Chap 3 Curve (PDF)

UNIT THREE CURVE 3.1 General Introduction to Curve Route Surveying includes all the surveying & mapping activates (the field and office work) required to plan, design, and layout (construct) and any "Long and narrow" transportation facility. This transportation facility could be road (high way), ra...

UNIT THREE CURVE 3.1 General Introduction to Curve Route Surveying includes all the surveying & mapping activates (the field and office work) required to plan, design, and layout (construct) and any "Long and narrow" transportation facility. This transportation facility could be road (high way), railways, pipeline, and power transmission lines. The "Shape" or geometry of any transportation route is called its alignment. This includes both its horizontal alignment (i.e. a plan view), and its vertical alignment (i.e., a profile view). The vertical alignment is also called the grade line. The centerline of highway, railroad, or canal consists of series of straight lines, (tangents) connected by curves for safe and comfortable ride. The plan view should include: Bearing of the tangents Geometric data for each horizontal curve Angle of intersection Topography adjacent to the center line Stationing Existing structures affected by the project The vertical alignment consists of straights (gradients) connected by vertical curves in a vertical plane. The profile view should include: tations (chainages) The existing ground surface Location of drainage structures Proposed route grade line Elevations of existing ground surface & Grades (slopes) of the tangents proposed grade lines Vertical curve data 1 3.2 Types of Curves and Their Uses Curves are provided in the line of communication like roads, railways, canals etc. to bring about the change of direction gradually. The two types of curves are horizontal curve and vertical curve. Fig. 3- 1: Types of Curves Horizontal Curve Horizontal curves are curves that are used to connect straight line called tangent in horizontal plane. The curves employed normally are circular although spiral curves may be used to provide gradual transitions to or from the circular curves. Their design & their construction are considered in the horizontal plane. The sharpness of circular curve may be described by their radius (R) or their degree of curve (D0). Therefore curves are designated either by either of them. E.g. a 750m curve or a 20 curve The followings are the types of horizontal curve 3.3 Simple Circular Curve A Simple Curve is a circular arc joining two intersecting tangents. The radius of the circle determines the sharpness or flatness of the curve. Elements of the simple circular curve Vertex (V) - the point of intersection (PI) of two intersecting tangents. Point of curvature (PC) - the point of tangency where the curve leaves the tangent. Point of tangency (PT) - the point of tangency where the curve meat the other tangent. Tangent distance (T) - the distance from the vertex to the PC or PT. Intersection angle (I) or - the angle by which the forward tangent deflects from the back tangent Radius (R) - the radius of the circle of which the curve is made. External distance (E) is the distance from the vertex to the mid- point of the circular curve. Long chord (C) - is the distance from the point of curvature (PC) to the point of tangency (PT). Middle Ordinate (M) -is the distance from the mid-point of the long chord to the mid- point of the circular curve. Degrees of curve (D0) - Curves are designated either by their radius (R) or their degree of curve (D0). 2  The degree of curvature (D0) is defined as the angle subtended at the center of the circle by an arc of 20m (Curve definition) D0 is the central angle subtended at the chord length of 20m (Chord basis). Length of curve (L) - length of the curve from PC to PT. Note: the sharpness of circular curve may be described by its radius or by the degree of curve (D0) subtending & standard arc length as defined above. Fig. 3- 2: Simple Circular Curve Important Relationship in circular curve (refer Fig. 3- 2: Simple Circular Curve) 1. Radius of curve (R) 6. The length of the curve (L) a) Arc basis By arc definition - the actual length of the D0 360 20  360 1145. 916 curve   R   m 20 2 R 2 D 0 D0 D0  20     L  b) Chord basis 20 L D0 D 0 10 10 By the chord definition - the total length as sin( )   R  2 R D 0 measured along the chords of an sin( ) 2 inscribed polygon. 2. Tangent distance (T) - take ∆PCOPI  T  6. Sub - Angle (d) - is the central angle. Less tan( )   T  R tan( ) 2 R 2 than degree of curve (D0) subtended by an 3. Length of long chord (C) - take PC-PT arc length less than 20m or chord length C  2  sin( 2 )  R  C  2 R sin( 2 ) less than 20m. 4. The external distance (E) - consider ∆PCOPI Arc definition,  R    D 0 d D 0  l cos( )   E  R  sec( )  1   d  2 E  R  2  20 l 20 5. The Middle ordinate (M) Chord definition,  R  M    lc cos( )   M  R  1  cos( ) d 2 d 2 R  2  sin( )   l c  2 R sin( ) 2 R 2 l  d  2 sin  1 ( c ) 2R Example: Two straight lines AB &BC intersect at chainage 10+020, the intersecting angle being 1400. It is desired to connect these two tangents by a simple C/ curve of 40. Calculate the elements of the curve & determine the chainage of the tangent PT. (arc basis) 3  Solution 1145. 916 1145. 916 20   R    286. 5 m D0 40 L  200.00m  D0 T  R tan( )  286. 48 * tan( 40 )  104. 28 Chainage of PI  10  020 2 2 C  2 R sin(  )  195. 98 m  T  0  104. 28 m 2  Chainage of PC  9  915.72   E  R  sec( )  1   18. 39 m  2   L  0  200. 00    M  R  1  cos( )   17. 28 m Chainage of PT  10  115.72  2  3.4 Compound Circular Curve It consists of two or more simple curves. It’s setting out involves setting out of two or more simple curves of different radii in continuation. (see Fig. 3- 3)  = total deflection angle 1 = deflection angle IMN  2 = deflection angle MNI Rs = radius of arc (small) T1C RL = radius of arc (large) T2C. Ts = total tangent length small), T1I Fig. 3- 3: Compound Circular Curve TL = total tangent length (large), T2I sin  2 Total Tangent Ts, Ts  t1  t1  t 2  Ts  T1M  MI = sin  sin 1 Total Tangent TL, = TL  t 2  t1  t2  TL  T2 L  LI sin  3.5 Reverse Circular Curve Reverse circular curve Consist of two consecutive circular curves. It may or may not have the same radii i.e.R1=R2 or R1≠R2 in Fig. 3- 4. The centres of the curve lie on opposite side of common tangent. 4  Fig. 3- 4: Reverse Circular Curve 3.6 Spiral Curve or Transition Curve Transition Curves are curves in which the radius changes from infinity to a particular value. The effect of this is to gradually increase the radial force P from zero to its highest value & thereby reduce its effect. For a Vehicle traveling from PC to PT, the force gradually increases from zero to its maximum on the circular curve and then decreases to zero again. This greatly reduces the tendency to skid & reduces the discomfort experienced by passengers in the vehicles. This is one of the purposes of transition curves; by introducing the radial force gradually and uniformly and uniformly they minimize passenger discomfort. If the transition curve is to introduce the radial force in a gradual & uniform manner it must have the property that the product of the radius of curvature at any point on the curve and the length of the curve up to that point is a constant value. K= RLT R- radius of the circular curve LT- length of transition curve Another purpose of transition curves is to gradually introduce super elevation Fig. 3- 5: Spiral Curve Clothoid (Euler Spiral) is the type spiral curve that the radius of the curvature decreases in proportion to the length i.e. the radius varies from infinity at that tangent end of the spiral to the radius of the circular arc at the circular curve end.  i  l i2 ( The basic equation) r a d ia n 2 R LT Where i =the angle between the tangent to the transition curve at i and the entry straight. li = distance along the transition curve of any point ( i ) from the point of tangency. LT = length of the transition curve. R = radius of curvature at the junction of the transition curve and simple curve. 5  LT At the end of the transition when l = L   radians 2R 3.7 Vertical Curve In the same way as horizontal curves are used to connect intersecting straights in the horizontal plane, vertical curves are used to connect intersecting straights (gradients) in the vertical plane. Gradients: these are usually expressed as percentages, for example, 1 in 50 = 2 percent 1 in 25 = 4 percent Fig. 3- 6: Vertical Curve [Crest or Valley) CREST CURVES: - Which can also be referred to as a summit or hogging curve, is one for which the algebraic difference is negative when calculated from right to left. Fig. 3- 7: Crest Curve SAG CURVES. A sag curve (valley or sagging) curve is one for which the algebraic difference of the gradients is positive when calculated from right to left. Fig. 3- 8: Sag/Valley curve A. Purposes of vertical curves There are two general purposes of Vertical curves: Adequate visibility (Sight distance)- For a vehicle to stop or to overtake safely, it is essential that on coming or any obstructions in the road can be seen clearly at good time. Passenger comfort & safety- As the vehicle travels along the curve a radial force acts on the vehicle in the vertical plan. 4 TYPES OF VERTICAL CURVES Vertical curves are usually parabolic since parabola curves provide constant rate of 6  change of curvature. I. Symmetrical parabola curve The curve lengths from PVC to PVI and from PVI to PVT are equal. Fig. 3- 9: Symmetrical Parabola Curve 5 Unsymmetrical parabolic curve L1 and l2 are not equal. Fig. 3- 10: Unsymmetrical parabolic curve 6 Elements of Parabolic Vertical Curve Vertex (V) - the point of intersection of the grade line (PVI) point of V. Curvature (PVC) - The pt of tangency where the parabolic V. curve leaves initial grade. Point of Vertical tangency (PVT)- The pt of tangency where the parabolic V. curve meets the forward grade. Length of vertical curve (L)- the horizontal distance b/n PVC to PVT. 7 Equations of symmetrical parabolic vertical curve Fig. 3- 11: symmetrical parabolic equation Let X and Y be the coordinates of point p on the curve w.r.t. point A. X= the vertical offset (m) from the vertex (PVI) to the middle of the curve. Y= the vertical offset (m) from the tangent to any point on the curve. 7 Method I Based On the Rate of Change of Grade (R) EPC = EPVC + g1X +(r/2)X2 Where, x = the distance of a point on curve from PVC g1 =the initial grade per station EPC= elevation of pts on curve EPT= elevation on tangent. EPVC= elevation of PV Method II Based on the rule of offsets The entire curve may be established by offsets from the initial tangent or the lst half may be referred to the 1st & the 2nd half to the second tangent  x  2 Where y offset from the tangent to the curve y  4e    L  Elevation On tangent (EPT) = EPVC +g1x Elevation On Curve (EPC) + EPT + y Where x = the distance of a point on curve from PVC L= the length of curve g1 =the initial grade per station EPC= elevation of pts on curve EPT= elevation on tangent. EPVC= elevation of PVC e = the offset from PVI to curve Example A vertical curve is to connect two tangents that intersect at station 5 + 000 & elevation 500m. The back tangent gradient is - 4 percent, the forward tangent gradient is 2 percent. The length of the length of the curve is 160m. setup a table that shows elevations at PVC,PVT and at full stations along the curve. The V.C. is symmetrical parabolic curve. Solution Station of PVI = 5+000 _____________________ -L/2 = -0+080 PVT = 5+080 ___________________ PVC = 4+920 Elevation of PVC = EPVI g1 L/2 + L = 0+ 160 8  4 160  2  4    160  =5000 + *  800  100 2 6 *160   1.2 = 500 +3.2 800 = 503.20m EPC  5 0 3.2 0  80  2   4 1.2   2 20 100   16   5 0 3.2 0  0. 8 0  0.0 7 5 Elevation On tangent  5 0 2.4 7 5 m EPT = EPVC + g1x  x  2 E P c 40  E P V C  g 1 x  4 e  Elevation on curve = Elevation on tangent + y  L   EPC = EPT +y 4e is c o n s ta n t fo r a cu rve 2 L 2 x 4 * 1.2  1.8 7 5 * 1 0 4 = EPVC + g1 x + 4e   1 6 0  2 L 4 4 2 at x= 20m, (at full station = 4+ 940) EPC 40  5 0 3.2 0  * 4 0  1.8 7 5 * 1 0 4 0  100 2  4   20   5 0 1.9 0 m EPC= 503. 20 +   20   4e    100   160  4 2 EPC 60  5 0 3.2 0  * 6 0  1.8 7 5 * 1 0 4 6 0  g 2  g1 100 but e  L  5 0 1.4 7 5 m 8  2 4  4 2    1 6 0 EPC 80  5 0 3.2 0  * 80    8 0  100 100  100 8  5 0 1.2 0 m EPC 100  5 0 1.0 7 5 EPC 120  5 0 1.1 0 0 EPC 140  5 0 1.2 7 5 EPC 160  5 0 1.6 0 0 EPT Station X in m Offset Y EPC Remark (EPVC+g1x) 4+920 0 503.20 0 503 PVC +940 20 502.40 +0.075 502.475 +960 40 501.60 +0.30 501.90 +980 60 500.80 +0.675 501.475 5+000 80 500.00 +1.20 501.20 +020 100 4.99.20 +1.875 501.075 +040 120 498.40 +2.70 501.100 +060 140 497.60 +3.675 501.275 +080 160 496.80 +4.80 501.600 PVT Example2: An upward 3% gradient meets a 2% downward gradient at station 1000.00 and elevation of 350m. If a vertical curve of length 200m is designed for the smooth transition of the gradients at 20m peg interval. Then determine: a. equation of the vertical curve b. elevations of the beginning of the vertical curve (PVC) c. elevations of the end of the curve (PVT): d. elevations of the 1st two stations after PVC and last two station before PVT 9  Solution: Where a = = , b= , C= Elevation of PVC and L=Length of the vertical curve which implies the distance between end points of the curve PVC and PVT. a. =0.03, =-0.02, L=200m which implies =100m.. = =-0.000125,b= =0.03,C=Elev. of PVC= − * =350-0.03*100=347m ∗ Then equation of the vertical curve = Y= -0.000125 +0.03x+347 b. Elevation of beginning point the PVC y(x=0)=347m c. Elevation of PVT y(x=200/20=10sta)= -0.00012510 +0.03*10+347=347.3m. d. Elevation of the curve at x=20m=1sta,x=40m=2sta after PVC are: y(1)= -0.0001251 +0.03*1+347=347.03, y(2)=-0.000125*22+0.03*2+347=347.05 Elevation of the curve at x=180m=9sta and x=160m=8sta before PVT are: y(9)= -0.000125∗ 9 +0.03*9+347=347.25, y(8)=-0.000125*82+0.03*8+347=347.23 3.8 Methods of Setting out (Horizontal Alignments) Prior to construction, it is normal practice for the Surveyor to prepare a detailed centre line survey. This centre line survey should normally consist of stakes located every 20 metres on straights and every 5 to 10 metres along curves. A mark is placed on each of these stakes defining the distance (up or down) to the finished formation level of the road surface. The followings are some of the method used to set out curves Deflection Angle Method (Angular Method) The method requires one theodolite and a tape or two theodolites. The formula used for the deflection angles is derived as follows. The angle formed between the back tangent and a line from PC to a point on the curve is the deflection angle to the point. This deflection angle, measured at the PC between the tangent and the line to the point, is one-half the central angle subtended between the PC and the point. From the geometry of circles, the angle b/n a tangent to a circular curve and a chord drown from that point of tangency to some other point on the curve equals one-half of the central angle subtended by the chord. Thus in fig. above the central angle d1 b/n the Pc and the first station  on the curves d1, and the deflection angle at PC is 1/2 d1. Similarly, the central angle b/n the PC and the second station  on the curve is d1+ D. and the deflection angle is 1/2 (d1 +D). The deflection angle to PT = 1/2 (d1+D+D+… +d2) = D/2 (This provides a check on the calculation. 10  Fig. 3- 12: Deflection Angle Central Angle & chord to first curve station. Let the difference in stationing b/n PC & the first full station on D 0 d1 the curve is C1. Then by direct proportioning, . 20 c1 The actual chord length b/n PC & the first full station on the curve is: C1 D 0  c1 d1 d d1  ; sin( )  2  C1  2 R sin( 1 ) 20 2 R 2 Central Angle & Chord from last curve station to PT If c2 denotes the difference in stationing b/n the last station on the curve & the PT, then the central angle d2 0 b/n these two points is: d 2  D  c 2 ; 20 The chord length C2 b/n the last curve station and the PT is: C 2  2 R sin( d 2 ) 2 Central Angle and Chord between Any Two Curve Points Let c be the difference in stationing b/n any two points on the curve, d be the central angle b/n the two points, D0  c and D be the degree of curve, either by chord or arc definition. Then, d  20 The actual chord distance C between the same two points, either by chord or arc definition, is given by the d relationship: C  2 R sin( ) 2 Field Procedures to Layout Circular Curve by Deflection Angle Method Case I When the theodolite is set at the PC. A. Set the theodolite at the PC with angle reading zero. On the initial tangent Points 1, 2, 3… etc are full stations to be established on the curve. 11  B. To locate point 1, turn the telescope until the angle reading is one half of d1, so that the line of sight is directed along the 1st sub-chord A1. Then when rod is held at the proper chord length and the tape is swing about A as a center until the rod is along the line of sight, point 1 on the curve is located. C. To locate point 2, turn the telescope till the angle reading equals the proceeding deflection plus D/2. Then the rod is held at the proper chord length for the full station and the tape is swing about point 1 as center until the rod is on the line of sight, point 2 on the curve is located. D. The remaining points on the curve are located in a similar manner. As a check the defection should be equal to D/2 when point PT is located Case II When the theodolite is set at PT properly oriented either by sighting on the PI angle reading D/2, or by sighting the PC angle reading zero, the same curve note can be used whether deflection are from PC or PT. Example: The PC of an arc definition curve is at station 8+345.25. The radius of the curve is 350m & Δ=35015’.Then Compute:  The first chord length (C1) & d1.  The last chord length (C2) & d2.  The deflection angles to each full  Successive station on the curve.  Chord length to layout each full station Solution Cumulative Deflection Station Arc length(M) Chord Length(M) Deflection Angle angle PC=8+345.25 - - 0 0 8+360 14.75 14.75 1012’26.3’’ 1012’26.3’’ 8+380 20 20 1038’13.28’’ 2050’39.58’’ 8+400 20 20 1038’13.28’’ 4028’52.86’’ 8+420 20 20 1038’13.28’’ 607’6.14’’ 8+440 20 20 1038’13.28’’ 7045’19.42’’ 8+460 20 20 1038’13.28’’ 9023’32.7’’ 8+480 20 20 1038’13.28’’ 11001’45.98’’ 8+500 20 20 1038’13.28’’ 12039’59.26’’ 8+520 20 20 1038’13.28’’ 14008’12.54’’ 12  8+540 20 20 1038’13.28’’ 15056’25.82’’ 8+560 20 20 1038’13.28’’ 17034’39’’ PT=8+560.58 0.58 0.58 002’50.91’’ 17037’30’’= Δ/2 Linear Method A ) Offsets from the tangent (back tangent) This method used when deflection angle is small, the length of the curve is short and the centre is inaccessible. In Fig. 3- 13 the line AB is drawn parallel to the tangent it cuts the radius. The length AT=Y & AO=(R-Y) In OAB, OA = (BO 2 - AB 2 ) R - Y = (R 2 - X 2 ) Fig. 3- 13: Offset from back Tangent Y = R - (R2 - X2 ) PROCEDURES FOR SETTNG OUT From PI, measure T along the tangents & drive pegs to the exact positions of PC & PT. Establish pegs at Xi intervals along the straights between PI & tangent points. Using an optical square set out the appropriate offset Y & drive in peg at each pt. Example: Given that the deviation angle  = 45o and radius R=60m. Calculate the offsets from the tangents at 5m intervals. Offset at 5m = 60 - (60 - 5 ) = 0.210m2 2 Solution: T=R tan  Offset at 10m = 60 - (60 - 10 ) = 0.84m 2 2 2 2 =60 tan 45o Offset at 15m = 60 - (60 - 15 ) = 1.905m 2 2 =24.85m Offset at 20m = 60 - (60 - 20 ) = 3.43m 2 2 Offset at 24.85m = 60 - (60 - 24.85 ) = 5.39m B) Offsets from the long chord It is the distance of a point on a curve from the long chord. It uses two tapes. Suitable for curves of small radius and useful when tangent lengths are inaccessible. See Fig. 3- 14 CD -major offset Y at mid-point C of the long chord. OC=a-Constant. In OPcC, a = (R2 - X2 ) 2 2 Therefore, Y = R - (R - X ), R  a  Y 13  In ABO, Any other offset Yn=AB-a 2 AB= (R2 - Xn )  a Yn  (R 2 - X n )  (R 2 - C ) 2 2 Fig. 3- 14: Offset from Long chord Yn  (R 2 - X n )  a ; 2 o Example1: Given that the deviation angle  = 45 and radius R=60m. Calculate the offsets from the tangents at 5m intervals. Yn  (R 2 -X 2 )  [R 2 - (C 2 ) 2 ] n Solution  (60 2 -X 2 )  (60 2 - 22.96 2 ) n 2 C =2Rsin(/2)  (3600 -X n )  (3600 - 3072.84) 2 =2*60*sin22.5o  (3600 -X n )  5 5. 43 =45.92m, Offset at 5m Y5 m  (3600 - 5 2 )  55.43  4.36m X(m) 0 5 10 15 20 22.96 Y(m) 4.57 4.36 3.73 2.66 1.14 0.0 Example2: Calculate the values of ordinates for the points on a circular curve having a long chord of 120m and a versed sine of 5m.The ordinates are to be measured on the long chord at an interval of 20m. Solution: =√ − - − ( ) (offsets from the long chord formula) The mid ordinate M= =5=√ −0 - −( ). This implies R=362.5m Then using offsets from the long chord formula: =5m (given), =√362.5 − 20 - 362.5 − ( ) =4.45 similarly, =2.79, =0 14