Surveying 9718510117

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an Surveying, Higher Surveying, Mine Surveying, Hydrographic urv ylng, Topographic Surveying, Astronomy, Simple Curves, Compound Curves, plral Curves, Reversed Curves, Parabolic Curves, Sight Distance, Earthworks, ass Diagram, Highway Engineering, Transportation Engineering, Traffic Engineering.

.>

DETI~

for CIVIL and

LieeQSure Exam Copyn~;t 1987

Venancio I. Besavilla, Jr. (BSCE, MSME, AS, F. (PICE)

-CIT (2nd Place) - August, 1969 Civil Engineer . Geodetic Engineer - CIT (7th Place) - July, 1966 Former Instructor: Cebu Institute of Technology Former Instructor: University of the Visayas Former Chairman: CE Dept. University of the Visayas Dean: College ofEngineering and Architecture, University of the Visayas Awardee: As an Outstanding Educator from the Phil. Veterans Legion on May 1984 Awardee: As Outstanding Alumnus in the Field of Education from CIT Alumni Association, Inc., Marcil 1990 Awardee: As Outstanding Engineering Educator from the CIT High School Alumni Association, December 1991 Member: Geo-Institute of the American Society of Civil Engineers (ASCE) Member: Structural Engineering Institute (ASCE) Member: American Concrete Institute (ACI) (Membership No. 104553) Member: American Society of Civil Engineers (ASCE) (Membership No. 346960) Member: PICE Delegation to the ASCE (Minneapolis, Minnesota, USA) (Oct. 1997) Head PICE Delegation to VFCEA, Hanoi, Vietnam (April 2009) Head PICE Delegation to JSCE, Fokuka Japan (September 2009) Head PICE Delegation to A SCE.~ Kansas City, USA (October 2009) Director: PICE National Board (1997-2008) Director: PICE Cebu Chapter 1991- 2008 ~lce President: PICE Cebu Chapter 2006, 2008 President: PICE Cebu Chapter 2009 Vice President: PICE National Board 2009 President: PICE National Board 2009 Chairman International Committee (PICE) Busan, Korea (February 2010) President: Cebu Institute ofTechnology Alumni Association (20OJ-up to the present) ISBN 971- 8510-11-7 Available at:

BISAVlllA Engineering Review Center

CEBU DAVAO 2nd Floor, Pilar-Bldg. 4th Floor, Cor. Osmena Blvd. Porras Bldg. & Sanciangko Sts. Magallanes St., Cebu City Davao City Tel. No. (032) 255-5153 Tel. No. (082) 222-3305 BAGUIO Lujean Chalet Bldg. Lourdes Grotto Dominican Road No. 68 San Roque SI. Baguio City Tel. Nos. (074) 445-5918

CAGAYAN DE ORO 3rd Floor, Ecology Bank Bldg. Tiano Bros. St. Cagayan De oro City Tel. (08822) 723-167(Samsung)

TACLOBAN Door 11.-303, F. Mendoza Commercial Complex 141 Sto. Nino Street Tacloban City Tel. No. (053) 325-3706

MANILA 2nd Floor, Concepcion Villaroman Bldg., P. Campa St., Sampaloc Melro Manila Tel. (02) 736-0966

GENERAL SANTOS RD. Rivera Bldg. Constar Lodge, Pioneer Ave. General Santos City Tel. No. (083) 301-0987

SURVEYING for CIVIL and GEODETIC Licensure Exam

Copyright 1984 by Venancio I. Besavilla, Jr. All Rights Reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher.

ISBN 971· 8510·11.7 •

•

~,.. ~;:,_;_~_,m., Punta Princesa, Cebu City. Tel. 272-2813

"~..'III.. I~ •• I~~ ~ '•• ~"I~I~XTS ~

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[email protected]!l$f@hd. l$bep:Qtf1~nJhe$t.Woyl1~.®()ut~.~.2@Q*:Itlm; ·.• ." . . . . . . .tl$ffi~ •. /••. . •. . . .

Total correction =0.014 -1.832 - 0.011 Total correction =-1.829 m. True length AB =458.65 - 1.829 True length AB =456.821 m.

Ii'~i:$.·of.:x·?··.····.·····

@•• • T\lij.~jstlil'lCElfrOm.P.Il)ei@fJi~@~d;.j$·.

• • • • • • • • 1§$.2.rl1.•••• lf.tM.$Q.m;t#p~~eaj~Q~Q1 • m;. tQP$lyjrt;Wh~!i$~#grf~W(ll$tM¢¢m rie? .. ... ... .... ...

Solution:

~~~I~~~!~~n~~~~i~I~~at~jl~_\\~~ .~~5f:o~~~~~~~Ii~~~~"'1~11;1~~l~'~.· ..' .. . .

CD Error in area: (99.962 _ (1002) A - 2.25 A=2.2482 hectares

Enur in area =2.25 - 2.2482~.l\ Errorin area =0.0018 hectares, Error in area = 0.0018 x10000 ~\ Error in area = 18 sq.m.

6~3,893,5dI9talj(18~S;&/

·.it$•. n()[llill~I.I~ngtb •.VjI'l~ll~IlPP()rtE!d.ptJly'~t

•.

~~~~.~f~~~ ~~4$t~I • _1~ ~.~6~~~

Unit weight of tape: _ 0.204 w...fAE PNPN- Ps

-J

.•

•.•. t@)~9hourit$ • leo91h.und~r.~.ll\at\!j~rd·PW · • •pt$·§.·kg•• wilh.•!J'tEl•. rnod~N~.9felastiglyfS .• ·~ • x.1~kgf(@fW"ldaieaOf.Q,06(:mf • • • • • •.• • ·•• •·

16 =0.204 w'./'--0.-05-(2-)1-06'

w=

=0.026 kglm

® Cross sectional area:

_ 0.204 w...fAE PN-J PN-Ps

·®•• • Aste~tapei$30:lt1 .•• ~ogu@era • ~nd~@.· •.• • • • • plJu • • ~f.9 • • • k~, • • • With•• • ~ • • • %J~~I~nt • • ett)$lh.

·(:.lilli'~_'1

18 =

O.204[~O.OO25)(40i...fAE ~

me¢M·P9IIlt~~ • tl)9~ls.W~.~ff®!·.()f$a~Wlll·

W=279.02

•.••.••.. ·.M.~I@i~W~pYlffil~l()l'lg~~9f;l§flIW¥lP~.

AE =77854.67 A = 77854.67 = 0039 rrf 2 x 106 . c

d~Mo.tI'lE!M~lj¢atl@Qft~I$IMdi~¢M~1

••.t6.t6.M....(feterlnio~lhe9~il·W~i9htcitt~e.

•• iaPE!•• j~ ~*lQ~Mtcnt.> taM·.MQ(I~ltl~ ()f•• [email protected]••

®Urid$ra$tandardpullQf~~g,Jhel@el tapej~4QlTl"()(lg,An9rm~IJ~n~i9Q:qf 1&.k9rll~~~s • ~I()ng~~qiJ9f.~h~ • • l~Il$·· ··()ff$e;t.t~~en~pt.pf.·~~·.· • • lfm~t~w(#Stm··

0.:

~

w=0.784 kg

m¢••

llqZ#·kgfril,•• •arld.E • =••z•)(•• j06•• • kglciril%)

Qeletrlline .• • it~ • • cr(Jss • • $ecll()nlll.··ar~.· • . in. s:(P::Il'l{

Solution: CD Normal tension: PN = 0.204

{Ai!

..,j Pw Pt" _ 0.204(0.84) ..,j'-O.-OO-(2-)1-06'

PN-

..,j Pw P1

_ 59.3608

PN-

..,j PN - 5.6

By trial and error: PN= 17.33 kg _ 59.3608 PN-

..,j 17.33 - 5.6

PN= 17.33 kg

.0) ..• Det~rminelli~ IEltlgll1 of the fine in meters if ... .there were 3 tallies'S phis aildthe last pin \ was9>·Iil,Jromtheend of tile Hne. The ...•..• tapeMectwas so. m.IOhg. .• . . lID A line was measured wilh a50 m. tape and fo~nd .to beJOO m. long. It was . .> ~• .• •.• • •.·•.~~I~i 6i• • • • • • • • •

•.••.•. '.•. •.•.•.•·.•.•. '.·.•L.• ..•·.•.•.1•• . ·• .E.•.·••.• •.•.•.•.•.•.•.• . ·•. ••. •. I.I.

•.••. : ..•

1.•11..• •.k1.•·. .•.• .•.•.L.•.•.• . E .•

l.·•.

U.

~~~:~»o5~6d2, • •.• ' \ @jyomputelh&l'll&YlnelElYaI160

pel>veenl\~nd8··) @'.Comput!3c.ftIe·.ElIEl\l()ti()n.of.~i •. Solution: G)

Dift. in elevation between A and B;

Miradorhilhvfth.an. elev9tionofp26rn,iSoll.a lifle•• ~Elwee8.A~ror() • ~ill • ~hlh$~· .• el~YWipry·.i~ 660.·m·.9nd•• Q()thedr()lhillna...i'19.1930 wheil the magn~lic declination WB$O'52' E. A$sumeB and C is on. the north eastsidji, . . .. . CD FInd the true bearing of A6.·••••. '. . ® .Find the length of AD with pOint Don the line Be and makillg the area of thetl'iangle ABD one third of the Whole area. . @ Compute the bearing of line AD.

69280 _ (AB)2 Sin 60' 2 AB=40m. 1 A1 = 3" (692.80) A1 =230.93 A 40 (x) Sin 60' 1

2

x= 13.3 m. (AD)2 = (40)2 + (13.3)2.2(40)(13.3) Cos 60' (AD)2 = 1245 . AD= 36.3m.

S-63

COMPASS SURVEYING @

Bearing of line AD: Sin f1I Sin 50' 13.3'=36:3 f1I = 20'08' Bearing of AD =49'37' +20'08' Bearing of AD = N 69'45' E

A triangUlar lot has for \Jne of its boundaries a.

nne 1500 rn, long Which runs due East «pm AThe eastern boundary is 900 m. long and the western boundClry 1200 m. long.. Astraight li~e

@

Bearing of line 8E: =90' • 53'0748" =N 36'52'12" W

@

Bearing of line DA: = 79'41'44" - 26'33'56" = S 53'07'48" W

In the defiectionaogie trClverse wilh atraO$il survey data below. Assume deflection T, T2 T3 and bearing Tl T2 is correct. .:.

cuts the western· bqundaiyal the iniddle point and meets Ihe easterly boundary E, 6{)O tit from the Sf. comerR· .

o

CD Find the bearing af line ED. @ Find the bearing of line BE @ Find the bearing of line DA.

Solution: CD Bearing of line ED:

T2 • Ta, Find the beClring of line T3• T4. Find the bearing of line T4 • Tl .

(1) Find the bearing af line

Angle ACB is a right angle having the ratio of its side as 3:4:5. . 900 Sin A = 1500 A = 36'52'12" B =90' - 36'52'12" = 53'OT48" 300 CotE= 600 E = 63'26'04" 0= 90' - 63'26'04" = 26'33'56" DE Sin 63'26'04 = 600 DE=670.83 Bearing of line ED =180' - (36'52'12" + 63'26'04") =S 79'41'44" W

@ @

Solution:

S-64

COMP'SS SURVEYING I Def. < S to the right R 96'42' R 176'33' R

IDef. \

Figure Adjustment Considering triangle ABC Angle 1 = 58'25'13" Angle 2 = 59'10'00" Angle 3 = 62'25'12" 180'00'25" Error =25"

Adjusted Angle 1 =58'25'05" - 08" =58'25'05" Adjusted Angle 2 =59'10'00" - 08" =59'09'52" Adjusted Angle 3 =62'25'12" - 09" = ~ 180'QO'00" Corrected value ofangle 3 = 62'25'03"

® Corrected value of angle 6: Considering triangle BCD Angle 4 = 60'29'05" Angle 5 = 59'25'12" Angle 6 = 60'05'09" 179'59'26" Error = 34"

Adjusted angle 4 = 60'29'05" + 12" = 60'29'17" Adjusted angle 5 =59'25'12" + 11" = 59'25'23" Adjusted angle 6 =60'05'09" + 11" = ~ 180'00'00" Corrected angle 6 =60'05'20" @

Corrected value of angle 9: Considering triangle COE: Angle 7 = 71'40'01" Angle 8 = 63'10'10" Angle 9 = 45'10'19" 180'00'30" Error =30" . =3 30" =10" CorrectIOn

Adjusted angle 7 = 71'39'51" Adjusted angle 8 = 63'10'00" Adjusted angle 9 = 45'10'09" 180'00'00" Corrected angle 9 =45'10'09"

AI::::..-...i:...------:-::-....l~

Baseline = 1420 m

111.ti'(lli'~~ Solution:

CD Constant F: O-C Constant F = 0

o = 10 (no. of directions observed forward and backward not including Ab)

C= (n' - s' + 1) + (n - 25 + 3) n' = no. of lines observed in both directions.

n'=6 s' = no. of occupied stations s'=4 n = total no. of lines in the figure including known lines

n=6 s = total no. of stations 5=4 C= (6 -4 + 1) +[6 - 2(4) +3) C=3+1 C=4 D-C F=O F= 10-4 10 F= 0.60

S-102

TRIUGUIlTION @

Strength of figure which gives the strongest route: Consider triangle ABC dand ACD with AC as common side. -AL_~

R =0.60(31.979) R = 19.19

AD

AB

Sin 90' = Sin 53' =AB Sin 90' AD Sin 53'

CD

Distance angles are 42' and 60' for triangle ABC 9.825510895 9.825513234 2339

AA =2.339 log Sine 60'00'00" log Sine 60'00'01"

hc) L (Ai + AA I.'>.a + I.'>.l)

Consider triangle ABO and ACD: with AD as common sides.

Sin 42' - Sin 60' AC - AB Sin 42' - Sin 60' CD AC Sin 41' =Sin 35' CD=ACSin41' S;n35' CD = AS Sin 42' Sin 41' Sin 60' Sin 35'

log Sine 42'00'00" log Sine 42'00'00"

R = (0

9.937530632 9,937531847 1215

AD

Sin 40' = Sin 104' CD =AD Sin 40' Sin 104' = AB Sin 90' Sin 40' CD Sin 53' Sin 104'

Distance angles are 53' and 90' for ABO log Sin 53'00'00" = 9.902348617 log Sin 53'00'01" = 9.902350203 1586 AA =1.586

Aa = 1.215

log Sin 90'00'00" = 0 log Sin 90'00'01" =0 Aa=O

(Ai + AA Aa + Art) =(2.339f + 2.339(1.215) + (1.215)2 (6} + AA Aa Art) =9.789

(Ai + AA Aa + Ai) = (1.586)2 + 0 + 0 (I.'>.i + AA Aa + Ai) = 2.51

+

Distance angles for triangle ACD are 41' and 35' log Sin 41'00'00" = log Sin 41'00'01" =

9.816942917 ~~

2422 log Sin 35'00'00" = 9.758591301 log Sin 35'00'01" = 9.758594308 3007

AA = 2.422 Aa = 3.007 (Ai + AA Aa + Ai) =(2.422)2 + (2.422)(3.007) + (3.007)2 (Ai + AA Aa + Ai) =22.19 L (Ai + AA Aa + Ai) =9.789 + 22.19 L (~i + AA Aa + Ai) :a: 31.979

Distance angles are 41' and 104' for triangle ACD log Sin 41'00'00" = 9.816942917 log Sin 41'00'01" = 9.816945339 2422 AA = 2.422 log Sin 104'00'00" = 9.986904119 log Sin 104'00'01" = 9.986903594 525 I.'>.a = 0.525

(Ai + I.'>.A I.'>.a + I.'>.il =(2.422)2 + 2.422(0.525) + (0.525)2 (Ai + AA I.'>.a + Ail = 7.41

L (Ai + I.'>.A Aa + I.'>.i) =2.51 + 7.41 L (Ai + AA I.'>.a + Ai) = 9.92

S-103

TRIANGULATION D-C) 2 2 R= ( C L(~A +~A~B+~B)

R = (0 ~.~ L (Ill + ~A t1B+ Ili)

R = 0.60(9.92) R= 5.952

R =0.60(5.96) R=3.58

Considering triangle ABC and BCD with BC as common side: Jin 78' - Sin 60' = AB Sin 78' BC Sin 60' CD BC Sin 48' = Sin 88'

Consider triangles ABO and BCD with BD as common side. BD AB Sin 37' = Sin 53' BD =AB Sin 37' Sin 53' CD BD Sin 48' = Sin 44'

- BC Sin 48' CD - Sin 88' = AB Sin 78' Sin 48' CD Sin 60' Sin 88'

CD _ Sin 48' - Sin 44' =AB Sin 37' Sin 48' CD Sin 53' Sin 54'

The distance angles are 60' and 78' for triangle ABC and 48' and 88' (or BCD

Distance angles of triangle ABO are 37' and 53' and for triangle BCD are 44' and 48'

~-~

log Sin 60'00'00" = log Sin60'OO'01" = ~A

9.937530632 9.937531847 1215

= 1.215

log Sin 78'00'00" = log Sin 78'00'01" =

9.990404394 9.990404842 448

Il B = 0.448 (Ili + ~A ~B + Ili) = (1.215)2 +(1.215)(0.448) + (O.44W (Ili + IlA IlB + Ili) = 2.22 log Sin 48'00'00" = log Sin 48'00'01" =

9.871073458 9;871075354 1896

IlA = 1.896 log Sin 88'00'00" = log Sin 88'00'01" =

9.999735359 9.999735432 073

Il B = 0.073 (Ill + IlA IlB + Ili) =(t.896)2 + (1.896)(0.073) + (0.073)2 (t1l + IlA Il B+Ili) =3.74 L (Ill + IlA IlB + Ili) = 2.22 + 3.74 L (Il/ + IlA IlB +Ili) = 5.96

eo

log Sin 37'00'00" = log Sin 37'00'01" =

9.779463025 9.779465819 2794

IlA = 2.794 log Sin 53'00'00" = log Sin 53'00'01" =

9.902348617 9.902350203 1586

~B=1.586

(Ill +IlA ~B + Ili) =(2.794)2 + (2.794)(1.586) + (1.586)2 (Ill + IlA IlB + ~i) = 14.75 log Sin 44'00'00"::: log Sin 44'00'01" =

9.841771273 9.841773454 2181

IlA =2.181 log Sin 48'00'00· = log Sin 48'00'01" = Il B=:.896

9.871073458 9.871075354· 1896

5-104

TRIANGUlaTION Solution:

(di + dA dB + di) =(2.181f + (2.181)(1.896) + (1.896)2 (di + dA dB + di) = 12.49 L (di + dA dB + di) =14.75 + 12.49 L (di + dA dB + di) = 27.24

CD Adjusted value of angle 4: ADJUST A ADJUST 8 L 1= 23'44' 38" 23'44' 37"

c0- C) L(di+dAdB+di)

R= (

R =0.60(27.24) R= 16.34 Relative strength of the quadrilateral R = 3,58 (smallest value) @

Length of check base CD: CO =AS Sin 78' Sin 48' Sin 60' sin SS' CO = 1420 Sin 78' Sin 48' Sin 60' Sin 88' CO = 1192.61 m,

ADJUSTC 23'44'35"

L 2 = 42'19' 09"

42'19' 08"

42'19' 06"

Z. 3 = 44'52' 01"

44 '52' 00"

44'52'00"

L 4 = 69'04' 21"

69'04' 20"

69'04' 19"

L 5 = 39'37' 48"

39'37' 47"

39'37' 49"

L 6 = 26'25' 51"

26'25' 50"

26'25' 52"

L 7 = 75'12' 14"

75'12' 13"

75'12' 13"

L 8 = 38'44' 06"

38'44' 05"

Sum = 360'00' 08"

360'00' 00"

3.8'44' 06" 360'00'00

Error =8" Correction =§. 8 Correction = 1" (sub)

L1+L2=L5+L6 23'44' 37" 42'19' 08" 66'03' 45"

• rfit~~ • foIIOWJdg"•• qM~drllatfjtal • ·.Wlth ¢or~~l;ljng·ClnSle~.·.~®lated.~h()wn.

a

39'37' 47" 26'25'50" 66'03' 37"

Error = 45·37 Error = 8"

c Correction =§. 4 Correction = 2" (subtract from L1 and L2 and add to L5 and L6)

L3+ L4 = L7 + L8 A O'-u....-

.-.L.~B

d) Compute the adjusted value of angle 4 by

appIyihg the artgleconditioll only.

® Compute the adjusted value of angle 7 by . applying the angleconditlon only. @

Compute the strength of figure factor.

44'52' 00" 69'04' 20" 113'56' 20"

75'12' 13" 38'44' 05" 113'56' 18"

Error = 20 - 18 =2" Correction = Add 1" to La and subtract 1" from L4

5-105

TRlAliGUlADON Check: L1 23'44' 35" L8 38'44' 06" L2 42'19' 06" L7 75'12'13" 180'00'00"

L1 L2 L3 L.4

23'44' 35" 42'19' 06' 44 '52' 00" 69'04'19" 180'00' 00"

L8 L.7 L.6 L.8 .

L.3 L4 L.5 L.6

38'44' 06" 75'12' 13" 26'25' 22" 39'37' 49" 180'00'00" 44'52' 00" 69'04'19" 39'37' 49" 26'25; 52" 180'00'00"

Angle 4 = 69'04'19"

Gtv~nthe~uCidrilateral.shownYJfjichh~~·qeerl adjU$t~d • u~lng·.IJrgle.C9ndi~()~, • • • ltl$rMUire~

1(l.aO,

ZD99mPutelheadju$te~~~glCl-~' ...i ®i.9omp\lte.ttmadjU!;ted.arl~Ie-.~ .• • • •. ·.· · · · . @PQmp\.ltetf)eadjllsted.al1gl~6.

L.1

.

=39'3749"

L.2 = 26'25' 52"

L3 = 75'12' 13" L4 = 38'44' 06"

@ @

Angle 7 = 75'12'13"

L5 =23'44' 35"

Strength of figure factor.

L6 = 42'19' 06"

D=10

L7 = 44'52' 00"

n'=6 n=6

L8 =69'04' 19" Sum = 360'00' 00"

s=4 s'=4 C =(n' - s +4) +(n - 2s +3) C = (6 - 4 + 4) + (6 - 8 + 3)

C=4 F=D-C

o

F= 10 -4 10 F =0,60 (Strength offigure factor)

Aif'-o"........................-

..........- -.........:.J..:.~B

Solution: CD Adjusted angle 3: . Sin L.2 Sin L4 Sin L6 Sin L8

1

~~-'-------=

Sin L1 Sin L3 Sin L5 Sin L.7 log Sin 26'25' 52" = 9.64847855 log Sin 38'44' 06" = 9.796379535 log Sin 42'19' 06" = 9.82817581 log Sin 69'04'19" = 9.970360677 9.243394570

.

5-106

TRIANGUlATION

log Sin 26'25' 52" log Sin 26'25' 53" log Sin 38'44' 06" log Sin 38'44' 07" log Sin 42'19' 06" fog Sin 42'19' 07" log Sin 69'04'19" log Sin 69'04' 20"

log Sin 39'37' 49" log Sin 75'12' 13" log Sin 23'44' 35" log Sin 44'52' 00"

log Sin 39'37' 49" log Sin 39'37' 50" log Sin 75'12' 13" log Sin 75'12' 14" log Sin 23'44' 35" log Sin 23'44' 36" log Sin 44'52' 00" log Sin 44'52' 01"

= 9.64847855 = 9.648482785 = 9.796379535 = 9.79638216 = 9.82817581 = 9.828178122 = 9.970360677 = 9.970361483

= = = =

= = = = = = = =

Diff. in 1" 4.2 2.62

Add 2" to all angles in the numerator: (smaller) Subtract 2" to all angles in the denominator: (bigger)

2.31

- L1 = 39'37' 49" +L2 26'25' 52" = .75'12'13" - L3 +L4 = 38'44'06i, - L5 = 23'44' 35" +L6 = 42'19' 06" - L7 = 44'52' 00" +L8 = 69'04' 19"

0.81 __ 9.94

9.804705675 9.985354379 9.604912331 9.848471997 9.243444381 Diff. in 1" 9.804705675 2.54 9.804708217 9.985354379 0.56 9,985354935 9.604912331 4.79 9.604917117 9.848471997 2.12 9.848474112 _ _ 10.01

• 02" +02" • 02" + 02" - 02" + 02" - 02" + 02"

39'37' 47" 26'25'54" 75'12' 11" 38'44'08" 23'44'33" 42'19'08" 44'51' 58" 69'04' 21" 360'00' - 00"

Adjusted angle 3 =75'12' 11'~ @

Adjusted angle 5 = 23'44' 33"

@

Adjusted angle 8 =69'04' 21" -----/

J

,c •••• "II1thE!figwe•• ~hpw§ • .~,.qy~drla~~r~IWl!htheit ~9q~$Pon(jin9, • • • • t*ngUItl.r• ' deSignated. "....,' ... .. ., ."•.• ','•, m~#$l)r~m~nls .'..'. ,', . . ,

Subtract: 9.243394570 - smaller 9.243444381' bigger 0.000049811 Add: 9.94 + 10.01 = 19.95

Difference =49.81 49.81 ' 0=-80=6.23 (J = 19.95

8 (J = 2.49

. 6.23 Correcllon =2,49 Correction = 2.5" say 2"

(j)

WhiCh. of InemilOWing equation dOes not ®ltsfy the figure shown. L2 + L3 =L7 + L6 b) .L1 + L8 =L4 + L5 c) L1 + L2 + L3 + L4 = 180' d) L1 + L8 + L6 + L7 = 180' a)

5·107

TRIANGUlATION

~ ~

c} d} ,

Sin L1 Sin L3 Sin L5 Sin L.7 Sin L8 Sin L2 Sin L4 Sin L6 Sin L2 Sin L4 Sin L6 Sin L8

=0 =1

Sin L3 Sin L5 Sin L7 Sin L1 Sin L1 Sin L3 Sin L5 Sin L1 Sin L2 Sin L4 Sin L2 Sin L4 Sin L6 Sin L8

@

n = total number of lines in figure, including the known side n=6

Sin L6 Sin L8 Sin L1 Sin L3

= ---=--....:..:...--=. Sin L5 Sin L7

.·WMt.Will • ~th~#~~Qr..• (.....F)l".~lvihg • me ..... . .

.

C----~::---~~~D

slt~ngtI'l9fflgur~.
} + ~A ~B + ~i) R = 0.625(5.02) R= 3.14

G

Solution:

CD Value of C:

n' =no. of lines observed in both directions including known side n'= 13 n =total no. of lines in figure unciuding known side n =13 s' =no. of occupied stations s' =7 s =total no. of stations s=7

C =(n' - s' + 1) + (n - 2s + 3) C=(13-7+1) + [13-2(7)+3] C=7+2 C=9

CD Compute the adjU$tedvalue of angle Aby @

@

diWibuting lhespherical excess and the remalning error equally. . Compute the adjusted value of angle B by distributing the spherical excess and the remaining error equally. . Compute the adjusted value of angle Cby distributing the spherical excess and the remaining error equally.

Solution: A

e=R2 Sin01" A =be Sin A 2 b Sin 56'10'30" b =33814.89

35965.47

= Sin 62'04'11"

5-109

SPHERICAl EXCESS A = be Sin A 2 33814.89 (35965.47) Sin 61'45'20" A= 2 A = 53568365b.2 m2

"

A

The interior angles in triangle ABC are

A "'. 57'30' 29", B ::: 65'17'27" • and C =57'12' 16". The distance from A to B is equal'to 180,420 m, ASsuming fh~ average radius Qf curvature is 6400 km. ',

e = R2 Sin 01" .' 535683650.2 e"= (6372000)2 Sin 01" [."=2.72"

CD Compute the area of fhe triangl~; @ Compute the second term of the spherical excess.

61'45'20" 56'10'30" 62'04'11"

,

'

@ Compute Ihe total spherical eXcess,

Solution:

180'-00'01" 180'-00'02.72"

G)

Area of triangle: B

Error =1.72" 1.72 1st Carr. = -31st Corr. =0.573 (added) 2.72 2nd Corr. =-3-

A

2nd Corr. =0.907 (to be subtracted)

.... c

~--~

Using Sine Law:

61'45'20" + 0.573" - 0.907 = 61'45' 19.676" 56'10'30" t 0.573" ·0.907 = 56'10' 29.676" 62'04'11" t 0.573" - 0.907 = 62'04' 10.676" 180'00' 00" B

b 180420 Sin 65'17' 27" = Sin 57'12' 16" b =194978.94 m. A bcSin A rea=-ZA rea

194978.94 (180420) Sin 57'30' 29" 2 6 Area = 14836 x 10 rrf

® Second term of the spherical excess: A

......... c

~-_

When the sides ofthe triangle are over 100 miles (160,000 m.) use the accurate formula for spherical excess. Area [ a2t b2t e" =R2 Sin 01" 1 t 24 R2

el) Adjusted value of angle A = 61'45'19.676"

=56'10'29.676"

c;"

Adjusted value of angle B

(;'t

Adjusted value of angle C =62'04'10.676"

The second term is 2 2 Area (a + b t R2 Sin 01" 24 R2

c2)

c2]

S-J10

SPIIIBICIl EXCESS a 180420 = Sin 57'30' 29" Sin 57'12' 16" a =181033.49 m. Second tent .

(j)()omPOle • the.adjlJ~tedvalue • ofMgle.A.PY 91~WputingmemmericaL~X(\{Jssandthe

~i:lrnlng:e@tequ~:"IY,>

·®·.Cp!l'lp~t~ • • the.9djl.lste9Wlue•• ofan~le~ • W• . . . .•··.·.4iS1riblJti/lg • lhe.spherj"al•• til)(ee$$~l'ld • • m~·

••.•..•..• @l)airyins~tI"QreqUEln}"

a2 =32773 x 1tl6

'.

.

• ~• • • Gompute•• t/)e.adJustedv~J~~Qf.~r~J~Cby . ••.•.•.•. •. . g~trll?~ljlls • . th~ • $P~~ig~I.·~~e;;san~ • the tElll'lafl'ling error eqU(lny. '. .. ..... .'

b2 =38017 x 106 c?- =32551 x 106 W- =40960000 x 106 Area = 14836 x 106

Solution: B

(a2

Area + ~ + c?-) 2nd term = R2 Sin 01" .24W-

14836 x 106 2nd term =40960000x1 06 Sin 01" A

(32773+38017+32551)106] [ 24(4096‫סס‬OO)106 2nd tenn

, = 74.7106"

(

103341 ) 24(40900000)

2nd term = 0.00785"

® Total spherical excess: Area [ a2+~+c2] e" =W- Sin 01" 1 + 24 We" =74.7106 + 0.OD7e5

d' =74.71845"

·~-_

....... c

C 5260 = Sin 52'03'17" Sin 88'33'05" C = 4149.3$ m. log m = 1.40658 ·10 m = 2.55023 x10-9 e"=mbc Sin A e' = 2.55023 x 10,9 (5260)(4149.35) Sin 39'23'40 e"= 0.035" A = 39'23'40" B = 88'33'05" C = 52'03'17" 180'· 00'02" 180',00,00.035" Error= 1.965 First Correction: 1,~5 ::; 0.655" Second Correction:

0,~35 =0,012

39'23'40" • 0.655" • 0,012" = 39'23' 39.333" 88'33'05"·0.655"·0.012" = 88'33' 4.333'.' 52'03'17"·0.655·0,012" = 52'03' 16,333" 180'00' 00" CD Adjusted value of angle A ::; 39'23'39,333" @ @

=88'33' 4,333" Adjusted value of angle C =52'03'16.333"

Adjusted value of angle B

S·lJl

SPHERICAl EXCESS m = 2.536 x 10.9 log m =1.40415·10

_b c_ Sin B - Sin C b 3012 Sin 63'44'59" = Sin 79'59'57" b = 2743.05

o @

m.

e" = 2 A

.I'fR'IS known. R Sin 01" e" = m bc Sin A if no radius is given

1 m- 2 R NArc 1"

CD Compute the adjusted value of aogle Eby .' ," 'distnbutingthe spherical excess and the remaining error equally. ,, @ ,Compute ,the adjusted value of angle Nby , distributing the spherical excess and the remaining ~rror equally. , ' ,'. , ® Cbmputeth,e adjusted value of angle L by distributing the spherical excess and the remaining error equally.

" 1t Arc 1 = 180(3600) N = 6376032 m.

At a given latitude e"=mbcSinA e" =2.536 x 10.9 (2743.05)(3012) Sin 36'15'07" e"= 0.012" (spherical excess) 79'59'27" 63'44'59" 36'15'07" 180',00'- 03" 180'- 00'· 0 012" Error = 2.988" (error of spherical triangle)

Solution: N

Correction for each angle 2.988 =-3-

= 0.966" (First Correction)

L

Second Correction 0.012 =-3-

~--""""-E

A e"= R2 Sin 01"

= 0.004" (spherical excess,subtracted from each angle)

Arc 1" =180 ~600)

79'59'57" ' 0.996" • 0.004" =79'59'56" 63'44'59",0.996" - 0.004" =63'44'58" 36'15'07"·0.995",0.004" =~ 180'00' 00"

bcSin A , A = - 2 - (area oftnangle)

e =m bcSin A 1 m= 2RNArc1" m=

CD Adjusted value of angle E = 79'59'56"

1 1t

2(6378160)(6376032) 180(3600)

® Adjusted value ofangle N = 63'44'58" @

Adjusted value of angle L

=36'15'06"

5-112

AREA Of ClOSED TIIAIERSE AREA OF CLOSED TRAVERSE In any closed traversed, there is always an error. No survey is geometrically perfect, until proper adjustment are made. For a closed traversed, the sum of the north and south latitudes should always be zero.

BALANCING A SURVEY Latitude of any line - is the projection on a north and south lines. It may be called as north or positive latitude and south or negative latitude. Departure of any line - is the projection on the east and west line. West departure is sometimes called negative departure and East departure is sometimes called positive departure.

C

DEPARTURE

B

1. Compass rule - the correction to be applied to the latitude or departure of any course is to the total correction in latitude or departure as the length of the course is to the length of the traverse. 2. Transit rule - the correction to be applied to the latitude or departure of any course is to the total correction in latitude or departure as the latitude or departure of that coUrse is to the arithmetical sum of all the latitudes or departures in the traverse without regards to sign. Error of closure = ."j LL2 + L02 RIf Error of closure eaIve error = Perimeter of all courses LL =error in latitude LO =error in departure

A

Line AB has its latitude AC and departure BC. The angle 0 is the bearing of the line AB. BC =AB.Sin 0 .

1 Area by Triangle Method Departure = Distance x Sin Bearing AC =AB Cos 0 Latitude = Oistance x Cos Bearing Dist = Latitud~ . Cos Bearrng O' t _ Departure IS - Sin Bearing

0

5-113

AREA OF ClOSED TRAVERSE or

A=A1+A2+A3 d1d2 Sin a A1 = 2 ·A - d3d4Sin {!, 22 - dsd s Sin l2I A3 2

4 Area by Double Meridian Distance

2 Area by Rule of Thumb Method

2A = [Y1 (X1 - X2) + Y2 (X1 - X3) + Y3 (X2 - ~) + Y4 (X3 - Xs) + Ys(~ - X1)]

Double Meridian Distance of line BC is the sum of meridian distances of the two extremeties.

3 Area by coordinates

Area ABCDE =Area BCEF + Area CDIF - Area EDIH • Area AEGH •Area ABGE

c

~ .'...x j....A~·

D.M.D. of BC = EB + FC Latitude of BC = EF

H······:':$·f,····

, !""I('»,.

Area of BCFE: _ (EB + FC) EF A2

- (X2 + x 3) (Y2 - Y3) + (X3 + X4) (Y3 - Y4) A2 2 (X4 + xs) (Ys • Y4) (xs + X1) (Y1 - Ys)

2

-

2

(1'1 + X2) (Y2 • Y1)

2A = (EB + FC) EF 2A = (D.M.D.)tatitude Double Area = Double Meridian Distance x Latitude

2 Simplifying 'this relation: CD 2A = • [Y1 (xs - X2) + Y2(X1 - X3) +Y3 (X2 • X4) + Y4 (X4 - Xs) + Ys (~ • X1)]

or @

2A = Yt X1+ Y3Y2 + X4Y3 + XSY4 + X1YS • XtY2 • X2Y3 - Y3Y4 . X4YS· XSY1

D.M.D. of the first course is equal to the departure of that course. 2. D.MD. of any other course is equal to the DMD of the preceding course, plus the departure of the preceding course plus the departure of the course itself: 3. D.M.D. of the last course IS numerically equal to the departure of the last course but opposite in sign. 1.

5-114

AREA OF ClOSED TRAVERSE Computing Area by D.M.D. Method: 1. Compute the latitudes and departures of all courses. 2. Compute the error of closure in latitudes and departures. 3. Balance the latitudes and departures by applying either transit rule or compass rule. 4. Compute for the D.M.D. of all courses. 5. Compute the double areas by mUltiplying each D.M.D. by the corresponding latitude. 6. Determine the algebraic sum of the double areas. 7. Divide the algebraic sum of the double area to obtain the area of the whole tract. Double Area =D.M.D. x Latitude

5

Area by: Double Parallel Distance

1. D.P.D. of the first course is equal to the latitude of that course. 2. D.P.D. of any other course is equal to the D.P.D. of the preceding course, plus the latitude of the preceding course, pius the latitude of the course itself. 3. D.P.D. of the last course is numerically equal to the latitude of the last course but opposite in sign. Double Area =Double Parallel Distance x Departure

Example: Area by Double Meridian Distance

Lines LAT. DEP.

DMD

-30

-30 -40 +40 +SO

1-2 2-3 3-4 4-1

+60 -20 -80 +40

+20 +60 -SO

Double Area ·30160\ =-1800 ·401- 20) =+800 +40(. 80\ =-3200 +SOI4O\- +2000 2A =- 2200 A=-1100m2

Area by Double Parallel Distance

Lines LAT. DEP. 1-2 2-3 3-4 4-1

+60 -20 - 80 +40

- 30 +20 +60 -SO

Double Area +60 601- 30\ =- 1800 +100 100120\ = +2000 d(6Q) = 0 0 -40 -401- SO\ = +2000 2A 2200 A=1100m 2 DPD

=-

•..••••..··· .·llOll,OOffi) 700.00fft·· .. 600.00trl. CD Compute the correction .. tiNES l. tAl1WP/l'·

DI1J'Af\r~ ...

CD

,

i

m+n

--V

3(300}2 + 2(150)2 x3+2 x= 251 m.

Anequilater~tl:1'iangufarlrack Of land has a

length of one '(jfits sides equal to 3600 m. Iori9' ·111$ requited to dIVide the lot into 3 equal .sharesoy aline parallel to one of Ihe 3sides.

CD COmpulethe whole area of Ihe lot jfl-£lCf8S. ® Compute the iength {If lhedivlding line on the portion near the vertex. . @ Compute the length of the dividing line on the se? = (9.55)2 + (14.02}2 - 2(9.55)(14.02) Cos 71'40' x= 14.27 m.

The centerline

of the' proposed service roM

Using Sine Law: 14.02 14.27 Sin", =Sin 71'40' 0=68'5" f3 = 180' - 71'40'·68'51' f!,=39'29' AB 14.27 Sin 50'31' = Sin 88'06' AB::: 11.02m.

cro~es at 9.55 m. from corner 4 along the line

3·4 and runs in adirection afN3'45' E. 1

@

S7J'Ji'W

Distance of other end of center line of proposed road from comer 1; AB B-1 Sin 50'31' = Sin 41'23' B-1 =11.02 Sin 41'23' Sin 50'31' B-1=9.44m.

S-166

SUBDIVISIOI @

Cost ofproperly to be expropriated:

tan 88'00' =~ X2

X2

=0.17

X1

5 = tan 88'06' = 0.17

X1

tan 20'14' =~ X4 = 1.84 m.

Xs = 1.84 m. CD =AB- x2 + X4 CD = 11.02 - 0.17 + 1.84 CD=12.69 EF = 11.02 -1.84 +0.17 EF=9.35

A =A 1 +A2 A1 = (12.69 +211.02) 5 = 59.275 m2

Line! 1- 2 2-3 3·4 4-5 5-1

Bearinas Distance N61'57' E 74.18 ........

--_ ..

S9'03'W 54.13 N68'21'W 55.43 N13'56'W 58.85

LAT +34.88 .......

DEP

+65.47

-

·53.46 -8.51 +20.45 - 51.52 +57.12 -14.17 +112.45 +65.47

::.lli2 :.l!2.Q

A = (11.02 ; 9.35) 5 = 50.925

+58.99

2

A = 59.275 + 50925 A = 110.20 m2

........

- 8.73

CD Length of (he boundary of the proposed

road on the sou/hem portion of the lot: Fortine 2- 3:

Total Cost =110.2 (2000) Total Cost = P220400,OO

tangent bearing =~

. 8.73 tangent beanng =58.99 bearing = S 8'25' E

h,!•• Vi~W • •9t•• lh~~@ffi • ·%.tll~9PY~mm@lJp relleyepet~~riial~ht¢ulararyqp~e~W~I1· ·.ttiif1ipcRmi~~ti(1~pf:a.city,a.dead.endrqligl~· .to•• b¢•• eXf~nded • arid·.coQ~tfUeted • .\Vl.th.~rroM

r1ghtqfw~YClf:2qgL • whi9h.iOter~s • ~.eYe~.1 Pl'jvat~IY • ()Wf\~dprpP~rtlM ..•• Qoe.• of·.ltieSe··lot

D'

t

dep

IS ance = Sin bearing

. 8.73 Distance = Sin 8'25' Distance = 59.64 m.

ha$tMfQlIOWI09f1e.I~Mttl~k . .:-:.:-:-:-:'.-:'"-:.:.', .• '.-.-.-.:-:.'.:.»:.'.",'.','.'.' .....:.:.:.:-,.:-:",-- ....

LINES··

SEARING

/. N6flm.s

.·····blStANCS··· lA

Complete Field Notes

LINES 1- 2 2-3 3-4 4-5 5-1

BEARING N61'57' E SS'25' E S9'03'W N68'2fW N13'56'W

DISTANCE 74.18 m. 59.64 m. 54.13 m. 55.43 m. 58.85 m.

S-167

SUBDIVISION X 68.50 Sin 0'45' = Sin 104'07' X=0.92m. y 68.50 Sin 75'08' =Sin 104'Or Y=68.27 h1 =0.92 Sin 75'53' h1 =0.89m. ~ =10- 0.89 ~=9.11 m.

2

tan 52'54' =!!.l

a

0.89 a=tan 52'54' a=0.67m.

4 Bearinqs Distance 3-4 S9'03'W 54.13 4-5 N68'21' W 55.43 ........ . -... 5··1

Line~

LAT -53.46 +20.45

DE? - 8.51 - 51.52

.... - ..

.. ......

- 33.01

- 60.03

tan 52'54' = & d d 9.11 =tan 52'54' d=6.89m.

tan 75'53' = 10

e

e= 2.51 m.

tan 70'2Z = 10 f f= 3.57 m. Distance ofline F - 5: 2

3

5

'

60.03

tangentb eanng =33.01 Bearing =N61'1 Z E 60.03 . . Distance =Sin 61'12' Distance = 68.50 m.

4

F - 5 =68.27· 0.67 +0.22 F- 5 =67.82 m. Distance DE: DE =67.82 - 7.86·6.89 DE=53.07

5-168

SUBDIVISION @

Lenght ofthe boundary ofhte proposal road on the northern portion of the lot: Distance AB: arotlljjrsABlldBare~Q'lhef:jtare.$idjjl1tiali()t

Whl~6·~h~II• • ~• dl~ed~llaIlY.lh~area • • ao~

•

.m~ lEl~liI\l'l.·~tl~,.Jt9~l~~.9butti99··~ • Nl1lipn'----o-J---.---.

:

\

I I

\

I I II

~----~----~----~ ~

-

~--

-~----~

----~----~

6. By transit and stadia.

'~l 6 II .

.stadia all boat ...-"'--'''''.-'-. '~ - .•~=:r---', .--0

I~ . ---- .--- ,&" I

~

5-196

HYBROGRAPHICSURIEYING 7. By intersection of fixed ranges.

Methods of Plotting Soundings: 1. By using the Two Polar Contractor 2. By using Two Tangent Protractors I

3. By the Tracin,!1 Cloth Method 4. By using the Three Arm Protractor 8. By a wire stretched along a river at known distances.

5. By the use ofPlolting Charts

Methods of Measuring Velocity in a Vertical Line:

Hydrographic maps - is similar to the ordinary topographic map but it has its own particular symbols. The amount and kind of informations shown on the hydrographic map varies with the use of the map. A hydrographic map contains the following informations: 1. Data used for elevation. 2. High and low water lines. 3. Soundings usually in feet and tenths, with a decimal point occupying the exact plolted location of the point. 4. Lines of equal depths, interpolated from soundings. On navigation charts the interval of line of equal depth is equal to one fanthom or six feet. 5. Conventional signs for land features as in topographic maps. 6. Light houses, navigation lights, bouys, etc., either shown by conventional signs or leIters on the map.

1. Vertical-velocity-culVe method: Measurements of horizontal velocity are made at 0.5 beneath the surface and at each tenth of the depth from the surface to as near the bed of stream as the meter will operate. If the stream is relatively shallow, measurements are taken at each one fifth of the depth. These measured velocities are plotted as abscissas and the respective'depths as ordinates. A smooth curve drawn through the plotted points defines the velocity at each point in the vertical. The are under this curve is equal to the product of the mean velocity and the total depth in that vertical line. This area may be computed by using aplanimeter or by Simpson's One Third Rule. The vertical velocity curve method gives us the most precise method of determining mean velocity but requires only too much time. 2. Two-tenths and Eight-tenths Method: The current meter is lowered downward at 0.2 and 0.8 of the total depth where observations are made. The mean of this two velocities is taken as the mean horizontal velocity in that vertical. 3 Six-tenths Method: Only one observation is made at a distance below the water surface equal to 0.6 the total depth of the stream. The velocity obtained at that particular depth is considered to be the mean velocity of vertical.

8-197

HYDROGRAPHIC SURVEYING 4. Integration Method: The cllrrent meter is lowere at a uniform rate down to the bed of the. tream and is raised also at the same rate up to the surface. The total time and the m mber of revolution during this interval con itute a measurement. ~ This method is based upon the th ry that all horizontal velocities in the ve 'cal have acted equally upon the meter w el thereby giving the average as the mea~ 01' all the velocity reading. I 5. Subsurface Method: In this particular method. the current meter is held at just sufficient depth below the surface usually 150 10m to 200 10m to avoid surface disturbance. The mean horizontal velocity is obtained by multiplying the sub-surface velocity by a coefficient. This coefficient varies with the depth and velocity of stream. This coefficient varies from 0.85 to 0.95.

Float Method of Measuring Stream Velocity From the figure shown, a base line AB is well selected and is established near the' bank of a river where no obstruction will interfere the line of sight during the observation period. Points Cand 0 are established on the opposite side of the river such that the sections AC and BO are' perpendicular to the line AB, hence they are parallel to each other. One transit is set up at A and the other at B. The transitman at B with vernier at zero, follows the float where it is being released at point E, at a distance of 15 m. above section BO. As the float approaches section BO, the transitman at A keeps the line of sight pointing at the float until the transitman at B shouts "shot" a the float passes section AB.

The transitman at A then clamps the lower plate, turns the line of sight to the signal station C and reads the angle 0. The transitman at Balso follows the float, until the transitman at A gives the "get ready' signal and by means of the upper tangent screw angle B is measured the moment the float passes the section AC. The time that the float. passes the section BO and AC is also recorded.

I

The base line AB is then measured accurately and the position C and 0 is then plotted. The path of the float is either scaled or computed using trigonometric principles. The distance divided by the time gives the mean velocity of the float. Three Distinct Methods of Determining the Flow Channels or in . Open Channels or Stream: 1. Velocity-Area Method: , The velocities at any vertical line is observed by using a current meter based on the five different method of velocity measurement using current meters. The area of a certain section is obtained by sounding, or by stretching a wire across the stream and marking the points where observations were made referred from an initial zero point. The depths at this particular points are also measured. The area of the section could then be computed by dividing the section into triangles and trapezoids. The product of the area and the mean velocity gives us the discharge of flow of a certain section. The sum of all the discharges at all sections gives us the total discharge or flow. 2. Slope Method: The 'slope method involves a detennination of the following: aj Slope of water surface. bj Mean area of channel cross-section cj Mean hydraulic radius d) Character of stream bed and the proper selection of roughness coefficient

I

S-198

,/

I

HYDROGRAPHIC SURVEYING Mean Velocity is computed by applying the Chezy Formula:

V=c{RS

3..)"'_00, ~il

A weir method is an obstruction place in a channel, over which water must flow. D' scharged of a stream using this method i valves the necessary information.

where V = mean velocity C = coefficient of roughness of stream

bed R = hydraulic radius

R=pA

c)

A :: cross-sectional area of stream P = wetted perimeter of stream

d) e) ~

Computing values of C by Kutter's Formula:

g) h)

Depth of water flowing over the crest of weir, H. Length of crest, L for rectangular or trapezoidal weir. Angle of side slopes if weir is triangular or trapezoidal. Whether flat or sharp crested. Height of crest above bottom of approach channel, P. Width and depth of approach channel Velocity of approach Nature of end contractions

a) English:

C=

41.65+ 0.00281 + 1.811 s n 1+ (41.65 + 0.0~281).

..JR

C = coefficient of roughness of stream

bed n =retardation factor of the stream bed R = hydraulic radius s = slope of water surface End Contracted Weir Q = 1.84 (L - 0.2H)

b) Metric:

H312

23 +0.00155 +1

c=

s

n

1 + ~ (23 + 0.00155)

{R

s

Computi~ values of C by

Manning's Formula: Rl/6

C=n Discharge = Area x Velocity

Suppressed Weir Q::: CLH3/2

8-199

HYDROGRAPHIC SURVEYING Three common types of floats used in measuring stream velocity: 1. Surface floats - it is designed to measure surface velocities and should be made light in weight and of such a shape as to offer less resistance to floating debris, wind, eddy currents and other extraneous forces. The use of surface float ;s the quickest and the most economical method of measuring stream velocity.

Triangular Weir Q = 1.4 H2.5

2. Sub-surface floats - this is sometimes called a double floats. It consists of a small surface float from which is suspended a second float slightly heavier than water. The submerged float. is a hollow cylinder, thus offering the same resistance in all directions and the minimum vertical resistance to rising currents.

Cipolletti Weir 1.86 LH3I2 o 1 when tan-::;:Q::;:

3. Rod float - the rod float is usually' a

2 4

cylindrical tube of thin, copper or brass 25 mm to 50 mm in diameter. The tube is sealed at the bottom and in weighted with shot until it will float in an upright position with 50 mm to 150 mm, projecting above the surface of the water.

Q ::;: 1.84 LH3/2 (Francis Formula Neglecting

Velocity of Approach) Q::;: 1.84 L [(H + hv)3/2 • hv 3/2 ] (Considering

Velocity of Approach)

Instruments used for measuring difference in level of water:

Discharge measurements are made for the following purposes: 1. To determine a particular flow without regard to stage of stream. 2. To determine flows for several definite gage readings throughout the range stage, in order to plot a rating curve for the station. From this curve the discharge for any subsequent period is computed from the curVe of water stage developed in the recording gage. 3. To obtain a formula or coefficient of dams, or rating flumes.

1. 2. 3. 4. 5. 6. 7.

Hook gauge 8taff gauge Wire-Weight gauge Float gauges Automatic gauges Piezometers Plumb bob

of Instruments used for measuring the velocit of flow:

1. Floats a) surface float b) sub-surface float c) rod float

5-200

HYDROGRAPHIC SURVEYING 2. Current meters a) Those which the revolving element is cup-shaped, or of the anemometer type and acts under differential pressure. Types of current meter: 1. 2. 3. 4. 5.

Price meter Ellis meter Haskell meter Fteley meter Ott meter

A. Measurement of dredged materials:

Measurement in place: Soundings of fixed section are taken both before and after dredging and the change in the cross-sectional area is obtained by calculation or by using a planimeter. The volume of the material removed is computed by using the borrow pit method or by the end-area method. 2. Scow measurement: Each scow is numbered and the capacity of each is carefully determined. When the scow is filled to the capacity the inspector records the full measurements. Materials is scow is sometimes measured by the amount displaced in loading.

B. Measurements of Surlace Current: Certain engineering problems require important information about the direction and velocity of currer)ts at all tidal stages. This. is done by locating the path and computing the velocity of floats from points whose locations are known and can be determined. Floats should be designed to give minimum wave resistance and to extend underwater to a sufficient depth to measure the current in question. The direction of the current may be determined by sextant angles from the boat between . known signals and the floats.

C. Wire drag or Sweep: This method is used in harbor or a bay Where corral reefs and pinnacle rocks are likely to occur. This consists of a wire of any length up to 120 m. which may be set at any desired depth. Depths are maintained by means of bouys placed at the wires and whose length can be adjusted. The drag is pulled through the water by means of a power launches, steering diverging forces to keep the drag taut. When an obstruction is met, the bouys are shown with the position of two straight lines intersecting at the obstruction. These intersection is located by sextant observations to reference points on the shore. Soundings are taken for the minimum depth.

D. Determination of stream slope: To determine surface slope, a gauge is installed on each side of the. stream at the end of the section. The zero's of the gauge are connected to permanent bench marcks on the shore. The gauges are read simultenaously every ten to fifteen minutes for six to eight hours. The mean of these elevation at that point of the stream. The difference in elevation between the ends of the section divided by the distance is the slope. Capacity of Existing Lakes or Reserviors: •

1. Contour Method: A traverse is run from a shore line and the desired shore topography are located by stadia. Take sufficient number of soundings by any method suited 'for the particular job and plot the sub-ageous contour. The area inclosed between contours are determined by planimeter. The average area of two consecutive contours multiplied by the contour interval gives the partial volume. The summation of the partial volumes gives the total volume.

S-lO]

HYDROGRAPHIC SURVEYING 2. Cross-Section Method: The outline of the water line is obtained as in the contour method. The water line is then plotted and divided into approximate trapezoids and tri-angles. Soundings are taken along the boundary lines between each station and are plotted on cross section paper. A perpendicular distances between sections are then obtained by the end area method. The summation of these partial volumes gives the total volume. . Two General Methods of Determining the Capacity of a Lake or Reservoir: 1. Contour Method: a) End-area method b) Prismoidal formula

The' areas A1, A2, etc. are determined by using a planimeter and h represents t~e contours interval. Area below A5 IS neglected. b) prismoid,1 Formula: L V =6(A +4A", +A2) In this case the midqle area Am is the Area A2 and ~ while L is equivalent to 2h.

2. Parallel Cross-Section Method: a) End-area method b) Prismoidal formula

a) End-area method:

a) End-area method:

Parallel ranges are laid out across the lake and soundings are then taken along the ranges. From the observed sounding the corresponding cross-sections could be plotted and its corresponding areas would then be computed.

5-202

HYDROGRAPHIC SURVEYING

F'rOll) • • the.culTent~~Wtlo~s • take~8r • th~ R>'h!itwillb~the)azirJ9lh.(>fth~$!Jl). ·@.wb~twm.~.·tne.(gimutnpffhEl.mal'k.

.

Solution: CD Corrected North Polar Distance: Horizontal Time Vertical Angle Angle 358'40' 54.7" Ave: 8:32:58.75 48'47' 08" (Ave) Oiff. in time = 8:32:58.75·8:00 Oiff. in time::: 00 • 32:58.75 Oiff. in time = 0.5496 hrs. Corrected for NPO = 0.5496(26.64} Corrected for NPO =- 14.64" Corrected NPO::: 68'22' 42.4" ,·14.64"

P = 68'22' 27.76" ® Azimuth of sun:

I it d 4.978" OI·ff.· , In ong u e::: -15" Oiff. in longitude ::: Oh 19m 54,8s Standard time:: 2:18:30.2 • 19:54,8 Standard time::: 1:58:35,4

H::: 48'47' 08" p::: 68'22' 27.76" L::: 14'20' 13.6" 28::: 13129' 49.3" 8::: 65'44' 5465"

S·243

PRACTICAl ASTRONOMY p = 68' 22' 27.76" Sop = 2'3T 33.11"

Solution: CD True bearing ofsun from the North:

S = 65'44' 54.65"

H:;: S-H:;: S:;: L= S-L:;:

+

48'4708" 16'5746.65" 65'44' 54.65" 51'24'41.05" 51'24' 41.05"

Cos ~ = ...j Sec SSec (SOP) Sin (S-H) Sin (S-L) ,

Z=1OI'2T50"

~:;:53.30' A = 106'35' 34.1" Azimuth of Sun :;: 360' - 106'35' 34:1" Azimuth of Sun =253'24' 25.9" @

Azimuth of Marie 358'40' 54.7" - 253'24' 25.9"

ex. = 105'16' 28.8"

SinD Cosl= Cos LCosh ·tanLtanh

B = 106'35' 34.1" -105'16' 28.8"

B:;: 1'19' 5.2"

Note: l =true bearing ofsun from the norlh NW if observed in the afternoon NE if observed in the morning 0= 20'52' 44" L=42'29' 30" h = 43'16' 48" SinD Cosl= Cos LCosh -tanLtanh

B = 1'19' 5.2"

Azimuth of Mark =253'24' 25.9" + 1'19' 5.2" Azimuth of Mark = 254'43' 31,2"

Anobsetvati\lli W8sniadetodeterminethe

Sin 20'52' 44" Cos l = Cos 42'29' 30" Cos 43'16' SO" Cosl=-0.19875 Z = 101'27' 50"

aiimuth btthellne .Ai3. by observing the allilode ot sun in theaftemoon. The follOwing

da~were~erve(t.


®11:iil.'~$$I~

@··.p()ttlpul~ • t~~e~~Pl!~¢l!Qntirli~9f

· •·•· · • ·.~~~t~i • ~• ~~Ii~jgl~~~. · · •· l· · ·i~1 ··· . ····~ehtth~d.3~.·····»·····

Solution: CD Length of sag curve: Assume: L > S AS2 L= 122 +3.5S A =2.98 - (- 2) A =4.98 _ 4.98 (115.32f L - 122 + 3.5 (115.32) L = 126m. > 115.32 ok @

L=AI/l

Stationing oflowest point of curve:

p.

Max. speed of the car:

395 126 =4.98 I/l 395 V=100kph @

Perception reaction time: S=Vt+

I/l

2gf V= 100000

3600 V= 27.78 mls

_ (27.78)2 115.32 - 27.78 t + 2 (9.81) (0.38) t= 0.42 sec.

S1--~ g1 - g2 S - - 0.015 (141) 1 - _0.015 - 0.025

S1 =52.875 m. from P.C. Stationing of lowest point of curve =(12 + 640.22) -17.625 = 12 + 622.595

AParap6I1C$a9Ctlry~ • ti~s • ~.sijltdi$tlllJ¢~(lf 11S·.rn;•• >Th~··tllllg~llt~@cl~(jflh~29iy~~r~ ·2%alld*3%. '.' . (i) CQmplitelheIM9thl)1th¢Cu~i< . . .@).. 9Qmp~~t~~m~x·speedthata~r·t;(l91d

• l'llClVeajoMthiS:@/"i~.@Jlr > ' . •. ai1dUre$isO.1~,

@.·.ROlllPIJ~trenead.lal11PS~hlcJjsl~®e

@

Max. velocity:

AVl

L=395 A=2-(-3) A=5 _ (5) Ij2 310.98 - 395 V= 156.74kph

.••••·• ·•.•

®¢O@Jmt~the~ogth.mln~$ag¢(lt· ~)99mptttt!.thel1'lex; • •Vel~ly.ofttiEl.tiSrtl'lat· COl1/dpa$sthruthesagcurve.inkph.

180

The length of the sag CllNe having gra4Elsof and +3;5% is e(JualtQ 310 m... , ....

, 2.5%

Solution: CD Head lamp sight distance:

0.25..' .. ' .

' .....

S-382

HEAD lAMP SIGHT DISTANCE Solution: G)

Head lamp sight distance:

.;r8••1t", ~~@Ms>················································

L =2 (h + SfIJ) fh.. gl 1.2n fIJ = 180 fIJ

=0.021 rad

Solution: . G)

L=A\f2

310 =2[0.80 + S (0.021)] (0.035)· (. 0.025) 0.80 + 0.021 S =9.3 S=404.76m. @

395 A =1.8· (- 3.2)

A=5

L = (5) (156f 395

L= 284.81 m.

Max. speed ofcar:

L=AV2

395 A=3.5 • (. 2.5)

@

6V2

V= 142.86 kph Pemepfion reaction time:

V2

2gf V= 142.86 (1000)

3600 V= 39.68 mls S=VT+

V2 2gf

(39.68f 404.76 =39.68 t + 2(9.81X025) t'= 2.11 sec.

V2

2gf V= 150 (1000) 3600 V=41.67m1s _ . (41.67)2 S - 41.67 (2.1) + 2(9.81) (0.40) S= 308.76

310= 395

S=VT+

Min. safe stopping distance: S=Vt+

A=6

@

Length ofsag curve:

@

Height ofhead lamp:

.

5-383

IUD lAMP SIGHT DISTANCE ® Height of head lamp above the pavement:

L=2(h+ $0)

L>S

gz- g,

_S2 (!l2 - g,)

_ 1.1 (1t) 1Il- 180 III = 0.0192 rad 284 81 =2 [h + 308.76 (0.0192)] . 0.018 - (- 0.032) h + 5.928 = 7.120

L- 2 (Sill +h) 200 = (164.69)2 (0.02 + 0.03)

.

2[(164.69) (~':b1t + h]

164.29:.9) 1t + h = 3.39 1

h= 1.19m

h=O.80m. @

Max. speed that a car could safely travel:

A1f2L=395 A=grg, A=2-(·3) A=5

2OO=51f2395 V=125.70kph

ilitC.! ~lilllmij;···

. . .• . . ):.• • .>

li.,i'iI~(lIll. Solution: CD Head lamp sight distance:

1f2S = lit + 2g (f + G) V= 80000

3600

s-

4

Solution:

CD Length of curve ofa sight distance of 130 m. h, = 1.5 m. ~= 100mm ~=0.10m.

V= 22.22 mfs _22.22 (3) +

ii.'"liiiBIB

(22.22f 2 (9.81)(0.20.0.03)

S= 164.69m. S=Sight Distance

1----L=Lmght of Curv,-

S-384

HEAD lAMP SIGHT DISTANCE Assume sight distance is lesser than the length of CUNe: A=2.8-(-1.6) A=4.4

AS2 ---=-=-----=--100(m1+~)2

L= L=

4.4 (13Of

100 ({3 + {Qi)2 L = 156.574m. > 130 m. (ok)

® Stationing of highest point of CUNe: S _.E.1..!:1-

S

Bevation ofhighest point of CUNe: Elev. B = Elev. A- Y Bev. A = 100.94 +0.016 (56.934) Bev. A = 101.85 m. Elev. B = 101.85·0.45 Elev. B = 101.40 m.

g1

-

g2

~ 0.028 (156.574) 1 -0.028

+ 0.016

S1 = 99.64 m.

·A.·yem~J30' circular cuMtlHhe stationing of the T,S. is 10 + 000, and lhegauge of beltact Qn the curve is 1.5rn. G)

@

--

Determine -the elevation -of the outer rail at the trti&pOiot;if Ih~ velocilY¢ the fastest train to pass OVer the ctltvels60 kph. ~~~:~rilhe spiral angle at the first

® Determine !he deflection angle at the end pOint _. @ Determine the offutMrom the tangent at the

second qualterpoinl

5-395

SPIRAl CURVE Solution:

@

CD Elevation of the outer rail:

Spiral angle at the first quarterpoint: L=20m. L2 180 s=-. 2Rc LeTt _ (20)2 (180) s - 2(176.3}(80) Tt 5=0.81' , s= 0'49'

® Deflection angle at the end point: At the end point Lc =80 m. _ Lc 180 sC-2RTt c _ 80 (180) Sc - 2 (176.3) Tt Sc = 13'

i=~ 3 . 13'

1=3

i = 4,33' deflection angle at the end point.

tan"

=-W-

@

Offset from the tangent at the second quarter point:

V"

tan" =-

fT V= 1000 K 3600 V=0.278K g = 9.8 m1sec2 e'

tan" =1" = e

e =(0.278 1\)2 9.8r 0.0079 K2 e= r R= 1145.916

D R= 1145.916 6.5 _ 0.0079 (60j2 (1.5) e176.30 e =0.241 (outer railj 0.241

e=-2

e = 0.1205 (at midpoint)

L = 40 m. at the second quarter point

_iL

XC -

6Rc

L3 X=Xc L 3 c

_

(80}2 Xc - 6 (176.30) Xc =6.05 X- 6.05(40f - (80}3 X= 0.756m,

5-396

SPIWCURVE

• t!l~·\MgM~·9f·#.·.~W~f#A~··Mi • ~ij]lAA~.m·.

·~ll'I'~gil.ltltl.~[1 • II~rJI~.~I.~·~~ll~.~ • ~~·. W.P~W:fulh~ttij@g~()@itl1Pl~#jjlY~;·.··

•.• • •·• • •

IllIi• •fli Solution: ":'

.

11111 Solution: CD Centrifugal acceleration: 80 C= 75 + V 80 C= 75+ 100 C = 0.457 nv'sec3

iT.

Sin 10.4' =

7.91 S.T. =Sin 10.4' S. T. =43.82 m.

8-402

SPIRAl CURVE Solution: G)

Solution:

Offset distance on the first quarter point:

CD Length ofspiral curve:

L3 x:::-6Rc Lc 1 L :::4"(80)

xc -k 6Rc

L:::20m. Lc :::80m.

p:::2tifc

p:::&

4 L2

_

(20)3 , x- 6 (280) (80) x::: 0.06m. @

Rc

5' Rc ::: 229.18 m.

Le2 1.02::: 24 (229.18)

Length ofthrow: L2 xc:::tR;

Lc ::: 74.90m.

-~

@

6(280) xc::: 3.81 m.

Xc -

Length of throw:::

CR

0.0215 Vl 74.90::: 0.50 (229.18)

Length ofthrow ::: 4

Length ofthrow::: 0.95

Velocity of car so as not to exceed the min. centrifugal acceleration: L ::: 0.0215 Vl c

4'x 3.81

@

::: 1145.916

V::: 73.63 kph @

Length of long tangent:

Max. velocity: ::: 0.036 J(3 Lc R c

80 ::: 0.036 J(3 280 K::: 85.37 kph .

s:::~

'tlle~h~I.l'#rve-otlln~sementcu~ispn.a

l~tI~I:.£oi.~~~:.~*~~$·,~··I~~ ··~""PQffiPUte-.ttie • reqijlte~lfm9thOftlffispiral

•••· • • • ~·· • • • ?

• • • ·.H • •·••••••••••.••••••••.••••.•••••••••••••••••••••••••••••••.•

~• • P~~~@i~lh~~¢!o/()f.tb~carpa$$in9 Jh~11:!i~9\ltV~$QJh8tjtwiO.nQt~xce~d

11.".

c 2Rc 1t S ::: 74.90 (180) c 2 (229.18)(n)

Sc= 9.36' Xc :::4 (1.02)

Xc =4.08m.

tan 9.36' =~ 4.08 tan 9. 36 ' =/1 h::: 24.75 Long tangent::: 73.60 - 24.75 Long tangent::: 48.85 m.

5-403

SPIRAl CURVE

_JL

11'_111.1

the••lell9lhof$pif1llcllrVe·f$ . .

Ihepa$$~llger$·· • •

aOmdCiog.

(I) • • CPmpme.·ijj~ • Y~@)ityQt.th~.apprQl@lng ~i~l\Pl1;< ®··.·.sdttipute.ltlareq~lfe~~iu$.Qtlb~centr~J· CQN~()flM • ~Setnerltc~f\I~ • t()•• timil·the

XC - 6R e _ (80f Xc - 6(266.61) Xc =4m. Ian 8.6' =~

h=26.45

LT= 79.30·26.45 LT= 52.85m.

cenlriftlgala~letation. ... . ...........•....•.• the.len9th•• of.tM.I()n~ • t;tn~~nt« tb~.spirlIl •.8Jrve•• if.lIJe•• ~istail~ • !llong • th~

@ •.••• C?ffiPllte••

tangentfrOm'T',S, IQS,CLis 79.301Tl;1(M'IQ,.· Solution: CD Veloctiy of approaching car:

80 C=75 + V

l"aditl$6f2()Qrildilttll~~¢l.lM!l'
~' .•.

5-412

EARTHWOIIIS Solution:

Assume a level section with an average value of cut and fill for each stretch.

CD Area section f + 020:

CD Determine the volume of cut. ® Determine the volume of fill. @ If the shrinkage factor is 1.2. determine the volume borrow or waste.

,,

3.0:

:

t--- ;;i---..+--4.S---I-----4.5-1-;--;.l

- 4(4.5) (4 + 2)(4.5) A1- 2 + 2

+

Solution: Average depth of cut:

(2 + 1.5){4.5) ~ 2 + 2

A1 = 31.50 m2 ® Area of section 1 + 040: - 7800 C - 850 C=9.18m.

_~

(5 + 4)(4.5)

A2- 2 +

2

(4 + 2)(4.5)

+

2

2(4.5)

+

2

A2 = 45.75m2 @

VOlUme between stations: (A 1 + A2) L V= 2

Average depth of fill: 8500 f= 1200 f= 7.08 1-----37.54.------1

V- (31.50 + 45.75)(20)

-

V=

2

772.5 m3

Side slope =1.5 : 1 Cut In determining the position of the balance line in the profile diagram, a horizontal grade line is drawn such that the length of the cut is· 850 m. and that of fill is 1200 m. The profile area between the ground line and the grade line in the cut is 7800 sj:l.m. while that of fill is 8500 sq.m. If the road bed is 10 m. wide for cut and' 8 meters wide for fill and if the side slope for cut is 1.5 : 1 while that for fill is 2 : 1.

_ (10 + 37.54) (9.18) A2 A = 218.21 sq.m.

S-412-A

EARTHWORKS CD Volume of cut: Vc :; 218.21 (850) Vc = 185,500 cU.m.

Solution:

(1)

Slope = ~.~ = 0.016

Side slope = 2 : 1 Fill _ (8 + 36.32) (7.08) A-. 2 A = 156.89sq.m. (2)

(2)

Distance in which the fill is extended: 0.048x = 1.2 + 0.016(50 - x) 0.064x'= 2 x = 31.25

@

Stationing of the point where the fill is extended: Sta. = (7 + 110) + (31.25) Sta. = 71 + 141.25

Volume of fill:

V, = 156.89 (1200) V, = 188,000 cU.m. @

Volume of borrow: Vol. of borrow == 188000 (1.2) - 185500 Vol. of borrow =40100 cU.m.

428-A CE Board May 200 ,h>i/'C;':;>:$"''''''

"kM.',"o/.. }·'·h';'>N'

",

"..

.. ,

'..' .

Solution: q=KJ.ls q= 22(40) q = 880 vehicles per hour (rate off/ow)

Solution: Severity ratio =. . Injury + Fatal Injury + Fatal + Prop. damage _ 318+14+(x+y) 0.26 - 318 + (14 + x +y) + 1006 _ 332 + (x+ y) 0.26 - 1338 + (x + y) 347.88 + 0.26 (x + y) = 332 + (x + y) 0.74 (x + y) = 347.88 - 332 x + ;r = 21.46 say 21

HoW·m~hY.~~j¢l$$P~$lhru • ll•• Cerl~IlPointl~

Compute the tate of flow in vehicles per hour if the space mean speed is 30 mph and the

dens~y

is 14 vehicles per km.

. .

Solution: K =14 vehicles per km J.ls =30 mph _ 30 (5280) J.ls - 3.281 (1000) J.ls = 48.28 kph

q=KJ.ls q = 14 (48.28) q = 675.92vehic/eslhour

a..• hi~nW~Y.·.~V~ry • • p~yr • • • jf•• • th~ • ·.#W~itY • • • is ~.~~~w!~,.~rQ.~p~~ 'JT1~~Il."~flllf#J • • iS

.•

Solution: q=K~s _

q- ~s

:>

50000 (3.28) 5280

~=31.06mph

q=K~s

q = 48 (31.06) q= 1490

The rate of flow at a point in lhe highway is 1200 vehicles per hour. Find the space mean speed if the density is 25 vehicles per mj(e.

Solution: q = K fJs 1200 =25 fJs fJs = 48 mph

5-436

TRANSPORTAnOI ENGINEERING

t.~,. Solution:

tMav~9~ • $PE!~~~ • efv~hiote$ • • i~• a•• sl!l9!e .~jghW~y.i$§Q.m;¢l!~f~rt9PEmler· • • J'tl~··YQl~m~ {)flt!1ffj~ • ~ • QO()•• v~h[c/e:sp~r.h()ur. . •. Delel1Tl~e

1b¢W~~g~ • s~q.Qt~e • cars.• u~ll~l~i$lare

illkPO·/

... .....

Solution: 800

No. of vehicles per hour = 40 No.

of vehicles

per hour = 20 (density)

. 0 f ve h'Ie/es = 20 1000 SpaCing Spacing of vehicles = 50 m.

Density = 100(J

50

Density = 20

600

Speed of car = 20 Speed of car = 30 kph

Delerrnine.theltPproPlial~.$pa¢log.9t\lehlcl~s 9~llt~r.toc~nterin • ~·.cerlaill.lp?rtICUlarlaneis Solution: 1000

10kpr~ndtO~'loIUrneoftraffic)~890

vehiclesPerh(lUt Solut/cm:

80 = Density .

1000

Denslty=80 Density = 12.5

No. of vehicles perkm = 8~g No. of vehicles per km =20 (average density)

.

1000

Vol. offraffic = 12.5 (50)

Spacing of vehicles = 20

Vol. of traffic = 625

Spacing of vehicles = 50 m. center to center

S-437

TRANSPORTATION ENGINEERING ® Capacity of single lane in vehicles/hr:

Compute

average

the speed inkphthat a passenger car should travel hfacertaiil ,treeWq¥ if the spaclngofl~e cars moVing in the same lane is 40 m, center 10 center. Volume of traffic at this instant is ,2000

vehicles per hour.

60000 C= 15.87 C = 3781 vehicleslhr ® Average density in vehicles/km: Vel = Vehicles/hr = km . Vehicles/km fr 3781 Vel. = Density

Solution: Density = 63 vehicles/hr.•

S . f h' I 1000 pacmg 0 ve Ie es = No. of vehicleslkm

40 =

1000 No. of vehicles/km . 1000 No. of vehicles per km = 40 No. of vehicles per km = 25 ~ I

.ty vehicles/hr e OCI = vehicles/km Velocity = km/hr Velocity =2000

25

Velocity = 80 kph

. ··In•• an•• qbservaijon.p(}st•• sho~s • thaf$V~h1Cles P~s •. through.the•• postal.,nlitryal$Of8•• 9•• • sec, • • •10•..•.sec.•••·•. 11 • seP • • ~od • •·13 • • • sec. r~~p~jv~ly·ThespeedS(lf ltle ...eblcle~wre 80kph,•• 1~kph.·.70' kpb,.~(}·kp~ • anli50.kph resp'ectiveJy. . ."

sec,

@i()ornpute.thetirneW='lfj.sp~d,.............i. @.• • cgrnputEl•• I~~ • $®q~ • • m~n • • $pe~if.tM. • 91~ta~~~.1~lbY.!Wl·.¥eri¢lflsjsf5Pm· @ AtthedenSjt¥of~affi9.1s • 2(lNl~niel~§~f

lffili•• 9PrnPlll~ • • lh~ • f~tfl9f.1I9W.9ftfaffjgm· ... ... .

vehipl~hdur:

··th~ • $p~a()fa.car.movingon • asiogl~Jao~.ts 60 kph/lfthelengthoflhecari$4.2in.an~ thfl·v!lltlepfthe~tinleis(H.~,

(i)•• • ¢9t'i1@!~ • the•• = 1550(1) h2 = 1550 Longest vehicle queue =1550·650 Longest vehicle queue = 900 vehicles

G) @ @

@

~·~taIConflict$,with54 being of rear.end . . .. .

COnflict type. . ..

Average hcitmyapproach vdume =1205 vehicles. Total ti.rrte to cOllision (TIC) ~verity =190 lor the 94 conflicts.· •. Tatal risk of colUsion(ROC} $llverily . 201 for the 94 conflicts.

.=

S-446

TRANSPORTlnON ENGINEERING

Solution: CD Average hourly conflict per thousand entering vehicles (AHC): AHC _ Total no. of conflict Number of OOseNation hours 94 AHC= 40 AHC = 2.35

Solution: CD No. of crashes prevented 10 yrs. from now:

AHC per thousand entering vehicles = 2.3~J~OO) AHC per thousand entering vehicles = 1.95 ® Total conflict severity (TCS): TCS = TTC + ROC TCS =190 + 201 TCS=391

N =(EC) (CRF) Forecast ADT baseADT EC =expected no. of crashes over a specified time CRF =crash reduction factor ADT = average daily traffic N= (11) (0.26) (~)

N = 3.5 per year ® Overall average conflict severity (OACS):

res OACS = Total conflict

® Total number of crashes prevented on the 3rd year:

OACS- 391

N= (ECi (CRF) (1 + r)0

OACS=4.16

N=(11) (0.26) (1.02}3 N= 3.035

-94

® Fatal benefit after 1year: Crashes prevented = (Ee) (eR?) (1 + r)0 Crash prevented after one year

The.~~llaPltys~gil'E!e~$taff.~elieY~s.lh~t in~tinlp.·.lin intersection of M8gsaysayAv~nueandQt.iirina Avenoeandslrucklibystanderandeontinued ul'\tll.lthit•• cme•• ofthe•• colVllln.sllpport.·pf•• the

$~y, • • a~$edqn.th~dam~g~ • tgt/)~fra~tof'

.mElcar; • • me.poliser~Plll'l.e~tilil~ledt!¥ltlhe.car

wa$d()ln~.8 • • kPI"laltM•• lJ1am~nt.()flmpaCt'on

IhElg)Il!mn·.• ~4PP()rt" • • • Th~l~mglb • Oflhe••·skid ma~$VJasreeor~dt9.M.40.1ll .•••.r~~.raa(j·has

ildpWnhiU•. 9@dll,oftS%, .• • Atest@r•• ~kid(jed ·.j~Tll .• oh.thesame • ~ctipn.·Af • ltte···roal:twren

Compute the passing slght distance·for·the following data: '..' .. ' .' . Speed of the pa$sing car :::~. kph Speed Of the overtaken vehicle; 88 kph n~ ofiili.~al maneuver =4.3 sec, . Average acceleration ~ 2.37 kphlsec,

TIme passing vehicle occupies the left lane

= 10.4 sec;

Oi5la~ between

!he passing Vehicle at the .end oHIs maneuyerand the opposing" ....

thr•• b@~~·isappli~.,ff:Pll'l.a.spe® •. qi~Okphto

vehiCle:;; 76 m.

the•• h~I~, •.• Deterrnine,1hEl.prob~bJe.s~d(lfthe Car InvQlved inJhe a¢eidentwnen lhe brakes

...

Solution:

'tVel"¢aPPljoolhkPh. ....

0ppOSlfli I'(n:c"!{ tlPpeat:> \n = 2 (181.41) = 362.82m.

.'

.'.

Solution:

,"

.C(jmpO.~• • llle•• ··reqUjr~d • • $~fe· • $t6ppiN~.· • $1~ht dl~~nqElf()r··~·.fMl • 'f/~Ylt::~:lh$tQadWay;··.·.·.·. Solution:

V= 65(1000) 3600 V=81. 06 m/s.

Vl

tane=-

gr

_ (18.06)2 tan e- 9.81(100) e=18.4'

tan

0 =

0'

= 31'

0.60

5-495

TRANSPORTlnOIi EIiGINEERIIiG V2 tan (0 + 8) =fT

V2

tan 49 . 4'=fT V2 = 9. 81 (100) tan 49. 4' V= 33. 83 mls

Power=PV 4000=300 V V= 13.33 mls V= 13.33 (3600) 1000

V=48kph

V= 33.83 (3600) 1000 V= 121. 79kph.

Solution:

w

Solution:

W=1200kN P = WSin 8 + 1doo (3500)

P = jOOO (0.03) + 0.004 (1‫סס‬OO) P=340kN

Power=PV 3500 =340 V

P=F+ WSin8 P=0.005 (12000) + 12000 (0.02) p=

~M

k'N

V= 10.29mts

5-496

TRANSPORTATION ENGINEERING SIGHT DISTANCE

Metric System:

-Metric System:

s>[

----~.s----_

where: L :: length of curve in meters S :: sight distance in meters A:: g1 - g2

·«······vtA >f~95

L :: length of curve in meters S :: length of sight distance in meters V :: velocity of car that could pass thru the curve in kph.

L

V :: velocity of car that could pass thru the curve in kph.

English System

English.System ··>S.~·L··>

-----s:---,--_

where: L :: length of curve in feet S :: sight distance in feet A:: g1 - g2

«v2A L=~···~··

····...,46,50

V :: velocity of car that could pass thru the curve in mph.

L :: length qf curve in feet S :: length of sight distance infeel V:: velocity of car thaI could pass thru the curve in mph.

5-497

TRANSPORTlnOI ENGINEERING

•

I

Metric or English System

A vertical sag curve has tan£j$nt .grapes of -1.5% and + 3.5%. Compute the following to have a minimum vl$ibilityofB9m.... (j) Length of sight distance In meters.

Length of curve in meters. ® Max. velocity of a car that could pass tnru 1M vertical sag curve inkph.·

@

Solution: where: L = length of crest of vertical summit in (m) orft S =sight distance (m) or ft. h1 = height of eye of average driver above roadway (m) or ft. h2 = height of object sighted above .roadway (m) or ft. A = algebraic difference in grades in percent.

CD Sight distance: S = 2 (89) S= 178m. ® Length of curve: Assume: S< L

AS2 L= 122 + 3.5S A =3.5 + 1.5 A =5 _ 5 (178)2 L - 122 + 3.5 (178) L = 212.64 m. ok as assumed

S>t> @

Max. velocity of car:

AV2

L =395

_V2 (5) 212.64 - 395 V= 129.6 kph

Metric or English System AS2

L = 1400 S>L

A descending curve has a downward grade of - 1.4% and an upward grade of +3.6%. The length of CUM is 220 m, long.· . (j)

~~:~~~~~~~~~~s?f minimum viSibility

@

Compute Ihe max. design speed of the car passing thru the sag curve in kph. What is the stationing of the lowesl polnl of Ihe curve if the p.e. is al station 12 + 12O.60?

L = 2 S. 1400

A where: L =length of curve in feet S = stopping sight distance in feet A= g1 - g2

5-498

TRUS'ORTlnOI EIOIIEERIIO Solution: G)

Minimum visibility of curve: Assume: SL ok

vQA

.

L =46.5 (Relation of L, Vand A)

· . 5 = vQ (4) 462 . 46.5

V= 73.33 mph

A=3.5 - (-1.5) A=5 S = 182.93 (3.28) S = 600ft. _ 5(600)2 L - 400 + 3.5(600) L = 720 ft. .

vQA L= 46.50

AilR

. r ... Solution:

720 = vQ(5) 46.50 V= 81.83 mph

¢(jlllPul~treC~pa¢lW6,fa#lrigl#I~Mjn

y~hjcl~$p~rMQ(ifJM$~ElgpttM¢ar

AssumeS< L A=f12-g1 A = 3- (- 2) A=5 AS2 L=122+3.5S

_. 5 (178)2 L -122 + 3.5 (178) L = 212.64m.

m~Vil'9l1l.the.W,\gl~.I@e.~ • $olq:ib>.~~S~Clt c:arJM.8rn.WithllI'll1l¢tioo til1le O.&$@./· .

Solution: Spacing of cars = V f + L

50000

S = 3600 (0.8) +4.8 S= 15.91 m. . 'f f' I Iane= 50000 Capaclyo smge 15.91

Capacity of single Jane =3142 vehicles/hour

5-502

TUIISPORTADOI EIGINEERING

It.I.111.'d1~1 Solution: Assume: SL

AS2 L= 3000 A=3- (- 3) A=6

S =160 (3.28) S= 524.8 ft. - 6 (524.8)2 L - 3000

L =175 (3.28) L =574 ft. A=2-(-2) A=4

3000 L=2S-A 3000 574=2S-

4

L=550.8 ft.

S = 662ft.

L= 167.94m.

S=201.83m.

tfnd•• ~h~.M#iMSi9ht9i$~~(,'tJ.9fa.·ggq.·m·

f)~lerflli~ • ft)~sig~t4i$~Ilt.e • ()f~>vertigal

.IQI'Ig·.,!~~lqll,t • • ~~m@t~H~ • ~yI119 • • ~'19$1lt g@dAAQf±2%an->:

=~

•

With dowels or tie ban: Purpose of dowel is to transmit the stresses due to the load from the adjacent pavement. At the edge of pavement:

92. - g1 n=-r _0.6 - (-1.2) n- 0.18 n = 10 stations

Length of curve = 10(20) Length of ClINe = 200 m.

~

M=-

2

6(~X

f--- 2x t1 2 t1

_f3W ='4 2i

5-506

TUlSPDITln.1 EIG.IEEI.I. At the center ofthe pavement: ~

M=T 6(~) f= 2x t22 t2 =

Cracks

. Cracks

~ (thickness at thecente~

By ratio and proportion: A1 A2

-'=(t+rl W

rr? T -;2= (t+r)2 1W

(t+~=:nT t+

r=o.564-vf

t =0.564 t1 =

-vf.

r

0.5641f- r

P T= K10 91O S P = wheel load S = sUbgrade pressure K = constant value from table

5-507

TIlANSPORTATION ENGINEEBIIIG

t _ expansion pressure - average pavement density t = thickness of pavement

contact area oftire

~D

Subgrade

_ (EB)1/3

SF. - E

p

EB = modulus of elasticity ofsubgrade

Ep = modulus of elasticity of pavement

w

SF. = stiffness factor

A d=-._..:..:...._D-E- (D·A) .

F d = bulk sp.gr. of core A= weight ofdry specimen in air [) = weight of specimen plus paraffin

_. ~[1.75

t- " W

GBR·

1-]1/2 rm

t =thicknessofpavemenUn em. W =wheel load in kg CBR = California Bearing Ratio p = tire pressure in kglcm 2

coating in air E = weight of specimen pIus paraffin coating in water F = bulk specific gravity of paraffin .

5·508

TRANSPORTAnON ENGINEERING

G=-WL Pc Pf

-+Gc Gf

G = absolute sp.gr. of composite aggregates

IG·r/\

V=~x100

G

v= percentage of voids G = theoritical or absolute sp.gr. d =bulk sp.gr.

Pc =percentages of course material by wt. Gc = sp.gr. of course material Pf = percentages of fine materials by weight Gf = sp.gr. of fine material

G-d n=-

G

n =porosity G= absolute sp.gr.

Gsb = bulk sp.gr. of total aggregate P1 = percentage of total weight of coarse aggregate . Pz = percentage of total weight of fine aggregate d1 =bulk sp.gr. of coarse aggregate d2 = bulk sp.gr. affine aggregate

d = bulk sp.gr.

d=~ wa - Ww d =bulk sp.gr. wa =weight of specimen in air Ww = weight of specimen in water

Gse =effective sp.gr. of aggregate Pmm = total loose mixture Gmm = max. sp.gr. of paving mixture Pb = asphaff (percentage by total weight) Gb = sp.gr. of asphalt

5-509

TRANSPORTATION ENGINEERING

Gse - GSb) Pba= 100 ( G G Gb sb se Pbe = asphalt absorption Gse = effective sp.gr. of aggregate Gsb = bulk sp.gr. of aggregate Gb = sp.gr. of asphalt

A flexible pavement carries a static wheel load of 53.5kN. The Circular contact area of thelirt'l Is 85806 mmhmd the trartsmiltedfoad is dIstributed acro$sa wide area of the subgrade at an angle of 45'. The sUbgrade bearing valuei$ 0.14 MPa, while that of the base Is 0,41 MPa. [)esign the thickness of pavement and that of the base. '. .

Solution:

Pbe

PbaPs

=Pb • 100"

Pbe = effective asphalt content Pb =% weight of fine aggregates Ps = sum of % weight of fine and coarse aggregates Pba =asphalt,absorption

VA = air voids Gmm = max. sp.gr. ofpaving mixture G-mIJ= bUlksp.gr. ofcoJIlfl~te!imix._ ...

VMA = 100 _ Gmb Ps Gsb VMA = percentage of voids in mineral aggregates Gmb = bulk sp.gr. of compacted mix Gsb = bulk -'p.gr. ofaggregates Ps =sum of %weight of fine and course aggregates

Flexible Pavement: A -W 1 - f1

A =nr2 A A1 -;2 = (t + ,2)

n,2

W1f1

7= (t+ 1)2

5-510

mlSPIRTlng EIIGIIIEERIIG

(t+r)2:0,564~ t =0,564

~ 5;,~~O

At the edge: (with dowels) r

-165

W 2 6M f -b~

M=-x

t= 184 mm

6(~X

A=nr2

f=-2 x /1 2

85806 =nr2 r= 165 mm

t1=0.564~r . t1 =0.564

-V 535M . 165 ~

t1 = 39,mm (thickness of pavement) t2 = t - t1 t2 = 184 - 39 t2 '= 145 mm (thickness of base)

t1

=-{iff /r'i

t ='" 3-(53-50-0-) .1 2(1.38) t1 = 241 mm

At the center:

M=~x 4

6M f= bd2

f- 6 (wf4)x - 2xt22

A.liQid.PilVelTleNiSto~!!p~m¢..fb,'e pavement: t =0.564 t =0.564

-{f

-v

1m r

54000 0.15 -165 l-3.5

t= 173.4mm

® Expansion pressure method: t = expansion pressure Ave. density of pavement t = 0.50 0.05 t=10cm. t= 100mm.

nDL x Bond Stress

~.

...-...-...-

...----------L

p

5-513

TRANSPORTATIOII ENGINEERING Consider one meter length ofslab. W= 0.18 (3.5)(1X2400) W=1512kg=N F=IJN F = 1.5(1512) F= 2268 kg

dLy

Solution:

•J.'.'. , ". , :."•. :. .

B /

-'_':-."'-';".'~.~ :.:~.-" .. .

:

. rf~·::···::· .

••.•

w

As fs = F As (1600) = 2268 As = 1.42 sq.cmlmeter

AS=~(1.6f As = 1.256 sq.cm.

. 1256 Spacmg = 1.42

Consider only half of the section (Using Principles of Mechanics)

Spacing = 0.88 mUse 80 em on centers

W = (~)

Note: The length ofbarmust be at least twice the computed value.

W=720L kg. N= 729L kg.

Length ofbars: As fs = (n DL) (Bondstress) 1.42 (1600) =n (1.6X24)L L = 18.83 say 19 mm ' Use L = 2(19) L=3Bcm

F=fJN F= 1.5(720L) F= 1080L kg.

i'1••'W

21ookg/cu,m.AlIoW~~leJeri~ll~stf~6f concrete•• isO.8.kgfcm2and.l&ato/ste~liS800

k9lCmZ·•• • unitwelght··9f:.*~llsT50q.kgtcu,m st~~lbars.~a¥jng • a•• diam~terof.1.~. @ ··Compple•• th~.P9ro~flY • 1:)flhe·.~mp'We~ • • specimen. . ..

p A

m

Stress=-

3.83 = 75.3

'!!:.cf2 4

Solution:

D=5cm.

CD Absolute sp.gr. of the bituminous mixture:

G

100

Es..ElEa + + Gs Gf Ga 100

ti¢.ITIBute•• tMll)odqI4S!ir·.~Wti¢ily.t)f.lhe

• $Ub'

grad~ifth~1'Jl99\:1lw;otellJ§ticityofJhe ·pav~Iii~~rf$·1~OMR~·WitliK$fitf#e[~]aCior()r

0.50("

.

...

® Bulk specific gravity:

d=~

Wa-Ww 1140 d= 1140 - 645

Solution: E 1/3 Stiffness factor = (r)

E

0.5 =(120)

d= 2.303

p

1/3

_b.

0.125 -120

Es =15 MPa

~.

@

Porosity: Porosity =(G - ~ 100

P

·t - (2.368 - 2.303)(100)

oroSI Y-

Porosity =2.74%

2.368

5-518

TUNSPORTATION OGINEERING

A plant mix is to be made usfnQlhe foil. percentages by weight Of !he lQtal mile •.,

••

··~ ~;~~~lc:etlti'e

••

$P'jf·•• • 0f•• •

th~'

(i)••·•••lf.thflteSI•.mePilT1~riw..~h$· •.t1~O • gr,ln.n~lti9D i$~9009'ThElYOluw~ptag~r~gate$ ~~q9dillg • ·tr~ \'pllJrlle•• pt@sprbed • • \',I!1ter•• i~

730Ctilh . • .

. '.' .

.'.

(i) • • • C9nJPYl~th~~pp!1rE!tlt.$Repiftc.graVlty • qf JtJ~saMPI~"lg9reg

@.•. .• (j.••..••.•. .•.•.••.,,:":::::-::::-;-:

PetSonalinjury=a.. ... . . Property damage only ~·1 ...

. ...

.

30 mph X

.

year at apartfcular slte, compute its severity

Solution: Severity number =12 (1) +3 (3) + 1 (5) Severity number = 26

The density of traffic in acerta:in obsElwation

point Oil a highway was recorded to be 30 .vehlcles per km. IfthEl space mean speed·of Ihe vehicle is 50 kph, haw many vehk:les will be passing every 30 seconds. Solution: K=!L fJ.s

30 =!L 50 q = 1500 vehicles/hr. . 1500 (30) No. of vehIcles per sec = 3600 No. of vehicles per sec = 12.5 vehicles

40 mph

45 mph

45 mph

• • • . •

IY

___.Vir.ctWn ofjlow

lfon~ fataLcrash,Spersonal trijuriesand 5 property damage crashes occurted durifjga nUmbeI"•..

.

15' 20' 30' 20' 35' 20' 85' 20' 45'

..

Fat~ity =12

..••.... .

l..----.----

~

300'

® CQmputethe~ty¢t~fflc. @iqoMW\Eltl1efiffi6.M~~spee~·

@Computethespa~mean~peed·

Solution: sil'l~lfl

d£!allr()r.lIlNW.~@$h·j~·~\1iVa~~tt9·.~ • prmmW ' .'. . .' d®lageCta~Ms;

&>•• • IUsobselYe~lhat4p~r~Ms~ttedon

.•..• • • ~ • • 1],?mIJ~~I¥#1~Il •. ~f:.a.h!9hW~Ylrjmle •.

.·.• .y~~r .• • •. Th~ .• aver~ll.~~¥Jr~ffl~.(APn.·Qn· '.' ·the.sec!i&iwas·5000vehicles) De.lerinine . .•• t~~ • • rat~·.qttQt~I~~~t#l$ • eer1pq.Il1111lo0 . yehjples!'I'11~,(BMVMtJ//" .. .

@ • 1l1ereilr~

• ~ • crashe$.·9ccurrlng.·i"a • 2q rnilesf.lclFqrt • of~ • • ~j9'HW~y.in .•. 9~~ • • Yf.llir. The.aVf.lrage•• daiIY•.• traffl9.·.on•• m~ $eCtlon was·6000vehj(:les;•• • ()etElrrninem~.rate • of • • fatal.crash~s • • mllll()r•• Ve~lcle • " milEls, • • if•.• S%·.·.9f .·ffi e. • ·¢rashf.lll·•. ir\VQlyed fatalitles,(RM\lMF).· .

•

per·1qq••

@ ·CalciJjaliitheiral¥icba$~lnYehiclespE!f

ktrl(ll3@
$~.

~jl.illli

2

Average delay per vehicle = 10 min. @

Longest time any vehicle spent in the queue:

,, ,, ,,,

@ .'Wfl