. Stresses in Pavements.1. Stresses in Fleible Pavements.1.1. Stresses in Homogeneous Mass Boussinesq formulated models for the stresses inside an elastic half-space due to a concentrated load applied on the surface. A half-space has an infinitely large area and an infinite depth with a top plane on which the loads are applied. The simplest way to characterize the behaviour of a fleible pavement under wheel loads is to consider the subgrade, the subbase, base, and the surfacing layers to form a homogeneous half-space. If the modulus ratio between the pavement and the subgrade is close to unity, as eemplified by a thin asphalt surface and a thin granular base, the Boussinesq theory can be applied to determine the stresses, strains, and deflections in the subgrade. rror! Reference source not found. Figure -1 shows a homogeneous half-space subjected to a circular load with a radius a and a uniform pressure q. The halfspace has an elastic modulus and a Poisson ratio, v. A small cylindrical element with centre at a distance z below the surface and r from the ais of symmetry is shown. Due to aisymmetry, there are only three normal stresses, z, r, and t, and one shear stress, τ rz, which is equal to τ rz. These stresses are functions of q, r/ a, and z/ a. Figure -1: Stresses under aisymmetric circular loading Foster and Ahlvin have developed charts as provided here from Figure - to Figure -6 for determining vertical stress z, radial stress r, tangential stress t, shear stress τ rz, and vertical deflection w, assuming the half-space is incompressible with a Poisson ratio of 0.5. After the stresses are obtained from the charts, the strains can be computed from [ ν( )] ε z 1 z r + [ ν( )] ε r 1 r t + t z 17
[ ν( )] ε t 1 t z + r If the contact area consists of two circles, the stresses and strains can be computed by superposition. Figure -: Vertical stresses due to circular loading (Foster and Ahlvin, 1954) Figure -3: Radial stresses due to circular loading (Foster and Ahlvin, 1954) 18
Figure -4: Tangential stresses due to circular loading (Foster and Ahlvin, 1954) Figure -5: Shear stresses due to circular loading (Foster and Ahlvin, 1954) 19
Figure -6: Vertical deflections due to circular loading (Foster and Ahlvin, 1954) When a wheel load is applied over a single contact area, the most critical stress, strain, and deflection occur under the centre of the circular area on the ais of symmetry, where τ rz 0 and r t, so z and r are the principal stresses. The stresses, strain, and deflection on the ais of symmetry of a wheel load applied to a pavement, which is similar to a load applied to a fleible plate with radius a and a uniform pressure q, can be computed by: z q 1 z 5 3 ( ) 1. a + z ε ε z r r ( + ν ) 3 q 1 z z 1 + ν + z 5 ( 1 + ν ) ( ) ( ) 0.5 1. a + z a + 3 q ν z z 1 ν + 5 ( 1 + ν ) q ( 1 ν ) ( ) ( ) 0.5 1. a + z a + z 3 z z 1 ν + 5 ( ) ( ) 0.5 1. a + z a + z ( + ν ) 0.5 [( a + z ) ] } 1 qa a 1 ν w + z 0.5 ( a + z ) a 0
When ν 0.5, the equation is simplified to 3qa w ( a + z ) 0. 5 On the surface of the loaded half-space, z 0, the deflection is ( 1 ) ν qa w0 If the load is applied on a rigid plate such as that used in a plate loading test, the deflection is the same at all points on the plate, but the pressure distribution under the plate is not uniform and is epressed as: q () r qa r ( a ) 0. 5 Figure -7: Differences between fleible and rigid plates The smallest pressure is at the centre and equal to one-half of the average pressure. The pressure at the edge is infinity. The deflection of the rigid plate is given by ( 1 ) π ν qa w0 All the above analyses are based on the assumption that the fleible pavement is homogenous, isotropic and semi-infinite, and that elastic properties are identical in every direction throughout the material. With these assumptions, Bousinesq theory has the following drawbacks: (1) Fleible pavements are multilayered structures each layer with its own modulus of elasticity. () The pavement layers and the subgrade soil are not perfectly elastic. (3) The assumption that the load is uniformly distributed may not be true. 1
.1.. Stresses in Layered Systems In actual case, fleible pavements are layered systems with better materials on top and cannot be represented by a homogeneous mass. Various multilayer theories for estimating stresses and deflections have been proposed. However, basic theories that utilize assumptions close to actual conditions in a fleible pavement are those proposed by Burmister. Burmister first developed solutions for a two-layer system and then etended them to a three-layer system with the following basic assumptions: 1. ach layer is homogeneous, isotropic, and linearly elastic with an elastic modulus and a Poisson ratio, ν.. The material is weightless and infinite in the lateral direction, but of finite depth, h, whereas the underlying layer is infinite in both the horizontal and vertical directions. 3. A uniform pressure q is applied on the surface over a circular area of radius a. 4. The layers are in continuous contact and continuity conditions are satisfied at the layer interfaces, as indicated by the same vertical stress, shear stress, vertical displacement, and radial displacement..1..1. Two-Layer Systems The eact case of a two-layer system is the full-depth asphalt pavement construction in which a thick layer of hot-mi asphalt is placed directly on the subgrade. If a pavement is composed of three layers (e.g., surface course, base course, and subgrade) the stresses and strains in the surface layer can be computed by combining the base course and the subgrade into a single layer. Similarly, the stresses and strains in the subgrade can be computed by combining the surface course and base course. Vertical stress: The stresses in a two-layer system depends on the modulus ratio 1 /, and the thickness-radius ratio h 1 /a. Figure -8a shows the effect of pavement layer on the distribution of vertical stresses under the centre of a circular loaded area when the thickness h 1 of layer 1 is equal to the radius of contact area, or h 1 /a 1 and a Poisson ratio of 0.5 for all layers. Figure -8b also shows the effect of pavement thickness and modulus ratio on the vertical stress, c, at the pavement-subgrade interface.
(a) (b) Figure -8 (a) Vertical stress distribution in a two-layer system (Burmister, 1958) and (b) effect of pavement thickness and modulus ratio on pavement subgrade interface vertical stresses (Haung, 1969) Deflection: Surface and interface deflections have been used as criteria of pavement design. The surface deflection, w 0, under a uniformly circular loaded area is given in terms of the deflection factor F as: 1.5qa w 0 F The deflection factor, F, can be obtained from 1 / and h 1 /a. Figure -9 for the corresponding Figure -9: Vertical surface deflection for two-layer system (Burmister, 1943) 3
If the load is applied on a rigid plate, then 1.18qa w 0 F The interface deflection, w, between the two layers is epressed in terms of the deflection factor F as: w qa F The deflection factor, F, is different from F and provided in Figure -10 as a function of 1 /, h 1 /a, and r/a, where r is the radial distance from the centre of loaded area. Figure -10: Vertical inteface deflection for two-layer systems (Haung, 1969) 4
Figure -11: (Continued) Critical tensile strain: The tensile strains at the bottom of the asphalt layer have been used as a design criterion to prevent fatigue cracking. The critical tensile strain, e, at the bottom of the first layer for a two-layer system can be determined by e q 1 F e Figure -1: Strain factor for single wheel (Haung, 1973) 5
Where, F e is the strain factor that can be obtained in Figure -1 as a function of 1 /, and h 1 /a. The critical tensile strain under dual wheels or dual-tandem wheels is obtained from the same equation, but the strain factor needs to be corrected..1... Three-Layer System With quick computational facilities available, the analysis of three or more layers is no more a difficult task. The three-layer system can be conceived as follows: 1. Top layer, representing all the bituminous layers taken together,. Second layer, representing the unbound base and subbase courses, and 3. Third layer, representing the subgrade. Figure -13 shows a three-layer system and the stresses at the interfaces on the ais of symmetry. z1 vertical stress at interface 1 z vertical stress at interface r1 radial stress at bottom of layer 1 r1 radial stress at top of layer r radial stress at bottom of layer r radial stress at top of layer 3 Figure -13: Stresses at interfaces of a three-layer system under a uniform circular load At the ais of symmetry, tangential and radial stresses are identical and the shear stress is equal to 0. Jones has developed a series of tables for determining the stresses in a three-layer system for the following dimensionless parameters : k 1 1 k 3 A a h h H h 1 6
Part of Jones s tables is presented here as Table.3, from which four sets of stress factors, ZZ1, ZZ, ZZ1-RR1, and ZZ-RR, can be obtained. The product of these factors and the contact pressure gives the stresses as: z 1 q( ZZ1) z q( ZZ ) ( ZZ1 1) z1 r1 q RR ( ZZ ) z r q RR From the continuity of horizontal displacement at the interfaces, r1 and r can be computed from ' r1 z1 z1 k 1 r1 ' r z z k r Once the stresses at the interfaces are calculated, strains can be computed from the equations of strains. 7
Table -1: Stress factors for three-layer systems (Jones, 196) 8
Table -1: (Continued) 9
.. Stresses in Rigid Pavements Stresses in rigid pavements result from variety of sources, of which the applied vehicle loads, changes in temperature of the slab, friction between the slab and the subgrade or base course are the most important. These factors tend to result in deformations of the concrete slab, which cause tensile, compression, and fleural stresses of varying magnitude...1. Stresses Due to Vehicle Loading Three methods can generally be used to determine the stresses and deflections in concrete pavements due to vehicle loading: 1. Westergaard s formulas. Influence charts 3. Finite element analysis Here, Westergaard s formulas derived to eamine three critical conditions of loading: corner loading, interior loading, and edge loading far from any corner are described. Westergaard s assumptions: The concrete slab acts as a homogenous isotropic, elastic solid in equilibrium. The reactions of the subgrade are vertical only and they are proportional to the deflections of the slab. The thickness of the concrete slab is uniform. The load at the interior and the corner is distributed uniformly over a circular area of contact and the circumference of the contact area at the corner is tangent to the edges of the slab. The edge loading is distributed uniformly over a semi-circular area, the diameter of the semi-circle being at the edge of the slab. Corner Loading: when a circular load is applied near the corner of the concrete slab, the stress, c, and the deflection, Δ c, at the corner are given by 0.6 3P a c 1 h l Δ c P kl a 1.1 0.88 l 3 h in which P is the load, l is the radius of relative stiffness defined as ( ) l, k is the 1 1 ν k modulus of the subgrade reaction, and a is the contact radius. The results obtained applying the finite element method of analysis are: 30
c 3P h 1 P c Δ c 1.05 0. 69 kl l where, c is the side length of the a square contact area, c 1.77a c l Modulus of subgrade reaction, k, is the constant that defines the subgrade in classical works of rigid pavements as shown in Figure -14 and defined as: p kδ where, p is the reactive pressure, and Δ is the deflection of the slab. The value of k is determined by means of the plate-loading tests. 0.7 (a) (b) Figure -14: (a) Definition of subgrade reaction, k, (b) plot of plate loading test to determine subgrade reaction Interior Loading: The formula developed by Westergaard for the stress in the interior of a slab under a circular loaded area of radius a is ( + ν ) 3 1 P l i ln + 0. 6159 πh b in which l is the radius of relative stiffness and b a when a 1.74h The deflection due to interior loading is b 1.6a + h 0. 675h when a < 1.74h Δ i P 1 a a 1 + ln 0. 673 8 kl π l l dge Loading: The stresses and deflections due to edge loading as formulated by Westergaard are: 31
For circular contact area e 3( 1+ ν ) ( 3 + ν ) P πh 3 h ln 100ka 4 4v 1 v 1.18 1 + 1.84 + + 3 l ( + v) a + 1.vP Δ e 3 h k For semicircular contact area Δ e e 3( 1+ ν ) ( 3 + ν ) P πh + 1.vP 3 h k ( 0.76 0.4v) + 1 l 3 h ln 100ka 4 a 4v + 3.84 + 3 ( 0.33 0.17v) + 1 l a ( 1+ v) When a load is applied over a set of dual tyres, the equations can be used after converting the contact area of the dual tyres into a radius, a, of equivalent circular contact area as: 0.851Pd S d pd a + qπ π 0.57q where, P d is the load on dual tyres, q is the contact pressure, S d is the spacing of the tyres.... Stresses Due to Curling Changes in temperature through the slab cause differential epansion or contraction between the top and bottom which results curling of the slab upward or downward. The weight of the slab restrains the slab from curling upward or downward. Consequently, stresses known as curling or warping stresses develop in the slap. During the day when the temperature on the top of the slab is greater than that of the bottom, the top tends to epand with respect to the neutral ais while the bottom tends to contract. Because the weight of the slab restrains the downward curling, compressive stresses are induced at the top while tensile stresses occur at the bottom. At night, when the temperature on the top of the slab is lower than that at the bottom, the effect is the reverse. The strain in the -direction in the infinite slab curled upward as shown in Figure -15 due to the stresses in the two directions can be determined by the generalized Hook s law as: 1/ l a 3
Figure -15: Upward curling of elastic slab due to temperature y ε ν where, is the strain in -direction due to, the stress in the -direction, and y ν is the strain in -direction due to y, the stress in the y-direction. and, y ε y ν When the slab is bent in the in the -direction, ε y 0, and y ν. Substituting this and solving for, gives: ε 1 ν (the stress in the bending direction), and y ν (the stress in the perpendicular direction to bending). When bending occurs in both directions as in the case of temperature curling, the stresses in both directions must be superimposed. Let Δ t represents the temperature differential between top and bottom of the slab, and α t represents the coefficient of thermal epansion of concrete. If the temperature at top is greater than at the bottom and the slab is completely restrained and prevented from moving, the strain developed at the top will be compressive and at the bottom tensile as shown in Figure -16 assuming the distribution of temperature is linear through out the slab depth. Figure -16: Strain due to gradient in temperature ε α tδ t ε y The stress in -direction due to bending in the in -direction is 33
α Δ t t ( 1 ) ν and the stress in the -direction due to bending in y-direction is να Δt ν The total stress in the -direction is then, t ( 1 ) α Δ να Δ + ν t t t t ( 1 ν ) ( 1 ) α tδ 1 ( ν ) t For a finite slab with length L and L y in the - and y-directions respectively, the total stress in the -direction can be epressed as: C α Δ C yνα tδ + ν t t ( 1 ν ) ( 1 ) α tδ t ( C + C yν ) ( 1 ν ) where C and C y are correction factors for a finite slab in the - and y-directions respectively. Similarly, the stress in the y-direction is y α tδ t ( C y + C ν ) ( 1 ν ) Based on Westergaard s analysis, Bradbury developed a simple chart shown as Figure -17 here for determining the correction factors depending on L /l and L y /l in the respective directions. In the above equations, and y are the maimum interior stresses at the centre of the slab. The edge stress at the midspan of the slab can be determined by Cα tδ t in which may be or y depending on whether C is C or C y. Unless actual field measurements are made, it is reasonable to assume a maimum temperature gradient of 0.055 to 0.077 0 C/mm during the day and about half of these values at night. Stresses due to curling may be quite large and cause concrete to crack when combined with loading stresses, but they are usually not considered in the thickness design of slabs. t 34
..3. Stresses Due to Friction Figure -17: Bradbury stress correction factor for finite slab The friction between a concrete slab and its foundation causes tensile stresses in the concrete, in the steel reinforcements, if any, and in the tie bars. It is the criteria for The spacing of plain concrete contraction joints Steel reinforcements for longer spaced concrete pavements The number of tie bars required as shown in Figure -18 Figure -18: Steel and joints in rigid pavements The volume change caused by the variation of temperature and moisture: Induces tensile stresses and causes the concrete to crack Causes the joint to open and decreases the efficiency of load transfer Figure -19 shows a concrete pavement subject to a decrease in temperature. Due to symmetry, the slab tends to move from both ends toward the centre, but the subgrade prevents it from moving; thus, frictional stresses are developed between the slab and the subgrade. The amount of friction depends on the relative movement, being zero at the centre where no movement occurs and maimum at some distance from the centre where the movement is fully mobilized, as shown in Figure -19b. The tensile stress in the concrete is greatest at the centre and can be determined by equating the frictional force per unit width of slab, to the tensile force as shown in Figure -19a. 35
Figure -19: Stresses due to friction γ c hlf ch γ Lf c c in which c is the stress in the concrete, γ c is the unit weight of the concrete, L is the length of the slab, and f is the average coefficient of friction between slab and subgrade, usually taken as 1.5. The spacing of joints in plain concrete pavements depends more on the shrinkage characteristics of the concrete rather than on the stress in the concrete. Longer joint spacing causes the joint to open wider and decrease the efficiency of load transfer. The opening of a joint can be computed approimately by ΔL CL (α t ΔT +ε) where, ΔL joint opening caused by temperature change and drying shrinkage of concrete; α t coefficient of thermal epansion of concrete, generally 9 to 10.8 10-6 / 0 C); ε drying shrinkage coefficient of concrete, approimately 0.5 to.5 10-4 ; L joint spacing or slab length; ΔT change in temperature; C adjustment factor for friction layers, 0.65 for stabilised and 0.8 for granular base. The design of longitudinal and transverse reinforcements and the tie bars across longitudinal joints is determined based on the stresses due to friction assuming that all tensile stresses are taken by the steel alone. Wire fabric or bar mats are used to increase the joint spacing and to 36
tie the cracked concrete together and maintain load transfers through aggregate interlock, but not to increase the structural capacity of the slab. h A c s f s A s γ chlf f s in which A s is the area of steel required per unit width and f s is the allowable stress in steel. The steel is usually placed at the middepth of the slab and discontinued at the joint. However, in actual practice the same amount of steel is used throughout the length of the slab. Tie bars are placed along the longitudinal joint to tie the two slabs together so that the joint will be tightly closed and the load transfer across the joint can be ensured. The amount of steel required for tie bars can be determined in the same way as the longitudinal or transverse reinforcements as: A s γ chl' f f s in which A s is the area of steel required per unit length of slab and L' is the distance from the longitudinal joint to the free edge where no tie bars eist. For two- or three-lane highways, L' is the lane width. If tie bars are used in all three longitudinal joints of a four-lane highway, L' is equal to the lane width for the two outer joints and twice the lane width for the inner joint. The length of tie bars is governed by the allowable bond stress. 37