Quadratic Equation Progression Trigonometric Identities Spherical Trigonometry Form: 2 AM β HM = (GM)2 Squared Identities: 2 2 Sine Law: Ax + Bx + C = 0 Arithmetic Progression: Roots: s 2 − 4AC −B ± √B x= 2A Sum of Roots: B x1 + x2 = − A x1 β x2 = + C A (x + y)n rth term: th = nCm x n−m y m where: m=r-1 cos π = cos π cos π + sin π sin π cos π΄ 1 πR3 E V = AB H = 3 540° 1 A = bh 2 1 A = ab sin C 2 Square: Case 1: Unequal rate rate = work time a+b+c 2 s= ο€ Clock Problems Trapezoid θ= 11M − 60H 2 Ex-circleIn-circle + if M is ahead of H - if M is behind of H 1 1 1 1 = + + π π1 π2 π3 Centers of Triangle INCENTER - the center of the inscribed circle (incircle) of the triangle & the point of intersection of the angle bisectors of the triangle. Ellipse a2 + b2 2 A = πab C = 2π√ 1 Area = n β R2 sinβ 2 1 Area = n β ah 2 β= 360° n 16 - hexadecagon 17 - septadecagon 18 - octadecagon 19 - nonadecagon 20 - icosagon 21 - unicosagon 22 - do-icosagon 30 - tricontagon 31 - untricontagon 40 - tetradecagon 50 - quincontagon 60 - hexacontagon 100 - hectogon 1,000 - chilliagon 10,000 - myriagon 1,000,000 - megagon ∞ - aperio (circle) 3 - triangle 4 - quad/tetragon 5 - pentagon 6 - hexagon/sexagon 7 - septagon/heptagon 8 - octagon 9 - nonagon 10 - decagon 11 - undecagon/ monodecagon 12 - dodecagon/ bidecagon 13 - tridecagon 14 - quadridecagon 15 - quindecagon/ pentadecagon Inscribed Circle: Cyclic Quadrilateral: (sum of opposite angles=180°) AT = rs A = √(s − a)(s − b)(s − c)(s − d) Escribed Circle: Ptolemy’s Theorem is applicable: AT = R a (s − a) AT = R b (s − b) AT = R c (s − c) ac + bd = d1 d2 diameter = opposite side sine of angle a b c = = sin A sin B sin C s= a+b+c+d 2 Non-cyclic Quadrilateral: A = √(s − a)(s − b)(s − c)(s − d) − abcd cos 2 Pappus Theorem Pappus Theorem 1: Prism or Cylinder Pointed Solid SA = L β 2πR V = AB H = AX L LA = PB H = Px L 1 V = AB H 3 v Pappus Theorem 2: Special Solids Truncated Prism or Cylinder: Sphere: 4 V = πR3 3 LA = 4πR2 Frustum of Cone or Pyramid: Spheroid: H (A + A2 + √A1 A2 ) 3 1 AB/PB → Perimeter or Area of base H → Height & L → slant height AX/PX → Perimeter or Area of crosssection perpendicular to slant height Spherical Solids V = AB Have LA = PB Have V= H V = (A1 + 4AM + A2 ) 6 1 2 Spherical Lune: Spherical Wedge: Alune 4πR2 = θrad 2π 3 Vwedge 3 πR = θrad 2π 4 2 Vwedge = θR3 Spherical Sector: 1 V = Azone R 3 2 V = πR2 h 3 Spherical Segment: For one base: about major axis 4 V = πaab 3 a2 + a2 + b2 ] LA = 4π [ 3 LA = PB L Azone = 2πRh V = πabb 3 a2 + b2 + b2 ] LA = 4π [ 3 Oblate Spheroid: LA = πrL Spherical Zone: 4 Prolate Spheroid: Prismatoid: Reg. Pyramid 3 V = πabc 3 a2 + b2 + c 2 ] LA = 4π [ 3 1 V = πh2 (3R − h) 3 For two bases: 1 about minor axis V = πh(3a2 + 3b2 + h2 ) 6 ε 2 Right Circ. Cone Alune = 2θR2 4 EULER LINE - the line that would pass through the orthocenter, circumcenter, and centroid of the triangle. Area = n β ATRIANGLE δ = 180° − γ abc AT = 4R NOTE: It is also used to locate centroid of an area. CENTROID - the point of intersection of the medians of the triangle. Deflection Angle, δ: Circumscribing Circle: V = A β 2πR ORTHOCENTER - the point of intersection of the altitudes of the triangle. (n − 2)180° n General Quadrilateral Triangle-Circle Relationship d= CIRCUMCENTER - the center of the circumscribing circle (circumcircle) & the point of intersection of the perpendicular bisectors of the triangle. A = ah A = a2 sin θ 1 A = d1 d2 2 A1 n ma2 + nb 2 = ;w = √ A2 m m+n 1 knot = 1 nautical mile per hour Polygon Names Rhombus: 1 A = (a + b)h 2 1 statute mile = 5280 feet Central Angle, β: Parallelogram: A = √s(s − a)(s − b)(s − c) Case 2: Equal rate → usually in project management → express given to man-days or man-hours γ= Rectangle: A = bh A = ab sin θ 1 A = d1 d2 sin θ 2 1 nautical mile = 6080 feet Interior Angle, Ι€: A = s2 A = bh P = 4s P = 2a + 2b d = √2s d = √b 2 + h2 1 sin B sin C A = a2 2 sin A ο€ Work Problems 1 minute of arc = 1 nautical mile n-sided Polygon 2 ο€ Age Problems → underline specific time conditions =0 = vt 180° sin 2A = 2 sin A cos A cos 2A = cos 2 A − sin2 A cos 2A = 2 cos 2 A − 1 cos 2A = 1 − 2 sin2 A 2 tan A # of diagonals: tan 2A = n 1 − tan2 A d = (n − 3) Common Quadrilateral →s Spherical Polygon: πR2 E E = spherical excess AB = E = (A+B+C+D…) – (n-2)180° Spherical Pyramid: Triangle →a cos π΄ = − cos π΅ cos πΆ + sin π΅ sin πΆ cos π Double Angle Identities: Worded Problems Tips ο€ Motion Problems Cosine Law for angles: sin (A ± B) = sin A cos B ± cos A sin B cos (A ± B) = cos A cos B β sin A sin B tan A ± tan B tan (A ± B) = 1 β tan A tan B r = a 2 /a1 = a 3 /a2 a n = a1 r n−1 a n = a x r n−x 1 − rn Sn = a1 1−r a1 S∞ = 1−r Form: Cosine Law for sides: Sum & Diff of Angles Identities: Geometric Progression: Binomial Theorem r d = a 2 − a1 = a 3 − a 2 a n = a1 + (n − 1)d a n = a x + (n − x)d n Sn = (a1 + a n ) 2 Harmonic Progression: - reciprocal of arithmetic progression Product of Roots: sin π sin π sin π = = sin π΄ sin π΅ sin π΄ sin A + cos A = 1 1 + tan2 A = sec 2 A 1 + cot 2 A = csc 2 A Archimedean Solids Analytic Geometry - the only 13 polyhedra that are convex, have identical vertices, and their faces are regular polygons. E= Nn 2 V= s Nn v where: E → # of edges V → # of vertices N → # of faces n → # of sides of each face v → # of faces meeting at a vertex Conic Sections General Equation: Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 Based on discriminant: B 2 − 4AC = 0 ∴ parabola B 2 − 4AC < 0 ∴ ellipse B 2 − 4AC > 0 ∴ hyperbola Based on eccentricity, e=f/d: π = 0 ∴ circle π = 1 ∴ parabola π < 1 ∴ ellipse π > 1 ∴ hyperbola Distance from a point to another point: d = √(y2 − y1 )2 + (x2 − x1 )2 Point-slope form: Distance from a point to a line: General Equation: General Equation: Ax 2 + Cy 2 + Dx + Ey + F = 0 Ax 2 − Cy 2 + Dx + Ey + F = 0 y − y1 m= x − x1 x 2 + y 2 + Dx + Ey + F = 0 Standard Equation: 2 (x − h) + (y − k)2 = r 2 Two-point form: y2 − y1 y − y2 = x 2 − x1 x − x 2 |C1 − C2 | √A2 + B 2 tan θ = 1 revolution = 2π rad = 360Λ = 400 grads = 6400 mills 2 H = a√ 3 √2 V=a 12 Elements: Eccentricity, e: e= 3 =1 LR = 4a Location of foci, c: c 2 = a2 − b2 Length of LR: 2b2 LR = a Versed cosine: covers A = 1 − sin A exsec A = sec A − 1 ο€ Inflation: ο€ Rate of return: annual net profit RR = capital πf = π + f + πf ο€ Break-even analysis: Annual net profit = savings – expenses – depreciation (sinking fund) 1 RR π¦" (+) minima (-) maxima Integral Calculus-The Cardioid r = a(1 − cos θ) r = a(1 + cos θ) CALTECH: Mode 3 2 x y (time) (BV) 0 FC n SV (1 + i)n − 1 −1 d = (FC − SV) [ ] π m (1 + i) − 1 Dm = d [ ] π n−m+1 dm = (FC − SV) [ ] ∑ years ∑nn−m+1 x Dm = (FC − SV) [ ] ∑n1 x ο€ Declining Balance (Matheson): BVm = FC(1 − k)m SV = FC(1 − k)n k → obtained Dm = FC − BVm where: F → future worth P → principal or present worth A → periodic payment i → interest rate per payment n → no. of interest periods n’ → no. of payments ο€ Perpetuity: P= where: FC → first cost SV → salvage cost d → depreciation per year n → economic life m → any year before n BVm → book value after m years Dm → total depreciation CALTECH: Mode 3 3 x y (time) (BV) 0 FC n SV n+1 SV k = 2/n k → obtained Dm = FC − BVm ο€ Service Output Method: A = F(1 + π)−n π ο€ Capitalized Cost: C = FC + OM RC − SV + π (1 + π)n − 1 AC = C β π AC = FC β π + OM + where: C → capitalized cost FC → first cost OM → annual operation or maintenance cost RC → replacement cost SV → salvage cost AC → annual cost (RC − SV)π (1 + i)n − 1 ο€ Single-payment-compound-amount factor: n (F/P, π, n) = (1 + π) ο€ Single-payment-present-worth factor: −n (P/F, π, n) = (1 + π) ο€ Equal-payment-series-compound-amount factor: CALTECH: Mode 3 6 x y (time) (BV) 0 FC n SV ο€ Double Declining Balance: BVm = FC(1 − k)m FC − SV Qn D = dQ m (1 + π)n − 1 ] π ′ ο€ Sinking Fund: d= ο€ Annuity: (1 + π)n − 1 P = A[ ] π(1 + π)n ο€ Sum-of-the-Years-Digit (SYD): π2 π¦ = y" = 0 ππ₯ 2 F = Pe ER = er − 1 ′ BVm = FC − Dm ρ= where: F → future worth P → principal or present worth i → interest rate per interest period r → nominal interest rate n → no. of interest periods m → no. of interest period per year t → no. of years ER → effective rate ο€ Continuous Compounding Interest: rt F = A[ Depreciation Radius of curvature: 3 [1 + (y′)2 ]2 where: m is (+) for upward asymptote; m is (-) for downward m = b/a if the transverse axis is horizontal; m = a/b if the transverse axis is vertical y − k = ±m(x − h) c e= a F = P(1 + π) r mt F = P (1 + ) m I r m ER = = (1 − ) − 1 P m 1 − cos A 2 FC − SV d= n Dm = d(m) Eccentricity, e: Eq’n of asymptote: ο€ Compound Interest: n Half versed sine: ο€ Straight-Line: c 2 = a2 + b2 I = Pπn F = P(1 + πn) vers A = 1 − cos A Exsecant: Same as ellipse: Length of LR, Loc. of directrix, d Eccentricity, e a d= e ο€ Simple Interest: Versed sine: hav A = Location of foci, c: Loc. of directrix, d: Engineering Economy Unit Circle RP = Point of inflection: A = 1.5πa2 P = 8a r = a(1 − sin θ) r = a(1 + sin θ) dd Length of latus rectum, LR: cost = revenue Maxima & Minima (Critical Points): ππ¦ = y′ = 0 ππ₯ df Elements: Elements: 2 Differential Calculus 3 (y′)2 ]2 (y − k)2 (x − h)2 − =1 a2 b2 m2 − m1 1 + m1 m2 General Equation: 2 π₯ 2 → π₯π₯1 π¦ 2 → π¦π¦1 π₯ + π₯1 π₯→ 2 π¦ + π¦1 π¦→ 2 π₯π¦1 + π¦π₯1 π₯π¦ → 2 y" (x − h)2 (y − k)2 + =1 b2 a2 - the locus of point that moves such that it is always equidistant from a fixed point (focus) and a fixed line (directrix). SA = a √3 Curvature: (x − h)2 (y − k)2 − =1 a2 b2 Parabola In the equation of the conic equation, replace: [1 + d= Standard Equation: (x − h)2 (y − k)2 + =1 a2 b2 Angle between two lines: x y + =1 a b Line Tangent to Conic Section To find the equation of a line tangent to a conic section at a given point P(x1, y1): Standard Equation: √A2 + B 2 Distance of two parallel lines: Point-slope form: Tetrahedron k= d= |Ax + By + C| (x − h) = ±4a(y − k) (y − k)2 = ±4a(x − h) General Equation: - the locus of point that moves such that the difference of its distances from two fixed points called the foci is constant. y = mx + b Slope-intercept form: Standard Equation: 2 - the locus of point that moves such that its distance from a fixed point called the center is constant. Hyperbola - the locus of point that moves such that the sum of its distances from two fixed points called the foci is constant. y + Dx + Ey + F = 0 x 2 + Dx + Ey + F = 0 Circle Ellipse ′ (1 + π)n − 1 (F/A, π, n) = [ ] π ο€ Equal-payment-sinking-fund factor: ′ −1 (1 + π)n − 1 (A/F, π, n) = [ ] π ο€ Equal-payment-series-present-worth factor: ′ where: FC → first cost SV → salvage cost d → depreciation per year Qn → qty produced during economic life Qm → qty produced during up to m year Dm → total depreciation (1 + π)n − 1 (P/A, π, n) = [ ] π(1 + π)n ο€ Equal-payment-series-capital-recovery factor: ′ (1 + π)n − 1 (A/P, π, n) = [ ] π(1 + π)n −1 Statistics Fractiles Transportation Engineering Traffic Accident Analysis ο«Measure of Natural Tendency ο€ Range = ππππππ π‘ πππ‘π’π − π ππππππ π‘ πππ‘π’π ο«Design of Horizontal Curve ο€ Mean, xΜ , μ → average → Mode Stat 1-var ο€ Coefficient of Range ο€ Accident rate for 100 million vehicles per miles of travel in a segment of a highway: → Shift Mode βΌs Stat Frequency? on ππππππ π‘ πππ‘π’π − π ππππππ π‘ πππ‘π’π = ππππππ π‘ πππ‘π’π + π ππππππ π‘ πππ‘π’π → Input → AC Shift 1 var xΜ ο€ Quartiles ο€ Median, Me → middle no. when n is even 1 n+1 2 1 n n = [( ) + ( + 1)] 2 2 2 Q1 = n Me th = Me th 4 2 3 Q2 = n Q3 = n 4 4 when n is odd Q1 = ο€ Mode, Mo → most frequent 1 1 1 (n + 1) ; Q1 = (n + 1) ; Q1 = (n + 1) 4 4 4 ο€ Interquartile Range, IQR ο«Standard Deviation = ππππππ π‘ ππ’πππ‘πππ − π ππππππ π‘ ππ’πππ‘πππ = Q3 − Q1 ο€ Population standard deviation → Mode Stat 1-var → Shift Mode βΌ Stat Frequency? on → Input → AC Shift 1 var σx ο€ Sample standard deviation → Mode Stat 1-var ππππππ π‘ ππ’πππ‘πππ − π ππππππ π‘ ππ’πππ‘πππ = ππππππ π‘ ππ’πππ‘πππ + π ππππππ π‘ ππ’πππ‘πππ Q − Q1 = 3 Q3 + Q1 ο€ Outlier → extremely high or low data higher than or lower than the following limits: NOTE: If not specified whether population/sample in a given problem, look for POPULATION. Q1 − 1.5IQR > x Q 3 + 1.5IQR < x ο«Coefficient of Linear Correlation or Pearson’s r ο€ Decile or Percentile → AC Shift 1 Reg r R= R → minimum radius of curvature e → superelevation f → coeff. of side friction or skid resistance v → design speed in m/s g → 9.82 m/s2 ο€ Centrifugal ratio or impact factor 2 Impact factor = v gR P = vR P → power needed to move vehicle in watts v → velocity of vehicle in m/s R → sum of diff. resistances in N ο«Design of Pavement ο€ Rigid pavement without dowels t=√ 3W 4f (at the center) ο€ Flexible pavement ο€ Population standard deviation ο€ Z-score or standard score or variate ο€ standard deviation = σ ο€ variance = σ2 → Mode Stat → AC Shift 1 Distr left of z → P( x−μ z= σ ο€ relative variability = σ/x ο«Mean/Average Deviation right of z → R( bet. z & axis → Q( → Input x → no. of observations μ → mean value, x Μ σ → standard deviation ο€ Mean/average value b 1 mv = ∫ f(x)dx b−a a b 1 RMS = √ ∫ f(x)2 dx b−a a Walli’s Formula ο€ Binomial Probability Distribution x n−x ∫ cosm θ sinn θ dθ = π 2 P(x) = C(n, x) p q 0 where: p → success q → failure f → fatal i → injury p → property damage ∑d n = ∑t ∑ 1 ( ) U1 ο€ Time mean speed, Ut: d ∑ U1 Ut = t = n n ∑ Ζ©d → sum of distance traveled by all vehicles Ζ©t → sum of time traveled by all vehicles Ζ©u1 → sum of all spot speed 1/Ζ©u1 → reciprocal of sum of all spot speed n → no. of vehicles q → rate of flow in vehicles/hour k → density in vehicles/km uS → space mean speed in kph ο€ Thickness of pavement in terms of expansion pressure ο€ Minimum time headway (hrs) = 1/q expansion pressure pavement density Es SF = √ Ep 3 q = kUs ο€ Spacing of vehicles (km) = 1/k ο€ Peak hour factor (PHF) = q/qmax s [(m − 1)(m − 3)(m − 5) … (1 or 2)][(n − 1)(n − 3)(n − 5) … (1 or 2)] βα (m + n)(m + n − 2)(m + n − 4) … (1 or 2) α = π/2 for m and n are both even α =1 otherwise ο€ Geometric Probability Distribution x−1 ) Tip to remember: Fibonacci Numbers ο€ Poisson Probability Distribution x −μ μ e x! an = Period, Amplitude & Frequency Period (T) → interval over which the graph of function repeats Amplitude (A) → greatest distance of any point on the graph from a horizontal line which passes halfway between the maximum & minimum values of the function Frequency (ω) → no. of repetitions/cycles per unit of time or 1/T Period 2π/B 2π/B π/B fβi fβiβp f1 → allow bearing pressure of subgrade r → radius of circular area of contact between wheel load & pavement NOTE: Function y = A sin (Bx + C) y = A cos (Bx + C) y = A tan (Bx + C) SR = ES → modulus of elasticity of subgrade EP→ modulus of elasticity of pavement Discrete Probability Distributions P(x) = p(q ο€ Severity ratio, SR: ο€ Rate of flow: ο€ Stiffness factor of pavement P(x ≥ a) = e−λa P(x ≤ a) = 1 − e−λa P(a ≤ x ≤ b) = e−λa − e−λb ο€ Mean value A → no. of accidents during period of analysis ADT → average daily traffic entering all legs N → time period in years W t=√ −r πf1 t= Exponential Distribution A (1,000,000) ADT β N β 365 Us = (at the edge) -1 ≤ r ≤ +1; otherwise erroneous ο«Variance ο€ Accident rate per million entering vehicles in an intersection: ο€ Spacing mean speed, US: 3W t=√ f t → thickness of pavement W → wheel load f → allow tensile stress of concrete Normal Distribution A (100,000,000) ADT β N β 365 β L A → no. of accidents during period of analysis ADT → average daily traffic N → time period in years L → length of segment in miles R= R → minimum radius of curvature v → design speed in m/s g → 9.82 m/s2 3W t=√ 2f NOTE: P(x) = v g(e + f) ο€ Rigid pavement with dowels m im = (n) 10 or 100 → Mode Stat A+Bx → Input R= ο«Power to move a vehicle ο€ Coefficient of IQR ο€ Quartile Deviation (semi-IQR) = IQR/2 → Shift Mode βΌ Stat Frequency? on → Input → AC Shift 1 var sx ο€ Minimum radius of curvature 2 Amplitude A A A 1 √5 n [( n 1 + √5 1 − √5 ) −( ) ] 2 2 x = r cos θ y = r sin θ r = x2 + y2 y θ = tan−1 x π₯2 − π₯ − 1 = 0 Mode Eqn 5 π₯= 1 ± √5 2 measure too long add too short subtract Measurement Corrections Due to temperature: Probable Errors C = αL(T2 − T1 ) Probable Error (single): (add/subtract); measured length (P2 − P1 )L C= EA (subtract only); unsupported length w 2 L3 24P 2 CD = MD (1 − ∑(x − xΜ ) n−1 ∑(x − xΜ ) Em = = 0.6745√ n(n − 1) √n E Proportionalities of weight, w: Due to slope: (subtract only); measured length π€∝ Normal Tension: 0.204W√AE 1 πΈ2 π€∝ 1 π π€∝π Area of Closed Traverse √PN − P Error of Closure: L H = (g1 + g 2 ) 8 L 2 x 2 ( 2) = L y H 1 Error of Closure Perimeter 1 acre = 4047 m2 from South D2 (h − h2 ) − 0.067D1 D2 D1 + D2 1 Stadia Measurement Leveling Horizontal: Elevπ΅ = Elevπ΄ + π΅π − πΉπ D = d + (f + c) π D = ( )s +C π D = Ks + C Inclined Upward: Inclined: Total Error: Reduction to Sea Level CD MD = R R+h error/setup = −eBS + eFS Subtense Bar Inclined Downward: error/setup = +eBS − eFS D = cot θ 2 eT = error/setup β no. of setups Double Meridian Distance Method DMD DMDππππ π‘ = Depππππ π‘ DMDπ = DMDπ−1 + Depπ−1 + Depπ DMDπππ π‘ = −Depπππ π‘ 2A = Σ(DMD β Lat) d [h + hn + 2Σh] 2 1 Double Parallel Distance Method DPD d A = [h1 + hn + 2Σhπππ + 4Σhππ£ππ ] 3 Relative Error/Precision: = h = h2 + Simpson’s 1/3 Rule: = √ΣL2 + ΣD2 Azimuth hcr = 0.067K 2 Trapezoidal Rule: A= Symmetrical: e ) TL Effect of Curvature & Refraction Area of Irregular Boundaries Lat = L cos α Dep = L sin α Parabolic Curves e ) TL D = Ks cos θ + C H = D cos θ V = D sin θ E=error; d=distance; n=no. of trials C 2 = S 2 − h2 PN = CD = MD (1 + Probable Error (mean): Due to sag: C= E = 0.6745√ too long too short (add/subtract); measured length Due to pull: lay-out subtract add Note: n must be odd Simple, Compound & Reverse Curves DPDππππ π‘ = Latππππ π‘ DPDπ = DPDπ−1 + Lat π−1 + Lat π DPDπππ π‘ = −Lat πππ π‘ 2A = Σ(DMD β Dep) Spiral Curve Unsymmetrical: H= L1 L2 (g + g 2 ) 2(L1 +L2 ) 1 g 3 (L1 +L2 ) = g1 L1 + g 2 L2 Note: Consider signs. Earthworks ππΏ 0 ππ ±ππΏ ±π ±ππ A= f w (d + dR ) + (fL + fR ) 2 L 4 T = R tan I m = R [1 − cos ] L = 2R sin L3 6RLs L (c − c2 )(d1 − d2 ) 12 1 VP = Ve − Cp L5 I Y=L− 2 π Lc = RI β 180° 20 2πR = D 360° 1145.916 R= D Prismoidal Correction: 40R2 Ls 2 Ls I + (R + p) tan 2 2 I Es = (R + p) sec − R 2 Ts = Ls = Volume (Truncated): 0.036k 3 R 0.0079k 2 R D L = DC Ls Σh = A( ) n e= A (Σh1 + 2Σh2 + 3Σh3 + 4Σh4 ) n Stopping Sight Distance Parabolic Summit Curve v2 S = vt + 2g(f ± G) a = g(f ± G) (deceleration) v (breaking time) tb = g(f ± G) f Eff = (100) fave L>S v → speed in m/s t → perception-reaction time f → coefficient of friction G → grade/slope of road x= 2 L VP = (A1 + 4Am + A2 ) 6 VT = θ Ls 2 ; p= 3 24R 2 Volume (Prismoidal): VT = ABase β Have i= I E = R [sec − 1] L Ve = (A1 + A2 ) 2 CP = L2 180° β 2RLs π 2 I Volume (End Area): θ= L= A(S)2 200(√h1 + √h2 ) 2 L<S 200(√h1 + √h2 ) L = 2(S) − A L → length of summit curve S → sight distance h1 → height of driver’s eye h1 = 1.143 m or 3.75 ft h2 → height of object h2 = 0.15 m or 0.50 ft 2 LT → long tangent ST → short tangent R → radius of simple curve L → length of spiral from TS to any point along the spiral Ls → length of spiral I → angle of intersection I c → angle of intersection of the simple curve p → length of throw or the distance from tangent that the circular curve has been offset x → offset distance (right angle distance) from tangent to any point on the spiral xc → offset distance (right angle distance) from tangent to SC Ec → external distance of the simple curve θ → spiral angle from tangent to any point on the spiral θS → spiral angle from tangent to SC i → deflection angle from TS to any point on the spiral is → deflection angle from TS to SC y → distance from TS along the tangent to any point on the spiral Parabolic Sag Curve Underpass Sight Distance Horizontal Curve L>S L>S L>S A(S)2 L= 122 + 3.5S A(S)2 L= 800H L<S L<S 122 + 3.5S L = 2(S) − A 800H L = 2(S) − A R= A → algebraic difference of grades, in percent L → length of sag curve S → sight distance A → algebraic difference of grades, in percent L → length of sag curve L<S L= A(K)2 395 H= C− h1 + h2 2 For passengers comfort, where K is speed in KPH R= S2 8M L(2S − L) 8M L → length of horizontal curve S → sight distance R → radius of the curve M → clearance from the centerline of the road Properties of Fluids Dams Pressure s Mg W= pπππ = pππππ + pππ‘π W Ι€= V p = Ι€h s. g.1 h2 = h s. g.2 1 M ; ρ= V pg Ι€ = ρg = RT V 1 s. v. = = M ρ Ι€ ρ s. g. = = Ι€π€ ρπ€ βP 1 ; β= βV EB V ππ¦ FT μ=τ = ππ L2 Ig AΣ― e= hΜ = Σ― (for vertical only) U2 = (h1 − h2 )Ι€B 2 RM OM & FSπ = ω2 x tan θ = g z1 + 2 2 2 y= ω x 2g V= 1 2 πr h 2 2 r x = h y ; Stresses/Hoops pD 2t 2T s= pD St = H. L. = f P1 v1 2 P2 v2 2 z1 + + + HA = z2 + + + H. L. Ι€ 2g Ι€ 2g H. L.T = H. L.1 + H. L.2 +. . . +H. L.n Fluid Flow Most Efficient Sections Q T = Q1 = Q 2 = Q n Q = Av Rectangular: Q → discharge → flow rate → weight flux b = 2d d R= 2 Q T = Q1 + Q 2 +. . . +Q n Constant Head Orifice Falling Head Orifice Without headloss: Time to remove water from h1 to h2 with constant cross-section: v = √2gh With headloss: v = Cv √2gh t= H. L. = v2 1 [ − 1] 2g Cv 2 H. L. = βH[1 − Cv 2 ] y= x CAo √2g (√h1 − √h2 ) Time to remove water from h1 to h2 with varying cross-section: h1 h2 Q = CA o √2gh C = Cc C v a Cc = A v Cv = vt 2As t=∫ As dh CAo √2gh Time in which water surfaces of two tanks will reach same elevation: t= (As1 )(As2 ) (√h1 − √h2 ) CAo √2g (As1 + As2 ) 2 Force on Curve Vane/Blade: Force on the Jet (at right angle): ∑ Fx = ρQ(v2x − v1x ) F = ρQv ∑ Fy = ρQ(v2y − v1y ) Force on Pipe’s Bend & Reducer: (same as on Curve Vane/Blade) BF = Ι€π€ Vπ Vbel Celerity (velocity of sound) L v2 D 2g Manning’s Formula: H. L. = 10.29 n2 L Q2 D16/3 Hazen William’s Formula: 1 atm = 101.325 KPa = 2166 psf = 14.7 psi = 760 mmHg = 29.9 inHg EB c=√ ρw EB c=√ E D ρw (1 + B ) Et Water Hammer βPmax = ρcv tc = 2L c A. TIME of closure: ο«rapid/instantaneous βP = βPmax ο«Slow Closure tc βP = βPmax ( ) t actual B. TYPE of closure: ο«Partial Closure (vf ≠ 0) βP = ρc(vi − vf ) ο«Total Closure (vf = 0) βP = ρcvi Open Channel x = y1 + y2 Specific Energy: 2 v E= +d 2g d R= 2 Triangular: v = C√RS b = 2d A = d2 θ = 90° Theoretically: Semi-circular: Kutter Formula: C=√ 8g Manning Formula: C= 1 1/6 R n Bazin Formula: C= 87 m 1+ √R f 1 0.000155 + 23 + n S C= n 0.000155 1+ (23 + ) S √R d = r (full) r R= 2 Circular: (rigid pipes) (non-rigid pipes) 0.0826 f L Q D5 Trapezoidal: Q max if d = 0.94D Vmax if d = 0.81D Hydrodynamics 2 4Cv 2 h H. L. = sg m A sg l tot sg m = V sg l tot Abel = BF = W 10.64 L Q1.85 H. L. = 1.85 4.87 C D Pump → Output & Turbine → Input volume flow rate → m3/s weight flow rate → N/s mass flow rate → kg/s vs I = VD sin θ VD 2 output QΙ€E efficiency = ; HP = input 746 H. L.T = H. L.1 = H. L.2 = H. L.n MBπ = Darcy Weisbach Eq’n: P1 v1 2 P2 v2 2 z1 + + − HE = z2 + + + H. L. Ι€ 2g Ι€ 2g Parallel Connection: B2 tan2 θ [1 + ] 12D 2 Major Losses in Pipes with pump: Series Connection: MBπ = Buoyancy St = tensile stress p = unit pressure D = inside diameter t = thickness of wall s = spacing of hoops T = tensile force P1 v1 2 P2 v2 2 + = z2 + + + H. L. Ι€ 2g Ι€ 2g Series-Parallel Pipes Use (-) if G is above BO and (+) if G is below BO. Note that M is always above BO. Rπ₯ with turbine: π 1 rpm = rad/sec 30 MG = MBπ ± GBπ RM or OM = Wx = W(MG sin θ) z = elevation head; P/Ι€ = pressure head; v2/2g = velocity head a tan θ = g MG = metacentric height μRπ¦ Bernoulli’s Energy Theorem Rotation: a p = Ι€h (1 ± ) g 2 NOTE: Δ§ = vertical distance from cg of submerged surface to liquid surface Horizontal Motion: Vertical Motion: ; B π = | − xΜ | 2 Rπ¦ B 6π π< ; q=− [1 ± ] 6 B B 2R π¦ B π> ; q= 6 3xΜ Rπ¦ B π= ; q=− 6 B 2R π¦ π = 0; q = B Fπ£ = Ι€V Relative Equilibrium of Fluids ah tan θ = g ± av FSπ = Fβ = Ι€hΜ A F = √ Fβ + Fπ£ 2 1 RxΜ = RM − OM Ι€Ig sin θ F On curved surfaces: pd σ= 4 4σcosθ h= Ι€d F2 = Ι€Ah2 = Ι€h2 2 h2 RM = W1 (X1 ) + W2 (X2 )+. . . +Wπ (Xπ ) + F2 ( ) 3 h 1 2 OM = F1 ( ) + U1 ( B) + U2 ( B) 3 2 3 F = Ι€hΜ A 2 1 ; U1 = Ι€h2 B On plane surfaces: μ L2 = ρ T Inclined Motion: 2 Hydrostatic Forces EB = − υ= F1 = Ι€Ah1 = Ι€h1 2 hπ€ = s. g.1 h1 e= Stability of Floating Bodies 1 TRAPEZOIDAL: For minimum seepage: b = 4d tan θ 2 If C is not given, use Manning’s in V: v= 1 2/3 1/2 R S n Volume Vv e= Vs Ww ω= Ws Vv n= V Vw S= Vv Gs = When S=0: Relative Compaction: Ι€d g = Gs (1 − n) W V WS Ι€d = V Ι€= 0<e<∞ n e= 1−n e n= 1+e (Gs − 1)Ι€w 1+e Gs Ι€w = 1 + Gs ω Ι€sub = Ι€zav Permeability PI = LL − PL Ι€d πππ₯ Dr (%) 0 – 20 20 – 40 40 – 70 70 – 85 85 – 100 βh v v = ki ; i = ; vπ = L n Q = vA = kiA qu PI ; St = und μ q u rem μ = % passing 0.002mm Cu = k= aL h1 ππ At h2 k eq = Casagrande: k = c β D10 2 k = 1.4e2 k 0.85 Kozeny-Carman: Samarasinhe: k = C1 β k eq = n 2 e 1+e k = C3 β e 1+e Stresses in Soil NOTE: Quick condition: Effective Stress/ Intergranular Stress: pE = 0 pE = pT − pw C hcr = eD10 pw = Ι€w hw h1 h2 h + +. . . + n k1 k 2 kn Flow Net / Seepage Isotropic soil: q = kH Capillary Rise: Pore Water Pressure/ Neutral Stress: Total Stress: Nf Nd ACTIVE PRESSURE: Nf q = √k x k z H Nd AT REST: k o = 1 − sin Ø cos β − √cos 2 β − cos 2 Ø cos β + √cos 2 β − cos 2 Ø For Horizontal: 1 − sin Ø ka = 1 + sin Ø If there is angle of friction α bet. wall and soil: 2 cos Ø sin(Ø + α) sin Ø ] cos α 2 k P = cos β cos 2 Ø cos β − √cos 2 β − cos 2 Ø For Horizontal: kP = Ø 2 TRI-AXIAL TEST: σ1 → maximum principal stress → axial stress β³σ → additional pressure → deviator stress → plunger pressure σ3 → minimum principal stress → confining pressure → lateral pressure → radial stress → cell pressure → chamber pressure ο€ Cohesive soil: For Inclined: − 1 + sin Ø 1 − sin Ø r x + σ3 + r c tan Ø = x sin Ø = ο€ Unconsolidatedundrained test: If there is angle of friction α bet. wall and soil: 2 c=r kP = ο€ Unconfined compression test: cos Ø cos α [1 − √ 9 Σ¨ → angle of failure in shear Ø → angle of internal friction/shearing resistance C → cohesion of soil r sin Ø = σ3 + r 1 pP = k P Ι€H 2 + 2cH√k P 2 cos β + 4 56 7 8 Nf → no. of flow channels [e.g. 4] Nd → no. of potential drops [e.g. 10] ο€ Normally consolidated: PASSIVE PRESSURE: √cos 2 β 3 Shear Strength of Soil θ = 45° + cos α [1 + √ 3 Compression Index, CC: Swell Index, CS: Cc = 0.009(LL − 10%) e − e′ Cc = βP + Po πππ Po 1 Cs = Cc 5 e − e′ H (for one layer only) 1+e Cc H βP + Po S= πππ 1+e Po S= 4 2 D75 D25 10 With Pre-consolidation pressure, Pc: when (β³P+Po) < Pc: For Inclined: ka = 1 D60 β D10 So = √ For normally consolidated clay: 2 Equipotential line ---- Non-Isotropic soil: 1 pa = k a Ι€H 2 − 2cH√k a 2 k a = cos β 1 Flow line ---- pT = Ι€1 h1 + Ι€2 h2 +. . . +Ι€n hn Lateral Earth Pressure r Q ππ 1 r2 k= 2πt(h1 − h2 ) H Cc = Compressibility of Soil Confined: for Perpendicular flow: Hazen Formula D60 D10 Sorting Coefficient: 3 1 1 Sn = 1.7√ + + 2 2 (D50 ) (D20 ) (D10 )2 r1 r2 k= π(h1 2 − h2 2 ) Q ππ h1 k1 + h2 k 2 +. . . +hn k n H Non-plastic Slightly plastic Low plasticity Medium plasticity High plasticity Very High plastic Suitability Number: Unconfined: for Parallel flow: Class AC < 0.7 Inactive 0.7 < AC < 1.2 Normal AC > 1.2 Active Description 0 1-5 5-10 10-20 20-40 >40 Coeff. of Gradation or Curvature: (D30 )2 Uniformity Coefficient: Pumping Test: Falling/Variable Head Test: Ac PI Sieve Analysis Constant Head Test: QL k= Aht State LI < 0 Semisolid 0 < LI < 1 Plastic LI > 1 Liquid LL − ω CI = LL − PI Description Very Loose Loose Medium Dense Dense Very Dense 1 − SL SR LI SI = PL − SL Ac = 1 GI = (F − 35)[0.2 + 0.005(LL − 40)] +0.01(F − 15)(PI − 10) ω − PL LL − PL LI = 1 1 − Ι€d πππ Ι€d Dr = 1 1 − Ι€d πππ Ι€d πππ₯ Stratified Soil Gπ = Atterberg Limits Relative Density/ Density Index: eπππ₯ − e Dr = eπππ₯ − eπππ Ι€sub = Ι€sat − Ι€w Ι€ Ι€d = 1+ω 0<n<1 R= (Gs + e)Ι€w = 1+e Ι€sat SL = Bulk Specific Gravity: When S=100%: Se = Gs ω Ι€s Ι€w (Gs + Gs ω)Ι€w Ι€= 1+e (Gs + Se)Ι€w Ι€= 1+e Gs Ι€w Ι€d = 1+e Weight m1 − m2 V1 − V2 − Ι€w m2 m2 e m2 SL = ; SR = Gs V2 Ι€w Specific Gravity of Solid: Unit Weight: sin(Ø − α) sin Ø ] cos α 2 σ3 = 0 S= Cs H βP + Po πππ 1 + eo Po when (β³P+Po) > pc: S= Cs H Pc Cc H βP + Po πππ + πππ 1+e Po 1 + e Pc Over Consolidation Ratio (OCR): OCR = pc ; po OCR = 1 (for normally consolidated soil) Coefficient of Compressibility: av = βe βP β³e → change in void ratio β³P → change in pressure Coefficient of Volume Compressibility: mv = βe βP 1 + eave Coefficient of Consolidation: Hdr → height of drainage path Hdr 2 Tv → thickness of layer if drained 1 side Cv = → half of thickness if drained both sides t Tv → factor from table Coefficient of Permeability: t → time consolidation k = mv Cv Ι€w DIRECT SHEAR TEST: σn → normal stress ο€ Normally consolidated soil: σs → shear stress σS tan Ø = σN ο€ Cohesive soil: tan Ø = σS c = x + σN x σS = c + σN tan ∅ Terzaghi‘s Bearing Capacity (Shallow Foundations) ο€ General Shear Failure (dense sand & stiff clay) Square Footing: qult = 1.3cNaSSSSSSSSSSSSSSSSSSSS c + qNq + 0.4Ι€BNΙ€ Circular Footing: Soil Stability ο€ Bearing Capacity Factor ο€ Analysis of Infinite Slope Ø Nq = tan2 (45° + ) eπ tan Ø 2 Factor of safety against sliding (without seepage) Nc = (Nq − 1) cot Ø qult = 1.3cNc + qNq + 0.3Ι€BNΙ€ NΙ€ = (Nq − 1) tan 1.4Ø Strip Footing: ο€ Parameters qult → ultimate bearing capacity qu → unconfined compressive strength c → cohesion of soil qult = cNc + qNq + 0.5Ι€BNΙ€ ο€ Local Shear Failure (loose sand & soft clay) Square Footing: ′ qu c= 2 ′ qult = 1.3c′Nc + qNq + 0.4Ι€BNΙ€ ′ q = Ι€Df qult = 1.3c′Nc ′ + qNq ′ + 0.3Ι€BNΙ€ ′ Strip Footing: qult = c′Nc ′ + qNq ′ + 0.5Ι€BNΙ€ ′ C tan ∅ + Ι€ H sin π½ cos π½ tan π½ β where: C → cohesion β → angle of backfill from horizontal Ø → angle of internal friction H → thickness of soil layer Factor of safety against sliding (with seepage) C Ι€′ tan ∅ FS = + Ι€π ππ‘ H sin π½ cos π½ Ι€π ππ‘ tan π½ ο€ Analysis of Finite Slope Factor of safety against sliding FS = (for no water table) qult Pallow qallow = = FS A qult − q qnet = FS Circular Footing: FS = EFFECT OF WATER TABLE: Ff + Fc W sin π β Maximum height for critical equilibrium (FS=1.0) Hcr = 4πΆ sin π½ cos ∅ [ ] Ι€ 1 − cos(π½ − ∅) Stability No.: Stability Factor: C m= Ι€H SF = θ where: Ff → frictional force; Ff = μN Fc → cohesive force Fc = C x Area along trial failure plane W → weight of soil above trial failure plane H H − = BC tan π tan π½ 1 m Capacity of Driven Piles (Deep Foundations) ο€ Pile in Sand Layer Case 1 Case 2 Case 3 q = Ι€(Df − d)+Ι€′d 3rd term Ι€ = Ι€′ q = Ι€Df 3rd term Ι€ = Ι€′ q = Ι€Df 3rd term Ι€ = Ι€ave for d ≤ B Ι€ave β B = Ι€d + Ι€′(B − d) NOTE: Ι€′= Ι€π π’π = Ι€ − Ι€π€ for d > B Ι€ave = Ι€ ο€ Alternate Equation for Group Efficiency (sand only) Group of Piles ο€ Group Efficiency (sand or clay) Eff = Q des−group Eff = Q des−indiv 2(m + n − 2)s + 4d mnπD where: m → no. of columns n→ no. of rows s → spacing of piles D → diameter of pile ο€ Pile in Clay Layer Q f = PAkμ where: P → perimeter of pile A → area of pressure diagram k → coefficient of lateral pressure μ → coefficient of friction 2 Q = C√2g L H 3/2 3 Considering velocity of approach: va 3/2 va 3/2 Q = m L [(H + ) 2g −( ) 2g Q des = ο€ Triangular (symmetrical only) Neglecting velocity of approach: 3/2 Francis Formula (when C and m is not given) Considering velocity of approach: va 3/2 va 3/2 ) 2g −( ) 2g Neglecting velocity of approach: 3/2 Q = 1.84 L′ H NOTE: L’ = L L’ = L – 0.1H L’ = L – 0.2H for suppressed for singly contracted for doubly contracted Time required to discharge: t= 2As 1 1 [ − ] mL √H2 √H1 8 θ C√2g tan H 5/2 15 2 Q = m H 5/2 Q= Q=mLH Q = 1.84 L′ [(H + where: W → channel width L → weir length Z → weir height H → weir head PARAMETERS: C → coefficient of discharge va → velocity of approach m/s m → weir factor ] ] When θ=90° Q = 1.4H 5/2 ο€ Cipolletti (symmetrical, slope 4V&1H) θ = 75°57’50” 3/2 Q = 1.859 L H ο€ with Dam: Neglecting velocity of approach: 3/2 Q = 1.71 L H where: c → cohesion Nc → soil bearing factor Atip → Area of tip QTIP Critical depth, dc: Loose 10 (size of pile) Dense 20 (size of pile) Q T = Q f + Q tip ο€ Rectangular Neglecting velocity of approach: Q tip = cNc Atip (AKA Qbearing) where: pe → effective pressure at bottom Nq → soil bearing factor Atip → Area of tip Q T = Q f + Q tip QT F. S. Q des = For all sections: NOTE: E is minimum for critical depth. For rectangular sections ONLY: Take note that it is only derived from the critical depth equation. NF = 1 NF < 1 NF > 1 Reynold’s Number NR = Dv Dvρ = υ μ 3 q2 2 dc = √ = Ec g 3 Q B v2 Eπ = + dπ 2g q= where: q → flow rate or discharge per meter width EC → specific energy at critical condition vC → critical velocity vc = √gdc Laminar Flow (NR ≤ 2000) 64 hf = NR Turbulent Flow (NR > 2000) 2 L v hf = f D 2g Hydraulic Jump Height of the jump: Power Lost: βd = d2 − d1 P = QΙ€E Length of the jump: L = 220 d1 tanh NF1 −1 22 Solving for Q: 2 hf = where: Q → flow rate m3/s g → 9.81 m/s2 AC → critical area BC → critical width Q2 Ac 3 = g Bc Q2 β B c NF = √ 3 Ac β g Critical Flow Subcritical Flow Supercritical Flow QT F. S. Critical Depth where: v → mean velocity (Q/A) g → 9.81 m/s2 dm → hydraulic depth (A/B) B → width of liquid surface va 3/2 −( ) ] 2g dc (AKA Qbearing) Froude Number v NF = √gdm Q = C√2g L [(H + ) 3 2g where: C → cohesion L → length of pile α → frictional factor P → perimeter of pile Qf Q tip = pe Nq Atip Weirs Considering velocity of approach: 2 va 3/2 Q f = CLαP Q 0.0826 f L Q D5 Boundary Shear Stress For all sections: P2 − P1 = Ι€Q (v − v2 ) g 1 τ = Ι€RS P = Ι€hΜ A Boundary Shear Stress (for circular pipes only) For rectangular sections ONLY: f τo = ρv 8 q2 1 = (d1 β d2 )(d1 + d2 ) g 2 Strength Reduction Factors, Ø Load Combinations → choose largest U in design Basic Loads: π = 1.4π· + 1.7πΏ With Wind Load: π = 0.75(1.4π· + 1.7πΏ + 1.7π) π = 0.9π· + 1.3π π = 1.4π· + 1.7πΏ With Earthquake Load: π = 1.32π· + 1.1π1 πΏ + 1.1πΈ π = 0.99π· + 1.1πΈ With Earth Pressure Load: With Structural Effects: π = 0.75(1.4π· + 1.7πΏ + 1.4π) π = 1.4(π· + π) Internal Couple Method: k= Factor j: n n+ 1 j= 1− k 3 fs fc Moment Resistance Coefficient, R: 1 R = fc kj 2 Moment Capacity: 1 Mc = C β jd = fc kdb 2 β jd = Rbd2 Ms = T β jd = As fs β jd Provisions for Uncracked Section: Values Over-reinforced: → concrete fails first → fs < fy (USD) → Ms > Mc (WSD) Choose Smaller Value/ Round-down → Moment Capacity → → Balance Condition: → concrete & steel simultaneously fail → fs = fy (USD) → Ms = Mc (WSD) Choose Larger Value/ Round-up → → 5 yrs + 12 mos 6 mos 3 mos 2.0 1.4 1.0 1.0 ο€ Solve for instantaneous deflection: 4 δi = 5wL 384Ec Ie (for uniformly distributed load) ο€ Solve for additional deflection: δadd = δsus β π δadd = (% of sustained load)δi β π Say, 70% of load is sustained after n yrs. δadd = 0.7δi β π ο€ Solve for final deflection: δfinal = δi + δadd fy = 230 MPa fy = 275 MPa fy = 415 MPa 424.3.2 for fy = 275 MPa; fs ≤ 140 MPa for fy = 415 MPa; fs ≤ 170 MPa Modular Ratio, n (if not given): n= Estronger Esteel 200,000 = = Eweaker Econcrete 4700√fc′ AyΜ above NA = AyΜ below NA x bx ( ) + (2n − 1)A′s (x − d′ ) = nAs (d − x) 2 x → obtained ο€ Solve transferred moment of inertia at NA: bx 3 INA = + nAs (d − x)2 ο€ Solve transferred moment of inertia at NA: bx 3 INA = + (2n − 1)A′s (x − d′ )2 + nAs (d INA INA 3 → obtained 3 → obtained ο€ Solve for Stresses or Resisted Moment: ο€ Solve for Stresses or Resisted Moment: For concrete: For tension steel: For concrete: fs Ms β (d − x) = n INA fc = Mc β x INA Solutions for Gross Section (Singly): Mc β x INA For tension steel: fs Ms β (d − x) = n INA − x)2 For comp. steel: fs′ Ms′ β (x − 2n = INA Solutions for Uncracked Section (By Sir Erick): ο€ Location of neutral axis, NA: AyΜ above NA = AyΜ below NA x d−x bx ( ) = b(d − x) ( ) + (n − 1)As (d − x) 2 2 x → obtained ο€ Location of neutral axis, NA: Ig = π 1 + 50π′ Structural Grade ASTM Gr.33 / PS Gr.230 Intermediate Grade ASTM Gr.40 / PS Gr.275 High Carbon Grade ASTM Gr.60 / PS Gr.415 AyΜ above NA = AyΜ below NA x bx ( ) = nAs (d − x) 2 x → obtained 409.6.2.4. For simply supported, Ie = Ie (mid) For cantilever, Ie = Ie (support) π= where: f’c → compressive strength of concrete at 28 days fy → axial strength of steel ο€ Location of neutral axis, NA: yt = 409.6.2.5. Factor for shrinkage & creep due to sustained loads: time-dep factor, ξ: fc = 0.25 f’c fs = 0.40 fy ο€ Location of neutral axis, NA: ο€ Solve for effective moment of inertia, Ie: Mcr 3 Mcr 3 Ie = ( ) β Ig + [1 − ( ) ] β Icr Ma Ma Ie mid + Ie support Ie = 2 fc = 0.45 f’c fs = 0.50 fy ο€ Vertical members (i.e. column, wall, etc.) Solutions for Cracked Section (Doubly): 409.6.2.3. if Ma < Mcr, no crack; Ig = Ie if Ma > Mcr, w/ crack; solve for Ie 3 ο€ Horizontal members (i.e. beam, slab, footing, etc.) 424.6.4 n must be taken as the nearest whole number & n ≥ 6 424.6.5 for doubly, use n for tension & use 2n for compression (for simply supported beam) ο€ Solve for inertia of cracked section: bx 3 Icr = + nAs (d − x)2 Allowable Stresses (if not given): Solutions for Cracked Section (Singly): fc = ο€ Solve for inertia of gross section, Ig. ο€ Solve for cracking moment, Mcr. ο€ Solve for actual moment, Ma: 2 wL Ma = 8 Design Conditions Under-reinforced: → steel fails first → fs > fy (USD) → Ms < Mc (WSD) π = 1.4π· + 1.7πΏ + 1.7π» π = 0.9π· π = 1.4π· + 1.7πΏ Factor k: (a) Flexure w/o axial load ……………………… 0.90 (b) Axial tension & axial tension w/ flexure .… 0.90 (c) Axial comp. & axial comp. w/ flexure: (1) Spiral ……………………………….………. 0.75 (2) Tie …………………….…………….………. 0.70 (d) Shear & torsion ……………………….………. 0.85 (e) Bearing on concrete ……………….…,……. 0.70 Working Strength Design (WSD) or Alternate Strength Design (ASD) h ; y → obtained 2 t ο€ Solve moment of inertia of gross section at NA: 3 bx 12 Ig → obtained ο€ Solve for cracking moment: Mcr β yt Ig → obtained fr = 0.7√fc′ = Mcr ο€ Solve transferred moment of inertia at NA: 3 3 bx b(d − x) + + (n − 1)As (d − x)2 3 3 → obtained INA = INA ο€ Solve for Stresses or Resisted Moment: For tension steel: For concrete: fc = Mc β x INA fs Ms β (d − x) = n INA d′) Ultimate Strength Design Steel Ratio ο€ Based in Strain Diagram: ο€ Ultimate Moment Capacity: εs 0.003 = d−c c d−c εs = 0.003 ( ) c d−c fs = 600 ( ) c Mu = ∅Mn Mu = ∅R n bd2 10 Mu = ∅fc′ bd2 ω(1 − ω) 17 fy ω=ρ ′ fc ο€a ο€ Coefficient of resistance, Rn: = β1 c a → depth of compression block c → distance bet. NA & extreme compression fiber Provisions for β1: * 1992 NSCP β1 = 0.85 − 0.008(fc′ − 30) * 2001 NSCP 0.05 7 * 2010 NSCP β1 = 0.85 − 0.05 7 − 10 17 (fc′ − 28) Maximum & Minimum steel ratio: 0.85fc′ 2R n [1 − √1 − ] fy 0.85fc′ Singly Reinforced Beam INVESTIGATION Singly Reinforced Beam DESIGN Computing MU with given As: Computing As with given WD & WL: ρmin Doubly Reinforced Beam (DRB) ρ > ρmax (rectangular only) As > As max (any section) Doubly Reinforced Beam Investigation if SRB or DRB: a = β1 c c → obtained (3rd) Solve for steel ratio, ρ: d−c fs = 600 [ ] c fs → obtained ρ= (4th) Solve for area of steel reinforcement, As and required no. of bars, N: C=T 0.85fc′ ab = As fs 0.85fc′ β1 cb = As β 600 [ d−c c ] c → obtained a = β1 c a → obtained (3rd) Solve for Moment Capacity: a Mu = ∅(C or T) [d − ] 2 Mu = ∅(0.85fc′ ab) [d a Mu = ∅(As fs ) [d − ] 2 a − ] 2 or As = ρbd As ρbd N= = 2 π Ab d 4 b If As < As max Solve the given beam using SRB Investigation procedure. If As > As max Solve the given beam using DRB Investigation procedure. Doubly Reinforced Beam DESIGN Computing As with given Mu: (1st) Solve for nominal M1: 0.85fc′ β1 600 fy (600 + fy ) ρmax = 0.75ρb As1 = 0.75ρb β bd ρb = M1 = (As1 fy ) [d − ] 2 (2nd) Solve for nominal M2: MU M2 = − M1 ∅ (3rd) Solve for As2: M2 = (As2 fy )[d − d′] As2 → obtained Doubly Reinforced Beam INVESTIGATION Computing MU with given As: (1st) Compute for a: Cc + Cs = T 0.85fc′ ab + As ′fs ′ = As fs 0.85fc′ ab + As ′fy = As fy a → obtained a = β1 c c → obtained d−c ] c fs → obtained fs = 600 [ If fs > fy, tension steel yields; correct a. If fs < fy, tension steel does not yield; compute for new a. c − d′ ] c fs ′ → obtained fs ′ = 600 [ If fs’ > fy, compression steel yields; correct a. If fs’ < fy, compression steel does not yield; compute for new a. (2nd-b) Recomputation: C=T 0.85fc′ ab + As ′fs ′ = As fs (4th) Solve for # of tension bars: NOTE: Use fs & fs’ as As As1 + As2 N= = π 2 Ab d 4 b fs = 600 [ (5th) Solve for fs’: c → obtained fs ′ = 600 [ a π (2nd) Check if assumption is correct: (2nd) Solve for given As & compare: (2nd-b) Recomputation: As πππ₯ = ρπππ₯ π bd a π = β1 cπ a π → obtained As max = 0.75As π If ρmin < ρ < ρmax, use ρ. If ρmin > ρ, use ρmin. If ρ > ρmax, design doubly. c − d′ ] c If fs’ > fy, compression steel yields; As’ = As2. If fs’ < fy, compression steel does not yield; Use fs’ to solve for As’. (6th) Solve for As’: As ′fs ′ = As2 fy (7th) Solve for # of compression bars: N= As ′ bd (assume tension steel yields fs=fs’=fy) C=T 0.85fc′ a π b = As π fy As π → obtained ρmin ≤ ρ ≤ ρmax ρπππ₯ π = 0.75ρb π + 600d 600 + fy cπ → obtained (2nd) Solve for Asmax: 0.85fc′ 2R n [1 − √1 − ] fy 0.85fc′ Check: If fs > fy, tension steel yields; correct a. If fs < fy, tension steel does not yield; compute for new a. d − cπ fs = fy = 600 [ ] cπ cπ = MU ∅bd2 As′ bd (1st) Compute for ab: Thus, (2nd) Solve for coeff. of resistance, Rn: (2nd) Check if assumption is correct: 75 mm → column footing → wall footing → retaining wall ρb π = ρb π + WU = 1.4WD + 1.7WL WU L2 (for simply supported) MU = 8 Rn = 40 mm → beam → column ρ < ρmax (rectangular only) As < As max (any section) (1st) Compute ultimate moment, Mu: a → obtained ρmin 20 mm → slab √fc′ = 4fy Balance Condition for Doubly C=T 0.85fc′ ab = As fs 0.85fc′ ab = As fy 1.4 = fy Singly Reinforced Beam (SRB) (1st) Compute for a: (assume tension steel yields fs=fy) Minimum Concrete Covers: ρmax = 0.75ρb As max = 0.75As b Singly or Doubly ? As ρ= bd ρ= 0.85fc′ β1 600 fy (600 + fy ) (choose larger between the 2) Mu Rn = ∅bd2 ο€ Combined ρ & Rn: (fc′ − 30) ρb = ω) ο€ Steel reinforcement ratio, ρ: 0.65 ≤ β1 ≤ 0.85 β1 = 0.85 − Rn = fc′ ω(1 Steel ratio for balance condition: As As′ = Ab π d 2 4 b d−c ] c c−d′ fs ′ = 600 [ c ] a = β1 c a → obtained (3rd) Solve for Moment Capacity: a Mu = ∅Cc [d − ] + ∅Cc [d − d′] 2 a Mu = ∅(0.85fc′ ab) [d − ] 2 + ∅(As ′fs ′)[d − d′] or a Mu = ∅T [d − ] 2 a Mu = ∅(As fs ) [d − ] 2 Design of Beam Stirrups (1st) Solve for Vu: NSCP Provisions for max. stirrups spacing: ΣFv = 0 Vu = R − wu d wu L Vu = − wu d 2 NSCP Provisions for effective flange width: NSCP Provisions for minimum thickness: i. Interior Beam: ii. exterior Beam: L bf = 4 L bf = bw + 12 s1 bf = bw + 2 bf = bw + 6t f Cantilever Simple Support One End Both Ends Slab L/10 L/20 L/24 L/28 Beams L/8 L/16 L/18.5 L/21 Factor: [0.4 + smax = d or 600mm 2 ] [1.65 − 0.0003ππ ] (for lightweight concrete only) Minimum Steel Ratio For one-way bending: k → steel ratio ii. when Vs > 2Vc, (3rd) Solve for Vs: smax = Vu = ∅(Vc + Vs ) Vs → obtained d or 300mm 4 i. fy = 275 MPa, k = 0.0020 ii. fy = 415 MPa, iii. & not greater than to: (4th) Theoretical Spacing: smax = n 3Av fy k = 0.0018 iii. fy > 415 MPa, n b k = 0.0018 [ Vs NOTE: 400 fy ] For two-way bending: ρ → steel ratio fyn → steel strength for shear reinforcement Av → area of shear reinforcement n → no. of shear legs Av = fy 700 i. when Vs < 2Vc, 1 Vc = √fc ′bw d 6 s= Thickness of One-way Slab & Beam s1 s2 bf = bw + + 2 2 bf = bw + 8t f 1 2Vc = √fc ′bw d 3 (2nd) Solve for Vc: dA v fy T-Beam ρmin = π 2 d βn 4 1.4 √fc′ ρmin = fy 4fy (choose larger between the 2) Design of One-way Slab LONGITUDINAL OR MAIN BARS (1st) Compute ultimate moment, Mu: (6th) Compute steel ratio, ρ: WU = 1.4WD + 1.7WL WU L2 MU = 8 ρ= (11th) Solve for As: As bd As = kbβ« h NSCP Provision for k: i. fy = 275 MPa, k = 0.0020 ii. fy = 415 MPa, k = 0.0018 iii. fy > 415 MPa, k = 0.0018 (400/fy) (7th) Check for minimum steel ratio: (2nd) Solve for slab thickness, h: See NSCP Provisions for minimum thickness. ρmin = (3rd) Solve for effective depth, d: d = h − cc − TEMPERATURE BARS/ SHRINKAGE BARS √fc′ 1.4 & ρmin = fy 4fy (12th) Determine # of req’d temp. bars: If ρmin < ρ, use ρ. If ρmin > ρ, use ρmin & recompute As. db 2 N= (8th) Determine # of req’d main bars: (4th) Solve for a: a As As = 2 π Ab d 4 b N= Mu = ∅(C) [d − ] 2 a Mu = ∅(0.85fc′ ab) [d − ] 2 a → obtained (13th) Determine spacing of temp. bars: s= (9th) Determine spacing of main bars: s= (5th) Solve for As: C=T 0.85fc′ ab = As fy As → obtained As As = Ab π d 2 4 b b N b N (14th) Check for max. spacing of temp. bars: smax = 5h or 450mm (10th) Check for max. spacing of main bars: smax = 3h or 450mm Design of Column TIED COLUMN SPIRAL COLUMN P = PC + PS P = 0.85fc′ (Ag − Ast ) + Ast fy PN = 0.8P PU = ∅0.8P ; ∅ = 0.7 PU = (0.7)(0.8)[0.85fc′ (Ag − Ast ) + Ast fy ] PN = 0.85P PU = ∅0.85P ; ∅ = 0.75 PU = (0.75)(0.85)[0.85fc′ (Ag − Ast ) + Ast fy ] ρ= Ast Ag No. of main bars: Thus, P Ag = ′ 0.85fc (1 − ρ) + ρfy 0.01Ag < Ast < 0.08Ag Design of Footing qA = qS + qC + qsur + qE qE = P A ftg ; qU = PU Aftg where: qA → allowable bearing pressure qS → soil pressure qC → concrete pressure qsur → surcharge qE → effective pressure qU → ultimate bearing pressure Ø = 0.85 Spacing of bars: Ast N= Ab ρs = 0.45 s = 16db s = 48dt s = least dimension N is based on Pu. NOTE: If spacing of main bars < 150mm, use 1 tie per set. fc′ Ag volume of spiral [ − 1] = fy Ac volume of core π (dsp )2 β π(Dc −dsp ) 4Asp s=4 π = Dc ρs (D )2 β ρs 4 c WIDE BEAM SHEAR PUNCHING/DIAGONAL TENSION SHEAR BENDING MOMENT VU1 = qU (B)(x) VU2 = PU − qU (a + d)(b + d) x MU = qU (B)(x) ( ) 2 VU1 ≤ ∅Vwb = ∅ τwb = VU1 ∅Bd τwb(allw) = √fc′ Bd 6 VU2 ≤ ∅Vpc = ∅ τpc = √fc′ 6 VU2 ∅bo d τpc(allw) = √fc′ 3 √fc′ b d 3 o ** design of main bars and temperature bars – Same as slab.