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Highway Design Engineering: Part 2 - Extra Width (Widening) Calculating Safety and Geometry etn oe lle (W,) Part 1 - Stopping Sight Distance (SSD) & Setback Se, widen Calculating Inner Lane Radius (R’) "Wo R’ = 200 - (3.5 / 2) = 198.25m /~ Mechanical Widening (W,,) = 0.5m pm y2 SSD = 0.278Vit + = 254(f +h) = : V = speed Psychological Widening vs. Speed t= reaction time f = friction Calculated SSD = 125m Based on a design speed (V) of 60 km/hr, a reaction time (£) of 2.4 seconds, and a friction Who, = 0.89 m Woy = 1.019 m i coefficient (/) of 0.355. - n - x Total Widening Required at Different Speeds (50m Radius) Parameter hr 80 km/hr ek ae Mechanical Widening (W,,) 05m 0.5m The central angle (a) is calculated Psychological Widening (W,.y) 0.89 m 1.019 m as 38.12", leading to a horizontal chord distance (x) of 188.48m. Engineering Analysis: Hydrostatic Forces on Dam Sector Gates Gate Geometry and Dimensions Radius R = 5m Angle @ = 60° Height y = 25m Vertical |) Offeet (y) 5sin(30°) =2.5m Horizontal Force Calculation (P,,) -——$—— Unit Length (L) = 1m Horizontal Pressure Formula: i Py=yA h ‘ C.6. y = specific weight (10) A= Projected Area (2y x 1 = 5m?) Key Finding: = depth to centroid (2y/2= 2.5m) hep >A (CP is lower Total Horizontal Force: than C.G.) 10x 52.5 = 125 kN Sector Area (Asector): TR?x 6 360 = 13.08 m? Vertical Force and ‘Imaginary Fluid’ (Fy) Fy is equivalent to the weight of the ‘imaginary fluid’ above the curved surface Imaginary Fluid a x s a i ie-——_—> x= 5c0s(30°) Triangle Area (Atiangle): Net Area (Ane1): 2x [0.5 x (Scos30°) x 2.5] Asector ~ Atriangle = 10.82? = 2.26 m? The Mechanics of Resistance: Practical Applications of Fluid Dynamics Fundamentals of Drag & Lift The Drag Force Equation Lift vs. Weight Drag Fiald Coefficient Density F, (Lift) Fo =3CppAV? ‘D ay DP. : Aerial & Aerospace Engineering Bie ACheactat canes Saar W (Weight) Parachute Diameter Calculation Aircraft Wing Surface Area Cian Ge Unoet eat ) ttt a, a (2) {r- Drag Coefficient (C,):1.34 Automotive & Industrial Efficiency | Power Requirements for Speed Wind Loads on Structures » Calculated Diameter: 3 uf Goaficent (G):04 Projected Area; Cy Traveling at: 4 ° 7.62 meters eso 0 2.6m? 0.3 120 km/h = 0) " 59 Mm = Drag Force: Wind tos 520N jn Rending Mement for Tal Requires: Structures (c.g., 20m 23.58 high) is calculated as Horaepower (HP) rag Force n (Height /2) to overcome air resistance Hydrodynamics & Particle Motio! 'yarody ——? Determining Flow Regimes 2 Laminar Turbulent Drag coefficients (C,) change with Revnolde number (Rc): Laminar uses S = 1.328/Ro; Turbulent uses C, = 0.672/(Rc)*? Stoke's Law & Rising Spheres in Viscous Liquids Sand Particles 6mm solid sphere Rising velocity: 1 em/sec Liquid Density: 900 kg/m? © ___ Falling Velocity: Viscosity: 0.7 20 cm/sec Net Torque results in 0) Rotational Speed: : 232.7 rpm Wind Speed: 60 km/h Drag Coefficients (C,): Conceve side (1.33) ve. < Convex side (0.24) * = bei Calculated Specific Weight: Critical Diameter of sand grain: 4,2 5.329 kg/m? 0.1mm Engineering Hydrology & Irrigation: Core Principles and Calculations Irrigation & Water Balance Infiltration Modeling (Horton's Equation) y Catchment and > | initial capacity f, Horton's Infiltration Formula: Effective Rainfall (ER) over Hyd rograph Geometry z ce a a 3 hours (f, = 6.8 + 8.72") 3 f,(t)= f+ ff.) @ = 2nd hr: | 4 = a 11.18 mm c| @ Fe = a] / é Constant eapacity f, 8 Be al Y 3 = | oat 5 7 to t Time = ; : Total Infiltration Loss Calculation Catchment Area ! Total loss is calculated by integrating the [Yt o> Horton equation over the duration of the Total ER: i Te ty Storm (t tot) eam Volume = nee is Rainfall () Unit Hydrograph (U.H.O) Principles ——— . DulGaicl Cah T See The Time Invariance — Linear Response (Superposition) E Height = 30 cumecs 3 Base = 18 hours aie la a ld ' a iA Base = 18 hours | Time Time | Rainfall-runoff relationship for a given 1-hour Unit Duplicate: Unit Required 2-hour catchment does not change over time Hydrograph Hydrograph Unit Hydrograph Seepage Loss Calculation: Hydrograph Coordination (Conversion Example) A 4.6 x 104 m? . I \ (Equivalent to 4.6 ha-m) ~ 6hours 6 hours Superposition for 12-hour DRH Adding coordinates of two 6-hour unit hydrographs lagged by 6 hours yields a 12-hour Dhect Runoff Hydrograph (ORH) with a peak of 225 cumecs Based on the water balance equation of the irrigation canculation in the volume of irrigation capacity. Engineering Gravity Dams: Principles of Stability and Design Applying Varignon's Theorem Eccentricity (e) of resultant force Must not exceed one-siath of the base width to prevent tension CALCULATING BASE WIDTH (B) Base Width for No-Tension Criterion LIMITING HEIGHT (Hyin) B H His dam height H ibs = — __ Gis specific gravity = ee TY VG—C Cisuplift coefficient w(G—C +1) Sliding vs. Uplift Criteria Required base width is the greater value between sliding criterion (B = H/u(G - C)) and no-tension criterion. THE EFFECT OF “NO UPLIFT” (C = 0) where f, is allowable principal stress (e.g, 7 N/mm?) J Height = Him LOW DAM HIGH DAM Low Dam if height < Him (approx. 209.8m); at the heel. pat otherwise, itis a High Dam. . = ; ve PRINCIPAL STRESS AT THE TOE (0) APPLYING VARIGNON'S THEOREM Ow vs. Uvs. Wa / o = P, sec? a Igebraic sum of all moments about aa ass the “middle one-third” must equal ‘“ —_— Stress at toe is function of the moment of the resultant force (R) If built on hard rock with no uplift pressure, vertical pressure and about that same point. formula simplifies to B = YG, significantly eometry, ensuring material UPLIFT PRESSURE AND DRAINAGE reducing required base width. IMPACT OF DRAINAGE GALLERIES: loes not crush under load. PARAMETERS FOR STABILITY EXAMPLE ALCULA L Incorporating a drainage gallery reduces. § “ ee rele Hida = \ tied uplift pressure (U,) at f allery location, Specie Ger fOueCnizels Ca eee Senate total \ significantly lowering total uplift force. Uplift Coefficient (C): ie th aa Lesulnltens 4) SU PERMISSIBLE FORCE LIMITS: in the | Height of Dam 4) 30m rovided case study, total calculated uplift | | Coefficient of Fricti 07 force of 10,500 kN was verified to be | ae a = a oy TN within permissible run-for-dam limits. | wable Principal Stress (F,): rm The Mechanics of Resistance: Practical Applications of Fluid Dynamics Fundamentals of Drag & Lift The Drag Force Equation Lift vs. Weight Drag Fiald Coefficient Density F, (Lift) Fo =3CppAV? ‘D ay DP. : Aerial & Aerospace Engineering Bie ACheactat canes Saar W (Weight) Parachute Diameter Calculation Aircraft Wing Surface Area Cian Ge Unoet eat ) ttt a, a (2) {r- Drag Coefficient (C,):1.34 Automotive & Industrial Efficiency | Power Requirements for Speed Wind Loads on Structures » Calculated Diameter: 3 uf Goaficent (G):04 Projected Area; Cy Traveling at: 4 ° 7.62 meters eso 0 2.6m? 0.3 120 km/h = 0) " 59 Mm = Drag Force: Wind tos 520N jn Rending Mement for Tal Requires: Structures (c.g., 20m 23.58 high) is calculated as Horaepower (HP) rag Force n (Height /2) to overcome air resistance Hydrodynamics & Particle Motio! 'yarody ——? Determining Flow Regimes 2 Laminar Turbulent Drag coefficients (C,) change with Revnolde number (Rc): Laminar uses S = 1.328/Ro; Turbulent uses C, = 0.672/(Rc)*? Stoke's Law & Rising Spheres in Viscous Liquids Sand Particles 6mm solid sphere Rising velocity: 1 em/sec Liquid Density: 900 kg/m? © ___ Falling Velocity: Viscosity: 0.7 20 cm/sec Net Torque results in 0) Rotational Speed: : 232.7 rpm Wind Speed: 60 km/h Drag Coefficients (C,): Conceve side (1.33) ve. < Convex side (0.24) * = bei Calculated Specific Weight: Critical Diameter of sand grain: 4,2 5.329 kg/m? 0.1mm Indian Railways: Historical Beginnings & Modern Rail Welding Technologies ia ——— The Birth of Indian Railways = Gp April 16, 1853 © Bombay to Thana Route CWR: Continuous Welded Rail | First railway in India oe Inaugural journey covered a distance of Meds ean Fea ara bs commenced operations. 21 miles, equivalent to 33.6 kilometers. 9 ‘switch expansion joints. a) jit . LWR: Long Welded Rail ZA These spans range from 200m to 1000m for Broad Gauge and 300m to 1000m for Meter Gauge. Welding Methods and Engineering Specs Thermit Welding (Chemical) Achemical welding process performed at the site, primarily for Long Welded Rails. The Thermit Reaction SWR: Short Welded Rail to produce Created by welding 3 rails together, + (zai) > (uo) Fe standard lengths are 3 x 13m for SS a Broad Gauge (BG) and 3 x 12m for Fe,03 2al Fe Meter Gauge (MG). (Iron Gxide) (Aluminum) deposited Fe Broad Gauge | Mater Gauge G 6) MG) ali i icati pall ype if 2h Hs wy Specialized Welding Applications ©@ swr 3x12m 32m 2 Ge) I> @ wr 200m to 300m to Expansion Gap Maintenance 1000m =~ 1000m Electric Arc Flash Butt Oxyacetylene To manage thermal expension. a liner =] CWR >1000m >1000m used forpoints for workshops for site cutting _—_with a specific thickness of mm is and crossings and changing used to maintain the expansion gap. Principles of Engineering Hydrology: Flood Routing & Statistical Analysis THE MUSKINGUM METHOD (FLOOD ROUTING) Iterative Flow Calculation Q2 = Cy + Colg + CQ) Outflow evolves from initial 10 cumec toa calculated peak of 74.33 cumec. = | | Se Whrs Whrs 24hrs Calculation of Coefficients CG =0.55 G=0.40 Co = 0,05 (=1-¢,-C) HYDROLOGIC RISK & RELIABILITY n R=1-[1-(/T)| Probability of “At Least Once” ee Occurrence <> Risk Analysis for a Barrage Lifetime n = 60 years, Design Flood T= 100 years Risk of Failure/Occurrence = 45.3% HYDROGRAPH CHARACTERISTICS Attenuation of Peak Flow | 100 (Inflow Peak) - 74.33 (Outflow Peak) = 25.67 cumec 6 hours Outflow Peak: 74.33 cumec (24 hrs - 18 hrs) Outflow Hydrograph Lag Time: Discharge 18 hrs 24 hrs Time Outflow curve is “flatter” and “shifted right" compared to the sharper, earlier inflow curve, GUMBEL’S DISTRIBUTION FOR PEAK FLOW Estimating Peak Discharge (Q,) Q = Q + Ko Gis mean flow ‘Pp K is frequency factor ais standard deviation Frequency Factor (K) derived from Gumbel's reduced variate, involving sample parameters y,, and S,, Calculated Peak Flow Example: Using a 90% non-occurrence probability over 100 years (T= 949.6 years), the predicted peak flow is calculated to be 1447.16 cumec. Engineering the Summit Curve: A Step-by-Step Highway Design Guide Establishing Design Parameters Safety & Length Calculations Geometric Offsets and Coordinates VPI (Vertical Point Step 1: Calculate Stopping Sight Distance (SSD) Tangent vs. Curve Offsets ' . of Intersection) ‘ F v2 Vertical difference between the VPC (Vertical Point — - Grades Meet VPT (Vertical Point SSD = 0,278VE+ peafemi Required theoretical meeting point of the of Curvature) of Tangency) Safe Stopping grades and actual curve. - Curve Begins - Curve Ends Distance = Curve Offset (y2) y = =3,31m hy if = 663m (i Step 3: Locate the VPC Coordinate VPC Horizontal Position = 1000m VPC Horizontal Coordinate (x) = 797.5m Step 2: Determine Summit Curve ; Length (L) +3%(0)) a a aS See Final VPC Elevation © Initial Design Constraints Grade Deviations Design Velocity (V) = 100 km/hr 7 Reaction Time (t) ; VPC Vertical (4 = 2.5 seconds (| f Coordinate (y) = 100m a Friction Coefficient | 93,925 m (f) = 0.35 Fluid Mechanics: Engineering Problem-Solving Guide Illustrating the application of core fluid dynamics principles—momentum, energy, and mass conservation—through solved engineering scenarios. Pipe Transitions & Momentum Calculating Force on Pipe Bolts Using Newton's 2nd Law of Motion (32 F =m x a), we determine the reaction force (R,) required to hold a owtale in place as water transitions from a 200mm pipe to a 100mm jet. ~~" 1.685 KN Reaction Force The final calculated force seting on the bolts (R,) is 1.685 kN, derived after calculating an inlet pressure (P,) of 84.07 kPa, 200mm Pipe Bernoulll’s Principle Application To find the internal pressure, the system applies the Bernoulll equation, accounting for velocity changes and head losses (0.1 x v2 /2g]. Open Channel Flow Dynamics Critical Depth and Energy Ina channel with a 3m width and 8 cumeo flow, specific energy (é) is used to determine haw the water reacts to obstacles lke humps. 0.202m & epth Maximum Hump Height Concitions For the given flow conditions, the maximum allowable hump height (Ap) to maiuitain the upstream depth is calculated as 0.202 meters. Humps vs. Contractions. The analysis explores two scenarios: a physical homp in the floor (Az) anda channel contraction (reducing width B,) to reach critical flow conditions, Channel > Contraction (B,) Physical Hump (Az) Jet Impact & Equilibrium Supporting Weight with Fluid Power A vertical water jet with a 75mm diameter and 12 m/s velocity is used to suspend a 40N object in mid-air. AON Object Q Bes v d % Momentum Flux Balance L ei is reached when ie upward force of the water jet (momentum change) exactly matches the downward weight of the object. 12 m/s A=731m 7.31 meters —samanve Water Jet Equilibrium Height Based on the deceleration of the jet due to gravity, the object will reach a stable floating height (n) of 7:31 meters. Tank Drainage (Unsteady Flow) Time-Dependent Discharge Calculating the time to empty a tank involves integration (0 x dt = dV) because the flow rate decreases as the water level drops. 3.49 tae Minutes to Drain Orifice Diameter 100m (im) ‘A.2m diameter tank discharging through Blseharge a10am orffice (8, © 0.92) takes exactly Coefficient (,) 209.92 seconds to reach the target ‘Total Time level. Key Variables 0.63 3.49 minutes, 
