Eng2Cg20survey1

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P P a a g g e e 1 1

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P P a a g g e e 2 2 E

Ellemementents s of of SiSimmplple e CCurvurvee

T

Thhe e ffoolllloowwiinng g aarre e tthhe e eelleemmeenntts s ffoouunnd d iin n a a sisimmpplle e ccuurrvvee::

P

P..C. C. ==PPooiint nt oof f CurCurvvaattururee R R = = RaRaddiius us oof f CurCurvvee

P

P..TT. . = = PPooiint nt oof f TTaangngeencncyy T T = = TTaangngeent nt DDiissttaancncee

P P..II. . = = PPooiint nt oof f IIntnteerrsseeccttiioonn D D = = DDeeggrreee e of of CCururvvee T Taannggeennt t II//2 2 = = TT//RR 1 1.T .T = = R R ttaannggeennt t II//22 22.. E E = = R R ((SSeec c II//2 2 – – II)) E E = = R R SSeec c II//2 2 – – RR M M = = R R – – R R ccoos s II//22 E

E = = extexternernaal l ddiiststanancece M = M = mmiiddddlle e oorrddiinnaattee 33. . M M = = R(R(1 1 – – CoCos s II//22)) LC LC = = LengtLength h of of CCuurrvve e SSiin n II/2 /2 = = CC/2R/2R 4 4. . C C = = 22R R SSiin n II/2/2 55. . LLcc//I I = = 2200//DD Lc Lc = = 220 0 II/D/D 6 6. . Lc/I Lc/I = = 101000/D/D Lc Lc = = 10100 0 II/D/D

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P P a a g g e e 2 2 E

Ellemementents s of of SiSimmplple e CCurvurvee

T

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P

P..C. C. ==PPooiint nt oof f CurCurvvaattururee R R = = RaRaddiius us oof f CurCurvvee

P

P..TT. . = = PPooiint nt oof f TTaangngeencncyy T T = = TTaangngeent nt DDiissttaancncee

P P..II. . = = PPooiint nt oof f IIntnteerrsseeccttiioonn D D = = DDeeggrreee e of of CCururvvee T Taannggeennt t II//2 2 = = TT//RR 1 1.T .T = = R R ttaannggeennt t II//22 22.. E E = = R R ((SSeec c II//2 2 – – II)) E E = = R R SSeec c II//2 2 – – RR M M = = R R – – R R ccoos s II//22 E

E = = extexternernaal l ddiiststanancece M = M = mmiiddddlle e oorrddiinnaattee 33. . M M = = R(R(1 1 – – CoCos s II//22)) LC LC = = LengtLength h of of CCuurrvve e SSiin n II/2 /2 = = CC/2R/2R 4 4. . C C = = 22R R SSiin n II/2/2 55. . LLcc//I I = = 2200//DD Lc Lc = = 220 0 II/D/D 6 6. . Lc/I Lc/I = = 101000/D/D Lc Lc = = 10100 0 II/D/D

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P P a a g g e e 3 3 FFiigugurre e 11. . A A SSiimmpplle e CCuurvrvee

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The deﬂection angles of two intermediate points A and B of a highway curve are 415’ ⁰ and 915’ ⁰ respectively. The chord distance between points A and B is 20.00 m. while the long chord is 120.00 m. Stationing of P.I. is 80 + 060. Find thestationingofP.C.andP.T.

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The bearing of the back tangent of a simple curve is N70°00’E while the forward tangent has a bearing of S82°30’E. the degree of curve is 4.5°. Stationing of PC is at 10+345.43. It is proposed to decrease the central angle by changing the direction of the forward tangent by an angle of 7°00’, in such a way that the position of the PT of the forward tangent and the direction of the back tangent shall remain the unchanged. Determine:

a)The new radius of the curve b)Stationing of new PC.

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WORKSHEET

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Two tangents intersecting at V with bearings N7512’⁰ E and S7836’⁰ E are connected with a 4 si⁰ mple curve. Without changing the direction of the two tangents and with the same angle of intersection, it is required to shorten the curve to 100.00 m. starting from the P.C.

a)By how much shall the P.T. be moved and in what direction? b)What is the distance between the two parallel tangents?

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It is required to layout a simple curve by deﬂection angles. The curve is to connect two tangents with an intersection angle of 32 and ⁰ a radius of 800 ft. Compute the deﬂection angles to each full stations on the curve, if the transit is set up at the P.C. which is at station 25 + 57.2. What is the station of P.T.?

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LESSON

2 –

COMPOUND

CURVE

Compound Curve consists of two or more consecutive simple curves having diﬀerent radius, but whose centers lie on the same side of the curve. In a compound curve, the point of the common tangent where the two curves join is called the point of compound curvature (PCC). Shown in Figure 2 are the elements of a Compound Curve.

Elements of Compound Curve

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P a g e 1 0 Example:

The long chord of a compound curve is 425.00 meters long and the angles that it makes with tangents of the curve were 20⁰ and 24⁰ respectively.

Find the radius R1 and R2 of the compound curve if the common tangent is paralleltothelongchord.

C1 / Sin 12 = ⁰ 425 / Sin 158⁰ C1 = 235.88 m. C2 / Sin 10 = ⁰ 425 / Sin 158⁰ C2= 197.01 m. Sin 10 = ⁰ 235.88 / 2R1 R1 = 679.15 m. Sin 12⁰ = C2 / 2R2 R2 = 473.78 m.

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P a g e

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COMPOUND

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The highway engineer of a certain road construction decided to use a radius of 100.00 m in laying out a simple curve having an angle of intersection of 3620’⁰ . The stationing of the vertex is 30 + 375.20 after verifying the actual conditions of the proposed route, it was found out that the PT should be moved out in a parallel tangent having a perpendicular distance of 10.00 meters with an angle of intersection remaining the same while the curve shall have the same PC. Determine:

1.The radius of the new curve.

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P a g e

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COMPOUND

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Two tangents intersect at Station 25 + 50. A compound curve laid on their tangents has the following data.

I1= 31⁰00’ I2= 36⁰00’ D1= 3⁰ 00’ D2= 5⁰ 00’ a)Compute the Stationing of the PC, PCC and PT of the curve.

b)If the PT is move 50.00 ft. out, compute the station of the PT with the PCC on the same point.

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P a g e

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COMPOUND

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A common tangent of a compound curve makes an angle with the tangents of the compound curve of 25⁰30’ and 30⁰00’ respectively. The

stationing of A of 10 + 362.42. The degree of curve of the ﬁrst curve is 4⁰30’

while that of the second curve is 5⁰00’. It is required to change this compound

curve with a simple curve that shall end at the same PT while the direction of the tangents remains the same. Find the radius of this simple curve and the stationing of the new PC.

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P a g e

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WORKSHEET

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COMPOUND

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Given a compound curve with a long chord equal to 135.00 meters forming an angle of 12⁰00’ and 18⁰00’ respectively with the tangents. The

common tangent is parallel to the long chord. Determine the radii of the compound curve.

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P a g e

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LESSON

3 –

REVERSED

CURVE

This kind of curve is formed by two circular simple curve with common tangents but lies on opposite side. Reversed curve is useful in laying out pipelines, ﬂumes, levees, and low speed roads and railroads. In canals, it is used with tremendous cautions since it makes the canal diﬃcult to navigate and contribute to erosion.

Elements of A Reversed Curve:

Figure 3. Reversed Curve

R1 and R2 =radiiofcurvature

D1an D2 = degee of curve

V1 and V2 =pointsofintersectionoftangents

θ = angle between converging tangents

I2– I1= θ

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P a g e 1 6 P.T.=pointoftangency

P.R.C. = Point of Reversed Curve

Lc = Lc1 + Lc2 = Length of reversed curve

P = distance between parallel tangents

Four types of reversed curve problems are:

1.Reversed curve with equal radii and parallel tangents. 2.Reversed curve with unequal radii and parallel tangents 3.Reversed curve with equal radii and converging tangents. 4.Reversed curve with unequal radii and converging tangents.

Method of Laying out

The method of latying out simple curve is applied. At the point where the curve reversed in its direction is called the Point of Reversed Curvature (PRC). After this point has been laid out from the P.C. the instrument is then

trnasferred to this point. With transit at P.R.C. and a reading equal to the total deﬂection anglefrom theP.C.totheP.R.C.,theP.C.isbacksighted.Iftheline of sight is rotated about the vertical axis untl the horizontal reading become zero, this line of sight falls on the common tangent. The next simple curve could be laid out on the opposite side of the tangent by deﬂection angle method.

Example:

From the ﬁgure shown, the two diverging tangents were connected

by a reversed curve with both arcs having a 5˚ curve. Determine the

Station of P.I. if I angle is 41˚, Determine also the Station of P.T. if Ts is

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P a g e 1 7 Solution:

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P P a a g g e e 1 1 8 8 P. P.II. . = = 220 0 + + 4400..330 0 + + 22,,775588..225 5 = = 47 47 + + 9988..5555 P. P.T. T. = = 447 7 + + 9988..555 5 – – 55550 0 = = 442 2 + + 4848..5555

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Thhe e ppeerrppeennddiiccuullaar r didissttaanncce e bbeettwweeeen n ttwwo o ppaarraalllleel l ttaannggeenntts s iis s eeqquuaal l ttoo 8

8..000 0 mmeteterers, s, cencenttrral al ananglgle e tto o 8˚8˚0000’’0000” ” anand d tthhe e rradadiiuus s of of cucurvrvatatuurre e of of tthhe e ﬁﬁrrstst cu

currvve e equequal al tto o 17175.5.00 m00 meetterers. s. FFiinnd d tthhe e rradadiiuus s of of tthhe e sesecond cond cucurrvve e of of tthhee r

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P P a a g g e e 1 1 9 9

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Twwo o ppaarraalllleel l rraaiillwwaay y 220000..000 0 mmeetteerrs s aappaarrt t wweerre e tto o bbe e ccoonnnneecctteed d bby y eeqquuaall t

tuurnrnououtts. s. IIf f tthhe e iintntermermedediiatate e ttanangengent t iis s 440000..000 0 metmeterers s anand d tthhe e rradadiiuus s of of cucurverve i

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P P a a g g e e 2 2 0 0

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Thhe e ppeerrppeennddiiccuullaar r didissttaanncce e bbeettwweeeen n ttwwo o ppaarraalllleel l ttaannggeenntts s oof f a a rreevveerrsseedd cu

currvve e iis s 3535..00 00 mmeetterers. s. ThThe e azaziimmuutth of h of tthhe e babacck k ttanangegent of nt of tthhe e cucurrvve e iiss 2

27700˚˚0000’’0000” ” anand d tthhe e aziazimmuutth h of of tthhe e comcommmon on ttangenangent t iis s 303000˚˚0000’’0000””. . IIf f tthe he rradadiiuuss of

of tthhe e bback ack curvcurve e iis s 115500..000 0 mmeteterers s anand d tthhe e ststatatiiononiing ng of of tthhe e P.P.R.R.CC. . iis s 110 0 + + 114400,, ﬁ

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REVERSED

CURVE

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A reversed curve connects two converging tangents intersecting at an angleof30˚00’00”.Thedistanceofthisintersectionfrom theP.I.ofthecurveis 150.00 meters. The deﬂection angle of the common tangent from the back tangent is 20˚00’00”R, and the azimuth of the common tangent is 320˚00’00”. The degree of curve of the second simle curve is 6˚00’00” and the stationing of

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P a g e

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the point of intersection of the ﬁrst curve is 4 + 450. Determine the stationing oftheP.C.,theP.R.C,andtheP.T.

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REVERSED

CURVE

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Two converging tangents have azimuth of 330˚00’00” and 90˚00’00” resectively while that of the common tangent is 350˚00’00”. The distance from

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the point intersection of the tangents to the P.I. of the second curve is 160.00 meterswhilethestationingoftheP.I.oftheﬁrstcurveisat10+432.24.Ifthe radius of the ﬁrst curve is 285.40 meters, determine the stationing of P.R.C and P.T.

LESSON

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SYMMETRI

CAL

PARABOLI

C

CURVES

Symmetrical parabolic curve is a vertical parabolic curve wherein the horizontal length of the curve from the PC to thevertexisequaltothe horizontal length from

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By Ratio and Proportion:

2) Using the squared property of parabola:

3) Location of Highest or Lowest Point of the Curve:

a) From the P.C. b) From the P.T.

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On a railroad a -0.80% grade meets a +0.40% grade at station 90+00 whose elevation is 100.00 ft. The maximum allowable change in grade per

station having a length of 100 ft. is o.20. It is desired to place a culvert to drain the ﬂood waters during heavy downpour. Where must be the location of the culvert? At what elevation must the invert of the culvert be set if the pipe has a diameter of 3.00 ft. and the backﬁll is 1.00 ft. high. Neglect the thickness ofthepipe. Figure: Solution: Length per station = 100ft. r=rateofchangeperstation Elevation of P.T. = 100 + 0.04(300) = 101.2 ft. AC = 101.8 – 100 = 101.8 ft. AB = BC = 0.90 ft. n = 6 stations H = 0.90 ft. L = 6(100) = 600 ft S1 = 400 ft. Stationing of P.C. = 90 + 00 – 300 = 87 + 00 Stationing of P.T. = 90 + 00 + 300 = 93 + 00

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P a g e 2 6 Elevation of P.C. = 100 + 0.008(300) = 102.4

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SYMMETRI

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PARABOLI

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A Symmetrical parabolic curve is designed to pass through point A at station 4 +50whoseelevation is76.20ft.TheP.C.ofthisparaboliccurveisatstation 2 + 75 whose elevation is 74 ft. The length of the parabolic curve is 400 ft. long

having a backward tangent grade of + 2.5%. It is required to determine the amount of cut and ﬁll at stations 2 + 75, 3 + 00, 4 + 00, 5 + 00, 6 + 00, and 6 +

75 if the ground elevations are as follows:

Station Ground Elevation Station Ground Elevation 2 + 75 74.00 5 + 00 76.925

3 + 00 75.881 6 + 00 75.825 4 + 00 76.815 6 + 75 70.940

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SYMMETRI

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A symmetrical parabolic curve connects two grades of + 6% and -4%. It is to pass through a point p the stationing of which is 35 + 280 and the elevation is 198.13 meters. If the elevation of the grade intersection is 200 meters with stationing 35 + 300 determine:

a)The length of the curve

b)Stationing and elevations of P.C. and P.T. c)The location of the highest point of the curve. d)Elevation of station 35 + 260 on the curve

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SYMMETRI

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A horizontal laid circular pipe culvert having an elevation of its top to be 85.26 ft. crosses the right angles under a proposed 400 ft parabolic curve. The pointofintersection ofthegradelinesisatstation12+80anditselevation is 88.50 ft while the culvert is located at station 13 + 20. The backward tangent has a grade of + 3% and the grade of the forward tangent is – 1.6%. Under this conditions, what will be the depth of cover over the pipe?

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SYMMETRI

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An underpass road crossing a reinforced concrete bridge along the Shaw Blvd. has a downward grade of – 4% meeting an upward grade of + 8% at the vertex V (elevation 70.00 m) at station 7 + 700, exactly underneath the center

line of the bridge having a width of 10.00 meters. If the required minimum clearance under the bridge is 5.00 meters and the elevation of the bottom of the bridge is 78.10 meters, determine the following:

a)Length of the vertical parabolic curve that shall connect the two tangents.

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A vertical highway curve is to pass through a railroad at grade. The crossing must be at Station 64 + 50 and at an elevation 724.00 ft. The initial grade of the highway is + 2% and meets a – 3% grade at station 62 + 00 at an elevation of 732.40 ft. The rate of change must not exceed 1% per station.

a)What length of curve will meet the condition?

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LESSON

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UNSYMMETRI

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PARABOLI

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CURVES

•Consist of a symmetrical parabolic curve from PC to PT. A,B

another symmetrical parabolic curve tangent to that point A and PT

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P a g e 3 2 Example: Given: g1= 7% g2 = -4% L1=160m L2 = 120m Required: a.Elevationofthe

Curve of the underpass

b. L2 if elevation of curve is 22.683m. c. Stationing of the HP of the curve for

question b.

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UNSYMMETRI

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PARABOLI

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An unsymmetrical parabolic curve has a forward tangent of -8% and a bacward tangent of +5%. The length of the curve on the left side of the curve is 40.00 meterslongwhilethatoftherightsideis60.00meterslong.IfP.C.isStation 6

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+ 780 and the elevation is 110.00 meters a) determine the height of ﬁll at the outcrop, b) Determine the height of curve at Sta. 6 + 820.

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UNSYMMETRI

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A forward tangent having a slope of -4% intersects the back tangent having a slope +7% at point V at Station 6 + 300 having an elevation of 230.00 meters. It is required to connect the two tangents with an unsemmetrical parabolic curve that shall pass through point A on the curve having an elevation of 227.57 meters at station 6 + 270. The length of curve is 60.00 meters on the side of the back tangent. A) It is required to determine the length of the curve on the side of the forward tangent. B) Determine the stationing and elevation of the highest point of the curve.

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P a g e 3 6

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In a certain road construction undertaken by DPWH it was decided to connect a forward tangent of 3% and a back tangent of -5% by 200 meter symmetrical parabolic curve. It was discovered that the grade intersection at 10 + 100 whose elevation is 100.00 meters fall on a rocky section with the exposed boulder at elevation 102.67 meters. To avoid rockky excavation, the project engineer decided to adjust the vertical arabolic curve in such a way that the curve will just clear the rock without altering the position of P.C. and the grade of the tangents. Determine the stationing and elevation of new P.T.

(38)

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A forward tangent of +6% was designed to intersect a back tangent of

-3% at a proposed underpass along EDSA so as to maintain a minimum clearance allowed under a bridge which crosses perpendicular to the underpass. A 200.00 meters curve lies on the side of the back tangent while a 100.00 meters curve lies on the side of the froward tangent. The stationing and elevation of the grade intersection is 12 + 530.20 meters and 100.00 meters respectively. The centerline of the bridge fall at station 12 + 575.20. The elevation of the underside of the bridge is 117.48 meters. Determine the minimum clearance of the bridge if it has a width of 10.00 meters.

(39)

P a g e 3 8

LESSON

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SI

GHT

DI

STANCE

Non-Passsing Sight Distance

Non Passing sight distance is the safe stopping distance of a vehicle running at design speed.

(40)

P a g e 3 9 Passing Sight Distance

Passing sight distance is the distance required to overtake safely another moving vehicle in the same traﬃc lane.

(41)

P a g e 4 0

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A vertical summit curve has a tangent grades of 2.8% and -1.6%. A motorist whom eye sight is 4.80 ft. above the road way sighted the top of a visible object 4.20 high at the right side of the summit. Calculate the length of

(42)

P a g e 4 1

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A grade ascending at the rate of 5% meets another grade descending at the rate of 4% at the vertex of elevation 20.00 m. and stationing 5 + 000. Solve fot the stationing and elevation of the summit of the vertical parabolic curve which will connect the grade lines for a safe distance of 150.00 m., the height

of the eyes of the drivers above the pavement at each end of the sight distance being 1.50 meters

(43)

P a g e

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LESSON

7 –

PASSI

NG SI

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FOR

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AT

UNDERPASS

(44)

P a g e

4 3

Using the previous relation of the parabolic curves. S > L

Where:

S = length of passing sight distance L = length of curve

h1 =heightofdriver’seye h2 =heightofobject

C = vertical clearance from the lowest point of under pass to curve

(45)

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4 4

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In a certain underpass, the vertical clearance of the 240.00 meter arabolic curve with P.C. at station 13 + 000 and elevation 30.00 meters is 14.50 ft. The height of the object at the instant of perception is 3.50 ft. while that of the driver’s eye is 4.50 ft. If the approach grade is -4% and the passing sight distance is 320.00 meters, what is the grade of the forward tangent? At

(46)

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4 5

what point (station and elevation) of the curve should the catch basin be installed?

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The vertical clearance of the parabolic sag curve of the newly constructed Balintawak Underpass is to be determined if the maximum height of the driver’s eye that could utilize such underpass measured from the pavement is

(47)

P a g e

4 6

4.50 ft., while that of the object at the instant of perception is 3.50 ft. The length of the parabolic curve is 1,152.00 ft. and that of the passing sight distance is 1,100.00 ft. The designed grade at the back tangent is -5% while the forward tangent has a designed grade of +3%.

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(48)

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A highway designed for speeds of 60 mph, is being constructed over a hill with a 3% ascendinhg grade and a 2% descending grade. The oint of intersection of the two grades is at elevation 100.00 ft. and at that station the elevation of the ground is 95.00 ft. What will be the depth of cut at the point where the two grades intersect if the vertical curve used is designed for a safe

passing sight distance of 2,100.00 ft. Height of observer’s eye from the pavement is 4.50 ft. and that of the object is also 4.50 ft. above the pavement.

(49)

P a g e

4 8

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A vertical parabolic summit curve was designed in order to have a clear sight distance of 120.00 meters. The grade lines intersect a station 9 + 100 at elevation 160.50 meters. The curve was so designed such that when the height of the driover’s eye is 4.50 ft. above the pavement it could just see an object whose height is 4.20 inches above the pavement. Determine then the maximum

speed that a car could travel along this curve. The grade lines has an upward grade of +5% and a downward grade of -3%.

(50)

P a g e

4 9

LESSON

8 –

SPI

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CURVE

A spiral curve is a transition curve inserted between a circular curve and the tangent to the curve. It is a curve of varying radius used to gradually

increase the curvature of the road or railroad used primarily to reduce

skidding and steering diﬃculties by gradual transition between straight-line and turning motion. It provides a method for adequately for super elevating curves.

(51)

P a g e 5 0 Elements of Spiral

TS = the point of change from tangent to spiral

SC = the point of change from spiral to circular curve CS = the point of change from circular curve to spiral ST = the point of change from spiral to tangent

(52)

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The symbols PC and PT, TS and ST, and SC and CS become transposed when the direction of stationing is changed.

a = the angle between the tangent at the TS and the chord from the TS to any point on the spiral A = the angle between the tangent at the TS and the chord from the TS to the SC

b = the angle at any point on the spiral between the tangent at that point and the chord from the TS  = the angle at the SC between the chord from the TS and the tangent at the SC

c = the chord from any point on the spiral to the TS C = the chord from the TS to the SC

d = the degree of curve at any point on the spiral ! = the degree of curve of the circular arc

f = the angle between any chord of the spiral "calculated when necessary# and the tangent through the TS $ = the angle of the deflection between initial and final tangents% the total central angle of the circular curve and spirals

& = the increase in degree of curve per station on the spiral

' = the length of the spiral in feet from the TS to any given point on the spiral

's = the length of the spiral in feet from the TS to the SC, measured in () e*ual chords

o = the ordinate of the offsetted PC% the distance between the tangent and a parallel tangent to the offsetted curve

r = the radius of the osculating circle at any given point of the spiral + = the radius of the central circular curve

s = the length of the spiral in stations from the TS to any given point S = the length of the spiral in stations from the TS to the SC

u = the distance on the tangent from the TS to the intersection with a tangent through any given point on the spiral

(53)

P a g e

5 2

 = the distance on the tangent from the TS to the intersection with a tangent through the SC% the longer spiral tangent

v = the distance on the tangent through any given point from that point to the intersection with the tangent through the TS

- = the distance on the tangent through the SC from the SC to the intersection with the tangent through the TS% the shorter spiral tangent

 = the tangent distance from the TS to any point on the spiral / = the tangent distance from the TS to the SC

y = the tangent offset of any point on the spiral 0 = the tangent offset of the SC

1 = the tangent distance from the TS to the offsetted PC "1 = /23, approimately#

= the central angle of the spiral from the TS to any given point = the central angle of the whole spiral

Ts = the tangent distance of the spiraled curve% distance from TS to P$, the point of intersection of tangents

4s = the eternal distance of the offsetted curve

Spiral Formulas

The following formulas are for the eact determination of the functions of the ()5chord spiral when the central angle, , does not eceed 67 degrees. These are suitable for the compilation of tables and for accurate fieldwor&.

"(# "3# "8#

"6# "7# A = " 28# 5 ).39: 8_seconds

(54)

P a g e 5 3 "4sec = ( Tan  " # "># / = C Cos A "9# 0 = C Sin A "()# "((# "(3# "(8# 1 = / 5 "+ Sin # "(6# o = 0 5 "+ -ers # "-ers = ( 5 Cos # "(7# Ts= "+ < o# Tan " $# < 1 "(;# 4s = "+ < o# 4sec " $# < o

"4sec " $# = Tan " $#"Tan"? $##

Empirical Formulas

@or use in the field, the following formulas are sufficiently accurate for practical purposes when does not eceed (7 degrees.

a = 28 "degrees# A = 28 "degrees# a = () &s3 _"minutes# _{S = () &S}3_"minutes# Spiral Lengths

!ifferent factors must be ta&en into account when calculating spiral lengths for highwa y and railroad layout.

Highways. Spirals applied to highway layout must be long enough to permit the effects of centrifugal force to be ade*uately compensated for by proper superelevation. The minimum transition spiral length for any degree of curvature and design speed is obtained from the the relationship 's = (.;-82+, in which 's is the minimum spiral length in feet, - is the design speed in miles per hour, and + is the radius of curvature of the simple curve. This e*uation is not

mathematically eact but an approimation based on years of observation and road tests. Table ( is compiled from the above e*uation for multiples of 7) feet. hen spirals are inserted between the arcs of a compound curve, use 's = (.;-82+ a. + a represents the radius of a curve of a degree e*ual to the difference in degrees of curvature of the circular arcs.

(55)

P a g e

5 4

Railroads Spirals applied to railroad layout must be long enough to permit an increase in

superelevation not eceeding ( ? inches per second for the maimum speed of train operation. The minimum length is determined from the e*uation 's = (.(: 4-. 4 is the full theoretical superelevation of the curve in inches, - is the speed in miles per hour, and 's is the spiral length in feet.

This length of spiral provides the best riding conditions by maintaining the desired relationship between the amount of superelevation and the degree of curvature. The degree of curvature

increases uniformly throughout the length of the spiral. The same e*u ation is used to compute the length of a spiral between the arcs of a compound curve. $n such a case, 4 is the difference between the superelevations of the two circular arcs.

Spiral elements are readily computed from the formulas given above. To use these formulas, certain data must be &nown. These data are normally obtained from location plans or by field measurements.

(56)

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5 5

The tangents of a spiral curve has azimuth of 226˚ and 221˚ respectively. The minimum length of the spiral is 40.00 meters. with a minimum super

elevation of 0.10 m/m width of roadway. The maximum velocity to pass over the curve is 70.00 km/hr. Assume width of the roadway to be 9.00 meters.

a)Determine the degree of simple curve.

b)Determine the length of the spiral at each of the simple curve. c)The super elevation of the ﬁrst 10.00 meters from the S.C. on the

spiral. Use e = 0.004 K2/R Solution: a) e = 0.004 K2/R 0.10 = 0.004(70)2/R R = 196.00 meters D = 1145.916/R D = 1145.916/196.00 D = 5.85˚ b) Lc = 0.0036K3/R Lc = 0.0036(70)3/196.00

Lc = 63.00 meters say 60.00 meter (use multiple of 10 m.)

c# e( = "(2;# ").()# = ).)(: m. "B () m. from TS on the Spiral# e3 = 7").)(:# = ).)>7 m. "B () m. from SC on the Spiral#

e = ).)>7"9# = ).:;7 m. "super elevation at () m. from SC on the spiral#

WORKSHEET

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A simple curve having a radius of 280.00 meters connects two tangents intersecting at an angle of 50˚00’. It is to be replaced by another curve having 80.00 meters spirals at ends such that the point of tangency shall be the same.

a)Determine the radius of new circular curve

b)Determine the distance that the curve will move nearer the vertex. c)Determine the central angle of the circular curve.

d)Determine the deﬂection angle at the end of the spiral.

(57)

P a g e

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f) Determine the distance along the tangent at the midpoint of the spiral.

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The two tangents of a simple curve has azimuths of 270˚00’ and 10˚00’ respectively. It has a radius of 320.00 meters. It is required to change this curve to a spiral curve that will have value of p = 2.50 meters and b = 30.00 meters as shown in the ﬁgure. Determine the distance on which the new curve must be moved from the vertex and its distance from T.S. to the P.C. of the simplecurve,ifDE isparalleltoh.

(58)

P a g e 5 7

LESSON

9 –

EARTHWORKS

EARTHWORKS – the construction of large open cuttings or excavations involving both cutting and ﬁlling of material other than rock.

(59)

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5 8

EXCAVATION – is the process of loosening and removing earth or rock from itsoriginalposition inacutandtransportingittoaﬁllortoawastedeposit.

EMBANKMENT – the term embankment describes the ﬁll added above the low points along the roadway to raise the level to the bottom of the pavement structurematerial for embankment commonly comes from roadway cuts or designated borrow areas

a.By Average End Areas

V = L/2 (A1+ A2)

Where:

V = Volume of Section of Earthworks between Sta. 1 and 2, m³ A1, A2 = Cross sectional area of end stations, m²

L = Perpendicular Distance between the end station, m

NOTES:

1.The above volume formula is exact only when A1= A2 but is approximate A1<> A2.

2.Considering the facts that cross-sections are usually a considerable distance apart and that minor inequalities in the surface of the earth between sections are not considered, the method of end areas is suﬃciently precise for ordinary earthwork.

3.By where heavy cuts or ﬁlls occur on sharp curves. The computed volume of earthwork may be corrected for curvature out of ordinarily the correction is not large enough to be considered

4.A.By Prismoidal Formula V = L/6 (A1+ AM + A2)

(60)

P a g e 5 9 Where:

V = Volume of section of earthwork between Sta 1 and 2 of volume of prismoid,m³

A1, A2 = cross – sectional area of end sections, m²

AM = Area of mid section parallel to the end sections and which will be computedas the averages of respective end dimensions, m³

NOTES:

1.A Prismoid is a solid having for its two ends any dissimilar parallel plane ﬁgures of the same number of sides, and all the sides of the solid plane ﬁgures. Also, any prismoid may be resolve into prisms, pyramids and wedges, having a common altitudes the perpendicular distance between the two parallel end plane cross – section.

2.As far as volume of earthworks are concerned, the use of Prismoidal formula is justiﬁed only if cross-section are taken at short intervals, is a small surface deviations are observed, and if the areas of successive cross-section cliﬀ or widely usually it yields smaller values than those computed from average end areas

PRISMOIDAL CORRECTION FORMULA

CD= L/12 (b1– b2)(h – 1 h2) Where:

CD =PrismoidalCorrection,Itissubtractedalgebraicallyfrom the volume as determined by the average and the areas method

to give the more nearly correct volume as determined by the Prismoidal formula, m³

(61)

P a g e

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L = Perpendicular distance between 2 parallel and sections, m b1 = Distance between slope stakes at end section ABC where the

altitudeish1, m

b2 = Distance between slope stake at end section DEF where the altitudeish2, m

h1= Altitude of end section ABC at Sta 1, m h2 = Altitude of end section DEF at Sta 2, m PROBLEM:

1.Given the following cross-section notes of a roadway with a base of 6m and SS of 1.25:1.00, between the volume of the prismoid between the two-end sections by the following methods: a) end area method; b) Prism oidal form ul;c)end area m ethod and

prism oidal correction form ua.

Station Cross Section Notes

10 + 00 +6.55 +2.84 +2.84 +6.55 +2.84 10 + 20 +7.55 +3.64 +1.85 +3.65 +0.52

SOLUTION:

Compute for the area at each station cross-section and at mid-section Figure

Check for Cut distances

DR1= DL1 = B / 2 + SHR= ½ (6m) + 1.25(2.84) = 6.55m

Area by method of triangle and rhombus A1= BC + SC² = 27.12m²

(62)

P a g e 6 1 Check for the distances DR2 = B/2 +SSHR2HR2= ½ (6) + 1.25(0.52)= 3.65m DL2 = B/2 + SHL2= ½ (6) + 1.25(3.64) = 7.55m Area by method of triangle

A2= A A + AL+ Ac+ Ad

= ½ (3)(3.64) + ½ (1.85)(7.55) + ½ (1.85)(3.65) + ½ (0.52)(3) A2 = 16.60m²

Compute for the dimensions of the mid sections

Figure DRm = ½ (DR1+ DR2) HRm= ½ (HR1+ HR2) = ½ (6.55 + 3.65) = ½ (2.84 + 0.52) DRm = 5.10 m. HRm = 1.68 m. DLm = ½ (DL1+ DL2) HLm = ½ (HL1+ HL2) = ½ (6.55 + 7.55) = ½ (2.84 + 3.64) DLm = 7.05 HLm= 3.24 HCm= 1 / 2 (HC1+ HC2)

(63)

P a g e 6 2 = 1 / 2 (2.84 + 1.85) HCm = 2.345m

Check for Cut distances DRm = B / 2 SHRm = 1 / 2 (6) + 1.25(1.68) DRm= 5.10m DLm = B / 2 SHLm = 1 / 2 (6) + 1.25(3.22) DLm = 7.05m

Area by method of triangle Am = Ae+ Af+ Ag+ Ah

= ½ (3)(3.24) + ½ (7.05)(2.345) + ½ (5.10)(2.345) + ½ (3)(1.68) Am = 21.68m

COMPUTE FOR THE VOLUME OF EARTHWORK VOLUME OF CUT IN BETWEEN THE TWO STATIONS

Figure 1. By End Area Method Ve = L/2 (A1+ A2) Where: L = (10 + 020) – (10 + 000)= 20m A1= 27.12m² A2 = 16.60m² Then, Ve= 20/2 (27.12 + 16.60)= 437.20m² 2. By Prismoidal Formula

Vp = L/ 6 (A1+ 4Am + A2) Where:

L = 20m; A1 = 27.12m²; A2 = 16.60m²; Am = 21.67m² Then

(64)

P a g e 6 3 Vp = 20/6 [27.12 + 4(21.67) + 16.60] = 434.13m³ 1.Prismoidal Formula for Correction

Cp = L/2 (A1+ A2)(b1– b2)

Note:Resolve the given prismoid into a series of triangular prismoid into a seriesoftriangularprismoid. Cp= Cpa+ Cpb+ Cpc+ Ppd Where: Cpa = 20/12 (2.84 – 3.64)(3-3) = 0 Cpb = 20/12 (2.84 – 1.85)(6.55 – 7.55) = -1.65m³ Cpc = 20/12 (2.84 – 1.85)(6.55 – 3.65) = 4.785m³ Ppd = 20/12 (2.84 – 0.52)(3-3) = 0 Then Cp = -1.65 + 4.785= 3.135m³ 2.Corrected Volum e Vc= Ve- Cp = 437.20 – 3.135 Vc = 434.065m

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Derive the prismoidal correction formula for a triangular end areas using the formula V = L/6 (A1+ AM + A2)

(65)

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Derive the prismoidal formula for determining volumes of regular solid. V = L/6 (A1+ AM + A2)

(66)

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Using the prismoidal correction formula, ﬁnd the corrected volume of cut between stations 80.00 meters apart if the area of irregular sections in cut at stations are 26.00 sq.m. and 68.00 sq.m. respectively. Base width = 8.00 meters, side slope 1:1.

(67)

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The side slope of a railroad cut shown is 1:1. The width of the roadway is 10.00 meters. Determine the corrected volume by applying the prismoidal correction. Distance between sections is 100.00 meters.

(68)

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Given the cross section notes below of the ground which will be excavated for a roadway, compute the volume of excavation between stations 47 + 00 and 48 + 00 by: a) end area method; and b) prismoidal formula. The road is 30.00 ft. wide with slopes of 1.5:1

(69)

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From the following cross section notes, compute the corrected volumes of cut and fill, the road bed being 20.00 ft. wide in cut and 16.00 ft. in fill. The sideslopeforbothcutandfillis1:1.Givetheresultsincubicyard.

(70)

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The following cross section notes are for a road passing a hilly country. The roads bed is 11.00 meters wide for thorough cut, 10.00 meters for side hill and 9.00 meters for fill. The slope of the cut is 1:1 and 1.5:1 for fill. Find the volume of cut and fill by the end area method.