Sums on Rigid Pavement Design

Problems regarding Rigid Pavement with solutions.

1. Example: Find the spacing between contraction joints for 3.8m slab width having thickness of 20 cm and f=1.5, for the following cases: (1) Plain cement concrete (11) for reinforced cement concrete with 1cm dia bars at 32cm spacing. ( i ) For plain cement concrete slab : Assume unit weight of concrete (W)= 2400kg/m 3 Allowable tensile stress in concrete = 0.8 Kg/cm 2 . Spacing between contraction joints= 2xS c x10 4 /W. f = 2x0.8x10 4 /2400x1.5 = 4.44m or 4.5m. (ii) For reinforced cement concrete slab: Spacing between contraction joints: L c = 200x S s xA s / bhwf Here A s = Area of steel in 3.8 m width of the slab in 32cm spacing = 3.14x 1x1 x 3.8 x 100/4 x32 = 9.32 cm 2 So, spacing of contraction joint = 200 x1750 x 9.32 / 3.8 x 20 x 2400 x 1.5 = 11.9 m (Here S s = allowable tensile strength of the steel is assumed as 1750 kg/cm 2 ), and over Example: Find radius of relative stiffness for a 20cm slab with E= 3x10 5 kg/cm 2 and Poisson’s ratio =0.15, over subgrade of modulus 5kg/cm 3 . E= 3x10 5 kg/cm 2 , 5kg/cm 3. Radius of relative stiffness = l = / 12K (1- } = = 79.98 cm.  

2. Example : Compute radius of relative stiffness for a 15cm slab with E= 2.1x10 5 kg/cm 2 and Poisson’s ratio =0.15, over subgrade of modulus (a) k= 3.5kg/cm 2 (b) k= 7kg/cm 2 . E= 2.1x10 5 kg/cm 2 , 3.5kg/cm 2. Radius of relative stiffness = l = / 12K (1- } = = 64.46 cm For k= 7.0 kg/ cm 2. Radius of relative stiffness = l = / 12K (1- } = = 54.2 cm . Example: Compute the equivalent radius of resisting section of 20cm thick slab, given that the radius of contact area wheel load is 16cm. Slab thickness h= 20cm. Radius of wheel load distribution, a =16cm. So a/h= 16/20 = 0.80 < 1.724 Therefore equivalent radius of the resisting section when ( a< 1.724 h) b = - 0.675h = - (0.675x20) = 14.95cm .  

3. Example: Using data given below determine: (a) Edge and Corner load stresses by Westergaard equation. Wheel load P= 5200kg, pavement thickness h= 18cm, Poisson ratio of concrete , radius of contact area ‘a’ = 15cm, Modulus of elasticity of concrete= E= 3.0x10 5 kg/cm 2 , modulus of subgrade reaction k= 6.0 kg/cm 2 Solution: Radius of relative stiffness; l = / 12K (1- } = = 70.6 cm. Equivalent radius of resisting section is given by a/h = 15/18 = 0.833 <1.724 Therefore equivalent radius of the resisting section= b = - 0.675h = - (0.675x18) = 14cm. Edge load stress, σ e = {4 log 10 (l/b) +0.359 } = {4 log 10 (70.6/14) +0.359} = 29.1 kg/ cm 2 Corner load Stress σ c = [ 1- ( ) 0.6 = [ 1- ( ) 0.6 ] = 24.75 kg/ cm 2  

4. Example: The spacing between the contraction joint of a CC pavement is 4.25m. Determine the tensile stress developed in the CC pavement due to contraction if the coefficient of friction between the bottom of the pavement and the supporting layer is 1.1 and unit weight of concrete 2400 kg /m 3 . Solution : Equating the total force( in Kg) developed in the cross section of the concrete pavement due to movement half of the length of slab ( L c /2) and the frictional resistance due to the restraint at the interface in half of the slab length, S f x h x b x 100 (kg) = b x ( L c /2) x h/100 x w x f (kg) = W L c f/ 2x10 4 Where S f = stress developed due to interface friction in cement concrete pavement /unit area. W= unit weight of concrete =2400 kg/m3 f= coefficient of friction at interface ( maxm . value =1.5) Lc = Spacing between contraction joint = slab length (m), b = slab width (m). Given L = 4.25m. f =1.1, W = 2400 kg/m3 Tensile stress developed in the pavement due to contraction S f = W L c f/ 2x10 4 = 2400 x 4.25 x 1.1/ 2 x 10 4 = 0.561 kg/cm 2 .  

5. Example : Design size and spacing of dowel bars at an expansion joint of concrete pavement of thickness 20 cm. Given the radius of relative stiffness of 90 cm. design wheel load 4000 kg. Load capacity of the dowel system is 40 percent of design wheel load. Joint width is 3.0 cm and the permissible stress in shear, bending and bearing stress in dowel bars are 1000,1500 and 100 kg/cm2 respectively. Given, P = 4000 kg, l = 90 cm, h = 20 cm, δ = 3 cm, Fs = 1000 kg/cm2, F f = 1500 kg/cm2 and F b = 100 kg/cm2; and assume d = 2.5 cm diameter. Step-1: length of the dowel bar Ld , Ld = 5×2.5 {100 (Ld +8.8×3)} = 12.5× Solving for Ld by trial and error, it is =39.5cm Minimum length of the dowel bar is Ld + δ = 39.5 + 3.0 = 42.5 cm, So, provide 45cm long and2.5cm φ. Therefore Ld =45−3=42cm. Step 2: Find the load transfer capacity of single dowel bar Ps = 0.785 × 2.5 2 × 1000 = 4906.25 kg, P f = 2×2.5 3 ×1500 / 42.0+8.8×3 = 685.307 kg, P b = 100×2.5×42.0 2 / 12.5 (42.0+1.5×3) = 758.71 kg Therefore, the required load transfer capacity (refer equation) max {0.4×4000/ 4906.25, 0.4×4000/ 685.307, 0.4×4000/ 758.71} = max {0.326, 2.335, 2.10} = 2.335 Step-3 : Find the required spacing: Effective distance of load transfer = 1.8 × l = 1.8 × 90 = 162 cm. Assuming 35 cm spacing, Actual capacity is 1+ + + + = 2.83 Assuming 40 cm spacing, Actual capacity is 1+ + + + = 2.52 So we should consider 2.52;2.335 as it is greater and more near to other value. Therefore provide 2.5 cm φ mild steel dowel bars of length 45 cm @ 40 cm center to center .  

6. Example: Using data given below, calculate wheel load stresses at:- ( a) Interior (b) Edge ( c) Corner regions of a cement concrete pavement using westargaard stress equation and also determine the probable location of crack is likely to develop due to corner loading. Wheel load =5200Kg. E= 3.0x10 5 kg/cm 2 , kg/cm 3 Radius of contact area= 15cm. Radius of relative stiffness = l = / 12K (1- } = = 70.6 cm. Equivalent radius of the resisting section= b = - 0.675h = - (0.675x18) = 14cm. ( a/h = 15/18 = 0.833 < 1.74) Interior load stress, σ i = {4 log 10 (l/b) +1.069 } = {4 log 10 (70.6/14) +1.069 }= 19.68kg/cm 2 . Edge load stress, σ e = {4 log 10 (l/b) +0.359 } = {4 log 10 (70.6/14) +0.359} = 29.1 kg/ cm 2 Corner load Stress σ c = [ 1- ( ) 0.6 = [ 1- ( )] 0.6 = 24.75 kg/ cm 2 Location where corner load cracks develop: location where the crack is likely to develop due to corner loading, the distance from the corner of the slab, x = 2.58 ( a.l ) 1/2 = 2.58x( 15x 70.6) ½ = 83.96 cm = 84 cm . Example: Determine the wrapping streses at interior, edge and corner of 30cm thick cement concrete pavement with transverse joint at 5m interval and longitudinal joints at 3.6 m interval. The modulus of subgrade reaction = kg/cm 3 and radius of loaded area is 15 cm. Assume maximum temperature differential during day to be 0.6 C per cm slab thickness ( for wrapping stresses at interior and edge) and maximum temperature differential at 0.4 C per cm slab thickness during night ( for wrapping stress at corner). Other data are E= 3.0x10 5 kg/cm 2 , -6 per C.  

7. Given Slab thickness h= 30cm. Modulus of Elasticity E= 3.0x10 5 kg/cm 2 , Poission’s ratio = -6 per C. a= 15 cm. L x = 500 cm, L y =360 cm. Temperature differential during day, t 1 = 0.6x 30 = 18 C Temperature differential during night, t 2 = 0.4x 30 = 12 C Radius of relative stiffness = l = / 12K (1- } = = 100 cm. L x /l= 500/100 = 5 , L y / l =360/100 =3.6 , Refer to Bradbury’s Chart for wrapping stress coefficients corresponding to L x /l = 5 , C x = 0.75, L y /l = 3.6 , C y = 0.4, Wrapping stress at interior region of the slab, during day S ti = ] = [ 0.75 + 0.15x 0.4/ (1- 0.15 2 )= 22.4 kg/cm 2 Wrapping stress at edge region of the slab, during day [ as it is higher than ] = = 20.25kg/cm 2 Wrapping stress at corner region of the slab, during night, ( a/l) ½ = ( 15/ 100) 1/2 = 5.47 kg/cm 2  

8. Example: Design a rigid pavement making use of wastergaard wheel load and wrapping stress equations at the edge region of the slab. The design data are given below: Design wheel load P= 7500Kg, Contact pressure p= 7.5Kg/cm 2 , spacing between longitudinal joints= 3.75 m , spacing between contraction joint= 4.2m. E= 3.0x10 5 kg/cm 2 Thermal coefficient of cc C -5 . Flexural strength of cc = 45 kg/cm 2 modulus of basecourse = kg/cm 3 , Maximum temperature differential at the location for pavement thickness value of 22,24,26 and 30cm are respectively14.8,15.6,16.2,16.8 C, Desired factor of safety with respect to load stress and wrapping stress at edge region is 1.1 to 1.2. Solution: P = 7500 kg, p = 7.5Kg/cm 2 , therefore radius ‘a’ = = = 17.84 cm Trial 1 , Assume pavement thickness h=25cm, Given K= kg/cm 3 , . E= 3.0x10 5 kg/cm 2 Radius of relative stiffness = l = / 12K (1- } = = 60.41 cm, a/h = 17.84/25 = 0.7136< 1.724, Radius of the resisting section= b = - 0.675h = - (0.675x25) = 16.80 cm. Edge load stress, σ e = {4 log 10 (l/b) +0.359 }] = 17.74 Kg/cm 2 For h= 25cm, temperature differential ( by interpolation) = (15.6+16.2) /2 = 15.9 C Lx= 4.2m = 420cm Ly=375cm, So, wrapping stress for higher ratio = Lx/l = 420/60.41 = 6.96.  

9. From Bradbury wrapping stress coefficient chart C x = 0.99, -5 0 C Wrapping stress at Edge region of the slab= = = 23.61 kg/cm 2 Total flexural stress S e + S te = (17.74+ 23.61) = 41.35 kg/cm 2 < 45 kg/cm 2 Factor of Safety = 45/ 41.35 = 1.1. As the factor of safety obtained is within the desired factor of safety between 1.1 to 1.2 so the adopted thickness for the rigid pavement 25 cm is Sufficient. Hence okay. Example: The design thickness of a CC pavement is 26cm considering a design axle load ( 98 th percentile load) of 13000kg on single axle and M-40 concrete with characteristics compressive strength of 400 kg/cm 2 .The relative stiffness is found to be 62.2 cm. If the elastic modulus of the dowel bar steel is 2x10 5 Kg/cm 2. Modulus of dowel concrete interaction is 41500 kg/cm 2 and joint width is 1.8cm, design the dowel bars for 40% load transfer considering edge loading. Solution: Radius of relative stiffness of pavement, l = 62.2 cm. Characteristics of compressive strength ( f ck ) = 400 kg/cm 2 , Elastic modulus of the dowel bar steel is 2x10 5 Kg/cm 2 , Design load ( 98 th percentile axle load of single axle) =13000kg, Therefore design wheel load for dowel bar design P = 6500kg, Total load to be sustained by the dowel bar group = 0.4P = 0.4x 6500 = 2600kg. Let the maximum load sustained by the first dowel bar near edge =P 1 kg. Assume diameter of dowel bar = 3 cm and the spacing S= 25cm in the 1 st Trail. Substituting the relevant values , Moment of inertia of dowel bar = I = ( b 4 / 64)= 3.976 cm 4  

10. Allowable bearing stress in concrete F b = f ck ( 10.16-b) / 9.525 = 400 (10.16-3) / 9.525 = 300.68 Kg/cm 2 . Total load transferred by the dowel group for the assumed dowel diameter and spacing in terms of maximum load carried:- Total load carried = P 1 {1+(l-25)/l + (l-50)/l } = 1.794P 1 1.794P 1 = 2600 Kg, So, P 1 = 1450Kg. Relative stiffness of dowel bars embedded in concrete = (M b/4EI ) 0.25 = 0.25 Check for maximum bearing stress between concrete and he dowel bar subs staining load, P 1 S bm = M. P 1 ( 2+ Z) / 4 E s I = 296.65 Kg/cm 2. ( Joint Width Z=1.8cm) As the maximum bearing stress between concrete and dowel bar (296.65 Kg/cm 2. ) < allowable bearing strength in concrete (300.68 Kg/cm 2 ) in the dowel bar design is safe and may be accepted . Rounded dowel bar of diameter 3 cm and spacing 25cm may be provided at expansion joints.  

11. Example: Design the dowel bars and their spacing from the following data: Wheel load=4000Kg, Modulus of subgrade reaction =6Kg/cm 2 , Modulus of Elasticity of concrete = 3x10 5 kg/cm 2 , Poisson’s ratio = 0.15, Slab thickness =20cm, Joint thickness= 18mm. Solution: Radius of relative stiffness = l = / 12K (1- } = = 76.42 cm, Allowable bearing stress in concrete F b = f ck ( 10.16-b) / 9.525 = 400 (10.16-3.2) / 9.525 = 292 Kg/cm 2 ( The Grade of concrete assumed M40 Grade and Dia of Dowel bar ( assumed) = 3.2cm) Assumed spacing between dowel bars = 32 cm and First dowel bar is placed at a distance=15cm from pavement edge. Assumed length of the dowel bar=50 cm. Dowel bars upto a distance of 1.0x radius of relative stiffness, from the point of load application are effective in load transfer. No of dowel bars participating in the load transfer when wheel load is just over the dowel bar close to the edge of the slab = 1+ (76.42/32) = 3 dowels Assuming that the load transfer by the first dowel is P 1 and assuming that the load on the dowel bar at a distance of l from the first dowel to be zero, the total load transferred by the dowel bar system = [ 1+ (76.42-32)/76.42 + ( 76.42-64)/76.42] . P 1 = 1.75 P 1 Load carried by the outer dowel bar, P 1 = 4000/1.75 = 2286 Kg.  

12. Check for bearing stress: Moment of Inertia of Dowel = I = ( b 4 / 64)= 5.147 cm 4 Relative stiffness of dowel bars embedded in concrete = ( Kb /4EI ) 0.25 =0.042 Check for maximum bearing stress between concrete and he dowel bar subs staining load, P 1 S bm = K P 1 ( 2+ Z) / 4 E s I = 62.2 Kg/cm 2. ( Joint Width Z=1.8cm) < 292 Kg/cm 2 Hence dowel bar spacing and diameter assumed are safe.  

13. Example