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# Civil Engr Analysis CIVL 1112

University of Memphis

GPA 3.54

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This 67 page Class Notes was uploaded by Dana Yundt on Friday October 23, 2015. The Class Notes belongs to CIVL 1112 at University of Memphis taught by Staff in Fall. Since its upload, it has received 22 views. For similar materials see /class/228419/civl-1112-university-of-memphis in Civil Engineering at University of Memphis.

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Date Created: 10/23/15

CIVL1112 Excel Solver Function Q What is the Excel Solver good for Q What must I do to use a solver Q How doI define a model Q What kind of solution can I expect Q What makes a model hard to solve Q Can you show me step by step Excel Solver 17 Excel Solver Function What is the Excel Solver good for Q Solvers or optimizers are software tools that help users find the best way to allocate scarce resources Q The resources may be raw materials machine time or people time money or anything else in limited supply Q The quotbestquot or optimal solution may mean maximizing profits minimizing costs or achieving the best possible quality Excel Solver Function What must I do to use a solver Q To use a solver you must build a model that specifies O The resources to be used using decision variable 0 The I mits on resource usage called constraints and o The measure to optimize called the objective Q The solver will find values for the decision variables that satisfy the constraints while optimizing maximizing or minimizing the objective Excel Solver Function What must I do to use a solver Q Spreadsheets such as Excel provide a convenient way to build a model Q Anyone who has used a spreadsheet is already familiar with the process 0 Cells on a worksheet can hold numbers labels or formulas that calculate new values such as the objective of an opt mization o Con milisare simply limits specified with lt or gt relations on formula cells 0 And the decision Variablesar39e simply input cells containing num ers Excel Solver Function How do I define a model Decision Variables Q A solver deals with numbers so you39ll need to quantify the various elements of your model dadsa variables constraints and the abjecriveu and their relationships Q Decision Variablesusually measure the amounts of resources such s time and money to be allocated to some purpose or the level of some activity Q For example the number of products to be manufactured the number of pounds or gallons of a chemical required for some process etc Excel Solver Function How do I define a model Decision Variables Q For example if you are shipping goods from 3 different plants to 5 different warehouses there are 3 x 5 15 different routes along which products could be shipped Q So you might have 15 variables each one measuring the number of products shipped along that route CIVL1112 Excel Solver FuncTion How do I define a model Decision Variables Q In addiTion you mighT also have 4 differenT producT Types and you mighT wanT To plan shipmenTs in each of The nexT 6 monThs Q So This mighT lead To 15 x4 x 6 360 variables Q This illusTraTes how a model can become large raTher quicklyl Q ParT of The arT of modeling is deciding how much deTail is really required Excel Solver 27 Excel Solver FuncTion How do I define a model Objective Q Once you39ve defined The decision Variables The nexT sTep is To define The objective which is a funcTion ThaT depends on The variables Q For example suppose you were planning how many uniTs To manufacTure of Three producTs TV seTs sTereos and speaker Excel Solver FuncTion How do I define a model Objective Q Your objecTive mighT be To maximize profiT Assume ThaT each TV seT yields a profiT of 75 each sTereo 50 and each speaker 35 Q Then your objecTive funcTion mighT be 75quot TV sets 50quot stereos 3 5quot speakers Excel Solver FuncTion How do I define a model Objective Q On a spreadsheeT where The number of TV seTs sTereos and speakers are in cells D9 E9 and F9 respecTively The formula would be 2 75D9 50E9 35F9 Q You39d be finished aT This poinT if The model did noT require any constraints Excel Solver FuncTion How do I define a model Constrants Q In mosT models con raints play a key role in v deTermining whaT alues can be assumed by The decision variables Q Constraints reflecT realworld limiTs on variables Q To define a constraint you firsT compuTe a value based on The decision variables Q Then you place a limiT lt2 or gt on This compuTed value Excel Solver FuncTion How do I define a model Constraints Senera Constraints Q For exam e The cell rane A1A5 conTains The percenTage of funds To be used To purchase 5 differenT Types of maTerial Q You could use cell B1 To calculaTe SUMA1A5 Q Then define a constraintof B1 1 so ThaT The percenTages allocaTed musT sum up To 100 CIVL1112 Excel Solver FuncTion How do I define a model Constrants Bounds on Variables Q Of course ou can also lace a limiT direch on a decision variable such as A1lt 100 Q Upper and lower bounds on The variables are efficienle handled by mosT opTimizers and are very useful in many problems Excel Solver 37 Excel Solver FuncTion How do I define a model Constraints Physical Constraints Q Many constraints are deTermined by The physical naTure of The problem Q For example if your decision Variabesmeasure The physical dimensions of an objecT negaTive values for These variables would make no sense Excel Solver FuncTion How do I define a model Constrants Physical Constraints Q This Type of nonnegaTiviTy consTrainT is very common Q Constraints such as A1 gt 0 musT be sTaTed explicile because The solver has no oTher way To know ThaT negaTive values are disallowed Excel Solver FuncTion How do I define a model Constraints Integer Constraints Q Advanced opTimizaTion sofTware also allows you To specify constraintsThaT require decision Variables To assume only inTeger whole number values aT The soluTion Q If you are scheduling a fleeT of Trucks for example a soluTion ThaT called for a fracTion of a Truck To Travel a cerTain rouTe would noT be useful 9 InTeger constraintsnormally can be applied only To decision Variables noT To quanTiTies calculaTed from em Excel Solver FuncTion WhaT kind of soluTion can I expecT Q A soluTion seT of values for The decision Variables for which all of The constraints in The Solver model are saTisfied is called a feasible solution Q MosT soluTion algoriThms firsT Try To find a feasible I m o 2 E m m g l I o I ampQ 5 n I m o m m m I m lt D l m o I m objecTive funcTion when maximizing or decreases iT when minimizing Q An optimal solution is a feasible soluTion where The objecTive funcTion reaches a maximum or minimum value Excel Solver FuncTion WhaT kind of soluTion can I expecT Q A globaly optimal solution is one where There are no oTher feasible soluTions wiTh beTTer objecTive funcTion values Q A localy optimal solution is one where There are no oTher feasible soluTions quotin The viciniTyquot wiTh beTTer objecTive funcTion values Q The Solver is designed To find optimal solution ideally The global opTimum buT This is noT always ssible CIVL 1112 Excel Solver 47 Excel Solver FuncTion WhaT makes a model hard To solve Q Three major facTors inTeracT To deTermine h w difficulT iT will be To find an opTimal soluTion To a solver model 0 The maThemaT cal relationships beTween the objective and constraints and The decision variable 0 The size of The model number of dentin yorbe and constraint 0 The use of nTeger variables memory and solution Time may rise exponentially as you add more nTeger vuria les Excel Solver FuncTion Can you show me sTep by sTep Consider The following problem Q Imagine ThaT you are managing a facTory ThaT is building Three producTs TV seTs sTereos and spea ers Q Each producT is assembled from parTs in invenTory and There are five types of parTs chassis picture tubes speaker cones power supplies and elecTronics uniTs Q Your goal is To produce The mix of producTs which will maximize profiTs given The invenTory of producTs on hand Excel Solver FuncTion Can you show me sTep by sTep Q From This descripTion we can see ThaT decision V l39l39dbGS are The number of producTs To build 0 The objective mam will be gross profit Q Assume ThaT you can sell TV seTs for a gross profiT of 75 each sTereos for a profiT of 50 each and speaker cones for 35 each 75quot TV sers 50srer39eas 35spearers Excel Solver FuncTion Can you show me sTep by sTep Q To assemble a TV seT you need 1 chassis 1picTure ube 2 speaker cones 1 power supply and 2 seTs of elecTronics Q To make a sTereo you need 1 chassis 2 speaker cones 1 power supply and 1seT of elecTronics Q To build a speaker all you need is 1 speaker cone and 1 seT of elecTronics Q The parTs you have on hand are 450 chassis 250 picTure Tubes 800 speaker cones 450 power supplies and 600 seTs of elecTronics Excel Solver FuncTion Can you show me sTep by sTep Q Before we implemenT This problem sTaTemenT in eiTher Excel leT39s wriTe ouT formulas corresponding To The verbal descripTion above Q If we Temporarily use The symbols 0 x for The number of TV seTs assembled o y for the number of stereos and o z for The number of speakers Q The ToTal profiT is Maximize 75 x 50 y 35 z ProfiT Excel Solver FuncTion Can you show me sTep by sTep Q Building each producT requires a cerTain number of parTs of each T e Q The number of parTs used depends on The mix of producTs builT consTrainT lefT hand side and The number of parTs of each Type on hand consTrainT righT hand side 1x 1y0 1lt 450 chassis 1x 1y 0 zlt 450 Power supplies 2 x 1y 1 1 lt 600 Electronics CIVL 1112 Excel Solver 57 Excel Solver Function Can you show me step by step 0 T e next step ls to create a worksheet where the formulas for t e kinh39n funk and the quotAshtins are calculated s 1h the worksheet ah the hexr sllde we have reserved cells 54 F4 aha em held our deelslsh varlables x y ml 1 the humber ofTV sets stereos and speakers to bulld Excel Solver Function Can you show me step by step Excel Solver Function Can you show me step by step l c The ublecllve Tuhclluh ls 75 r 50 y r 5 ll c The Excel formula ls EA EIAVFA FMoBV BM Excel Solver Function Can you show me step by step l c The Excel formula fur lhe humber ufchassls used ls E739E4oF739F4o673964 c The Excel formula fur lhe humber qulclure lubes used ls EabEbwasrsmsabew c The remaln hg Used values are cumpmed h slmllclr ways Excel Solver Function Can you show me step by step l c The cuhslrelms are c Number ufused perls musl be less lhen hvehlury ur m b7 on 67611 c Number ufTVsels lereus ers musl be hum Excel Solver Function Can you show me step by step c TurunSulver cllck uhlhe om lab h lhe Amiysis gruup cllck Solver CIVL 1112 Excel Solver 67 Excel Solver Function Can you show me step by step v r l o Sellhemrgelcell urlhe value uflhe ablealve funcllurl n we caselhe Tmal Pruflt eell Em Excel Solver Function Can you show me step by step Excel Solver Function Can you show me step by step l 0 Set the By Chang rig Cells urlhe value unlle dm sm variaka l nmls ease the W lere Excel Solver Function Can you show me step by step 0 Sellhe nvemury cunslralm D7DllltC7Cll 5 Excel Solver Function 39 a Can you show me step by step r l o Sellhenunrnegallve cunslralm E464gt0 Excel Solver Function Can you show me step by step Excel5ulver l5 ready la run 7 0 quotli Just clle ml Solve CIVL 1112 Excel LOOKUP Functions 14 Excel Lookup FuncTions Q LookupType funcTions can reTurn parTicular informaTion from a series of a Table of daTa Q The Two mosT common lockup funcTions are VLOOKUP for verTical lcckup HLOOKUP for hcrizchal lcckup Q BoTh These funcTion lookup a parTicular value or TexT enTry in aTable and reTurn The relaTed informaTion Excel Lookup FuncTions Q The VLOOKUP funcTicn moves verTically down The rows of a lockup Table looking for maTching informaTicn in The firsT column of The Tab e Q The HLOOKUP funcTicn moves horizchally across The columns of a lockup Table looking for maTching informaTicn in The firsT row of The Table Excel Lookup FuncTions Q The formaT of The VLOOKUP funcTicn is VLOO KUPlcckLl pvalLleTablearray cclind egtltnle rangelcckLl p loohwwv is the value to be found in the first column the arm tooeomyis the Table of informaTicn n which daTais locked up ulildxlmis the column number in Tableiarray from which The maTch ng value musT be reTurned Ml 00kbw is a logical value that specifies wheTher you wanT VLOOKUP Tc ind an exac m r an approximaTe maTc Excel Lookup FuncTions VLOOKUPlcckup cclln exnumrangelcckup valueTablearray d Q The looruLVaue can be a value a reference or a TexT sTring Q The fabearray is a reference To a range Q A COL deLullll of 1 reTurns The value in The firsT column in fabearray a caLndeLIum of 2 reTurns The value in The second column in fabearray and so on Excel Lookup FuncTions VLOOKUPlcckupvalueTablearray cclindexnumrangelcckup Q If co39Idexnun is less Than 1 VLOOKUP reTurns The VALUE error value Q If co39Idexnun is greaTer Than The number of columns in fabearray VLOOKUP reTurns The REF error value Excel Lookup FuncTions VLOOKUPlcckupvalueTablearray cclindexnumrangelcckup Q If I39allgeoarup is TRUE or omiTTed an approximaTe maTch is reTurned Q In oTher words if an exacT maTch is noT found The nexT largesT value ThaT is less Than looruLVaue is reTurned Q If rangeoarup is FALSE VLOOKUP will find an exacT maTc Q If one is noT found The error value NA is reTurned CIVL1112 Excel LOOKUP Functions 24 Excel Loo kup FuncTions VLOOKUPlookupvalueTablearray co i ndexnumrangelookup 0 If ralyeookup is TRUE The values in The firsT column of labearray musT be placed in ascending or er 1 U 1 L AL FALSE IRUE oTherwise VLOOKUP may noT give The correcT value 0 If ralyeookup is FALSE Inbearray does noT need To be sorTed Excel Lookup FuncTions VLOOKUPlookupvalueTablearray colindexnumrangelookup 0 You can puT The values in ascending order by choosing The Sort command from The Dafa menu and selecTing Ascending o The values in The firsT column of fabearraycan be TexT numbers or logical values 0 Uppercase and lowercase TexT are equivalenT VLOOKUP Examples m X 2 3 19 n9 8 7 7 15 3 A 1 21 22 8 1E 24 g 25 3 1E 31 23 31 E 28 94 24 48 2D no 5m VLOOKUP1 E19H24 1 TRUE reTurns O 9 VLOOKUP1 E19H24 1 FALSE reTurns NA VLOOKUPO 8 E19H241 reTurns NA VLOOKUP Examples VLOOKUP3 E19H24 3 reTurns 15 6 VLOOKUP3 E19H24 3 FALSE reTurns NA VLOOKUPZ 2 E19H24 4 reTurns 234 VLOOKUP Examples F c n I J Mm oggn nh six in i K AIP Cor M VLOOKUP 3 To 4quot B4J6 4 reTurns 340 VLOOKUP l To 2quot B4J6 9 reTurns 190 VLOOKUP l To 7quot B4J6 1 reTurns 1 To 2quot MATCH looku valuelookul arraymaTchType Excel MATCH FuncTion O The MATCH funcTion reTurns The relaTive posiTion of an iTem in an array ThaT maTches a specified value in a specified order laakzpvaueis The value you wanT To maTch n The IanmLmy IanmLmyis a conTiguous range of cells conTa n ng possible lockup values MarcLfmeis The number 1 O or 1 CIVL 1112 Excel LOOKUP Functions 34 Excel MATCH FuncTion MATCHlookupvalue lookuparray maTchType 0 The ookupvaue is The value you wanT To maTch in ookuparray For example when you look up someone39s number in a Tele hone book ou are usini The erson39s name as The lockup value buT The Telephone number is The value you wanT o Lookupvalue can be a value number TexT or logical value or a cell reference To a number TexT or logical ue val Excel MATCH FuncTion MATCHlookupvalue lookuparray maTchType o If quotmicLI is 1 MATCH f nds The lurgesT value ThuT is less Than or equuyrfo lookupivulue Lookupiurray musT be placed in ascending order 10 12 AZ FALSE TRUE o If mereUm is o MATCH f nds The firsT value ThaT is exuchy equal To lookupivulue Lookupiurruycun be n an order o If maleLiypeis 1MATCH finds The smullesT value ThaT is greuTer Than or equal To lookupivulue Lookupiurray musT be placed n descending order TRUE FALSE ZA2 1 o 1 and so on o If matsLink is omiTTed H is assumed To be 1 Excel MATCH FuncTion MATCHlookupvalue lookuparray maTchType o MATCH does noT disTinguish beTween uppercase and lowercase leTTers when maTching TexT va Lies 0 If MATCH is unsuccessTul in Tinding a maTch iT reTurns The NA error value 0 If MarcLfype is O and ooupvaue is TexT ooupvaue can conTain The wildcard characTers asTerisk and quesTion mark 0 An asTerisk 0 maTches any sequence of chowder511 quesTion mark 2 maTches any s ngle characTer Excel MATCH Examples MATCH4 O F19F241 reTurns 2 MATCH4 0 F19F24 0 reTurns NA MATCH4 0 F19F24 1 reTurns NA Excel MATCH Examples MATCH4 8 F19F241 reTurns 3 MATCH4 8 F19F24 O reTurns 3 MATCH4 8 F19F24 1 reTurns NA Excel MATCH FuncTion 7 i 1 d 1 1 L 1 M 1 17 TB X 2 3 o 7197 AB 2 n 1EE am i 31 E 2 E 9 A A 25 3 1E 31 g 22 E 1B 2 A g 15 3 A 1 a n9 8 7 7 25 0 Suppose you wanT To know which row The value of x3 exceeds a value of 20 The value in row quotampMATCH20L19L241amp exceeds 20 o This funcTion reTurns The value in row 2 exceeds 20quot CIVL 1112 Surveying Angles and Directions 16 Angles and DirecTions Angles and DirecTions Q Surveying is The science and arT a Q Evidenceof surveying and recorded measuring disTances and a ESon or far mahon 3 05 fro I H f f H h ive ousan ears in aces suc as near a sur ace 0 2 ea China India Ebb Ion ar39i d Em IT Q IT is an orderl recess of acquirin daTa reaTing To Theyprhysical characTerisETics of The word WigEcon from The Lahquot The an in parTicuIar The reaTive word MENquot5 mammg 1 5quotquot posiTion of poinTs and The magniTude o erquot Angles and DirecTions Angles and DirecTions Tak Q In surveying The direcTion of a line is Q The Term quotmeridianquot comes from The described by The horizonTaI angle ThaT iT makes wiTh a referenc in LaTin ierdies meaning quotmiddayquot Q The sun crosses a given meridian midway Q This reference line is called a meridian beTween The Times of sunrise and sunseT on ThaT meridian o The same Latin slem givzs V lSZ lo The terms A M Ame Meridian and PM Post Meridian used is disambiguate hours of ihz day when using ih212rhour l Angles and DirecTions Angles and DirecTions Q A meridian or line of longitude is an imaginary arc on The Q The meridian that passes 39 E f through Greenwich England 1quot 5 5 quot 3 zsmbhshzsihz mzahh of from The NorTh Pole To The SouT Pole ThaT connecTs a locaTions running along iT wiTh a given longiTude Q The posiTion of a poinT on The meridian is given by The aTiTude CIVL 1112 Angles and Direc rions Surveying Angles and Directions Q There are Hiree fypes of meridians 26 culled geodetic nor Magnetc direction tuk needie at observer s posit conveni smred arbitrary direction taken for ence OASII39MWIIc direction determined fro the shape of the earth and gravity also en by a magnetic 39 ion 1 is divided fu he Eu Yiiniiins and is stiii asid in mi ciiginatid Win tn aneiint Saniiiiins int t d giis and guanine eccidinatis Sexagesinai renew is unaniiai semi Win sixty as tn as Angles and Direc rions fr ggt f gquot 1 Q Mefhods for expressing Hie magnifude of plane angles are sexagesma cenfesmal radials and mils SergeantSystem Tne ci cumfererice of circies iYiiO 360 parts degrees each deg N 7 divided iYilO minuies arid seconds b r ennnsac wastiansn inn h ittid ti tn Mains tci msniing m Angles and Direc rions Q Babylonian mafhemafics Sexagesi system rnaias ed naneient Mempma niawm nct apu idaee ntne sense tnat it didn t use an dist nci ymhai tci it a a digit Appmximations 10 i 20 Tres tne ab Angles and Direc rions Q Mefhods for expressing Hie magnifude of plane angles are sexagesmal cenfesmal radials and mils 1 is appruximciiely tne Widtn ufci iittie finger at arm s iengtn s appruximaiely tne Widtn ufciciused tist at arm s is appruximaiely tne Widtn ufci nan s cinciiarm s uve snauid be treated as r e measuremenis clearly depend an tne individual su u engtn engtn bieet and n appr Ximaiiuns wily Angles and Direc rions 39s Q Mefhods for expressing Hie magnifude of plane a gles are sexagesm cenfesl39lnal radius and mils Cznizsina System The circumference of circies is di ided inio 400 paris caiied gun previously caiie grads Angles and Direc rions Q Mefhods for expressing Hie magnifude of plane angles are sexagesm cenfesmal radials and mils Radany There are 27 radians Wi a circle 1 radian 57 295E or 57 173945quot CIVL 1112 Surveying Angles and Directions 36 Angles and DirecTions Angles and DirecTions 7 Azimuth QMeThod for expressmg The magnITude Q A common Terms used for designang The direcTIan of of plane angles are sexageSImal a iine is the azimuth cem esi 0 radans and i I I Q from Arabic aIsumuT from al The samT way Q The azimuTh of a line is defined as The clockwise angle from The narTh end or souTh end of The reference meridian MI The circumference of a circle is divided inTa 6400 parTs used in miIiTary science Q AzimuThs are usually measured from The norTh end of The meridian Angles and DirecTions Angles and DirecTions Azimuth Nam Azimuth Q Every line has Two azimuTh forward and back and Their values differ by 180 Q AzimLITh are referred To asTronomic magneTic or assumed meri ian Angles and DirecTions Angles and DirecTions Azimuth 1 Bearing For example The famard azimurhof line AB is 50 The back azimurwr azimuTh of BA is 230 Nor Nari1 Q AnoTher meThod of describing The direcTion of a line is give iTs beary Q The bearing of a line is defined as The 505 smallesT angle which ThaT line makes wiTh The gt3 reference meridian A B Q A bearing cannoT be greaTer Than 90 bearings are measured in relation to the north or souTh end of 230 the meridian NE NW SE or s CIVL 1112 Surveying Angles and Directions 46 Angles and DirecTions Angles and DirecTions Nari1 Bearings Bearings NEDDE Q IT is convenT To say N90 E is due EasT N759 B 590 W is due WesT D Q UnTiI The IasT few decades American surveyors favored The use of bearings over azimuTh EasT Q However wiTh The advenT of compuTers and calcuIaTors surveyors generally use azimuTh Today insTead of bearings Traverse and Angles Traverse and Angles Q A fmverse is a series of successive sTraighT Q An exfer or crye is one ThaT is noT enclosed lines ThaT are connecTed TogeTher A by The sides of a closed Traverse Q A Traverse is dosed Q An infer390quot crye is one enclosed by sides of a such as in a boundary Closed Traverse survey or open as for a B highway E I ExTerIor K D C C B Traverse and Angles Traverse and Angles Q An angle 7 0 12 rigIf is The clockwise angle Q A de edon crye is The angle beTween The beTween The preceding line and The nexT line preceding line and The presenT one of The a Traverse Angle lo the righl Angleioiherighi m B 4quot c gle 0 The righT A A D c 23 2539 Angle To The righT CIVL1112 Traverse and Angles Q A de edon angle is The angle beTween The preceding line and The presenT one C Angle To The lefT g B 39 65 1539 Angle To The le 39 Surveying Angles and Directions 56 Traverse and Angles Traverse Compufafions Q If The bearing or azimuth of one side of Traverse has been deTermined and The angles beTween The sides have been measured The bearing or azimufrs of The oTher sides can be compuTed Q One Technique To solve mosT of These problems is To use The deflecTion angles Traverse and Angles Q Traverse Example From The Traverse shown below compLITe The azimLITh and bearing of side BC N 30 3539 E E5 14 n Deflection angle 180 85 1439 94 4639 Traverse and Angles Q Example CompLITe The inTerior angle aT B B 5 75 15 E N 62 2039 E C U a Q Example CompLITe The inTerior angle aT B Nari1 N 62 2039 E InTerior ABC 62 203975 1539 D 137 3539 CIVL 1112 Detention Pond Part 1 19 Detention Ponds o A detention basin is an artificial flow control structure that is used to contain flood water for a limited period of a time o Detention basins are best management practices BMPs used to mitigate the effects of storm water runoff with the primary purpose to control storm water runoff provide temporary storage and control storm water release to downstream areas Detention Ponds o Detention basins are for short term storage of storm water o The objective of a detention facility is to regulate the runoff from a given rainfall event and to control discharge rates to 39 pact on downstream storm water systems 0 Detention basins are permanent storing of water indefinitely Detention Ponds slon that urbanizutlo increases runoff from rainstorms and snow melt o It is generally acknowledged by the stormwater management profes 39 39 n o For example studies in Denver show that during a utine summer afternoon rainsiiower an acre of pavement will produce the same amount of runoff as would occur from several square miles of native rangeland Detention Ponds o A retentiondetention lao facility can be eitner apond an undergraundlank or vault or an nfittration system specifically designed to capture store and tnen slowty release stormwdier runoff downstream or nto tne surround ng ground Detention Ponds o How do RD Facilities Work 0 A detention pond stores accumulated stormwater runoff and slowly releases it downstream o A flow control structure regulates ttie release rate of tine stored water 0 some detention ponds are combined witn water quality treatment ponds i e wetponds wnicn are intended to nave some water in nem on apermanent basis Detention Ponds Wlnat are tine Benefits of RD Facilities o In additiontoneip ng prevent fioodin and erosion Rl e t nt excess nutr ents and luxic cnemicais o In some cases ponds can provide feeding nesting breeding and niding places for many species of fisn birds and reptiles CIVL 1112 Detention Pond Part 1 Detention Ponds Detention Ponds Q Why is stormwater runoff a problem Q Why is stormwater runoff a problem Q As we cut woodlands clear land pave road d WY 5 parking lots and construct houses and buildings we 39 ear 0quotquot change the permeability of the ground Q Fallin rain has fewer places to soak in gradually R no f on hard surfaces occurs faster and in greater Discharge Q Increased stormwater runoff can worsen flooding erosion and water pollution and destroy stream habitat Mln um Detention Ponds Detention Ponds Fc an Urban stormwater discharges cause no significant harm to our lakes rivers and oc Q Planning for Detention Ponds Q It is not sufficient to address only hydrology and Facf39 hydraulics Urban Smrmwmer is one of The m 9 cum sources Q According to ASCE successful detention facilities of pollution in our nation39s rivers lakes and estua 39es also have strong recreational or other community According to t e U 5 Environmental Protection Agency uses urban stormwater is the second largest source of water The daler T39Oquot Peel 395 o equot Consldere j second lry quality damage in estuaries and a significant contributor by T 5 res39dems 39 The are Plannmg 0quot dalemmquot To The damage To lakes rivers and buys needs also to consider the social enVIronmental a d recreational needs of each communit Detention Ponds Detention Ponds Q Planning for Detention Ponds b Q Planning for Detention Ponds Q Effects on the Landscape Aesthetics Q Pond Environment Q Recreational Opportunities Q Removal of Pollutants Q Detention in Natural Lakes Q Safety Q Layout of Detention Ponds CIVL 1112 Detention Pond Part 1 39 Detention Ponds Detention Ponds 0 Effects on the Landscape Aesthetics 0 Effects on the Landscape Aesthetics n 0 As an m2ng part aftne eammunity it seryes adetentian pund needs ta biend intatne iandsea e and intatne eammunit Tun aften detentian punds are instaiied merety as a naie in tne gruund sametimes referred ta as an HIBwiihuui any redeem rig landscape features 0 5i e yet inexpensive measures suen as gentie side slupes pianting aftrees and snrubs and diner landscaping features ean transfarm an HIE nta an dtraetiye amenity far tne neighburhuud Detention Ponds Detention Ponds o Pond Environment 0 Recreational Opportunities 0 Funds tnat naye apermanem puul uf water affer many attraetiye eny ranmentai passibiiities o betentian basins and pundswilh ar wnnaut permanent puuls a water a er man reereaiianai a artunities in an urban se t rig rbanizaiian aeeu s h bquot 39 5 h h bquot Sim sg uss W M f db 2 o5tarmwaier detentian ean be ncurpurmed neety ntatne a a W a a is MP W W WW2 WW 5 V 5i landscape and prayide reereaiianai appartunities far iaeai and trees tnat affer nabitat far seieet smaii b rds and animais residents between starms suen as squirreis 0 These natural enyiranmentai packets are eansidered by many 39 ifg39jj ffjbfglfxm xfeed39 eny dweiiers ta be atreasure in an ulherwise densety urbanized P 39 WWW reereatianai uses mast aftne time Detention Ponds Detention Ponds 0 Recreational Opportunities Skyline Park Denver Colorado 0 Recreational Opportunities Skyline Park Denver Colorado CIVL 1112 Detention Pond Part 1 Defen rion Ponds o Bioretention Defen rion Ponds Blurelem lElrl ar u s alsa o Bioretent reu re erred la as blarelenllan fmers er 0 BlElrelerlllElrl areas alsu referredlu as blarelenllan fmers er ralrl gardens are sl u ural slarrnwaler eanlrals lnal eaplure ralrl gardens are slrudural slarrnwaler eanlrals lnal eaplure and lempurarW slare lne waler auallly vulume usng sells and andlernpararlly slare lnewaler qualle vulume us lng sells and vegelallun lrl snallaw baslrls er landseaped areas la remuve vegefaflElrll snallaw aslrls er landseaped areas la remuve pElHularlls frum slarrnwaler runElff rmHlnnnN frnm slnrrnwnler rllrlnff o Tne fulluwlrlg ls a llsl ElfsElme anne bellest er blarelenllan areas and ralrl gardens pelleaele e snall dranage a 0 seed far nlgnle lmpervlau areas sudn as park ng m o nelaneele 1aw mamenallce requlremem 0 can ee planned as an aeslnene feamre Defen rion Ponds Defen rlon Ponds o Bioretention ur o Bioretention ur O BlElrelemlElrl areas a1sEl referred U as blElrelerlflElrl leers Elr O BlElrelerlflElrl areas a1sEl referred fEl as blElrelerlflElrl leers Elr raln gardens are slrudural slarrnwaler eanlrals lnal eaplure raln gardens are slrudural slarrnwaler eanlrals lnal eaplure and lernpararl slare lne waler quallly vulume usln salls and an lern ararl y slare lnewaler qualny vulume us lng salls vegelall n n snallaw baslrls ar landscaped areas la remuve vegelallun ln snallaw baslrls ar 1anascape areas la Elve pallulanls frum slarrnwaler ru pallulanls frum slarrnwaler r n f O The fElHElerlg ls a1lsl ufsElme Elffhe benefns Elf blElrelerlflElrl O The fElHElerlg ls a1lsl Elf sume Elf Me hmlfaYlElrls Elf areas and ran gardens blarelenllan areas andr lngard s o Applleaele le srnall dramage areas o cann l ee used lelreal large dranage areas o eeed far nlgnly lmpervlau areas sudn as park ng lels u eepl ele le elegglng byxedl enl o R21QHVB1Y1DW rnanlenanee requlreme s o Terd le eensurne spaee aeeul 6af he dranage area o can ee planned as an aeslnelle feamre o wnen eernpared le elner delenl len epllens lne eenslruellen eesl anee nlgn Defen rion Ponds o Bioretention urea Defen rion Ponds 0 Removal of Pollutants o Deenlen baslns and pends le eause suspended sellds e sel He o Slnce rnan af ne pelluans are allaened e suspende sellds pends le rerneve serne af ne 0 Ha lrl1 w rnuen ls rerneved le depend en pend velurne e and eule eenflguramen pend depn and shape and e lrne lne serrnwaer resldes ln lne end CIVL 1112 Detention Pond Part 1 Detention Ponds 59 o Detention in Natural Lakes when the downstrearn reci ient ot urban storrn unott is a naturai iake awater suppiv reservoir or a recreationai reservoir each of them can provide peak ow attenuation othe ow routing advantages of these water bodies however can extract a price in the form of water quaiitv deterioration and adverse impacts on their naturai or designated uses Detention Ponds o Detention in Natural Lakes utrient enrichment and excessive aigae ieveis can depiete oxvgen and cause tish kiii o Deposits of sediments containing heavy rnetais and attached petroieuni product wiii occur in the bottom Detention Ponds o Detention in Natural Lakes o It sait is used to controi street icing increases in take saiinitv can occur o If acid rain is of concern the increased surtace runott trorn urbanizatio av increase the aciditv of the receiving water bo Detention Ponds 0 Safety Satetv issues inciude the structurai integritv ot the o tining embankment the outiet works peopie using the taciiitv tor recre othe iatter inciudes the need to protect peopie when the pond is storing runott i e operating and during the periods between storms Detention Ponds o Layout of Detention Ponds o when pianning adetention basin try to iav it out so that it fits the surrounding iatidscape and the com unitv o Detention ponds shouid be iaid out to insure that the ow entering the Vid is distributed across the pond so that stagnant zones do not deveiop in the pond Detention Ponds o Layout of Detention Ponds CIVL 1112 Detention Pond Part 1 69 Defen rion Ponds 0 Technical Configuration 0 In ow swucmre r Canngummn bf Pond Bo 39om r Slope Prmcmn o Ou e smcme 0 Trash Racks 0 prHway vs Embankmen Overfoppmg r Embankmen Loss Analys s Defen rion Ponds o Inflow Structure 0 Eroson and sedwmenf depaswian problems can deveiop aH39He mfiow To The defen on basm o Anhaugh 1 15 posswble w deswgn m aw swcmres n mmrmxe eroson depasman mum be prevenfed OTa mimmze mamrenance ems However 1 is a good deafo iocahxe much of ne deposmon where 1139 can removed be easwly Defen rion Ponds o Inflow structures should have the following features 0 Dwsswpme uw energy aube mbe o Pruwde prulemmn against emsmn o Pruwde mmmenance access fur the repcl rs m we met and fur we remuval uf semmenvs o Incurpumle safely femures m prulecl we publ e 1 e genne slupes fencmg up rmlmg av yernem faces uflne smaure 0 Be unublruswe yume pubhc eye by biendmg the met mu we surruund rig lerrmn Defen rion Ponds o Inflow Structure Defen rion Ponds o Inflow Structure Defen rion Ponds o Inflow Structure CIVL 1112 Detention Pond Part1 79 Defen rion Ponds 0 Configuration of Pond Bottom 0 We can gum an of fne pond ba 39am um depend o a arge degree on unefner fne pond um nave a perrnanenf poo1 of wafer o Ba 39ams um perrnanerd poo1s are m some respeefs easwer Po desng The areas above fne permanenf poo1 can nen be graded re a ve y naf kep dean unfn 1ess effaM39 o The permanenf poo1 ean a1so aef as ne se 39hng basm Defen rion Ponds o Can gum an of Pond Ba 39am Defen rion Ponds o Can gum an of Pond sworn o A defen an basm PnaP s139ares waer an y durmg sforrns snou1d nave afrmk e ow dwen befween fne mwef and The ou ef 0 Law non dnenes are sued o carry we frequen y accurr mg runoff and fruwe ows 0 wnen arge ransforrns oeeur fne capacn39y of fnese dnenes 15 exceeded and fne wafer noods fne adJacenf pond ba 39am Defen rion Ponds o Can gum an of Pond Ba 39am Defen rion Ponds o Can gum an of Pond Ba 39am 0 The mas successfu msmHa ans of 1m ow and ka e ow enanne1s have a eoneree ba 39am o Concrefe hmng faemfafes se1fe1eansmg of ne dwen a mmr 39enance o Dramage of Pne pond baH39am be139ween sarms nas o oeeur ef men y M n 15 gmng o be used for recrea an 0 Ta achweve st The ba 39am Has a be crasss1aped a no ess fnan 2 foward ne ka e enanneKs Defen rion Ponds o Slope Protection 0 Swde dupes of The pond w1Hend a erade whenever fne defamed wc 39er surface ucmc 39es fre uenH or en Pnere 15 wave act on 0 Good vegem an um He p Ta profeef fne swde dope agamsf eraswan reds of men ve acmes and wave a ack sfrucfum measures are needed o supp emenf vegem an CIVL 1112 Detention Pond Pan 1 89 Defen rion Ponds o slope Protection Defen rion Ponds o slope Protection Defen rion Ponds o slope Protection Defen rion Ponds o slope Protection Defen rion Ponds Q Ouflef Sfrucfure c The configuration ofo pond outlet determines the type of pond wet or dry the storage volume and the control the pond provides the storm runoff c lvlany detention ponds are designed to control runoff rordirrerent rainstonns Defen rion Ponds Q Ouflef Sfrucfure c Deslgn fur lne cunlnul uf lwu ur innee leyels uf aw fur example 27 and lo ear c Pruvlde maintenance access in lne uullel c If possible use no muying pants ur pumps n an uullel Use m si e cumpunenls to reduce damage frum yandalism c Pruvlde eruslun protection at lne inlet and outlet ends uf lne uullel p pe c Pruvlde cuanse grayel pocklng to screen out debris o Always design Wlln maintenance and aeslnel cs 1quot mind CIVL 1112 ACI Mix Design 19 ACI Mix Design I The mosT common meThod used in NorTh America is ThaT esTablished by ACI Recommended PracTice 211 1 I Any mix design procedure will provide a firsT approximaTion of The proporTions and musT be checked by Trial baTches I Local characTerisTics in maTeriaIs should be considered I The following sequence of sTeps should be followed 1 deTermine The job parameTers aggregaTe properTies max mum aggregaTe size slump wc raTio admixTures 2 calculaTion of baTch weighT and 3 adjusTmenTs To baTch weighTs based on Trial mix ACI Mix Design WaTercemenT raTio w c raTio Theory sTaTes ThaT for a given combinaTion of maTerials and as long as workable consisTency is obTained The sTrengTh of concreTe aT a given age depends on The wc raTio The lower The wc raTio The higher The concreTe sTrengTh Whereas sTrengTh depends on The wc raTio economy depends on The percenTage of aggregaTe presenT ThaT would sTill give a workable mix The aim of The designer should always be To geT concreTe mixTures of opTimum sTrengTh aT minimum cemenT conTenT and accepTable workabiliTy ACI Mix Design 50 51 Iill 21 Airentrained concrete ii 23day mmprsssius strength MPH m D i Nanaairientrainsd concrete 39 if 5 28day G f39 ipi E SSWIE s39irsngiht 1WD psi 390 l J a J I a I I 113 14 35 ISLE ULquot 113 Water in ssrnsniiiiaus rnaisrisls rails I1219 ACI Mix Design Once The wc raTio is esTablished and The workabiliTy or consisTency needed for The specific design is chosen The resT should be simple manipulaTion wiTh diagrams and Tables based on large numbers of Trial mixes Such diagrams and Tables allow an esTimaTe of The required mix proporTions for various condiTions and permiT predeTerminaTion on small unrepresenTaTive baTches ACI Mix Design Resistance to weathering and chemicals I The f lowcharT is a represenTaTion Of The pr nC pal pr oper l39ies 0f Appropriatetementtwe Low wc ratio 900d H CO ncr39e le Proper curing Use of superplasticizers or 9 ix ures I I wc raTio I aggregaTe I cemenT pasTe and aggregaTe I mix ng I placemenT and handl ng of fresh concreTe Good quality of paste Low wcp ratio I Optimal oemenv contem Sound aggregate grading nd vibration Low air content Large maximum aggregate size Efficient grading Minimum slump Minimum cement content Optimal automated plant operation Admixtures and entrained air Quality assurance and control ACI Mix Design Basic ConsideraTions I Economy The maTerial cosTs are mosT imporTanT in deTermining The relaTive cosTs of differenT mixes I The labor and equipmenT cosTs excepT for special concreTes are generally independenT for The mix design I Since cemenT is more expensive Than aggregaTe iT is clear ThaT cemenT conTenT should be minimized I This can be accomplished by 1 using The lowesT slump ThaT will permiT handling 2 using a good raTio of coarse To f ne aggregaTe and 3 pass ble use of admixTures CIVL 1112 ACI Mix Design 29 ACI Mix Design Basic Considerations I WorkabiliTy A good mix design musT be capable of being placed and compacTed wiTh minimal bleeding and segregaTion and be finishable I WaTer requiremenTs depend on The aggregaTe raTher Than The cemenT characTerisTics I WorkabiliTy should be improved by redesigning The morTar facTion raTher Than simply adding more waTer ACI Mix Design Basic ConsideraTions I TrengTh and DurabiliTy In general The minimum compressive sTrengTh and a range of wc raTios are specified for a given concreTe mix I FOSSIble requiremenTs Tor resnsmnce To freezeThaw and chemical aTTack musT be considered I Therefore a balance or compromise musT be made beTween sTrengTh and workabiliTy ACI Mix Design I A measure of The degree of consisTency and exTenT of workabiliTy is The slump I In The slump TesT The plasTic concreTe specimen is Tormea mm a conical meTaI m0Ia as described In ASTM STandard C143 I The mold is lifTed leaving The concreTe To I39slump quot ThaT is To spread or drop in heighT ACI Mix Design I Process of measur ng The slump of fresh concreTe ACI Mix Design I Process of measuring The slump of fresh concreTe ACI Mix Design I This drop in heighT is The slump measure of The degree of workabiliTy of The mix CIVL 1112 ACI Mix Design 39 ACI Mix Design I Here are some examples of different measures of slump ACI Mix Design If slump is not given decide the slump use Table 31 I Decide the maximum size of aggregate choose the maximum possible size using the following guidelines maximum size 2lgt 15 narrower dimension between forms I The flowchart is a representation gt13demhofsia9 Zlgt 34 of clear spacmg between reinforcmg bars of the prmcupal properties of ACI i m X n Decide the amount of water and air Table 323 I workability slump I wafer and a r39 CO nle n1 laculate cement content cw X wt of wateil I wc ratio l I Choose the amount of coarse aggregate Table 34J I cement Calculate the amount or fine aggrngratglusiansgphed esim dwihoffr nrete an I the kntowiiuiaiveightgs f wet ciirisnt andacogrse aggregate I fine aggregate 39 39 inc reandfin a r te I maisture ad J ustment 5 e 99 93 I trial mix ACI Mix Design Mix Design Procedures 1 Required material information sieve analyses of both fine and coarse aggregates unit weight specific ravities and absor39 tion ca acities of areates 2 Choice of slump Generally specified for a particular job However if not given an appropriate value may be chosen from Table 1 As a general rule the lowest slump that will permit adequate placement should be selected ACI Mix Design Table 1 Recommended Slumps for Various Types of Construction Slump in Slump mm Types of Construction Maxb Min Max Min Reinforced foundation walls and 3 1 80 20 footings Plain footings caissons and 3 l 80 20 substructure walls Beams and reinforced walls 4 l 100 20 Building columns 4 l 100 20 Pavements and slabs 3 l 80 20 Mass concrete 3 l 80 20 ACI Mix Design Mix Design Procedures 3 Maximum aggregate size The largest maximum aggregate size that will conform to the following limitations I MaXmum size should not be larger than 5 the minmum dimension of struc tural members 3 the thickness of a slab or 34 the clearance be tween reinforcing rods and forms These restrictions hmit maXmum aggregate size to 15 inches except in mass applications I Current thought suggests that a reduced maximum aggregate size for a given wc ratio can ach eve higher strengths Also in many areas the largest available sizes are 34 in to 1 in ACI Mix Design Mix Design Procedures 4 Estimation of mixing water and air content An estimation of the amount of water required for air entrained and nonairentrained concretes can be obtained from Table 2 One major disadvantage of concrete is its susceptibility to damage by s ngle or multiple freezethaw cycles However concrete can be made frostresistant by using a r entraining admixtures Concrete is routinely airentrained n the Northern US and Canada CIVL 1112 ACI Mix Design 49 ACI Mix Design Mix Design Procedures Appruximme mixingwmer ibyd and cm cement fur differem siumps and nummai maximum sizes uf aggregmes stumpttn its 2 Nonek irrEmr amzd Concrete ACI Mix Design Mix Design Procethres Appruximme mixingwmer ibyd and cm cuntent fur differem siumps and nummai max mum sizes uf aggregmes A irrEmr amzd Concrete ACI Mix Design Mix Design Procedures ACI Mix Design Mix Design Procethres 5 Watercement ratio 7 Tnts component ts governed by strengtn and durabiiiiy requtrements e5tnengtn W thnuut strengtn vs we mttu dam fur a certain mmeriai acunservmwe est imme can be made fur tne accepted 23er cumpresswe strengtn tmn Tobie 3 b bumbiiiiyw Iftnere are severe expusure cundniuns sucn as freezm unetnawtngexpusunetuseewatenur suitetestne wc mttu requiremems may have tn be e41 ustee ACI MIx DesIgn Mix Design Procedures Reimiunsnip between wmercemem mtm and cumpresswe strengtn uteunenete 2ampde Cumpresswe Sirenlin s ACI Mix Design Mix Design Procethres o Cale Iation of cement content W Once the water c u onte owever amtntmum cement content ts requtred to ensure good ttntsnabtitty workabiiiiy and strengtn Weth of 6571er M W CIVL 1112 ACI Mix Design 59 ACI Mix Design Mix Design Procedures 7 Estimation of coarse aggregate content The percent of coarse aggregate to concrete for a given maximum size and fineness modulus is given by Table 4 The value from the table multiplied by the dryrodded unit weight the ovendry OD weight of coarse aggregate required per cubic foot of concrete To convert from OD to saturated surface dry SSD weights multiply by 1 absorption capacity AC ACI Mix Design Mix Design Procedures Volume of dryrodded coarse aggregate per unit volume of concrete for different coarse aggregates and fineness moduli of f ne aggregates Fineness Modulus Max Aggregate in 24 25 26 27 28 29 3 H f r quot 0391quotquot 3 Ag 39 39 A r r 0500 7 0750 1000 quot 39 1500 2000 3000 6000 ACI Mix Design Mix Design Procedures Volume of dryrodded coarse aggregate per unit volume of concrete for different coarse aggregates and fineness moduli of fine himTrial l g g39lgl39r39urr l augireg l 553E rI If EU 1 LI I d5 3 LE Wr flu 39II39UUIFI U 2 394 3 tr 1 anal a F39rluriLss mL dulm EJ Frannmis mEhjuius P E Friarr155 l39ltL39h39JulLI l nnnsr mrrnriulut 39I III ram 39e39CIli 39lT lrnctmn Euquot marsh nggrngnlri I 1 l 39 i lk i I l I I I u 45 Lari ME39WH39QI naazrl urraggmgalu 55 mm ACI Mix Design Mix Design Procedures 8 Estimation of fine aggregate content There are two standard methods to establish the fine aggregate content the mass method and the volume method We will use the quotvolu Wequot Method I quotVolumequot Method This method is the preferred method as it is a somewhat more exact procedure I The volume of f ne aggregates is found by subtracting the volume of cement water air and coarse aggregate from the total concrete volume ACI Mix Design Mix Design Procedures 9 Adjustment for moisture in the aggregate The water content of the concrete will be affected by the moisture content of the aggregate o Q 9 Q a Q 00 0 96 a 0 a 0 o O Q a Q Ovendry Airdry Saturated Wet surface dry Absorption capacity 39 Effective Surface absorption mousture ACI Mix Design Mix Design Procedures 9 Adjustment for moisture in the aggregate The water content of the concrete will be affected by the moisture content of the aggregate o Q 9 Q a Q 00 0 96 a 0 a 0 o O Q a Q Ovendry Airdry Saturated Wet surface dry l A c E333 544 A CIVL 1112 ACI Mix Design 69 ACI Mix Design Mix Design Procedures 9 AdjusTmenT for moisTure in The aggregaTe The waTer conTenT of The concreTe will be affecTed by The moisTure conTenT of The aggregaTe o Q 9 Q a Q 00 0 96 a 0 a 0 o O Q a Q Ovendry Airdry Saturated Wet surface dry Mm39sfure com em M6 A6 514 I ACI Mix Design Mix Design Procedures 10 Trial baTch Using The proporTions developed in The preceding sTeps mix a Trial baTch of concreTe using only as much waTer as is needed To reach The desired slump buT noT exceeding The permissible wc raTio 7 ACI Mix Design Mix Design Procedures 10 Trial baTch The fresh concreTe should be TesTed for slump uniT weighT yield aIr conTenT and HS Tendencnes To segregaTe bleed and finishing characTerisTics Also hardened samples should be TesTed for compressive and flexural sTrengTh ACI Mix Design Example I ConcreTe is required for an exTerior column locaTed above ground where subsTanTial freezing and Thawing may occur The 28day compressive sTrengTh should be 5 OOO Ibin2 The slump should be beTween land 2 in and The maximum aggregaTe size should noT exceed in I The properTies of The maTeriaIs are as follows I CemenT Type I specific graviTy 315 I Coarse AggregaTe Bulk specific graviTy SSD 270 absorpTion capaciTy 1 dryrodded uniT weighT 100 IbfT3 surface moisTure 0 I Fine AggregaTe Bulk specif c graviTy SSD 265 absorpTion capaciTy 13 fineness modulus 270 Surface moisTure 3 ACI Mix Design Example I Tep 1 Required maTeriaI informaTion already given I Tep 2 Choice of slump The slump is given consisTenT wiTh Table 1 Slump in Slump mm Types of Construction Maxquot Min Max Min Reinforced foundation walls and 3 1 80 20 footings Plain footings caissons and 3 l 80 20 substructure walls Beams and reinforced walls 4 l 100 20 Building columns 4 l 100 20 Pavements and slabs 3 l 80 20 Mass concrete 3 l 80 20 ACI Mix Design Example I Tep 4 EsTimaTion of mixing waTer and air conTenT Since freezing and Thawing is imporTanT The concreTe musT be airenTrained From Table 2 The recommended air conTenT is 6 The waTer requiremenT is 280 Ibyd3 Maximum aggregaTe size in lumpin 0375 05 075 1 15 2 3 6 1 To 2 3 To 4 6 To 7 Air ConTen Mild Moderate Extreme CIVL1112 ACI Mix Design 79 ACI Mix Design Example I Step 5 WaTercemenT raTio From Table3 The esTimaTe for required wc raTio To give a 28day sTrengTh of 5 000 lbin2 is 04 ACI Mix Design Example I Step 6 CalculaTion of cemenT conTenT Based on sTeps 4 and 5 The required cemenT conTenT is wetth ofcemem 3 28 2yd 700b de ACI Mix Design Example I Step 7 EsTimaTion of coarse aggregaTe conTenT InTerpolaTing Table 4 for The fineness modulus of The fine aggregaTe of 2 70 5 Midulus ACI Mix Design Example I The coarse aggregaTe will occupy 3 063x27 Ada 1701 f 3 Value from Table 4 I The OD weighT of The coarse aggregaTe 3 7 lb 1701fa3gtlt100 Ib fg 7 1701 Ada DryrRodded UnliWZlgl f ACI Mix Design Example I Step 8 EsTimaTion of fine aggregaTe conTenT by The absoluTe volume meThod 39I39Illlllllltllll39F nunslmllln39 60 62 368 65 62 337 70 62 302 75 62 261 80 62 216 85 62166 ACI Mix Design Example Step 8 EsTimaTion of fine aggregaTe conTenT by The absoluTe volume meThod I WuTer I CemenT I Course AggreguTe I Air 280 lb624 lbfTa 449 H3 700 lb315 x 624 lbfTi 356 H3 1701lb270 x 624 lbfTa 1010 Hi 6 x 27fT3yda 162 H3 Total CIVL1112 ACI Mix Design Example 27 3 e 1977 3 723 26 3 I The OD weighT of The fine aggregaTe is 723 3 gtlt 265x 624b f3 1196b Speclf c 6mva ofF ne Aggregale I Therefore The fine aggregaTe musT occupy a volume of ACI Mix Design Example I Step 9 AdjusTmenT for moisTure in The aggregaTe The weighT of aggregaTe from The sTock pile is wagIran Weighfoo1 MC The change in The weighT waTer due To The moisTure of The aggregaTe from The sTock pile is A Weighfwd Weighfoo 514 A dus39red Wegh TM Wegh TM 7 A Wegh fwd ACI Mix Design 89 ACI Mix Design Example I Step 9 AdjusTmenT for moisTure in The aggregaTe I Since The moisTure level of The fine aggregaTe in ur sTorage bins can vary we will apply a simple rule To adjusT The waTer required Decrease The amounT of waTer required by surface moisTure conTenT of The weighT of The fine aggregaTe I Increase The amounT of aggregaTe by The amounT equal To The surface moisTure ACI Mix Design Example I Step 9 CompuTe sTockpile weighT based on moisTure conTenT I Fine aggregaTe required from The sTockpile is 1196 lb 1 0 04312474lbyd3or1247lbyd3 Molslure Comenl 43 Molslure Conlenl 1 1 om The sTockpile is I Coarse aggregaTe requ39 1701 lb 1 001 1718 Ibyd3 ACI Mix Design Example I Step 9 AdjusT The amounT of waTer based on moisTure conTenT The required mixing waTer required is 280 lb 1196 lb 0 043 v V10 6 fine aggregaTe Molslure Comem 4 3 Absorpllon Capaclly13 1718 lb 0 01 0 01 6 coarse aggregaTe 2441lbyd3 or 244 lbyd3 ACI Mix Design Example I Thus The esTimaTed baTch weighTs per yd3 are WaTer 244 lb CemenT 700 lb Coarse aggregaTe 1718 lb Fine aggregaTe weT 1247 lb I Total 3909 lbydSI 1448 lbfts CIVL 1112 ACI Mix Design Group Problem 2 13 ACI Mix Design Example I The 28day compressive strength should be 7 000 lbin2 The slump should be between 3 and 4 in and the maximum aggregate size should not exceed in I The properties of the materials are as follows I Cement Type I specif c gravity 315 I Coarse Aggregate Bulk specific gravity SSD 265 absorption capacity 05 dryrodded unit weight 100 lbft3 surface moisture 1 I Fine Aggregate Bulk specific gravity SSD 260 absorption capacity 11 f neness modulus 270 Surface moisture 3 ACI Mix Design Example I Step 1 Required material information already given I Step 2 Choice of slump The slump is given consistent with Table 1 Slump in Slump mm Types of Construction Maxquot Min Max Min Reinforced foundation walls and 3 1 80 20 footings Plain footings caissons and 3 l 80 20 substructure walls Beams and reinforced walls 4 l 100 20 Building columns 4 l 100 20 Pavements and slabs 3 l 80 20 Mass concrete 3 l 80 20 Step 3 Maximum aggregate size Given inches ACI Mix Design Example I Step 4 Estimation of mixing water and air content From Table 2 the recommended air content is 2 the water requirement is 340 lbyd3 Maximum aggregate size in ACI Mix Design Example I Step 5 Water cement ratio From Tabe3 the estimate for required wc ratio to give a 28day strength of 7 000 lbin2 is 0 33 28day Compressiv NonAE AE Strength psi 2000 3000 4000 5000 6000 7000 ACI Mix Design Example and 5 the required cement content is 340by0 3 033 weight of cement I Step 6 Calculation of cement content Based on steps 4 1030bya 3 ACI Mix Design Example I Step 7 Estimation of coarse aggregate content Interpolating Table 4 for the fineness modulus of the fine aggregate of 2 70 Fineness Modulus Max Aggregate in 0375 t a 7 r CIVL1112 ACI Mix Design Group Problem 2 23 ACI Mix Design Example I The coarse aggregaTe will occupy 3 ff3 063x27 Ada 1701 yds Valuefrom lable 4 I The OD weighT of The coarse aggregaTe s 7 lb 1701fa3gtlt100b f3 7 1701 g DryrRodded um Weight ACI Mix Design Example I Step 8 EsTimaTion of fine aggregaTe conTenT by The absoluTe volume meThod I Water 340 lb624 lb a 545 Ha I Cement 1030 lb315 X 624 lb a 524 Ha I Course Aggregate 1701lb265 X 624 lb i 1029 3 I Air 2 X 27fT3yda 054 a Total ACI Mix Design Example I Therefore The fine aggregaTe musT occupy a volume of 27 g 7 2152 26 3 548 26 3 I The SSD weighT of The fine aggregaTe is 548 3 gtlt 260x 6247 3 889 lb Speclf c 6mva of F ne Aggregale Class ACI Mix Design Example I Step 9 AdjusTmenT for moisTure in The aggregaTe The weighT of aggregaTe from The sTock pile is WeyhrMM Weyhfw 1 MC o The moisTure of s The change in The weighT waTer due T The aggregaTe from The sTock pile A Weyhfw wag12104544 z el39 A dus39red Way fwd Way TM 7 A 14237 TM Class ACI Mix Design Example I Step 9 CompuTe sTockpile weighT based on moisTure conTenT I Fine aggregaTe required from The sTockpile is 889 lb 1 0041 9251114ch3 Molslure Conlenl 41 I Coarse aggregaTe requir MolslureConlenl 15 a om The sTockpile is 1701 lb 1 0 015 1727 lbyd3 Class ACI Mix Design Example I Step 9 AdjusT The amounT of waTer based on moisTure conTenT The required mixing waTer required is 340 lb 889 lb 003 6 fine aggregaTe Surface molslure 3 1701 lb 0 01 6 coarse aggregaTe CIVL1112 Reinforced ConcreTe Beam Analysis I LeT39s use The failure models To predicT The ulTimaTe sTrengThToweighT SWR of one of our reinforced concreTe beams from lab I Consider a beam wiTh The following characTerisTics Concrete strength PC 4000 psi Steel strength Y 60000 psi The tension reinforcement will be 1 we rebars The shear reinforcement will be 3 rebars bent in a U shepe with a 3 inch specing Use the minimum width to accommodate the reinforcement Reinforced Concrete Beam Example 1 15 Reinforced ConcreTe Beam Analysis I Reinforcing bars are denoTed by The bar number The diameTer and area of sTandard rebars are shown below I b n Diana 0 As an 3 0375 011 4 0500 020 5 0625 031 6 d e 0750 044 7 0575 060 l 5 1000 079 6 9 1125 100 10 1270 127 11 1410 15s Reinforced ConcreTe Beam Analysis I Based on The choice of reinforcemenT we can compuTe an esTimaTe of ban of b 21O75in 2o75 in 2o375 in l b l Reinforced ConcreTe Beam Analysis I If we allow a minimum cover under The rebars were can esTimaTe d ibi d6eeo 5 70375 l 6 Reinforced ConcreTe Beam Analysis I We now have values for b d and AS l I The A5 for one 6 rebars is k 4 044 in2 Reinforced ConcreTe Beam Analysis I CompuTe The momenT capaciTy A MA5fy d 059 7 2 7 044quot260ksi 7044 60k545m 059 4kS3n 845 kI39n 3 CIVL1112 Reinforced ConcreTe Beam Analysis I LeT39s check The shear model If PM i4 24fde S 2 2l 3911 2ll6239 045 2 4000psi 3in45in 43015 Since Pram PSMWTherefore PMS conTrols Reinforced Concrete Beam Example 1 25 Reinforced ConcreTe Beam Analysis I LeT39s check The reinforcement raTio cf39c 2085 P Ag 7 T0 compuTe p firsT we need To esTimaTe l Reinforced ConcreTe Beam Analysis I The heighT of The sTress box a is defined as a percenTage of The depTh To The neural axis 2quot 4000 psi A 085 f 2 4000 psi 31 0857 005 2 065 4000 4 4000 5 7 085 4005139OT Reinforced ConcreTe Beam Analysis I Check The reinforcemenT raTio for The maximum sTeel allowe 4ks oks pmm 08513137 Ir 0a50a50375 d 3 00181 c 7quot 4ks 085 085 0600 d fy 60 ksi 00289 0853l pmymmn Reinforced ConcreTe Beam Analysis I Check The reinforcemenT raTio for The maximum sTeel allowed A 044 072 5 00326 p bd 3m45m p gt WW 0289 I LeTs consider compression failure in overreinforced concreTe beam Reinforced ConcreTe Beam Analysis I FirsT leT define an equaTion ThaT given The sTress in The Tensile sTeel when concreTe reaches iTs ulTimaTe sTrain f d 87000psd 5 i 52 C I LeT39s compuTe values for aand c Asfy 044n260ks 085752 0854ks3n 2395939quot i 3o5n 31 085 CIVL1112 Reinforced ConcreTe Beam Analysis I FirsT leT define an equaTion ThaT given The sTress in The Tensile sTeel when concreTe reaches iTs ulTimaTe sTrain 52 87000 mg C 6 45077310507 41l360PS 30507 c I If fsyeelt fyThen and7gt0600 87000 psi Reinforced Concrete Beam Example 1 35 Reinforced ConcreTe Beam Analysis I FirsT leT define an equaTion ThaT given The sTress in The Tensile sTeel when concreTe reaches iTs ulTimaTe sTrain 52 87000 mg C C 45077 310507 413601575 305m C I If feelt fyThen and7gt0600 2 pm mm M M 45 259 quot 87000 P 4 30511 2 1458I39ps 87000 ps Reinforced ConcreTe Beam Analysis I The minimum force conTrols 8mm 211 1475 5 4302 kips eumpres iun 1458605 Reinforced ConcreTe Beam Analysis I In summaryThis reinforced concreTe beam will fail in Tension 4in l 3 P1458kps 45 in 6 14 580b 6 R 30 l o l 5W 4795b This beam should fail in compressionl Reinforced ConcreTe Beam Analysis I An esTimaTe of The weighT of The beam can be made as W 7 3in6in30in 145m 1728n3fr3 f1 044In230in 49g 145ij 1728in3fT3 H3 4531b 264b Reinforced ConcreTe Beam Analysis I The cosT of sTeel may be esTimaTed as follows 7 ILL A 896 for Cost ofsfee71728i 7 490 aIifon 172000 lb ff I where A is The cross secTional area of sTeel rebars L is The lengTh of The sTeel rebars and 490 lbTT3 is The LIniT weighT of sTeel CIVL1112 Reinforced ConcreTe Beam Analysis I For example if one 6 rebar incplaced in The beam The sTeel casT is esTimaTe as Cost of sizz 7044in230in 490i 896 m 1728m 73 m 2000b 168 Reinforced Concrete Beam Example 1 45 Reinforced ConcreTe Beam Analysis I Consider The following mix for a yd3 of cancreTe developed using The ACI mix design procedure Reinforced ConcreTe Beam Analysis I The casT of The cancreTe required for a 3quot by 6quot b 39 d s y 30quot beam Is esTimaTe a somemom 5531 102 m c r r r quotS mm 17287 27731 m 2000b o33 30in6in30in1641b 14 m C r f 99 r 7 7 7quot as a nurse a ran 2 172877 27 ffa 1quot1239000 w 013 Reinforced ConcreTe Beam Analysis I The casT of The cancreTe required for a 4quot by 6quot by 30quot beam is esTIma ed as 30quot6ill30ill 14311 10 m c r ff 9 r quot5 w my 17287 27er 1mm 2000b 008 The casT cancreTe is esTimaTed as 054 The casT reinforced cancreTe beam is esTimaTed as 222 Reinforced ConcreTe Beam Analysis I The casT adjusTmenT for The reinforced cancreTe beam is If casT lt 300 Then Casf Facfar 1 If casT gt 300 Then 605139 Facfor 605139 Reinforced ConcreTe Beam Analysis I If The unadjusTed SWRfar a beam is 304 and The casT is 222 Then The casT adjusTed SWR is SWRA ymf SWR x 605139 Facfor SWRMWW 304 x1 304 CIVL1112 Excel IF Function 13 Excel IF FuncT ion Q The logical lencTions in Excel are a small group consisTing of six lencTions Q These lencTions are noTed for Their black or whiTe resul s Q A logical lencTion can reTLlrn only one of Two values TRUE or FALSE Excel IF FLlncTion Q The mosT common and powerful of The logical lencTions in Excel is The IF lencTion Q This lencTion is parTicLllarly powerful because iT can TesT for a parTicLllar condiTion in The worksheeT and use use one value is The on condiTion is TRUE and anoTher is The condiTion is FALSE Excel IF FuncT ion Q The formaT of The IF lencTion is IFlogicalTesT valueifTrue valueiffalse 0 Log calitest is any value or expression that can be evaluated to TRUE or FALSE 0 Valueiifitrue is thevalue that is returned if log calitest is TkUE o Valueiififalse is the value that is returned if log calitest is FALSE 0 Valueiifitrue andor Valueiififalse can be another formula Excel IF FLlncTion Q The operaTors in The logicalTesT of The IF lencTion is may e Equals To ltgt NoT Equals To gt GreaTer Than gt GreaTer Than or equal To lt Less Than lt Less Than or equal To The logical TesT 1000100 returns FALSE 3quot6gt2quot3 returns TkUE Excel IF FuncTion Q Consider The following IF lencTion IFBZgt100010050 o If the Log caLtest is TkUE or in other words if the value of 32 is greater than or equal to 1000 than the function returns avalue o 00 o If the Log caLtest is FALSE or in other words if the value of 32 is not greater than or equal to 1000 than the function returns avalue o 5 Excel IF FLlncTion Q Consider The following IF lencTion IFBZgt1000BZO1BZ005 o If the Logicalitest is TkUE or in other words if the value of 32 is greater than or equal to 1000 than the function returns 10 of 32 o If the Logicalitest is FALSE or in other words if thevalue of 32 is not greater than or equal to 1000 than the function returns 5 of 32 CIVL 1112 Excel IF Function 23 Excel IF Function Q Consider the following IF function IFBZgt1000quotA 1000 or betterquot Less than a grandquot o If the value of 52 is greater than or equal to 1000 than the function returns the string quotA 1000 or betterquot 0 If the value of 32 is not greater than or equal to 1000 than the function returns the string Less than a grandquot When you use text as the valueiifitrue or the valueiififalse argurnents you must enclose the text in a pair of double quotation marks H Excel IF Function Q Consider the following IF function IFA1o1oo5UMl35B15quotquot o If the value of A10 is equal to 100 than the function returns the sum of the values in cells ES to 315 o If the value of A10 is not equal to 100 than the function returns a blank string Excel IF Function Q Suppose you want to assign letter grades to numbers referenced by the name Score See the following table If Score is Then return B From 70 1079 C D F Less than 60 IFScoregt89 quotAquot IFScoregt79 quotBquot IFScoregt69 IFScoregt59 quotDquot quotFquot Excel IF Function IFScoregt89 quotAquot IFScoregt79 quotBquot IFScoregt69 quotCquot I FScoregt59 o If the first log caLtest Scoregt89 is TRUE quotAquot is returned 0 If the first log calitest is FALSE the second IF staternent is evaluated 0 If the second log calitest Scoregt79 is TRUE quotBquot is returned 0 If the second lo calitest is FALSE the third IF statement is evaluated an so on 0 Up to seven IF functions can be nested together in one function Excel AND Function Q The format of the AND function is ANDlogical1 logical2 Q Returns TRUE if all the logical arguments are TRUE Q Returns FALSE 39 one or more arguments is FALSE Q Up to 30 conditions you want to test that can be either TRUE or FALSE 7 Excel AND Function Q Suppose you want to display B4 if it contains a number strictly between 1 and 100 and you want to display a message if it is not IFAND1ltB4 B4lt100 B4 quotThe value is out of range quot Q If B4 contains 104 then the value of the IF function is quotThe value is out of range quot Q If B4 contains 50 then the value of the IF function is 50 CIVL1112 Reinforced ConcreTe Beam Analysis I LeT39s use The failure models To predicT The ulTimaTe sTrengThToweighT SWR of one of our reinforced concreTe beams r m ab I Consider a beam wiTh The following characTerisTics ConcreTe sTrengTh PC 6000 psi Steel sTrengTh Y 60000 psi The Tension reinforcemenT will be 2 4 rebars The shear reinforcemenT will be 3 rebars insTuIIed verTicuIIy an 4 inch spacing Use the minimum concrete cover of 1 inch and a bar spacing of 075 nches Reinforced Concrete Beam Example 2 13 Reinforced ConcreTe Beam Analysis I Reinforcing bars are denoTed by The bar number The diameTer and area of sTandard rebars are shown below b hr 3 Dimkr in As ilz O 375 O 11 4 O 500 O 20 5 O 625 O 31 6 O 750 O 44 6quot d 7 O 575 O 60 4 B 1 000 O 79 9 1 12B 1 00 i 10 1 270 1 27 11 1 410 1 56 Reinforced ConcreTe Beam Analysis I Based on The choice of reinforcemenT we can compuTe an esTimaTe of ban d Hi b id 1 w 2205 n21039Ii O7539n Reinforced ConcreTe Beam Analysis I If we allow a minimum cover under The rebars were can esTimaTe d i b i a 05 T d 6 e 7 e 10 6quot d i 4 l d Reinforced ConcreTe Beam Analysis I The A5 for Three 4 rebars is 4 2o2o m2 040 m2 i b i a 7 46 T 085f395b 6quot d 04072605139 4 l a O856ksi375in 13926 Reinforced ConcreTe Beam Analysis I We now have values for a d and AS we can compuTe The momenT capaciTy a M A5fyd Ej i b i T M o 417960 000ps4 75m1 226 6quot d i 4 l M 98880 bm CIVL1112 Reinforced ConcreTe Beam Analysis I Use The long formula To compuTe The momenT capaciTy A57 M Asfyd 059707 2 04n260ks475n 7 059 W 6ks375n 9888 k m lP g 2472 kips I Reinforced Concrete Beam Example 2 23 Reinforced ConcreTe Beam Analysis I LeT39s check The shear model fd PM 4 V 2lfbd S 2 2 6000 psi 375n475in 201112 60000 psi475in 36869 Since PMSam PMEWTher39efor39e Fwy conTrols Reinforced ConcreTe Beam Analysis I LeT39s check The reinforcemenT raTio A5 p bd p085 1 fc 0 T0 compuTe p firsT we need To esTimaTe 51 Reinforced ConcreTe Beam Analysis I The heighT of The sTr39ess box a is defined as a percenTage of The depTh To The neural axis g 4000 psi 81 085 f39f4000 f 2 4000 pg 81 0857 005W 2 065 600074000 0857005 0 5 8 l 1000 j Reinforced ConcreTe Beam Analysis I Check The reinforcemenT r39aTio for The maximum sTeel allowed c 7quot 6ks 085 085 075 0375 Pym1m 51 d 7 60 ks 00239 c 7quot 6ks 085 085 075 0600 pcampressa 51 d fy 60 ks 00383 Reinforced ConcreTe Beam Analysis I Check The reinforcemenT r39aTio for The maximum sTeel allowed A 4 72 5 00224 p bd 375m475m plt pmm ar39 00224lt 00239 I The beam should fail in Tension CIVL 1112 Water Treatment Floculation and Sedimentation 19 Coagulant Chlorine BR TS M F S GF Solution cw Raw I fquot T ws l 1 quotSP Water U i at T to Distri button System FP Flltratc MI Sludge Treatment Processes O BR Bar Rack l F Flocculatlon CW Clear Well TS traveling Screen Cake 5 Settling HSP ngh Service Pumps M Mlxlng GF Granular Filtration FP Fllter Press Sedimentation 4 Flocculation 4 Coagulation Rapid Mixing I Prior to entering a flocculation tank the water flows Through a rapid mixing basin Chemicals are injected just before the first mixing impeller and baffles throughout the basin to minimize shortcircuiting Rapid Mixing I The rapidmixing characteristics of a flocculation basin are investigated using highperformance computational models Rapid Mixing I The rapidmixing characteristics of a flocculation basin are investigated using highperformance computational models Rapid Mixing Ti39Wuquot Emilia iotaH1354 Rapid Mixing I The rapidmixing characteristics of a flocculation basin are investigated using highperformance computational models CIVL 1112 FlocculaTion I FlocculaTion is genTle mixing To speed The agglomeraTion of colloidal maTerials I The waTer enTers a small Tank or secTion of a Tank in which paddles are Turning slowly Their movemenT i causes The small parTicles To collide and sTick TogeTher fasT or vigorous Oe of Water Treatment Floculation and Sedimentation 29 t FlocculaTion I The genTe mixing combined wiTh The reducTion in repulsive surface charges allow The parTicles To join rapidly I The seTTling velociTy of a parTicle is proporTional To The square of The diameTer I Doubling The parTicle diameTer increases iTs seTTling velociTy by a facTor of 4 I The waTer Then flows i nTo The sedimenTaTion basin where The solids seTTle To The boTTom and are removed mixing would separaTe combined 6 a parTicles 2 F FlocculaTion I In rap39dmXhg inTense mixing or agiTaTion is required To disperse The chemicals uniformly I In The faccua on The fine microfloc begins To agglomeraTe i nTo larger foc parTicles I The Types of devices usually used To furnish The agiTaTion required in boTh rapid mixing and foccuaTion may be generally classified as 1 mechanical agiTaTors such as paddles 2 pneumaTic agiTaTors and 3 baffle basins 395 Var1 caI Padde Whee HorizonTal Paddle Wheel FlocculaTion Graphic courTesy of Jim Myers amp Sons CharloTTe NC VerTical Paddle Wheel Walking 360m 1 FlocculaTion I To improve The efficiency of a foccuaTor good mixing is required aT The sTarT and musT be followed by a long residence Time To allow The mosT seTTling To occur I The mixing characTerisTics of a foccuaTion basin are i nvesTigaTed using highperformance compuTaTionaI models CIVL 1112 Water Treatment Floculation and Sedimentation 39 i i i FlocculaTIon I The degree of mixing is based on The power imparTed To The waTer which is measured by The veacIyyraa l39enr Change I39ll VeocfyDI39sfarce I For example The velociTy gradienT of Two fluid parTicles which are 0 05 TT 0 01524 m aparT and have a velociTy relaTive To each oTher of 2 0 fps feeT per second 0 6096ms is oo5f f r SedimenTaTion I SedimenTaTion is The downwards movemenT of an objecT relaTive To iTs surrounding medium due To The force of graviTy Sedimentarimz 6L Particles Em Iliumminim IIMIJZD Zli olumu Iranian 013 SedimenTaTion I Dissolved air floTaTion DAF is a meThod whereby bubbles are produced by The reducTion of pressure in a waTer sTream saTuraTed wiTh air mum uAF maTALLAnan nuouuc v nonme SedimenTaTion I The purpose of sedimenTaTion is To remove preexisTing solids as well as The precipiTaTes formed in coagulaTion and flocculaTion 51m am u mm SedimenTaTion I The purpose of sedimenTaTion is To remove preexisTing solids as well as The precipiTaTes formed in coagulaTion and flocculaTion SedimenTaTion I Model of a circular seTTlemenT Tank wiTh sludge scrapers was used To esTimaTe The disTribuTion of parTiculaTe concenTraTion over Time H r CIVL 1112 Water Treatment Floculation and Sedimentation Sedimentation Sedimentation Sedimentation Sedimeniaiion is ihe accumuiaiion ihrough graviiy of pariicuiaie maiier ai ihe boiiom of afiuid Sedimentation I Dimer Individuai pariicies seiiie independeniiy neiiher aggiomeraiing nor inierfering Wiih ihe seiiiing of ihe oiher pariicies preseni This occurs in waier I This naiurai process is frequeniiy used io separaie Wiih a iow concentration of Pariicizs coniaminanis from air waier and wasiewaier I Flaccua rr Pariicie conceniraiions are high enough I Wm M four pr of gaming ihai aggiomeraiion occurs This reduces ihe number of d pariicies and increases average pariicie mass T e 39 WW heavier pariicies Sink fasier I fioccuiam I hindered I compression Sedimentation Sedimentation I Hi dll39ldr Pariicie COYiCBYiiV aiiOYi iS sufficieni ihai pariicies inierfere Wiih ihe seiiiing of oiher V iiCiBS I Camprlssa r In ihe iower reaches of ciar ifier s where pariicie conceniraiions are highesi pariicies can seiiie oniy by compressing ihe mass of pariicies beiow 49 CIVL1112 Water Treatment Floculation and Sedimentation Sedimentation I If the ip gt lyythen settling can occur 7 0 Sedimentation I The horizontal velocity V of a particle can be approximated by considering the flowrate Q and the crosssectional flow area of the tank owlt m 470 WZ 59 Sedimentation I If the iplt lyythen shor39tcircuiting can occur V L Sedimentation I The residence time of water in the sedimentation tank can be approximated as Q lh E 3 sz Q f Sedimentation I Estimate of the residence time of water in a small sedimentation tank where Q1litermin L 1 in w 3 in and 210 in dimensions of a tank in the lab 7 sz 127339n1039n Q 1000 quot rm 360m3 min 1639m 1000m m3 Sedimentation I Discrete settling can be analyzed by calculating the settling velocity of the individual particles contained within the water I The forces acting on a particle are I gravity in the downward d rection I drag acting n the upward direction as the particle settles I upward buoyancy due the water displaces by the particle CIVL 1112 Water Treatment Floculation and Sedimentation 69 Sedimentation The forces aciiiig on a seiiiiiig pariicie are Sedimentation I The grauiiaiiowai force can be expressed as Fb Fd F mpg I Using ihe dehsiiy and voiume of ihe pariicie yieids 9 pp 119 FE is ihe farce due ia grawy F Fa is the drag force 9 where p is ihe dehsiiy of ihe pariicie ibrmaSS a a is ihe buoyahi force Qgis ihe voiume of ihe pariicie n3 and F Fd Fb 31 ihe grauiiaibhai cahsiam ns2 Sedimentation I The d rag on ihe pariicie can be caicuiaied by ihe drag equation om uid mechanics 1 F it39d391 sz Sedimentation I The buoyant force aciihg on ihe pariicie is Fbm7 I Subsiiiuiihg ihe pariicie vaiume arid dehsiiy of water yizids where Cdis ihe drag coefficiehi dimensioniess F pullPg A is ihe pariicie crosseseciiahai area pw is ihe dehsiiy of waier ibemas 3 m vacw Wm where pwis ihe dehsiiy ofwaier ibrmaSS a Sedimentation Sedimentation I By baiahcihg ihe forces aciihg on a seiiiihg pariicie and I Soivmg for ihe seiiiihg veiociiy r resuiis in using ihe reiaibhships for FE ihe force due is grauiiy Fd ihe drag force and a ihe buoyahi force ihe foiiowmg reiaibhship can be d eveioped 1 PM 541 my I If ihe pariicie is assumed is round and for ar the formuias ea andvoiume of a d sphere are use where a is ihe diameier of ihe pariicie CIVL 1112 Water Treatment Floculation and Sedimentation 79 Sedimentation I At low Reynolds numbers for NEE lt 1 Cd can be approximated by 2 d NR2 I For Reynolds Numbers is transition flow 1lt Ngelt 10 000 the drag coefficient for spheres is 7 24 3 ilVee I For turbulent flow NR2 gt10 000 the relationship for the drag coefficient for spheres is 4 04 034 Cd Sedimentation Vd I The Reynolds Number is NM FT where u is the absolute viscosity of the water lb forcesecfi392 at 50 F u 2 7310395 lbsecft2 For NR2 lt 1 the particle settling velocity can be estimated as a function o t e ro erties of the particle and water and the particle diameter or memw V 18u Sedimentation I This relationship is known as Stokes39 law and the velocity is known as the Stokes velocity mAW W 1 The vertical velocity of water in a settling basin is often described as the aveflow rare OFR I It is usually expressed as galftZday m3m2day Sedimentation I The overflow rate is calculated in the following way Q 0R F A where OFR is the overflow rate galftZday Q is the flowrate galday and A is the clarifier area ft2 Sedimentation Example I Estimate the settling velocity of sand p 2650 kgm3 with a mean particle diameter of 0 21 mm I Assume the sand is approximately spherical I Using a safety factor of 14 to account for inlet and outlet losses estimate the area required for a chamber to remove the sand if the flowrate is 010 m3sec 1000 liters 1m3 Sedimentation Example I The density of water at 20 C is 998 kgm3 and the viscosity of water at 200C is 10110393 Nsm2 Newton kgmsz The Stokes settling velocity is pp 7 mfg VP 7 V5 7 0F 7 T k k 2650937 99821gtlt10394m2981 j 18101x10 3kij I775 0039 IIIS 39 cIns CIVL 1112 Water Treatment Floculation and Sedimentation 89 Sedimentation Example I Knowing the overflow rate the area required is 3 Q 701omA 7 2 A F 7 14 736 0526 0039l m where 5F is the safety factor 14 Sedimentation Example Group Problem Estimate the settling velocity of of the floc particles we have seen in lab especially thejar test results Use Stokes39 law to estimate the settling velocity What are quotgoodquot estimates of the particle density and diameter How does your estimate compare to what you have seen in the lab Sedimentation Example Group Problem I What are quotgoodquot estimates of the particle density and diameter I Let39s assume the following values Part cle density 1100 kgm3 Part cle diameter 1039 m Sedimentation Example Group Problem 7 7 7 pp WWW VP 7 V5 7 OFR e T mooL937 9981x104m298112 7 m m 5 18101x104k 9 ms 55 x 10394 IIIS 0055 cIns Sedimentation Example Group Problem OFA 55 x 10394 IIIS 0055 cIns 378541 cmquot 1 ff 539 one WJWWT day For ferric chloride typical OFRs are in the 700 1000 nd2 Sedimentation Example Group Problem If the settling velocity of of the floc particles is 0 055 cms determine the area of the sedimentation tan Assume a factor of safety of 13 Assume the system flowrate can varying from 1000 mlmin to 2 000 mlmin How does your estimate compare to what you have seen in the lab CIVL 1112 Water Treatment Treatment Cost Example 17 a Project 1 Treatment Cost The objective of this project is to utilize within given constraints a prototype water filter system to design a fullscale system The effectiveness of the filter design will be evaluated by the yearly apera bna and Inaquotreliance cans Project 1 Treatment Cost Group Problem Treatment Cost Consider a prototype system with the following eristics coagulant dosage of 25 mgl flowrote 1000 mlmin 7 qu 1 m2 of 60 mlnutes 2 inches of anthracite and 4 inches of filter sand replace filter material once ayeor 4 prototype sedimentation tanks n omemm Compute the total yearly cost of this system Project 1 Treatment Cost Coagulation and Flocculation Cost The weight of coagulant kg required per gallon of treated water is estimated as my ky 25 mg 3785 liters kg 9 5 liter galon 10quot my 7 k9 79462x10 5 A Project 1 Treatment Cost Coagulation and Flocculation Cost The number of coagulation and flocculation units NCF required are tEIYo Factorot Safety 1019Pd NCF7W x11 NCF 44 units or 5 units Project 1 Treatment Cost Coagulation and Flocculation Cost The total yearly cost of the coagulation and flocculation system for 1 lltloo is 5000 year 9462x105k9 totgo sesdoys 39 go day year 9 Custh 5E Project 1 Treatment Cost Sedimentation System Cost Step 1 Compute the prototype sediment tank MilIrina fink tP gallon 231iquot3 VolumeW 360 in 156 gallons 4156 gallons 23 62 mln ml liter gallon 100 minute1000m3785 liter CIVL 1112 Water Treatment Treatment Cost Example 27 a Project 1 Treatment Cost Sedimentation System Cost Step 2 The fullscale treatment ourate 711 257 per sedimentation tanks is 7 1000 gallons an 23am 4234gpm Project 1 Treatment Cost Sedimentation System Cost Step 3 The effective ourate Wm 25 in a sedimentation tank is 7 4234 gpm X 60 minutes 7 7 60 minutes 7 Qse 4234 gpm Project 1 Treatment Cost Sedimentation System Cost Step 4 The number of fullscaled sedimentation tanks NS required to handle the daily volume is estimated as 71000000gpd day Ns i 4234gpm 1440min1391 1804 tanks or 19 tanks Project 1 Treatment Cost Sedimentation System Cost The operation and maintenance costs per tanks is 2 OOOtanks Tlie yearly costs per sediment tank is 2000 CastS 19 tanks cln k t Project 1 Treatment Cost Filtration System Cost Step 1 Convert the average flowrate through the prototype filter the 3 5 inch diameter prototype filter has an area of o 0668 ft into a prototype lter oddity rate QF9Pm 2 Q7 1000m liter gallon 1 39 minute 1000m 3785liter oooosft2 3959pmft2 Project 1 Treatment Cost Filtration System Cost Step 2 The fullscale treatment ourateQris QT 3959Pm1hr2X 100ft2 395 gpm CIVL 1112 Water Treatment Treatment Cost Example 37 a Project 1 Treatment Cost Filtration System Cost Step 3 Considering that each filter is inoperable during backwashing the effective ourafe 25 is 7 395 gpmx O minutes 60 minutes 395 gpm st Project 1 Treatment Cost Filtration System Cost Step 4 The number of fullscaled filters NF required to handle the daily volume is estimate as tux animalquot NF 1000000gpd day I X 395gpm 1440mln 193filters or 2filters Project 1 Treatment Cost Filtration System Cost The yearly cost per filter is CostF 2 filters539oooj 1 000 filter Project 1 Treatment Cost Filtration System Cost nuances 2 in Costh 9395 ft100 ft2gtltNF Fta 12 inches The yearly cost for filter sand is 7 590 4in 2 Costm57 Ha Ilainchesft 100 ft xNF Cost 317 393 710 w 1 Project 1 Treatment Cost Total Treatment System Cost Total Cost 59I538 coagulation 38000 sedimentation 1oooo nitration 710 Filtration Media 1 Project 1 Treatment Cost Total Treatment System Cost Total Cost 108248 CIVL 1112 Water Treatment Treatment Cost Example 47 a Project 1 Treatment Cost Group Problem Treatment Cost How would your cost change if you decreased your flowrate to 950 mlmin and remaining variables were coagulant dosage ot 25 mgl Yl llme of 60 m YlLllBS a 2 inches ot anthracite and 4 nches ot tilter sand 4 replace tilter tnaterial once ayear 5 4 prototype sedimentation tanks Compute the total yearly cost of this system Project 1 Treatment Cost Coagulation and Flocculation Cost The weight of coagulant kg required per gallon of treated water is estimated as k5 25 mg 3785 lifer kg mt I 9 9 7 Inter galon 10quot my 9462 x104 ka Project 1 Treatment Cost Coagulation and Flocculation Cost The number of coagulation and flocculation units NCF required are tEIYo Factor ot Satety 1019Pd NCF7W x11 NCF 44 units or 5 units Project 1 Treatment Cost Coagulation and Flocculation Cost The total yearly cost of the coagulation and flocculation systemtortlllo i 5000 year 9462X105k9 to ga sesdoys 39 go day year 9 Custh 5E Project 1 Treatment Cost Sedimentation System Cost Step 1 Compute the prototype sediment tank retention time tP Volumemk 360 in3 e 1 56 gallons in 4156 gallons I H e 2486 min m iter ga on 950 minute1000mlj3 785 liter Project 1 Treatment Cost Sedimentation System Cost Step 2 The fullscale treatment ourate win 257 per sedimentation tanks is 7 1000 gallons QST m 4022 gpm CIVL 1112 Water Treatment Treatment Cost Example 57 a Project 1 Treatment Cost Sedimentation System Cost Step 3 The effective ourate 77 25 in a sedimentation tan is 7 4022 gpmgtlt60 minutes 7 7 60 minutes 7 QSE 4022 gpm Project 1 Treatment Cost Sedimentation System Cost Step 4 The number of fullscaled sedimentation tanks NS required to handle the daily vo ume is estimate as NS 7 1000000gpd day X 7 4022 9pm 1 440min 39 1899 tanks or 19 tanks 1 I Project 1 Treatment Cost Sedimentation System Cost The operation and maintenance costs per tanks is 2 OOOtanks The yearly costs per sediment tank is 2000 CostS 19 tanks Tank Project 1 Treatment Cost Filtration System Cost Step 1 Convert the average flowrate through the prototype filter the 5 5 inch d39ameter prototype filter has an area of 00668 ft into a prototype fiter low rate QF9Pm 2 Q 7 950ml liter gallon 1 minute 1000m 3785liter ooeeaft2 3 76 gptnft2 Project 1 Treatment Cost Filtration System Cost Step 2 The fullscale treatment ourate Qris QT 3769me2x100ft2 376 gpm A Project 1 Treatment Cost Filtration System Cost Step 3 Considering that each filter is inoperable during backwashing the effectbe owrate 25 is 7 376 gpmx 60 minutes 60 minutes 376 gpm Q E CIVL 1112 Water Treatment Treatment Cost Example 67 a Project 1 Treatment Cost Filtration System Cost Step 4 The number of fullscaled filters NF required to handle the daily volume is estimate as iii whammy NF 1000000gpd day I X 376gpm 1440mln 203 filters or 3 filters Project 1 Treatment Cost Filtration System Cost The yearly cost per filter is 5000 filter CostF 3 filters Project 1 Treatment Cost Filtration System Cost The yearly cost for anthracite is amni i e p m 950 Eliquot 2 C t ft 100ft NF W fta 12inches X The yearly cost for filter sand is 7 590 4in 2 Costm e W Mmft 100 ft xNF CastFM 475 590 Project 1 Treatment Cost Total Treatment System Cost Total Cost 59I538 coagiiiaiioi 38000 saiimahhion 15ooo aimquot 1065 Filiraiioiilvleaia I Project 1 Treatment Cost Total Treatment System Cost Total Cost 113603 A Project 1 Treatment Cost Comparison of Treatment System Cost CIVL 1112 Surveying Traverse Calculations 114 Surveying Traverse M Introduction Q AlmosT all surveying requires some calculaTions To reduce measuremenTs inTo a more useful form for deTermining disTance earThwork volumes land areas Tc Q A Traverse is developed by measuring The disTance and angles beTween poinTs ThaT found The boundary of a siTe Q We will learn several differenT Techniques To compuTe The area inside a Traverse DisTance Traverse E Methods of Computing Area Q A simple meThod ThaT is useful for rough area esTimaTes is a graphical method QInThismeThod The irriiir r eBeeee Traverse is ploTTed To scgle n graph paper data1 I an e nu er o t r 71 squares inside The 39 39 39 Traverse are counTed A 5 G l 074939 5quot l DisTance Traverse Methods of Computing Area B b Area ABC acsina DisTance Traverse Methods of Computing Area Area ABD gar sina C Area 360 bcsin Area ABCD Area ABD Area 360 DisTance Traverse Methods of Computing Area C Area ABE aesina Area 605 god sin Q To compuTe Area 360 more daTa is required Surveying Traverse Balancing Angles Q Before The areas of a piece of land can be compuTed iT is necessary To have a closed traverse Q The inferior angles of a closed traverse should ToTaI I7 2180 where n is The number of sides of The Traverse CIVL1112 Surveying Traverse Calculations 214 Surveying Traverse M Balancing Angles Error of closure Angle conTaining misTuke Surveying Traverse E Balancing Angles Q A surveying heurisTic is ThaT The ToTaI angle should noT vary from The correcT value by more Than The square rooT of The number of angles measured Times T e precision of The insTrumenT Q For example an eighTsided Traverse using a 139 TransiT The maximum error 395 i1g i28339 i Surveying Traverse Balancing Angles Q If The angles do noT close by a reasonable amounT misTakes in measuring have been ma e Q If an error of 139 is made The surveyor may correcT one angle by 139 Q If an error of 239 is made The surveyor may correcT Two angles by 139 each Q If an error of 339 is made in a 12 sided Traverse The surveyor may correcT each angle by 33912 or 15quot Surveying Traverse Latitudes and Departures Q The casure of a Traverse is checked by compuTing The IaTiTudes and deparTures of each of iT sides N N Bean 4 DeparTure CD E Bearing 4 Departure is LuTiTude CD 5 Surveying Traverse Latitudes and Departures Q The latrudeof a line is iTs projecTion on The norTh souTh meridian N B Q The depan ur39e of a line is iTs projecTion on The easT Luhmde 3 wesT line E Q A norTheasTerIy bearing has a IaTiTude and DepurTure Ag deparTure Bearing 4 A Surveying Traverse Error of Closure Q Consider The following sTaTemenT 39Tf Slur at on corur of 1 closed traverse and wit it lim mi you rcle to your starting pail you m39l have Mike 1 for I a you mind south and 1 for can a you have winked mmquot Q Therefore Z IaTiTudes 0 and Z deparTures 0 CIVL1112 Surveying Traverse M Error of Closure Q When IaTiTudes are added TogeTher The resuITing error is called The error I39ll latrude E Q The error resuITing from adding deparTures TogeTher is called The error I39ll departures ED Surveying Traverse Calculations 314 Surveying Traverse E Error of Closure Q If The measured bearings and disTances are pIoTTed on a sheeT of paper The figure will noT close because o E an 59 Error of closure B Eb E 2 2 l 5W 51 ED A E C Precisionamp perimeTer o Typlcal preclslon 15000 for rural land 17500 for suburban land and 110000 for urban land Surveying Traverse Latitudes and Departures Example A Surveying Traverse Latitudes and Departures Example N 7w 18953 ffsin6 1539 72063 ft w A E s 6 1539 w LuTITude 75 18953 ffcos6 1539 718840 ff B Surveying Traverse Latitudes and Departures Example N 5 17518 ffsin29 3839 8662 ft w B E 17518 Latitude EC 75 17518 ffcos29 3839 715227 ff Surveying Traverse Latitudes and Departures Example 93966 1 39 l Edwe kc quotf ED2 leoo792 701632 0182 5mm 7 0181 7 1 perimeTer 7 93946 TT 7 5176 Precision CIVL 1112 Surveying Traverse Calculations 414 Surveying Traverse Group Example Problem 1 A Surveying Traverse E Group Example Problem 1 o Compute the latttudes departures EMUquot and the prectstoh tor the tollowmg traverse Lgng hm Lawn Surveying Traverse m Group Example Problem 1 Em ltsf 459 quot06002 471110 1262 f 5 262 f 1 P mw 4300 10quot 7 quotmmquot perimeter 2 6294 ft X 2033 Surveying Traverse M Balancing Latitudes and Departures o Balahcthg the latttudes and departures of atraverse ls to attempt to obtam more probable values for the locattohs of the corners of the traverse A popular method tor balanclng errors ls called the eerpess or the sewetiteh rue o The preterm mk39was devlsed by Nathan el Bowatteh surveyur navlgatur aha mathemattetah as aproposea soluttohto the problem or cumpass traverse adJustmem vm eh was posea m the merlcanJuuNlal m Artery m 1807 Surveying Traverse Balancing Latitudes and Departures The eerpessmethod assumes 1 angles and dlstatlces have same error 2 errors are accldental The rule states quotThe error in laii iiide departure afa lire is to the total error in latitude departure as iergth 01 the ire is the perimeter 01 the fratuse Surveying Traverse m Balancing Latitudes and Departures CIVL1112 Surveying Traverse Latitudes and Departures Example Q Recall the results of our example problem Surveying Traverse Calculations 514 Surveying Traverse Balancing Latitudes and Departures N 75 18953 ffcos6 1539 718840 f Correction nLcitAg 7 1M 7 perimeter E Correction n LcitAg M per meter 0079 ff18953 ff Correction in Letw Surveying Traverse Balancing Latitudes and Departures N ew 18953 ffsin61539 72063 ft E W A Correction nDeIoAg 7 4w 5 6 1539 W ED perimeter 1E9 5339 5 Correction n DeliaAE perimeter 0033fr Surveying Traverse Balancing Latitudes and Departures N 75 17518 frcos293839 715227 ff w B E Correctionin Lat 7 41 7 perimeter 4 51 perimeter 0015fr Correction in Lat Surveying Traverse Balancing Latitudes and Departures 5 17518 frsin293839 8662 ft Correction in pep 4 E perimeter E Correction in pep perimeter Common in mpg 7 0163 fr17518 fr 5 93946 ft Surveying Traverse Balancing Latitudes and Departures CIVL1112 Surveying Traverse Balancing Latitudes and Departures Length R Latltude Depamre Latitude Latitude Surveying Traverse Calculations 614 Surveying Traverse Balancing Latitudes and Departures Length R Latltude Depamre Latltude Latitude a l a V 39 i i i i Corrected iatitudes and departures No error in corrected iatitudes and departures Surveying Traverse Balancing Latitudes and Departures correct ions gives quotm39 WWW m W Q Combining the latitude and departure calculations with Surveying Traverse Group Example Problem 2 Q Balance the latitudes and departures for the following averse Surveying Traverse Group Example Problem 2 Q Balance the latitudes and departures for the following traverse Surveying Traverse Group Example Problem 3 Q In the survey of your assign site in Project 3 you will have to balance data collected in the following form N 69 5339E B A CIVL1112 Surveying Traverse Calculations 714 Surveying Traverse M Group Example Problem 3 Q The firsT sTep is To compuTe The bearings for each side N 69 5339E B Surveying Traverse E Group Example Problem 3 Q Find The bearing of side DA on 180 69 5339 51 2339 on 58 4439 DA N58 4439W Surveying Traverse Group Example Problem 3 Q Find The bearing of side BC on 105 3939 69 5339 on 35 4639 BC S 35 4639E Surveying Traverse Group Example Problem 3 Q Find The bearing of side CD on 180 78 1139 35 4639 on 66 0339 CD S 66 0339W Surveying Traverse Group Example Problem 3 Q In The survey of your assign siTe in ProjecT 3 you will have To balance daTa coecTed in The following form m quotquot quotquotquot m W Surveying Traverse Group Example Problem 3 Q In The survey of your assign siTe in ProjecT 3 you will have To balance daTa coecTed in The following form CIVL1112 Surveying Traverse Calculations 814 Surveying Traverse M Calculating Traverse Area Q The besT known procedure for calcuIaTing land areas is The double meridian distance DMD meThod Q The meridian distance of a line is The easT wesT disTance from The midpoinT of The line To The reference meridian Q The meridian distance is posiTive To The easT and negaTive To The wesT Surveying Traverse E Calculating Traverse Area N A 142 BQ keference N12 2439 W D N81 1839W Surveying Traverse Calculating Traverse Area Q The mosT wesTerIy and easTerIy poinTs of a Traverse may e found using The deparTures of The Traverse Q Begin by esTainshing a arbiTrary reference line and using The deparTure values of each poinT in he Traverse To deTermine The far wesTerIy poinT Surveying Traverse Calculating Traverse Area 720601 B A B C D 7195470 C 7305515 D E 159974 A PoinT E is The farThesT To The wesT Surveying Traverse Calculating Traverse Area keference Meri ian E N 12 2439 W D N81 1839W Surveying Traverse DMD Calculations Q The meridian disTance i N of line EA 5 N E e Reference Meridian D C E DMD of line EA is The deparTure of line CIVL1112 Surveying Traverse Calculations 914 Surveying Traverse M DMD Calculations Q The meridian disTance of line Meridian distance AB is equa a Dr line AB N The meridian disTance of EA The deparTure of line EA deparTure of AB Q The DMD of line AB is Twice The meridian disTance of line AB Surveying Traverse E DMD Calculations Meridian dismm The 0440 of My Side 5 Wu we to Me DMD of le lasr side plus rhe departure of Me lasr side plus rhe departure of Me present 539 N Surveying Traverse DMD Calculations Balanced Surveying Traverse DMD Calculations Q The DMD of line AB is deparTure of line AB Q The DMD of line BC is DMD of line AB deparTure of line AB The deparTure of line BC Surveying Traverse DMD Calculations Surveying Traverse DMD Calculations Q The DMD of line CD is DMD of line BC deparTure of line BC The deparTure of line CD Q The DMD of line DE is DMD of line CD deparTure of line CD The deparTure of line DE CIVL1112 Surveying Traverse Calculations 1014 Surveying Traverse M DMD Calculations Latitude Surveying Traverse E DMD Calculations Latitude HillIII 159974 Q The DMD of line EA is DMD of line DE departure of line DE the departure of line EA Q Notice that the DMD values can be positive or negative Surveying Traverse Traverse Area Double Area Q The sum of the products of each points DMD and latitude equal twice the area or the double area Surveying Traverse Traverse Area Double Area Q The sum of the products of each points DMD and latitude equal twice the area or the double area Latitude Departure DMD Babette um 8 64 ll Q The double area for line AB equals DMD of line AB times the latitude of line AB Q The double area for line BC equals DMD of line BC times the latitude of line BC Surveying Traverse Traverse Area Double Area Q The sum of the products of each points DMD and latitude equal twice the area or the double area Latitud Departure DMD Double A Surveying Traverse Traverse Area Double Area Q The sum of the products of each points DMD and latitude equal twice the area or the double area Latitud Departure DMD DoubleA as Q The double area for line CD equals DMD of line CD times the latitude of line CD Q The double area for line DE equals DMD of line DE times the latitude of line DE CIVL1112 Surveying Traverse Calculations 1114 Surveying Traverse M Traverse Area Double Area Q The sum of The producTs of each poinTs DMD and laTiTude equal Twice The area or The double area Q The double area for line EA equals DMD of line EA Times The laTiTude of line EA Surveying Traverse E Traverse Area Double Area Q The sum of The producTs of each poinTs DMD and laTiTude equal Twice The area or The double area Latimde 7188 388 139 080 171 627 1 acre 43 560 T T2 Surveying Traverse Traverse Area Double Area Q The sum of The producTs of each poinTs DMD and laTiTude equal Twice The area or The double area L d Departu 7188 388 7152 253 29 933 139 080 171 627 1 acre 43 560 HZ Surveying Traverse Traverse Area Double Area Q The word quotacrequot is derived from Old English aecer originally meaning quotopen fieldquot cognaTe To Swedish quot39ikerquot German Acker LaTin ager and Greek Hypo tyros e acre was selecTed as approximaTely The amounT of land Tillable by one man behind an ox in one ay Q This explains one definiTion as The area of a recTangle wiTh sides of lengTh one chain and one furlong Surveying Traverse Traverse Area Double Area Q The word quotacrequot is derived from Old English aecer originally meaning quotopen fieldquot cognaTe To Swedish quot39ikerquot German Acker LaTin dyer and Greek Hypo NEWS Q A long narrow sTrip of land is more efficienT To plough Than a square ploT since The plough does noT have To be Turned so ofTen Q The word quotfurlongquot iTself derives from The facT ThaT iT is one furrow org Surveying Traverse Traverse Area Example 4 Q Find The area enclosed by The following Traverse 1 acre 43 560 T T2 CIVL1112 Surveying Traverse Calculations 1214 Surveying Traverse M Traverse Area Example 4 Q Find The area enclosed by The following Traverse Latitude D 1 acre 43 560 H2 Surveying Traverse E DPD Calculations Q The same procedure used for DMD can be used The double parallel distances DPD are mulTipIied by The balanced deparTures Q The parallel distance of a line is The disTance from The midpoinT of The line To The reference parallel or easT wes line Surveying Traverse Rectangular Coordinates Q RecTanguIar coordinaTes are The convenienT meThod available for describing The horizonTaI posiTion of survey poinTs Q WiTh The applicaTion of compuTers recTanguIar coordinaTes are used frequenle in engineering projecTs Q In The US The X axiscorresponds To The easT wesT direcTion and The y aXsTo The norTh souTh direcTion Surveying Traverse Rectangular Coordinates Example Q In This example The Ien h of AB is 300 fT and bearing is shown in The igure below DeTermine The coordinaTes of poinT LaTiTude AB 300 fT cos42 30 221 183 fT N 42 3039 E DeparTure AB 300 fT sin42 30 202 677 H xB 200 202 667 402 667 fT x CunrdinalesufPuinlA 300 yB300221183521183fT 200 Surveying Traverse Rectangular Coordinates Example Q In This example iT is assumed ThaT The coordinaTes of p i e w and we wanT To calcuIaTe The aTiTude and deparTure for line AB Cuurdinalesuf PumlA 100 300 v LaTiTude AByB yA LaTiTude AB 400 H A DeparTure AB xB xA DeparTure AB 220 fT B Cuurdinales ur Puml B 320 7100 Surveying Traverse Rectangular Coordinates Example Q Consider our previous example deTermine The Xand y coordinaTes of all The poinTs CIVL1112 Surveying Traverse M Rectangular Coordinates Example A Q xcoordinates E 0 H A E 159 974 159 974 ff BA 20601139373 6 B86648 226021 79 DC 19547030551 ED 305510 Surveying Traverse Rectangular Coordinates Example v A 159974 340640 139373152253 00 169013 E 30551 29933 c 226020 00 X Surveying Traverse Calculations 1314 Surveying Traverse E Rectangular Coordinates Example A Q ycoordinates c 0 ft D c 29 933 E D 139080 169 013 AE171627340640 79 B A 188 388 152 252 CB 152 2520 Surveying Traverse Group Example Problem 5 Q Compute the x and y coordinates from the following balanced Surveying Traverse Group Example Problem 5 Q Compute the x and y coordinates from the following balanced Bearm Surveying Traverse M Area Computed by Coordinates Q The area of a traverse can be computed by taking each ycoordinate multiplied by the difference in the two adjacent xcoor inates using a sign convention of for next side and for last side

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