Road Drainage - Term Paper Example

Download file to see previous pages Road Drainage Design Report Name: Course: Instructor: Institution: Date of Submission: 1.0. Introduction Road drainage is an important factor during road design projects, as some of the designs are presented in the document, which differ depending on the situation at hand. Road drainage design is important since it guarantee that the pavements on the road perform their activities effectively. Other important benefits of road drainage is through prevention of flooding on the road, avoidance of ponding on road surfaces, protection of the pavements and subgrade materials, and also ensures erosion of the side slopes is avoided. The three main principles that drainage designs on the roads must consider include the open, French, and piped drain systems. The system are selected for design depending on characteristics such as urban or rural road, fill or cut roads, and the ground water conditions of the roads. Open drain systems transport surface water and carry subsoil water, thus facilitates early visual detection of possible blockages on the road. However, it is a restricted system since it does not consider safety considerations, risk assessments and roadside space. Piped drains are highly recommended for urban roads, which ensures road water and shoulders do not reach the embankment. The French drain is common since it is frequently used for all new roads in the rural areas. The benefit of the French drain is the provision of subgrade drainage and surface water despite limited spaces. 2.0. Design Discharge for Longitudinal Table Drainage Figure 1. Cross section of the road and table drain The design discharge of the above picture is calculated by attaining the peak discharge. The rational method is used in estimating the design discharge of sized catchments as provided in the figure above. The rational method formula is given as: Qy = 0.278 CY tcIY.A Where Qy = the design discharge for an ARI in y years ARI = the average value between the exceedances of a given rainfall intensity Cy = the runoff coefficient, which is dimensionless for ARI of y years Y = Years tcIy = The rainfall intensity at an average given in mm/h, it is also the depth of rainfaill per unit time tc = concentration time which denotes the time water flows from a remote point to the outlet point. It is used as the duration of design rainfall intensity. L = location A = the area of the catchment given in km2 Based on figure 1: which aims to drain the water from a 485 m section of a MAJOR ROAD in Mt Nathan (27.987 deg. South, 153.269 deg. East) Major road (ARI) = 50 years Road width = 12 m Shoulder width = 2 m Concentration time = 5 minutes. 2.1. Rainfall Intensity tcIY tc = 5minutes y = 50 years 5I50 = Is attained using the Interpolate IFD data That is; 60I50 = 88.73mm/hr: 90I50 =69.33mm/hr Therefore, 5I50 = 69.33 + (88.73 - 69.33) * (90 – 5) / (90 -60) = 5I50 = 69.33 + 19.4 * 85 / 30 5I50 = 69.33 + 19.4 * 2.8 5I50 = 69.33 + 54.32 5I50 = 123.65 mm/hr 2.2. Runoff Coefficient C50 = using table 5.9.1 C50 = (0.3 * 77.46 + 4 + 0 + 10 + 40) 100 C50 = 0.77 2.3. Area Km = m / 1, 000 Km = 485 m / 1000 Km = 0.485km A = L * W A = 0.485 * 12 * 2 A = 11.64 2.4 Design Discharge of the major road Qy = 0.278 CY tcIY.A Qy = 0.278 *0.77 * 123.65 * 11.64 Qy = 308.0936 3.0. Table Drain The table drains are located on the outer edges of the shoulders in cuts, with the design water level been below the level of the subgrade on the pavement of the road. Slope = 1V:2H Vegetated soil with clean bottom and brush on sides = 70% Design Discharge = 11.7 m3/s Longitudinal slope / gradient = 2% Best trapezoidal channel is a half-hexagon = X= 0.58 3.1. Designing the Table Drain Since we have the design discharge and bed slope, we have to calculate the manning value and the side slope channel. Design discharge = Q = 11.7 m3/s Bed slope = S0 Manning value = n Channel slope = X = 0.58 The design is founded on the method of maximum permissible velocity through the formula below. Average velocity < max permissible velocity The channel gradient = 2% Thus, the maximum permissible velocity = 2.5m/s, based on table 8.8.1.3 One must first determine, Q, S0, n, X, and Vmax Q = 11.7 m3/s S0 = 0.5% (0.005 m/m) The minimum slope S0 for the hard lined channels = 0.25% (0.0025 m/m) Vegetated channels =0.5% (0.005 m/m) n = 0.14 based on table 8.4.3b X = 0.58 Vmax = 2.5 A = Q / Vmax A = 11.7 / 2.5 A = 4.68 Manning’s Discharge Equation Q = 1/n AR2/3 S01/2 = 1/n A (A / P) 2/3 S0 ½ The wetted perimeter = P = (A5/3 S0 ½ / Qn) 3/2 P is also attained through P = 2y X2 + 1 + A/ Y –Yx P = (A5/3 S0 ½ / Qn) 3/2 P = [4.68 5/3 * 0.0051/2 m/m / (11.7 * 0.14) 3/2] P = [4.68 5/3 * 0.0051/2 m/m / 2.1974] P = 709.2245 * 2.5 -03 / 2.1974 P = 709.2245 * 0.3457 P = 245.1843 Manning’s equation = 1/n A (A / P) 2/3 S0 ½ = 1/0.14 of 4.68 (4.68 / 245.1843) 2/3of 0.0051/2 = 1/0.14 of 4.68 (0.0191)2/3 of 2.5-03 = 1/0.14 of 5.6910-04 of 2.5-03 = 6.8095-03 of 2.5-03 = 2.0269-04 Where R = A/P R = 4.68 / 245.1843 R = 0.0191 X = 0.58 Finding y in the equation b = 2y 1 + X2 –X Calculating the flow of depth, which is y b = 2y 0.7564 b = A/y –yX Since real roots for Y are not available, we use the best hydraulic section, so we can design the channel geometry to be determined with the freeboard added. It allows the process to calculate the Froude number to perceive if the answer is acceptable for the design. Q -1/nAR2/3 S1/2 = nQ / S1/2 –AR 2/3 y2 [2 1 + X2 – X (Y/2)2/3 Re-arranging = Y – {1.59Qn/s1/2 (2 1+ x2 –x]}3/8 y – {1.59 * 11.7 * 0.14 / 0.0051/2 [2 1 + 12 -1]}3/8 y = 2.60442 / (0.005 ½ (2 1 + 12 – 1) y = 2.60442 / (0.51/2 * 100–1/2) . (1) y = 2.60442 / 0.070710678 y = (36.83206286)3/8 y = (8 36.83206086 )3 y = 1.5695611933 y = 3.8667 b = A/ y – yX b = 4.68/3.8667 – 3.8667 * 0.58 b = 1.2103 – 2.2467 b = -1.0324 The negative symbol is ignored. Thus, b = 1.0324 A = 4.68m2 V = Q/A V = 11.7/4.68 V = 2.5m/s B = b + 2Xy = 1.0324 + 2 (0.58 * 3.8667) = 5.5178m Fr = Q B/ gA3 = 11.7 5.5178 / 2 * 4.683 = Fr = 5.5178/2 *4.68^3 = 0.16406 *11.7 Fr = 1.91949 Freeboard = max (0.3m,0.2y, V2/2g) = 0.3 * 0.2 * 3.8667/4 = 0.06 * 3.8667/4 0.232/4 = 0.058 Freeboard = 0.058m 4.0. Estimating the Peak Design Discharge of the Cross Drainage Structure The rational method formula is given as: Qy = 0.278 CY tcIY.A Where Qy = the design discharge for an ARI in y years ARI = the average value between the exceedances of a given rainfall intensity Cy = the runoff coefficient, which is dimensionless for ARI of y years Y = Years tcIy = The rainfall intensity at an average given in mm/h, it is also the depth of rainfall per unit time tc = concentration time which denotes the time water flows from a remote point to the outlet point. It is used as the duration of design rainfall intensity. A = 0.58km2 70% - open forest area 30% dense vegetation Qy = 0.278 CY tcIY.A From the first equation Qy = 0.278 * 0.77 * 123.65 * 0.58 Qy = 15.3517m3/s 5.0. Culvert Design In designing culverts, extra energy is required, which assists overcome the losses and pass the discharge though the culverts. The damming of the flow raises the water level by generating the extra energy. The water increase is based on the maximum culvert entrance in the upstream. Design discharge =12.4m3/s Allowable Headwater (AHW) = 100mm Inlet invert = 122.85m Outlet invert 122.75m Trapezoidal section bed width = 8.0 m Bed slope = 0.65% Side slopes 1V:2H Manning’s value of n = 0.035 Maximum allowable velocity = 2.5 m/s Culvert = 15.6 metres long Reinforced concrete with a maximum height of 1.4m Culvert entrance = 0% and square edged flare For downstream outlets, the protection is provided by calculating the batter slope based on the length of the wingwalls. The wingwalls should be shorter for protection against any possible scour or erosion that is exposed to embankments for increasing the design and durability of normal wingwalls. Protection at the culvert outlets should be compared to the stream where the culvert should have a lesser width and greater depth. Therefore, the protection should be estimated trough an improved controlling and operating condition. Thus, protection is based on ensuring the inlet and outlet control where the culvert capacity is controlled and culvert flow as well gets controlled through considering the outlet conditions. For protection, hydraulic conditions must be attained to ensure the most conservative answer of the controlling condition must be attained. Inlet control of the Head Water (HWi) Inlet invert = 122.85m Oulet Control of Head Water (HW0) Outlet invert 122.75m In inlet control = maximum discharge is 12.4m3/s HW = 100mm Square Culvert of the wing wall flare = 00 D = B = 0.1 m, Q = 12.4m3/s and a 00 flare Q/B 12.4/0.1 = 124m3/s/m 00 flare (scale 3) HW / D = 3 HW = 3 * 0.1 HW = 0.3m For the outlet portion, the energy loss must be considered That is; h0 + V2/2g + hL + he = S0 * L + HW hL = friction loss he = entrance loss hL = Se * L = V2n2/ 4/3 * L he = ke * V2 / 2g Outlet Control = h0 + V2/ 2g + V2n2/ R4/3 * L + ke* V2/2g = S0 * L + HW V = mean velocity of flow in the culvert barrel m/s = 2.5m3/s 2g = acceleration due to gravity = (2 *2) = 4 n = manning’s roughness coefficient = 0.035 L = length of culvert m = 15.6m S0 = Slope of culvert barrel = 0.65% R = Hydraulic radius = 1.872m Ke = entry loss coefficient = 0.5 based on table 9.10.2 h0 = adopted tail water depth = 4.7 h0 = is equal to TW based on manning’s equation 4.7 TW > D it stipulates the outlet is submerged TW = (D + dc) / 2 dc = culvert critical depth (m) 100m D = diameter and height of the culvert (m) 1.4m TW = (8.0 + 1.4) / 2 TW = 4.7 Thus, to design the culvert with the above dimensions the culvert should be as follows Allowable HW level = 100m H 100.24m TW 4.7 ho Vmax 2.5 L= 15.6m Q = 12. 4m3/s V = q / A A = Q / Vmax A= 12.4 / 2.5 A = 4.96 V = Q / A V = 12.4 / 4.96 V = 2.5 Since 2.5 < > 2.5 m3/s Thus, the erosion possibility is minimum, showing the culvert does not need to be modified since it provides sufficient protection. h0 + H = S0 * L + HW H = (V2 / 2g) + V2n2/ R 4/3 * L + Ke V2/ 2g V2/2g = 2.5/ 19.6 = 0.13 Ke = 0.5 he = 0.5 * 0.13 = he = 0.065m n = 0.035 r = A/p = 4.68 / 2.5 R= 1.872m hL = V2n2/R3/4 * L hL = 2.52 * 0.0352 / 1.8723/4 * 15.6 hL = 6.25 * 1.225-03 / 8.4302 * 15.6; hL = 3.3999 / 131.5111; hL = 0.0259m H = 0.13 + 0.0259 + 0.065m; H = 0.2209 h0 = TW = 2.1 > D (1.4m) h0 = 2.1m HW = h0 + H – S0 * L HW = 4.7 + 0.2209 – 0.3 * 15.6 = HW = 4.7 + 0.22 – 0.3 * 15.6 ; HW = 4.7 + 0.22 – 4.68 = HW = 0.24 Water elevation inlet = 100m + 0.24 where 100.24 > 100m Thus, the culvert has to be bigger to accomodte the 0.24m longer than the expectced AHW ...Download file to see next pages Read More