Topic: Simple Curve
Problem #1 The angle of intersection of a circular curve is 36° 30'. Compute the radius if the external distance is 12.02 m. a) b) c) d)
203.74 m 253.72 m 226.94 m 214.67 m
Solution:
[ C ] answer
Problem #2
The angle of intersection of a circular curve is 36° 30'. Compute the radius if the external distance is 20.00 m.
A. 453.74 m B. 377.61 m C. 214.67 m D. 367.93 m
SOLUTION
cos½ I = R/R+E cos18°15’ + = R/R+20.00 Rcos18°15’ + 20.00cos18°15’ = R R - Rcos18°15’= 20.00cos18°15’ R(1-cos18°15’) = 20.00cos18°15’ R = 20.00cos18°15’/(1-cos18°15’) R=377.61 m Answer is Letter B
Problem # 3 Given the following elements of a circular curve: middle ordinate = 2 m; length of long chord = 70 m. Find its degree of curve, use arc basis. a.
4.5°
b.
5.3°
c.
2.9°
d.
3.7°
Apply Pythagorean theorem to find the radius:
Degree of curve (arc basis):
TOPIC: Compound Curves Problem 1. The common tangent AB of a compound curve is 76.42 m with an azimuth of 268° 30′ . The vertex V being inaccessible. The azimuth of the tangents AV and VB was measured to be 247° 50′ and 262° 50′ respectively. If the stationing of A is 43 + 010.16 and the degree of the first curve was fixed at 4 based on the 20 m chord. Using chord basis. What is the stationing at P.T.?
a. 43 + 109.65
c. 42 + 109.65
b. 43 + 108.65
d. 43 + 100.1
Solution: Stationing at P.C. I1 = 268° 30′ − 247° 50′ I1 = 20° 40′ I2 = 282° 50′ − 268° 30’ I2 = 14° 20′
D1 = 4 Sin D1/2 = 10/R1 R1 = 286.56
T 1 = R1 tan I1/2 T 1 = 286.56 tan 10° 20′ T 1 = 52.25 m
P.C. = (43 + 010.46) – 51.25 P.C. = 42 + 958.21
Stationing at P.C.C.
T1 + T2 = 76.42 T2 = 76.42 – 52.25
T2 = 24.17
T2 = R2 tan I2/2 24.17 = R2 tan 7° 10′ R2 = 192.233 m
Sin D2/2 = 10/R2 Sin D2/2 = 10/192.23 D/2 = 2°59′ D2 = 5°58′ Lc1 = 20°40′ (20) Lc1 = 103.34
P.C.C. = (42 + 958.21) + 103.34 P.C.C. = 103.34
Stationing at P.T.
Lc2 = I2 (20) / D2 Lc2 = 14°20′ 20 / D2 Lc2 = 48.10 P.T. = (43 + 06.55) + 48.10 P.T. = 43 +109.65
Problem 2. The long chord from the P.C. to the P.T. of a compound curve is 300 meters long and the angles it makes with the longer and shorter tangents are 12° and 15 respectively. If the common tangent is parallel to the long chord. Find the Radius of the first curve.
a. 800.1 m
c.709.03
b. 802.36 m
d. 801.33
Solution: Radius of the first curve: I1 = 12 I2 = 15 Considering Triangle ABC: 300 𝐵𝐶 = 𝑆𝑖𝑛 166°30′ 𝑆𝑖𝑛 6° 𝐵𝐶 =
𝑆𝑖𝑛 6° 𝑆𝑖𝑛 166° 30′
𝐵𝐶 = 134.33 𝑚
300 𝐴𝐶 = 𝑆𝑖𝑛 166°30′ 𝑆𝑖𝑛 7°30′ 𝐴𝐶 =
300 𝑆𝑖𝑛 7°30′ 𝑆𝑖𝑛 166°30′
𝐴𝐶 = 167.74 𝑚
Sin I1/2 = AC/2R1 R1 = 167.74 /2 Sin 6 R1 = 802.36 m
Problem 3. The locating engineer a railroad curve runs a 6° curve to the P.C.C., 300 m long from the P.C. of the compound curve, thence from the P.C.C., a 1°40′ curve was run towards to the P.T. 600 m long. (Use Arc basis) Determine the length of the long chord connecting the P.C. and P.T.
a. 713.13
c. 700.33
b. 739.67
d. 719.76
Solution: Lc1 = 300 D1 = 6 Lc1 = 20 I1 /D1 I1 = 300(6) / 20 I1 =90 Lc2 = 600
D2 = 1°40′ Lc2 = 20 I2 / D2 I2 = 600(1.667) / 20 I2 = 50 R1 = 1145.916/D1 R1 = 1145.916/6 R1 = 190.99 m
Sin 45° = C1 / 2R1 C1 = 2R1 Sin 45 C1 = 2(190.99) Sin 45 C1 = 270.10 m
R2 = 1145.916 /D2 R2 = 1145.916 / 1.66 R2 = 687.55 m
Sin 25 = C2 / 2R2 C2 = 2R2 Sin 25 C2 = 2(687.55) Sin 25 C2 = 581.14 m L2 = (270.10)2 + (581.14)2 - 2(270.10)(581.14) Cos 110 L = 719.76 m
Topic: Reversed Curve
Problem #1 Two tangents intersecting at an angle of 46o40’ (at PI) are to be connected by a reversed curve. The tangent distance from PI to PT of the reversed curve is 48.6 m and from PI to PC is 360.43. The radius of the curve through the PC is 240 m. Sta. of PI is 25 + 863.2. What is the stationing of PT? a.
25 + 814.60
c.
26 + 061.97
b.
26 + 167.98
d.
25 + 934.21
Solution:
In triangle OAD: tan α =
240 360.43 o
α = 33.658 𝑂𝐷 =
2402 + 360.432
In right triangle DFO: OF = OD cos β OF = 433.024 cos 13.009o OF = 421.91 m DF = OD sin β DF = 433.024 sin 13.009o
β = 46 o40’ – 33.658 o
β = 13.009
DF = 97.48 m [CF = OF – OE]
CF = 421.91 – 48.6 CF = 373.31 m
In right triangle DGE:
Lc2 =
(240 + R)2= (CF)2 + (R - DF)2 2
2
𝜋𝑅2 𝐼2 180°
Lc2 =
𝜋(135.21)(84.23°) 180°
Lc2 = 198.77 m 2
(240 + R) = 373.31 + (R – 97.48)
57600 + 480R + R2 = 139, 360 + 194.96R + 9502
Lc1 =
674.96R = 91,262
Lc1 = 157.34 m
𝜋𝑅1 𝐼1 180°
Lc1 =
𝜋(240)(37.563°) 180°
R = 135.21 m = R2 Stationing of PC = (25 + 863.20) – 360.43 𝐶𝐹
tan I2 = 𝑅−𝐷𝐹
373.31
tan I2 = 135.21−97.48
Stationing of PC = 25 + 502.77
I2 = 84.23o Stationing of PT = (25 + 863.20) + Lc1 + Lc2 γ = 90 – I2 = 5.77o
Stationing of PT = 26 + 061.97
Problem #2
Given broken line AB = 57.6m, BC = 91.5m and CD = 91.5m arranged as shown. A reverse curve is to connect these three lines thus forming the center of the common radius of the reverse curve. Find the length of the common radius of the reverse curve and the total length of the reverse curve. a.
111.688m; 167.642m
c.
134.642m; 155.453m
b.
112.421m; 135.744m
d.
105.232m; 187.893m
Solution:
T1 = Rtan11˚
P.C. to P.R.C = L1
T2 = Rtan32˚
L1 = 111.6888 (22˚) 180
T1 + T2 = 91.5
L1 = 42.885m
R (tan11˚ + tan32˚) = 91.5 R = 111.688m
𝜋
L2 = 124.757m Total Length of the reverse curve L1 + L2 = 167.642m
Problem #3
The perpendicular distance between two parallel tangents is equal to 8 meters, central angle equal to 8˚ and the radius of curvature of the first curve is equal to 175m. Find the radius of the second curve of the reversed curve. a.
647 m
c.
669 m
b.
635 m
d.
753 m
Solution:
a = R1 – R1 cos8˚ a = 175(1 – cos8˚) a = 175 (0.00973) a = 1.70
a+b=8
b = R2 (1 – cos8˚)
b = 8 – 1.70
b = R2 (0.00973)
b = 6.3 6.3
R2 = 0.00973 R2 = 647 meters
TOPIC: SYMMETRICAL CURVE PROBLEM NO. 1 A symmetrical parabolic curve 120m long passes through point X whose elevation is 27.79m and 54 m away from PC. The back tangent of the curve has a grade of +2%. If PC is at elevation 27.12, what is the elevation of the summit? a. 27.18m b. 28.81m
c. 29.57m d. 27.83m
Solution:
elev A = 27.12 + 54(0.02) Y1 = 28.2 – 27.79 Y1 = H X12 (L/2)2 0.41/ (542) H
= 28.20 = 0.41
= H / 602 = 0.5062
2H = L/2 (g1 – g2) L/2 L 2 (0.5062) = 60 (0.02 – g2) 60 120 g2 = -0.0137
From PC S1 = g1 L / (g1 – g2) = (0.02 x 120) / 0.02 – 0.0137 = 71.22 Y2 = H X22 (L/2)2 Y2 / 48.782 Y2
= 0.506 / 602 = 0.334
Elev of Summit = [27.12 + (0.02 x60)] – (11.22 x 0.0137) – 0.334 Elev of Summit= 27.83m
Problem no. 2 A descending grade of 6% and an ascending grade of 2% intersect at Sta 12 + 200 km whose elevation is at 14.375 m. The two grades are to be connected by a parabolic curve, 160 m long. Find the elevation of the first quarter point on the curve. A. B. C. D.
16.082 m 15.575 m 17.175 m 13.936 m
Horizontal distance from the lowest point to point Q:
Grade at point Q by ratio and proportion of triangles:
Solution
Elevation of PC:
Difference in elevation between PC and Q:
Elevation of the first quarter point Q:
From the grade diagram:
[ C ] answer
Problem no. 3 A grade line AB having a slope of +5% intersect another grade line BC having a slope of –3% at B. The elevations of points A, B and C are 95 m, 100 m and 97 m respectively. Determine the elevation of the summit of the 100 m parabolic vertical curve to connect the grade lines. a) b) c) d)
98.32 m 99.06 m 97.32 m 96.86 m
Solution Horizontal distance between A and B = (100 - 95)/0.05 = 100 m Horizontal distance between B and C = (100 - 97)/0.03 = 100 m
The figure above place the parabolic curve at the middle-half:
Distance from PC to the summit:
Elevation of the summit:
[ B ] answer
TOPIC: UNSYMMMETRICAL PARABOLIC CURVE Problem No. 1 An unsymmetrical parabolic curve has a forward tangent of -8% and a backward tangent of +5%. The length of the curve on the left side of the curve is 40m long while that the right side is 60m long. At station 6+780 and at elevation 110m. Determine the height of fill at the outcrop.
a.
2.221m
c.
1.074m
b.
3.462m
d.
4.366m
Solution: 2𝐻 𝐿2 (𝑔1 + 𝑔2) = 𝐿1 𝐿1 + 𝐿2 𝐻=
40(60)(0.05 + 0.08) 2(40 + 60)
H= 1.56 𝐻 𝐿1 2 1.56 40 2
𝑦 = 2 𝑥 𝑦
= 20 2 y= 0.39
El. Of A = 110 + 20(.05) El. Of A = 111m El. Of B = 111 –0.39 El. Of B = 110.60m Height of fill = 110.61 – 108.40 Height of fill = 2.221m
Problem No. 2 A forward tangent having a slope of -4% intersects the back tangent having a slope +7% at point V at stations 6 + 300 having an elevation of 230m. It is required to connect two tangents with an unsymmetrical parabolic curve that shall pass through point A on the curve having an elevation of 227.57m at station 6+270. The length of the curve is 60m on the side of the back tangent. Determine the stationing of the highest point of the curve.
a.
6 + 105.47
c.
6 + 228.65
b.
6 + 125.33
d.
6 + 315.76
Solution: El. of B = 230-30(0.07) El. of B = 227.90m y=227.90-227.57 y=0.33 𝑦 𝐻 = 2 2 𝑥 𝐿1 0.33 𝐻 = 2 (30) (60)2
𝑆2 =
𝑔2𝐿2 2 2𝐻
2𝐻 𝐿2 (𝑔1 + 𝑔2) = 𝐿1 𝐿1 + 𝐿2 2(1.32) 𝐿2 (0.07 + 0.04) = 60 60 + 𝐿2 2.64(60+𝐿2 ) = 60 (0.11)𝐿2 158.4 + 2.64𝐿2 = 6.6𝐿2 𝐿2 = 40m 𝐿1 𝑔1 60(0.07) = = 2.1 > 𝐻 2 2 Therefore the highest point of the curve is on the right side.
0.04(40)2 𝑆2 = 2(1.32) 𝑆2 = 24.24 m from P.T. Stationing of highest pt. = (6+340) – 24.24 Highest point of curve = 6 + 315.76
Topic: Sight Distance (S < L) Problem #1 A 5% grade intersects a -3.4% grade at station 1 + 990 of elevation 42.30 m. Design a vertical summit parabolic curve connecting the two tangent grades to conform with the following safe stopping sight distance speciifications. Design velocity = 60kph Height of driver’s eye from the road pavement = 1.37 Height of an object over the pavement ahead = 100 mm Perception – reaction time = ¾ seconds Coefficient of friction between the road pavement and the tires = 0.15 DETERMINE: a. Stopping sight distance b. Station of P.C and P.T.
a. b. c. d.
S = 83.29 m ; Sta of P.C = 1 + 924.06 ; Sta. of P.T = 2 + 055.91 S = 82.69 m ; Sta of P.C = 1 + 924.00 ; Sta. of P.T = 2 + 055.92 S = 83.29 m ; Sta of P.C = 1 + 964.06 ; Sta. of P.T = 2 + 045.94 S = 83.29 m ; Sta of P.C = 1 + 924.06 ; Sta. of P.T = 2 + 059.96
Solution: a. Stopping sight distance S
𝑉2
𝑉𝑡
S = 3.6 + 2𝑔 (𝑓+𝐺)3.62 S=
60(3 4) 3.6
(60)2
+ 2(9.81)(0.15+0.05)3.62
S = 83.29 m
b. Stationing of P.C and P.T S
h1
h2
P.C
P.T L
S
𝐴𝑆 2 100 ( 21 +
2
22)
A = g1 – g2 A = 5 – (-3.4) A = 8.4 2h1 = 2(1.37) = 2.74 2h2 = 2(0.10) = 0.20 L=
8.4(83.29)2
100 ( 2.74 +
2
0.20)
L = 131.82 > S (OK) Sta. of P.C = (1+990) -
131.82 2
Sta of P.C = 1 + 924.06 Sta of P.t = (1 + 990) +
131.82
Sta. of P.T = 2 + 055.91
2
Problem #2 A vertical curve has a descending grade of -12% starting from the P.C and an ascending grade of +3.8% passing thru the P.T. The curve has a sight distance of 180 m. 1. Compute the length of the vertical curve. 2. Compute the max. velocity of the car that could pass thru the curve. 3. Compute the distance of the lowest point of the curve from the P.C a) 53.01m b) 51.09m
c) 52.40m d) 51.70m
Solution 1. Length of the curve
L=
𝐴𝑉 2 395
S
5𝑉 2
AS 2
L = 122+3.5 S A = g2 – g1 A = 3.5 – (-1.2) A=5 5 (180 2 )
215.43 = 395
V = 130.46 kph 3. Distance of lowest point of curve from P.C. S=𝑔
𝑔1 𝐿 1 −𝑔2
L = 122+3.5 (180)
S=
L = 215.43
S = 51.70m
2. Max. Velocity
−0.012(215.43) −0.012−0.038
Problem #3 A vertical curve has a descending grade of -12% starting from the P.C and an ascending grade of +3.8% passing thru the P.T. The curve has a sight distance of 180 m. 1. Compute the length of the vertical curve.
a) 220.5 b) 216.9
c) 214.79 d) 215.43 A=5 5 (180 2 )
Solution
L = 122+3.5 (180)
1. Length of the curve S
L = 122+3.5 S A = g2 – g1 A = 3.5 – (-1.2)
L = 215.43
SIGHT DISTANCE ON VERTICAL SUMMIT CURVES (S > L)
Problem #1 A vertical summit curve has tangent grades of +2.5% and -1.5% intersecting at station 12 + 460.12 at an elevation of 150m. above sea level. If the length of curve is 190m., compute the length of the passing sight distance. a. 206 m b. 208 m c. 209 m
Solution: Passing sight distance: h1 = 1.14 m. h2 = 1.14 m.
Assume S > L 𝐿 = 2𝑆 −
200( 1 + 2 𝐴
A = g1 – g2 A = 2.5 – (–1.5) A=4 190 = 2𝑆 −
200( 1.14 + 1.14 𝐴
2S = 190 + 228 S = 209 > 190 ok
d. 2 10 m
SGIHT DISTANCE FOR A SAG VERTICAL CURVE Problem no. 2: A highway has a 55 mph designed speed. There is a negative 1% grade followed by a 2% grade. Refer to the following table: Assumed Speeds Designed
Passed
Passing
Minimum Passing
Speed
Vehicle
Vehicle
Sight Distance (ft)
(mph)
(mph)
(mph)
50
41
51
1,840
60
47
57
2,140
65
50
60
2,310
70
54
64
2,490
What is the required length of vertical curve to satisfy AASHTO stopping sight distance? a. b. c. d.
316ft 585 ft 200 ft 1460 ft
Solution: From the table, stopping sight distance=550 ft. A= (-1)-2= -3 or 3 For a sag vertical curve, S>L: 𝐿 = 2𝑆 − 𝐿 = 2(550) −
400 + 3.55𝑆 𝐴 400 + 3.55(550) 3
L=315.83
Problem no. 3 Determine how far from each other will the two drivers with a height of 1.5 meters from eye level to the ground see each other when they are approaching each other through a vertical curve having an ascending grade of 5% and a descending grade of 3.5%. The total length of the curve is 150 meters. Sight distance is greater than the length of the curve. a. 144.59 b. 146.49 Solution: 𝐿=
2𝑆 𝑔1 − 𝑔2 − 8 𝑔1 − 𝑔2
By transposition: 𝑆=
𝐿(𝑔1 − 𝑔2) − 8 2(𝑔1 − 𝑔2)
𝑆=
150 0.05 + 0.035 − 8 ∗ 1.5 2(0.05 + 0.035)
S= 145.59 m.
c. 145.59 d. 145.69
TOPIC: EARTHWORKS (Volume by End Area Method) Problem No. 1 Given the cross section notes below of the ground which will be excavated for a roadway, compute the volume of excavation between station 4 + 000 and 4 + 020 by end area method. The roadway is 9 m. wide with sides slope of 1.5: 1. Station 4 + 000 7.38 + 1.92
0 +3.15
12.285 + 5.19
Station 4 + 020 8.82 + 2.88
0 + 2.52
10.08 + 3.72
a.
859.22
c.
853.48
b.
856.34
d.
855.13
Solution: A1=
1.92+3.15 7.38 2
−
1.92 2.88 2
+
3.15+5.19 12.285
+
2.52+3.72 10.08
2
−
5.19 7.785 2
A1= 46.969875 m2 ≈ 46.97 m2 A2=
2.88+2.52 8.82 2
A2=38.664 m2 V= V=
𝐿(A 1+A 2 ) 2 20(46.97+38.664) 2
V=856.34 m2
−
2.88 4.32 2
2
−
3.72 5.58 2
Problem 2 Find x if the end areas shown below have a prismoidal correction of 8.75m3and are 120 meters apart.
0.75m
0.75m
0.75m 1.5m
3m
3m
1.5m
x 0.75m
0.5m 0.1m
a. 3.10m b. 2.5m
3m
3m
c. 2.25m d. 1.98m
1.5m
Solution: D
C
D
C
1m
C1 = 1.5 + 3 + 3 +1.5 = 9 C2 = 1 + 3 + 3 + 1.5 = 8.5 Cp = L (C1 – C2) (D1 – D2) 12 8.75 = 120 (x – 0.75)(9 – 8.5) 12 105 = 120 (x – 0.75) (0.5) 105 = 60x – 45 60x = 150 x = 2.5 m (b)
TOPIC: EARTHWORKS (Volume using Prismoidal Formula) Problem No. 3 Given the following cross section notes of an earthwork on a rolling terrain. STA. 5+000 +10 +5 +3 31 0 13.5
STA. 5+020 +14 +7 +4.5 41 0 17.25 The width of the road is 12m. and the side slope is 2.5:1. Compute the volume using Prismoidal formula. a. 4055.37
c. 4069.55
b. 4050.42
d. 4063.79
Solution: Use average values of dimensions of A1and A2 to get Am (mid-section).
A1
𝐴1 =
10(6)
𝐴2 =
6(14)
2 2
A2
+
5(31)
+
7(41)
2 2
+
5(13.5)
+
7(17.25)
𝐴𝑚 =
𝑉𝑜𝑙. =
2 2
+ +
6(3) 2
= 150.25𝑚2
6(4.5) 2
Am
= 259.375𝑚2
12(6) 6(36) 6(15.375) 3.75(6) + + + = 201.375𝑚2 2 2 2 2
𝐿 20 𝐴1 + 4𝐴𝑚 + 𝐴2 = 150.25 + 4 201.375 + 259.375 = 4050.42𝑚3 6 6
ASSEMBLY OF PRISM Problem #1 Determine the volume of earth to be removed in the figure shown.
a. 516,000 m3 b. 561,000 m3 c. 521,000 m3 d. 511,000 m3
Solution: Corner Points: 6.0 + 3.4 + 4.0 + 2.0 = 15.4 m Border Points: 5.8 + 5.2 + 4.6 + 3.0 + 2.8 + 3.0 + 3.5 + 4.8 + 4.8 + 5.5 = 43.0 m Interior Points: 5.0 + 4.6 + 4.2 + 3.6 + 4.0 + 4.9 = 26.3 m Average Depth:
1 15.4 + 2 43.0 +4 26.3 48
= 4.30 𝑚
Area: 400m x 300m = 120000 m2 Volume: 120000 m2 x 4.30 m = 516000 m3 ANS: The volume of earth to be removed is 516000 m3.
TRUNCATED TRIANGULAR PRISM Problem #2 Calculate the lateral area, surface area and volume of the truncated square pyramid whose larger base edge is 24 cm, smaller base edge is 14 cm and whose lateral edge is 13 cm. a. 912 cm 2 , 1684 cm 2 ,4029.43 cm 3 b. 910 cm 2 , 1784 cm 2 ,4059.66 cm 3 c. 812 cm 2 , 1900 cm 2 ,4439.44 cm 3 d. 956 cm 2 , 1543 cm 2 ,3429.82 cm 3
Solution: 12 2 =h 2 +5 2 Ap= 132 + 52 = 12cm P=24 x 4 = 96cm 𝑝𝑡 =14 x 4 =56cm 𝐴𝑙 =
96+59 2
x 2 = 912cm 2
A = 24 2 =576cm 2 𝐴𝑡 = 14 2 = 196cm 2 𝐴𝑇 = 912+576+196 = 1684cm 2 V=
10.91 3
x (576+196+ 576 𝑥 196 )=4029.43cm 3
TRUNCATED RECTANGULAR PRISMS Problem # 3 An area is to be excavated where a house is to be built. The elevation of the area’s corner is indicated in meters. The area is desired to be excavated with the same elevation which is 40. Determine the volume of excavation in cubic meters using truncated prism. a. 6150 m 2 b. 6105 m 2 c. 6015 m 2 d. 6115 m 2 SOLUTION: A = 30(40) = 1200 m 2 V = 1200(3.7 + 4.6 + 5.3 + 6.9)/4 V = 1200(20.5)/4 V = 6150 m 2
