Westergaards theory for rigid pavement PDF

Preview

Summary

Download Westergaards theory for rigid pavement PDF

Description

Design of Rigid pavement The design of rigid pavements by the Westergaard method is still commonly used today for rigid pavements loaded by non-highway vehicles. The Westergaard method for rigid pavement design involves a calculation of the stresses acting in the pavement under the wheel load. This stress is then compared to the strength of the pavement slab to determine whether the slab is sufficiently strong to accommodate the proposed loadings.

Westergaard's modulus of sub-grade reaction: k is directly proportional to displacement. If p is pressed induced in N\mm2 in a rigid plate of diameter 750mm with a deflection of 1.25mm then k=p/Δ

Temperature Gradient: It is the temperature difference between top fiber and bottom fiber divided by thickness of concrete slab.

Δ t = (Ttop-Tbottom)/thickness

WESTERGAARD'S ANALYSIS: 1. Rigid pavement is made of cement concrete which may or may not have reinforcement. 2. Develops slab action and load distribution over wider area. 3. Stress below subgrade is greatly reduced and 4. Failure is due to overstressing of the concrete or not by overstressing of the sub-grade. 5. Stress in concrete slab due to wheel loads are determined using Westergaard's theory. 6. Rigid pavement as a thin elastic plate resting on liquid foundation or Winkler foundation (Emil Winkler was first to formulate and solve a problem of elastic beam on deformable foundation. The model of a beam on elastic foundation which assumes linear force-deflection relationship is known as Winkler Foundation. 7. Pressure deformation characteristics of a rigid pavement depend upon the relative stiffness of the slab and the sub-grade. 4

𝐸ℎ 3

Radius of relative stiffness(l)=√ 12(1−𝜇2)𝑘 E= modulus of elasticity of cement concrete

H= thickness of slab 𝜇 = 𝑝𝑜𝑖𝑠𝑜𝑛𝑠 𝑟𝑎𝑡𝑖𝑜 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒(.12 𝑡𝑜 .15) 8. Intensity of stress induced depends upon the location of wheel load on the road. Following are the locations which are critical: Interior, Corner and Edge.

According to Westergaard’s the concrete slab is homogeneous, thin elastic plate with sub-grade reaction being vertical and proportional to the deflection. Interior loading (Si) = Edge loading(Se) =

0.316𝑝 ℎ2

𝑙

[4 log ( )] + 1.069 𝑏

𝑙 0.572𝑝 [4 log ( )] + ℎ2 𝑏

Corner loading(Sc) =

3𝑝

ℎ2

𝑎√2 𝑙

[1 − (

0.6

0.359

) ]

Where, a= radius of wheel distribution load b= radius of resisting section(cm) b=a when a>=1.724h; when a that in interior. Critical combination= load stress + warping stress – frictional stress at edge region 2. During winter, same as above but frictional resistance is tensile during contraction. Critical combination= load stress + warping stress + frictional stress

at corner region

3. At corner region, when slabs warps upward no frictional stress. 4. Stress due to warping/curling: a. Interior stress: 𝐶𝑥 𝐸𝛼𝑡 𝐷𝑡 𝐶𝑦 𝐸𝑚𝛼𝑡 𝐷𝑡 + 𝜎𝑥 = 2(1 − 𝜇2 ) 2(1 − 𝜇2 )

𝜎𝑦 =

𝐶𝑦 𝐸𝛼𝑡 𝐷𝑡 𝐶𝑥 𝐸𝑚𝛼𝑡 𝐷𝑡 + 2(1 − 𝜇2 ) 2(1 − 𝜇2 )

Based on Brad bury chart, values of Cx and Cy are computed on the basis of lx/l and ly/l.

b. Edge stress: 𝝈 𝑚𝑎𝑦 𝑏𝑒 𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝑡ℎ𝑒 𝐶𝑥 𝑜𝑟 𝐶𝑦. 𝜎=

𝐶𝐸𝛼𝑡 𝐷𝑡 2

c. Corner stress 𝐸𝛼𝑡𝐷

𝑎 𝑙

𝜎 = 3(1−𝜇𝑡2) √

5. Stress due to friction: 𝑊𝐿𝐹

𝜎 = 2𝑋104

Where, W= density of concrete(1400kg/m3) F= coefficient of sub-grade restraint (1.5) L= length of slab in meters

TYPES OF CONCRETE PAVEMENTS

   

Jointed plain concrete pavement Jointed reinforced concrete pavement Continuous reinforced concrete pavement Pre-stressed concrete pavement

Design wheel load = 7500kg, Contact pressure = 7.5kg/cm2, Spacing between longitudinal joints = 3.75m and contractions joints is 4.2m, Elastic modulus of the pavement materials/CC is 3×10^5kg/cm2, Poisson ratio = 0.15, Modulus of sub grade reaction = 30kg/cm^3. Thermal coefficient of cc per ∘C=1×10−5/CC∘C=1×10−5/CC, flexural strength of CC = 45kg/cm2, Max. temperature differential at the location for pavement thickness values of 24.26 and 30 cm are respectively 15.6, 16.2 and 16.8 ∘C∘C. Calculate the desired factor of safety with respect to load stress and warping stress at edge region.

L/1

C

L/1

C

L/1

C

1

0.00

5

0.720

9

1.080

2

0.04

6

0.920

10

1.075

3

0.175

7

1.030

11

1.050

4

0.440

8

1.077

12

1.000

To find radius (l) of relative stiffness: P=Design wheel load=7500kgP p=Contact pressure=7.5kg We know, a=Radius of load=√

=17.24cm

𝑃

𝑝𝜋

TRIAL 1: Assume pavement thickness (h) as 25 ∴h=25cm Given: subgrade reaction,K=30k/cm3, E=3×105kg/cm2, μ=Poisson's ratio=0. We know, Radius of relative stiffness,l=[Eh312K(1−μ2)] =[3×105×(15)312×30(1−0.152)]=[3×105×(15)312×30(1−0.152)] ∴l=60.41cm

2.

To find radius of resisting section(b) a/h=17.24/25=0.69...