ARCHES & CABLES

Structural Theory

Maria Cristina V. David, PhD

Chair, Department of Civil Engineering
Arch Structures
Dvortsovaya Ploshchad (Palace Square) in
St. Petersburg was built to commemorate the
1812 Russian victories over Napoleon (Top).
The Gateway Arch is a 630-foot-tall
monument in St. Louis, Missouri, United
States.
Types of Arches
Three-Hinged Arch
A three-hinged arch is a geometrically stable and statically
determinate structure. It consists of two curved members
connected by an internal hinge at the crown and is supported by
two hinges at its base. Sometimes, a tie is provided at the support
level or at an elevated position in the arch to increase the stability
of the structure.
Derivation of Equations for the Determination of Internal Forces in a
Three-Hinged Arch
Consider the section Q in the three-hinged arch shown below. The three internal
forces at the section are the axial force, NQ, the radial shear force, VQ, and the
bending moment, MQ. The derivation of the equations for the determination of
these forces with respect to the angle φ are as follows:

Bending moment at point Q.

𝑀𝜑 = 𝐴𝑦𝑥 − 𝐴𝑥𝑦 = 𝑀𝑏(𝑥) − 𝐴𝑥𝑦
where
𝑀𝑏(𝑥) = moment of a beam of the same span as the arch.
y = ordinate of any point along the central line of the arch.
 Consider the entire arch
Sample Problem No. 1
A three-hinged arch is subjected to two concentrated
loads, as shown in the figure below. Determine the
support reactions of the arch.
Consider Segment CE
Sample Problem No. 2
A parabolic arch with supports at the same level is
subjected to the combined loading shown below.
Determine the support reactions and the normal thrust
and radial shear at a point just to the left of the 150 kN
concentrated load.
Sample Problem No. 3

A parabolic arch is subjected to a

uniformly distributed load of 600 lb/ft
throughout its span, as shown in the
right. Determine the support
reactions and the bending moment at
a section Q in the arch, which is at a
distance of 18 ft from the left-hand
support.
Due to symmetry:

Taking the moment about point C of the free-body diagram suggests the following:

Consider the free-body diagram of the entire arch.

Bending moment at point Q: To find the bending moment at a point Q, which is located 18 ft from support A, first
determine the ordinate of the arch at that point by using the equation of the ordinate of a parabola.

The moment at Q can be determined as the summation of the moment of the forces on the left-hand portion of the
point in the beam, as shown below, and the moment due to the horizontal thrust, Ax. Thus, MQ = Ay(18) – 0.6(18)(9)
– Ax(11.81)
 Golden Gate
Suspension Bridge
Cables
The Golden Gate Bridge is a
Cables are flexible suspension bridge spanning
structures that support the the Golden Gate, the one-
applied transverse loads by mile-wide (1.6 km) strait
the tensile resistance connecting San Francisco Bay
and the Pacific Ocean.
developed in its members.

Jiaxing-Shaoxing Sea
Bridge (Jiashao Bridge)
Other Applications
Jiashao is a multi-pylon cable-stayed
bridge, supported by six 227m pylons. It Cables are used in
carries a two-way eight-lane expressway suspension bridges, tension
with six lanes dedicated to traffic. With leg offshore platforms,
the main span of 2,680m, the bridge transmission lines, and
runs to a total distance of 10,138m. The several other engineering
55.6m-wide bridge features a steel box applications.
girder structure with a pillar, six towers
and four cable faces.
Sample Problem 1.

The suspension bridge in the figure below is constructed using the two stiffening
trusses that are pin connected at their ends C and supported by a pin at A and a rocker
at B. Determine the maximum tension in the cable IH. The cable has a parabolic shape
and the bridge is subjected to the single load of 50 kN.
Free-Body Diagram
To determine the value of an assumed uniform distributed loading, 𝑤𝑜

To obtain the maximum tension in the cable

 Sample Problem No. 1

A cable supports two concentrated loads at B and C, as shown in Figure a. Determine the
sag at B, the tension in the cable, and the length of the cable.
 SOLUTION:
Support reactions. The reactions of the cable are determined by applying the equations
of equilibrium to the free-body diagram of the cable shown in Figure b.
 Sag at B. The sag at point B of the cable is determined by taking the moment of B, as shown in the free-body
diagram in Figure c.

• Tension in cable.

Tension at A and D.
 • Tension in segment CB

Length of cable. The length of the cable is determined as the algebraic sum of the lengths of the segments. The
lengths of the segments can be obtained by the application of the Pythagoras theorem.

Adrian Jade Hipolito

Carlo M. Pabustan
Erica P. Pingul
Jovi Sese
Sample Problem No. 2

A cable supports three concentrated loads at B, C, and D, as shown in Figure a. Determine

the sag at B and D, as well as the tension in each segment of the cable.
 SOLUTION:
Support reactions. The reactions shown in the free-body diagram of the cable in Figure
b are determined by applying the equations of equilibrium, which are written as follows:
 Sag. The sag at B is determined by summing the moment of B, as shown in the free-body diagram in Figure c, while
the sag at D was computed by summing the moment of D, as shown in the free-body diagram in Figure d.

Sag at B.

• Tension.
Sag at D. Tension at A.

Tension at E.
 • Tension B.

• Tension D.

• Tension C.

Angel Nicole Garcia

Dianne Joy Guevarra
Francesca Kate Gatus
Sean Klaude Santos
 Sample Problem 3
A cable subjected to a uniform load of 240 N/m is suspended between two supports at the same level 20
m apart, as shown in Figure 6.12. If the cable has a central sag of 4 m, determine the horizontal reactions
at the supports, the minimum and maximum tension in the cable, and the total length of the cable.

Acacio, Jelica Mae

Figure 6.12 Cable
Caballes, Keisser Lois
Supan, Patricia
Solution
Horizontal reactions. Applying the general cable Figure 6.12c, the minimum tension is as follows:
theorem at point C suggests the following:

when x= L/2 , h=4

Therefore, From equation 6.15, the maximum tension is found, as follows:s follows:

The total length of cable:

Answers:
Horizontal reactions at the support
Ax = 9000 N, Bx = 9000 N

Minimum Tension in the cable

3000 N

Maximum Tension in the cable

3841.87 N

Total length of the cable

21.93 meters
 Sample Problem 4

Determine the tension in each segment of the cable shown in the figure, what is
the dimension h?

Catap, Jahir Mae

David, Kimberly May
Dayrit, Sherina B.
Valenzuela, Kim Nicole