1 - Purdue University
CE361 Introduction to Transportation Engineering |HW Posted: Mon. 6 November 2006 | |
|Homework 9 Solutions (revised and expanded) |Due: Wed. 15 November 2006 |
PAVEMENT DESIGN
1. ESAL calculations. 40,000 containers per year on 3-S2 trucks. Empty containers weigh 5.25 tons and full containers can contain up to 46,000 lbs.
|Truck type |Unloaded |Max. Gross Weight |Steering axle |Max. number of | |
| |weight | |load |containers | |
|3-S2 |30,000 lbs |80,000 lbs. |12,000 lbs. |1 | |
A. (10 points) Show how you calculate the load for each axle on the truck, for both empty and loaded container conditions.
Empty containers weigh 5.25 tons * 2000 lb/ton = 10,500 lbs.
3-S2 with empty container weighs 30,000 + 10,500 = 40,500 lbs.
Steering axle load is 12,000 lbs.
A truck’s load is distributed equally over all non-steering axles.
Load on each tandem axle set is (40,500 – 12,000)/2 = 14,250 lbs.
3-S2 with full container weighs 30,000 + 46,000 = 76,000 lbs.
Load on each tandem axle set is (76,000 – 12,000)/2 = 32,000 lbs.
B. (10 points) How many ESALs would be applied to the special truck road during the next 15 years if it is rigid pavement?
(9.1) ESALSingle = [pic]= 0.1975;
(9.3) ESALTandem = [pic]= 0.0530;
(9.3) ESALTandem = [pic]= 1.4825
G = (1+0.031)15-1)/0.031 = 18.74.
Sample calc: ESAL(1)/life = 80,000*0.1975*18.74 = 296,092
| | |Axle |Design lane | | |Compounded |
C. (10 points) How many ESALs would be applied to the special truck road during the next 15 years if it is flexible pavement?
(9.3) ESALTandem = [pic]= 0.0339;
(9.3) ESALTandem = [pic]= 0.8631
Sample calc: ESAL(2)/life = 80,000*0.0339*18.74 = 50,883
| | |Axle |Design lane | |Growth |Compounded |
2. Flexible pavement design. Lifetime ESALs = 2.0 million, R = 95 percent, S.D. = 0.35, ΔPSI = 1.9, MR values of 21,750 psi (base), 12,200 psi (subbase), and 4200 psi (subgrade).
A. (15 points) Using the design chart (see FTE Figure 9.15 below), SN3 = 4.7, SN2 = 3.2, and SN1 = 2.6, or values close to these. If you failed to show the values at each scale, 5 points will be deducted.
[pic]
B. (15 points) a1 =0.41, a2=0.24, a3=0.13 in (9.6).
Using the design chart solution: SN1 = a1 * d1; 2.6 = 0.41 * d1; d1 = 6.34 “ ( 6.5”. SN2 = a1d1 + a2d2; 3.2 = (0.41*6.5) + (0.24 * d2); d2 = (3.2-2.67)/0.24 = 2.21” ( 2.5” ( 6.0” by Table 9.7. SN3 = a1d1 + a2d2 + a3d3; 4.7 = (0.41*6.5) + (0.24 * 6.0) + (0.13 * d3); d3 = (4.7 – 2.67 – 1.44)/0.13 = 4.53” ( 5.0”.
C. (10 points) Cost per lane-mile-inch for each layer in Part B. Use the materials in Table 9.8a with a values closest to those given in the problem. Sample calc for tons/la-mi/in for Hot Asphaltic Concrete: 2.65 * 62.4 lb/ft3 = 165.36 lb/ft3; lane-mi = 12 ft * 5280 ft = 63,360 ft2; volume of la-mi one inch thick = 63,360 ft2 * 1/12 ft thick = 5280 ft3; wt of Hot A.C. = 5280 ft3 * 165.36 lb/ft3 * 1 ton/2000 lbs = 436.55 tons
|Material | |S.G. |lbs/cu ft |tons/la-mi/in. |$/ton |$/la-mi/in. |
|Hot A.C. |2.65 |165.36 |436.55 |90.00 |39289.54 |
|Asphaltic stabilized base |2.40 |149.76 |395.37 |40.00 |15814.66 |
|Coarse Aggreg. |2.30 |143.52 |378.89 |7.00 |2652.25 |
|Excavation | | |$/cu yd: |11.80 |2307.56 |
Total pavement cost for one lane-mile, with excavation and earthwork costs at $11.80/CY. Using the d values from the design chart solution, …
|Layer | |SN(i) |a(i) |D(i) to use |$/la-mi |
|Surface > 3.5" |2.67 |0.41 |6.50 | $255,382 |
|Base > 6" |1.44 |0.24 |6.00 | $ 94,888 |
|Subbase |4.70 |0.13 |5.00 | $ 13,261 |
|Excavation | | |17.5 | $ 40,382 |
| | | |Total: | | $403,913 |
D. (10 points) Using the minimum thicknesses of d1 and d2 allowed in FTE Table 9.7, show how to find d3. How much would this pavement design cost to construct? Include excavation and earthwork costs at $11.80/CY.
Because W18=2.0*106, min d1=3.5” and min d2 = 6.0”.
Design chart d3 = (4.7-(3.5*0.41)-(6.0*0.24))/0.13=14.04”(14.5”
|Layer | |SN(i) |a(i) |D(i) to use |$/la-mi |
|Surface > 3.5" |2.60 |0.41 |3.50 | $137,513 |
|Base > 6" |3.20 |0.24 |6.00 | $ 94,888 |
|Subbase |4.70 |0.13 |14.50 | $ 38,458 |
|Excavation | | |24.00 | $ 55,381 |
| | | | |Total: | $326,240 |
This design is $77,673 cheaper than the one found in Part B.
3. (20 points) Rigid pavement design. S’c = 786 psi, J = 3.2, Ec = 3.7*106 psi, k = 66 pci, and Cd = 1.0. Using the design charts (see below), the concrete slab must be (approximately) 7.6” thick. Round up to 8”. If you failed to show the values at each scale, 5 points will be deducted.
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