CE Board Exam Randomizer

⬅ Back to Subject Topics

Highway Engineering

Highway engineering is a branch of civil engineering that focuses on the planning, design, construction, operation, and maintenance of roads and highways to ensure safe, efficient, and sustainable transportation. It involves geometric design, pavement materials and structural design, traffic engineering, drainage, and safety analysis, while considering economic, environmental, and social factors. The goal of highway engineering is to provide reliable roadway systems that accommodate current and future traffic demands while minimizing accidents, travel time, and maintenance costs.

Simple Curves

A simple curve is a horizontal curve formed by one continuous circle, used to change the direction of an alignment gradually instead of abruptly at an intersection of straight lines.

Problem: Situation Involving Multiple Unknowns and Given the Bearings of Two Tangent Lines

A simple curve connects two tangents, AB and BC, with bearings of N76ºE and S63ºE, respectively. If the stationing of the vertex is 3+988 and the stationing of PC is at 3+866, determine the stationing of a point on the curve which intersects with the line making a deflection angle of 6º with the tangent passing through the PC.

Highway Engineering – Problem: Situation Involving Multiple Unknowns and Given the Bearings of Two Tangent Lines – Diagram
Highway Engineering – Problem: Situation Involving Multiple Unknowns and Given the Bearings of Two Tangent Lines – Diagram

Problem: Offset Distance

A simple curve connects two tangents, AB and BC, with bearings of N76ºE and S63ºE, respectively. If the stationing of the vertex is 3+988 and the stationing of PC is at 3+866, determine the offset distance of a point on the curve which intersects with the line making a deflection angle of 6º with the tangent passing through the PC.

Highway Engineering – Problem: Offset Distance – Diagram
Highway Engineering – Problem: Offset Distance – Diagram

Problem: Arc Basis and Fillet of Curve

If the degree of simple curve whose central angle is 26º is 6º, determine the area of the fillet of curve in m2. Use arc basis and S.I. units.

Highway Engineering – Problem: Arc Basis and Fillet of Curve – Diagram
Highway Engineering – Problem: Arc Basis and Fillet of Curve – Diagram

Problem: Radius of Curve Given the Angle of Intersection and Middle Ordinate

A simple curve has an angle of intersection of 30º. If the middle ordinate is 16m, determine the radius of the simple curve.

Highway Engineering – Problem: Radius of Curve Given the Angle of Intersection and Middle Ordinate – Diagram
Highway Engineering – Problem: Radius of Curve Given the Angle of Intersection and Middle Ordinate – Diagram

Problem: Chord Basis with English Units

A simple curve has a degree of curve of 5º. If English Units are to be used, determine the length of chord (in ft) if I=36º. Use chord basis.

Highway Engineering – Problem: Chord Basis with English Units – Diagram
Highway Engineering – Problem: Chord Basis with English Units – Diagram

Problem: Offset Distance from PT

The offset distance of the simple curve from the PT to the tangent line passing through the PC is equal to 126.2m. The stationing of PC is at 2+978. The simple curve has an angle of intersection of 26º. Determine the stationing of PT.

Highway Engineering – Problem: Offset Distance from PT – Diagram
Highway Engineering – Problem: Offset Distance from PT – Diagram

Compound Curves

Compound curves are horizontal curves in highway and railway engineering that consist of two or more simple circular arcs of different radii, joined together and curving in the same direction, with a common tangent at their point of connection. They are used when site conditions or alignment constraints make a single-radius curve impractical, allowing a smoother transition between tangents while fitting the roadway within limited right-of-way or difficult terrain.

Problem: Length of Common Tangent and Common Chord

Two simple curves are connected together as physical obstructions do not permit the use of a single circular curve. The radius of the first curve is 400m while that of the second is 300m. The angles of intersection of the first and second curve are 36º and 50º, respectively. The stationing of PC is 4+509. Determine the following:
a. Length of the common tangent
b. Stationing of PT
c. Stationing of the vertex
d. Length of the common chord

Highway Engineering – Problem: Length of Common Tangent and Common Chord – Diagram
Highway Engineering – Problem: Length of Common Tangent and Common Chord – Diagram Highway Engineering – Problem: Length of Common Tangent and Common Chord – Diagram Highway Engineering – Problem: Length of Common Tangent and Common Chord – Diagram Highway Engineering – Problem: Length of Common Tangent and Common Chord – Diagram

Problem: Compound Curve where the Common Tangent is Parallel to the Long Chord

The long chord from the PC to the PT of a compound curve is 300 meters long and the angles it makes with the longer and shorter tangents are 12º and 15º respectively. If the common tangent is parallel to the long chord.
a. Find the radius of the first curve.
b. Find the radius of the second curve.
c. If the stationing of PC is 10+204.30, find the stationing of PT.

Highway Engineering – Problem: Compound Curve where the Common Tangent is Parallel to the Long Chord – Diagram
Highway Engineering – Problem: Compound Curve where the Common Tangent is Parallel to the Long Chord – Diagram Highway Engineering – Problem: Compound Curve where the Common Tangent is Parallel to the Long Chord – Diagram

Problem: Compound Curve Given Three Tangents with Respective Azimuths

A compound curve connects three tangents having azimuths of 254º, 270º, and 280º, respectively. The length of the chord is 320m long measured from the PC to the PT of the curve, and it is parallel to the common tangent having an azimuth of 270º. If the stationing of PT is 6+520:
a. Determine the total length of the curve.
b. Determine the stationing of the PCC.
c. Determine the stationing of PC.

Highway Engineering – Problem: Compound Curve Given Three Tangents with Respective Azimuths – Diagram
Highway Engineering – Problem: Compound Curve Given Three Tangents with Respective Azimuths – Diagram

Problem: Compound Curve Given the Length of the Common Tangent and Azimuths | Chord Basis

The common tangent AB of a compound curve is 76.42m with an azimuth of 268º30', the vertex V being inacessible. The azimuths of the tangents AV and VB were measured to be 247º50' and 282º50', respectively. The stationing at A is 10+010.46 and the degree of the first curve is 4º based on the 20m chord. Use chord basis.
a. Compute the stationing of PCC
b. Compute the radius of the second curve
c. Compute the stationing of PT
d. Compute the length of the common chord.

Highway Engineering – Problem: Compound Curve Given the Length of the Common Tangent and Azimuths | Chord Basis – Diagram
Highway Engineering – Problem: Compound Curve Given the Length of the Common Tangent and Azimuths | Chord Basis – Diagram Highway Engineering – Problem: Compound Curve Given the Length of the Common Tangent and Azimuths | Chord Basis – Diagram

Reversed Curves

A reversed curve is a horizontal alignment consisting of two consecutive circular curves that bend in opposite directions and are joined directly without an intervening tangent; its engineering significance lies in the abrupt reversal of centrifugal force acting on vehicles, which can reduce riding comfort and compromise safety if not properly designed. Because drivers must immediately adjust steering in the opposite direction, reversed curves are generally avoided on high-speed facilities unless adequate transition curves, proper superelevation development, and sufficient sight distance are provided. When unavoidable due to terrain or right-of-way constraints, careful geometric design is essential to minimize operational hazards and ensure smooth vehicle maneuvering.

Reverse Curve with Parallel Tangents

Concept ${i};

Reverse Curve with Converging Tangents

Concept ${i};

Reverse Curve with Intermediate Tangent

Concept ${i};

Problem: Reversed Curve with Parallel Tangents

The perpendicular distance between two parallel tangents is equal to 8 meters. These tangents are connected by a reversed curvature such that the central angle is 8º and the radius of curvature of the first curve is equal to 175m. Find the radius of the second curve of the reversed curvature.

Highway Engineering – Problem: Reversed Curve with Parallel Tangents – Diagram
Highway Engineering – Problem: Reversed Curve with Parallel Tangents – Diagram

Problem: Reversed Curve with Parallel Tangents and Connected by Intermediate Tangent

Two parallel tangents have directions of due east and are 200 meters apart. They are connected by a reversed curve having the same degree of 1.4º. PC of the curve is on the upper tangent while the PT of the curve is at the lower tangent. If the horizontal distance parallel to the tangent from the PC to the PT of the reversed curve is 800m, determine:
a. The distance of the intermediate tangent between the two curves.
b. The distance between the centers of the reversed curvature.
c. The stationing of PT if the stationing of PC is 10+200.

Highway Engineering – Problem: Reversed Curve with Parallel Tangents and Connected by Intermediate Tangent – Diagram
Highway Engineering – Problem: Reversed Curve with Parallel Tangents and Connected by Intermediate Tangent – Diagram

Problem: Reversed Curve with Converging Tangents

A reverse curve connects two converging tangents with an angle of intersection of 32º. The distance from PI of the second curve to this point of intersection is 180m. The deflection angle of the common tangent from the back tangent is 22º and its degree of curve is 5º.
a. Determine the radius of the 2nd curve.
b. Determine the tangent distance of the 2nd curve.
c. Determine the degree of curve of the 1st curve by arc basis.
If the stationing of PC is 3+600, determine the stationing of PT.

Highway Engineering – Problem: Reversed Curve with Converging Tangents – Diagram
Highway Engineering – Problem: Reversed Curve with Converging Tangents – Diagram

Vertical Curves: Symmetrical Parabolic Curve

Highway Engineering – Vertical Curves: Symmetrical Parabolic Curve – Diagram

In highway practice, an abrupt change in the vertical direction of moving vehicles should be avoided. To provide a gradual change in vertical direction, a parabolic vertical curve is adopted because its slope varies at a constant rate with respect to horizontal distance.

  1. The vertical offsets from the tangent to the curve are proportional to the squares of the distances from the point of tangency.
  2. The curve bisects the distance between the vertex and the midpoint of the long chord.
  3. If the algebraic difference in the rates of grade of the two slopes is positive, i.e., $(g_1-g_2)>0$, we have a summit curve. Otherwise, we have a sag curve.
  4. The length $L$ of a vertical parabolic curve refers to the horizontal distance from the PC to the PT.
  5. The stationing of vertical parabolic curves is measured not along the curve, but along a horizontal line.
  6. For a symmetrical parabolic curve, the number of stations to the left must be equal to the number of stations to the right of the intersection of the slopes (forward and backward tangents).
  7. The slope of the parabola varies uniformly along the curve.
  8. The maximum offset $H$ is $\tfrac{1}{8}$ of the product of the algebraic difference between the two rates of grade and the length of curve.

    Summary of Formulas in Symmetrical Parabolic Curves

    Location of the highest point from PC:

    $$s_1=\frac{g_1L}{g_1-g_2}$$

    Location of the highest point from PT:

    $$s_2=\frac{g_2L}{g_2-g_1}$$
    $$H=(g_1-g_2)\cdot\frac{L}{8}$$
    $$\text{rate of change of grade}=\frac{g_1-g_2}{n}$$
    n = no. of stations
    1 station = 100ft in English Units
    1 station = 20m in S.I. Units

    Grade Diagram for Vertical Parabolic Curves

    A grade diagram is a graphical representation of how the slope (grade) varies along the horizontal distance of the curve. Since the slope of a parabolic curve varies uniformly, the grade diagram is a straight line.

    Steps in Constructing the Grade Diagram

    1. Determine the algebraic difference in grade:
      Compute
      $$ A = g_2 - g_1 $$
      where $g_1$ is the initial grade and $g_2$ is the final grade.
    2. Determine the length of curve:
      Let $L$ be the horizontal length from PC to PT.
    3. Compute the rate of change of grade:
      $$ r = \frac{g_2 - g_1}{L} $$
      This represents the constant rate at which the slope changes per unit horizontal distance.
    4. Establish horizontal axis:
      Lay out the horizontal distance from PC (0) to PT ($L$).
    5. Plot the initial and final grades:
      At $x = 0$, ordinate = $g_1$. At $x = L$, ordinate = $g_2$.
    6. Join the two grade points with a straight line:
      Since the slope varies uniformly, the grade diagram is linear.

    Important Interpretations of the Grade Diagram

    • The ordinate of the grade diagram at any horizontal distance $x$ represents the slope (grade) of the curve at that point.
    • The area under the grade diagram between two points represents the difference in elevation between those two points.

    Mathematically, since grade is defined as the rate of change of elevation:

    $$ g = \frac{dy}{dx} $$

    The elevation difference between two points is therefore:

    $$ \Delta y = \int g \, dx $$

    Thus, the grade diagram is essentially a graphical integration tool — the area under the grade line gives the vertical change in elevation.

Problem: Symmetrical Vertical Summit Curve with Allowable Rate of Change of Grade

A symmetrical vertical summit curve has tangents of +4% and -2%. The allowable rate of change of grade is 0.3% per meter station. The stationing and elevation of PT is 10+020 and 142.63m, respectively.
a. Compute the length of curve.
b. Compute the distance of the highest point of the curve from the PC.
c. Compute the elevation of the highest point of the curve.
d. Determine the stationing of the highest point of the curve.
e. Determine the elevation of a point at station 9+720.

Highway Engineering – Problem: Symmetrical Vertical Summit Curve with Allowable Rate of Change of Grade – Diagram
Highway Engineering – Problem: Symmetrical Vertical Summit Curve with Allowable Rate of Change of Grade – Diagram

Problem: Sag Vertical Curve with English Units

A symmetrical sag vertical curve is shown. Determine the elevation of the lowest point of the curve.

Highway Engineering – Problem: Sag Vertical Curve with English Units – Diagram
Highway Engineering – Problem: Sag Vertical Curve with English Units – Diagram

Problem: Symmetrical Sag Vertical Curve | Invert Elevation and all Possible Cases of the Grade Diagram

A symmetrical sag vertical curve has a back tangent grade of -3.2% and a forward tangent grade of 2.4%, intersecting at the point PVI whose stationing is 2+200 at an elevation of 700.28m. The stationing of PVC is 1+900.
a. At what station should the cross-drainage pipes be situated?
b. What is the elevation of the lowest point?
c. If the overall outside dimensions of the reinforced concrete pipe to be installed is 95cm, and the top of the culvert is 0.3m below the subgrade, what will be the invert elevation (elevation of the lowest point of the pipe)?
d. Determine the elevation at station 2+120.
e. Determine the elevation at station 2+270.

Highway Engineering – Problem: Symmetrical Sag Vertical Curve | Invert Elevation and all Possible Cases of the Grade Diagram – Diagram
Highway Engineering – Problem: Symmetrical Sag Vertical Curve | Invert Elevation and all Possible Cases of the Grade Diagram – Diagram Highway Engineering – Problem: Symmetrical Sag Vertical Curve | Invert Elevation and all Possible Cases of the Grade Diagram – Diagram Highway Engineering – Problem: Symmetrical Sag Vertical Curve | Invert Elevation and all Possible Cases of the Grade Diagram – Diagram Highway Engineering – Problem: Symmetrical Sag Vertical Curve | Invert Elevation and all Possible Cases of the Grade Diagram – Diagram

Unsymmetrical Parabolic Curves

An unsymmetrical parabolic curve is a vertical curve in which the lengths of the curve on either side of the point of vertical intersection (PVI) are not equal. Unlike symmetrical curves—where the PVI lies at the midpoint—an unsymmetrical curve has different distances from the PVI to the beginning (PC) and to the end (PT) of the curve. These curves are commonly used when site constraints, existing structures, drainage requirements, or design limitations prevent equal curve lengths on both sides. Although the curve remains parabolic and the rate of change of grade is still constant, the vertex does not lie at the midpoint of the curve.

Highway Engineering – Unsymmetrical Parabolic Curves – Diagram Highway Engineering – Unsymmetrical Parabolic Curves – Diagram

Let:

1. Maximum Offset (Unsymmetrical Curve)

$$ H = \frac{(g_1 - g_2)L_1 L_2}{2L} $$

2. Location of Highest/Lowest Point (HP)

Grade at HP:

$$ g_3 = 0 $$

Interpretation:

3. Relationship of Intermediate Grade

$$ g_3 = g_1 - \frac{2H}{L_1} $$
OR
$$ g_3 = g_2 + \frac{2H}{L_2} $$

Consider algebraic signs of grades.

4. Station to Highest/Lowest Point

If $g_3 < 0$:

$$ S_1 = \frac{g_1 L_1^2}{2H} $$

If $g_3 > 0$:

$$ S_2 = \frac{g_2 L_2^2}{2H} $$

Problem: Unsymmetrical Parabolic Curve — Summit Curve

An unsymmetrical parabolic curve has a forward tangent of -8% and a back tangent of +5%. The length of curve on the left side of the curve is 40m long while that of the right side is 60m long. The PC is at station 6+780 and has an elevation of 110m.
a. Determine the elevation of the highest point.
b. Determine the elevation at 6+840.

Highway Engineering – Problem: Unsymmetrical Parabolic Curve — Summit Curve – Diagram
Highway Engineering – Problem: Unsymmetrical Parabolic Curve — Summit Curve – Diagram

Step 1: Compute Maximum Offset $H$

$$ H = \frac{(g_1 - g_2)L_2 L_1}{2L} $$ $$ H = \frac{(0.05 - (-0.08))(60)(40)}{2(100)} $$ $$ H = 1.56 \text{ m} $$

Step 2: Locate the Highest Point (HP)

$$ g_3 = g_1 - \frac{2H}{L_1} = 0.05 - \frac{2(1.56)}{40} = -0.028 $$

OR

$$ g_3 = g_2 + \frac{2H}{L_2} = -0.08 + \frac{2(1.56)}{60} = -0.028 $$

Since $g_3 < 0$, the highest point is located to the left of the PVI.

$$ S_1 = \frac{g_1 L_1^2}{2H} = \frac{0.05(40)^2}{2(1.56)} = 25.64 \text{ m} $$

Step 3: Compute Offset at HP using SPP

$$ \frac{y_1}{(25.64)^2} = \frac{1.56}{40^2} $$ $$ y_1 = 0.64 \text{ m} $$

Step 4: Elevation of the High Point (Answer to Part a)

Using: $\text{El. HP} = \text{Elev. PC} + g_1 S_1 - y_1$

$$ \text{El. HP} = 110 + 0.05(25.64) - 0.64 $$ $$ = 110.642 \text{ m} $$

Step 5: Elevation at Station 6+840 (Answer to Part b)

First compute elevation of PVT:

$$ \text{El. PVT} = 110 + 0.05(40) - 0.08(60) $$ $$ = 107.2 \text{ m} $$

Compute offset using SPP:

$$ \frac{y_2^2}{40^2} = \frac{1.56}{60^2} $$ $$ y_2 = 0.693 \text{ m} $$

Using: Elevation = El. PVT + $g_2(40) - y_2$

$$ \text{El. } 6+840 = 107.2 + 0.08(40) - 0.693 $$ $$ = 109.707 \text{ m} $$

For the grade diagram in unsymmetrical parabolic curves, we do not connect the back tangent grade and the forward tangent grade using a straight line. Instead, we divide the diagrams into two. First, we plot the ordinate of the back tangent (+5% or +0.05), then we plot the ordinate of PVI (g3=-2.8% or -0.028), and we connect these two points with a straight line. Next, we plot the ordinate of the forward tangent (-8% or -0.08), and connect this to the ordinate of PVI using another straight line. The area under the diagram still defines the difference in elevation between two points, just like in symmetrical parabolic curves. Here, we compute the elevation of the highest point and the elevation at 6+840 using the grade diagram and verify that we still obtain the appropriate values. The small discrepancy is only due to rounding off error.

Highway Engineering – Problem: Unsymmetrical Parabolic Curve — Summit Curve – Diagram
Scroll to zoom