ĐỒ án ĐƯỜNG 2 Đại Học Xây Dựng

© NATIONAL UNIVERSITY OF CIVIL ENGINEERING HIGHWAY AND TRAFFIC DEPARTMENT BRIDGES AND ROAD FACULTY ROADBED AND PAVEMENT DESIGN PROJECT CHAPTER I: CALCULATION OF ROADBED SLOPE STABILITY I Some method for slope stablity Mechanical roadbed slope stability Soil block will slide 1.1.Calculation method: i Sliding surface i Ti Pi Pi.cos Considering the equilibrium condition of a slide i on its slide surface we have: Sliding forces : T = Q sin i i i  Holding force, obstructing the soil sliding: Ni=Q Cos tg  + c di i i i cos i  In which :  Q i - Weight of slice ( Qi  di * hi *  i *1m )   , c,  : density, cohesion of soil, angle of internal friction Equilibrium conditions: Ti  Ni or tg i  tgi  c  hi cos 2 i To put it simply Professor Maslov added safety factor K: tg i  c (tg  ) K  hi LECTURER : NGUYEN DUC NGHIEM STUDENT : SU DUY LINH - 59CDE - 588859 © NATIONAL UNIVERSITY OF CIVIL ENGINEERING BRIDGES AND ROAD FACULTY HIGHWAY AND TRAFFIC DEPARTMENT ROADBED AND PAVEMENT DESIGN PROJECT 1.2 Comment: - Advantage:  Simple calculation, easy to understand -Disadvantage:  Less accuracy and reliability  Scope of application is small, rarely used in practice Classical Method (Ordinary Method) 2.1 Calculation method Considering the plane problem , circular cylinder sliding mass divided into slices as shown, assuming the soil block will slide at a time Thus each slices will subject to the effects of self weight Pi ; Pi divided into two parts:  Tangential force at the sliding surface Ti  Pi sin  i (sliding force)  Normal force Ni  Pc i os i (causing friction N i tg i with cohesion cili is the holding force)  Considering the effect of the earthquake, each piece was subjected to a sliding forces Wi , the lever arm compared with O is Zi LECTURER : NGUYEN DUC NGHIEM STUDENT : SU DUY LINH - 59CDE - 588859 © NATIONAL UNIVERSITY OF CIVIL ENGINEERING HIGHWAY AND TRAFFIC DEPARTMENT BRIDGES AND ROAD FACULTY ROADBED AND PAVEMENT DESIGN PROJECT Factor of Safety: i n K=  M igiu i 1 i n n = �( Nitgi  cili )  M ilat i 1 n Z (Ti  Wi i ) � R n  �( Pcos tg i i n i  cili ) �( Pi sin i  Wi Zi ) R 2.2 Comment: -Advantage:  Results calculated with relative accuracy  Apply popular in practice -Disadvantage:  Relatively complex calculations  Assuming no exact reality (the pieces not interact with each other) Bishop Method (Simplified) 3.1 Calculation method - Same as Fellenius Method except that it includes normal forces E i+1 and Ei-1 along sides of slices - Factor of Safety : n Ptg ( i i  ci li )mi � cos i K  n1 with mi  (1  tgi tg i ) 1 Z K �( Pi sin  i  Wi i ) R 3.2 Comment -Advantage:  Method for the most accurate results of the three methods  Assumptions and calculation close to reality -Disadvantage:  Methods of finding groping Kmin  Calculations quite long LECTURER : NGUYEN DUC NGHIEM STUDENT : SU DUY LINH - 59CDE - 588859 © NATIONAL UNIVERSITY OF CIVIL ENGINEERING HIGHWAY AND TRAFFIC DEPARTMENT BRIDGES AND ROAD FACULTY ROADBED AND PAVEMENT DESIGN PROJECT 4.Conclusion - In the methods mentioned above: � Selection classical method Fellenius used to calculate II Detailed calculation Classical method Fellennius 1.Input data - Check stabilize of soft ground - Geological data:  ( g / cm ) 1.95 Embankment c (kg/cm2)  (˚) 1.2 29  ( g / cm ) 1.65 Subgrade c (kg/cm2)  (˚) 0.96 17 Ground Embankment 10m 2.Checking method 2.1- Determining the monolithic stability factor Kođ - The monolithic stability factor is determined follow formula:  K1 * K * K * K * K * K PP Kođ In which: o K1 - the reliability of the data on the mechanical characteristics of the soil (such as cohesion c and friction angle) K1=1.0 – 1.1 � choose K1=1.1 o K2 - factor consider the significance of roadbed , depending on road level; K 2=1.0 – 1.03 � K2=1.02 o K3 - factor consider extent of causing damage to the national economy if road is undermined disrupting traffic; K3=1.0 – 1.2 � K3=1.15 o K4 - factor consider degree of fit between the calculated diagram with hydrogeological at road construction site K4=1.0 – 1.05 � K4=1.05 o K5 - factor consider type of soil and its work in roadbed structure K 5=1.0 – 1.05 � K5=1.05 o Kpp - factor consider degree of reliability of calculation methods � Kpp=1.01 � The monolithic stability factor: Kođ=1.1x1.02x1.15x1.05x1.05x1.01=1.4 LECTURER : NGUYEN DUC NGHIEM STUDENT : SU DUY LINH - 59CDE - 588859 © NATIONAL UNIVERSITY OF CIVIL ENGINEERING HIGHWAY AND TRAFFIC DEPARTMENT BRIDGES AND ROAD FACULTY ROADBED AND PAVEMENT DESIGN PROJECT 2.2- Using classic methods to find Kmin - Finding Kmin based on finding the position sliding center of the most dangerous sliding surface based on the experience locus sliding center - Detailed calculations:  Determine the the experience locus sliding center  On experience locus sliding center , take a point , at that point draw a slip with different radii, each radii will give stability coefficient K according to the formula: i n K=  M igiu i 1 i n M ilat i 1 n = �( Nitgi  cili ) n Z (Ti  Wi i ) � R n  �( Pcos tg i n i i  ci li ) Z ( Pi sin  i  Wi i ) � R  Each sliding center gives Kmin value Made with various sliding center will find the coefficient Kmin  After that define Kmin required in (the value of each center have found Kmin)  Value Kmin newfound corresponding position which has the most dangerous sliding surface 2.3 Assessment and conclusions: -From Kmin compared with Kođ if:  Kmin > Kođ � roadbed slope stability  Kmin < Kođ � roadbed slopes unstable , should propose measures to strengthen Detailed calculations  The calculation is done with sliding center O1, O2, O3 with different radii  In the calculation process, taking into account vehicle weight following II.4.3 [2]  Loads are considered the load of maximum heavy vehicles at the same time can park around the roadbed distributed over 1m lengths The load is equivalent to a embankment whose height is determined by the formula hx hx  n.G  B.l LECTURER : NGUYEN DUC NGHIEM STUDENT : SU DUY LINH - 59CDE - 588859 © NATIONAL UNIVERSITY OF CIVIL ENGINEERING BRIDGES AND ROAD FACULTY HIGHWAY AND TRAFFIC DEPARTMENT ROADBED AND PAVEMENT DESIGN PROJECT  In which : o G- The heaviest weight of a vehicle (T) � G=13T o n- The maximum number of vehicles put on the road, n=2 o  - Soil density of embankment �   1.7 T/m o l- vehicle distribution range in the longitudinal direction m � l = 4.2 m o B- horizontal distribution width of vehicles, according to figures � B = 6.9m � hx  0.51 m  Detailed calculations are shown in appendix III Conclusion Result: Safety factor K B/2(O1) B/4(O2) B/8(O3) 2.572 3.38 3.603 The calculation process with the center O1, O2, O3 determining Safety factor Kmin= 2.350 Thus Kmin=2.350 > Kođ=1.2 � soft ground stability LECTURER : NGUYEN DUC NGHIEM STUDENT : SU DUY LINH - 59CDE - 588859 © NATIONAL UNIVERSITY OF CIVIL ENGINEERING BRIDGES AND ROAD FACULTY HIGHWAY AND TRAFFIC DEPARTMENT ROADBED AND PAVEMENT DESIGN PROJECT LECTURER : NGUYEN DUC NGHIEM STUDENT : SU DUY LINH - 59CDE - 588859 © NATIONAL UNIVERSITY OF CIVIL ENGINEERING BRIDGES AND ROAD FACULTY HIGHWAY AND TRAFFIC DEPARTMENT ROADBED AND PAVEMENT DESIGN PROJECT LECTURER : NGUYEN DUC NGHIEM STUDENT : SU DUY LINH - 59CDE - 588859 © NATIONAL UNIVERSITY OF CIVIL ENGINEERING BRIDGES AND ROAD FACULTY CHAPTER II: HIGHWAY AND TRAFFIC DEPARTMENT ROADBED AND PAVEMENT DESIGN PROJECT FLEXIBLE PAVEMENT DESIGN Pavement is directly component subjected to the regular destruction of vehicles and elements of the natural environment , it affects directly the quality of the operation and exploitation of roads as well as construction costs So pavement design to ensure driving smoothly, economic, satisfies the technical requirements shall require meticulous and precise Therefore pavement must:  There is enough strength to resist the shear stress deformation, bending and stability of strength  Pavement must be flat to make vehicles running smoothly  Pavement must enough roughness to enhance the sticking coefficient between the wheels and the pavement  Minimize the amount of dust caused by the pavement, avoiding pollution, pavement must have good abrasion resistance Suitable with the construction capability of local area The design standards: - Highway − Specifications for design TCVN 4054-2005 [1] - Flexible pavement design standard 22TCN211-06 [2] - Pavement Material Specifications [3] - Highway design volume IV, [4]; I Loads and Times (22TCN 211 – 06) [2] - Standard axle loading :  a single axle, double-wheel grove  P=100kN  p=0.6Mpa  D=33cm Table 1: The characteristics of standard axle loading Standard axle loading Pressure on the pavement P (kN) p (Mpa) 100 0.6 120 0.6 Wheel streak diameter D (cm) 33 36 LECTURER : NGUYEN DUC NGHIEM STUDENT : SU DUY LINH - 59CDE - 588859 © NATIONAL UNIVERSITY OF CIVIL ENGINEERING HIGHWAY AND TRAFFIC DEPARTMENT BRIDGES AND ROAD FACULTY ROADBED AND PAVEMENT DESIGN PROJECT - Calculation time of pavement structure is taken by overhaul time of the most sustainable surface layer, with class A1 layer asphalt mediating particles T overhaul = 15 years, so the calculation time of pavement structure is 15 years II Traffic flow and vehicle components Traffic flow crossing points A27-B27 in the year 15th : 1500 veh/day + Vehicle component: 25% passenger car (Volga) 25% light truck (Gaz-51) 30% medium truck (Zil-150) 20% heavy truck Traffic volume growth rate : q = 10% III Converted to calculate traffic flow 100kN: The formula for calculating the trafic flow: Nt = N0*(1+q)t Nt � N0  = 584 (veh/day) (1  q)t Table 1: Table of traffic volume over the time Year Ntt(veh/day) 257 414 10 668 15 1075 Passenger car 64 104 167 269 Light truck 64 104 167 269 Medium truck 77 124 200 323 26 41 67 108 26 41 67 108 Heavy truck with axles Heavy truck with axles Converted to calculate flow 100 kN Traffic flow of different vehicles have converted to the standard axle loading through cross-section of road at the end of the exploitation by the formula: N   =  C1.C ni  Pi  i 1  Ptt  k tk 4 (axles/day) (3-1) 10 LECTURER : NGUYEN DUC NGHIEM STUDENT : SU DUY LINH - 59CDE - 588859 © NATIONAL UNIVERSITY OF CIVIL ENGINEERING HIGHWAY AND TRAFFIC DEPARTMENT BRIDGES AND ROAD FACULTY ROADBED AND PAVEMENT DESIGN PROJECT kb : coefficient that consider stress distribution properties in the pavement under the effect of the calculation load is double or single wheel I use the term should double wheels grove kb = 0.85;  ku : bending tensile stress unit; p : tire pressure of standard loading axle , p = 0.6 (MPa) H 12   0.3636  D 33    Looking up diagram 3.5 [1] :  ku = 1.6(MPa) E1 1683  8.89 E ch,m 189.31  So :  ku   ku p.kb = 1.6x0.6x0.85 = 0.815(MPa) + For wearing course layer : H1 = 5cm ; E1=1800 Mpa Value E’tb of under layers is shown in following Table : Table 20 Ei (Mpa) Layer Graded-aggregate type II Graded-aggregate type II AC binder course t E2 E1 250 300 1600 hi (cm) K h2 h1 32 300 1.20 250 1600 5.986 267.30 18 18 0.5625 32 0.140 50 Htb E’tb (cm) (MPa) 32 250 50 267.30 57 355.94 H 57 �H �  1.727 Consider the adjustment coefficient   f � �với D 33 �D � Looking up Table 3-6 [1] :  = 1.196 So that Etbdc   Etb' = 1.196x355.94 = 425.70 (MPa) H 57   1.727  Ech.m D 33  Looking up Kogan diagram � dc = 0.467  E0 42 Etb  0.0987  dc  425.70 Etb dc Thus : Ech.m = Etb x0.467 = 425.70 x 0.467 = 198.80 (MPa) 31 LECTURER : NGUYEN DUC NGHIEM STUDENT : SU DUY LINH - 59CDE - 588859 © NATIONAL UNIVERSITY OF CIVIL ENGINEERING HIGHWAY AND TRAFFIC DEPARTMENT BRIDGES AND ROAD FACULTY ROADBED AND PAVEMENT DESIGN PROJECT H   0.1515  D 33   Looking up diagram 3.5 [1] :  ku = 1.94 (MPa) E1 1800  9.05  E ch.m 198.80  ku =  ku p k b Thus  (3-10)  ku =1.94 0.6 0.85=0.989(Mpa) + Determine R ttku : Calculation bending tensile strength of monolithic materials is determined by formula: Rttku  Rku k1.k2 In which: Rku : Bending tensile strength limit at calculations temperatures in calculation perfusions under the effect of load is applied times k2 : factor considering the declining strength over time compared with the agent of climate and weather With asphalt concrete type I take k2 = 1,0; k1 : actor considering the strength decline due to material fatigue under the effect of loads, k1 is taken according to the formula below: k1  11,11 11 11  =0.524 , 22 Ne (1.065 10 ) 0, 22 Thus: Calculation bending tensile strength of AC binder course layer: Rttku  Rku k1.k2 = 2.0x0.524x1.0 = 1.048 (MPa) Calculation bending tensile strength of AC wearing course lay er: Rttku  Rku k1.k2 = 2.8x0.524x1.0 = 1.468 (MPa) ku Coefficient K cd = 0.94 (Table 3.7[1]) for road level III with realiability 0.9 + Check for AC binder course :  ku 0.815  Rttku 1.048  1.115 ( Mpa) � OK 0.94 k dcku + Check for wearing course:  ku 0.989  Rttku 1.468  1.562( Mpa) � OK 0.94 k dcku Conclusion : Asphalt concrete layers ensure the bended-tensile stress condition 32 LECTURER : NGUYEN DUC NGHIEM STUDENT : SU DUY LINH - 59CDE - 588859 © NATIONAL UNIVERSITY OF CIVIL ENGINEERING HIGHWAY AND TRAFFIC DEPARTMENT BRIDGES AND ROAD FACULTY ROADBED AND PAVEMENT DESIGN PROJECT General conclusion: To upgrade pavement for the next 10 years, we add AC wearing course 5cm and AC binder course 7cm CHAPTER : DESIGN DETAIL CULVERT 7.1 DESIGN CULVERT Terrain culvert is located at positions which alignment cut through stream, standing water slit When it rains, they make flow Structure culvert is located to drain water on the suface and slope of the road Culvert is perpendicular with center of the road Due to road bed width is quite large So, we will use diameter of culvert that is larger than or equal to 1m In order to maintenance is convenient Diameter of culvert is chosen with flow regulation is not pressure Set alignment is as perpendicular with flow as possible Shoulder must be larger than the highest level of water front of culvert 0.5m (for diameter �2m, no pressure) Embankment width above culvert is larger than or equal to 0.5 m or must be located pavement width if pavement width larger than 0.5m Using concrete pile culvert, due to it is cheap, convenient for mechanization Hydrology and hydraulic calculation basis is design flow rate with design flood frequency 1% 7.1.1 DETERMINE FLOW RATE Maximum flow volume flows drainage construction (bridge, culvert) with design flood frequency 1% is calculated by Viet Nam standard 22TCN 220-95 of Viet Nam Ministry of transportation: Q p  A pH p F Where: + Ap: Flood flow modulus with design frequency +  : flood flow coefficient (look up table 9-7) + Hp: daily rainfall (mm) with design frequency P%, (see appendix 15) + F: basin area km2 +  : volume reduction coefficient cause by pond, lake (determine table 9-5) 33 LECTURER : NGUYEN DUC NGHIEM STUDENT : SU DUY LINH - 59CDE - 588859 © NATIONAL UNIVERSITY OF CIVIL ENGINEERING HIGHWAY AND TRAFFIC DEPARTMENT BRIDGES AND ROAD FACULTY ROADBED AND PAVEMENT DESIGN PROJECT 7.1.1.1 Determine culvert waterway and culvert length Determine culvert waterway and hydrologic factors of culvert Using pile culvert have m diameters, sewer manhole is normal type – type The flow regulation is not pressure Look up appendix 16 – TKD F3, we can determine rising water level is H=1.93 m and flow velocity is V=3.55 m/s Determine drainage capacity of the culvert: Due to flow regulation of the culvert is not pressure; drainage capacity of culvert is calculated by formula: Qc  c  c 2g H  hc   Where: +  c : velocity coefficient when culvert is not pressure, make equal to 0,85 + c : pile culvert area (look up in figure 10-2 TKĐ F3) + hc: rising water level in culvert at narrow area Between hc and H: H hc  vc2 or hc=0,9hk g c2 + hk: critical depth + g: gravity acceleration, make equal to 9,81 m/s2 From the result of calculation, we have Q c>Qtt so, drainage capacity of culverts are satisfied Determine culvert length If we don’t consider grade of the culvert, the culvert length is determined as diagram: B 1:m Hd LECTURER : NGUYEN DUC NGHIEM STUDENT : SU DUY LINH - 59CDE - 588859 1:m Hn D Lc 34 © NATIONAL UNIVERSITY OF CIVIL ENGINEERING BRIDGES AND ROAD FACULTY HIGHWAY AND TRAFFIC DEPARTMENT ROADBED AND PAVEMENT DESIGN PROJECT Figure 8-1: Culvert Lc=B+2x(Hn-D)xm Where: B: road bed width, B=7 m Hn: road bed level, m, depending on cross section m: grade of slope D: culvert diameter, m Lc: Length of culvert, m Hd: rising water level at inlet, m The grade of culvert depend on cross section which is located culvert, I c=0,5%3% Determine length of reinforcement after culvert After culvert, the flow have high velocity, velocity increase about 1.5 times at after construction So, we have to design downstream of construction with flow velocity is V = 1,5Vc and at the end of the reinforcement must have skew line in order to resist scour Lgc b Figure 8-2: Reinforcement after culvert diagram h Hscour 0,5m t The length of reinforcement lgc after culvert is equal to three times culvert waterway and is calculated from out door  lgc = 3.b The foot of wall of resistance scour depth is calculated: hw  hscour + 0,5m Where: hxói : calculation scour depth 35 LECTURER : NGUYEN DUC NGHIEM STUDENT : SU DUY LINH - 59CDE - 588859 © NATIONAL UNIVERSITY OF CIVIL ENGINEERING BRIDGES AND ROAD FACULTY Hscour = 2H HIGHWAY AND TRAFFIC DEPARTMENT ROADBED AND PAVEMENT DESIGN PROJECT b b  2,5l gc With: b : waterway of culvert H : resing water level at inlet 36 LECTURER : NGUYEN DUC NGHIEM STUDENT : SU DUY LINH - 59CDE - 588859 © NATIONAL UNIVERSITY OF CIVIL ENGINEERING HIGHWAY AND TRAFFIC DEPARTMENT BRIDGES AND ROAD FACULTY ROADBED AND PAVEMENT DESIGN PROJECT The method of reinforcement after culvert: Base on flow velocity at the downstream of construction, and in order to execution of the work is simple, whole reinforcement after culverts have the same structure: as figure 8-3 Reinforcement width hgc = 25cm Upper layer uses rubble stone 15cm thickness Lower layer uses chipping 10cm thickness 15cm 10cm Figure 8-3: Structure of reinforcement 7.1.2 DETAIL DESIGN OF PILE CULVERT D 1,0 M 7.1.2.1 Data to design: The calculated discharge Qtt=1.427 (m3/s) Culvert type : reinforced concrete pipe culvert  = 1,0 m Number of culverts : Valley’s area: F=0,027 Km2 Main stream lenght: L=0.15 Km Side gradient : isd= 194.7 % Roughness coefficient of stream : mls=7 Roughness coefficient of valley : mlv=0.2 Consider the cross-section of stream as triangle: Left side slope 1:12 Right side slope 1:12 7.1.2.2 Culvert design procedures a Determine the natural water flow depth of the stream h: The discharge determination formula of Sedi-Pavlopski: 37 LECTURER : NGUYEN DUC NGHIEM STUDENT : SU DUY LINH - 59CDE - 588859 © NATIONAL UNIVERSITY OF CIVIL ENGINEERING HIGHWAY AND TRAFFIC DEPARTMENT BRIDGES AND ROAD FACULTY ROADBED AND PAVEMENT DESIGN PROJECT 1 n   y Q =   R R.i  Where : R – Hydraulic radius R= : Water flow section : Wettedg perimeter y – The coefficient in formula, it depends on y= f(n,R) 1 4 1 6 y=    base on hydraulic radius Take y= n – Roughness coefficient of the stream n=7 i – Stream bed slope, i=0,016 The depth of water flow on stream bed are supposed as : 0,1; 0,245; 0,3m; …and with each water depths we can calculate the stream discharge use formula SediPavlopski above The result of discharge calculations are presented in this table : h (m) h1 =0,1 m h2 =0,245 m h3 =0,3 m (m ) 0,18 1,08 1,62 (m) 3,61 8,83 10,82 R(m) 0,049 0,122 0,150 Q(m /s) 0,053 0,722 1,306 Use the interpolation method we have the water flow depth of stream in nature h=0,2489 m b Inlet flow velocity: The formula to determine: VCV= Q r With the entrance loss coefficient : VCV= 4.Q   d The number of culvert is reinforced pipe culvert  = 1,0 m so the flow discharge for culvert is Qc = 1.427 (m3/s) 38 LECTURER : NGUYEN DUC NGHIEM STUDENT : SU DUY LINH - 59CDE - 588859 © NATIONAL UNIVERSITY OF CIVIL ENGINEERING HIGHWAY AND TRAFFIC DEPARTMENT BRIDGES AND ROAD FACULTY ROADBED AND PAVEMENT DESIGN PROJECT The outlet velocity: VCV= 4.Q 4.1.427  = 2.8 (m/s)   d 0,65.3,14.12 Vcv=2.8 (m/s) < V0xả = (m/s) (V0xả of pipe concrete) c Determine the critical depth hk: The critical depth ( hk ) is the depth of a cross section where the flow energy is minimum We have the ratio hk depends on the ratio d Qc2 1.472   g d 9,81.1 0.208 Refer the table 10-3 in High way design III we have: Qc2 0,208 h g d  k = 0,66 hk= 0,66m d So h=0,2489 m < 1,3.hk= 1,3.0,66 =0,858 m (the water flows inside the culvert with free surface) d The critical gradient ik: Consider the pipe culvert, the ik can be calculated as follow : ik = Qc Kk Where : Kk = kCk Rk characteristic flow discharge, refer table 10.3 in Highway design with Qc2/g.d5 ; d is the diameter of the culvert Qc2 1.472   K 9,81.1 0,208 refer the table : k =0,83 We have g.d Kd then Kd =24.d8/3=24.18/3=24 So : Kk = 0,83.24 =19.92 39 LECTURER : NGUYEN DUC NGHIEM STUDENT : SU DUY LINH - 59CDE - 588859 © NATIONAL UNIVERSITY OF CIVIL ENGINEERING HIGHWAY AND TRAFFIC DEPARTMENT BRIDGES AND ROAD FACULTY ik = Qc Kk ROADBED AND PAVEMENT DESIGN PROJECT 1.472  = 0,005 19.92 So the critical gradient ik =0,5% We choose the gradient to lay the culvert ic = 5% e Determine the headwater H With the flow discharge Q =1.472 (m/s) we choose the unpressured culvert Because of the culvert gradient ic so we can know the drainage capability of the unpressured culvert as follows : Qc =c.c 2.g  H  hc  Where: + c: velocity coefficient in unpressured condition, c =0,85 + c: water flow section at narrow place + hc: water flow depth in culvert at narrow place, usually take hc=0,9 h k + hk: the critical depth hk = 0,69 m + g: acceleration due to gravity, assume g=9,81 (m/ss) We have: hc= 0.9.0.66= 0.594m Refer Highway design III : H  2*hc= 2*0,594 =1.188 (m) H =1.188  1,2*hcv=1,2*1 =1,2 m So the unpressured work condition is satisfied f Determine the Tail-water h0: We have the ratio h0 K depends on the ratio d d Where: K0, Kd : the characteristic flow discharge K0 = Qc i  1.472 6.58 0.05 Kd =24.d8/3=24.18/3=24 40 LECTURER : NGUYEN DUC NGHIEM STUDENT : SU DUY LINH - 59CDE - 588859 © NATIONAL UNIVERSITY OF CIVIL ENGINEERING HIGHWAY AND TRAFFIC DEPARTMENT BRIDGES AND ROAD FACULTY ROADBED AND PAVEMENT DESIGN PROJECT Refer table 10-3, Highway design No.3, we have: K0 h  0, 274  = 0,362 h0=0,362 m Kd d g Outlet flow velocity V0: Formula to define: V0 =W0 i Where: W0 is characteristic velocity Refer table 10-3 Highway design III we have: W K0  0, 274  = 0,831 W0=0,831*30,5.12/3=25.34 Kd Wd with Wd=30,5.d2/3 =30,5.12/3= 30,5 so V0 =25.34 * 0, 05 =5.67 (m/s) < V0xãi = (m/s) h Culvert outlet protection In the free flow condition case, the culvert outlet velocities are high They will be 1,5 times compare with the velocity of the flow in culvert, we have the outlet velocities: So Vgia cè=1,5.V=1,5x5.67 =8.5 (m/s) Refer the appendix of Highway design III with V= 8.5 (m/s) and h=0.362 (m) with the treated material by cement mortar masonry with V0xãi =5.67 (m/s) The length of the outlet protection lgc  (34)d, choose lgc =3.d =3*1 =3(m) Determine the scour depth at the treated part of downstream by the formula: hx(1)=2.H b b  2,5l gc (1) where: + Headwater: H =0,9(m) + Construction width, b=1 (m) + lgc the length of protection, lgc =3 (m) We have: hx(1)=2.0,9 =0,61 m  2,5.3 So the scour depth is hx(1)= 0,61 m 41 LECTURER : NGUYEN DUC NGHIEM STUDENT : SU DUY LINH - 59CDE - 588859 © NATIONAL UNIVERSITY OF CIVIL ENGINEERING BRIDGES AND ROAD FACULTY HIGHWAY AND TRAFFIC DEPARTMENT ROADBED AND PAVEMENT DESIGN PROJECT 7.2 DETAIL OF SIDE DITCH 7.2.1 PRINCIPLE AND FACTORS DESIGN Side ditch is located at embankment that its high is smaller than 0.6m, at excavation, one part is embankment and other is excavation Side ditch dimension is designed by structure, it doesn’t need hydraulic calculation Ditch section is trapezoidal section The bottom edge of ditch width is 0.4 m, the depth of ditch is 0.4m from natural ground, slope of ditch is 1:1 The grade of side ditch is designed to not only ensure drainage capacity but also it doesn’t accumulate silt The grade of side ditch is usually the same the grade of red line At positions which their grade of side ditch is large, scour of bottom ditch phenomenon is occur, so we must have suitable reinforcement methods Water at embankment doesn’t flow to excavation (unless length of excavation