BUREAU OF DESIGN AND ENVIRONMENT
SURVEY MANUAL
May 2001
Illinois CURVES May 2001
A-1
APPENDIX - A
CURVES
The radius can be used to define a curve in lieu of the degree of curve definition to describe
horizontal curves.
For projects designed or constructed in English units of measure the arc definition of degree of
curve is primarily used. The arc definition is defined as the change of direction of the central
angle per an arc length of 100 feet. The relationship between the degree of curve and the
radius in feet is stated by the following equation:
R = 5729.58
D
Some of the various definitions and relationships of a simple curve are illustrated in Figure A.1,
page A-9 and given as follows:
(Δ) = Deflection Angle
Tangent distance (T) = R(Tan Δ/2)
External distance (E) = R(Sec Δ/2 -1) = R(Exsec Δ/2) = T (Sin Δ/2) - M
Middle Ordinate (M) = R(1-Cos Δ/2) = R(Vers Δ/2) = R – R (Cos Δ/2)
Long Chord (L.C.) = 2R(Sin Δ/2) = 2T(Cos Δ/2)
Length of Curve (L) = Δ(100) = Δ2πR = ΔR
D 360° 57.2958
The deflection angle (Δ) is measured in decimal degrees. For English units D is measured in
decimal degrees and R is in feet.
In addition to simple curves, compound curves, reverse curves and vertical curves are also
used in highway work.
Compound curves are a combination of two or more simple curves and their use should be
avoided where a simple curve can be used. However, due either to right-of-way problems or
topographic considerations, a compound curve may occasionally be necessary. The curve
must not be compounded at a ratio greater than 2:1. If the adjacent curves differ by more than
2º-00', a transition curve should be used.
Illinois CURVES May 2001
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A reverse curve is a combination of two simple curves of opposite curvature with a common
tangent. A tangent adequate in length to provide the superelevation transition required by the
Design Policies should be provided between the curves. If the reverse curves do not contain
any superelevations, a tangent between the curves is not required.
A “Broken-back” curve is a term used to denote two curves in the same direction separated by
a short tangent or by a flat curve whose radius is greater than twice the radius of either of the
two initial curves. This layout is particularly objectionable on highways and should be avoided
by using one simple curve or a compound curve if necessary.
The curves described above are illustrated in Figure A.2, page A-10.
In the past, there was a period when transition or spiral curves were extensively used in
connection with pavement widening at curves. As the curves became flatter and the pavement
wider, it became unnecessary to widen at curves in this manner. It remains, however, that
vehicular paths entering and leaving circular curves follow a spiral curve. For this reason or for
reasons of “fitting” an alignment into a problem area, transition curves may occasionally be
used.
Vertical curves are used to transfer the smoothly change from one slope to another along an
alignment to account for the change in terrain. See Figure A.3, page A-11 for a diagram of a
vertical curve.
CALCULATIONS FOR A HORIZONTAL CIRCULAR CURVE
Given:
P.I. Sta. 107+67.90, Δ = 11° 00' 00", D = 2° 30' 00"
Calculate the Radius
R = 5729.58/D
R = 5729.58/2° 30' 00"
R = 2291.83'
Tangent Distance
T = R(Tan Δ/2)
T = 2291.83(Tan 11° 00' 00"/2)
T = 220.68'
Length of Curve
L = 100 (Δ/D)
L = 100 (11° 00' 00"/ 2° 30' 00")
Illinois CURVES May 2001
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L = 440.00'
External Distance
E = T(Tan Δ/4)
E = 220.68(Tan 11° 00' 00"/4)
E = 10.60'
CALCULATE P.C. AND P.T. STATIONS
P.C. Station
P.C. = P.I. Station – Tangent Distance
P.C. = 107+67.90 – 220.68
P.C. = 105+47.22
P.T. Station
P.T. = 105+47.22 + 440.00
P.T. = 109+87.22
CALCULATE DEFLECTION ANGLES
Deflection for 100' of Arc
100' Arc = D/2
100' Arc = 2° 30' 00"/2
100' Arc = 1° 15' 00"
Deflection for 50' of Arc
50' Arc = D/4
50' Arc =2° 30' 00"/4
50' Arc =0° 37' 30"
Deflection for 25' of Arc
25' Arc = D/8
25' Arc =2° 30' 00"/8
Illinois CURVES May 2001
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25' Arc = 0° 18' 45"
Deflection for 1' of Arc
1' Arc = D/200
1' Arc = 2° 30' 00"/200
1' Arc = 0° 00' 45"
CALCULATE CHORD LENGTHS
Chord Length for 100' of Arc
100' Arc = (2)(R)(Sin of Deflection Angle) Deflection Angle = D/2
100' Arc = (2)(2291.83)(Sin 1° 15' 00")
100' Arc = 99.99'
Chord Length for 50' of Arc
50' Arc = (2)(R)(Sin of Deflection Angle) Deflection Angle = D/2
50' Arc = (2)(291.83)(Sin 0° 37' 30")
50' Arc = 50.00'
CALCULATE THE DEFLECTION FOR THE FIRST STATION FROM THE P.C. OR ANY ODD
STATION ALONG THE CURVE
1. Take the distance from the last point with a known deflection to the station you are
calculating.
2. Multiply this distance by the deflection of a 1' Arc (D/200), this will give you the
deflection between these two points.
Example: Find the deflection angle for Sta. 108+55.
(108+55 - 105+47.22) = 307.78'
307.78'(0° 00' 45") = (3° 50' 50") Note: Use decimal degrees for this calculation.
Illinois CURVES May 2001
A-5
EXAMPLE OF A FIELD BOOK SETUP FOR A HORIZONTAL CURVE
Deflections For Curve #1
Chord Deflection Total
Sta. Distance Distance Angle Deflection
105+00
+47.22 0 0 0 0 P.C.
+50 2.78' 2.78' 0° 02' 05" 0° 02' 05"
106+00 50.0 50.00 0° 37' 30" 0° 39' 35"
+50 50.0 50.00 0° 37' 30" 1° 17' 05"
107+00 50.0 50.00 0° 37' 30" 1° 54' 35"
+50 50.0 50.00 0° 37' 30" 2° 32' 05"
108+00 50.0 50.00 0° 37' 30" 3° 09' 35"
+50 50.0 50.00 0° 37' 30" 3° 47' 05"
109+00 50.0 50.00 0° 37' 30" 4° 24' 35"
+50 50.0 50.00 0° 37' 30" 5° 02' 05"
+87.22 37.22 37.22 0° 27' 55" 5° 30' 00" P.T.
Calculated by (Initials) (Date)
Checked by (Initials) (Date)
Δ = 11° 00'00" D = 2° 30' 00"
P.C. Marked by P.K. Nail
Deflection Angles Chord Lengths
100' of Arc = D/2 = 2° 30' 00" 100' of Arc = (2)(R)(Sin of Deflection
Angle)
= 1° 15' 00" = (2)(2291.83)(Sin 1° 15' 00")
50' of Arc = D/4 = 2° 30' 00"/4 = 99.99'
= 0° 37'30" 50' of Arc = (2)(R)(Sin of Deflection
Angle)
1’ of Arc = D/200 = 2° 30' 00"/200 = (2)(2291.83)(Sin 0° 37' 00")
= 0° 00' 45" = 50.00'
P.T. (Note: Total Deflection should equal Δ/2)
CALCULATIONS FOR VERTICAL CURVE
Please refer to Figure A.3 on page A-11 of this appendix for a graphic view of the
components of a vertical curve. Following are several definitions of the elements of a
vertical curve and a set of sample field notes that have been prepared for field use to stake
out a vertical curve.
Definitions:
PVC: “Point of vertical curve”. Station on centerline where the vertical curve starts.
PVI: “Point of vertical intersection”. Station at which the two tangent grade lines intersect.
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PVT. “Point of vertical tangency”. Station on centerline where the vertical curve ends.
LVC. Length of vertical curve.
OFFSET: The vertical distance from the tangent grade line to the vertical curve.
e: a mathematical constant whose value is determined by the grades of the two intersecting
tangents and the length of the vertical curve.
e: Grade #2(%) – Grade#1 (%) x (LVC) (Stations)/8 (Constant) = (G2-G1)(LVC/100)/8
CL ELEV: = Tangent Elevation + offset.
Highpoint/low point locations.
Distance from PVC = G1 x (Stations)/G2 –G1
Sample set of field notes for a vertical curve.
Tangent Elevation
Station (X) Distance Elevation (Y)offset on Curve
45+50 429.34
46+00 50.00 428.84 0.03 428.87
47+00 150.00 427.84 0.26 428.10
48+00 250.00 426.84 0.71 427.55
49+00 350.00 425.84 1.40 427.24
49+50 400.00 425.34 1.83 427.17
50+00 450.00 424.84 2.31 427.15
50+50 500.00 424.34 2.86 427.20
51+00 550.00 423.84 3.46 427.30
52+00 650.00 422.84 4.83 427.67
52+50 700.00 422.34 5.60 427.94
53+00 650.00 423.44 4.83 428.27
54+00 550.00 425.64 3.46 429.10
55+00 450.00 427.84 2.31 430.15
56+00 350.00 430.04 1.40 431.44
57+00 250.00 432.24 0.71 432.95
58+00 150.00 434.44 0.26 434.70
59+00 50.00 436.64 0.03 436.67
59+50 437.74 437.74
e = (G2-G1)LVC/8 = (2.2-(-1.0))14/8 = 3.2(14)/8 = 5.60 Vertical offset at
VPI.
Y-Offset @ 46+00 & 59+00 = (50/700)2(5.60) = 0.03
47+00 & 58+00 = (150/700)2(5.60) = 0.26
48+00 & 57+00 = (250/700)2(5.60) = 0.71 Y = e(X/L)2
49+00 & 56+00 = (350/700)2(5.60) = 1.40
49+50 = (400/700)2(5.60) = 1.83
50+00 & 55+00 = (450/700)2(5.60) = 2.31
50+50 = (500/700)2(5.60) = 2.86
51+00 & 54+00 = (550/700)2(5.60) = 3.46
52+00 & 53+00 = (650/700)2(5.60) = 4.83
52+50 = (700/700)2(5.60) = 5.60
When using this method the elevation difference is calculated from both the tangent
gradients at specified distances from the PVC and PVT and then applied to the tangent
Illinois CURVES May 2001
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elevations determined for the two gradients. The value of (e) is the offset at the VPI and
each individual station’s offset value is determined as a percentage of the external at the
VPI. The vertical offsets from a tangent to a parabola are proportional to the squares of the
distances from the point of tangency.
STATION AND ELEVATION OF LOW POINT OF VERTICAL CURVE
X (In Station from VPC) = G1L/(G2-G1) = 1.0(14)/(2.2-(-1.0)) = 1.0(14)/3.2 = 4.375 Sta. or
437.50'.
Station of Low Point = 45+50 + (4+37.5) = 49+87.50
X-Distance = 437.50 Tangent Elev. = 429.34 – 437.50(0.01) = 424.96'
Y-Offset (437.50/700)2(5.60) = 2.19'
Elevation of Low Point = 424.96 + 2.19 = 427.15'
ALTERNATE METHOD OF CALCULATING CURVE ELEVATIONS
An alternative method of calculating the elevations of a vertical curve is as follows: calculate
the value of (a) using the following formula: The tangent elevations are computed using the
elevation of the PVC and the slope of the forward tangent. An elevation is computed for
each station needed. Then compute the offsets from the forward tangent to the curve. The
offset equals aX2. Apply the offset values to the tangent elevation to obtain the curve
elevation.
a = 100(G2-G1)/2L = 100(2.2-(-1.0))/2*1400 = 0.114
2a = 0.228
Station Tangent Offsets from AV=aX2 Curve. Check
El. =G1X Elev. 1st Diff. 2nd Diff
45+50 429.34 429.34
46+00 428.84 -.114*0.502 =0.03 428.87
47+00 427.84 -.114*1.52 =0.256 428.10 -0.77
48+00 426.84 -.114*2.52 =0.71 427.55 -0.55 0.22
49+00 425.84 -.114*3.52 =1.40 427.24 -0.31 0.24
49+50 425.34 -.114*4.02 =1.83 427.17
50+00 424.84 -.114*4.52 =2.31 427.15 -0.09 0.22
50+50 424.34 -.114*5.02 =2.86 427.20
51+00 423.84 -.114*5.52 =3.46 427.30 +0.15 0.24
52+00 422.84 -.114*6.52 =4.83 427.67 +0.37 0.22
52+50 422.34 -.114*7.02 =5.60 427.94
53+00 421.84 -.114*7.52 =6.43 428.27 +0.60 0.23
54+00 420.84 -.114*8.52 =8.26 429.10 +0.83 0.23
55+00 419.84 -.114*9.52 =10.31 430.15 +1.05 0.23
56+00 418.84 -.114*10.52 =12.60 431.44 +1.29 0.24
57+00 417.84 -.114*11.52 =15.11 432.95 +1.51 0.22
58+00 416.84 -.114*12.52 =17.86 434.70 +1.75 0.24
59+00 415.84 -.114*13.502 =20.83 436.67 +1.97 0.22
59+50 415.34 -.114*14.002 =22.40 437.74
Illinois CURVES May 2001
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One method of checking your calculations is to calculate the first and second differences of
the curve elevation between the full stations. The second difference should be the same
and is equal to 2a, which is the percent of constant change in slope per station.
If other methods of calculating the station vertical offsets are desired, see a survey textbook
for the procedures.

A-9
Figure A.1

A-10
Figure A.2

A-11

METHOD FOR PULLOUT TEST FOR PILES

Pile to be tested shall be chipped off and dressed to natural horizontal plane till sound concrete is met or till up to cut off level. Arrangement shall be made to fix the four dial gauges / LVDT set on periphery of the pile.

Reinforcement bars or special anchor bars shall be used for anchoring of girder with test pile as per design.

Supports shall be built as per design &requirement at both sides of test piles. The reaction to pull out will be generated from these supports (as per design), which shall be at least 3 D away from the test pile periphery, where D is the diameter of test pile.

Girder of sufficient length & section (as per design) shall be placed over test pile with center coinciding with center of test pile. Both ends of girder shall be temporarily supported to accommodate the hydraulic jacks & bearing plates kept over reaction supports.

Girder shall be anchored with reinforcement bars or anchor bars with test pile as per design.

Datum bars parallel to girder, rested on immovable support shall be fixed at a distance of 3D (subject to minimum 1.5 m) from the edge of the pile on both sides, where D is the diameter of pile. Datum bars should not be disturbed during Test period.

Four nos, of Dial Gauges of 0.01 mm sensitivity to measure the upward deflection of the pile shall be fixed with datum bars and needle of gauge shall be touched to the arrangement made on periphery of pile. Small glass piece shall be placed below needle of dial gauge to get smooth surface.

Hydraulic jack shall be placed over reaction supports at both sides of test pile. Bearing plate of sufficient size & thickness shall be placed at bottom & top of Hydraulic jacks. Hydraulic jack shall be connected to manifold which in turn shall be connection to Hydraulic pump fitted with calibrated pressure Gauge to read the applied pressure.

After completion of all above arrangement, load shall be applied in following manner;

The test shall be carried out by applying a load in stages, each increment being 20% of safe load on the pile (exact percentage of increment of load depends on ram diameter of jack & least count of Pressure Gauge).    

For each stage of loading, the load shall be maintained till rate of displacement is either 0.1mm in first 30 minutes or 0.2mm in first one hour or till 2 hours whichever occurs first. The settlement shall be noted at each stage of loading. 

The full test load shall be maintained for 24 hours and settlement shall be measure at every one-hour interval. 

Unloading of test load shall be started after 24 hours of observation. Unloading shall be done in same and rebound shall be measured at each stage of unloading.

VISCOSITY TEST OF BITUMEN EMULSION BY SAYBOLT FUROL VISCOMETR (IS-3117-2004)

Test for Bitumen Emulsion: Viscosity by Saybolt Furol Viscometer

Purpose

Viscosity indicates the resistance to flow due to its internal friction. Higher the viscosity, lower the rate of flow. The test is conducted to see if it is in the specified range. The Test result is expressed in seconds for the flow of 60 cc of the emulsion sample through a 3.8 mm diameter orifice under the specified conditions of the test.

Apparatus

• Saybolt Furol Viscometer
• Stop watch

Saybolt Furol Viscometer – used for viscosity test of bitumen emulsion

Procedure

1. Clean the oil tube with a solvent, such as benzene, and remove excess solvent from the gallery. Pass the entire material through a 150 micron wire strainer before introducing into the oil tube.
2. After the tube is cleaned, pour into the tube a quantity of the material to be tested, sufficient to wet the entire surface of the tube. Allow to drain out.
3. The plunger commonly supplied with the viscometer shall never be used on instruments maintained as standards. Insert the cork stopper not less than 6.0 mm and not more than 9.5 mm into the lower end of the air chamber at the bottom of the oil tube, taking care that the cork fits tightly enough to prevent the escape of air, as tested by the absence of oil on the cork after it is withdrawn.
4. If the test temperature is above that of the room, heat the material to not more than 1.5⁰C below the temperature of test.
5. Pour the material into the oil tube until it ceases to overflow into the gallery. Keep it well stirred with the oil tube thermometer, care being taken to avoid touching the outflow tube. Adjust the bath temperature until the temperature of the material remains constant.
6. After thermal equilibrium has been attained, no further adjustment shall be made in the bath temperature. The test results shall be discarded if the indicated bath temperature varies by more than +0.03⁰C.
7. After the temperature of the material in the oil tube has remained constant within +0.02⁰C of the desired temperature for one minute with constant stirring, withdraw the oil tube thermometer and remove the surplus liquid quickly from the gallery by means of the withdrawal tube so that the level of the material in the gallery is below the level in the oil tube proper. Insert the tip of the withdrawal tube at one point in the gallery.
8. The test shall be started over again if the tip of the withdrawal tube touches the overflow rim. Under no condition shall the excess liquid be removed by rotating the withdrawal tube around the gallery.
9. Place the receiving flask in position so that the stream of liquid from outlet tube strikes the neck of the flask, care being taken that the graduation mark on the receiving flask is not less than 10cm, not more than 13cm, from the bottom of the bath. Snap the cork from its position and at the same instant start the timer. Stop the timer when the bottom of the meniscus of the liquid reaches the mark on the neck of the receiving flask.
10. Time in seconds as determined by the prescribed procedure, with the proper calibration correction, is the Saybolt Furol viscosity of the material at the temperature at which the test is made.
11. Report the results to the nearest 0.1 second for viscosity values below 200 seconds and to the nearest whole second for values 200 seconds or above.

Reporting of Result

Time in seconds as determined by the above described procedure, with the proper calibration correction, is the Saybolt Furol Viscosity of the material at the temperature at which the test is made.
Report the results to the nearest 0.1 s for viscosity values below 200s and to the nearest whole second for values 200s or above.

Reference

IS 3117 – 1965

Requirement Criterion

As per IS: 8887 -2004, the acceptance limits of viscosity for different types of emulsion are as follow.

Type of Emulsion	Acceptance Limits at 500C
Rapid Setting (RS-1)	20 – 100
Rapid Setting (RS-2)	100 – 300
Medium Setting (MS)	50 – 400
Slow Setting (SS-1)	20 – 100
Slow Setting (SS-2)	30 – 150

What is Fineness Modulus and how to calculate it

Definition of Fineness Modulus :

The fineness modulus is define in ACI terminology as “A factor obtained by adding the total percentages of material in the  sample that are coarser than each of the following sieves (cumulative  percentages retained), and dividing the sum by 100: 0.15 mm (No. 100),0.30 mm (No. 50), 0.60 mm (No. 30), 1.18 mm (No. 16),2.36 mm (No. 8), 4.75 mm (No. 4), 9.5 mm (3/8in.), 19.0 mm (3/4 in.), 37.5 mm  (1-1/2in.), 75 mm (3 in.), 150 mm (6 in.).”

Fineness modulus gives the idea about how course or fine the sand is. Lower FM indicates fine sand and higher FM indicates courser sand.Fine sand requires more cement paste due to high surface area and tendency to develop crack. On the other hand course sand produces a concrete mixture that is harsh and difficult to finish and will cause segregation.Therefore recommended FM range is 2.3 to 3.1.
Following table shows fineness modulus for different types of sand

Type of Sand	Fineness Modulus Range
Fine Sand	2.2 – 2.6
Medium Sand	2.6 – 2.9
Coarse Sand	2.9 – 3.2

How to calculate Fineness Modulus

Procure to determine Fineness Modulus

1. Sieve the aggregate using the appropriate sieves (80 mm, 40 mm, 20 mm, 10 mm, 4.75 mm, 2.36 mm, 1.18 mm, 600 micron, 300 micron & 150 micron)
2. Note down the weight of aggregate retained on each sieve.
3. Calculate the cumulative weight of aggregate retained on each sieve.
4. Find out the cumulative percentage of aggregate retained on each sieve.
5. Sum the cumulative weight of aggregate retained and divide the sum by 100.
   FM = ∑ Cumulative % retain on each Sieve  / 100

Calculation sheet to calculate Fineness Modulus

Sieve of Size (in mm)	Weight of sand Retained (g)	Cumulative weight of sand retained (g)	Cumulative percentage of sand retained (%)
80 mm	–
40 mm	–
20 mm	–
10 mm	0	0
4.75 mm	10	10	2
2.36 mm	50	60	12
1.18 mm	70	130	26
600 micron	90	220	44
300 micron	160	380	76
150 micron	80	460	92
Pan	40	500
Total	500	1760	252

Fineness Modulus = 252/100 = 2.52

Test Procedure for FINENESS MODULUS OF FINE AGGREGATE

This method determines the fineness modulus

of concrete fine aggregate used in

evaluation of natural and manufactured

sands for portland cement concrete.

1.2

The values given in parentheses (if provided)

are not standard and may not be exact

mathematical conversions. Use each system

of units separately. Combining values from

the two systems may result in nonconformance with the standard.

2. APPARATUS

2.1

Apparatus,

specified in Tex-401-A.

2.2

Standard U.S. sieves,

meeting the requirements of Tex-907-K, in the following sizes:



4.75 mm (No. 4)



2.36 mm (No. 8)



1.18 mm (No. 16)



600

μ

m (No. 30)



300

μ

m (No. 50)



150

μ

m (No.100).

3. PROCEDURE

3.1

Determine particle size distribu

tion in accordance with Tex-401-A.

4. CALCULATIONS

4.1

Calculate Cumulative Percent Retained:

Cumulative Percent tained Cumulative Mass tained Mass of Total Samp

Calculate the Fineness Modulus (

FM

):

FM Cumulative percent retained

=



( ) / 100

4.3

Table 1 lists example amounts for discussion purposes.

Table 1—Amounts for Example Calculations

Sieve Size Cum. Mass Retained Cum. % Retained

4.75 mm (No. 4) 31.5 g 6.3

2.36 mm (No. 8) 99.1 g 19.8

1.18 mm (No. 16) 195.6 g 39.1

600

μ

m (No. 30)

306.7 g 61.3

300

μ

m (No. 50)

367.2 g 73.4

150

μ

m (No. 100)

482.8 g 96.5

(Dry Weight of Original Sample = 500.3 g)

Fineness Modulus = (6.3 + 19.8 + 39.1 + 61.3 + 73.4 + 96.5) / 100 = 2.964

Fineness Modulus = 2.96

5. REPORT

5.1

Report Fineness Modulus to the nearest 0.01.

Specific Gravity & Water Absorption of Coarse Aggregate

https://www.youtube.com/watch?v=hqXFPq676iM&ebc=ANyPxKppNh_89mtQrZiiHj_IlBMTTSfd_hKvtnH_BXMyIbBfi78pSWSTeqQwOei8LqN7dqFS_jWNUd6xqAMdJo6w7gNnQpFT0A