Civil Engineering Equations

7/25/2019 Civil Engineering Equations

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CIVIL ENGINEERING EQUATIONSTopic ...................................................................................................... Page #Environmental.................................................................................................3Wastewater Flow Rates ....................................................................................3Unit Processes ..................................................................................................3

Treatment Process: ..........................................................................................3Biochemical Oxygen Demand ..........................................................................4Chemical dosage rate.......................................................................................4Ultimate BOD....................................................................................................5Sludge Digestion...............................................................................................6Amount of Heat Produced ................................................................................7Stream Purification ............................................................................................9Water Demand................................................................................................10Landfills...........................................................................................................10Number years use of a Landfill .......................................................................11Groundwater....................................................................................................11

Hydrology......................................................................................................12General Equations...........................................................................................12Force on an inclined plane ..............................................................................13To find determine measuring fluid in manometer ............................................13Horsepower.....................................................................................................14Bernoulli Equation ...........................................................................................14Friction Head Loss ..........................................................................................14Darcy Equation................................................................................................15Hazen Williams................................................................................................15Pipes in Parallel...............................................................................................15Open Channel Flow.........................................................................................16

Depth of flow in a channel ..............................................................................19Total Head Loss in Pipe..................................................................................19Minimum Pipe Size.........................................................................................19Rainfall Runoff.................................................................................................19Hydraulic Jump................................................................................................20Find water depth upstream or downstream of critical depth ...........................20Water depth downstream of hydraulic jump....................................................20Energy Loss in a Hydraulic Jump ...................................................................20Geotechnical .................................................................................................20Earthwork ........................................................................................................20Volumetric Properties ......................................................................................22

Soil Classification ............................................................................................24General Size Classes...............................................................................26

Soil Stress .......................................................................................................26Settlement.......................................................................................................29Consolidation...................................................................................................30Retaining Walls ...............................................................................................32Retaining Wall Analysis...................................................................................33Pile Analysis....................................................................................................35


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Structural.......................................................................................................38Reinforced Concrete Footings.........................................................................38Surveying/Transportation ............................................................................39General Equations...........................................................................................39Vertical Curves................................................................................................41

Minimum length required to make a transition................................................41Elevation on curve @ Sta. X...........................................................................41Elevation at high point on curve......................................................................41Passing Sight Distance...................................................................................42Bearings..........................................................................................................42Horizontal Curves...........................................................................................43Vertical Curves ...............................................................................................44

Asphalt Properties...........................................................................................49Economics ....................................................................................................50General Equations...........................................................................................50Ranking Alternative Projects...........................................................................51

Annual Costs ..................................................................................................51Rate of Return ................................................................................................51Perpetual Lives...............................................................................................52Depreciation and Taxes..................................................................................52Ranking Alternatives.......................................................................................53Concrete Mix Design.......................................................................................54Critical Path Analysis.......................................................................................55Metric Conversions ......................................................................................59Revision History ...........................................................................................73


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CIVIL ENGINEERING EQUATIONSEnvironmental

Wastewater Flow Rates

o Approx. 70-80% of water supply returns as wastewatero Sanitary sewer sizing assumed to average 100 to 125 gpcdo Infiltration limited typically to 500 gpd/mi-in

o Modern piping materials reduce infiltration to 200 gpd/mi-in or lowero Infiltration can also be estimated to be 3 to 5% of the peak hourly

rate or 10% of the average rate

Ratio of peak hourly flow to average hourly flow can be calculated from thefollowing:

18

4

peak

average

Q P

Q P

+=

+

Organic Loading of a treatment plantExpressed in BOD (lb/day)

/ /

,1000

6

( )(8.345 )

10 (1000 ) 0.2

mg L gal day

eqiuvalent

lb LBOD Q

MG mg P

gal lbpersons

MG person day

=

Sewer Velocitieso

Sewers should be 2 ft/sec at a minimum to be self-cleaningo Minimum slope needed is listed on table 28.8 on Pg 28-4 CERM

Sewer Sizingo Use Mannings Equation to size gravity sewerso Depth of flow at design flow should be less than 70 to 80% of pipe

diameter

Street Inlets/Gutterso Use Mannings Equation to size gutters

Unit ProcessesTreatment Process:(1) Pretreatment of wastewater large solids removed (debris)(2) Grit Removal heavier organic solids settle out(3) Primary Clarifier water circulates to remove sludge that settles out and

floats on surface(4) Aeration air is bubbled into tanks to supply a maximum amount of

oxygen to the microorganisms that come in the wastewater. These


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microorganisms provide the biological treatment of the wastewater bydecomposing the biological compounds. Retention time is typically 8hours.

(5) Secondary Clarifier Heavier solids such as dead microorganisms orundigestible material settle out.

(6) Chlorination/dechlorination/aeration chlorine is added to kill pathogensby destroying coliform bacteria and associated pathogens (but notsterilization). Sulfur dioxide (SO2) is added to dechlorination. Finalaeration is to raise the dissolved oxygen levels before discharge.

(7) Discharge

Biochemical Oxygen Demand

5i f

sample

sample dilution

DO DOBOD

V

V V

=

+

Where: DO = concentration of DissolvedOxygen (mg/L)V = volume of sample & dilutionamount (mL)

BOD at time tis known as the BOD exertion.

(1 10 )dK tt u

BOD BOD = where Kd= deoxygenation rate contact(approx. 0.05 to 0.1 day-1)t = time (days)

Volume of tank needed to treat wastewatergiven daily flow and detention time

tan detentionkV flow rate time=

Convert flow from MGD to ft3/day (3 31

7.48

ft ftMGD

gal day = )

Therefore volume is tank is:3

tan

160 sec

86,400 seck

dayftV

day=

Chemical dosage rateGiven a daily flow and a dose rate, how many pounds of stuff must be applied toachieve a required concentration?

8.345 lb liters(dose in mg/l)(daily flow in MGD) # per day

mg MG

=

To determine the velocity difference between a paddle and the water.Find paddle velocity:

2

60p

rnV

=

Where: r = paddle radius (ft)n = paddle rotationspeed (rpm)

The velocity differential is:

(1.0 fractional difference between paddle & water velocities)d pV V=


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To determine horsepower required for turning paddles:Find power reqd:

3

2d dC ADVP

g=

Where: Cd= coefficient of dragD = Density (62.4)g = gravity constant (32.2)

A = Area (LW on rect tank)Divide P by 550 to get Hp

To find hydraulic detention time:Tank Volume

Flowd

VT

Q= =

and convert to min via:3

3

1440 min7.48

1 1

ft gal

MGD ft day

To find mean velocity gradient (G)

tank

PG

V=

Where P = power

= dynamic viscosityV = Tank Vol.

Mixing Opportunity Parameter:dt d

G G t=

To find the surface overflow rate of a sedimentation basin for a given flow:

Surface Overflow RateSurface Area

Q=

Surface Area = LW (for rectangular tank)=

2

(for circular tank)4

d

To determine weir overflow rate:

Weir Overflow Rateweir overflow length

Q=

Ultimate BODo BODultimateis the total oxygen used by carbonaceous bacteria if the test is

run for a long time (20 days)

51.47uBOD BOD

When lime is added to Carbon Dioxide in water: Calcium Carbonate & water areproduced.

Add Soda Ash to remove CaSO4


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When Magnesium Sulfate reacts with lime: Calcium Sulfate does not precipitateout.

Total Hardness = Ca+2+ Mg+2

Carbonate Hardness =3HCO = alkalinity

Non-carbonate hardness = total hardness carbonate hardness

Lime and Soda Ash Softening

To determine how much of each multiply equivalent weight of required lime orsoda ash by quick lime (CaO = 28) or Soda Ash (Na2CO3= 53) by milli-equivalents to get mg/L of the substance needed

Added Lime = Ca + Mg + CO2+ excess

Added Soda Ash = Non-carbonate hardness

To calculate Lime add up meq of CO2 + Ca + Mg then multiply by equivalentweight of quick lime (CaO = 28) to get mg/L of CaO needed and multiply by 8.34to get lb/million gallons

Sludge Digestion

Amount of solids removed by the primary clarifier:

( ) ( )

mg 8.34 Lb liters

Flow in MGD SS conic in % SS Primary Clarifier RemovesL mg MG

l

d

= Total Amount Solids producedby WW plant:

o Find Amount solids removed by Primary Clarifier (see above)o Find Amount solids removed by Secondary Clarifier (mg/l):

(Solids going in + Solids Already In) Removal Rate

o Solids Going In:

inSolids Raw Water Solids (mg/l) (1 - % removed by Primary Clarifie=

o Solids Already In (Produced Inside)

removed Removed Primary

5 5 5 5BODIN effluent BOD BOD BOD=

o To find total amount, convert mg/l to lb/day by multiplying by flowand conversion factor.

o Total Solids = Primary Weight + Secondary Weight

DetermineAmount of volatile Solids sent to digester:


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volatile solids total solidsWt = Wt % Volatile Solids

Weight of Volatile Solids Digested:

volatile digestedWt Digestion RatevolatileWt=

Amount of Heat Produced3

total

solids 3

ft BTUHeat Produced=Wt given rate in heat rate in

ft

BTU

lb day

=

Amount of Heat Required to heat sludgefrom x temp to y temp:

solidsWt BTU 1 dayHeat Req'd= Temp BTU rate (in )% Solids in sludge lb F 24 hrs

Total BTU

hr =

o

Amount of Heat Required to maintain temperaturegiven a certain loss in BTU/hr

and efficiency (E):Heat Lost + Heat Temp

E=Bacteria Heat + Added Heat

Bacteria heat is the total heat produced (convert from BTU/day to BTU/hr)Rest should be given except for Added Heat, solve for that (answer in

BTU/hr or Millions BTU/hr)

Primary clarifier Suspended Solids removal:= amt. of W.W. treatment plant handles * Suspended Solids * removal rate ofS.S. * (8.34)(conv. factor)

BOD5removed by Secondary Clarifier = BOD5amt. Removed by Prim. - BOD5not removed

S.S. produced by Sec. Clarifier = BOD5removed * plant production of S.S.

To determine depth of two parallel clarifiers with a given overflow rate and peakoverflow rate and a hydraulic detention time.

o Find depth: d

2 clarifiers

daily flow t

surface areadepth

=

o Find surface area for average and peak flows, use the larger number:

odaily flow

average overflowavgSA =

opeak flow

peak overflowpeakSA =

o3

ddaily flow t 1 day 1 ft

24 hrs 7.48depth

SA gal

=


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To determine diameter of two parallel filters using the National Research CouncilEquation:

o Use the equation for surface area to find diameter:2

4

dSA

=

o Use the National Research Equation to find the volume:

1

1 0.0085

EW

VF

=

+

Where: E = Filter efficiencyW = #/day BOD goinginto filterV = Tank VolumeF = (see below)

o W (BOD loading)= pounds of BOD5coming into filter after passingthrough clarifiers (which remove % given of BOD5)

mg 8.345 lb liter dose in (1 % ) = lbm/day going into filter

mg MGW MGD given

l

=

Solving equation for V Yields:

2

2

0.0085

11

WV

FE

=

o To find filter efficiency, we need to know how much BOD5goes intofilters and how much comes out. Amount coming out should be given,concentration of BOD5coming is given as well.

o Subtract % of BOD5removed from concentration going ino Filter efficiency is:

= =

in out

(concentration (1-% removed) - concentration ) (250 (1 0.35)

(1 % removed) 250 (1 0.3inE concentration

o Since the water does not re-circulate F = 1o Now find V using the formula above (units: acre-feet)o Convert to cubic feet (43,560 ft3= 1 acre-foot) then divide by

number of tankso Since depth of tank should be given use this value for the

Volume & solve Area=Volume/Depth to find the Area then use

that to find the diameter of the tanks(note area circle: 24

d )

To determine diameter of two filters using the Velz Equation:

1010 p

DdL

L

= Where: L = removable portion of incoming

BODuLd= Portion of BODu that remainsafter filtrationDp= Required depth of filter to


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remove BODuNOTE: This equation uses BODuNOTBOD5K = rate constant for high or lowtrickling filters

o L is found by multiplying incoming BODuby the % that will be removed(i.e. 90%)

o Ldis the portion that remains after filtration (effluent concentration - %BODuthat remains)

o Plug in values and solve Velz equation for Dp: dL

logL

K D

=

o Now find the surface area in the following equation:

u p#/ (BOD Loading Rate)(Surface Area)(D )day=

(NOTE: Surface area will be for the total number of filters

given)o Plug in surface area for one filter into 2

4SA d

= and solve for d

To find removal efficiency:o Given % removal of primary clarifier and % removal of trickle filtero % Primary removed is x%-o IF secondary filters are in parallel, use entire percentage for removal, if

filters are in series, then each filter gets a share of the overall percentage(i.e. for 3 filters, each one gets 1/3 of the removal percentage).

o Total efficiency:

secondary% Primary removed + % efficiency (% reamining after primary)

Stream Purification

Flow from treatment plant: Q = VA

Temperature after dischargegiven discharge flow and temp and flow andtemperature of stream.

(weighed average)discharge discharge

discharge

stream stream

mix

stream

T Q T QT

Q Q

+=

+(note Temps in C)

Find BOD5 in stream after discharge:o Use weighted average like above but BOD5instead of Temp.

Determine BODu in stream after discharge.o Find BODufor stream and dischargeo Use weighted average like above


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Given stream velocity and aeration constant, find BOD level x miles belowdischarge.

o Use Streeter-Phelps equation to find Dt:

( )10 10 10mix d r rd u k t K t k t t ar d

k BODD D

k k

= +

distance(in miles)

(in ft/sec)t

velocity=

Where: kd= deaeration coeffkr= reaeiration coeff

BODu= ult. BODt = time (days)Da= DOsat-DOmix

(DOmixis weighted avglike above)

Water Demand

Total water use per capita:gpd

population(units gal per capita per day)

Average Annual Daily Flow:daily flow for each month

12AADF

=

To find seasonal difference in AADF:

o Find ADF for season ( )seasonADF ADF multiplier=

o Find difference in ADFs: 1 2season seasonADF ADF ADF =

To find maximum demand: Demand ADF multiplier=

To find future demand: Future Demand = Max Demand growth(i.e. use 1.5 for a 50% increase)

Fire Fighting demand is based on STORAGE capacity NOT treatment capacity

Water Demand = AADF * Max. daily flow multiplier

Landfills

o Typical Daily Cover Depth is 6 to 12o Typical landfill caps are around 24 (Pg 31-3 CERM, Subtitle D Landfills)o Glass recycling can reduce waste by 4% - 16% (Pg 31-2 has typical

percentages of different types of waste in municipal landfills)ko Waste is compacted in landfill to 1000 1250 pcy for design

Refuse Generated: population .refuse generated per capita

For Size of a Landfill:


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o Find Volume of lost minus buffer zone (remember to subtract buffer fromall sizes on width)

o If Landfill is a berm (above ground), volume will consist of 1 cube (inmiddle), 4 triangles and 4 pyramids.

o Vcube= LWH

o triangles 1V (# )2

bhl Sides=

o pyramids1

V (# )3

baseA h Sides=

o totalV V=

Number years use of a Landfillo Find yearly demand based on population and per capita refuseo Convert to acre-feet and then factor in volume of cover. (i.e. if refuse in

24 lifts with 6 cover then find yearly total demand as follows:

= demand (Refuse lift thickness + cover thickness)V Refuse lift thicknesstotal

in placeV

=in placeRefuse generated

Refuse densityV

o availabletotal

demand

VYears =

V

Remaining life of landfillgiven change (population, refuse generation, etc.)o Find volume already usedo Subtract that volume from total volumeo

Find new yearly demand based on new rate (i.e. for a 5% reductionmultiply old rate by 0.95)o Divide volume remaining by new yearly volume demand to get years

remaining

GroundwaterDupuit Equation:

=

2 2

0( )

( )

w

O

w

K h hQ

rLn

r

Where: ho= aquifier thickness

hw= ho drawdowndrawdown = ho - hw

ro= radius of influencerw= radius of pipe in ft.K = Coefficient of Permeability

NOTE: Be sure to convert Q from gpm to cfs

If an aquifer is free to move up and down, it is referred to as an unconfinedaquifer. The phreatic zone (water table) is the fluctuating upper boundary.


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Water trapped in a confined or artesian aquifer is trapped by rock or dense soiland does not freely move up and down.

To find the coefficient of permeability (K) of an aquifer (assuming an unconfinedaquifer), use the Dupit Equation and solve for K:

2 2

0 1( )

log oew

K h hQ

r

r

=

Where: Q = flow

ho= aquifer depth (ft)hw= drawdown (ft)ro= well radius of influcencerw= pipe radius of influence

2 2log

( )o

o w w

rQK

h h r

=

To find the drawdown of a well with a given flow solve the Dupit Equation for hwThen subtract that from the aquifer depth (ho) to get the drawdown

Hydrology

General EquationsAbsolute Pressure: Pabs= Pg* Patm

where Patm= 2116.8' at sea level

Absolute Force: Fabs= Pabs* Area

Atmospheric Pressure in terms of Head of Mercury (Hg): hHg= Patm/ (Hg* g)

where Hg= 13.6(sp.grav.) * 1.94 and g = 32.2Gage Pressure: Pg= * g * hwhere = 1.94, g = 32.2, h = height

Bernoullis Equation:2 2

1 1 2 21 2

2 2f t p

P V P V z z h h h

g g g g + + = + + + +

Cross-Sectional Area: 2

4A d

=

Darcy-Weisach Equation (head loss):2

2f

fLVh

Dg=

Energy Grade Line: 2 2 2EGL E z = +

Force at Fx: 1 1 1 2( )xF P A Q V V = +

Force at Fy: 2 2 1 2( )yF P A Q V V gH = + +

Form Loss (head of mercury): atmHgHg

Ph

g=

Form Loss:2

2m

Vh K

g

=


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Faroude Number: cV

Fgy

=

Hazen-Williams Equation (form loss):1.85

1.85 1.165

3.012f

V Lh

C D=

Horsepower:550

pgh QHPe

=

Hydraulic radius:4

h

Dr =

Manning Equation (flow):2

31.49

hQ Ar Slope

n=

Manning Equation (form loss):2 2

432.22

f

h

V n Lh

r=

Pressure head, hydraulic grade line:2

p

Ph

g

=

Rational Formula (design flow from rainfall): Q=CIA

Reynolds Number:VD

R

=

Specific Energy:2

22 2

2

VE y

g= +

Specific Gravity:

2

xx

H O

SG

=

Time of Concentration:0.77

0.385

0.0078c

LT

Slope

=

Velocity Head:2

2v

Vh

g=

Force on an inclined plane1. Find pressure at top of plane (Ptop)2. Find pressure at bottom of panel (Pbottom)

3. Find average force2

top bottomP P+

4. Find the force (F=PaverageA)

To find determine measuring fluid in manometer1. The pressure in the water will be equal to the measuring fluid:

2

2

0

0

H fluid

H fluid

P P

gh gh

=

= solve for fluid


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2. Find the specific gravity of the fluid:

2

fluid

H O

SG

= , look up SG to find fluid

Horsepower

Q = A * V

=2

4pipe

dA

=2

2

vVelocity Head

g where g= 32.2

Bernoulli Equation

+ + = + + + + 2 2

1 1 2 21 2

2 2

f t p

P V P V z z h h h

g g g g

when determining HP of pump, you can assume hp= z2- z1

Horsepower of a pump:

=550

pgh QHP

e

where e = efficiency of pump and hpis rated horsepower of pump

Turbine Horsepower:

=550

p

turbine

gh QEHP

Friction Head Loss

Reynolds # =

V diameter

Where = coskinematic vis ity

Temp (F)

Kinematic Viscosity

510 / secft 32 1.931

40 1.664

50 1.410

60 1.21770 1.059

80 0.930

90 0.826

100 0.739


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Darcy Equation

Relative Roughness (E/D) = roughness / dia.use Moody Diag. (17-6)Begin on Right side & go left along the curve until you meet Refrom the bottom,

then go straight left to get the friction factor.

Friction Head Loss(hf) =

2( )

(2 )

friction factor length of pipe V

diameter g

Pump Head (hp) = elev2 elev1+ hf

Hazen Williams

Friction Head Loss:

=

1.85

1.85 1.165

3.022 ( )

( )f

v Length of pipeh

C pipe diameter

where C = Hazen-Williams Roughness Coefficient

FORM LOSS (minor loss) = (K * V2) / (2 * g) ????Page 3 of original notes

situation K valueordinary pipe inlet 0.5gate valve fully open 0.25sudden contractiond2 / d1 = 3/4 0.25

d2 / d1 = 1/2 0.43d2 / d1 = 1/4 0.49

Pipes in Parallel

1. total a bQ Q Q= + 2. Assume HL6= HL123. Use Head Loss Equation (Darcy, Hazen-Williams or Manningdepending on givens) with velocities unknown, solve for one of theunknown velocities (Vaor Vb)

Pipe A

Pipe B

A B


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4. Plug velocity from #3 into total a a b bQ V A V A= + and solve for V (i.e. if Vafound in #3 then find Vb)5. Find Q for that pipe (Q=AV)

Open Channel Flow

TablesPg A-18 to A-23 Pipe DimensionsPg A-27 Roughness values for Hazen WilliamsPg A-28 to A-31 Darcy Friction Factors

Hydraulic Parameters of Basic Channel Sections

Section Area (A) WettedPerimeter (P)

Hydraulic Radius(R)

dw 2d+w2

dww d+

tan

db d

+

2sin

db

+

2sin cos

sin 2

bd d

b d

++

2

tan

d

2zy

2

sin

d

2 22 ( )y zy+

cos

2

d

22 1

zy

z+

( )

( )

21 sin8

in radians

D

1

2D

1 sin1

4D

For pipe flowing fullor full 4

d

Open Channel Flow Chapter 19 CERM


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Uniform Flow Slope of water surface is same as slope of channel so assumethe depth does not change

Mannings Equation

Q=AV 23

2

3

2

3

1.49

1.49

1

Q A R S n

V R Sn

Q A R S n

=

=

=

Energy Equation: 22 2

1 1 21 2

2 2L

PP V Vz h z

g g + + = + +

In an open channel:2 2

1 21 2

2 2L

V Vz h z

g g+ = +

Essentially Zero dueTo roughness

Specific Engergy

Flow Characteristics

LaminarTurburlent

SupercriticalSubcritical

Reynolds Number (Rn, Nr)H

R

VRVDN

= =

Laminar Flow Reynolds Number < 500 for Open ChannelsTurbulent Flow Reynolds Number > 2000 for Open Channels

Froude Number (Fr) Rh

vF

g D=

Dh= Hydraulic DepthFor a rectangular

channel: Dh= Depth

For everything else:

h

areaD

width= @top

Subcritical Flow Faroude # < 1

(US)

(SI)

Reynolds Number

Faroudes Number


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Supercritical Flow Faroude # > 1Critical Flow Faroude # = 1

Conjugate Depths are depths where thesame energy is possible

Critical Depth will occur when Faroude # = 12

3R

Q TF

gA=

Example:

Given a channel cross section and flow, find normal depth of flow. Find the flowregime (laminar, turbulent, sub/supercritical) and find the critical depth.

Use Trial & Error or Solver

2

31.49

Q AR S n

= RW

= (W is wetted perimeter, Table on Pg 19-3)

Example:

Given this channel with the following

parameters:Q = 10 m

3/sec

n= 0.012S = 0.0016 m/mFind Normal Depth (dn), Critical Depth (dc),Critical Velocity (vc) and Critical Slope (Sc)

Area = 3y2 2 22 ( )WP y zy= +

2

31.49

Q AR S n

= 22

2 3

2

1 310 (3 )( ) 0.0016

0.012 2 10

yy

y= y = 1.205 m

A = 3y2T = 2zy = 2(3y) = 6y

A

P


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Critical Flow(Faroude # = 1)22

3 3

(10) (6 )1 1.178

(9.81)(3 )

cc

c

yQ TF y m

gA y

= = =

Critical Velocity:2

2

103 2.4 /

3(1.178)c c c c

c

QV where A y V m s

A= = = =

Critical Slope:2

31.49

Q AR S n

= 22

2 31 3(1.178)

10 (3(1.178) )( ) 0.00180.012 2 10

c cS Sy

= =

Depth of flow in a channel1. Find Area & Hydraulic Radius using y for depth2. Plug into Mannings Equation and use trial & error to find given flow

Total Head Loss in Pipetot inlet length outlet

L L L Lh h h h= + + Typically use Form Loss for Inlet & outlet and eitherMannings or Hazen-Williams for Length dependingon givens.

Minimum Pipe Size1. Find Elevation Change (this is the minimum head loss) from upstream to

down stream (be mindful of freeboard elevation and water depth downstream)

2. Use Total Head Loss in Pipe equation using y for depth in V equations3. Simplify and use trial & error until head loss is less than minimum headloss (be about pipe sizes being given in inches)

Rainfall Runoff

Time of Concentration:

=0.77

0.385

0.0078C

LengthT

Slope

where Length = Dist. Water travels in a diagonal direction

Peak Dischargevia Rational Method:Qp= CIAd Qp = Peak Discharge

I = Intensity of rainfall (in/hr)

Ad= Area of discharge (acres)C = Coefficient of discharge (App A-34)Intensity can be found from curves on Page 20-5


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Hydraulic Jump

Find water depth upstream or downstream of critical depthTo find water depth at either Points #1 or #3

1. Find info at Point #2 (Faroude # = 1), Q, V, etc.2. Find Specific Energy at Point #23. Find EGL at point #2

4. SubstituteQ

byfor V in and set EGL1= EGL2

5. Use Trial & error changing y to match EGL2

Water depth downstream of hydraulic jump

1. ( )22

344 30

2 2

yyF gb Q V V

= = +

2. Isolate all variables for points 3 & 422

344 3

2 2

yygb QV gb QV + = +

3. Plug in known variables & simplify4. Use trial and error solution varying unknown y to required EGL

Energy Loss in a Hydraulic Jump1. Energy Loss: EL = EGL3-EGL42. EGL3can be found using above methods, EGL4= E4+z

3.2

44 4

2

VE y

g= + , substitute

Q

byfor V4 and find E4, use this to find

EGL44. Find Energy Loss (see #1)

Geotechnical

Earthwork

End Area Method:

+ =

1 2( )2

27cut

LA A

V


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+ = +

1 2( )2 (1 % )

27fill

LA A

V shrinkage

Cut > Fill = surplusCut < Fill = needed

Balance Point:Mass Ordinate = Ordinsteprev. + Vcut- Vfill the first sta. Ordinate = 0

Mass - Haul DiagramSta. Cut(+) Fill(-) Adj. Fill* Cumulative** Haul Vol.**** (fill*shrinkage factor)** (add/subt. Cut and Adj. Fill)*** [(A1+A2)/2]*(sta.)

Find Overhaul in cubic stations Lfreehaulis the length of stations CMembankment& CMexcavagtionis Center of mass for the embankment and

excavation respectfully

Loverhaulis the length of overhaul: Loverhaul= CMembankment CMexcavationLfreehaul

HLordinateis the volume between the CMembankment& CMexcavation Voverhaul = HLordinate* Loverhaul


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Volumetric Properties***NOTE: For volumetrics/soil properties.work with the phase diagram!! ***

Air

Solid

Wtotal

Wwater

Wsolid

Vtotal

Vair

Vwater

Vsolidl

Property Saturatedsample (ms,mw, SG areknown)

UnsaturatedSample (mw,mw, SGt, vtareknown)

Supplementary formulas andcomputed factors

Volume Components

Vs VolumeSolids ( )

s

w

m

SG Vt-(Vg+Vw) Vt(1-) (1 )

t

V

e+ v

V

e

Vw VolumeWater *

w

w

m

Vv-Vg SVv

1

SVe

e+ SVse

Vg Volumeof gas orair

ZeroVt (Vs+

Vw)

Vv Vw (1-S)Vv(1 )

1

S Ve

e

+

(1-

S)Vse

Vv VolumeVoids *

w

w

m

( )s

t

w

mV

SG Vt-Vs

1SV

1

Ve

e+ Vse

Vt Total

VolumeVs+ Vw

MeasuredVg+ Vw+ Vs

Vs+Vg+Vw 1SV

Vs(1+e)

(1 )v

V e

e

+

Porosityv

t

V

V 1 v

t

V

V 1

( )s

t water

m

SG V

1

e

e+

e VoidRatio

v

s

V

V 1v

S

V

V

( )1t water

s

SG V

m

( )w

S

m SG

m S

Mass for specific sample

ms Mass

solids Measured 1tm

w+ ( ) (1 )t wSG V ( )

Wm SG

eS

( )S wV SG

mw Masswater Measured wms SwVv

sem S

SG Vtdw

mt Masstotal

ms+ mw ms(1+w)

Mass for sample of unit volume (density)


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Property Saturatedsample (ms,mw, SG areknown)

UnsaturatedSample (mw,mw, SGt, vtareknown)

Supplementary formulas andcomputed factors

d DryDensity

s

s W

m

V V+ s

g w s

m

V V V+ +

(1 )t

t

m

V w+ ( )

1wSG

e

+ ( )

( )1

wSG

w SG

S

+

1 w

+

WetDensity

s w

s w

m m

V V

++

s w

t

m m

V

+

t

t

m

V

( )

1wSG Se

e

++

(1 )

1ww

w

S SG

+

+

(1 )d w +

sat Saturateddensity

s w

s w

m m

V V++

s v w

t

m V

V

+ 1

sw

t

m e

V e

+

+

( )

1

wSG e

e

+

+

(1 )

1ww

wSG

+

+

b Buoyant(submer-ged)density

*

sat w

*1

1s

w

t

m

V e

+ *1

1w

SG e

e

+ +

*

11

1 wSG

wSG

+

Combined relations

wWatercontent

w

t

m

m 1t

s

m

m

Se

SG

* 1w

d

SSG

SDegree ofsaturation

100%w

v

V

V *

w

v w

m

V

( )w SG

e

* 1w

d

w

SG

w

g w

V

V V+

SGSpecificgravity ofsolids

s

s w

m

V

Se

w

wis the density of water. Where noted with an asterisk (*), use the actualdensity of water. In other cases usee 62.4 lb/ft3or 1000 kg/m3.

Total density: t w s

t g w s

m m m

V V V V

+= =

+ +

Specific Gravity: s

w

SG

=

PI = LL - PL


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Volume of borrow required given a required density(i.e. 95%):

d borrow d fillborrow fill V V = solve for Vborrow

Amount water to be added to borrow to get optimum m.c.:

w

s

www

= ,opt

water opt

s

www

= & ,w

w reg

s

wWW

=

, , ,w opt w reg w required w w w = (in pcf)

to convert to gal/ft3:3

3 3

27

8.34

ftgal gal

ft yd lb

Moisture content for saturated soil:

sSe wG= Where S = 1.0 (saturated)

Gs& e should be given, then solve for w (%)

Number of trucks to complete a fill:o Find volume required first

d truck d filltruck fill V V = Solve for Vtrucko Then divide Vtruckby number of yd

3per truck to obtain number oftruckloads required

NOTE: These are drydensities, filld should be the minimum spec

limit for compaction (i.e. 95%). Watch units, fill volume may be inyd

3and densities are given in pcf.

Soil Classification

To find classification using AASHTO System (including Group Index):o Find Plasticity Index (PI = LL PL) (LL = Wl& PL = Plon above chart)o Use chart to find soil group


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o % passing #40, #40 & #200o LL/PI limitso Find Group Index = (F200- 35) * [0.2 + 0.005(LL - 40)] +

0.01(F200- 15)(PI - 10)o Round Group Indexes to the nearest whole number.

To find soil classification using the Unified Soil Classification System (USCS)o % passing #200 > 50%

o If yes, fine grained soil % passing #4

o If no, coarse grained soil


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LL

General Size ClassesPASSING #200 = FINES

PASSING #4 = SANDALL ELSE = GRAVEL1. %gravel = 100-%passing #42. %sand = 100-%fines-%gravel3. %fines = %passing #200

Unit Weight (dry)

(1 )total

dryw

=

+

Where: totalis total unit weightw is moisture content

Use a Vtof 1ft3if no volume given

MIN. gd of compacted fill: gdmin = (% compacted) * (gdmax)

Moisture Content if fill is saturated:Se = wGs Where S = degree of saturation (if fill is saturated, S = 1.0)

e = void ratiow = moisture contentGs= specific gravity of solids

Contraction factor = ,

mod

s borrow

roctor

w

w

Contraction factor =3(1 )

(% )s ft

compaction

Volume of fill you actually get =Total Volume

Contraction Factor

Soil Stress

Vertical Stress (v):

Total vertical stress (v) @ a point = (depth of layer1* 1) +(depth of layer topointn* n)

IF UNDER WATER TABLE (v) = (depth of layer1* 1) - w


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Pore Water Pressure (u): u = wh if there is NO seepage where h = dist. fromwater table to point

Vertical Effective Stress ('v): ('v) = v- u

If water rises above the ground surface:

Total vertical Stress = (total vertical stress @ the point) + w(change in waterlevel)

Change in Vertical Effective Stress = (v ) - unew then take change in vertical

effective stress

If water table is at ground surface, treat pore water pressure as another layer andadd to total vertical stress. NOTE: Changes in water table will not effect stresson an embankment as long as the water table stays below the embankment.

For a load (i.e. fill) placed on soil, must find stress increase due to load and thenadd to vertical stress.

Stress Increase: = totalv v

I

I comes from Boussinesq Charts

= totalv

h

( )

(1 )

sand wsat

Gs e

e

+=

+

(1 )

1( )

clay wsat

w

wGs

+=

+

( )

(1 )w

total

Gs Se

e

+=

+where

w GsS

e

= or S=1.0 if saturated

Passive Earth Pressure:

2tan (45 )2p

K = 1pa

KK

=

Active Earth Pressure:

2

tan (45 )2aK

= + 1

aK K=

Lateral Forces

Total Active Force: 21

2a aP H K=

v= Hh= vKa


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with cohesion: h= vKa-2C( aK )

Overturning moment of a wall about point a

3o a

HM P=

effective friction angle w/out Cohesion = 12

sin

2

v h

v h

+

where 'v v u = &'

h h u =

Cyclic Stress Ratio = 'v

T

'

2000 ( )vT CS Ratio= where T = shear stress & 'v = overburden pressure

Factor of SafetyFS against liquification:

FS = (Available T) / (Required T) where you solve for available

FS against sliding:

( tan )sliding

a

N PFS

P

+=

where N = object weight = A (FS>1.4 is good)

FS against overturning:

( )roverturning

OT a

M resisting moment object wt dist FS

M overturning moment P dist

= = =

ignore Ppand sum the moments about the toe

Bearing Pressure FS = (q / qmax)

max6( )(4 ( )dNq

B B= where: 1

2f qq b N BN = +

N = AB = Base or widthD = height


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Settlement(1) Initial vertical stress @ midpoint of layer:

vo= (thickness of sat. layer above)(g) + (1/2 layer thickness)(g)(2) Initial pore pressure:

(thickness of sat. layer above + 1/2 layer)(w)

(3) Initial vertical effective stress: (1) (2)(4) Final total vertical stress = (thickness)(g) +.(1/2 layer thickness)(g)

(5) Final pore pressure: (thickness of layer below water table + 1/2 layer)(w)(6) Final vertical effective stress: (4) (5)

Settlement () =(6)

log1 (3)

c

i

CH

e

+

Settlement () =' '

0

'

0 0

log1

c sC H

e

+

+

where: Cc= compression indexei= initial void ratioH = Height of added layerHs= Height of layer that

will settle'

v =effective vertical

stress from fill

Ultimate Bearing Capacity (qult):

0.5ult c f qq cN D N BN = + + (us)

0.5ult c f qq cN gD N gBN = + + (si)

where: B = width of strip footingDf = depth of footing belowsurface

Nc, Nq, N= Bearingcapacity factors (Terzaghior Meyerhof & Vesic)(CERM pg 36-3)

NcBearing capacitymultipliers for varouslvalues of B/L

NBearing capacitymultipliers for varouslvalues of B/L


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B/L Multiplier B/L Multiplier

1.0(square)

1.25 1.0(square)

0.85

0.5 1.12 0.5 0.90

0.2 1.05 0.2 0.95

0.0 1.00 0.0 1.00

1 (circular) 1.2 1 (circular) 0.70

Net Bearing Capacity:The ultimate bearing capacity is corrected by the overburden giving the netbearing capacity.

qnet= qult- Df

Allowable Bearing CapacityIs the net bearing capacity divided by a factor fo safety:

neta

qqFS

=

Bearing capacity of sand:The cohescion of an ideal said is zero. Therefore the ultimate bearing capacityof a sand is:

( ) 0.5ult q f qq p D N BN = + +

The net baring capacity when there is no surface surcharge (pq= 0):

( 1) 0.5net ult f f qq q D D N B N = = +

c = 0 for sand

if concentric vertical loading, Ss= Si= Sqs= Sqi= 1.0

D= tD

Ultimate End-Bearing Capacity (Terzaghi)(qult) = [gDfNq/2000] = [sv'tipNq/2000][Area]

Ultimate Side-Friction Capacity (Mohr-Coulomb)

T = tan Q=T*AT = c+tan

Consolidation

Cv= compression indexCc= coefficient of consolidation

Total unit weight (of a saturated sand)


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( )1

s w

sat

G e

e

+=

+

Total unit weight of a sand above ground water level:

( )

1

s w

t

G Se

e

+=

+

where swG

Se

=

Total unit weight of a saturated clay:

(1 )

1w

sat

s

w

wG

+=

+

Void ratio of a saturated clay:

swGeS= (S = 1.0 because it is saturated)

Find total settlement of a clay layer:

Settlement for normally consolidated clay ( )' 'm fv v

= : f

0

'

v

'

0 V

H log1

cC

e

= +

Find initial stresses at mid-depth in clay layer: '

2

clay

v sand sand clay

hh = +

0 wu h=

0 0

'

0v v u = Find final stresses at mid-depth in clay:

fv fill fill moistsand moistsand satsand satsand clay clay h h h h = + + +

f wu h= '

0f fv vu =

Now plug variables into and solve

Pore pressure after placement of a fill layer:Pore pressure = static + initial excess pore pressures(u = ui+ u0)

=l wu h

0 v fill fill u h = =

Time for settlement (consolidation)Time it takes for a single layer to reach a given consolidation:


32/73

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2

v d

v

T Ht

C=

Where Cvis the coefficient ofconsolidation (ft2/day) as:

0(1 )v

v water

K eC

a

+=

and av is the coefficient ofcompressiblity found from the voidsratio and effective stress from any twoloadings:

( )2 1' '

2 1

v

e ea

p p

=

For Tv, time factor, it depends on degree of consolidation Uzand can be derivedfrom the following formula or table for Uz>0.6

[ ]21

0.604

v z zT U U= <

Uz Tv

0.10 0.008

0.20 0.031

0.30 0.071

0.40 0.126

0.50 0.197

0.55 0.238

0.60 0.287

0.65 0.340

0.70 0.403

0.75 0.477

0.80 0.567

0.85 0.684

0.90 0.858

0.95 1.129

0.99 1.781

1.00

Retaining Walls

For granular soils: will use '

&v For cohesive soils: will use & undrained shear strength

Granular Soils:

Lateral Earth Pressure = 'vK

Value of K will depend on friction angle and mode of movement


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Rankine Pressure Coefficients (3 modes):

At Rest: = 0 1 sin( )K

Active:

=

+1 sin( )

1 sin( )aK

Passive:

+=

1 sin( )

1 sin( )pK

Use Rankine only when: Backfill is horizontalNo friction between wall and backfill

Must use Coulomb if either of these exist, however, Rankine can be usedin the Passive Condition (Coulomb incorporates sloping backfill and wallfriction)

Cohesive Soils:

Active Pressure: 2v c

Passive Pressure: + 2v c

Other loads to consider:

Granular soils include pressure due to unbalanced water pressure Undrained cohesive soils Already dealing with total stress so loads are

already included

External Loads:o Line Loads use graph to estimate increase in line load pressureo Surcharges Uniform surcharge (q) is the resulting horizontal

stress = qK

Retaining Wall Analysis

3 Modes: Overturning, Sliding, and Bearing CapacityWhen designing, Factor of Safety should be determined for each.

To find FSoverturning:

1. Find overturning moment (Mo): = 3

o a

HM P

2. Find resisting moment (Mr):a. Divide retaining wall and backfill into areas

Soil

Ret. Wall

1

2

3

4

AForce Arm Length

b. Construction and fill in the following table


34/73

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# Dimensions

Area(ft2)

(pcf)Force(kips)

(area . )

Forcearm (ft)

Moment (kip-ft)(force .force arm)

NOTE: Force arm is length from reference point (A in the example) to

the center of the area in the x direction.c. Sum up the moment column to get Mr.

3. = ro

MFS

M

To find FSsliding:

1. Find s: = ( ) tan( )y sS F

2. Find Pa:21

2a aP H K=

3. Find Factor Safety Sliding: =SLa

sFSP

NOTE: to find Fy, see eccentricity; s friction angle for shear alongthe wall

To find FSbearing capacity:1. Find resisting moment (Mr) (See FSoverturning)2. Find overturning moment (Mo) (See FSoverturning)

3. Find Fy (See FSoverturning, sum up the Force column)4. Find eccentricity5. Find effective footing width (B): B=B-2e

6. Find Qult: = + + 0.5 'ult f qQ cNc D N B N

7. Find FSbc: =

ultbc

y

QFS

F

NOTE: The equation for Qultwill not give the exact answer for Qult,this is an approximation using the equation for ultimate bearingcapacity. It is estimated that this will yield an answer within 10% ofthe correct answer using the shape factors.

To find eccentricity (e):1. Find resisting moment and overturning moment (see FSoverturning)

2. Sum the Force column to get Fy

3. Find (X):

=r o

y

M MX

F

4. Find (e): = 2

Be x


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Pile Analysis

Piles derive their capacity from side friction between the pile and the soil and theend bearing capacity of the tip.

Side friction (fs) increases up to critical depth (Dc), after Dc, theside friction does not increase.D

c

Fs

D

Ultimate capacity = skin friction + end bearing: Qult= Qs+ Qt

Allowable capacity = ultimate capacity / Factor of Safety: ultall QQFS

=

Critical Depth:Based on soil type

Dc= 10B for loose sands ( 30 o )

Dc= 15B for medium dense sands (30 36>


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To find side/skin friction capacity at a given depth (in a granular soil):1. use equation for side friction, if d and k are not given, solve the

equation for k tan().Find the effective stress from the test pile since you have the fs for

that value and solve for k tan().2. Find the critical depth of the pile3. Find the side friction at Dc, use the formula for side friction with the

value for k tan() as found in part 1.4. Find the ultimate capacity (Qs) using the following formula:

1( )

2 12 12s c s c s

D DQ D f D D f

= +

To find the ultimate bearing capacity at a given depth:If depth (d) is > critical depth, then use values for Qsand Qtfrom the testpile, otherwise:

To find Qt:1. Use the formula for tip capacity to find Nqusing the Qtfrom the test

pile.2. Use the value for Nqin the formula now for the vertical stress at the

given depth.

To find Qs, use the method as described above.

Seepage

Vertical effective stress below the piezo level (i.e. Elev = 75ft) Treat just like

vertical effective stress below the water table ( 'v v u = )

Vertical effective stress above the piezo level (but below water table) (i.e. Elev =90ft):

'

v v u =

1. Find vertical stress ( v )

2. Pore water pressure: w pu h=

Where t eh h h= he= ElevA


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t top Ah Elev h=

To find head loose, use Darcys Equation (Q=kia)Use A = 1 ft2& QA=QBMaking QA & QB equal and plugging in 1 for A yields:

A A B Bk i k i =

with: &A BA BA B

h hi i

H H

= =

kA, HA, KB& HBwill be given so plug them into above Darcys to

get: A BA BA B

h hk k

H H

=

Solving for Ah yields:B B A

A

A A

k h Hh

k h

=

Plug this into A Bh h H + = and solve for &A Bh h

Now find the returning head (ht): t top Ah Elev h= t eh h h =

Now find pore water pressure

3. Effective stress can now be found: 'v v

u =

To find rate of vertical seepage:Use Darcys Law

Area should be given (i.e. per foot)

For i use: nnn

hi

H

=

Find Q (seepage rate)

Flownets

Example diagram:Nf= # flow paths (4 in diagram)Nd= # equipotential drops (12 indiagram)K = coefficient of permeability

H = total head loss due to flow

Rate of Seepage:f

d

NQ k H

N

=

To find head loss at a point (i.e. Point A):

# Drops

Total # Drops

AT T A

A

h h h

h H

=

=


38/73

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NOTE: # drops is from the water sideto where point is (6 in the diagram)

Uplift Pressure at Point A:

Where: (see head loss at Pt A)

w p

p t e

t

e A

u h

h h h

hh Elev

=

=

=

Pressure head at a point under a dam(depth from top of water to the point) - (#drops to the point along the flowline)(DHH20/Nd)

Structural

Reinforced Concrete Footings

Square R.C. Footings - smallest dimension B given qall

2

61

1all

P Mq

B B

+ =

Where P is applied forceM is applied momentB is dimension

For a wall using a 1.0 ft. x-section:2

61

1all

P Mq

B B

+ =

31where I=12

MxBH

L

Shear Force'

'

c

0.85(2) ( )(length or B is square)

1000

where f is usually 3000

cf d

capacity=

Punching Shear'0.85(2) ( )(length or B is square)

1000

2

c

c

u s y

f dV

aM A f d

=

=

Stirrup Spacing

s y

s

A f dV

S

=

Where S = stirrup spacingAs= area of steel = 2 bar areas

min 50sV bw d = Where bw = width of webb


39/73

7/31/2004 Page 39

calculate minVsand plug back intos y

s

A f dV

S

=

and solve for S

'max 8s c

V f bw d =

Surveying/Transportation

General Equations

Central angle ofcircular arc:

(2 )c s = Where is angle

between tangents & sis spiral angle

Cut Volume:( )1

2 27

i iA A Lvol =

Where Aiis cut atstation I, L is the length

between stations &

dividing by 27 convertsft3to yd3

Deflection Angle:2

arc

L

=

Where arc is segment ofarc length, L is length of

circular arc & is thedeflection angle

between 2 tangents

Deflection Angle:

2

1

3s

s

L

L

=

Where L is length ofspiral curve, Lsis length

of spiral & sis spiralangle

Elevation of P ( )2

12

p

p pvc p

R XY Y g X

= + +

Where Ypvcis elevationat point of vertical

curvature, g1is gradient1, Xpis horizontal

distance from PVC to P,R is range of change of

grade

Elevation of PVC 12

PVC PVI

LElev Elev g

=

Where ElevPVCiselevation at PVC, g1isgradient 1 & L is length

of curve

Elevation at X2 12

X P

WElev Y

=

Where is cross slope,W is width of pavement

and dividing by 12converts from feet to

inches


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Elev at StaX

2

1

2

x 1

or2

Elev2

x PVC

PVC

RXElev Elev g X

g XElev g X

L

= + +

= + +

Where X = Stax- StaPVC,R is curve radius

Fill Volume( )1 112

2 27 100

i iB B LVol +

=

Where Bi is fill area atStai, L is length betweenstations & 12% is added

for shrinkageHorizontal

Distance fromPVC to X

1

1 2

p

g LX

g g

=

Where g1is gradient 1,g2is gradient 2, L is

length of vertical curve

Length of circulararc

100c

c

LD

=

Where c is centralangle of arc, Dmaxismaximum degree of

curvature

Length of Curve L KA=

Where K is rate ofcurvature & A is total

change in grade of thecurve

Length ofoverhaul ohaul emb exc free

L CM CM L=

Where CMembis centerof mass of

embankment, CMexcisctr mass of excavation &

Lfreeis free haul dist

Length of

Subtangent

min tan

2

T R

=

Where Rminis min. curve

radius & is deflectionangle

Max Degree ofcurvature

5729.578D

R=

Minimum CurveRadius

min

max

5729.578R

D=

Minimum CurveRadius*

2

min15 ( )

VR

e f=

+

Where V is velocity, e issuperelevation & f is

side friction factor

Minimum Lengthof Spiral*

33.15

c

VL

R C=

V is design speed, C israte of increase of accel,

Rcis curve radiusOffset distancefrom initial point

to spiral0.017426P L= L is length of spiral

Radius of Curvetan

2

TR=

T is len of subtangent, is deflection angle

between subtangents


41/73

7/31/2004 Page 41

Rate of change ofgrade

1 2g g

RL

=

Sight distance forcurve*

55.61 (if S


42/73

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2.2

1 1where

2 2pt PVC PVC PVI

rx LElev Elev g x Elev Elev g = + + =

therefore2

1 12 2

pt PVI

L rxElev Elev g g x

= + +

because x and L are

unknowns ( )2

PVI x Lx Sta Sta= and 1 2g gR L=

3. Plug back into equation and solve

Passing Sight DistanceS < L S = Sight Dist.S > L L = Length of Curve

1st, assume S < L ( )1 255.61 whereL

S A g g A

= = , if the calculated S is less

than L, assumption was correct.

CREST (pg. 271) SAG (pg. 275) ALWAYS use SSD not PSDS>L S>LS


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Law of Sines ( ) ( ) ( )Sin A Sin B Sin C

a b c= =

Law of cosines 2 2 2

2 2 2

2 cos( )

2 cos( )

a b c ab A

c a b ab C

= +

= +

3 different types of intersection problemsbearing bearingbearing distancedistance distance

Bearing Bearing(vectors are tail to tail on bearing-bearing problems)

Bearing Distance

Distance Distance

1. Find Dist C2. Use Law of Cosines 3 times

3. Add up Angles to check for 180

Horizontal Curves


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Arc Definition

Magic Number -5729.578

RD

=o

(Arc Definition)

( )2

IT R Tan= (Be sure to Decimal

degrees, not degrees, min, sec)

100 ( )I

LD

=

2 ( )2

ILC R Sin=

1tan( ) ( )

4cos( ) 1

2

IE T R

I= =

cos( ) (1 cos( ))2 2

I IM E R= =

Chord Distance: 2 sin( )ch R =

Given and D or R:1. Find R or D using Magic Number2. Find Length (T)3. Find L4. Find Long Chord (LC)5. Find External Distance (E)6. Find Middle Ordinate (M)

Stapc= Stapi TStapt= Stapc+ L

To Find Tangent Offset:1. Find Deflection Angle

()2. Find Chord Distance3. Use Trig to find

Tangent Offset

Vertical Curves


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L is in stationsG1& G2are in %

22 11

2X PC

G GElev x G x Elev

L

= + +

Stahi/low 11 2

G Lx

G G=

Break into 2 curves, one being 600,the other 400

Stopping Sight DistanceD = 1.47 V T + V2 / [30(f g)]

where (1.47VT) = Dist. During reaction time (1.47 = conv. factor)

(V2 / [30(f g)] = Dist. During deceleration (if f is not given, f=(11.2 / 32.2))

V = speed in mphT = time in sec. (2.5 sec.)f = friction factor (0.348)g = uphill or gownhill grade (decimal)

D = Stopping Dist.

Trip Generation by Regression EquationResidential:Ln(T) = 0.920 Ln(X) + 2.707where T = # trips

X = # houses

Commercial:Ln(T) = 0.643 Ln(X) + 5.866where T = # trips

X = floor area in 1000 sq.ft. incriments

Using Average Trip Rate:Residential:T = (Trip Rate) (# Units) where Trip Rate = 9.57

Commercial:


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T = (Trip Rate) (sq.ft/1000's) where Trip Rate = 42.92

Queue

Utilization Factor(p) = Arrival Rate / Service Rate

Avg. Queue Length = E(m) = p2 / (1-p)Avg. # of veh. In sys. = E(n) = p / (1-p)95th percentile # veh. In sys. = P(n) = pn (1-p)

plug and chug n=0, 1, 2, etc. until the cumulative

answer reaches 0.95.

Traffic Signal TimingVolume to Saturation Flow Rate = (V/S)

Min. Cycle Length:C = LX / [X - (V/S)1 - (V/S)2 - (V/S)n]where L = lost time (multiply by # of signals)

(i.e., 2 streets in one direction = 2 signals)where X = V/S for specified street

Optimum Green Time:G = (C - L)(V/S)1 / [(V/S)1 + (V/S)2 + (V/S)n]

Given ATR readings, what is the daily volume?

# Axles = (# axles)*(Trucks)+(2 axles)(Cars) % Trucks = (# Trucks)/(# Trucks + # Cars) Solve for # Trucks and substitute back into first equation

Maximum Service Ratio: At maximum capacity when the volume to capacityratio (V/C) is 1.0

Median Speed:

Find Middle Speed Find cumulative frequency on table that contains the middle speedfrequency

Assume a median speed within that group

85thpercentile speed:

85thpercentile speed = total frequency * 0.85 Find Cumulative frequency that contain the nth speed above


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Assume a speed within that range

Mode Speed is the assumed speed group with highest frequency

Pace of vehicle speeds is the 10 mph speed group with the highest frequency

Distance to Stop: Formula in green book (Distance traveled during reaction &deceleration)

If skid distance given but not reaction time or initial speed, what was speed.

Perform iterations of the distance traveled during deceleration formula inGreen Book solving for coefficient of friction until you get a frictioncoefficient that is approximately what you see in the given table.

If asked how far skid would continue if no object would have been hit:

Add skid distance to distance traveled during deceleration using speed

vehicle was traveling when it hit the object

Initial speed of car

Find skid distance then use distance during deceleration varying speed tomatch up with friction value like in above problem.

GREEN BOOK

Rural Freeway - Sight Distance1. Select type of sight dist. (Decision, Passing, Intersection)

2. Select speed based on condition (usually 70mph)

3. Choose avoidance maneuver if dsd (pg. 116)

4. Choose sight dist.

Rural two-lane hwy.1. Select type of sight dist. (Decision, Passing, Intersection)

(pg. 123+)

2. Select speed based on condition3. Choose sight dist.

ISD (Intersection Dight Distance) (pg. 664)

ISD = 1.47*Vmajor*tg where tg = time gap


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Rail Road Intersection Equations (pg 738 - 740)

M = R(1-cos[(28.65*S) / R] (pg. 231)where S = SSD, M = Middle Ordinate, R = Radius

Rmin = [V2 / 15*(0.01emax + fmax)] (pg. 143)f = pg.145 E 3-14 Rural & High Speed Urbanpg.197 E 3-41 Low Speed Urbanpg.201 E 3-43 Intersection Curves (Turning)

SPIRALS (pg. 177)L = (3.15*V3) / RC where C = 1 - 3 (usually use 2)

RUNOUT AND RUNOFF (pg. 170 - 176)

Runout = [(rise per lane / max. gradient(dec.))] * (multiplier for the # of lanes)

(pg. 170) (pg. 172)Runoff = [(lane width)*(e(dec.)) / max. gradient(dec.)] * (multiplier for the # oflanes)

M.U.T.C.D.

Ped. walk speed = 4 fpsInitial walk signal time = about 7 sec.Vehicle Length = 18 - 20 ft.Traffic Signal tpr = 1.0 sec.Deceleration = a = -10 ft/sec2Lost time = about 3 - 4sec / f

Time for yellow and red:time = tpr + V/2a + (W +L) / V where V(ft/sec)or(mph*1.467), tpr(sec), L(ft),a(ft/sec2)

yellow redtime = tpr + V85/2a + (W +L) / V15

M.U.T.C.D. (pg. 40-15) says min yellow = 3 sec., max yellow = 6 sec.

I.T.E. = 5 sec. for yellow

HWY. CAPACITY MANUAL

TRAFFIC FLOW


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q = kVsmswhere q = flow (veh./time)where k = density (veh./dist.)where Vsms = velocity (dist./time)

TMS = (dist./time) / # veh.

SMS = (dist.) / (avg. travel time) = (# veh.) / (S(1/Vel.))

TMS > SMS except for uniform speedsTMS = SMS + (s2 / SMS)

FREEWAYS

Vp = (vol.) / (PHF*N*fHV*fp)

where Vp = V15min, peakwhere fp = population factor (usually 1.0)where N = # lanes in ONE directionwhere fHV = (1) / [1+PT(ET - 1) + PR(ER - 1)]where PHF = V(vph) / (4*V15)

# veh. flowing in peak direction during peak 15 min. of day.PHF = [(AADT)*(K)*(D)] / (4*V15)where AADT = daily flowwhere K = PK hr / daywhere D = Direction%

Flow Rate in terms of Passenger Car Equivalents (PCE)(Hourly Vol. Veh.)*(% non Trucks) + (Hourly Vol. Veh.)*(%Trucks)*(# Veh. asTrucks)

Given: FFS and other info.Q: determine LOS

1. fHV (HCM E23-9 & E23-10) for Freeways2. Vp = (vol.) / (PHF*N*fHV*fp)

3. Go to tables in HCM and choose LOS based on flow. (E23-2) forFreeways

Asphalt Properties

1 mb

mm

GAirVoids

G

=

Gmb= Bulk Specific GravityGmm= Theor. Max SG (Rice)Pb= % Binder


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100 mb seff

se

G PVMA

G

=

100

100

bse

b

mm b

PG

P

G G

=

Correction Factoractual eff VMA VMA=

100VMA AirVoids

VFAVMA

=

Ps= % StoneGse= Effective SG if Stone

Economics

General Equations

Capitalized cost:

Annual Cost

Initial Cost + Yearly Interest Rate

Direct-reduction Loan Payment: ( )AA = P P

Effective interest rate for compound period (cash-flow period>compounding

period): (1 )ki = + where is the effective interest rate for the compoundingperiod and k is the number of compounding periods per cash flow period.

Effect interest rate for compounding period (compounding period < yearly):

1 1

m

effrim

= +

, where r is the nominal yearly interest rate and m is the number

of compounding periods per year.

Equivalent uniform annual cost: ( )AEUAC = initial cost , , + Annual Costsi nP ,where i is the yearly interest rate and n is the number of years.

Future value of present sum: ( ) nF : F = P(1+i)P

Future value of a uniform series: ( )n(1+i) -1

: F = Ai

FA

Internal rate of return: The effective interest rate for a cash flow that makes apresent value = 0

Present value of future sum: ( ) -n: P = F(1+i)PF


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Present value of a uniform series: ( )nA (1+i) 1

: P =(1 )n

PA i i

+

Straight line (fixed percentage) depreciation: nj C SDn= , where C is the initial

cost, Snis the depreciation in year n

Uniform series for future value ( )( )

i: A=F

1+i 1n

AF

Uniform series for present value: ( )nP i(1+i)

: A=(1 ) 1n

AP i

+

Uniform Gradient to Present Worth: ( )n

2

(1+i) -1: P=G

i (1 ) (1 )n nnP

G i i i

+ +

Uniform Gradient to Future Worth: ( )n

2

(1+i) 1: F=G

nFG i i

Uniform Gradient to Uniform Series: ( ) 1: A=Gi (1 ) 1n

nAG i

+

(A Uniform Gradient is a uniformly increasing cost or cash flow)

Present Cost: for each alt., convert all cash flows to (P), then calc. total (P).

Select alternative with the smallest total (P).P = F(P/F, i, n) where n = done in "n" yearsP = A(P/A, i, n)

Ranking Alternative ProjectsPresent Worth --- convert all to (P) where initial cost is a negative face valuePW = PW of yearly income - initial costInternal Rate of Return: change the i (trial & error) until the PW = 0

Annual CostsAnnual Cost: Equivalent Uniform Annual Cost:

Convert all to (A) the add all.initial = A/P, ann. cost + yearly income = A/G, salvage = A/F (negative value)for fixed costs, use their face value

Rate of Return1. Calc. Profit2. Calc. Tax on profit


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3. Calc. Profit after tax --- profit - tax

4. Express Rate of Return:Profit after tax

ROR =Construction Cost

Effective Interest Rate 1 1effr

i mm

= +

, where ieffis effective interest rate, r

is nominal interest rate and m is the number of compounding periods per year.

Effective interest per compounding period: ( )1 1cp cfpi i m= + , where icpis theeffective interest rate for compounding period, icfpis the effective interest rate forthe cash flow period and m is the number of compounding periods per cash flowperiod.

( )( )( )

1 1

1

cfp

cfp cfp

i mPW A

i i m

+ = +

Perpetual Livesgiven initial cost, annual cost, i, which one can be sustained forever w/initialinvestment x:(1) Capitalized Cost = Annual Cost / i(2) Total Cost = Cap. Cost + initial costProject can be sustained if total cost < or = to x

Depreciation and Taxes

1. Calculate the amount of initial cost not funded by the loan. That is thepresent worth of the initial cost

2. Calculate the yearly loan payments.3. For each year:

a. Depreciate 1/n of the initial cost (i.e. 1/3 for a 3 year project)b. Calculate the portion of the loan payment that is interestc. Calculate taxd. Calculate after-tax cash flowe. Calculate the present worth of the after tax cash flow

4. Calculate the present worth of salvage after tax5. Calculate the present worth of all cash flows. If that total is 0, then the

project should proceed.

EOY BTCF Dep Loan Pmt Prin Int Tinc Tax ATCF P/F PW

0

1


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n

Dep: Depreciation per year = initial cost/number of yearsLoan:

Pmt: Payment = Loan Amount (A/P, i%, n)Principle = Payment (P/F, i%, n)

Interest = Payment Principle(Note: Find last payment by itself, smaller interest amount)Tinc: Taxable Income = Before Tax Cash Flow Depreciation Loan Interest

Tax: Tax = Taxable Income Tax RateAFTC: After Tax Cash Flow = BTCF + Loan Loan Payment Tax

Present Worth: AFTC (P/F,i%,n)Total Present Worth = PW for years 0 to nTotal Present Worth > 0, then project should proceed.

Ranking AlternativesFor project with higher present worth:

1. For each projecta. Take Initial Costb. Calculate present worth of year income with P/A multiplierc. Add results of parts (a) and (b) to obtain present worth

2. Select the project with the larger total present worth

Projects

A B

MARR

Project Life

Initial Cost

Income per yearP/A multiplier

PW income

PW total

MARR: Minimum Acceptible Rate of Return (if given, use this for i)

Initial Cost: PW = Initial Cost (P/F,i%,n)Income: PW = Yearly Income (P/A,i%,n)Salvage: PW = Salvage Value (P/F,i%,n)Total: PW

For higher internal rate of return:1. For each project

a. Calculate total present worthb. Adjust interest rate until total present worth is zeroc. The interest rate obtained in part (b) is the internal rate of return

2. Select project with the larger internal rate of return

To see which project should be funded:


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1. total present worth is always correct2. Internal rate of return can be misleading3. If you must use rate of return, use incremental rate of return.

a. Rank projects from lower to higher initial costb. Calculate all cash flows as increment from B to A (for example)

c. Calculate the total present worth of these incremental cash flowsd. If the total incremental present worthies positive, the return on theadditional investment is greater than the MARR, so the project withthe higher initial cost should be funded.

Concrete Mix Design1. Abs. Vol. Method (work with 1.0 cy.) (see chapter 77 for method)2. Weight Method (work with 3900#) conc. weighs 145#/cf(27cf/cy) =3915#/cy

SG cement = 3.15SG aggregate = 2.65

SG water = 1.00 Vol(cf) = [Wt.(lb)] / [SG(w)]Wt. Water = 62.4#/cf

1 gall. Water = 8.34 lb Wt.(lb) = Vol.(cf) * SG * (w)Vol. 1 bag cement = 1.0cfWt. 1 bag cement = 94#1.0cf concrete = 145#Wt. SAND = 3900# - water - cement - rockVol. SAND = 27cf - water - cement - rock - air

STRUCTURES

TRUSSES CH. 41SFx = 0SFy = 0SM = 0


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Critical Path AnalysisSlides taken from Project Management Course on Critical Path


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Metric Conversions

Conversions ProceduresApproximateConversions

From ToMultiplynumber

of

byTo getnumber

of

(Actual answers will bewithin 25% of

approximate answer.)acres hectares (ha) acres 0.4047 ha 1 acre = 2.471 ha

acressquare feet(ft2)

acres 43,560 ft2 1 acre = 40,000 ft2

acressquarekilometres(km2)

acres 0.004047 km2 1 acre = 0.004 km2

acressquare metres(m2)

acres 4047 m2 1 acre = 4000 m2

acressquare miles(mi2)

acres 0.001563 mi2 1 acre = 0.0015 mi2

acressquare yards(yd2)

acres 4840 yd2 1 acre = 5000 yd2

acre-feet (acre-ft)

cubic feet (ft3) acre-ft 43,560 ft3 1 acre-ft = 40,000 ft3

acre-feet (acre-ft)

cubic metres(m3)

acre-ft 1233 m3 1 acre-ft = 1000 m3

acre-feet (acre-ft)

gallons (gal) acre-ft 325,851 gal 1 acre-ft = 300,000 gal

centimetres(cm)

feet (ft) cm 0.03281 ft 1 cm = 0.03 ft

centimetres(cm)

inches (in) cm 0.3937 in. 1 cm = 0.4 in.

centimetres(cm)

metres (m) cm 0.01 m --

centimetres(cm)

millimetres(mm)

cm 10 mm --

centimerers persecond (cm/sec)

metres perminute (m/min)

cm/sec 0.6 m/min --

cubiccentimetres(cm3)

cubic feet (ft3) cm3 0.00003531 ft3 1 cm3 = 0.00004 ft3


cubicinches(in.3)

cm3 0.6102 in.3 1 cm3 = 0.06 in.3


cubic metres(m3)

cm3 0.000001 m3 --

cubiccentimetres

cubic yards(yd3)

cm3 0.000001308 yd31 cm3 = 0.0000015yd3


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ofby

To getnumber

of


approximate answer.)

(cm3)cubiccentimetres(cm3)

gallons (gal) cm3 0.0002642 gal 1 cm3 = 0.003 gal


litres (L) cm3 0.001 L --

cubit feet (ft3)acre-feet (acre-ft)

ft3 0.00002296 acre-ft 1 ft3 = 0.00002 acre-ft

cubit feet (ft3)cubiccentimetres

(cm3)

ft3 28,320 cm3 1 ft3 = 30

cubit feet (ft3)cubic inches(in.3)

ft3 1728 in3 1 ft3 = 1500 in.3

cubit feet (ft3)cubic metres(m3)

ft3 0.02832 m3 1 ft3 = 0.03 m3

cubit feet (ft3)cubic yards(yd3)

ft3 0.03704 yd3 1 ft3 = 0.04 yd3

cubit feet (ft3) gallons (gal) ft3 7.481 gal 1 ft3 = 7 gal

cubit feet (ft3) kilolitres (kL) ft3 0.02832 kL 1 ft3 = 0.03 kL

cubit feet (ft3) litres (L) ft3 28.32 L 1 ft3 = 30L

cubit feet (ft3) pounds (lb) ofwater

ft3 62.4 lb ofwater

1 ft3 = 60 lb of water

cubic ft persecond (cfs)

cubic metresper second(m3/sec)

cfs 0.02832 m3/sec 1 cfs = 0.03 m3/sec

cubic feet persecond (cfs)

million gallonsper day (mgd)

cfs 0.6463 mgd 1 cfs = 0.6 mgd


gallons perminute (gpm)

cfs 448.8 gpm 1 cfs = 400 gpm

cubic feet perminute (cfm)

gallons persecond (gps)

cfm 0.1247 gps 1 cfm = 0.1 gps


litres persecond (L/sec)

cfm 0.4720 L/sec 1 cfm = 0.5 L/sec

cubic inches(in.3)


in.3 16.39 cm3 1 in.3 = 15 cm3

cubic inches(in.3)

cubic feet (ft3) in.3 0.0005787 ft3 1 in.3 = 0.0006 ft3


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ofby

To getnumber

of



cubic inches(in.3)

cubic metres(m3)

in.3 0.00001639 m3 1 in.3 = 0.000015 m3

cubic inches(in.3)

cubicmillimitres(mm3)

in.3 16,390 mm3 1 in.3 = 15,000 mm3

cubic inches(in.3)

cubic yards(yd3)

in.3 0.00002143 yd3 1 in.3 = 0.00002 yd3

cubic inches(in.3)

gallons (gal) in.3 0.004329 gal 1 in.3 = 0.004 gal

cubic inches(in.3)

litres (L) in.3 0.01639 L 1 in.3 = 0.015 L

cubic inches(in.3)

acre-feet (acre-ft)

m3 0.0008107 acre-ft 1 m3 = 0.0008 acre-ft

cubic inches(in.3)


m3 1,000,000 cm3 --

cubic inches(in.3)

cubic feet (cf3) m3 35.31 cf3 1 m3 = 40 ft3

cubic inches(in.3)

cubic inches(in.3)

m3 61,020 in.3 1 m3 = 60,000 in.3

cubic metres(m3)

cubic yards(yd3)

m3 1.308 yd3 1 m3 = 1.5 yd3

cubic metres(m3)

gallons (gal) m3 264.2 gal 1 m3 = 300 gal

cubic metres(m3)

kilolitres (kL) m3 1.0 kL --

cubic metres(m3)

litres (L) m3 1000 L --

cubic metres perday (m3/day)

gallons per day(gpd)

m3/day 264.2 gpd 1 m3/day = 300gpd

cubic metres persecond (m3/sec)


m3/sec 35.31 cfs 1 m3/sec = 40cfs

cubic millimetres(mm3)

cubic inches(in.3)

mm3 0.00006102 in.3 1 mm3 = 0.00006 in.3

cubic yards(yd3)


yd3 764,600 cm3 1 yd3 = 800,000 cm3

cubic yards(yd3)

cubic feet (ft3) yd3 27 ft3 1 yd3 = 30 ft3

cubic yards cubic inches yd3 46,660 in.3 1 yd3 = 50,000 in.3


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ofby

To getnumber

of



(yd3) (in.3)cubic yards(yd3)

cubic metres(m3)

yd3 0.4646 m3 1 yd3 = 0.8 m3

cubic yards(yd3) gallons (gal)

yd3 202.0 gal 1 yd3 =200 gal

cubic yards(yd3) litres (L)

yd3 764.6 L 1 yd3 =800 L

feet (ft)centimetres(cm)

(ft) 30.48 cm 1 ft = 30 cm

feet (ft) inches (in.) (ft) 12 in. --

feet (ft) kilometres (km) ft 0.0003048 km 1 ft = 0.0003 km

feet (ft) metres (m) ft 0.3048 m 1 ft = 0.3 m

feet (ft) miles (mi) ft 0.0001894 mi 1 ft = 0.0002 mi

feet (ft)millimetres(mm)

ft 304.8 mm 1 ft = 300 mm

feet (ft) yards (yd) ft 0.3333 yd 1 ft = 0.3 yd

feet (ft) ofhydraulic head

kilopascals(kPa)

ft of head 2.989 kPa 1 ft of head = 3 kPa


metres (m) ofhydraulic head

ft of head 0.3048m ofhead

1 ft of head = 0.3 m ofhead


pascals (Pa) ft of head 2989 Pa 1 ft of head = 3000Pa

feet (ft) of waterinches ofmercury (in.Hg)

ft of water 0.8826 in. Hg1 ft of water = 0.9 in.Hg

feet (ft) of waterpounds persquare foot(psf)

ft of water 62.4 psf 1 ft of water = 60 psf

feet (ft) of waterpounds persquare inchgage (psig)

ft of water 0.4332 psig 1 ft of water = 0.4 psig

feet per hour(fph)

metres persecond (m/sec)

fph 0.00008467 m/sec 1 fph = 0.00008 m/sec

feet per minute(fpm)

feet per second(fps)

fpm 0.01667 fps 1 fpm = 0.015 fps


kilometres perhour (km/hr)

fpm 0.01829 km/hr 1 fpm = 0.02 km/hr

feet per minute metres per fpm 0.3048 m/min 1 fpm = 0.3 m/min


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ofby

To getnumber

of



(fpm) minute (m/min)feet per minute(fpm)

metres persecond (mps)

fpm 0.005080 m/sec 1 fpm = 0.005 m/sec


miles per hour(mph)

fpm 0.01136 mph 1 fpm = 0.01 mph



fps 60 fpm --


kilometres perhour (km/hr)

fps 1.097 km/hr 1 fps = 1 km/hr


metres perminute (m/min)

fps 18.29 m/min 1 fps = 20 m/min


metres persecond (m/sec)

fps 0.3048 m/sec 1 fps = 0.3 m/sec


miles per hour(mph)

fps 0.6818 mph 1 fps = 0.7 mph

foot-pounds perminute (ft-lb/min)

horsepower(hp)

ft-lb/min 0.00003030 hp1 ft-lb/min = 0.00003hp


kilowatts (kW) ft-lb/min 0.00002260 kW1 ft-lb/min = 0.00002kW


watts (W) ft-lb/min 0.02260 W 1 ft-lb/min = 0.02 W

gallons (gal)acre-feet (acre-ft)

gal 0.000003069 acre-ft1 gal = 0.000003 acre-ft

gallons (gal)cubiccenimetres(cm3)

gal 3785 cm3 1 gal = 4000 cm3

gallons (gal) cubic feet (ft3) gal 0.1337 ft3 1 gal = 0.15 ft3

gallons (gal)cubic inches(in.3)

gal 231.0 in.3 1 gal = 200 in.3

gallons (gal) cubic metres(m3)gal 0.003785 m3 1 gal = 0.004 m3

gallons (gal)cubic yards(yd3)

gal 0.004951 yd3 1 gal = 0.005 yd3

gallons (gal) kilolitres (kL) gal 0.003785 kL 1 gal = 0.004 kL

gallons (gal) litres (L) gal 3.785 L 1 gal = 4 L

gallons (gal) pounds (lb) of gal 8.34 lb of 1 gal = 8 lb of water


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ofby

To getnumber

of



water watergallons (gal) quarts (qt) gal 4 qt --

gallons percapita per day(gpcd)

litres per capitaper day (Lpcd)

gpcd 3.785 Lpcd 1 gpcd = 4 Lpcd


cubic metresper day(m3/day)

gpd 0.003785 m3/day 1 gpd = 0.004 m3/day


litres per day(L/day)

gpd 3.785 L/day 1 gpd = 4 L/day

gallons per dayper foot (gpd/ft)

square metres

per day(m2/day)

gpd/ft 0.01242 m2/day 1 gpd/ft = 0.01 m2/day

gallons per dayper foot (gpd/ft)

squaremillimetres persecond(mm2/sec)

gpd/ft 0.1437 mm2/sec1 gpd/ft = 0.15mm2/sec

gallons per dayper square foot(gpd/ft2)

millimetres persecond(mm/sec)

gpd/ft2 0.0004716 mm/sec1 gpd/ft2 = 00005mm/sec

gallons per hour(gph)


gph 0.001052 L/sec 1 gph = 0.001 L/sec



gpm 0.002228 cfs 1 gpm = 0.002 cfs



gpm 0.06309 L/sec 1 gpm = 0.06 L/sec

gallons perminute persquare foot(gpm/ft2)

millimitres persecond(mm/sec)

gpm/ft2 0.6790 mm/sec1 gpm/ft2 = 0.7mm/sec

gallons persecond (gps)


gps 8.021 cfm 1 gps = 8 cfm

gallons per

second (gps)

litres per

minute (L/min)gps 227.1 L/min 1 gps = 200 L/min

grains(gr) grams (g) gr 0.06480 g 1 gr = 0.06 g

grains(gr) punds (lb) gr 0.0001428 lb 1 gr = 0.00015 lb

grains(gr) grains (gr) gr 15.43 gr 1 gr = 15 gr

grams (g) kilograms (kg) g 0.001 kg --

grams (g) milligrams (mg) g 1000 mg --


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ofby

To getnumber

of



grams (g) ounces (oz),avoirdupois

g 0.03527 oz 1 g = 0.04 oz

grams (g) pounds (lb) g 0.002205 lb 1 g = 0.002 lb

hectares (ha) acres ha 2.471 acres 1 ha = 2 acres

hectares (ha)square metres(m2)

ha 10,000 m2 --

hectares (ha)square miles(mi2)

ha 0.003861 mi2 1 ha = 0.004 mi2

horse power(hp)


hp 33,000 ft-lb/min 1 hp = 30,000 ft-lb/min

horse power(hp)

kilowatts (kW) hp 0.7457 kW 1 hp = 0.7 kW

horse power(hp)

watts (W) hp 745.7 W 1 hp = 700 W

inches (in.)centimetres(cm)

in. 2.540 cm 1 in. = 3 cm

inches (in.) feet (ft) in. 0.08333 ft 1 in. = 0.08 ft

inches (in.) metres (m) in. 0.02540 m 1 in. =

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Civil Engineering Equations