Stresses in Rigid Pavements - KFUPM
Transcript of Stresses in Rigid Pavements - KFUPM
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Chapter 3
Stresses in Rigid Pavements
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E, µ, h
WESTERGARD’S THEORY
SUBRADE REACTION MODULUS - k
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STRESSES IN RIGID PAVEMENTS
1- TRAFFIC (Single Tire, Multiple Tires)
2- TEMPERATURE
3- FRICTION
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TRAFFIC STRESSSingle Tire
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MULTIPLE TIRES
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31 in
60 in
P=160,000 lb
p=150 psi
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31
60
Joint
Edge
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P =160,000 / 4 = 40000 lb , A = 40,000 / 150 psi = 267 in2
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TEMPERATURE
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εt = strain/ 1.0 F = 5* 10-6 in / F Δt = Total temp. differenceμ = 0.15E = 4*106 psiC = C1 = Max of C1 or C2
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FRICTION
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1-Thermal Reinforcement
( 144 lb/ft3 )( h/12)
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2-Tie Bars
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3 - Dowel Bars
Step 1: Find Joint Opening (z)
Example: slab length= 50ft, Δt= 60oF
Z= 50 (12)[0.000005 X 60+ 0.00005] = 0.21 in
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Load transfer = 50 % of the applied load max.
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LANE WIDTH = 12 ft , 6 ft - Axle Step 2: Find critical Dowel load
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P=15500 lbP=15500 lb
Given: Z= 0.25, Assume dowel diameter = 1 in,Load transfer = 50 % of the applied load
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= 3000 psi
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155
Allowable stress =[(4-1)/3] X 3000= 3000 psi
Transferred load =(50/100)(15500)[(1/5)+(0.44/8.3)]
= 1763 lb
Bearing stress =(155/100) X 1763 =2732 psi < 3000 OK.
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