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Mastering Surveying Calculations: A Field Note Guide TAC! METRIC SURVEYING Tacheometer Constants: Multiplying (fA) = 100; Additive (f+d) = 0.1m Horizontal Distance (D) = (is cos?6 + (f+d)cos® ————— COMPASS SURVEYING & POLYGON GEOMETRY Solving Inaccessible Distance (QR): Using Law of Cosines (cos @ = (a? + b?- c*)/2ab) with Dpg = 32.1m, Dpg = 21.1m, and angle = 61° 30°30". Distance QR = 78m A B Interior Angles (Regular Hexagon): Sum = (2n — 4) x 90° F ¢ = 720° (6 sides); Each Angle = 120°. 3 D B A oe Bearing AB= 10° B Cc Ww C Determining Fore Bearing (F8) of BC: Based on 120° Interior angle, FB of BC = 70° (90°- 20°). Ss TRIANGULATION & LOCAL ATTRACTION P A ae are Well-conditioned Ill-conditioned (Stations P, S): (Stations Q, R): Angles between Angles outside 30° & 120° 30°-120° range. ® Identifying Local Attraction: Difference between Fore Bearing (es and Back Beary BB) is NOT exactly Stations Affected yy Interference: R (diff 182°), § (Giff 183°), (PQ, Tunaffected) RECIPROCAL LEVELING ar=1.35m b: 55m Overcoming Obstacles: Finds true elevation difference across obstacles like rivers, eliminating curvature & refraction arrors. True Difference in Level () qe a) (be 8 Sas Staff Readings (50m wide river): Case (Inst. at A): | a1=1.35m | br=1.55m_ Apparent Diff=8.20m Case Il (inst. at B); | ay=B.55m | by=0.75m ‘Apparent Diff=0.20m Core Principles of Geotechnical Engineering: From Site Selection to Stability Analysis EARTHWORK ECONOMICS & SITE SELECTION Borrow Plt 1 KEY FINDING: Constant Volume of Solids (V, = 342.8 x 10° m2) Most Economical: Borrow Pit 2 Despite higher void ratio (e,=1.7), lower overall cost. Borrow Pit 2 Required Embankment: 600,000 m° (e = 0.75) Unit Cost: 5/m? SLOPE STABILITY ANALYSIS Circular Slip Surface FOS = 0.867 Infinite Slope with Seepage FOS = 1.48 For a 5m thick clay layer on a 15° slope with seepage; uses saturated unit weights and pore water pressure effects Unstable: FOS < 1.0 (Likely to fail) SOIL STRENGTH & LABORATORY TESTING: SKEMPTON'S THEORY UNCONFINED COMPRESSION a, Skempton's Undrained Shear Unconfined Le Strength (C,) = 39.78 kPa i Compression % Calculated using effective stress (0) iS Test (UCT) and Plasticity indes (LL - PL) = | q,, = 23 kPa i : Calculating Excess Pore = Water Pressure (AU,) CLTTTTLSTISSSTTA. Me cctaits Av, fon pea Determining Shear Strength (A=0.6, B=0.8) embankment raising Parameters (C & ) Vl Cohesion (C) Angle of Internal Factors in Cohesion (50 kPa), effective stress, and internal friction angle (20°) Friction (ip) = 10° SEEPAGE & FLOW NET ANALYSIS Sheet Pile Hydraulic Calculating Seepage Discharge Structure (d,,9p) = 0.3825 m*/day per meter run Using a flow net with 4 flow channels (N,) and 8 head drops (Nz) Piping Fallure Protection (FOS = 2.6) Calculated using Specific Gravit (G,=2.7) and Void Ratio (e=0.83) Us I -— head loss per drop (1.06251 f Bu [ pressure head at P is 13.873m, Mastering Traffic Engineering: Signal Timing and Flow Calculations Core Traffic Signal Parameters Traffic Flow Inputs (IIT Kharagpur Problem) The Webster's Cycle Time Formula Direction Example Calculation 1 7 Saturation Fl a (IISc Bangalore) poumy ow | 2500 | 2500 | 3000 | 3000 Total Lost Time L = 10s; Flow ratios: y,=0.3, y.=0.25, Flow Ratio (y) y=0.25. Resulting Cycle Time Cis exactly Flow Ratio (y) Total Lost Time (L) $100 seconds. Calculated as the ratio of actual Represents time not used by any traffic flow (q) to the saturation phase (e.g., | = 12s for 2-phase, _ N-S flow ratio = 1000/2500 = 0.4). or 4s per phase in multi-phase). Resulting Cycle Ranges b Typical cared cycle lengths . range from $80 seconds to $100 Calculate Optimum fe Cycle Length (C) ak Saturation Flow Maximum hourly rate vehicles pass Ge: through an intersection; typically 1800 to 3000 PCU/hr per lane. Green Time Distribution G=[(C-L)xul/dXy Against Traffic (q2) Multi-Phase Results (for 100s Cycle) SS 92= 9+ kV, Allocating Green Time (G): B B Traffic Density (k) Calculated Density Green time is directly proportional to the Defined as vehicles per Solving simultaneous fiow flow ratio (y) of that phase; higher traffic Phase 1 (y=0.3): Phase 2 (y=0.25): Phase 3 (y=0.25): kilometer (vehvkm), representing equations yields a density (k) of volume = longer green duration. 33.75s 28.125s 28.125s concentration, 3 veh/km (IISc Bangalore case). Environmental Engineering: Key Calculations and Atmospheric Stability Electrostatic Precipitators (ESP) n=1-e cA Electrostatic Precipitators remove Suspended Particulate Matter (SPM) larger than 1 micron from hot gases using the efficiency formula. Calculating the Environmental ELR vs. Lapse Rate (ELR) > Based on a temperature drop from 15.5°C at 10m to 14.5°C at 110m, the ELR is determined to. MSW per day Every 120g of Municipal Solid Waste produces 60g of Greenhouse Gases, comprised of 20g of Methane (CH,) and 40g of Carbon Dioxide (CO,). Ay = 6100 Nm? Relationship of ‘c' and ‘A’: The effici the collection area (A) and the precipitation rate constant (c). Efficiency vs. Surface Area Calculation 96% Efficiency Atmospheric Lapse Rates & Stability Adiabatic Lapse Rate (ALR) 1o°c/km = 9.8°C/km = 6.56°C/km be 10°C/km. Calculated Dry ELR ALR Total Daily GHG Production $500 tonnes $250 tonnes Greenhouse Gasesdaily 98% Efficiency Az = 7413 Nm? icy of the ESP is an exponential function of Super-Adiabatic and Unstable Conditions: Because ELR > ALR (lorc > 9.8°C), the atmosphere is in a ‘Super-Adiabatic’ state, leading to unstable environmental conditions, 6° Wet ALR $250g Per Capita GHG Emission For a population of 1 million people, the per capita Greenhouse Gas emission is calculated at 250g per person per day. Traffic Flow Engineering: Solving the Greenshields Linear Model FUNDAMENTAL FORMULAS SS Speed-Density Relationship: V = u; x (1- Ky (%) —— rl Flow Equation: qzkxv P==4 = fy* Headway Conversion: H, = 3600/q = Ga Gn PROBLEM 1: |..Sc BANGALORE (GATE) PROBLEM 2: |.I.T BOMBAY MODEL Input Parameters lit __ u,=8Okm/hr — (Free-flow speed) k, = 100 veh/km (Jam density) Given Linear Equation U = 70 - 0.7K Where U is speed, K is density Calculating Critical Speed & Density Deriving Maximum Flow Ge] dq/dk = 0 for optimal densit Cr " a, est, Set = 0 for optimal density 7 ue uf2 = 40 km/hr dk (Gay —- Gai —- lain —- eli ke kf2 =50 veh/km : Flow (q) Optimal Density (K) = 50 veh/km 2000 1750 ~2.1 t veh/hr sec/vehicle veh/hr sec/vehicle MAXIMUM FLOW (qx) TIME HEADWAY MAXIMUM FLOW (dmax) TIME HEADWAY max = (Uy * ky) 14 derived from 1/ Qmax Capacity found using K= 50 Calculated as 3600/1750 in flow equation Highway Traffic Engineering: Sal Calculating Road Capacity via Stopping Sight Distance This infographic outlines the fundamental calculations used to determine traffic flow (q), beginning cae . . with Stopping Sight Distance (SSD), then determining vehicle spacing (S), and finally calculating Step 2 a Determining Vehicle Spacing (S) the theoretical road capacity in vehicles per hour. Step 1 - Calculating Stopping Sight Distance (SSD) REACTION DISTANCE BRAKING DISTANCE Vehicle moving at constant Vehicle braking to a halt t=2.5 speed during driver reaction time f=04 seconds V=65 km/h Distance = 0.278 * V*t =a itcn \eeeelie Reaction Time (t): 2.5 seconds Center-to-Center Spacing (S) is the sum of vehicle Vehicle Length ie length and the safe gap. assumed to be 5 meters S (L/2) + SSD + (L/2) (5/2) + 86.76 + (5/2) Friction Coefficient (F}: 0.4 Total Spacing = 91.76 Meters = Adding the 5-meter vehicle length to the SSD Total SSD = 86.76 Meters results in the total center-to-center spacing. The required safe stopping distance based on the variables. Step 3 - Calculating Total Traffic Flow (q) Conversion to hourly volume < ay Capacity ~ 710 Vehicles/Hour q = u * k u= Speed 65 x 1000 The final calculated flow is 708.37, rounded to k = Density (1000/5) 91.76 approximately 710 vehicles per hour per lane. Structural Mechanics: Bea alysis and Weld Capacity Guide PHASE 1-4: BEAM ANALYSIS PHASE 3: STRESS ANALYSIS & DISTRIBUTION ———— Bending Stress (o) at Fibers Shear Stress (t) at Fibers Moment of Inertia ul , S0x8 Toll ye=7.73x10mm* total Shear Capacity .*, 393 KN Bottom Fiber (Point D): ls = (t+ Zz x6). 7.73 x 10° mm4 a= 80 MPa (Tensile) 4 =0.207s 1 f H 4 ey Top Fiber (Point A): a 1 2 Simple Beam w=30kN/m Hl 0 = 53.33 MPa E i (Compressive) ' Maximum Shear Stress | i rh 7 u i \ L=6m S ! Fr x i Bottom Fiber (Point D): =90k R=90KN Section Geomet ! o=50 MPa Revue Step 1: Support Reaction (R) is sind piblaal ' (Tensile) Re Me. 306 - 99 kN = lle ike ae tober PHASE 4: COMBINED STRESSES & MOHR’S CIRCLE Ie? gy = qq = 225 * 10 mmé | + FG ! | Principal Stress Equation: 6, 2 = “Ss atts File.- 6 oy)? +47? | f Step 2: Shear Force (V,,) Step 3: mae! Moment (M,,) ' Point B Stress State (at Neutral Axis) Point D Stress State (at Bottom Fiber) Vg, = R Wx = 90 - (30x 2) = 30 KN My. = Rx - # C een oa t +1.5 MPa i ee 1.5, ; = 1.5. MP; rae ee C (15,0) 4, a(T) ) = "1 ' 9, =-1.5 MPa(C) 0, = -1.5MPa (T) a =0 (80, 0: hx i aos : i (0,-1.5) Bi Zero Shear, Max Tensile Stress: Bending Stress —_Top Fiber (Point A): ; i Principal Stresses: ¢ 1.5 MPa Principal Stress = Bending Stress Bending Stress (0) o = $3.33 MPa (Compressive) ' 5 ' PHASE 5: WELD CAPACITY ANALYSIS ' Weld Design P: tel i i ANAL Step 5: Composite Moment af inertia ip Hs arameters : Final shear Capacity ) Total |, = 7.73 x 108 mm4 ' gs Step 5: Composite Permissible Shear Stress = 100 MPa i f 1 ' 1 1 Geotechnical Stability and Stress Analysis: Comprehensive Base Pressure Structural Calculation Guide Ashes pistiburicn Calculating the Sliding Resistance Deriving Shear Stress from Principal Stress je- MN 4 Se =H =100 MN c= << Base Width (B) = 140m = ant Resisting Moment (Mj) = 2 x 10° kNm; Overturning Moment (M,) = 1* 10° kNm Heel Pressure (Phcoi) xBtulVv Toe Pressure (Proe) = -408.16 kPa srFF= H = 1122.45 kN/m? (Tension) 5 ghar each 5 Calculated Toe Shear Stress yiltueee ceefltgcapiti en a Nin? = 1.879 MPa Principal vs, Shear Stress at the Toe Calculated Shear Factor of Safety Calculated via the formula a ey ote alt Calculated safety factor exceeds the 3.5 threshold, = 9% tand_ providing specific ‘= r i ‘ indicating a stable design against silding, ~ 1+ tan2a ' intensity at the toe. Using the vertical pressure at the toe. Kinetics of Disinfection and Sludge Characterization in Environmental Engineering Chick's Law & Disinfection Kinetics Sludge Physical Properties Theoretical Formulas Chick's Law of Bacterial Kill The rate of bacterial kill is directly proportional to the number of bacteria remaining at any given time. Sludge Composition by Weight: 98% Water, 2% Solids (Specific Gravity G. = 2.2) . il Ki Reaction Constant (K): = -Kt = e N(t) = No *e K=3x 102 I Combined Specific | per second alll Gravity (Greombinea) © E | Gycombinea) = 1.011 Contact Time & Efficiency (Based on weighted percentages) (™) (‘remaining Unit Weight of Sludge (¥jsuag0)) Visiudge) = Gisiuagey X Yw Determining Contact Time (t) x “Tot i ht t=1-in__No K No - Nxxitles)) Theoretical Formulas for Multi-Component Sludge Theoretical Volume and Discharge Theoretical Specific Gravity (Gieq)) Limit Relationship To reach 100% G(eq) = ty 00 Calculating Remaining Bacteria: kill rate Volume (L x B x H) q r(P;/G) N(Remaining) = Ny * e°Kt aaa Ne) an is equivalent to : i ; =No Infinity duration H il Where P, = Percentage of each component, is required Discharge (Q) xTime 1) Gj= ae gravity of each component Problem 32: Pile Foundation Capacity in Dense Sand Ultimate Capacity Calculation (Q,) Qu=Qp+Q, Since cohesion C = 0 (sandy soil), total ultimate load = 1643.87 kN Factor of Safety (FOS) = 2.5 Qsate = Qu / FOS Final Safe Load = 657.55 kN 5 Critical Depth, L, = 15x D= 4.5m Dense Sand Friction Angle, d = 40° Lateral Earth Pressure, K = 2 — Bearing Capacity Factor, Nj = 137 Mastering Geotechnical Engineering: Pile Capacity and Soil Compaction “a Problem 33: Vibro-Flotation & Soil Compaction o 8 e Loose Sand = 0 W=70kN @ sa 0 (os oe Initial Unit Weight y= 14 kN/m3 Initial Soil Profile (Before Compaction) Final Settlement & Thickness Reduction Total Reduction, AH = H - H" = 5m - 4.119m = 0.881m Target Relative Density (RD) Target RD = 0.85 (85%) for dense state Calculating Max Dry Unit Weight Using RD formula and void ratio relationships, Max Dry Unit Weight, ya(mag = 16.99 kN/m? 4 =4.119m Final Soil Profile (After Compaction) Parameter | Symbol | Value Weight of Sol == = W | 70KN Target Relative Density) RD | 0.85 _ Initial Thickness _ a ee | _FinalThickness =H” 4.119 m Total Reduction AH 0.881m Rigid Pavement Engineering: Stress Analysis and Structural Design Wheel Load Stresses + Warping (6) + Friction (4) = 42 kgficm? ae Load Wheel (32) Summer Corner: 9 Edge: 7 kgficm? RESULTANT STRESS COMBINATIONS (The Critical Value) TYPES OF PAVEMENT STRESSES Warping Stresses (Thermal) Winter Corner: 8 Edge: 6 kgflcm? Corner Sitges Analysis ier Wheel (30) + Warping (8) = 38 kgficm? Frictional Stresses a Yi ubgrade Critical Design Value Setitical = 42 kgflem? Engineers adopt the greatest value to ensure design safety under demanding conditions. CONTRACTION JOINT SPACING (L,) STRUCTURAL REINFORCEMENT (Tie Bar Design) val Calculating Total Tie Bar Length: Total Length = (Development Length L, * 2) + Central a For a 12mm diameter bar, calculated length is 640mm Ie | Steel Area and Spacing For a 3.5m slab width, Required Area of Steel A, = 87.6 mm? Calculated Spacing S = 1291mm | Transveres Reinforcement Constants Yee= 24,000 Nim? fe12 DATA TABLE Max. ee spacing Z, is . 05m [Variable Value Description _ aes P5100 kgf Design Wheel Load E 5x 10®kgficm? | Modulus of Elasticity of Concrete #1015 Poisson's Ratio a —10*109°C _ Coeft. of Thermal Expansion At |15°C Temperature Differential _h__ | 25cm Slab Thickness Spacing Without Dowel Bars Spacing With Dowel Bars a ae Relying on concrete “ ay (0.8 kgficm?), Adding dowel bars (e.g., 8 or 10 bars of 12mm) | increases permissible spacing to approx. 20m Radius of Relative Stiffness (2) 1 = Function of(Z, k, #) = 90.3cm Foundational value for joint design, determined by slab thickness (4=25cm), modulus (£), and subgrade reaction (k). Structural Analysis Masterclass: 3-Hinged Arches and Hinged Beams Analysis of the 3-Hinged Arch (0) Bs A f L=20m 1 Calculating Maximum Horizontal Thrust (H) L Vmax = 1.25 (=) Focus 40kN H = (80 x 1.25) + (40 1.0) 1 | = 140KN ] ILD Maximum Positive 20kN Maximum Negative Bending Moment _ * Bending Moment x WL ! Miye= a ! Mmnax = 38.48kNm Mer .16 =a j x= 4.22m (0.2111) Ww Beam with Internal Hinge Analysis 4OkN Reaction Force at Hinge (R,) Equating Deflections R, = 11.11kN i. =-255uhmn DM = 222248) Visualizing Forces (BMD & SFD) Moment Results for Propped Cantilever BMD Characteristics: Sagging Moment Moment must always Calculation 7 be zero by definition at internal hinge. 40kN g SFD: Illustrates step-changes in internal shear. Meag = 22.22kNm 
