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Analysis of Rigid Pavement Built with Multi-layer Concrete Slabs Resting on Pasternak Foundation Sanjeev Kumar Suman1 and Sunita Kumari2 1

Assistant Professor, Department of Civil Engineering, National Institute of Technology Patna, Bihar, India, [email protected] 2 Assistant Professor, Department of Civil Engineering, National Institute of Technology Patna, Bihar, India, [email protected]

Abstract: Rigid pavement construction is emerging due to its long term performance and less maintenance cost. Its rigidity and high modulus of elasticity tend to distribute load over a wide area of foundation, resulting in very low deflections under traffic and environmental loading. Thus a major portion of the structure capacity is imparted by slab itself. Multi-layer slab is the composition of several superposed concrete layers of different quality. An idea is that lower quality material can be sandwiched in the neutral axis zone. This paper deals a structural analysis of the three layered structure resting on the Pasternak foundation subjected to traffic loading. The three layered structure is replaced by an equivalent single layered structure. An assumption was taken that the deflection at the surface of the upper layer may be considered to be equal with the deflection at the bottom of the lower layer. The mechanical characteristics of the equivalent structure depend on the mechanical characteristics of the different layers and the various adhesion conditions at the interfaces. Adhesion condition are full slip at each of the interface, full friction at the first interface as well second interface, full friction at both interfaces and partial friction at all interfaces. Analytical solutions are arrived and structural responses are investigated. It has been found that the structural responses for full friction case showed significant results. This study is made towards enhancing current design method and economic construction. Keywords: Multi-layer slab, Pasternak foundation, Flexural strength INTRODUCTION Rigid highway pavements are normally constructed of Portland cement concrete and may or not have a bottom course between the subgrade and the concrete surface. Rigid pavements have some flexural strength that permits them to sustain a beamlike action across minor irregularities in the underlying material. Properly designed and constructed rigid pavements have long service lives and usually are less expensive to maintain than the flexible pavements [Abusharar S.]. This statement may be true in the contest of rate of deterioration is comparatively much slower than flexible pavement. However, as suggested by the Indian road congress (IRC): 58, the minimum grade of concrete to be used is M40 and thickness of concrete slab varies from 16 cm to 36 cm. Thus providing the same grade of concrete throughout the large depth of concrete slab, which can be mitigated by using higher to lower grade of concrete from top to bottom to reduce the cost of construction indirectly without affecting structural capacity. All pavements derive their ultimate support from the underlying subgrade. Subgrade models such as Winkler, Filenenko-Borodich, Pasternak, Vlasov, Reissner, Hetenyi, etc., play a very important role in geotechnical engineering.

Generally, the foundation is described either as a continuum for which linear stressβstrain relations are assumed valid, or as a system of continuously distributed independent springs which offer resistance in the direction of their axis only. Winkler (1867) assumed that the reactive forces of the foundation were proportional at every point to the deflection of the beam at that point. Pasternak (1954) assumed that shear interactions existed between the springs. This was accomplished by assuming the tops of the equivalent springs to be connected by a mechanical element [Patil et al.]. Two-lift concrete paving involves the placement of two wet-on-wet layers instead of the homogenous single layer commonly placed in concrete paving. The two-lift process has great potential as a sustainable paving solution. It reduces the cost of materials and materials transportation, reduces the environmental impact caused by quarrying and importing aggregates, and increases social benefits by maintaining friction and low noise levels and providing a long-lasting pavement that does not need frequent reconstruction [Map Brief]. Two lift pavements consisted of two layers placed, wet on wet, with the top layer consisting of a special surface mix. Coarse aggregate was placed in the surface layer and built as an exposed aggregate surface to provide a quiet, but rough surface texture for durability and friction resistance [James K.]. The objective of this paper is to analyse the multi-layer slab concrete pavement resting on a Pasternak foundation model taking a model of three layered system with different cases of interaction among different interfaces and comparing the results with a single layered structure of equivalent thickness. Four cases have been considered namely: full slip at each of the interface, full friction at first interface, full slip at the second interface and full friction at both interfaces. BEAM ON AN ELASTIC FOUNDATION WITH PASTERNAK MODEL A single-layer elastic foundation of finite thickness is considered as shown in FIG.1. The subject matter is restricted to the problems where the horizontal displacement is negligible. In addition, it is assumed that the shear stress at the interface between the compressible layer and the rigid bottom equals zero. For a relatively thin compressible layer of foundation, the variation of the normal stresses along the depth may be small and therefore can be considered as constant with depth. Based on the conventional stress-strain and straindisplacement relationships, the equilibrium of an infinitesimal element of the foundation under an externally distributed load over the contact area is shown in FIG. 1. Concrete pavement with modulus Pasternak shear layer with modulus G

Figure 1. Concrete slab on Pasternak foundation. The final differential equation using equation of equilibrium is given in Eqn.1 [Patil et al.]. π4 π€ ππ₯ 4

πΊ

π

π

- πΈπΌ π» 2 w + πΈπΌw = πΈπΌ

(1)

Where w = deflection at any point in beam, G = shear modulus of elasticity of the subgrade, EI or D=flexural rigidity of beam, k = modulus of subgrade reaction and p=distributed load intensity. MULTI-LAYER SLAB MODEL Consider a pavement consisting of three superposed concrete layers (of different quality). It can be designed using the models developed for a single slab as far as the total thickness of the layers can still be considered as small against the whole structure. Mathematically, one must assume that the radius of curvature remains important, in other words, that the deflection at the surface of upper layer may be considered to be equal with the deflection at bottom of lower layer. The mechanical characteristics of the equivalent structure depend of the mechanical characteristics of the different layers and the adhesion condition at their interfaces. 2a p E1

Concrete slab 1

E2

Concrete slab

q1

E3

Concrete slab

q2

G

p1

h1

p2

h2

p3

h3

q3

Pasternak shear layer with modulus G

k

Figure 2. Multi-layer slab on Pasternak foundation with loading. The three layered structures (FIG.2) are replaced by an equivalent single layered structure under following conditions. Case 1: Full Slip at Each of the Interfaces [Cauwelaert, F.V] The equilibrium equations for first, second and third layer are Eqns.1, 2 and respectively. πβπ π π2 π βπ π» 4 π€1 = π· 1 (2), π» 4 π€2 = 1β (3), π» 4 π€3 = 2π· 3 (4) π· 1

2

3

By hypothesis, w=w1=w2=w3, Hence, π = π β (π·1 + π·2 + π·3 ) The equilibrium equation of equivalent slab is πβπ πβπ π» 4 π€1 = β π· = π

(5)

π

Where D =β π·π

is the equivalent stiffness of the structure, πΊ

π

π

Consider a Pasternak subgrade hence, π» 4 π€ β πΈπΌ β2 π€ + πΈπΌ π€ = πΈπΌ

(6)

The moment in x is given by M = -D*( ππ₯ 2 ) The bending stresses at the bottom of the layers are

(7)

π2 π€

ππ = -

6π·π (βπ )

2

π2 π€

( ππ₯ 2 ) =

6π·π π (βπ )

2

π·

(8)

Case 2: Full Friction at the First Interface, Full Slip at the Second Interface [Cauwelaert, F.V] Two upper layers are replaced by an equivalent layer whose modulus is equal with the modulus of the upper layer. The transverse section of the equivalent layer is a I- section as given in FIG.3. The width of the vertical bar of the I-

section is equal with the ratio of moduli: b= E2/E1.Hence the moment of inertia of the transformed section remains equal with the moment of inertia of the initial section.

Figure 3. Multi-layer slab with friction at first interface, slip at the second. The position of the neutral axis is given by c=

β1 2 +π(2β1 β2 +β2 2 )

(9)

2(β1 +6β2 )

The moment of inertia of T-section is 3β π 2 β3β 2 +β 3

3β (β βπ)2 +3(β1 βπ)β2 2 +β2 3

1 1 πΌ12 = 1 +π 2 1 3 The stiffness of equivalent section is πΈ1 .πΌ12 π·12 = (1βπ 2)

3

(10) (11)

1

The equilibrium equations are π βπ2

π»4π€ =

π·12

π2 βπ

π»4π€ =

π·3

(12) (13)

Therefore we can write π βπ

π» 4 π€ =π·

=

πβπ

π· 12 +π·3 πΊ 2 ππ€ π π» w+ π· =π· π·

π»4w Now the bending stresses are: β βπ π· At the bottom of upper layer, ππ₯ =ππ₯ 1π π·12 At the bottom of mid layer,

ππ₯ =ππ₯

6ππ₯ π·3

At the bottom of lower layer, ππ₯ = β

3

2

π·

(15) (16)

β1 +β2 βπ π·12 πΈ2 π

(14)

π· πΈ1

(17) (18)

Case 3: Full Friction at Both Interfaces [Cauwelaert, F.V] Three layers are replaced with an equivalent layer whose modulus is equal with the modulus of the upper layer. The equivalent transverse is a section with decreasing width as shown in the FIG.4. The width of the sub sections is equal with the ratios of moduli.

Figure 4. Multi-layer slab with friction at both interfaces. The position of neutral axis is given by: π=

β1 2 +π1 (2β1 β2 +β2 2 )+π2 (2β1 β3 +2β1 β3 +β3 2 )

(19)

2(β1 +π1 β2 +π2 β3 )

The second moment of area about neutral axis at c from top: I123 = h1 c 2 β h12 c +

h31 3

+ b1 (h2 (h1 β c)2 + (h1 β c)2 h2 +

b2 (h3 (h2 + h1 β c)2 + h23 (h2 + h1 β c) +

h33 3

)

At the bottom of lower layer, ππ₯ =ππ₯

)+

(21)

123

β1 +β2 βπ πΈ2

ππ₯ =ππ₯

3

(20)

The bending stresses are: β βπ At the bottom of upper layer, ππ₯ =ππ₯ πΌ1 At the bottom of mid layer,

h32

πΌ123 πΈ1 β1 +β2 +β3 βπ πΈ2 πΌ123

πΈ1

(22) (23)

ANALYSIS OF MUTISLAB RIGID PAVEMENT The analysis of multi-layer slab rigid pavement is carried out by considering a rigid slab resting on Pasternak foundation model. The slab is considered with free edges. The basic equilibrium equation will remain the same in all cases, only the properties of the slab will change according to the interaction among the interfaces between different layers. The basic differential equation is nonhomogeneous linear differential equation. The programs for computational algebra such as MATLAB are high-level language program systems with built-in functions for algebraic computations including manipulations with singularity functions. The suggested approach for an analytical solution is a combination of the transformation of the loading as singularity functions with the resources of the modern computer algebra systems for a solution of ordinary differential equations. The procedure for structural analysis of a beam on an elastic foundation are transformation of the load as a singularity functions, computation of the general solution of the given boundary-value problem and definition of the boundary conditions at the beamβs ends and calculation of the constants of integration. Boundary conditions are presented in Eqn. -EI -EI

π3 π€ ππ₯ 3

π2 π€ ππ₯ 2

(0)=0

(0) + Gb -EI

-EI

π2 π€ ππ₯ 2

π3 π€ ππ₯ 3

ππ€ ππ₯

(24)

(0) = 0

(L) =0

(L) + Gb

(25) (26)

ππ€ ππ₯

(L) = 0

(27)

MATLAB coding for deflection computation is w(x)=dsolve('(EI)*D4w(x)G*D2w(x)+K*w(x)==F*dirac(x-1.75)','(EI)*D2w(0)==0',' (EI)*D3w(0)+G*Dw(0)==0','-(EI)*D2w(3.5)==0',' (EI)*D3w(3.5)+G*Dw(3.5)==0','x'); The considered problem as shown in FIG.2 is solved by using data, Modulus of subgrade reaction (k) = 42000 kN/m3, Shear modulus (G) =20000 kN/m. The beam is 3.5m long (L), 1m wide (b) and has elastic modulus of E1=30000MPa; E2=20000MPa and E3=10000 MPa. Thickness are h1=8cm, h2=9cm, h3=12cm and sum of h1+h2+h3=h=29cm. It is loaded by dual wheel with tire pressure of 0.7MPa and equivalent wheel load of 60kN. Second moment of area is calculated by using formula Ii=bhi3/12. Maximum computed structural response is presented in Table 1 under differnt conditions. FIG.5 to 8 is given for first case representation of analysis. Paramter mentioned in the tables are named as x=distance from either of edges in m, w(x) =deflection on top surface at a distance x in m, R(x) =subgrade reaction in kN, S(x) =shear force at a distance x in kN, M(x)=Bending moment at a distance x in kNm,Ο1(x)=Bending stress at bottom of upper layer at a distance of x in kN/m2, Ο2(x)= Bending stress at bottom of middle layer at a distance of x in kN/m2 Ο3(x)= Bending stress at bottom of lower layer at a distance of x in kN/m2. Table 1. Maximum computed responses in different cases. π2 (π₯) Conditions x w(x) R(x) S(x) M(x) π1 (π₯) Case 1 Case 2

1.749 1.749

0.43 0.41

18.4 17

30 30

14 26

Case 3

1.749

0.41

17.44

30

23

Case 4# 1.749 0.412 17.32 #Single layer of equivalent thickness

30

18

-354.9 -1.94 103.35 -74.1

DEFLECTION CURVE

x 10-4

Case1 y min y max

4.3

4.1 4 3.9 3.8

0

0.5

1

1.5

2

2.5

3

3.5

Distance(in m)

Figure 5. Deflection curve. REACTION CURVE 18.5 Casea1 y min y max 18

Reaction(in kN)

Deflection(in m)

4.2

17.5

17

16.5

16 0

0.5

1

1.5

2

2.5

3

Distance(in m)

Figure 6. Reaction curve.

3.5

π3 (π₯)

-354.9 14.26

-236.6 5.06

74.81

133.27

14.8

133.34

SHEAR FORCE DIAGRAM 30 Case1 y min y max

Shear force(in kN/m 2 )

20

10

0

-10

-20

-30 0

0.5

1

1.5

2

2.5

3

3.5

Distance(in m)

Figure 7. Shear force diagram. MOMENT DIAGRAM 14

case1 y min y max

12 10

Bending moment

8 6 4 2 0 -2 0

0.

1

1.

2

2.

3

3.

Distance(in m)

Figure 8. Bending moment diagram. DISCUSSION The bending stress variation in each of the case at the bottom of each layer is drawn for comparison as shown in FIG.9-11.In the Case-1 tensile maximum bending stress is equal at bottom of upper and middle layer but at the bottom of lower layer reduced. The nature of maximum bending stress is changed from tensile to compressive in the second and third case but the value of compressive stress is more in the third case when compared to the second case. When overall equivalent thickness is considered maximum bending stress is changing from tensile to compressive. Table 2 represents the comparison of results obtained by the Cauwelaert F.V. and authors. Authors have considered three layers whereas literature author has considered only two layers under different input parameters. However data gives an idea about the proximity of the results. Table 2. Comparison of results. Bending stress at Bending stress at Condition Observed by bottom of first bottom of second layer layer Cauwelaert, F.V. 0.02MPa 0.15MPa Full friction Authors 0.10MPa 0.075MPa Cauwelaert, F.V. 0.47MPa 0.23MPa Full slip Authors 0.35MPa 0.35MPa Full friction at both interfaces creates bond which mitigate the bending action. Hence bending stress reduces in this case.

BENDING STRESS AT BOTTOM OF UPPER LAYER 50 0

case2 case4

Stress(in kN/m2)

-50 -100

case3 case1

-150 -200 -250 -300 -350 -400 0

0.5

1

1.5

2

2.5

3

3.5

Distance in m)

Figure 9. Bending stress at BUL. BENDING STRESS AT BOTTOM OF MIDDLE LAYER 100 50

case3

0

case4

case2

Stress(in kN/m

2)

-50 -100

case1

-150 -200 -250 -300 -350 -400 0

0.5

1

1.5

2

2.5

3

3.5

Distance(in m)

Figure 10. Bending stress at BML. BENDING STRESS AT BOTTOM OF LOWER LAYER 150

100 Case4

Case3

50

2 Stress(in kN/m )

Case2 0

-50 Case1 -100

-150

-200

-250 0

0.5

1

1.5

2

2.5

3

3.5

Distance (in m)

Case1 refers to full slip at each interface Case 2 refers to full friction at first and full slip at second interface Case 3 refers to full friction at both interfaces Case 4 refers to single layer of equivalent thickness BUL: Bottom of upper layer BML: Bottom of middle layer BLL: Bottom of upper layer Figure 11. Bending stress at BLL.

CONCLUSION Multi-layer slab having full friction at both interfaces and the equivalent single slab are behaving in the same manner. Hence it is suggested that to use multilayer slab having full friction at both interfaces to reduce the uniform high quality of material throughout the higher thickness. Further this work may be carried out by using finite element method. ACKNOWLEDGMENTS We would like to thank our graduate students for research support. The writers are in debt to International Engineering Foundation for giving opportunity to present a paper. REFERENCES Abusharar, S. (2015-16)."Design of rigid pavement, βTraffic and Highway Engineering," Applied Engineering and Urban Planning, University of Palestine, 1-63, ocw.up.edu.ps. Cauwelaert, F.V. (2003). "Pavement design and evaluation, Chapter 16, The Multislab," The required mathematics and its applications, Federation of Belgian Cement Industry, B-1170, Brussels, Rue Volta 9:163-166. Dobromir, D. (2012). "Analytical solution of beam on elastic foundation by singularity functions," Engineering Mechanics, Vol. 19(6):381-392. Map Brief (2010). "Two-lift concrete pavingβ, National Transportation Library,https://ntl.bts.gov.1-4 Patil, V.A., Sawant, V.A. and Deb, K. (2010). "Use of finite and infinite elements in static analysis of pavement," Interaction and Multiscale Mechanics, Vol. 3(1):95-110. Pasternak, P. L. (1954). "Fundamentals of a New Method of Analysis of Elastic Foundation by Means of Two Foundation Constants (in Russian)", Gosuedarstvennoe Izadatelstvo Literaturi po Stroitelstvui Arkhitekture, Moscow IRC: 58-2002, Guidelines for the design of plain jointed rigid pavements, The Indian road congress, second edition, New Delhi, reprinted Feb 2008:1-68. James K. Cable and Daniel P. Frentress (2004) Two-Lift Portland cement Concrete Pavements to Meet Public Needs, Final Report, Portland Cement Concrete Pavement Technology, Federal Highway Administration, U.S. Department of Transportation,1-14. Winkler, E. (1867). "Die Lehre von der Elastlzitiit and Festigkeit." Prague. 182.

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