Highway Pavement

A pavement must carry traffic loads for its design life without excessive cracking, rutting or roughness. The factors affecting the design (IRC:37-2018 for flexible and IRC:58-2015 for rigid pavements; Nepal: DoR Pavement Design Guidelines / Nepal Road Standard 2070, based on CBR charts) are:

1. Design wheel/axle load

• Magnitude of load and axle configuration (single, tandem, tridem). The standard axle is 80 kN (8.16 t) single axle with dual wheels.
• Tyre pressure and contact area govern the surface stress; the contact pressure of about 0.56 MPa is used in IRC:37.
• Repetition of loads: damage increases with the number of passes.

2. Traffic

• Present commercial traffic (vehicles of more than 3 t laden weight) in each direction, expressed as CVPD.
• Growth rate $r$ and design life $n$ (15 years for NH/SH, 20 years or more for expressways).
• Vehicle damage factor (VDF), the equivalent standard axles per commercial vehicle, and lane distribution factor (LDF).
• Cumulative standard axles:

$$N_s = \frac{365\left[(1+r)^n - 1\right]}{r} \times A \times D \times F$$

where $A$ = initial traffic at opening, $D$ = lane distribution factor and $F$ = VDF. The pavement is designed for $N_s$ in million standard axles (msa).

3. Subgrade strength

• Soaked CBR of the subgrade (at the expected field density and moisture), or the resilient modulus $M_R$ (IRC:37-2018: $M_R = 10.4\,CBR^{0.64}$ for CBR $\le$ 5 % and $17.6\,CBR^{0.64}$ for CBR > 5 %); for rigid pavements the modulus of subgrade reaction $k$.
• Soil type, plasticity, compaction, depth of water table and swelling potential.

4. Pavement materials

• Strength and stiffness of granular sub-base, base (WBM/WMM), bituminous layers (modulus, fatigue, rutting) or cement concrete (flexural strength, modulus of elasticity).
• Availability and cost of local materials.

5. Climate and environment

• Temperature (bitumen stiffness, thermal stresses in concrete), rainfall, and moisture variation of the subgrade; in the hills, frost action and landslides.
• Drainage and ground-water level: poor drainage greatly reduces the subgrade strength.

6. Other factors

• Design period, required reliability (80-90 percent), and failure criteria (fatigue cracking, rutting).
• Construction quality, method and maintenance level.
• Economy: initial cost and life-cycle cost; stage construction.
• In rigid pavements: joints and load transfer, slab thickness, and the temperature differential.

A cement concrete (rigid) pavement slab resting on the subgrade is subjected to stresses from wheel loads and from changes in temperature and moisture. The types of stress (Westergaard's analysis, IRC:58-2015 for design) are:

1. Wheel load stresses

Caused by the load of the vehicle, treated by Westergaard's theory (slab on a Winkler foundation). Three critical positions:

$$\begin{aligned} \text{Interior:}\ \sigma_i &= \frac{0.316P}{h^2}\left[4\log_{10}\frac{l}{b} + 1.069\right]\\ \text{Edge:}\ \sigma_e &= \frac{0.572P}{h^2}\left[4\log_{10}\frac{l}{b} + 0.359\right]\\ \text{Corner:}\ \sigma_c &= \frac{3P}{h^2}\left[1 - \left(\frac{a\sqrt{2}}{l}\right)^{0.6}\right] \end{aligned}$$

where $P$ = wheel load, $h$ = slab thickness, $a$ = radius of the loaded area, $b = a$ if $a \ge 1.724h$ and $b = \sqrt{1.6a^2 + h^2} - 0.675h$ otherwise, and $l$ = radius of relative stiffness:

$$l = \left[\frac{E\,h^3}{12(1-\mu^2)\,k}\right]^{1/4}$$

($E$ = modulus of elasticity of concrete, $\mu$ = Poisson's ratio, $k$ = modulus of subgrade reaction.) The edge and corner stresses are the highest, so these control the design (corner cracks develop at the joint corners).

2. Temperature (warping) stresses

The top and bottom of the slab are at different temperatures. In the day the top is hotter and tends to expand, so the slab curls down at the edges; at night it curls up. The self-weight and subgrade restrain this, giving tensile stresses:

$$\sigma_{te} = \frac{C\,E\,\alpha\,t}{2}\ \ \text{(edge)},\qquad \sigma_{ti} = \frac{E\alpha t}{2}\left[\frac{C_x + \mu C_y}{1-\mu^2}\right]\ \ \text{(interior)}$$

where $\alpha$ = coefficient of thermal expansion ($10 \times 10^{-6}$ per $^\circ$C), $t$ = temperature difference between top and bottom, and $C$, $C_x$, $C_y$ = Bradbury's coefficients (functions of slab length/$l$). Hot-afternoon edge warping stress is tension at the bottom, night curling gives tension at the top.

3. Frictional stresses

When the slab contracts due to a fall in temperature or shrinkage, friction with the subgrade resists the movement and causes tension at the middle of the slab:

$$\sigma_f = \frac{W\,L\,f}{2 \times 10^4}\ \text{kg/cm}^2$$

where $W$ = unit weight of concrete (2400 kg/m$^3$), $L$ = slab length (m) and $f$ = coefficient of friction (about 1.5). It determines the joint spacing and the steel in the slab.

4. Other stresses

Moisture-warping and shrinkage stresses, stresses due to subgrade settlement and swelling, and stresses at joints.

Combination for design (IRC:58): the critical stress is the sum of the load stress and warping stress at the edge (day: load + warping, night: load + reverse warping) and the corner stress, and it is compared with the flexural strength of concrete with a factor of safety (fatigue is checked).

Factors affecting pavement design

The main factors that control the design of a pavement are:

1. Design wheel load - magnitude of the load, tyre (contact) pressure, contact area, axle configuration (single, tandem, tridem) and repetitions. Legal and standard axle load (80 kN) are used to convert mixed traffic to ESAL/msa.
2. Traffic - initial commercial vehicles per day (CVPD, vehicles above 3 t laden weight), growth rate, design life, lane and directional distribution, and the vehicle damage factor (VDF).
3. Climate - rainfall and drainage (moisture changes the subgrade strength), temperature (stiffness of bituminous layers; warping and expansion in concrete slabs), frost and freeze-thaw in hills.
4. Subgrade soil - strength (CBR, modulus of subgrade reaction k or resilient modulus M_R), volumetric change, drainage and the water-table depth. Strength is taken at soaked condition.
5. Pavement component materials - strength, modulus, Poisson's ratio, fatigue and rutting resistance of bituminous mix, granular layers, and the flexural strength and elastic modulus of concrete.
6. Environment and local factors - ground water, terrain, availability of materials, construction quality and maintenance level.
7. Failure criteria and reliability - allowable serviceability loss (PSI), fatigue cracking, rutting limit (20 mm) and the reliability chosen.

Steps of IRC design of rigid pavement

The design follows IRC:58 (Guidelines for the Design of Plain Jointed Rigid Pavements for Highways) based on the Westergaard/Bradbury/Pickett-Ray approach.

1. Collect data - design wheel load (5100 kg for highways, 4100 kg for small roads), tyre pressure, radius of contact area, flexural strength (modulus of rupture) and E, Poisson's ratio and thermal coefficient of concrete; modulus of subgrade reaction k (on top of sub-base); traffic (CVPD), growth rate and design life.
2. Design traffic - find the commercial vehicles at the end of construction, classify the traffic (A to G) and note the thickness adjustment.
3. Assume a trial slab thickness h and the joint spacing (transverse L, longitudinal width B).
4. Radius of relative stiffness $l=\left[\dfrac{Eh^3}{12(1-\mu^2)k}\right]^{1/4}$.
5. Load stresses - edge load stress (Pickett and Ray chart) and corner load stress (Kelly's formula $\sigma_c=\frac{3P}{h^2}\left[1-\left(\frac{a\sqrt2}{l}\right)^{0.6}\right]$).
6. Temperature (warping) stresses - from temperature differential between top and bottom: Bradbury's formula $\sigma_{te}=\dfrac{C\,E\,\alpha\,t}{2}$ with coefficient C from L/l and B/l; corner warping stress $\sigma_{tc}=\dfrac{E\alpha t}{3(1-\mu)}\sqrt{a/l}$.
7. Frictional stress $\sigma_f=\dfrac{WLf}{2\times10^4}$ (W in kg/m3, L in m).
8. Check critical combinations - (a) summer midday: $\sigma_e+\sigma_{te}-\sigma_f$, (b) winter midday: $\sigma_e+\sigma_{te}+\sigma_f$ (different $\Delta T$), (c) night: warping + friction, and corner: $\sigma_c+\sigma_{tc}$. Each must not exceed the permissible flexural stress (modulus of rupture divided by a factor of safety, or by the fatigue stress-ratio in IRC:58-2002/2015).
9. Revise thickness by trial until the stresses are safe; apply the traffic correction (A to G) to the thickness.
10. Joints and steel - design contraction, expansion and longitudinal joints, dowel bars at transverse joints and tie bars at longitudinal joints (IRC:58 / IRC:15).

Functions of pavement layers

A flexible pavement is a layered system; load spreads down so that stress on each layer reduces with depth.

The thickness of each layer is fixed so that the stress reaching the layer below is within its capacity.

Joints in rigid pavement

Joints are provided in a cement concrete slab to control cracking from temperature change, shrinkage and warping, and to allow construction in stages.

1. Contraction joint (transverse) - a weakened plane (sawn groove 1/4 to 1/3 of slab depth) that lets the slab shrink on cooling and creates a controlled crack. Dowels are provided. Spacing is usually 4.5 m (IRC:58), up to 5 m for unreinforced slabs.
2. Expansion joint - a full-depth gap (20-25 mm) filled with compressible filler (bitumen-impregnated fibre board or cork) and sealant, provided at bridges, culverts, intersections and where slab meets a structure so that the slab can expand freely. Smooth dowel bars with a cap on one end carry the load. IRC:58 permits omitting them in long slab runs.
3. Construction joint - placed where concreting stops at the end of the day or a break. Designed as a butt joint with dowels, preferably at a contraction-joint location.
4. Longitudinal joint - between adjacent lanes (about every 3.5 m) to control longitudinal cracking from warping. Tie bars (12 mm deformed bars) hold the slabs together and prevent lane separation.
5. Warping joint (hinged joint) - in older practice, placed in the middle of a slab to relieve warping stress; a longitudinal type with tie bars is used.

Design method

Four-lane divided highway, so IRC:37-2012 (Guidelines for the Design of Flexible Pavements) is used with its CBR-based catalogue; Nepal DoR adopts the same approach (cumulative msa and subgrade CBR). Traffic of 8000 is taken as commercial vehicles per day (CVPD) in both directions.

Step 1 - Design subgrade CBR

For a major highway the 90th percentile CBR is used (value at which 90 percent of results are equal or greater):

The 90th percentile lies between 5 percent (100 percent of values) and 6 percent (71.4 percent of values), so the design CBR is taken as the lower value, CBR = 5 %.

Step 2 - Design traffic

• Traffic count was one year ago and construction takes 2 years, so $x = 1+2 = 3$ years.
• Traffic at completion (both ways): $A_{total}=8000(1.07)^3 = 9800$ CVPD.
• Heavier direction 55 percent: $A = 0.55\times9800=5390$ CVPD.
• Lane distribution factor for a dual two-lane carriageway (IRC:37-2012): $D = 0.75$ of the directional traffic.
• VDF $F = 4.0$, design life $n = 15$ years, $r=7\%$.

$$\begin{aligned} N_s &= \frac{365[(1+r)^n-1]}{r}\,A\,D\,F\\ &= 365\times\frac{(1.07)^{15}-1}{0.07}\times 5390\times0.75\times4.0 \\ &= 365\times25.129\times5390\times0.75\times4.0 = 148.3\times10^{6}\ \text{standard axles} \end{aligned}$$

Design traffic is about 150 msa (the upper limit of the IRC:37-2012 catalogue).

Step 3 - Pavement composition (IRC:37-2012 plate for CBR 5 %, 150 msa)

  +-----------------------------+
  | BC               50 mm      |
  | DBM             170 mm      |
  +-----------------------------+
  | WMM base        250 mm      |
  +-----------------------------+
  | GSB sub-base    300 mm      |
  +-----------------------------+
  | Subgrade (CBR 5 %)          |
  +-----------------------------+

Answer: design traffic $\approx 148$ msa (150 msa); pavement = 50 mm BC + 170 mm DBM + 250 mm WMM + 300 mm GSB = 770 mm total. The sub-base should extend 0.5 m beyond the wearing surface for drainage.

Given data and assumptions

• Current traffic (both directions) 1850 ESA/day, growth 6 %, design period 15 years, construction period 16 months (1.33 years).
• Four-lane divided carriageway: design lane carries 45 percent of the total two-way traffic (AI table for 4 lanes).
• Moduli: asphalt concrete 2500 MPa, bituminous treated base 1200 MPa, granular sub-base 125 MPa.
• Method: Asphalt Institute (AI) thickness design as taught for IOE (and used in the Nepal DoR Pavement Design Guidelines); design reliability 90 percent, terminal serviceability 2.5.

Step 1 - Subgrade strength

CBR at 2.5 mm = $\frac{68.5}{1370}\times100=5.00\ \%$; CBR at 5.0 mm = $\frac{80}{2055}\times100=3.89\ \%$ (standard loads 1370 kg and 2055 kg). The 2.5 mm value is higher, so CBR = 5.0 % is adopted.

Resilient modulus of subgrade: $M_R = 10\times CBR = 10\times 5.0 = 50$ MPa.

Step 2 - Design traffic (ESAL)

Traffic at opening: $1850\times0.45\times(1.06)^{1.33} = 899.8$ ESA/day (design lane).

$$\begin{aligned} ESAL &= 365\times 899.8 \times \frac{(1.06)^{15}-1}{0.06} \\ &= 365\times899.8\times23.276 = 7.644\times10^6 \end{aligned}$$

Design ESAL = 7.644 $\times 10^6$

Step 3 - Thickness required over each layer

The AI charts give the thickness of asphalt concrete (full depth) needed to protect a layer of a given modulus for the design ESAL. The same requirement is expressed here as a structural number $SN$ (AASHTO-93 relation with R = 90 percent, $S_0$ = 0.45, $\Delta PSI$ = 1.7), which closely approximates the AI chart readings (if the chart is supplied in the exam, read the thickness from it and use the same layer-coefficient steps below).

Step 4 - Layer coefficients

Using $a_1=0.323\log_{10}E-1.42$, $a_2=0.249\log_{10}E-0.977$, $a_3=0.227\log_{10}E-0.839$ with $E$ in psi:

• Asphalt concrete (2500 MPa): $a_1 = 0.376$
• bituminous treated base (1200 MPa): $a_2 = 0.328$
• granular sub-base (125 MPa): $a_3 = 0.128$

Step 5 - Layer thicknesses

• Asphalt concrete: $D_1 \ge SN_{base}/a_1 = 1.42/0.376=96$ mm, so provide 100 mm (minimum 50 mm). $SN_1 = 1.48$.
• Bituminous treated base: $D_2 \ge (SN_{sub}-SN_1)/a_2 = (3.47-1.48)/0.328 = 154$ mm, so provide 160 mm. $SN_2 = 2.07$.
• Granular sub-base: $D_3 \ge (SN_{sg}-SN_1-SN_2)/a_3 = (4.82-1.48-2.07)/0.128 = 255$ mm, so provide 260 mm. $SN_3 = 1.31$.

Check: provided $SN = 1.48+2.07+1.31 = 4.85 \ge 4.82$ (required) - safe.

Answer: asphalt concrete 100 mm, bituminous treated base 160 mm, granular sub-base 260 mm; total pavement thickness = 520 mm.

   +-------------------------------------+
   | Asphalt concrete             100 mm |
   +-------------------------------------+
   | Bituminous treated base      160 mm |
   +-------------------------------------+
   | Granular sub-base            260 mm |
   +-------------------------------------+
   | Subgrade, CBR 5.0 %                 |
   +-------------------------------------+

Given data

Dowel diameter $d=2.5$ cm, slab $h=25$ cm, joint width $\delta=2.0$ cm, wheel load $P=5000$ kg, load transferred by dowels $=0.4P=2000$ kg, $F_s=1000$, $F_f=1400$, $F_b=100$ kg/cm2, $E=2.45\times10^5$ kg/cm2, $k=8$ kg/cm3, $\mu=0.15$. (Method of Bradbury as in Khanna and Justo; IRC:58 detailing.)

Step 1 - Radius of relative stiffness

$$l=\left[\frac{Eh^3}{12(1-\mu^2)k}\right]^{1/4}=\left[\frac{2.45\times10^5\times25^3}{12(1-0.15^2)\times8}\right]^{1/4}=79.9\ \text{cm}$$

Step 2 - Capacity of one dowel

• Shear: $P_s=0.785\,d^2F_s=0.785\times2.5^2\times1000=4906$ kg
• Bending: $P_f=\dfrac{2d^3F_f}{L_d+8.8\delta}=\dfrac{2\times2.5^3\times1400}{L_d+17.6}=\dfrac{43750}{L_d+17.6}$
• Bearing: $P_b=\dfrac{F_bL_d^2d}{12.5(L_d+1.5\delta)}=\dfrac{100\times2.5\,L_d^2}{12.5(L_d+3)}=\dfrac{20L_d^2}{L_d+3}$

Step 3 - Length of dowel

Equating $P_f=P_b$ and solving by trial gives $L_d=40.5$ cm. Adopt dowel length $L_d=45$ cm (rounded up; this is within the usual 45-50 cm).

With $L_d=45$: $P_f=699$ kg, $P_b=844$ kg, $P_s=4906$ kg. The least value governs: $P_d=699$ kg per dowel.

Step 4 - Spacing

Required capacity of the dowel group $=2000$ kg, so the effective number of dowels (in terms of the dowel under the load) must be $\ge 2000/699=2.86$.

The dowel under the load takes the full capacity; one at distance $x$ takes $(1-x/1.8l)$, where $1.8l=143.9$ cm. For spacing $s$ the sum is $\Sigma=1+2\sum(1-js/1.8l)$:

Spacing up to about 50 cm satisfies the load; for IRC:58 practice adopt 30 cm centre to centre ($\Sigma=4.83$, capacity $=3375$ kg $>2000$ kg).

Answer: 25 mm diameter dowels, length 45 cm, spacing 30 cm c/c (first dowel 15 cm from the slab edge), smooth bars with one half greased and capped at the free end.

Factors affecting pavement design

1. Traffic and loading - axle loads, axle configuration, repetitions, tyre pressure, growth and design life.
2. Subgrade - strength and volume change of the roadbed soil.
3. Pavement materials - strength, stiffness and durability of each layer.
4. Climate and environment - rainfall, drainage, temperature and frost.
5. Serviceability, reliability and failure criteria - the performance level expected during the life.

How AASHTO (1993) considers them - the design is based on the structural number (SN) equation

$$\log_{10}W_{18}=Z_RS_0+9.36\log_{10}(SN+1)-0.20+\frac{\log_{10}\left[\frac{\Delta PSI}{4.2-1.5}\right]}{0.40+\frac{1094}{(SN+1)^{5.19}}}+2.32\log_{10}M_R-8.07$$

Given data and assumptions

$h=20$ cm, $k=15$ kg/cm3 (the unit given as kg/cm2 is taken as kg/cm3), $\Delta T=18^\circ$C, $L=4.5$ m (transverse joint spacing), $B=3.5$ m, $P=5100$ kg, $a=15$ cm, $E=3\times10^5$ kg/cm2, $\mu=0.15$, $\alpha=10\times10^{-6}$/$^\circ$C, $f=1.5$, unit weight of concrete $W=2400$ kg/m3. Method: IRC:58 (Westergaard, Bradbury, Pickett and Ray).

Step 1 - Radius of relative stiffness

$$l=\left[\frac{3\times10^5\times20^3}{12(1-0.15^2)\times15}\right]^{1/4}=60.8\ \text{cm}$$

Step 2 - Edge load stress

Since $a=15 < 1.724h=34.5$ cm: $b=\sqrt{1.6a^2+h^2}-0.675h=\sqrt{1.6\times15^2+20^2}-13.5=14.07$ cm.

The IRC edge load stress chart is based on the Westergaard edge equation (Teller and Sutherland form), which is used here for the value:

$$\sigma_e=\frac{0.572P}{h^2}\left[4\log_{10}\frac{l}{b}+\log_{10}b-0.359\right]=\frac{0.572\times5100}{20^2}\left[4\log_{10}\frac{60.8}{14.07}+\log_{10}14.07-0.359\right]=24.3\ \text{kg/cm}^2$$

Step 3 - Warping stress at the edge (Bradbury)

$L/l=450/60.8=7.40$; from the Bradbury chart/table $C=1.049$.

$$\sigma_{te}=\frac{C E\alpha\Delta T}{2}=\frac{1.049\times3\times10^5\times10\times10^{-6}\times18}{2}=28.3\ \text{kg/cm}^2$$

Step 4 - Frictional stress

$$\sigma_f=\frac{WLf}{2\times10^4}=\frac{2400\times4.5\times1.5}{2\times10^4}=0.81\ \text{kg/cm}^2$$

Step 5 - Critical combinations at the edge

• Summer midday (top hotter; bottom tension adds to load tension; friction compressive): $\sigma=\sigma_e+\sigma_{te}-\sigma_f=24.3+28.3-0.81=51.8$ kg/cm2
• Night (top cooler, top tension; load stress is not additive): $\sigma_{te}+\sigma_f=28.3+0.81=29.1$ kg/cm2 (no wheel load)

Answer: the worst combination is the summer-midday case, with a total edge stress of about 51.8 kg/cm2 (load 24.3 + warping 28.3 - friction 0.81). This must be compared with the flexural strength of concrete; a stress this high shows the 20 cm slab is not adequate for 5100 kg unless the concrete is very strong, so a thicker slab is needed.

Given and assumptions

Surface 5.3 in ($E_1=365$ ksi), base 8.6 in ($E_2=30$ ksi, $m_2=1$), subgrade $E_3=9$ ksi $=M_R=9000$ psi. Reliability, standard deviation and $\Delta PSI$ are not given for this part; the values of the companion problem are used: $R=90\%$ ($Z_R=-1.282$), $S_0=0.5$, $\Delta PSI=2.0$. Two-lane two-way road: design lane carries 50 percent of the total.

Step 1 - Layer coefficients and SN

$$\begin{aligned} a_1&=0.323\log_{10}(365000)-1.42=0.377\\ a_2&=0.249\log_{10}(30000)-0.977=0.138\\ SN&=a_1D_1+a_2D_2m_2=0.377\times5.3+0.138\times8.6\times1=3.18 \end{aligned}$$

Step 2 - ESALs the section can carry (AASHTO-93)

$$\log_{10}W_{18}=Z_RS_0+9.36\log_{10}(SN+1)-0.20+\frac{\log_{10}[\Delta PSI/2.7]}{0.4+1094/(SN+1)^{5.19}}+2.32\log_{10}M_R-8.07$$

Substituting $SN=3.18$, $M_R=9000$: $\log_{10}W_{18}=5.954$, so $W_{18}=900251$ ESALs.

Step 3 - Initial commercial vehicles per day

Growth factor for 10 years at 6 percent: $\dfrac{(1.06)^{10}-1}{0.06}=13.181$.

$$W_{18}=365\times A\times0.5\times VDF\times GF \ \Rightarrow\ A=\frac{900251}{365\times0.5\times2.5\times13.181}=149.7\ \text{CVPD}$$

This is the traffic in the year the road opens. The construction period is 2 years, so the traffic counted at the start is

$$A_0=\frac{149.7}{(1.06)^2}=133.2\ \text{CVPD}$$

Answer: the pavement was designed for an initial traffic of about 133 commercial vehicles per day (at the time of the survey), which grows to about 150 CVPD at opening.

Types of rigid pavement

Rigid pavements are classified by the way joints and steel are used.

1. Jointed Plain Concrete Pavement (JPCP) - no reinforcement except dowel and tie bars; transverse contraction joints every 4.5 m (up to 5 m). This is the most common type (IRC:58).

 |<-4.5m->|<-4.5m->|   joints with dowels
 +--------+--------+
 |  slab  |  slab  |
 +--------+--------+
     sub-base (DLC/GSB)

2. Jointed Reinforced Concrete Pavement (JRCP) - slabs 10 to 30 m long with distributed steel mesh (0.1 to 0.2 percent) to hold cracks tight; dowels at joints.

 |<-------- 15 m -------->|
 +--==== mesh ====---------+
 |         slab           |
 +------------------------+

3. Continuously Reinforced Concrete Pavement (CRCP) - continuous longitudinal steel (0.6 to 0.7 percent) and no transverse contraction joints; fine transverse cracks are held closed by the steel.

 ---==== steel bars ====---
 |   continuous slab      |
 +------------------------+

4. Prestressed Concrete Pavement (PCP) - post-tensioned slabs, thinner and long (100 to 200 m) with few joints. Costly; used mainly at airports.
5. Fibre-reinforced concrete pavement - steel or synthetic fibres added to control cracking and give thinner slabs.

Importance of standard axle load

Importance of the standard axle load in flexible pavement design

1. Traffic contains axles of many different weights. The standard axle (80 kN or 18 kip single axle with dual tyres, AASHO Road Test) is a common reference, so all axles are converted into equivalent standard axle repetitions (ESAL or msa).
2. Pavement damage increases roughly with the fourth power of axle load: $\text{EALF}=(P/80)^4$. A 100 kN axle is 2.4 times and a 120 kN axle 5.1 times as damaging as an 80 kN axle, so overloaded trucks dominate the design.
3. It gives a single design variable (msa) with which thickness catalogues are prepared (IRC:37, Asphalt Institute charts, Nepal DoR design charts).
4. It helps to compare designs and to estimate the effect of overloading and the need for axle-load control.
5. Axle-load survey data can be summarised into one VDF and used with growth rate and design life.

Differences between flexible and rigid pavement

Radius of relative stiffness (l)

The radius of relative stiffness $l$ is a measure of how stiff the slab is compared with the subgrade support. It defines the area of slab that takes part in resisting a load (Westergaard).

$$l=\left[\frac{E\,h^{3}}{12\,(1-\mu^{2})\,k}\right]^{1/4}$$

where $E$ is the modulus of elasticity of concrete (kg/cm2), $h$ slab thickness (cm), $\mu$ Poisson's ratio and $k$ modulus of subgrade reaction (kg/cm3). The unit of $l$ is cm.

• A thicker slab or higher $E$ gives a larger $l$ (stiffer slab spreads load over a wider area); a higher $k$ gives smaller $l$.
• Typical values for 20-30 cm slabs are about 60-100 cm.
• It is used in all stress equations (Westergaard, Kelly, Bradbury) and in the ratio $L/l$ and $B/l$ for the warping coefficient, and to decide joint spacing and the spread of dowel load transfer (about $1.8\,l$).

Modulus of subgrade reaction (k)

The modulus of subgrade reaction $k$ is the pressure required to cause unit deflection of the subgrade (Winkler's foundation assumption):

$$k=\frac{p}{\Delta}\ \ (\text{kg/cm}^3)$$

It is found by a plate-load test: a rigid 75 cm diameter plate is loaded in increments, and $k$ is taken at 0.125 cm deflection (so $k = p_{0.125}/0.125$). When the test is on the subgrade, $k$ is corrected for saturation and for the sub-base:

• Correction for the plate size: $k_{75}= k_{30}/2$ for a 30 cm plate on clay (bending of the subgrade).
• A sub-base (e.g. lean concrete or granular) raises the effective $k$; IRC:58 gives tables of $k$ over granular or cement-treated sub-base.
• Roughly, IRC:58 relates k to the soaked CBR of the subgrade: about 2 kg/cm3 for CBR 2 and 5 to 6 kg/cm3 for CBR 10; granular or cement-treated sub-bases raise it further.

Rigid pavement stresses depend only weakly on $k$ (through $l$), so approximate values are acceptable.

Given data

Surface $D_1=5.3$ in, $a_1=0.401$; base $D_2=8.6$ in, $E_2=30$ ksi, $m_2=1$; subgrade $M_R=E_3=9$ ksi $=9000$ psi; $R=90\%$, $Z_R=-1.282$, $S_0=0.5$, $\Delta PSI=2.0$.

Step 1 - Layer coefficient of base

$$a_2=0.249\log_{10}(E_2)-0.977=0.249\log_{10}(30000)-0.977=0.138$$

Step 2 - Structural number

$$SN=a_1D_1+a_2D_2m_2=0.401\times5.3+0.138\times8.6\times1.0=2.125+1.185=3.310$$

Step 3 - ESALs from the AASHTO equation

$$\begin{aligned} \log_{10}W_{18}&=Z_RS_0+9.36\log_{10}(SN+1)-0.20+\frac{\log_{10}\left[\frac{\Delta PSI}{4.2-1.5}\right]}{0.4+\frac{1094}{(SN+1)^{5.19}}}+2.32\log_{10}M_R-8.07\\ &=(-1.282)(0.5)+9.36\log_{10}(4.310)-0.20+\frac{\log_{10}(2.0/2.7)}{0.4+1094/(4.310)^{5.19}}+2.32\log_{10}(9000)-8.07 \end{aligned}$$

Terms: $Z_RS_0=-0.641$; $9.36\log_{10}(SN+1)=5.939$; fatigue term $=-0.136$; $2.32\log_{10}M_R=9.174$.

$$\log_{10}W_{18}=6.066\ \Rightarrow\ W_{18}=10^{6.066}=1163446$$

Answer: the pavement can carry about 1.16 $\times10^6$ ESALs (1.16 million) at 90 percent reliability.

The design follows IRC:58 (Guidelines for the Design of Plain Jointed Rigid Pavements for Highways) based on the Westergaard/Bradbury/Pickett-Ray approach.

1. Collect data - design wheel load (5100 kg for highways, 4100 kg for small roads), tyre pressure, radius of contact area, flexural strength (modulus of rupture) and E, Poisson's ratio and thermal coefficient of concrete; modulus of subgrade reaction k (on top of sub-base); traffic (CVPD), growth rate and design life.
2. Design traffic - find the commercial vehicles at the end of construction, classify the traffic (A to G) and note the thickness adjustment.
3. Assume a trial slab thickness h and the joint spacing (transverse L, longitudinal width B).
4. Radius of relative stiffness $l=\left[\dfrac{Eh^3}{12(1-\mu^2)k}\right]^{1/4}$.
5. Load stresses - edge load stress (Pickett and Ray chart) and corner load stress (Kelly's formula $\sigma_c=\frac{3P}{h^2}\left[1-\left(\frac{a\sqrt2}{l}\right)^{0.6}\right]$).
6. Temperature (warping) stresses - from temperature differential between top and bottom: Bradbury's formula $\sigma_{te}=\dfrac{C\,E\,\alpha\,t}{2}$ with coefficient C from L/l and B/l; corner warping stress $\sigma_{tc}=\dfrac{E\alpha t}{3(1-\mu)}\sqrt{a/l}$.
7. Frictional stress $\sigma_f=\dfrac{WLf}{2\times10^4}$ (W in kg/m3, L in m).
8. Check critical combinations - (a) summer midday: $\sigma_e+\sigma_{te}-\sigma_f$, (b) winter midday: $\sigma_e+\sigma_{te}+\sigma_f$ (different $\Delta T$), (c) night: warping + friction, and corner: $\sigma_c+\sigma_{tc}$. Each must not exceed the permissible flexural stress (modulus of rupture divided by a factor of safety, or by the fatigue stress-ratio in IRC:58-2002/2015).
9. Revise thickness by trial until the stresses are safe; apply the traffic correction (A to G) to the thickness.
10. Joints and steel - design contraction, expansion and longitudinal joints, dowel bars at transverse joints and tie bars at longitudinal joints (IRC:58 / IRC:15).

Design method and assumptions

IRC:37-2012 (also followed by Nepal DoR for road pavements) is used.

• Two-lane single carriageway: design lane carries 50 percent of the two-way commercial traffic as per IRC:37-2012; a more conservative 75 percent (IRC:37-2001) is used here to cover heavier lane loading on a two-lane road.
• Each commercial vehicle is taken to impose its quoted load on one critical axle (10 000 kg and 9000 kg) and the damage of that axle is counted; standard axle = 80 kN (8160 kg); $g=9.81$ m/s2.
• Traffic is counted now; the road opens after 2.5 years; design life $n=10$ years; $r=7\%$.

Step 1 - Design CBR

The CBR value at which 90 percent of results are equal or greater lies between 4 % (100 percent) and 5 % (75 percent), about 4.4 %. The design subgrade CBR is taken as 4 % (conservative; on the safe side).

Step 2 - Vehicle damage factor (fourth power law)

• 10 000 kg axle: $(10000\times9.81/1000\,/\,80)^4=(98.1/80)^4=2.261$
• 9000 kg axle: $(88.29/80)^4=1.483$
• Weighted: $VDF=0.6\times2.261+0.4\times1.483=1.950$

Step 3 - Design traffic

Traffic at opening: $A=340(1.07)^{2.5}=402.7$ CVPD.

$$N_s=\frac{365[(1.07)^{10}-1]}{0.07}\times A\times D\times F=365\times13.816\times402.7\times0.75\times1.950=2.97\ \text{msa}$$

Step 4 - Pavement composition

The lowest design traffic in the IRC:37-2012 catalogues is 10 msa, so for the subgrade CBR of 4 % the 10 msa section is adopted (it is safe for 3 msa; a thinner section can be designed using the IRC:37-2001 plates or IRC:SP:72 if economy is required).

Answer: design traffic $\approx 2.97$ msa; section = 40 mm BC + 80 mm DBM + 250 mm WMM + 330 mm GSB over subgrade of CBR 4 %.

Vehicle Damage Factor

Vehicle Damage Factor (VDF) is the number of standard axle load repetitions (80 kN single axle) that cause the same damage as one pass of a commercial vehicle. It converts mixed traffic into equivalent standard axles.

• By the fourth power law, the damage factor of one axle is

$$\text{EALF}=\left(\frac{\text{axle load}}{80\ \text{kN}}\right)^{4}$$

• For a vehicle, VDF is the sum of the EALFs of all its axles; for a traffic stream, VDF = $\dfrac{\sum n_i\,\text{VDF}_i}{\sum n_i}$ (weighted average). Values are obtained from an axle-load survey at the site. If no survey is possible, IRC:37-2012 gives indicative values (for plain terrain 1.5, 3.5 and 4.5 for 0-150, 150-1500 and over 1500 CVPD respectively).
• Design traffic: $N_s=\dfrac{365\,[(1+r)^n-1]}{r}\,A\,D\,F$, where $F$ is VDF, $A$ the initial CVPD, $r$ the growth rate, $n$ the design life and $D$ the lane distribution factor.
• Overloading in Nepal is common, so a site axle-load survey is preferred to default values.

Design of the 4-lane divided pavement (IRC:37-2012)

Assumptions: each commercial vehicle is represented by its quoted axle load on the critical axle; directional split 50:50; the 4000 CVPD is the traffic in the year of completion, so no construction-period growth is needed.

Step 1 - Design CBR.

The 90th percentile lies between 4 % (100 percent) and 5 % (85.7 percent), about 4.3 %, so the design CBR is taken as 4 %.

Step 2 - VDF (80 kN standard axle):

• 100 kN: $(100/80)^4=2.441$; 80 kN: $1.000$; 70 kN: $(70/80)^4=0.586$
• $VDF=0.30\times2.441+0.50\times1.000+0.20\times0.586=1.350$

Step 3 - Design traffic. Dual two-lane carriageway: directional share 0.5 and lane factor 0.75 (IRC:37-2012), $r=8\%$, $n=15$ years:

$$N_s=365\times\frac{(1.08)^{15}-1}{0.08}\times4000\times0.5\times0.75\times1.350=365\times27.152\times1500\times1.350=20.06\ \text{msa}$$

Step 4 - Section from the CBR 4 % catalogue at 20 msa (the design traffic of 20.1 msa falls on the 20 msa column):

Answer: VDF = 1.350, design traffic $\approx$ 20.06 msa (adopt 20 msa); pavement = 40 mm BC + 110 mm DBM + 250 mm base + 330 mm sub-base = 730 mm.

Temperature stresses in rigid pavement

Temperature stresses in a concrete slab are of three kinds:

1. Warping stress - arises from the difference of temperature between the top and bottom of the slab (day: top hotter, bottom in tension; night: top in tension). The curling is resisted by the self-weight of the slab. Bradbury's formula for the edge is

$$\sigma_{te}=\frac{C\,E\,\alpha\,\Delta T}{2}$$

where $C$ is the coefficient from $L/l$ (chart), $E$ modulus of concrete, $\alpha$ coefficient of thermal expansion ($10\times10^{-6}$ per $^\circ$C) and $\Delta T$ the temperature differential across the slab. At the corner, $\sigma_{tc}=\dfrac{E\alpha\Delta T}{3(1-\mu)}\sqrt{\dfrac{a}{l}}$. 2. Frictional stress - due to overall expansion or contraction of the slab restrained by the sub-base:

$$\sigma_f=\frac{WLf}{2\times10^4}$$

3. Expansion stress - if free movement is blocked at joints (for example by sand or stones in joints), a compressive stress $E\alpha\,\Delta t$ develops and the slab may buckle or blow up, so expansion joints and sealants are provided.

Critical combinations: summer midday $\sigma_e+\sigma_{te}-\sigma_f$, winter midday $\sigma_e+\sigma_{te}+\sigma_f$, and night $\sigma_{te}+\sigma_f$.

Design procedure of dowel bars at expansion joints

Dowel bars transfer load across the expansion joint (smooth round mild steel bars, one end greased or sleeved with a cap). Design procedure (Bradbury, as in Khanna and Justo, with IRC:58 detailing):

1. Given data - wheel load $P$, joint width $\delta$, dowel diameter $d$, permissible stresses in shear ($F_s$), bending ($F_f$) and bearing ($F_b$), slab thickness, $E$, $k$, $\mu$.
2. Radius of relative stiffness $l=\left[\dfrac{Eh^3}{12(1-\mu^2)k}\right]^{1/4}$.
3. Load to be transferred by the dowel system is taken as 40 percent of the design wheel load.
4. Capacity of a single dowel (least of the three governs):
   • Shear: $P_s=0.785\,d^2F_s$
   • Bending: $P_f=\dfrac{2d^3F_f}{L_d+8.8\delta}$
   • Bearing: $P_b=\dfrac{F_b\,L_d^2\,d}{12.5\,(L_d+1.5\delta)}$
5. Length of dowel $L_d$ is obtained by equating the bending and bearing capacities; adopt a practical length.
6. Spacing - the dowel under the load transfers its full capacity; a dowel at distance $x$ transfers the proportion $(1-x/1.8l)$ and dowels beyond $1.8l$ transfer nothing. For a trial spacing $s$ add the proportions on both sides: $\Sigma = 1+2\sum(1-ks/1.8l)$.
7. Check $P_d\times\Sigma\ge 0.4P$; adjust spacing (usually 30-45 cm) until satisfied.
8. Detail the bar: one half coated with bitumen or greased, and a cap leaving an expansion gap at the free end.

Approach

The structure is given, so its capacity (ESAL) is found with the layer-coefficient (structural number) method of AASHTO, then divided by the yearly traffic to get the life. Assumed: reliability 90 % ($Z_R=-1.282$), $S_0=0.5$, $\Delta PSI=2.0$, drainage coefficients $m=1$, single lane so the design lane takes 100 percent of traffic.

Step 1 - Subgrade

CBR values at 7 locations: 11, 9, 7, 10, 8, 6, 4. The 90th percentile value (90 percent equal or greater) lies between 4 % (100 percent) and 6 % (85.7 percent) and is about 5.4 %; adopt CBR = 5 %, $M_R=10\times5=50$ MPa $=7252$ psi.

Step 2 - Layer coefficients (E in psi)

$$\begin{aligned} a_1&=0.323\log_{10}(2400\times145.04)-1.42=0.370\\ a_2&=0.249\log_{10}(500\times145.04)-0.977=0.233\\ a_3&=0.227\log_{10}(275\times145.04)-0.839=0.205 \end{aligned}$$

Step 3 - Structural number of the section

$$SN=0.370\times\frac{75}{25.4}+0.233\times\frac{120}{25.4}+0.205\times\frac{145}{25.4}=1.092+1.102+1.172=3.37$$

Step 4 - ESALs the pavement can carry

With $SN=3.37$ and $M_R=7252$ psi in the AASHTO-93 equation: $\log_{10}W_{18}=5.896$, so $W_{18}=786452$ ESAL ($\approx0.79$ million).

Step 5 - Traffic per year

VDF from truck factors: $0.30\times5.6+0.30\times0.6+0.40\times0.1=1.90$. Traffic at opening (18 months of construction): $A=140(1.08)^{1.5}=157.1$ CVPD.

$$W_{18}=365\,A\,F\,\frac{(1.08)^n-1}{0.08}\ \Rightarrow\ \frac{(1.08)^n-1}{0.08}=\frac{786452}{365\times157.1\times1.90}=7.22$$ $$(1.08)^n=1+0.08\times7.22=1.577\ \Rightarrow\ n=\frac{\ln1.577}{\ln1.08}=5.92\ \text{years}$$

Answer: the pavement will last about 5.9 years after opening (CBR 5 %, SN = 3.37, capacity 0.79 million ESAL).

Assumptions

Design method: AASHTO-93 structural number with layer coefficients from the given moduli. Not stated in the problem, so assumed: design life 20 years (new alignment, no construction period), 50:50 directional split, design-lane factor 0.8 for two lanes in one direction, reliability 90 % ($Z_R=-1.282$), $S_0=0.45$, $\Delta PSI=4.2-2.5=1.7$, drainage coefficients 1.0. Standard axle 80 kN, fourth power law for ESAL.

Step 1 - ESAL per day (first year, both directions)

Step 2 - Design ESAL

$$W_{18}=196.7\times0.5\times0.8\times365\times\frac{(1.04)^{20}-1}{0.04}=196.7\times0.4\times365\times29.778=855345$$

Step 3 - Layer coefficients (E in psi) and required SN

• Subgrade $M_R=10\times6.5=65$ MPa $=9427$ psi.
• $a_1=0.323\log_{10}(2400\times145)-1.42=0.370$, $a_2=0.249\log_{10}(500\times145)-0.977=0.233$, $a_3=0.227\log_{10}(275\times145)-0.839=0.205$.
• Required $SN$ over base (protecting $M_R$ = 500 MPa): $SN_1=1.38$; over sub-base (275 MPa): $SN_2=1.77$; over subgrade: $SN_3=3.10$.

Step 4 - Thicknesses

• AC is specified as 7.5 cm (2.95 in): $SN_{AC}=0.370\times2.953=1.09$. A layer-wise check over a 500 MPa base asks for $SN_1=1.38$ (about 94 mm of AC); because the problem fixes 7.5 cm, this small shortfall is covered by the stiff base and the overall SN is checked instead.
• Base: $D_2\ge(SN_2-SN_{AC})/a_2=(1.77-1.09)/0.233=2.90$ in $=74$ mm; this is below the practical minimum, so provide 15 cm (5.91 in): $a_2D_2=1.38$.
• Sub-base: $D_3\ge(SN_3-SN_{AC}-a_2D_2)/a_3=(3.10-1.09-1.38)/0.205=3.08$ in $=78$ mm; provide the minimum practical 15 cm: $a_3D_3=1.21$.
• Check: $SN=1.09+1.38+1.21=3.68\ge3.10$ - safe. The subgrade CBR of 6.5 % supports this because traffic is light.

  +--------------------------+
  | Asphalt concrete  7.5 cm |
  | Base (500 MPa)   15.0 cm |
  | Sub-base (275)   15.0 cm |
  | Subgrade CBR 6.5 %       |
  +--------------------------+

Answer: AC 7.5 cm, base 15 cm, sub-base 15 cm (total 37.5 cm), for a design traffic of about 0.86 million ESAL. The passenger cars and light trucks cause very little damage; the 3-axle trucks produce most of the ESAL.

Westergaard theory: assumptions and analysis

Westergaard (1926) analysed a concrete slab resting on an elastic subgrade, treating the slab as a thin plate on a liquid (Winkler) foundation.

Assumptions

1. The slab acts as a homogeneous, isotropic elastic plate of uniform thickness.
2. The reaction of the subgrade is vertical and proportional to deflection ($p=k\,\Delta$), where $k$ is the modulus of subgrade reaction (a "dense liquid"; no shear transfer in the subgrade).
3. The slab is in full contact with the subgrade at all times.
4. The load is applied on a circular area of radius $a$ with uniform pressure; the thickness is small compared with the other dimensions and the slab is treated as infinite or semi-infinite for the three cases.
5. The load is static; temperature, joint load transfer and friction are not included in the basic analysis.

Analysis - three loading positions

• Interior load (far from edges):

$$\sigma_i=\frac{0.316\,P}{h^2}\left[4\log_{10}\frac{l}{b}+1.069\right]$$

• Edge load (at the free edge):

$$\sigma_e=\frac{0.572\,P}{h^2}\left[4\log_{10}\frac{l}{b}+\log_{10}b-0.359\right]$$

• Corner load (Kelly's modification):

$$\sigma_c=\frac{3P}{h^2}\left[1-\left(\frac{a\sqrt2}{l}\right)^{0.6}\right]$$

Here $P$ is the wheel load, $h$ thickness, $a$ contact radius, $b=\sqrt{1.6a^2+h^2}-0.675h$ for $a<1.724h$ (else $b=a$) and $l=\left[\dfrac{Eh^3}{12(1-\mu^2)k}\right]^{1/4}$ the radius of relative stiffness. For the same load the edge stress is the largest bending stress and the interior stress is the smallest; the corner load gives the largest deflection.

Development of warping and friction stresses

Warping stress

Warping stress comes from the temperature difference between the top and bottom of the slab.

• By day, the top is hotter than the bottom, so the top tries to expand more and the slab edges curl downwards. The weight of the slab and the subgrade resist this, so compression develops at the top and tension at the bottom.
• At night the top is cooler, the edges curl upwards and tension develops at the top.

Bradbury gave the edge stress for a finite slab:

$$\sigma_{te}=\frac{C\,E\,\alpha\,\Delta T}{2}$$

and for the interior $\sigma_{ti}=\dfrac{E\alpha\Delta T}{2}\left(\dfrac{C_x+\mu C_y}{1-\mu^2}\right)$. The coefficient $C$ depends on $L/l$ and $B/l$ (Bradbury chart), $E$ is the modulus of concrete, $\alpha$ the thermal coefficient and $\Delta T$ the top-bottom temperature difference.

Frictional stress

When the whole slab expands or contracts with an average temperature change, the movement is resisted by friction between the slab and the sub-base. At the centre of the slab the tension from contraction is

$$\sigma_f=\frac{W\,L\,f}{2\times10^{4}}\ \ (\text{kg/cm}^2)$$

with $W$ the unit weight of concrete (2400 kg/m3), $L$ the joint spacing (m) and $f$ the coefficient of friction (about 1.5). It is tensile in winter or at night (contraction) and compressive in summer. It is generally small and fixes the spacing of contraction joints.

Traffic and loading factors decide the thickness and the strength of the pavement layers. The main ones are:

1. Design wheel load and axle load - the magnitude of the load on one wheel or axle. Damage rises about as the fourth power of the axle load, so heavy and overloaded trucks control the design. Axles are converted to the standard 80 kN axle (IRC:37).
2. Axle configuration - single, tandem and tridem axles spread load differently; the number of wheels per axle (dual tyres) changes the stress under the surface.
3. Contact (tyre) pressure and area - the contact radius and tyre pressure (about 0.5-0.8 MPa) decide the stress at the surface and the upper layers; for rigid slabs the contact radius enters the Westergaard equations.
4. Repetition of loads - the cumulative number of standard axles (msa) in the design life, found from initial CVPD, growth rate, design period, lane distribution factor and vehicle damage factor. Fatigue cracking and rutting depend on it.
5. Traffic volume and growth - only commercial vehicles above 3 t laden weight count. Counts from the last survey are grown by the rate $r$: $A=P(1+r)^x$.
6. Speed of vehicles and wheel position - slow and standing loads (at junctions, climbing lanes) cause greater deformation of bituminous layers; lateral wander in the lane reduces the damage at one point.
7. Impact and dynamic effects - increase the effective load on rough surfaces.
8. Legal limits - the allowed maximum axle load and enforcement (axle-load control and weigh stations) fix the upper bound of loading.

Lane distribution factor (D) is the fraction of the total commercial traffic that uses the most heavily loaded (design) lane. Traffic is not shared equally between lanes and directions, so the design lane carries more than an equal share.

IRC:37-2012 recommends:

If the directional split is not 50:50 (for example 55:45), the heavier direction is used.

Vehicle damage factor (F) is the number of standard (80 kN) axle passes equal in damage to one commercial vehicle. It is found by the fourth power law $F=\sum (\text{axle load}/80)^4$ over all axles and averaged over the traffic from an axle-load survey. The cumulative standard axles for design are

$$N_s=\frac{365\,[(1+r)^n-1]}{r}\,A\,D\,F$$

Method

IRC:58 (jointed plain concrete pavement) - trial thickness is checked for the critical stress combinations by Westergaard, Bradbury and Pickett-Ray analysis, then the traffic adjustment is applied.

Assumptions

Pavement 7 m wide = two lanes of 3.5 m with a central longitudinal joint ($B=3.5$ m); transverse contraction joint spacing $L=4.5$ m; unit weight of concrete 2400 kg/m3; friction coefficient 1.5; permissible (design) flexural stress = modulus of rupture 40 kg/cm2; the edge stress chart is replaced by the Westergaard edge formula (same basis as the IRC chart). The summer and winter air temperatures (50 and 15 degrees C) affect the joint gap only, while the given differential $\Delta T$ is used for warping.

Step 1 - Traffic classification

Traffic at completion: $A=1050(1.075)^3=1304$ CVPD, which is in class E (450-1500 CVPD) and the thickness adjustment is 0 cm. (Even the present 1050 CVPD is class E.)

Step 2 - Frictional stress

$\sigma_f=\dfrac{WLf}{2\times10^4}=\dfrac{2400\times4.5\times1.5}{2\times10^4}=0.81$ kg/cm2.

Step 3 - Trial thickness: edge load and warping stress

$l=\left[\dfrac{Eh^3}{12(1-\mu^2)k}\right]^{1/4}$, edge load stress $\sigma_e=\dfrac{0.572P}{h^2}\left[4\log_{10}\dfrac{l}{b}+\log_{10}b-0.359\right]$, warping $\sigma_{te}=\dfrac{CE\alpha\Delta T}{2}$ with $C$ from the given table at $L/l$ (interpolated).

All stresses in kg/cm2. For h = 20 cm the combined stress is too high; for h = 25 cm it is within the 40 kg/cm2 limit.

Step 4 - Corner check (h = 25 cm)

Kelly's corner load stress: $\sigma_c=\dfrac{3P}{h^2}\left[1-\left(\dfrac{a\sqrt2}{l}\right)^{0.6}\right]=\dfrac{3\times5100}{25^2}\left[1-\left(\dfrac{15\sqrt2}{90.3}\right)^{0.6}\right]=14.2$ kg/cm2. Corner warping: $\sigma_{tc}=\dfrac{E\alpha\Delta T}{3(1-\mu)}\sqrt{a/l}=\dfrac{3\times10^5\times10\times10^{-6}\times20.3}{3(0.85)}\sqrt{15/90.3}=9.7$ kg/cm2. Total corner stress $=14.2+9.7=23.9$ kg/cm2 $<40$, safe.

Step 5 - Result

Thickness from analysis = 25 cm; traffic adjustment (class E) = 0 cm.

Answer: provide a 25 cm thick plain cement concrete slab (modulus of rupture 40 kg/cm2) over a granular/DLC sub-base, with 4.5 m contraction joints, a longitudinal joint with tie bars at the 3.5 m lane edge and dowel bars at the transverse joints as per IRC:58.