Curves
October 2, 2022 | Author: Anonymous | Category: N/A
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INTENSE SANITARY ENGINEERING REVIEW CENTER ENGINEERING SURVEYS
SIMPLE CURVE -
A circular arc extending from one tangent to the next.
Elements of Simple Curve: PC = = Point of curvature. It is the beginning of curve. = Point of tangency. It is the end of curve. PT = PI = = Point of intersection of the tangents. Also called vertex T = = Tangent distance from PC to to PI and and from PI to to PT R = Radius of simple curve, or o r simply radius. C = = Length of chord from PC to to PT to PT . Lc = Length of curve from PC to E = = External distance, the nearest distance from PI to to the curve. m = Middle ordinate, the distance from midpoint of curve to midpoint of chord. I = == Angle intersection x = offset of distance from tangent to the curve. Note: x is is perpendicular to T . between PI and and any point in the curve θ = offset angle subtended at PC between D = Degree of curve: English system, one station station = 100 ft SI, one station = 20 m
Formulas:
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Engineering Surveys | Prepared by: Engr. Michael Delos Santos 1
INTENSE SANITARY ENGINEERING REVIEW CENTER ENGINEERING SURVEYS
Problems: 1. A simple curve has a central angle of 36° and a degree of curve 6°. a. Find the nearest distance from the midpoint of the curve to the point of intersection of tangents. 9.83m b. Compute the distance from the midpoint of the curve to the midpoint of the long chord joining the point of curvature and point of tangency. 9.35m c. If the stationing of the point of curvature is at 10+020, compute the stationing of a point on the curve which intersects with the line making a deflection angle of 8° with the tangent through the PC. 10+073.33 2. The tangent distance of a 3° simple curve is only half its radius. a. Compute the angle of intersection in the curve. 53.13° b. Compute the length of the curve. 354.20m c. Compute the area of fillet of the curve. 5304.04 sqm. 3. If the radius of simple curve is R, the length of the chord for calculating offsets by the method of chords produced, should not exceed, a. R/10 b. R/15 R/20 c. R/20 d. R/25 4. For a curve of radius 100m and a normal cord 10m, the Rankine’s deflection angle is is °25’15” b. 0°35’15” 0°35’15” c. 1°25’53” 1°25’53” d. 2°51’53” 2°51’53” a. 0°25’15” 5. The offset distance of the simple curve from the PT to the tangent line passing through the PC is equal to 120.2m. The stationing of PC is at 2+540.26. The simple curve has an angle of intersection of 50°. a. Compute the degree of curve. 3°24’ b. Compute the external distance. 34.79m c. Compute the length of long chord. 28.41m 6. A simple curve has a radius of 286.48m. Its distance from PC t PT along the curve is equal to 240m. a. Compute the central angle of the curve. 48° b. Compute the distance from the midpoint of the long chord to the midpoint of the curve. 24.76m c. Compute the area bounded by the tangents and the portion outside the central curve in Acres. 0.53 acre 7. For setting out a simple curve, using two theodolites, a. Offsets from tangents are required. b. Offsets from chord produced are required. c. Offsets from long chord are required. d. Deflection angles from Rankine's formula are required. e. None of these.
Engineering Surveys | Prepared by: Engr. Michael Delos Santos 2
INTENSE SANITARY ENGINEERING REVIEW CENTER ENGINEERING SURVEYS COMPOUND CURVE -
Consists of 2 or more circular curves having different radius, but whose centers lie on the same side of the curve, likewise any two consecutive curves must have a common tangent at their meeting point.
Elements (in addition to shown in simple curve) of Compound Curve: R1 = Radius of curve 1 R2 = Radius of curve 2 T1 = Tangent distance of curve 1 T2 = Tangent distance of curve 2 I1 = Intersection angle of curve 1 I2 = Intersection angle of curve 2 I = Angle of Intersection of the compound curve V1V2 = Length of Common Tangent = T 1 + T2 PCC = Point of intersection of tangents of the compound curve Problems: 1. The common tangent AB of a compound curve is 76.42m with an azimuth of 268°30’. The vertex is being inaccessible. The azimuth of the tangents AV and VB was measured m easured to be 247°50’ and 282°50’, respectively. If the stationing of A is 43+010.46 and the degree of degree of the first curve was fixed at 4° based on the 20m chord. Using chord basis, a. Determine the stationing of the PC. 42+958.21 b. Determine the stationing of the PCC. 43+061.55 c. Determine the stationing of the PT. 43+109.65 2. The long chord from the PC to the PT of a compound curve is 300m long and the angle it makes with the longer and shorter tangents are 12° and 15°, respectively. If the common tangent is parallel to the long chord, cho rd, a. Find the radius of the first curve. 802.36m b. Find the radius of the second curve. 514.55m Engineering Surveys | Prepared by: Engr. Michael Delos Santos 3
INTENSE SANITARY ENGINEERING REVIEW CENTER ENGINEERING SURVEYS c. If station of PC is 10+204.30, find the stationing of PT. 10+507.06 3. A compound curve has a common tangent of 84.5m long which makes angles of 16° and 20° with the tangents of the first curve and the second curve, respectively. The length of the tangent of the first curve is 38.6m. What is the radius of the second curve? a. 260.3m 260.3m b. 145.2m c. 250.3m d. 301.6m 4. The common tangent of a compound curve makes an angle of 14° and 20 ° with the tangent of the first curve and the second curve, respectively. The length of chord from the PC to PCC is is 73.5m and that from the PCC to PT is 51.3m. a. Find the length of the chord from PC to PT if it is parallel to the common tangent. 124m b. Find the radius of the first curve. 301.55m c. Find the radius of the second curve. 147.71m
REVERSED CURVES -
Formed by two simple curves having a common tangent but lies on opposite sides.
Elements (in addition to shown in compound curve) of Reversed Curve: PRC = Point of Intersection of the Reversed curve Problems: 1. Two parallel tangents 10m apart are connected by a reversed curve. The chord length from the PC to the PT equals 120m. a. Compute the length of tangent with common direction. 60.17m b. Determine the equal radius of the reversed curve. 359.78m c. Compute the stationing of the PRC if the stationing of A at the beginning of the tangent with common direction is 3+420. 3+449.99 2. In a railroad layout, the centerline of two parallel tracks is connected with a reversed curve of unequal radii. The central angle of the first curve is 16° and the distance between parallel tracks is 27.6m. Stationing of the PC is 15+420 and the radius of the second curve is 290m. a. Compute the length of the long chord from PC to PT. 198.31m b. Compute the radius of the first curve. 422.47m c. Compute the stationing of PT. 15+618.96 Engineering Surveys | Prepared by: Engr. Michael Delos Santos 4
INTENSE SANITARY ENGINEERING REVIEW CENTER ENGINEERING SURVEYS SPIRAL CURVES
Elements of Spiral Curves: = Tangent to spiral TS = SC = = Spiral to curve CS = = Curve to spiral ST = = Spiral to tangent LT = = Long tangent ST = = Short tangent R = Radius of simple curve T s = Spiral tangent distance T c = Circular curve tangent L = Length of spiral from TS to to any point along the spiral Ls = Length of spiral PI = = Point of intersection I = = Angle of intersection I c = Angle of intersection of the simple curve p = Length of throw or the distance from tangent that the circular curve has been offset X = = Offset distance (right angle distance) from tangent to any point on the spiral X c = Offset distance (right angle distance) from tangent to SC = Distance along tangent to any point on the spiral Y = Y c = Distance along tangent from TS to to point at right angle to SC E s = External distance of the simple curve θ = Spiral angle from tangent to any point on the spiral θs = Spiral angle from tangent to SC i = Deflection angle from TS to to any point on o n the spiral, it is proportional to the square of its distance is = Deflection angle from TS to to SC Engineering Surveys | Prepared by: Engr. Michael Delos Santos 5
INTENSE SANITARY ENGINEERING REVIEW CENTER ENGINEERING SURVEYS D = Degree of spiral curve at any point Dc = Degree of simple curve
Formulas:
External Distance, Problems:
1. The tangent of a spiral curve has azimuths of 226° and 221°, respectively. The minimum length of spiral is 40m with a minimum superelevation of 0.10m/m width of roadway. The maximum velocity to pass over the curve is 70 kph. Assume width of roadway to be 9m. a. Determine the degree of simple curve. 5.85° b. Determine the length of spiral at each end of simple curve. 60m c. Determine the superelevation of the first 10m from SC on the spiral. 0.765m 2. A simple curve having a radius of 280m connects two tangents intersecting at an angle of 50°. It is to be replaced by another curve having 80m spirals at its ends such that the points of tangency shall be the same. a. Determine the radius of the new circular curve. 192.84m b. Determine the new central angle. 26.26° c. Determine the deflection angle at the end of the spiral. 3.96° d. Determine the offset from tangent at the end of the spiral. 5.53m e. Determine the distance along the tangent at the midpoint of the spiral. 39.99m 3. The tangents of a spiral curve forms I=25° at station 2+058. Design speed is 80 kph. Radius of central curve is 300m and the length of spiral is 52.1m. a. Find the stationing at the point where the spiral starts. 1+965.36 b. Find the stationing of the start of central curve. 2+017.46 c. Find the length of the central curve. cu rve. 78.8m
Engineering Surveys | Prepared by: Engr. Michael Delos Santos 6
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