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) 203.74 m

b) 253.72 m

c) 226.94 m

d) 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

𝑆𝑖𝑛 166°30′=

𝐵𝐶

𝑆𝑖𝑛 6°

𝐵𝐶 = 𝑆𝑖𝑛 6°

𝑆𝑖𝑛 166° 30′

𝐵𝐶 = 134.33 𝑚

300

𝑆𝑖𝑛 166°30′=

𝐴𝐶

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

α = 33.658o

𝑂𝐷 = 2402 + 360.432

β = 46 o40’ – 33.658

β = 13.009o

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

DF = 97.48 m

[CF = OF – OE] CF = 421.91 – 48.6

CF = 373.31 m

In right triangle DGE:

(240 + R)2= (CF)

2 + (R - DF)

2

(240 + R)2= 373.31

2 + (R – 97.48)

2

57600 + 480R + R2 = 139, 360 + 194.96R +

9502

674.96R = 91,262

R = 135.21 m = R2

tan I2 = 𝐶𝐹

𝑅−𝐷𝐹 tan I2 =

373.31

135.21−97.48

I2 = 84.23o

γ = 90 – I2 = 5.77o

Lc2 = 𝜋𝑅2𝐼2

180° Lc2 =

𝜋(135.21)(84.23°)

180°

Lc2 = 198.77 m

Lc1 = 𝜋𝑅1𝐼1

180° Lc1 =

𝜋(240)(37.563°)

180°

Lc1 = 157.34 m

Stationing of PC = (25 + 863.20) – 360.43

Stationing of PC = 25 + 502.77

Stationing of PT = (25 + 863.20) + Lc1 + Lc2

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˚

T2 = Rtan32˚

T1 + T2 = 91.5

R (tan11˚ + tan32˚) = 91.5

R = 111.688m

P.C. to P.R.C = L1

L1 = 111.6888 (22˚) 𝜋

180

L1 = 42.885m

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

b = R2 (1 – cos8˚)

b = R2 (0.00973)

a + b = 8

b = 8 – 1.70

b = 6.3

R2 = 6.3

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 c. 29.57m

b. 28.81m d. 27.83m

Solution:

elev A = 27.12 + 54(0.02) = 28.20

Y1 = 28.2 – 27.79 = 0.41

Y1 = H

X12 (L/2)

2

0.41/ (542) = H / 60

2

H = 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 = 0.506 / 60

2

Y2 = 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. 16.082 m

B. 15.575 m

C. 17.175 m

D. 13.936 m

Solution

From the grade diagram:

Horizontal distance from the lowest point to

point Q:

Grade at point Q by ratio and proportion of

triangles:

Elevation of PC:

Difference in elevation between PC and Q:

Elevation of the first quarter point Q:

[ 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) 98.32 m

b) 99.06 m

c) 97.32 m

d) 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

𝐿1=

𝐿2 (𝑔1 + 𝑔2)

𝐿1 + 𝐿2

𝐻 = 40(60)(0.05 + 0.08)

2(40 + 60)

H= 1.56

𝐻

𝐿12 =

𝑦

𝑥2

1.56

402 =𝑦

202 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

𝑥2=

𝐻

𝐿12

0.33

(30)2=

𝐻

(60)2

𝑆2 =𝑔2𝐿2

2

2𝐻

2𝐻

𝐿1=

𝐿2 (𝑔1 + 𝑔2)

𝐿1 + 𝐿2

2(1.32)

60=

𝐿2 (0.07 + 0.04)

60 + 𝐿2

𝐿1𝑔1

2=

60(0.07)

2= 2.1 > 𝐻

2.64(60+𝐿2 ) = 60 (0.11)𝐿2

158.4 + 2.64𝐿2 = 6.6𝐿2

𝐿2 = 40m

Therefore the highest point of the curve is

on the right side.

𝑆2 =0.04(40)2

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. S = 83.29 m ; Sta of P.C = 1 + 924.06 ; Sta. of P.T = 2 + 055.91

b. S = 82.69 m ; Sta of P.C = 1 + 924.00 ; Sta. of P.T = 2 + 055.92

c. S = 83.29 m ; Sta of P.C = 1 + 964.06 ; Sta. of P.T = 2 + 045.94

d. S = 83.29 m ; Sta of P.C = 1 + 924.06 ; Sta. of P.T = 2 + 059.96

Solution:

a. Stopping sight distance

S

S = 𝑉𝑡

3.6 +

𝑉2

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 < L

L= 𝐴𝑆2

100 ( 2𝑕1 + 2𝑕2)2

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 + 0.20)2

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

2

Sta. of P.T = 2 + 055.91

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

S<L

L = AS2

122+3.5 S

A = g2 – g1

A = 3.5 – (-1.2)

A = 5

L = 5 (1802)

122+3.5 (180)

L = 215.43

2. Max. Velocity

L = 𝐴𝑉2

395

215.43 = 5𝑉2

395

V = 130.46 kph

3. Distance of lowest point of curve from

P.C.

S = 𝑔1𝐿

𝑔1−𝑔2

S = −0.012(215.43)

−0.012−0.038

S = 51.70m

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

Solution

1. Length of the curve

S<L

L = AS2

122+3.5 S

A = g2 – g1

A = 3.5 – (-1.2)

A = 5

L = 5 (1802)

122+3.5 (180)

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

d. 2

10 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

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

Speed

(mph)

Passed

Vehicle

(mph)

Passing

Vehicle

(mph)

Minimum Passing

Sight Distance (ft)

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. 316ft

b. 585 ft

c. 200 ft

d. 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 −400 + 3.55𝑆

𝐴

𝐿 = 2(550) −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 c. 145.59

b. 146.49 d. 145.69

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.

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−

5.19 7.785

2

A1= 46.969875 m2

≈ 46.97 m2

A2= 2.88+2.52 8.82

2−

2.88 4.32

2 +

2.52+3.72 10.08

2−

3.72 5.58

2

A2=38.664 m2

V=𝐿(A1+A2 )

2

V=20(46.97+38.664)

2

V=856.34 m2

Problem 2

Find x if the end areas shown below have a prismoidal correction of 8.75m3and are 120

meters apart.

a. 3.10m c. 2.25m

b. 2.5m d. 1.98m

x

0.75m

1.5m 1.5m

0.75m

0.75m 0.5m

0.1m 1.5m

0.75m

3m 3m

3m 3m

C

Solution:

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)

D

D

C

1m

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

31

+5

0

+3

13.5

STA. 5+020

+14

41

+7

0

+4.5

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 = 10(6)

2+

5(31)

2+

5(13.5)

2+

6(3)

2= 150.25𝑚2

𝐴2 = 6(14)

2+

7(41)

2+

7(17.25)

2+

6(4.5)

2= 259.375𝑚2

𝐴𝑚 = 12(6)

2+

6(36)

2+

6(15.375)

2+

3.75(6)

2= 201.375𝑚2

𝑉𝑜𝑙. = 𝐿

6 𝐴1 + 4𝐴𝑚 + 𝐴2 =

20

6 150.25 + 4 201.375 + 259.375 = 4050.42𝑚3

A2 A1 Am

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 m

3

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 cm2, 1684 cm

2,4029.43

cm3

b. 910 cm2, 1784 cm

2,4059.66

cm3

c. 812 cm2, 1900 cm

2,4439.44 cm

3

d. 956 cm2, 1543 cm

2,3429.82 cm

3

Solution:

122=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 = 242

=576cm2

𝐴𝑡= 142

= 196cm2

𝐴𝑇= 912+576+196 = 1684cm2

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 m2

b. 6105 m2

c. 6015 m2

d. 6115 m2

SOLUTION:

A = 30(40) = 1200 m2

V = 1200(3.7 + 4.6 + 5.3 + 6.9)/4

V = 1200(20.5)/4

V = 6150 m2