# Rigid pavement design

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3 Warping stress The warping stress at the interior, edge and corner regions, denoted as σ ti, σ te, σ tc in kg/cm 2 respectively and given by the equation σ ti = Eɛt ( ) Cx µc y 2 1 µ 2 (29.6) ( Cx Eɛt σ te = Max, C ) yeɛt 2 2 (29.7) σ tc = Eɛt a 3(1 µ) l (29.8) where E is the modulus of elasticity of concrete in kg/cm 2 ( ), ɛ is the thermal coefficient of concrete per o C ( ) t is the temperature difference between the top and bottom of the slab, C x and C y are the coefficient based on L x /l in the desired direction and L y /l right angle to the desired direction, µ is the Poisson s ration (0.15), a is the radius of the contact area and l is the radius of the relative stiffness Frictional stresses The frictional stress σ f in kg/cm 2 is given by the equation σ f = W Lf (29.9) where W is the unit weight of concrete in kg/cm 2 (2400), f is the coefficient of sub grade friction (1.5) and L is the length of the slab in meters Combination of stresses The cumulative effect of the different stress give rise to the following thee critical cases Summer, mid-day: The critical stress is for edge region given by σ critical = σ e σ te σ f Winter, mid-day: The critical combination of stress is for the edge region given by σ critical = σ e σ te σ f Mid-nights: The critical combination of stress is for the corner region given by σ critical = σ c σ tc 29.5 Design of joints Expansion joints The purpose of the expansion joint is to allow the expansion of the pavement due to rise in temperature with respect to construction temperature. The design consideration are: Provided along the longitudinal direction, design involves finding the joint spacing for a given expansion joint thickness (say 2.5 cm specified by IRC) subjected to some maximum spacing (say 140 as per IRC) Introduction to Transportation Engineering 29.3 Tom V. Mathew and K V Krishna Rao

4 Filler h/2 No bond Full bond Figure 29:2: Expansion joint Contraction joints The purpose of the contraction joint is to allow the contraction of the slab due to fall in slab temperature below the construction temperature. The design considerations are: The movement is restricted by the sub-grade friction Design involves the length of the slab given by: L c = S c W.f (29.10) where, S c is the allowable stress in tension in cement concrete and is taken as 0.8 kg/cm 2, W is the unit weight of the concrete which can be taken as 2400 kg/cm 3 and f is the coefficient of sub-grade friction which can be taken as 1.5. Steel reinforcements can be use, however with a maximum spacing of 4.5 m as per IRC. Filler h/2 Full bond Full bond Figure 29:3: Contraction joint Dowel bars The purpose of the dowel bar is to effectively transfer the load between two concrete slabs and to keep the two slabs in same height. The dowel bars are provided in the direction of the traffic (longitudinal). The design considerations are: Introduction to Transportation Engineering 29.4 Tom V. Mathew and K V Krishna Rao

6 Solution: Given, P = 5000 kg, l = 80 cm, h = 25 cm, δ = 2 cm, F s = 1000 kg/cm 2, F f = 1400 kg/cm 2 and F b = 100 kg/cm 2 ; and assume d = 2.5 cm diameter. Step-1: length of the dowel bar L d L d = = (L d 1.5 2) 100 (L d 8.8 2) 14 (L d 3) (L d 17.6) Solve for L d by trial and error: put L d = L d = put L d = L d = put L d = L d = Minimum length of the dowel bar is L d δ = = 42.5 cm, So, provide 45 cm long and 2.5 cm φ. Therefore L d = 45 2 = 43 cm. Step 2: Find the load transfer capacity of single dowel bar P s = = 4906 kg P f = = 722 kg P b = ( ) = 804 kg Therefore, the required load transfer capacity { } max,, max {0.41, 2.77, 2.487} = 2.77 Step-3 : Find the required spacing: Effective distance of load transfer = 1.8 l = = 144 cm. Assuming 35 cm spacing, Actual capacity is = 2.57 < 2.77 (the required capacity) Therefore assume 30 cm spacing and now the actual capacity is = 2.92 > 2.77 (the required capacity) Therefore provide 2.5 cm φ mild steel dowel bars of length cm center to center Tie bars In contrast to dowel bars, tie bars are not load transfer devices, but serve as a means to tie two slabs. Hence tie bars must be deformed or hooked and must be firmly anchored into the concrete to function properly. They are smaller than dowel bars and placed at large intervals. They are provided across longitudinal joints. Introduction to Transportation Engineering 29.6 Tom V. Mathew and K V Krishna Rao

7 Step 1 Diameter and spacing: The diameter and the spacing is first found out by equating the total sub-grade friction tot he total tensile stress for a unit length (one meter). Hence the area of steel per one meter in cm 2 is given by: A s S s = b h W f A s = bhw f 100S s (29.16) where, b is the width of the pavement panel in m, h is the depth of the pavement in cm, W is the unit weight of the concrete (assume 2400 kg/cm 2 ), f is the coefficient of friction (assume 1.5), and S s is the allowable working tensile stress in steel (assume 1750 kg/cm 2 ). Assume 0.8 to 1.5 cm φ bars for the design. Step 2 Length of the tie bar: Length of the tie bar is twice the length needed to develop bond stress equal to the working tensile stress and is given by: L t = d S s 2 S b (29.17) where, d is the diameter of the bar, S s is the allowable tensile stress in kg/cm 2, and S b is the allowable bond stress and can be assumed for plain and deformed bars respectively as 17.5 and 24.6 kg/cm 2. Example A cement concrete pavement of thickness 18 cm, has two lanes of 7.2 m with a joint. Design the tie bars. (Solution:) Given h=18 cm, b=7.2/2=3.6m, S s = 1700 kg/cm 2 W = 2400 kg/cm 2 f = 1.5 S b = 24.6 kg/cm 2. Step 1: diameter and spacing: Get A s from A s = = 1.33 cm 2 /m Assume φ = 1 cm, A = cm 2. Therefore spacing is = 59 cm, say 55 cm Step 2: Length of the bar: Get L t from L t = [Ans] Use 1 cm φ tie bars of length of cm c/c = 36.0 cm 29.6 Summary Design of rigid pavements is based on Westergaard s analysis, where modulus of subgrade reaction, radius of relative stiffness, radius of wheel load distribution are used. For critical design, a combination of load stress, frictional stress and warping stress is considered. Different types of joints are required like expansion and contraction joints. Their design is also dealt with. Introduction to Transportation Engineering 29.7 Tom V. Mathew and K V Krishna Rao

8 29.7 Problems 1. 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/cm 2 respectively. 2. Design the length and spacing of tie bars given that the pavement thickness is 20cm and width of the road is 7m with one longitudinal joint. The unit weight of concrete is 2400 kg/m 3, the coefficient of friction is 1.5, allowable working tensile stress in steel is 1750 kg/cm 2, and bond stress of deformed bars is 24.6 kg/cm Solutions 1. Given, P = 4000 kg, l = 90 cm, h = 20 cm, δ = 3 cm, F s = 1000 kg/cm 2, F f = 1500 kg/cm 2 and F b = 100 kg/cm 2 ; and assume d = 2.5 cm diameter. Step-1: length of the dowel bar L d, L d = = (L d 1.5 3) 100 (L d 8.8 3) 15 (L d 4.5) (L d 26.4) Solving for L d by trial and error, it is =39.5cm Minimum length of the dowel bar is L d δ = = 42.5 cm, So, provide 45 cm long and 2.5 cm φ. Therefore L d = 45 3 = 42 cm. Step 2: Find the load transfer capacity of single dowel bar P s = = kg P f = = kg P b = ( ) = kg Therefore, the required load transfer capacity (refer equation) { } max , , max {0.326, 2.335, 2.10} = Step-3 : Find the required spacing: Effective distance of load transfer = 1.8 l = = cm. Assuming 35 cm spacing, Actual capacity is = 2.83 Introduction to Transportation Engineering 29.8 Tom V. Mathew and K V Krishna Rao

9 Assuming 40cm spacing, capacity is, = 2.52 So we should consider as it is greater and more near to other value. Therefore provide 2.5 cm φ mild steel dowel bars of length cm center to center Given h=20 cm, b=7/2=3.5m, S s = 1750 kg/cm 2 W = 2400 kg/cm 2 f = 1.5 S b = 24.6 kg/cm 2. Step 1: diameter and spacing: A s = = 1.44 cm 2 /m Assume φ = 1 cm, A = cm 2. Therefore spacing is = cm, say 55 cm Step 2: Length of the bar: L t = = 36.0 cm [Ans] Use 1 cm φ tie bars of length of cm c/c Introduction to Transportation Engineering 29.9 Tom V. Mathew and K V Krishna Rao

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