflection of a point of subgrade element:
16. uee N K  u ee (3.54) v Nv K  wee N K dw d  In this, NNNeee ,, are vectors that have 8 compositions – u v wd interpolation functions following (3.52). Expantion : N Ne , z e ro 1,16  uu (3.55) e Nvv ze ro 1,8 , N , z e ro 1,8  N ze ro 1,16 , N e wwdd  Deflection of a point in element is becoming: ue  N K  u e (3.56) v  Nv  K  we N K dw d  From (3.56), could define stress and relations based on Symbolic of Matlab. And zd - Functional subgrade:  Z   dV   dV   dV xy  dV yz  dV zx  dV 0(3.58) dxx yy zz xy yz zx VVVVVVGGGo o o 111 e ee e e e e e e u vwd  u  v  v  w d   u  w d  i  j  j xy  j j yz  j j zx  j j dxdydz 0 xx y  y z  zGyxGzyGzx       111 o o o (3.59) Have done this work with i = 1,2, 8 , could define subgrade stiffness matrix: Ce 24 24 . 3.1.7. General stiff matrix: Functional ZZZ dt min , when:
17. 11      Mxx My 2 M xy Qee Q y dxdy x y xy x y 00 ui  u i  u i  u i  u i i 1,2, 28 e ewwee e e e uj  v jddjj xy  u j  v j yz  v j x   y   z    111 x  y  z G  y  x G  z  y oo dxdydz 0 e e 111  u w d zx i i G  z x o j 1,2, 8 (3.60) Element of Plate-Foundation of plate on elastic half-space foundation is like plate-foundation, and based on external load, and with boundary conditions, and continuous conditions, could establish general stiffness matrix for plate-foundation elastic systems. BLOCK DIAGRAM
18. To dec l a r e Plate Element Size Load Subgrade Model Mo ving l o a d t o no de WINKLER FOUNDATION / ELASTIC HALF-SPACE Unnknown of Subgrade Stiffness matrix of Subgrade Pl a t e Unknown of Plate Parameters of Plate Interpolation function Stiffness matrix of Plate Est a bl ish g ener a l st if f ness ma t r ix Load Boundary conditions Continuous conditions so l ve r esul t s Deflection Curly moment on edge Internal values of Plate Subsidence and stress of subgrade c hec k (f ) Bending-tractive stress of plate < = US kBending-tractive stress of material (t ) t he end Fig3.10. Block diagram of program 3.2. Example Table 3.8: Square plate- P converged load in center plate h/ a 1/ 20; E 1000 daN / cm2 ; 0,35; l 90.3 cm oo -13 Mat vong x 10 Bieu do mo men xoan tren bien x=0 10 8 0 6 -0.005 4 -0.01 -0.015 2 -0.02 0 Gia Gia tri vong,do cm -0.025 -2 Gia Gia tri mo men, daNcm -0.03 15 -4 15 10 10 -6 5 5 -8 0 0 0 1 2 3 4 5 6 7 8 9 10 Kich thuoc tam, 50cm x 10 Kich thuoc tam, 50cm x 10 Kich thuoc tam, 50cm x 10
19. Bieu do mo men Mx giua phan tu dat tai trong Bieu do mo men My giua phan tu dat tai trong 1400 1400 1200 1200 1000 1000 800 800 600 600 Gia Gia tri mo men, daNcm Gia Gia tri mo men, daNcm 400 400 200 200 0 0 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 Kich thuoc tam, 50cm x 10 Kich thuoc tam, 50cm x 10 Bieu do ung sua dat nen Bieu do do lun tam+nen va cua rieng nen (net dut) 0 0 -0.5 -0.5 -1 -1 -1.5 -1.5 -2 -2 -2.5 -2.5 -3 -3 Do sau,Do 50cm x 5 sau,Do 50cm x 5 -3.5 -3.5 -4 -4 -4.5 -4.5 -5 -5 -6 -5 -4 -3 -2 -1 0 1 2 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05 0.055 Gia tri ung suat, daN/cm2 Gia tri do lun, cm 3.4. Conclude chapter 3 By the way considering effects of shear deformation and solve plate on elastic foundation program, good resuts: - Satisfying three boundary conditrions on edge plate. - Not only define redistribution of internal in the plate, but also define contemporaneous stress and displacements of plate and foundation. Change of stress and deformation status of plate: - - Curly moment in free edge of the plate is not zero while ignore effects of shear deformation. And else, curly moment is very small ( < 10-11) while considering effects of shear defomation, can be zero. - - Deflection maximum value is 5.5% difference-while ratio h/a 1/20 . While 1/15 h/a 1/5 , this difference increase 8.8 21% .
20. - Having redistribution of internal in the plate. Even, having reversal of moment-graph. This change is very more clear when thickness plate is more increased, specially when the place be effected of load in corner or in edge plate, so, 32.1% difference. - About moment value, when ratio (h/a) the more increasing, the more moment and deflection value is disparity, comparing that be in edge and corner to center plate. Plate-Foundation model that Post-Graduate have choosed: - Plate on elastic foundation-4 free edges. - Most adversary status: P – Converged load in midle length-edge of plate. Have to reseach about effect of shear deformation, when: - Ratio: h/a ≥ 1/15. - Converged Load are in edge and corner of plate. - Thick plate. Have better not do plate what has b/a > 1.3. And, Post-Graduate propose using coefficient K = 1.32.2 be considered increasion moment. CHAPTER 4 AN APPLICATION OF TC2BRP & TC32RP PROGRAME TO DESIGN CONCRETE PLATE PAVAMENT In this: Establish nomogram to define thickness of plate Calculate steel to reinforce for edge and corner of plate. Calcuate foundation. Comparison result of programe with some the other methods. 4.1. Establish nomogram. Result 1:
21. h 2 D daN/cm 0.8 q= P 0.7 R 2 25.0 20.0 0.6 15.0 0.5 10.0 5.0 0.41 0.4 0.3 -2 Eb x10 0 1.97 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 Ech Fig 4.1.Nomogram to define thickness of concrete pavament plate Based on model: plate on elastic half-space foundation Result 2: h(cm) daN/cm2 45 40 35 30 q= P R 2 25 25 20 21.15 20 15 10 15 3.35 5 0 K 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10 11 12 13 14 15 (daN/cm3 ) Fig 4.1.Nomogram to define thickness of concrete pavament plate Based on model: plate on elastic Winkler foundation
22. The nomogram is established for the aim: - Based on parameters input,such as: External load, material, k, Eo, , put into the nomogram to define rudiment thickness plate (h). - Put (h) and other parameters into TC2BRP or TC32RP programe, have calculated internal forces and stress and displacement of plate and of foundation under plate. And then: + Calculate steel to reinforce for plate based on chart moment. + Based on chart of stress and displacement of foundation to calculate and design foundation layer material. 4.2. Calculate steel to reinforce for plate 4.2.1. Calculate steel to reinforce for edge of plate Mat vong tam chu nhat Bieu do mo men My tren bien x = 0 2500 0 2000 -0.01 1500 -0.02 1000 -0.03 Gia Gia tri vong,do cm 500 -0.04 Gia tri mo men, daNcm 15 0 8 10 6 -500 5 4 2 Kich thuoc tam b=1.3a, cm 0 0 -1000 Kich thuoc tam a, cm 0 1 2 3 4 5 6 7 8 9 10 Kich thuoc tam b = 1.3a, cm - Use the chart to calculate torque reinforcement plate edge enhancement, especially for panels on sewer lines, underground lines of life. Mesh bottom plate placed parallel to the edges, the mesh plate placed perpendicular to the edge plate. - Seeing is 3 to 4 sheets of edge element has deflection most, Post- Graduate proposals to strengthen reinforced this domain. Based on the particle size, plate size and characteristics of elastic plates, Post-Graduate choose range enhance reinforcement from the back edge of the plate, by: Max (120cm, a / 3; 1.3L), (L is the elastic characteristics of the plate). Fig 4.3. Basic solution to calculate steel to reinforce for edge of plate
23. 4.3. Calculate and design foundation layer material. Based on the charts and graphs of stress subsidence of the ground, from which the decision of the foundation layer material. With each work load, plate material and plate thickness, we establish the relationship between the bending tensile stresses in the plate section is reviewed and elastic modulus of the ground / foundation. If the fixed plate material, thickness of nails beneath, to satisfy the conditions of bending tensile stresses in the plate, we completely determine the elastic modulus for subgrade requirements, and vice versa. 2  (daN/cm ) 2 Plate: h=22cm; Eb = 315000daN/cm Foundation: 18cm; E m Ratio b/a: 1.3 Upright load: 55kN 56.3 22 24 45.96 26 2 Plate: h=24cm; Eb = 315000daN/cm i Foundation: 18cm; E m Ratio b/a: 1.3 Upright load: 65kN 37.1 24.6 12.5 3 2 10.1 E ( X10 daN/cm ) 3.2 m 0 0.3 0.5 E 1.0 2.0 5.0 10.0 50.0 100.0 mi Fig 4.8. Relationship between the stress of plate with module of foundation According to calculations by graduate students, in the same condition as the other, then: When cement concrete sheet thickness increased / decreased by 1 cm, modulus of elasticity of foundation under a reduced / increased by 40-50daN/cm2. When the work load increase / decrease of about 10kN, thick concrete slabs increase / decrease of about 2cm. 4.4. Some comparisons
24. Post-Graduate have compared the results of their calculations with analytical formulas of Westergaard and Shekter-Gorbunov-Pasadov, R805FAA software, KenSlabs-2003 The above comparison purposes only difference comes when and when not taking into consideration the effects of transverse shear strain in the plate, rather than specifying the value and stress torque plates to design, because not to mention other factors, such as considering the impact factor of the load, for the same load, the effect of boundary links, the influence of temperature, humidity, The biggest difference here is TC32RP program of Post-Graduate allows to define simultaneously determine the stresses and displacements of the plate and the subgrade. The authors above, who found the stress and displacement of foundation by other ways. CONCLUSIONS AND RECOMMENDATIONS 1.Conclusions: Thesis research allows us to identify the state of stress and deformation of the plate and of the foundation simultaneously, directly satisfy the boundary conditions on the edges 3 plates. The new study results of the thesis, as follows: a/ Theoretical:  First: By considering the effects of transverse shear strain and solving elastic plate-foundation system by the way of comparison method based on Gauss principle extreme theory, Post-Graduate have completed another step computation theory " plate on elastic foundation ": The plates on Winkler elastic foundation, without considering the effect of transverse shear strain, just have to solve a eq (2.10), implicitly define a single deflection w x, y of the plate, thereby determining the value of the internal force of plate.
25. While considering the impact horizontal shear strain induced by shear forces,z 0 we must solve theab 3 equations (2.58), (2.59) and (2.60) to define 3 unkown functions: , Qx x, y and Qy x, y . The flexural plate on the elastic half-space: - When and in scope of the plate, established 5 balance equations (2.71), (2.72), (2.73) and (2.74) to define 5 unknown functions: w,,,, u v Qxy Q . - When and out side of the plate, has (2.66). Solve the above equation, not only to identify the components of stress and displacement of the plate, but also simultaneously have defined components of stress and displacement of the ground.  Second: z 0 Develop a program and solved the plate on elastic foundation. Stress-strain state of the plate and make the following changes: Simultaneously, define stresses and displacements of the plate and the subgrade. To satisfy 3 boundary conditions on the free edge of plate. Point out the need to consider the effects of horizontal shear strain in the plate on elastic foundation, when ratio: ha/ 1/15. Specify the model is the model most unfavorable 4 free edges ofplate on elastic foundation and load placement detrimental for weight plate is positioned between the effects long side of the plate. b/ In terms of applications to design concrete pavament plate:  First: - Establish 2 nomogram to define thickness plate - Proposals of Post-Graduate: Have better not use a rectangular sheet size difference of the two sides passingw x1.3, y And can be used to increase the moment coefficient: K =1.3 2.2  Second:
26. Calculate and layout steel to have reinforced edges and corners strengthened concrete slabs are made more accurate through the bending moment diagram when the load placed on the corner and edge plate.  Third: Identify and stresses and displacements in the plate and in the function, lets build nomogram 4.8, investigated the effect of subgrade to calculate the thickness of concrete slabs on motorways and airports. Know the internal forces in the plate, stresses and displacements of the foundation, help choose thickness plate and foundation layer material more reasonable. 2. Recommendations: Thesis made several recommendations in the calculation sheet, contributing more complete theoretical calculations plate on elastic foundation. To TC2BRP and TC32RP program is more practical applications, higher reliability in computing concrete pavement plate for highwway and airports, need for meticulous research and comprehensive than the other complex issues that fact requires, such as the influence of dynamic loads and loads for the same; influence of environmental effects of contact between the plate and foundation, the influence of plate thickness in the calculation plate and calculate background, relationship betwween the size and the various links of the plate boundary, the friction between the bottom plate to the foundation, etc, 3. Directions for further research: Research on the method of elastic sheet, taking into account the dynamic loads, horizontal thrust, temperature changes and the influence of the other factors.
27. CATALOGUE OF WORKS BY AUTHOR 1. Nguyen Anh Tuan,(2009),“Calculate plate on the elastic Winkler by finite difference method”, Vietnam Bridge-Road Journal, (11), p. 17-26. 2. Nguyen Anh Tuan,(2010), “ Research the plate on the elastic foundation and considering the effects of transverse shearing deformation” Science topic of the UCT, Ha Noi. 3. Nguyen Anh Tuan,(2010),“Solve plate on the elastic Winkler by Maurice Lévy method”, Vietnam Bridge-Road Journal, (10), p. 16-22. 4. Nguyen Anh Tuan, (2012), “Analysing the plate on elastic half-space by new theory” , The Transport Journal, (8), p.20-21,33. 5. Nguyen Anh Tuan, (2012), “ Solve the plate on elastic half-space by finite element method”, The Transport Journal, (9), p.16-18.