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This Document Contains Chapters 15 to 16 Chapter 15 Geometric Design of Highway Facilities 15-1 A rural collector highway located in a mountainous terrain is designed to carry a design volume of 800 veh/day. Determine the following: (a) a suitable design speed, (b) lane and usable shoulder widths, (c) maximum desirable grade. From Table 15.1 (for mountainous terrain) minimum design speed = 30 mi/h Lane width = 12 ft, usable shoulder width = 10 ft From Table 15.3 maximum desirable grade = 10% 15-2 Repeat Problem 15-1 for an urban freeway in rolling terrain. From Table 15.2, for urban freeway minimum design speed = 50 mi/h Lane width = 12 ft, usable shoulder width = 12 ft From Table 15.3, for freeway in rolling terrain with design speed of 50 mi/h maximum grade = 5% 15-3 Given: A rural collector is to be constructed in rolling terrain with an ADT of 650 veh/day. Determine: minimum design speed recommended lane width preferable shoulder width maximum grade From Table 15.1, for rolling terrain, minimum design speed = 40 mi/h Lane width = 12 ft, Shoulder width = 10 ft From Table 15.3 maximum desirable grade = 8% 15-4 A +2% grade intersects with a –1% grade at station (535+24.25) at an elevation of 300 ft. If the design speed is 65 mi/h, determine: the minimum length of vertical curve using the rate of vertical curvature Then, using the length found in part (a), the stations and elevations of the BVC and EVC the elevation of each 100-ft station the station and elevation of the highpoint From Table 15.4, K = 193 L = KA = 193 (2 – (–1)) = 579 ft Station of BVC = (535+24.25) – (579 ft)/2 = 532+34.75 Station of EVC = (535+24.25) + (579 ft)/2 = 538+13.75 Elevation of BVC = 300 – (0.02)(579/2) = 294.21 ft Elevation at any station on the leading tangent can be found in a similar manner. The elevation on the curve can be found by subtracting the elevation on the leading tangent by the offset, which can be found using Equation 15.15, Y = A x2 200L Using this procedure, the following table, which tabulates the elevation at 100 ft stations on the curve, can be generated.
Station Dist. from BVC Tangent Elev. Offset Curve Elev.
532+34.75 0 294.21 0 294.21
533+00 65.25 295.52 0.11 295.41
534+00 165.25 297.52 0.71 296.81
535+00 265.25 299.52 1.82 297.70
536+00 365.25 301.52 3.46 298.06
537+00 465.25 303.52 5.61 297.91
538+00 565.25 305.52 8.28 297.24
538+13.75 579.00 305.79 8.69 297.10
The distance from the BVC to the high point can be found as: xhigh = LG1 / (G1 – G2) = (579)(2)/(2 – (–1)) = 386 ft The station of the high point is (532+34.75) + (386 ft) = 536+20.75 The difference between the elevation of the BVC and the elevation of the high point can be found as: yhigh = LG12 / (200(G1 – G2)) = (579)(2)2 / (200)(2 – (–1)) = 3.86 ft Therefore, the elevation of the high point is 294.21 + 3.86 = 298.07 ft 15-5 Determine the minimum length of a crest vertical curve, using the minimum length based on SSD criteria if the grades are +4% and –2%. Design speed is 70 mi/h. State assumptions used. Assumptions used include: perception-reaction time is 2.5 seconds, deceleration rate is 11.2 ft/sec2, and the case is sight distance is less than the length of the curve. Determine required stopping sight distance using Equation 3.27: SSD =1.47ut + u2 =1.47(70)(2.5) + (70)2 a 11.2 30( ± G) 30( − 0.04) g 32.2 SSD = 788 ft Since S < L, use Equation 15.5 to calculate the minimum length of the curve. Lmin = AS 2 = (6)(788)2/2158 = 1727 ft 2158 Therefore, for the given design conditions, the minimum length of the curve is 1,727 ft. 15-6 Determine the minimum length of a sag vertical curve if the grades are –4% and +2%. Design speed is 70 mi/h. State assumptions used. Consider the following criteria: stopping sight distance, comfort, and general appearance. Assumptions used include: perception-reaction time is 2.5 seconds, deceleration rate is 11.2 ft/sec2, and the case is sight distance is less than the length of the curve. For the sight distance criterion: Determine required stopping sight distance using Equation 3.27: SSD =1.47ut + u2 =1.47(70)(2.5) + (70)2 a 11.2 30( ± G) 30( − 0.04) g 32.2 SSD = 788 ft Since S<L, use Equation 15.9 to calculate the minimum length of the curve. AS2 (6)(788)2 L= = = 1179 ft 400 + 3.5S 400+ 3.5(788) For the comfort criterion, use Equation 15.10: = Au2 = (6)(70)2/46.5 = 633 ft L 46.5 For the general appearance criterion, use Equation 15.11: L = 100 A = (100)(6) = 600 ft Therefore, for the given design conditions, the minimum length of the curve is 1,179 ft. 15-7 Show that the offset y of any point of the curve above the BVC is given as: 100y = g x1 − (g1 − g2)x2 2L From the properties of a parabola, Y = ax2 , where a is constant The rate of change of the slope can be written as: d y2 = 2a 2 dx If L is the total length of the curve in feet and the total change in slope is A then, 2a= A 100L The equation for the curve can now be written as: Y = A x2 200L At any point on the curve, the vertical offset, y, can be determined by the equation: y= g x1 −Y = g x1 − A x2 = g x g1 − 1 −g2 x2 100 100 200L 100 200L = 1 − (g1 −g x2) 2 100y g x 2L 15-8 Given a sag vertical curve connecting a –1.5% grade with a +2.5% grade on a rural arterial highway, use the rate of vertical curvature and a design speed of 70 mi/h to compute the elevation of the curve at 100 ft stations if the grades intersect at station (475+00) at an elevation of 300.00 ft. Identify the station and elevation of the low point. From Table 15.5, K = 181 L = KA = 181 (1.5 – (–2.5)) = 724 ft Station of BVC = (475+00) – (724 ft)/2 = 471+38.00 Station of EVC = (475+00) + (724 ft)/2 = 478+62.00 Elevation of BVC = 300 + (0.015)(724/2) = 305.43 ft Elevation at any station on the leading tangent can be found in a similar manner. The elevation on the curve can be found by adding the elevation on the leading tangent to the offset, which can be found using Equation 15.12, Y = A x2 200L Using this procedure, the following table, which tabulates the elevation at 100 ft stations on the curve, can be generated.
Station Dist. from BVC Tangent Elev. Offset Curve Elev.
471+38 0 305.43 0 305.43
472+00 62 304.50 0.11 304.61
473+00 162 303.00 0.72 303.72
474+00 262 301.50 1.90 303.40
475+00 362 300.00 3.62 303.62
476+00 462 298.50 5.90 304.40
477+00 562 297.00 8.72 305.72
478+00 662 295.50 12.11 307.61
478+62 724 294.57 14.48 309.05
The distance from the BVC to the low point can be found as: xlow = LG1 / (G1 – G2) = (724)(1.5)/(1.5 – (–2.5)) = 271.50 ft The station of the low point is (471+38) + (2+71.50) = 474+09.50 The difference between the elevation of the BVC and the elevation of the low point can be found as: ylow = LG12 / (200(G1 – G2)) = (724)(1.5)2 / (200)(1.5 – (–2.5)) = 2.04 ft Therefore, the elevation of the low point is 305.43 – 2.04 = 303.39 ft 15-9 A crest vertical curve connects a +4.44% grade and a –6.87% grade. The PVI is at station 43+50.00 at an elevation of 1240.00 ft. The design speed is 30 mi/h. Determine: The length of the vertical curve using the AASHTO method (“K” factors) The station of the BVC (c) The elevation of the BVC (d) The station of the EVC (e) The elevation of the EVC The station of the high point (g) The elevation of the high point (h) The elevation of station 44+23.23 From Table 15.4, K = 19 L = KA = (19)|–4.44-6.87| = (19)(11.31) = 214.89 ft Station of BVC = (43+50) – (214.89 ft)/2 = 42+42.55 Elevation of BVC = 1240.00 – (0.0444)(214.89 ft/2) = 1235.23 ft (d) Station of EVC = (43+50) + (214.89 ft)/2 = 44+57.45 Elevation of EVC = 1240.00 – (0.0687)(214.89 ft/2) = 1232.62 ft Station of the high point = (42+42.55) + (214.90)(4.44)/(11.31) = 43+26.91 (214.90)(4.44)2 Elevation of the high point = 1235.23+ = 1237.10 ft (200)(11.31) Elevation of station 44+23.23 = 1235.23+ (0.0444)(180.68) − (11.31)(180.68)2 =1234.66 ft (200)(214.90) 15-10 A crest vertical curve connects a +4.90% grade and a –1.85% grade. The PVI is at station 77+00.00 at an elevation of 770.00 ft. The design speed is 60 mi/h. Determine: The length of the vertical curve using the AASHTO method (“K” factors) The station of the BVC The elevation of the BVC The station of the EVC The elevation of the EVC The station of the high point The elevation of the high point The elevation of station 75+30.00 Also, provide a tabulation showing the elevations at each full station on the curve, including the distance from the BVC for each full station, the elevation on initial tangent, the offset, and the elevation on the vertical curve. This can be in the form of a computer printout from Microsoft Excel or another program. From Table 15.4, K = 151L = kA = (151)|–4.90–1.85| = (151)(6.75) = 1019.25 ft Station of BVC = (77+00) – (1019.25)/2 = 71+90.38 Elevation of BVC = 770.00 – (0.0490)(1019.25/2) = 745.03 ft Station of EVC = (77+00) + (1019.25)/2 = 82+09.63 Elevation of EVC = 770.00 – (0.0185)(1019.25/2) = 760.57 ft Station of the high point = (71+90.38) + (1019.25)(4.90)/(6.75) = 79+30.28 (1019.25)(4.90)2 Elevation of the high point = 745.03+ = 763.16 ft (200)(6.75) Elevation of station 75+30 = (6.75)(339.62)2 745.03+ (0.0490)(339.62) − = 757.85ft (200)(1019.25) Elevation of BVC = 770.00 – (0.0490)(1019.25/2) = 745.03 ft Elevation at any station on the leading tangent can be found in a similar manner. The elevation on the curve can be found by subtracting the elevation on the leading tangent by the offset, which can be found using Equation 15.12, Y = A x2 200L Using this procedure, the following table, which tabulates the elevation at 100 ft stations on the curve, can be generated.
Station Dist. from BVC Tangent Elev. Offset Curve Elev.
71+90.38 0 745.03 0.00 745.03
72+00 9.62 745.50 0.00 745.50
73+00 109.62 750.40 0.40 750.00
74+00 209.62 755.30 1.45 753.85
75+00 309.62 760.20 3.17 757.03
76+00 409.62 765.10 5.56 759.55
77+00 509.62 770.00 8.60 761.40
78+00 609.62 774.90 12.31 762.60
79+00 709.62 779.80 16.67 763.13
80+00 809.62 784.70 21.70 763.00
81+00 909.62 789.60 27.40 762.20
82+00 1009.62 794.50 33.75 760.75
82+09.63 1019.25 794.97 34.40 760.57
15-11 A horizontal curve is to be designed for a two-lane road in mountainous terrain. The following data are known: Intersection angle: 40 degrees, tangent length = 436.76 ft, station of PI: 2700+10.65, fs = 0.12, e = 0.08. Determine: design speed station of the PC station of the PT deflection angle and chord length to the first 100 ft station From the given horizontal curve data, the radius can be calculated, from which design speed for the curve can be derived. The radius can be found by rearranging Equation 15.22, R = T / (tan Δ/2) = 436.76 / tan (40°/2) = 436.76 / 0.3640 = 1200 ft The design speed can then be found: R = u2 / [15(e + fs)] (1200) = u2 / [ 15 (0.08+0.12) ] u2 = 3600 u = 60 mi/h Station of the PC can be found by subtracting the tangent length from the station of the PI. PC = (2700+10.65) – (4+36.76) = 2695+73.89 The length of the curve can be found using Equation 15.26, L = πRΔ/180 = (3.1415926)(1200)(40)/180 = 837.76 ft Station of the PT can be found by adding the length of the curve to the station of the PC. PT = (2695+73.89) + (8+37.76) = 2704+11.65 Deflection angle and chord length to the first full station To find the deflection angle to the first full station, use Equation 15.27, δ1/2 = 180 l1 / 2πR = (180)(100–73.89)/(2)(3.1415926)(1200) = 0.6233° The chord to the first full station can be found using Equation 15.28, C1 = 2R sin (δ1/2) = (2)(1200) sin (1.24666/2) = 26.11 ft 15-12 A proposed highway has two tangents of bearings N 45º54’36” E and N 1º22’30” W. The highway design engineer, attempting to obtain the best fit for the simple circular curve to join these tangents, decides that the external ordinate is to be 43.00 ft. The PI is at station 65+43.21 Determine: The central angle of the curve The radius of the curve The length of the tangent of the curve The station of the PC The length of the curve The station of the PT The deflection angle and chord from the PC to the first full station on the curve Δ = 45.91° + 1.375° = 47.285° R = 43.00 = 469.30 ft 1 cos(47.285/2) −1 T = 469.30 tan (47.285/2) = 205.44 ft Station of the PC = (65+43.21) – 205.44 ft = 63+37.76 (e) L = π(469.30)(47.285)/180 = 387.31 ft (f) Station of the PT = (63+37.76) + 387.31 ft = 67+25.07 (g) Deflection angle from the PC to the first full station: DA1 = δ1Δ/2L = (100 – 37.76)(47.285)/(2)(387.31) = 3.7993° Chord from the PC to the first full station: C1 = 2(469.30) sin (3.7993) = 62.19 ft 15-13 Given a simple circular curve with the following properties: D = 11°, bearing on incoming (back) tangent is N 89°27’25” E, bearing on outgoing (forward) tangent is S 60°10’05” E, the station of the PI = 22+69.77 Determine: The intersection angle (Δ) The radius (R) The tangent (T) The external distance (E), The middle ordinate (M), The long chord (C), The length of the curve (L) Station of the PC Station of the PT Δ = 180° – (89.45694° + 60.16806°) = 30.375° R = 5729.58 / 11 = 520.87 ft T = R tan (Δ/2) = (520.87) tan (30.375/2) = 141.39 ft E =520.87[ −1]= 18.85 ft M = 520.87 [1 – cos (30.375/2)] = 18.19 ft C = 2(520.87) sin (30.375/2) = 272.91 ft L = (30.375)(520.87)π/180 = 276.14 ft Station PC = (22+69.77) – 141.39 ft = 21+28.38 Station PT = (21+28.38) + 276.14 ft = 24+04.52 15-14 A simple circular curve exists with a degree of curve D=12º and e=0.08. A structure is proposed on land on the inside of curve. Assume the road is on level grade. Determine: The radius of the curve The current maximum safe speed of the curve The minimum distance allowable between the proposed structure and the centerline of the curve such that the current maximum safe speed of the curve would not need to be reduced Using Equation 15.20: R = 5729.58 / 12 = 477.47 ft Next, determine the maximum safe speed for the curve, which can then be used to determine the required stopping sight distance. u= (15)(477.47)(0.08+ 0.16) = 41.4 mi/h (41.4)2 SSD = (1.47)(2.5)(41.4) + =315 ft (30)(0.35− 0) Then, determine the minimum offset distance using Equation 15.43: −cos(28.65)(315)= 25.8 ft m = 477.47 1 477.47 15-15 Given a circular curve connecting 2 tangents that intersect at an angle of 48°. The PI is at station (948+67.32) and the design speed of the highway is 60 mi/h. Determine the point of the tangent and the deflection angles from the PC to full stations for laying out the curve. First determine the radius of the curve: R = u2 / [15(e + fs)] For u = 60 mi/h, from Table 3.4, fs = 0.12, e = 0.08 R = (60)2 / [15(0.08+0.12)] R = 1200 ft The length of the tangent, T, can be found using Equation 15.22, T = R tan(Δ/2) = 1200(tan(48°/2)) T = 534.27 ft The length of the curve, L, is given by Equation 15.26: L = RΔπ / 180 = 1200(48)(3.1415926) / 180 L = 1005.30 ft Station of the PC can be found by subtracting the tangent length from the station of the PI. PC = (948+67.32) – (5+34.27) = 943+33.05 Station of the PT can be found by adding the length of the curve to the station of the PC. PT = (943+33.05) + (10+5.30) = 953+38.35 First full station is located at 944+00 δ1 / Δ = l1 / L δ1 / 48 = 66.95 / 1005.3 δ1 = 3.197° The first chord can be found using Equation 15.28, C1 = 2R sin (δ1/2) = 2(1200) sin (3.197° / 2) C1 = 66.94 ft The first deflection angle = δ1/2 = 1.5985° Last full station is located at 953+00 δ2 / Δ = l2 / L δ2 / 48 = 38.35 / 1005.30 δ2 = 1.831° The last chord can be found using Equation 15.28, C2 = 2R sin ( δ2/2) = 2(1200) sin (1.831°/ 2) C2 = 38.35 ft For other deflection angles between full stations: δ/48 = 100/1005.3 δ = 4.775° δ/2 = 2.3875° The chords between full stations can be found using Equation 15.28, C = 2R sin (δ/2) = 2(1200) sin (2.3875°) = 99.97 ft
Station Deflection Angle Chord Length
PC 943+33.05 0 0
944+00 1.598 66.94
945+00 3.986 99.97
946+00 6.373 99.97
947+00 8.760 99.97
948+00 11.148 99.97
949+00 13.535 99.97
950+00 15.922 99.97
951+00 18.310 99.97
952+00 20.697 99.97
953+00 23.084 99.97
PT 953+38.35 24.000 38.35
15-16 Given two chords xy and yz of 135 ft each marked on an existing curve to determine the radius. The perpendicular distance between y and the chord xz is 15 ft. Determine the radius of the curve and the angle set out from xz to get a line to the center of the curve. y C Since it is known that the chord xy = 135 ft and yw = 15 ft, The length of xw can be found using the Pythagorean theorem as follows: xy2 = xw2 + yw2 xw = xy2 − yw2 = 1352 −152 = 134.16 ft Solve for the radius, R, using the Pythagorean theorem, knowing the length of xw and that xc = R and wc = R – 15. xc2 = xw2 + wc2 R2 = 134.162 + (R-15)2 R2 = 18000 + R2 – 30R + 225 30R = 18225 R = 607.50 ft Solve for the deflection angle from xz to the center of the curve, cos δ = (xw/xc) = (134.16/607.50) = 0.220846 δ = 77.2412˚ 15-17 Given a compound circular curve with radii of 600 ft. and 450 ft. designed to connect two tangents that intersect at an angle of 75°, determine the central angles and the corresponding chord lengths for setting out the curve if the central angle of the first curve is 45° and the PCC is at station (675+35.25). Δ = Δ1 + Δ2 t1 = R1 tan (Δ1/2) t2 = R2 tan (Δ2/2) 75 = 45 + Δ2 t1 = 600 tan (45/2) t2 = 450 tan (30/2) Δ2 = 30 t1 = 248.53 t2 = 120.58
L1 = RΔ1 (π/180) L2 = RΔ2 (π/180)
L1 = 600 (45) (π/180) L2 = 450 (30) (π/180)
L1 = 471.24 L2 = 235.62
PC = PCC – L1 PT = PCC + L2 PC = (675 + 35.25) – (4 + 71.24) PT = (675 + 35.25) + 2 + 35.62) PC = 670 + 64.01 PT = 677 + 70.87 For curve 1 (R = 600 ft), using Equation 15.20: D = 5729.6/R = 5729.6 / 600 D = 9.5493° D/2 = 4.7747° L1 = 671 – (670 + 64.01) = 35.99 δ1/l1 = Δ/L δ1 = (35.99)(45)/471.24 δ1 = 3.4368° l2 = 35.25 δ2/l2 = Δ/L δ2 = (35.25)(45)/471.24 δ2 = 3.3661° For curve 1:
Station Deflection Angle Chord Length
670+64.01 0 0
671 1.7184 35.99
672 6.4931 99.89
673 11.268 99.89
674 16.042 99.89
675 20.817 99.89
675+35.25 22.50 35.25
For curve 2 (R = 450 ft), using Equation 15.20 D = 5729.6/R D = 5729.6 / 450 D = 12.732° D/2 = 6.3662° L1 = 676 – (675+35.25) = 64.75 δ1/l1 = Δ/L δ1 = (70.87)(30)/235.62 δ1 =8.2442 δ1/2 = 4.1221º l2 = 70.87 δ2/l2 = Δ/L δ2 = (70.87)(30)/235.62 δ2 = 9.0234º For curve 2:
Station Deflection Angle Chord Length
675+35.25 0 0
676+00 4.1221 64.69
677+00 10.488 99.79
677+70.87 15.00 70.80
15-18 Given an arterial road is to be connected to a frontage road by a reverse curve with parallel tangents. The distance between the centerline of the two tangent sections is 60 ft. The PC of the curve is located at station (38+25.31) and the central angle of each of the component curves is 25° Determine: the station of PT. Solve for the radius of the reverse curves, R, R= d = 60 = 320.20 ft 2(1−cos )Δ 2(1−cos25 )° Solve for the length of each of the reverse curves using Equation 15.26, L = RΔπ / 180 = 320.20(25)(3.1415926) / 180 = 139.71 ft The station of the PT can be found by adding 2 times the length of each reverse curve to the station of the PC, PT = (38+25.31) + (2+79.42) PT = 41+04.73 15-19 For the data provided in Problem 15-1, determine (a) minimum radius of horizontal curvature, assuming a superelevation rate of 8%, (b) minimum length of crest vertical curves, and (c) minimum length of sag vertical curves, using the rate of vertical curvature method, when joining two segments at maximum grade. Find the minimum radius R = u2 / [ 15(e + fs)] R = (302/15) (1/(e + fs)) R = 60 / (emax + fs) Assuming e = 0.08 and fs = 0.16 Rmin = 60 / (0.08+0.16) = 250 ft Using Equation 15.12, L= kA From Table 15.4, K = 19 = 19 (10 – (–10)) L = 380 ft Using Equation 15.12, L = kA From Table 15.5, K = 37 = 37(10 – (–10)) L = 740 ft 15-20 Repeat Problem 15-19 for an urban freeway in rolling terrain. Find the minimum radius R = u2 / [ 15(e + fs)] R = (502/15) (1/(e + fs)) R = 166.7 / (e + fs) Assuming emax = 0.08 and fs = 0.14 Rmin = 166.7 / (0.08 + 0.14) = 758 ft Using Equation 15.12, L = kA From Table 15.4, K = 84 = 84 (5 – (–5)) L = 840 ft Using Equation 15.12, L = kA From Table 15.5, K = 96 = 96 (5 – (–5)) L = 960 ft 15-21 For the data provided in Problem 15-3, determine (a) minimum length of crest vertical curves, and (b) minimum length of sag vertical curves, using the rate of vertical curvature method, when joining two segments at maximum grade, (c) maximum superelevation, and (d) maximum degree of curve (use fs = 0.15) (Degrees-minutes-seconds). Using Equation 15.12, L= kA From Table 15.5, k = 44 = 44 (8– (–8)) L = 704 ft Using Equation 15.12, L = kA From Table 15.6, k = 64 = 64 (8– (–8)) L = 1024 ft Maximum superelevation Maximum values of superelevation can be as high as 0.12 but vary from state to state. In Virginia, emax = 0.08 for rural roads. Maximum degree of curve Find the minimum radius R = u2 / [15(e + fs)] R = (402/15) (1/(e + fs)) R = 106.7 / (e + fs) Assuming emax = 0.08 and fs = 0.15 Rmin = 106.7 / (0.08+0.15) = 463 ft 15-22 Determine the distance required to transition pavement cross-slope from a normal crown section with a normal crown cross-slope of 2% to superelevation of 6% on a two-lane highway with a design speed of 50 mi/h. First, determine the length of superelevation runoff required to transition from a cross-slope of zero on the outside lane to full superelevation, using Table 15.11. Lr = 144 ft Next, determine the tangent runout required using Equation 15.40, Lt =eNC Lr = 0.02(144) = 48 ft ed 0.06 Then, the determine superelevation transition length required by summing the superelevation runoff and tangent runout. Required transition = 144 + 48 = 192 ft. 15-23 A building is located 19 ft from the centerline of the inside lane of a curved section of highway with a 400 ft radius. The road is level; e = 0.10. Determine the appropriate speed limit (to the nearest 5 mi/h) considering the following conditions: stopping sight distance and curve radius. To determine sight distance past the building located on the inside of the curve, use Figure 15.26 and a rearrangement of Equation 15.43. S = R cos [−1 R m− ] = 400 cos [−1 400−19]= 248 ft 28.65 R 28.65 400 Use Equation 3.27 for stopping sight distance to determine the appropriate speed. Assume a = 11.2 ft/sec2, perception-reaction time = 2.5 sec, and that the road is level. u2 2 SSD =1.47ut += 248 = (1.47)(2.5)u + u / (30)(0.35) a 30( ± G) g 0.095238 u2 + 3.675 u – 248 = 0 u = 35.3 mi/h Speed limits are posted in increments of 5 mi/h; the appropriate speed is 35 mi/h. To determine the appropriate speed based on curve radius, use Equation 15.24, R = u2 / [15(e + fs)] Assume fs = 0.15 (appropriate for u = 40 mi/h) 400 = u2 / [15(0.10+0.15)] u = 38.7 mi/h Speed limits are posted in increments of 5 mi/h; the appropriate speed is 35 mi/h. 15-24 Describe the factors that must be taken into account in the design of bicycle paths. The design criteria for bicycle paths are somewhat similar to those for highways, but some of these criteria are governed by bicycle operating characteristics, which are significantly different from those of automobiles. Important design considerations for a safe bicycle path include the path width, the design speed, the horizontal alignment, and the vertical alignment. The minimum width specified by AASHTO for a 2-way path is 10 ft. and 5 ft. for a 1-way path. Uniform graded shoulders of at least 2 ft should be provided on either side. AASHTO recommends a design speed of 20 mph for paved paths or 15 mph for unpaved paths. Design speeds should be increased for grades greater than 4%. Superelevation rates for bicycle paths vary from 2% to 5% and coefficients of side friction should vary from 0.30 to 0.22 for paved paths and 0.15 to 0.11 for unpaved paths. Grades should not exceed 5%. 15-25 For a bicycle path with an average speed of 20 mi/h, a maximum superelevation of 2%, and a total change in grade of 10%, determine the minimum radius of horizontal curvature and the minimum length of the vertical curve. The radius can be found: R = u2 / [15(e + fs)] Assume fs = 0.30 R = (20)2 / [15(0.02 + 0.30)] R = 83.33 ft From Figure 15.32, for a design speed of 20 mi/h, L = 180 ft 15-26 Given an available area that is 400 ft by 500 ft, design a suitable parking lot layout for achieving each of the following objectives: Provide the maximum number of spaces. Provide the maximum number of spaces while facilitating traffic circulation by providing one way flow on each aisle. Provide the maximum number of spaces (accomplished with spaces at a 90° angle). Assume space dimensions: Stall width = 8.5 ft Stall length = 18 ft Aisle width = 24 ft Aisle + row width = 24 + 18 + 18 = 60 ft Allow 25 ft at each row end for traffic circulation, Number of aisles = 400 ft / 60 ft = 6.67 There can be 6 aisles with 2 rows per aisle Number of spaces per row = [500 ft – (2)(25 ft)] / (8.5 ft/space) = 52 Total number of spaces = (6 aisles)(2 rows/aisle)(52 spaces/row) = 624 spaces For optimum traffic circulation, spaces are at a 45° angle (herringbone design). Allow 25 ft at each row end for traffic circulation, Number of aisles with 2 rows each = 500 ft / 45 ft = 11.11 There can be 11 aisles with 2 rows per aisle Number of spaces per row = [400 ft – (2)(25 ft)] / (14 ft/space) = 25 Total number of spaces = (11 aisles)(2 rows/aisle)(25 spaces/row) = 550 spaces Chapter 16 Highway Drainage 16-1 What are the two sources of water a highway engineer is primarily concerned with? Briefly describe each. The highway engineer is concerned primarily with two sources of water. The first source, surface water, is precipitation that occurs as rain or snow. Some of this is absorbed into the soil, and the remainder remains on the surface of the ground and should be removed from the highway pavement as surface drainage. The second source, ground water, is that which flows in underground streams; this is referred to as subsurface drainage. This may become important in highway cuts or at locations where a high water table exists near the pavement structure. 16-2 Briefly describe the main differences between surface drainage and subsurface drainage. Surface drainage is the system that drains surface water away from the surface of the pavement. This surface water consists of the precipitation in the form of rain, ice and snow, less that which is absorbed into the soil. Surface drainage system features are incorporated into the overall design of the highway with the objective of ultimately directing all surface runoff to natural waterways, including: Transverse slopes to facilitate the removal of water from the pavement surface in the shortest possible time. Longitudinal slopes to facilitate the provision of adequate slopes in the longitudinal channels. Longitudinal channels or ditches to collect the surface water that runs off from pavement surfaces, subsurface drains, and other areas of the highway right-of-way. Curbs and gutters. Drainage structures such as bridges and culverts. Subsurface drainage is the system that drains groundwater from the highway pavement structure. The groundwater may be in one or more of the following forms: (1) surface water that has permeated through cracks and joints in the pavement to the underlying strata, (2) water moving upward through the underlying strata due to capillary action, and (3) water existing in the natural ground below the water table, usually referred to as groundwater. A subsurface drainage system usually consists of: Longitudinal drains usually consisting of pipes laid in trenches within the pavement structure and parallel to the center line. Transverse drains placed below the pavement and in a direction perpendicular to the center line. Horizontal drains in cuts and embankments to relieve pore pressure. A drainage blanket, which is a layer of material with a high coefficient of permeability beneath or within the pavement structure placed beneath or within the pavement structure to facilitate the flow of subsurface water away from the pavement. 16-3 What are the two main disadvantages of using turf cover on unpaved shoulders? The two main disadvantages of using turf cover on unpaved shoulders are that turf cover cannot resist continued traffic and that it loses firmness under conditions of heavy rain. 16-4 Briefly describe the three properties of rainfall that primarily concern highway engineers. Highway engineers are primarily concerned with three properties of rainfall: intensity, duration, and frequency. The rate of fall (typically expressed in inches per hour) is known as intensity. The length of time for a given intensity is known as duration. The probable number of years that will elapse before a given combination of intensity and duration will be repeated is known as frequency. 16-5 What is meant by a: (a) 10 year storm, (b) 50 year storm, (c) 100 year storm, and (d) 500 year storm? A 10-year storm is a storm of given intensity and duration for which the probability that it will occur in a one year period is 1 in 10. A 50-year storm is a storm of given intensity and duration for which the probability that it will occur in a one year period is 1 in 50. A 100-year storm is a storm of given intensity and duration for which the probability that it will occur in a one year period is 1 in 100. A 500-year storm is a storm of given intensity and duration for which the probability that it will occur in a one year period is 1 in 500. 16-6 Define the following: (a) drainage area, (b) run-off coefficient (C), (c) travel time (ti), and (d) time of concentration (tc). Drainage area is that area of land that contributes to the runoff at the point where the channel capacity is to be determined. This area is normally determined from a topographic map. Runoff coefficient (C) is the ratio of runoff to rainfall for the drainage area. The runoff coefficient depends on the type of ground cover, the slope of the drainage area, storm duration, prior wetting, and the slope of the ground. For small drainage areas, typically only type of ground cover and slope of the drainage area are considered in determination of runoff coefficients. (c) Travel time (ti) is the ratio of flow length to average flow velocity for a specific watershed. (d) Time of concentration (tc) is the time required for runoff to flow from the most distant point, along hydraulic channels, to the point of interest in the watershed and is therefore the sum of travel times for the various elements within the watershed. 16-7 A 196-acre rural drainage area consists of four different watershed areas as follows: Steep grass covered area = 35% Cultivated area = 15% Forested area = 40% Turf meadows = 10% Using the rational formula, determine the runoff rate for a storm of 100-year frequency. Use Table 16.2 for runoff coefficients. Assume that the rainfall intensity curves in Figure 16.2 are applicable to this drainage area and the following land characteristics apply. Use Figure 16.4 to calculate average velocity using "fallow or minimum tillage cultivation" ground cover. Overland flow length = 0.5 miles. Average slope of overland area = 2% Determine the runoff rate for a 100-year storm. Determine the runoff coefficients for each of the four groundcover-type areas from Table 16.2.
Type of Cover % Area Coefficient
Steep grass covered 35 0.60
Cultivated area 15 0.30
Forested area 40 0.20
Turf meadow 10 0.25
Calculate the weighted run-off coefficient using Equation 16.1. n C Ai i Cw = i=1n Ai i=1 Cw = 196 [(0.35)(0.60) + (0.15)(0.30) + (0.40)(0.20) + (0.10)(0.25)] / 196 Cw = 0.36 Estimate the average velocity using Figure 16.4. This area can be considered as Fallow or minimum tillage cultivation while average slope is 2%. v = 0.65 ft/sec Determine travel time using Equation 16.2, Ti = L / 3600 V Ti = (0.5)(5280) / (3600)(0.65) Ti = 1.128 h Since only one segment is being investigated, Ti = Tc. Determine the rainfall intensity, using Figure 16.2. I = 2.75 in/h Determine the runoff flow rate using the rational formula (Equation 16.4). Q = CIA Q = (0.36)(2.75)(196) Q = 194.04 ft3/sec Therefore, the runoff flow rate for this watershed will be 194 ft3/sec. 16-8 A 210-acre urban drainage area consists of three different watershed areas as follows: Flat residential (30% impervious area) = 58% Moderately steep residential (50% impervious area) = 28% Flat commercial (90% impervious area) = 14% Using the rational formula, determine the runoff rate for a storm of 100-year frequency. Use Table 16.2 for runoff coefficients. Assume that the rainfall intensity curves in Figure 16.2 are applicable to this drainage area and the following land characteristics apply. Use Figure 16.4 to calculate average velocity using "fallow or minimum tillage cultivation" ground cover. Overland flow length = 0.4 miles. Average slope of overland area = 3% Determine the runoff rate for a 100-year storm. Determine the runoff coefficients for each of the four groundcover-type areas from Table 16.2.
Type of Cover % Area Coefficient
Flat residential (30% impervious area) 58 0.40
Moderately steep residential (50% impervious area) 28 0.65
Flat commercial (90% impervious area) 14 0.80
Calculate the weighted run-off coefficient using Equation 16.1. n C Ai i Cw = i=1n Ai i=1 Cw = 210 [(0.58)(0.40) + (0.28)(0.65) + (0.14)(0.80)] / 210 Cw = 0.526 Estimate the average velocity using Figure 16.4. This area can be considered as Fallow or minimum tillage cultivation while average slope is 3%. v = 0.8 ft/sec Determine travel time using Equation 16.2, Ti = L / 3600 V Ti = (0.4)(5280) / (3600)(0.8) Ti = 0.733h Since only one segment is being investigated, Ti = Tc. Determine the rainfall intensity, using Figure 16.2. I = 3.6 in/h Determine the runoff flow rate using the rational formula (Equation 16.4). Q = CIA Q = (0.526)(3.6)(210) Q = 397.66 ft3/sec Therefore, the runoff flow rate for this watershed will be 398 ft3/sec. 16-9 Compute rate of runoff using the rational formula for a 256-acre rural drainage area consisting of two different watershed areas as follows: Steep grass area = 15% Forested area = 30% Cultivated fields = 55% If the time of concentration for this area is 2.4 hours, determine the runoff rate for a storm of 50-year frequency. Use the rainfall intensity curves in Figure 16.2. Use Table 16.2 for runoff coefficients. Determine the runoff coefficients for each of the two groundcover-type areas from Table 16.2.
Type of Cover % Area Coefficient
Steep grass covered 15 0.60
Forested area 30 0.20
Cultivated fields 55 0.30
Calculate the weighted run-off coefficient using Equation 16.1, n C Ai i Cw = i=1n Ai i=1 Cw = 256 [(0.15)(0.60) + (0.30)(0.20) + (0.55)(0.30)] / 256 Cw = 0.315 Determine the rainfall intensity, using Figure 16.2. I = 1.45 in/h Determine the runoff flow rate using the rational formula (Equation 16.4). Q = CIA Q = (0.315)(1.45)(256) Q = 116.9 ft3/sec Therefore, the runoff flow rate for this watershed will be 117 ft3/sec. 16-10 Using the TR-55 method, determine the depth of runoff for a 24-hour, 100-year precipitation event of 9 inches if the soil can be classified as group B and the watershed is contoured pasture with good hydrologic condition and an antecedent soil condition III. Determine depth of runoff for a 24-hour, 100-year, precipitation of 9 inches. First determine CNII value from Table 16.3 for contoured pasture with good hydrologic condition and soil group B. CNII = 35 Since the antecedent moisture condition of the soil is condition III, now find the CNIII value for this soil condition from Table 16.5. CNIII = 55 Determine the potential maximum retention after runoff begins, using Equation 16.7. S = (1000 / CN) – 10 = (1000 / 55) – 10 S = 8.18 in Determine depth of runoff using Equation 16.5. = (P −0.2 )S 2 ((9) − (0.2)(8.18))2 h = P + 0.8S 9 + (0.8)(8.18) h = 3.49 in Therefore, the depth of runoff will be 3.5 inches. 16-11 Determine the depth of runoff by the TR-55 method for a 24-hour, 100-year precipitation of 9 inches for an antecedent moisture condition III, if the following land uses and soil conditions exist. Area Fraction Land/Use Condition Soil Group 0.33 Wooded/fair condition D 0.27 Small grain/straight row/good condition D 0.12 Pasture/contoured/fair condition D 0.23 Meadow/good condition D 0.05 Farmstead D Since different land uses are present, a weighted CN value must be calculated. The CNII values can be found in Table 16.3, and the corresponding CNIII values in Table 16.5, as shown below.
Land-use Type % Area CNII CNIII
Wooded/fair 0.33 79 93.4
Small grain/straight/good 0.27 87 97.4
Pasture/contoured/fair 0.12 83 95.8
Meadow/good 0.23 78 92.8
Farmstead 0.05 86 97.2
Weighted CNIII = (0.33)(93.4)+(0.27)(97.4)+(0.12)(95.8)+(0.23)(92.8)+(0.05)(97.2) Weighted CNIII = 94.82 Determine the potential maximum retention after runoff begins, using Equation 16.7. S = (1000 / CN) – 10 S = (1000 / 94.82) – 10 S = 0.546 in Determine depth of runoff using Equation 16.5. h = (P − 0.2 )S 2 = [9−(0.2)(0.546)]2 P + 0.8S 9+ (0.8)(0.546) h = 8.38 in Therefore, the depth of runoff will be 8.38 inches. 16-12 Determine the peak discharge that will occur for the conditions indicated in Problem 16-11 if the drainage area is 0.67 mi2, and the time of concentration is 1.4 hours. From Problem 16-10, the depth was found to be h = 8.38 inches. From Figure 16.5, for a time concentration of 1.4 hours: q'p = 260 ft3/sec/mi2/in Determine the peak discharge using Equation 16.6. qp = q'p (A)(h) qp = (260)(0.67)(8.38) qp = 1459.8 ft3/sec Therefore, the peak discharge that will occur will be approximately 1500 ft3/sec. 16-13 What is the difference between supercritical and subcritical flow? Under what conditions will either of these occur? Flows in channels can be tranquil or rapid. Flow is considered to be critical when the depth of flow is the critical depth (i.e. the depth at which the flow in a channel changes from tranquil to rapid). At this depth the specific energy is minimized. When the flow depth is less than the critical depth, the flow is known as supercritical. This type of rapid, turbulent flow is prevalent in steep flumes and mountain streams. When the flow depth exceeds the critical depth, the flow is subcritical, characterized by slower velocity and tranquil flow. Subcritical flow is typically found in shallow broad channel of nearly flat slopes. 16-14 A trapezoidal channel of 2:1 side slope and 5 ft bottom width, discharges a flow of 275 ft3/sec. If the channel slope is 2.5% and the Manning coefficient is 0.03, determine (a) flow velocity, (b) flow depth, and (c) type of flow. Using the graphical solution of Manning’s equation for a trapezoidal channel with a side slope of 2:1 and bottom width of 5 ft, provided in Figure 16.6, Flow velocity = 11 ft/sec Flow depth = 2.6 ft Type of flow: The intersection of the discharge flow and the 2.5% slope lies above the critical curve, thus, the flow is supercritical. 16-15 A 6 ft wide rectangular channel lined with rubble masonry is required to carry a flow of 300 ft3/sec. If the slope of the channel is 2% and n = 0.015, determine (a) flow depth, (b) flow velocity, and (c) type of flow. Using the graphical solution of Manning’s equation for a rectangular channel of a width 6 ft and n = 0.015, provided in Figure 16.7, Flow depth = 2.8 ft Flow velocity = 18 ft/sec Type of flow: The intersection of the discharge flow line and the 2% slope line lies above the critical curve, thus, the flow is supercritical. 16-16 For the conditions given in Problem 16-15, determine the critical depth and the maximum channel slope at which subcritical flow can occur. Using Figure 16.7, the critical flow curve intersects the vertical line for a flow of 300 ft3/sec at a depth of approximately 4.25 ft. The associated channel slope is approximately 0.007 (0.7%), which is the maximum slope at which subcritical flow will occur. 16-17 Determine a suitable rectangular flexible lined channel to resist erosion for a maximum flow of 20 ft3/sec if the channel slope is 2%. Use channel dimensions given in Problem 16-15. Using Figure 16.7, the normal depth of flow = 0.45 ft and the flow velocity = 7 ft/sec. To allow a free board of 1 ft, the revised channel dimension is: 1.45 ft × 6 ft Determine if a channel lined with jute mesh is suitable to prevent erosion. Using Figure 16.9, determine dmax dmax = 0.62 ft Determine the hydraulic radius, R R = a /p R = (0.62)(6.0) / [(2)(0.62) + (6.0)] R = 0.51 ft Determine the flow velocity, V, using Figure 16.13. V = 61.53 R1.028 S00.431 V = 61.53(0.51)1.028(0.02)0.431 V = 5.75 ft / sec Determine the flow. Q = VA Q = (5.75)((0.62)(6.0)) Q = 21.4 ft3 / sec Since the allowable flow rate for this material is slightly greater than the flow required, the channel design using jute mesh is adequate. 16-18 A trapezoidal channel of 2:1 side slope and 5 ft bottom width is to be used to discharge a flow of 220 ft3/sec. If the channel slope is 1.5% and the Manning coefficient is 0.015, determine the minimum depth required for the channel. Is the flow supercritical or subcritical? Using the graphical solution of Manning’s equation for a trapezoidal channel with a side slope of 2:1 and bottom width of 5 ft, provided in Figure 16.6, Qn = flow × manning coefficient Qn = (220)(0.015) Qn = 3.3 From Figure 16.6, depth of flow = 1.8 ft. The intersection of the discharge flow line and the 1.5% slope line lies above (barely) the critical curve, thus, the flow is supercritical. Allowing a free board of 1 ft, the minimum depth required for the channel is 2.6 ft. 16-19 Determine whether a 5 ft × 5 ft reinforced concrete box culvert with 45o flared wingwalls and beveled edge at top of inlet carrying a 50-year flow rate of 200 ft3/sec will operate under inlet or outlet control for the following conditions. Assume ke = 0.5. Design headwater elevation (ELhd) = 105 ft Elevation of stream bed at face of invert = 99.55 ft Tailwater depth = 4.75 ft Approximate length of culvert = 200 ft Slope of stream = 1.5% n = 0.012 Determine the required conditions for inlet control. Calculate the flow rate per ft of width. Q / NB = 200/5 = 40.0 ft3/sec/ft Using Figure 16.16, draw a line connecting 5 ft to 40 ft to obtain the headwater depth at culvert face (HW / D). HW / D = 1.19 ft/ft Calculate the required headwater, HW. HW = (HW / D) × depth of culvert HW = 1.19 × 5 HW = 5.95 ft Neglect the approach velocity head in this problem. Therefore; HWi = 5.95 ft Calculate the required depression. Use Equation 16.13 to determine the design headwater depth, HWd. HWd = ELhd – ELsf HWd = 105 – 99.55 HWd = 5.45 ft Determine the fall using Equation 16.14. Fall = HWi – HWd Fall = 5.95 – 5.45 Fall = 0.50 ft Calculate the culvert invert elevation invert elevation = 99.55 – 0.50 = 99.05 ft Determine the required conditions for outlet control. Determine the critical depth with Q / B = 200 / 5 = 40.0, from Figure 16.23. dc = 3.7 ft Calculate depth from outlet invert to hydraulic grade line and determine if TW is greater. (dc + D) / 2 = (3.7 + 5) / 2 = 4.35 ft Since TW is not greater, use ho = 4.75 ft Determine the total head loss (H) from Figure 16.21 H = 2.2 ft Calculate the required outlet headwater elevation using Equation 16.23. ELho = Elo + H + ho ELho = (99.05 – (0.015)(200)) + 2.2 + 4.75 ELho = 103.00 ft The required outlet headwater elevation (103 ft) is less than the design headwater elevation (105 ft); therefore the 5 ft × 5 ft culvert is acceptable and inlet control governs. 16-20 Repeat Problem 16-19 using a 6.5 ft diameter circular pipe culvert with ke = 0.5. Determine the required conditions for inlet control. Using Figure 16.16, draw a line connecting 78 in to 200 ft3/sec to obtain the headwater depth at culvert face (HW / D). HW / D = 0.88 ft/ft Calculate the required headwater, HW. HW = (HW / D) × depth of culvert HW = (0.88)(6.5 ft) HW = 5.72 ft Neglect the approach velocity head in this problem. Therefore; HWi = 5.72 ft Calculate the required depression. Use Equation 16.13 to determine the design headwater depth, HWd. HWd = ELhd – ELsf HWd = 105 – 99.55 HWd = 5.45 ft Determine the fall using Equation 16.14. Fall = HWi – HWd Fall = 5.72 – 5.45 Fall = 0.27 ft Calculate the culvert invert elevation invert elevation = 99.55 – 0.27 = 99.28 ft Determine the required conditions for outlet control. Determine the critical depth with Q / B = 200 / 6.5 = 30.8, from Figure 16.24. dc = 3.75 ft Calculate depth from outlet invert to hydraulic grade line and determine if TW is greater. (dc + D) / 2 = (3.75 + 6.5) / 2 = 5.13 ft Since TW is greater, use ho = 5.13 ft Determine the total head loss (H) from Figure 16.22 H = 1.05 ft Calculate the required outlet headwater elevation using Equation 16.23. ELho = Elo + H + ho ELho = (99.28 – (0.015)(200)) + 1.05 + 5.13 ELho = 102.46 ft The required outlet headwater elevation (102.46 ft) is less than the design headwater elevation (105 ft); therefore the 6.5 ft diameter circular culvert is acceptable and inlet control governs. 16-21 Determine the ground water infiltration rate for a new two-lane pavement with the following characteristics: Lane width = 12 ft Shoulder width = 10 ft Length of contributing transverse cracks (Wc) = 20 ft Rate of infiltration (Kp) = 0.05 ft3/day/ft2 Spacing of transverse cracks = 30 ft Determine the width of granular base. W = (number of lanes) × (lane width) + (shoulder width) × 2 W = (2 × 12) + (10 × 2) W = 44 ft Since this is a new pavement, Nc = N + 1 Nc = 3 Use Equation 16.24 to determine infiltration rate (assume Ic = 2.4): qi = Ic (Nc + Wc )+K p W WCs qi = (2.4) [(3 / 44) + (20 / (44)(30))] + 0.05 qi = 0.25 ft3 / day / ft2 Therefore, the ground water infiltration rate will be 0.25 ft3/day/ft2 for this new pavement section. 16-22 Determine the ground water infiltration rate for a new four-lane pavement with the following characteristics: Lane width = 12 ft Shoulder width = 6 ft Length of contributing transverse cracks (Wc) = 40 ft Rate of infiltration (Kp) = 0.05 ft3/day/ft2 Spacing of transverse cracks = 40 ft Determine the width of granular base. W = (number of lanes) × (lane width) + (shoulder width) × 2 W = (4 × 12) + (6 × 2) W = 60 ft Since this is a new pavement, Nc = N + 1 Nc = 5 Use Equation 16.24 to determine infiltration rate (assume Ic = 2.4): qi = Ic (Nc + Wc )+K p W WCs qi = (2.4) [(5 / 60) + (40 / (60)(40))] + 0.05 qi = 0.29 ft3 / day / ft2 Therefore, the ground water infiltration rate will be 0.29 ft3/day/ft2 for this new pavement section. 16-23 In addition to the infiltration determined in Problem 16-21, ground water seepage due to gravity also occurs. Determine the thickness of a suitable drainage layer required to transmit the net inflow to a suitable outlet. Thickness of subgrade below drainage pipe = 12 feet Coefficient of permeability of native soil = 0.35 ft/day Height of water table above impervious layer = 21 feet Slope of drainage layer = 2% Permeability of drainage area = 2,000 ft/day Length of flow path = 60 ft Calculate the amount of drawdown. H – Ho = 21 – 12 = 9 ft of drawdown Determine the radius of influence. Li = 3.8 (H – Ho) Li = (3.8) (9) = 34.2 ft Compute the average inflow rate due to gravity drainage: ((Li + 0.5W) / Ho) = (34.2 + ((0.5) × (60))) / 12 ((Li + 0.5W) / Ho) = 64.2 / 12 ((Li + 0.5W) / Ho) = 5.35 W/Ho = 60/12 = 5.00 From Figure 16.34, determine: k × (H – Ho) / 2q2 = 1.9 q2 = k × (H – Ho) / (2 × (1.9)) q2 = (0.35 × 9) / 3.8 q2 = 0.83 qg = q2 / 0.5W qg = 0.83 / (0.5 × 60) qg = 0.028 ft3/day/ft2 Calculate the net inflow. From Problem 16-20, qi = 0.29 ft3 / day / ft2 qn = qi + qg qn = 0.29 + 0.028 qn = 0.318 p = qn / kd p = 0.318 / 2,000 p = 1.59 × 10–4 From Figure 16.38: L / Hm = 200 Hm = L / 200 Hm = 60 / 200 Hm = 0.30 ft = 3.6 in; use 4.0 inches Therefore, the suitable drainage layer thickness should be 4.0 inches. Solution Manual for Traffic and Highway Engineering Nicholas J. Garber, Lester A. Hoel 9781133605157
