Arches and Their Structural Behavior
Arches
An arch is a curved structural element that
spans an opening and primarily carries loads through compression along
its curved path, transferring forces to the supports (abutments).
Arches and efficiency
- Arches
use material efficiently because applied loads mostly cause axial
compression.
- In a
special shape—the funicular shape—all sections are in direct
compression (zero bending moments).
Load considerations
- Dead
load is usually the major load on arches.
- Funicular
shape is based on dead load distribution.
- Live
loads differ in distribution, creating moments.
- These
live load moments usually produce small bending stresses compared to axial
stresses → overall net compression remains.
Applications of arches
- Used
in long-span bridges (say 400–1800 ft).
- Ideal
for large column-free buildings (airplane hangars, field houses,
convention halls).
Types of Arches
By number of hinges / base construction:
1.
Three-hinged arch
o Statically
determinate → easiest to analyze & construct.
o Not
affected by temperature changes, support settlements, or fabrication errors.
o More
flexible due to extra hinge.
(a)
three-hinged arch, stable and determinate
2.
Two-hinged arch
o Statically
indeterminate.
o Requires
advanced analysis (flexibility method or computer programs).
(b)
two-hinged arch, indeterminate to the first degree
3.
Fixed-ended arch
o Statically
indeterminate.
o Often
built of masonry/concrete with massive foundations.
(c)
fixed-end arch, indeterminate to the third degree.
4.
Tied Arch (or Bowstring Arch)
- Purpose:
- Used
when large foundation abutments are not feasible.
- Suitable
where clearance is not a problem.
- Construction:
- Supports:
pin and roller connections.
- Tie
rod connects supports.
- Behavior:
- Tie
rod carries the horizontal thrust, so abutments don’t need to
resist it.
- Structure
acts as a rigid unit.
- Insensitive
to settlement of supports.
5.
Special Types
- Barrel
arch (short-span bridges, often in parks/roads with rock sidewalls):
- Constructed
of masonry or reinforced concrete.
- Wide,
shallow arch with heavy compacted fill → ensures compression dominates
over tension.
- Stress
levels: typically, low (300–500 psi) due to large cross-sectional
area.
- Proven
durability: 19th-century masonry barrel arches still outperform modern
steel/concrete bridges (resist corrosion, carry heavier loads).
(a) Barrel
arch resembles a curved slab;
(b) barrel
arch used to support a compacted fill and roadway slab.
Three-Hinged Arches
To explain certain characteristics of arches, here we
consider how the bar forces vary as the slope θ of the bar’s changes in the
pin jointed arch shown in figure below.
Vector Diagram (b, c)
Members carry axial load only and this configuration
represents the funicular shape for an arch. Here the arch supporting a single
concentrated load P at midspan.
Reactions: For Symmetry → reactions at A and C each
carry P/2 vertically.
Slope angle: θ = slope of bars AB & CB.
Bar Forces:
This equation shows that
- As θ
increases (0° → 90°):
- Force
decreases (∞ → P/2).
- But
bar length increases → more material needed.
Volume of Bar Vs Slope
To establish the slope that produces the most economical
structure for a given span L, we will express the volume V of bar
material required to support the load P in terms of the geometry of the
structure and the compressive strength of the material.
Volume of two bars: V
= 2ALB
where A is the area of one bar and LB is the length of a
bar.
To express the required area of the bars in terms of load P,
divide the bar forces by the allowable compressive stress σallow.
The bar length LB can be written in terms of
the slope angle θ and the span L.
Substituting A and LB in equation of volume
Volume of Material vs. Arch Slope (θ)
Figure: Variation
of volume of material with slope of bars
Observations from Plot (Fig. 6.12)
- Minimum
volume → at θ=45∘.
- Shallow
arches (θ≤15∘) → require large volume.
- Deep
arches (θ≥75∘) → also require large volume.
- Moderate
slopes (30°–60°) → volume not very sensitive to θ.
- Designer
can vary shape within this range without major effect on weight or cost.
Curved Arches (Distributed Load)
- Similar
trend: material volume not sensitive to arch depth within a certain
range.
- Extremes
(too shallow or too deep) → more costly.
Practical Design Considerations
- Site
profile.
- Location
of solid foundation material.
- Architectural
+ functional requirements.
Funicular Shape of an Arch
Definition
- A funicular
arch is the profile of an arch that carries loads entirely in direct
compression (no bending moments).
- Material
usage is minimized when the arch axis follows the funicular shape.
How to Determine Funicular Shape
- Imagine
the loads supported by a flexible cable.
- The
cable naturally assumes the funicular shape for those loads.
- Invert
(flip) the cable → gives the funicular arch profile.
Load Considerations
- Dead
loads (self-weight, permanent) are usually greater than live loads
(traffic, variable).
- Designers
often base the funicular shape on dead loads, since they dominate the
structural form.
Design Insights
- Funicular
arches minimize bending → reduce material requirement.
- Used
when efficiency and economy of material are important.
- Geometry
must also fit:
- Site
profile
- Foundation
conditions
- Architectural/functional
needs
Establishing Funicular Shape:
Figure: Establishing the shape of the funicular arch:
(a) loads supported by arch applied to a cable whose sag h3 at
midspan equals the midspan height of the arch; (b) arch (produced by
inverting the cable profile) in direct stress.
E x a m p l e
Establish the shape of the funicular arch for the set of
loads acting on the trussed arch in Figure a. The rise of the arch at
midspan is set at 36 ft.
Load → Cable analogy
- Imagine
the given set of vertical loads applied to a cable spanning the
same length as the intended arch.
- Sag
of the cable = rise of the arch (here: 36 ft at midspan).
End loads (at supports)
- The
30-kip loads at each support act through the supports.
Apply General Cable Theory
- Replace
the cable with an imaginary simply supported beam of equal span.
- Apply
the loads to this beam.
Shear & Moment Diagrams
- Construct
shear (V) and moment (M) diagrams for the simply supported beam.
By the general cable theorem
Hhz = Mz
where M =
moment at an arbitrary point in the beam
H = horizontal
component of support reaction
hz = cable sag
at an arbitrary point
h = 36 ft at midspan and M = 8100 kip⋅ft
H = M/h = 8100/36 = 225 kips
Now, at x1 = 30 ft
h1 = M/H = 4500/225 = 20 ft,
At x2 = 60 ft
h2 = M/H = 7200/225 = 32 ft
If the cable profile is turned upside down, a funicular arch
is produced. When the vertical loads acting on the cable are applied to the
arch, they produce compression forces at all sections equal in magnitude to the
tension forces in the cable at the corresponding sections.
E x a m p l e
The three-hinged arch truss shown in Figure a has a
bottom chord that is same as funicular shape. To demonstrate one benefit of
using a funicular arch shape, (a) analyze the truss assuming the applied
loads represent the dead load of the structure, and (b) analyze the
truss with a single concentrated live load of 90 kips applied at joint L.
Member KJ is detailed with a slotted connection at one end so that it
cannot transmit axial force. Assume joint D acts as a hinge.
Due to symmetric pattern, the vertical reactions at A and
G are equal to 180 kips
Consider the free body of the arch to the left of the hinge
at D (Figure b), and sum moments about D
- Start
the method-of-joints analysis at support A and progress through the
truss (results sketched in Fig. b).
- Because
the arch rib follows the funicular shape for the top-chord
loading, it carries the axial force path for those loads.
- Aside
from the rib, only the vertical columns carry load (they transfer
the vertical loads down into the arch).
- The diagonals
and top-chord members remain unstressed (zero-force) for this
specific loading pattern.
- Those
members become stressed only when the applied load pattern departs from
the funicular shape — then the force flow must redistribute and
diagonals/top chord will pick up axial forces.
Truss under Single 90-kip Live Load
- When
the uniform funicular loading is applied → only the rib + verticals
carry force (many other members are zero-force).
- But
with a single concentrated 90-kip live load:
- The
loading pattern no longer matches the funicular arch shape.
- The
rib alone cannot carry the load in pure compression.
- Diagonals
and top-chord members now develop significant axial forces to
redistribute the unbalanced load.
- Result:
much fewer zero-force members compared to the funicular load case.
Funicular Shape for an Arch That Supports a Uniformly Distributed Load
- A
common load on arches is the dead load.
- This
dead load often has a uniform or nearly uniform distribution across
the entire span.
- A typical example is the weight of a bridge's
floor system, which generally has a constant weight per unit length.
·
The
analysis focuses on a uniformly loaded, symmetric, three-hinged arch.
·
The funicular shape is the specific form an arch must
take to carry loads using direct compression only (no bending moment).
·
The
analysis uses a symmetric
three-hinged arch as
the model.
·
The
arch has a span of L and
a height (or rise) of h.
·
Due
to symmetry:
o The vertical reactions at both supports are equal.
o Each vertical reaction has a magnitude
of wL/2 (half the total load).
o RA = RC = wL/2
To
find the horizontal thrust (H) at the base of the arch-
·
Consider
the free-body diagram of one half of the arch (e.g., the right side from the
crown hinge at B).
·
Apply
the equilibrium equation of summing moments about the center hinge (B) and set them equal to zero.
To
establish the equation of the axis of the arch-
·
A
coordinate system is set up with its origin (O) at the center hinge (B).
·
The vertical
y-axis is positive in the downward direction.
·
The
goal is to find the funicular shape (parabola) that ensures bending
moment = 0 everywhere.
·
Consider
section at D (distance x from the center hinge B).
·
To
do this, the bending moment (M) at an arbitrary point D on the
arch is analyzed.
·
The
moment is determined by considering the free-body diagram of the arch segment from point D
to the support at C.
·
Vertical reaction
at C = wL/2
·
Load on segment
(length L/2−x) = w(L/2−x)
·
Resultant of this
load acts at centroid (distance ½ (L/2 – x ) from
section D).
·
For an arch to be
a funicular shape, the bending moment (M) must be zero at
every section along its axis.
·
This condition (M
= 0) is substituted into the moment equation.
·
Solving the
resulting equation provides the specific mathematical relationship
between the coordinates y and x that defines the arch's curve.
This equation represents the equation of a
parabola.
·
Funicular
Shape for Uniform Load: A
parabolic arch is the funicular shape for a uniformly distributed load.
·
Implication
for Stress: This means that a
uniformly loaded parabolic arch, even a fixed-ended, will experience direct
(axial) stress only at all sections, with no bending moment.
·
Horizontal
Thrust: The horizontal thrust (H)
is constant at every section of the arch and is equal to the horizontal
reaction at the support.
·
Total Axial
Thrust: The total axial force (T)
at any section can be calculated using the horizontal thrust (H) and the
slope (θ) of the arch at that specific section.
·
Formula for
Thrust: The relationship is
given by the equation:
T = H / cos θ
To
evaluate cos θ, first differentiate parabolic equation above with
respect to x to give
·
The
value of tan θ is represented graphically by a right triangle
in the figure c.
·
Using
this triangle and the Pythagorean theorem, the length of the
hypotenuse (r) can be calculated.
·
The
formula for the hypotenuse is: r² = x² + y² (where x and y are
the legs of the triangle).
·
From
the relationship between the sides of the triangle in Figure c and the
cosine function –
Substituting this Equation into axial thrust Equation
This
Equation shows:
·
Maximum
Thrust: The
largest axial thrust (T) in the arch occurs at the supports.
·
Reason: This is because the thrust is
proportional to the horizontal distance (x), which is maximum at the
supports (x = L/2).
·
Design
Implication: For
large arches (with large w or L), the thrust force
can be very high at the supports.
·
Tapering
Solution: To
manage this, a designer can taper (vary) the cross-section of
the arch.
·
Design Goal: The cross-sectional area is made proportional to the
thrust force T. This creates a constant stress throughout the arch, leading to an
efficient use of material.
E x a m p l e
Establish
the shape of the funicular arch for the uniform loading acting on the
three-hinged arch in Figure below. To achieve economy, the arch is tapered
along its length. Determine the minimum cross-sectional area at three locations
(x1 = 17.5 ft, x2 = 35 ft, and x3 = 52.5 ft) if the
maximum allowable compressive stress is 2000 lbs/in2.
Solution
Since
this arch is uniformly loaded, the funicular shape is given by
As
the maximum allowable compressive stress is 2000 lb/in.2 and the
funicular shape of the arch, following Equation can be used to determine the
minimum cross-sectional area:
The
required cross-sectional area A :
·
If,
A key design decision is to hold the depth of the arch cross-section
constant at 10 inches.
·
The width of
the cross-section is then varied to create an economical,
constant-stress design.
·
The
required width is calculated at different points along the arch to ensure the
stress does not exceed the allowable limit.
·
The
resulting economic widths at three specific points are:
o 15.2 inches
o 17 inches
o 19.6 inches
·
This
demonstrates that the widest section (19.6 in) is required where the
thrust is greatest, typically at the supports.
Home Work
For the parabolic
arch in Figure 1,
plot the variation of the thrust T at support A for values of h
= 12, 24, 36, 48, and 60 ft.
For the
arch and loading in Figure 2, compute the reactions and determine the height of
each point. The maximum height permitted at any point along the arch, hzmax, is
20 m.
Determine
the reactions at supports A and C of the three-hinged circular arch.
The arch
shown in Figure 4 has a pin support at A and a roller at C. A tension rod
connects A and C. Determine the reactions at A and C and the tension in rod AC.
Compute
the reactions at supports A and E of the three-hinged parabolic arch in Figure
5. Next compute the shear, axial load, and moment at points B and D, located at
the quarter points.
The
three-hinged parabolic arch in Figure 6 supports 60-kip load at the quarter
points. Determine the shear, axial load, and moment on sections an
infinitesimal distance to the left and right of the loads. The equation for the
arch axis is y = 4hx2/L2.
Compute
the support reactions for the arch in Figure 7. (Hint: You will need two moment
equations: Consider the entire free body for one, and a free body of the
portion of truss to either the left or right of the hinge at B.)
Compute
the horizontal reaction at A of the arch in Figure 8.
From
Figure 9 (a) Determine the reactions and all bar forces of the three-hinged,
trussed arch for the following cases:
·
Case A: Only the 90-kN force at joint D acts.
·
Case B: Both the 90-kN and 60-kN forces at
joints D and M act.
(b)
Determine the maximum axial force in the arch in Case B.
Establish
the funicular arch for the system of loads in Figure 10, if the maximum
allowable compressive force in the arch is 85 kips.
Determine
the load P such that all the members in the three-hinged arch in Figure 11 are
in pure compression. What is the value of y1?
If the
arch rib ABCDE in Figure 12 is to be funicular for the dead loads shown at the
top joints, establish the elevation of the lower chord joints at B and D.
For the
arch rib to be funicular for the dead loads shown in figure 13, establish the
elevation of the lower chord joints B, C, and E.
The
power transmission cable weighs 150 N/m. If the resultant horizontal force on
tower BD is required to be zero, determine the sag h of cable BC.
