Question

Questions 14 to 15 are based on the description below: The drainage layer has a thickness of 10 m., a cross slope of 3%, a porosity of 25%, and permeability of 12,000 ft/day. The drainage length is 15 ft. 14. The steady-state capacity of the drainage layer is most nearly: A. 420 t/day/ft B. 578 f/dayft C. 390 ft day/ft D. 665 ft/day/f 15·The time required for 50% drainage is most nearly: A. 0.0018 day B. 0.0014 day C. 0.0012 day D. 0.0010 day Questions 16 to 18 are based on WELSEA software and the following information. A pavement structure is designed as follows. The traffic load is single, 8000 lb, and with tire pressure of 120 psi. The traffic repetition is 20,000 Laver Material HMA ranular Soil Thickness, in. Modulus. ksiPoisson's Ratio 350 0.35 0.45 15 16. The critical strain at the bottom of the asphalt layer for fatigue analysis is most nearly A. 554.6 x 106 B. 703.22 x 10-6 C. 817.92 x 106 D. 1025.32 x 10% 17. The damage caused by rutting is most nearly A. 0.84 B. 6.93 C. 23830 D. 2886 18. In order to reduce rutting damage by approximately 20%, the most possible method among the following is A. Increase HMA layer to be 4 in. B. Increase granular layer to be 8 in. C. Increase the modulus of granular layer to be 50 ksi D. Increase HMA layer's modulus to be 500 ksi

Answer 1

Ans 14) The steady state capacity (q) of drainage layer can be given by :

q = k H ( S + H/2L )

where k = permeability

H = thickness of layer

S = slope

L = Length of drainage layer

q = 12000 (10/12) ( 0.03 + 10/( 2 x 12 x 15) )

q = 577.77 ft³/day /ft

or q = 578 ft³ /day/ft

Ans 15) Time required for 50% drainage (T₅₀) ,

T₅₀ = n L² / 2 k(H + SL)

where n = porosity

T₅₀ = 0.25 x 225 / [(2 x12000)( (10/12) + (0.03 x 15)]

= 56.25 / 30792

= 0.0018 day

Ans 16) We know,

N_f = 0.000221 x (1/$\epsilon$)^3.89 x (1/E)^0.84

where N_f = repetition

$\epsilon$ =strain at bottom of asphalt layer

E = modulus of elasticity

Putting values,

$\epsilon$^3.89 = 0.000221 x (1/350000)^0.854

     = 0.00021 x 1.84232 x 10^-5

= 7.03 x 10^-4

= 703 x 10^-6

Ans 17) Nr = 4.1656 x 10^-8 x (1 / $\epsilon$ ) ^4.5337

2.25 = 4.1656 x 10^-8 x (1 / $\epsilon$ ) ^4.5337

     $\epsilon$ = 0.84

Ans 18) To reduce rutting damage by 20 % , increase the modulus of granular layer to be 50 ksi