2. A rigid pavement for a rural collector is to be designed using the 1993 AASHT
ID: 1714126 • Letter: 2
Question
2. A rigid pavement for a rural collector is to be designed using the 1993 AASHTO guide procedure to carry a design ESAL of 5 x 10. The soil is considered to be dry and well drained. The pavement cross-section will include asphalt shoulders with dowels. · = 4,000,000 MR-13,500 psi . S-900 psi If a 25-year design life with a 4% growth factor is to be used, determine the required slab thickness. Assume a reliability level of 85%, a standard deviation of 0.34, an initial serviceability index of 4.2, and a terminal serviceability index of 2.5Explanation / Answer
For calculating the thickness of rigid pavement AASHTO guidelines provide the following equation
log( W18) = Zr x So + 7.35 log (D+1) -0.06 +(log ((PSI/3)))/(1+(1.624*10000000)/(D+1)^8.46) + (4.22-0.32Pt ) x log ( (Sc Cd ( D^0.75 - 1.132))/(215.63 J ( D^0.75 -(18.42/ (Ec/ K )^0.25)))
Where W18 is design EASL of 80 KN ; Zr is standard normal deviate corresponding to given reliability value ; So is standard deviation ; D is pavement thickness in inches ; PSI is difference between initial and terminal serviceability conditions ; Pt is terminal serviceability index ; J is load transfer coeffecients ; Sc is modulus of rupture of PCC ; Cd is drainage coeffecient ; Ec is elastic modulus and K is subgrade modulus
For reliability of 85% Zr = - 1.037
For JRCP condition with HMA J =3.2
Assuming no sub base thus K = Mr / 18.8 = 13500/18.8 = 718.085
Thus from AASHTO equation we get
6.698 = -0.3528 + 7.35 log ( D+1) -0.06 -( 0.246/ ( 1+ (1.624*10000000)/(D+1)^8.46)))+ 3.42log ( 1080(D^0.75 -1.132)/ ( 690.016 (D^0.75 -2.132)))
By hit and trial equating LHS = RHS for various values of D
For D = 5 RHS = 6.81
For D = 4 RHS = 6.7
For D = 3 RHS = 7.718
For D = 4.2 RHS = 6.701
For D = 4.5 RHS =6.73
For D =3.9 RHS = 6.718
FOR D= 3.7 RHS = 6.748
Thus D=4 inches is integral value that would be an approximate answer
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