Cooper Rivet Bridge in Charleston, South Carolina

The nell\ll Ooarier Rf\/_en Brfdge is a cao/e-stay_ed 6rfdge witfi a main span of 1,5416 ff~.
Completeo ,lo 2006, Ifie bridge prol!Jides a 1,000-f~ navigaltional cbannel wiith a m,lnimum
Leet, K.L, Uang, C.M. and Gilbert, A. (2011). Fundamentals of struc
~ertica/ clearance of 186 £t above tbe river Each !J74-ft tall, oiamono-sharieo concrete
tower is protected from sfiip r:ol!iisian oy a rock fs/aod surrounding the base ofl tfie rower.

analysis (4th ed.). Boston: McGraw-Hill. Chapters 6 and 7.

 H A- ! I

fhapter Objectives · ·
'I Study the characteristics and behavior of cable structures. Since cables are flexible, they carry tension
J forces only and compression and bending stresses equal zero along the cable.
·• Analyze determinate cable structures and calculate support reactions by two methods, namely by
! equations of static equilibrium and by the general cable theorem, as well as determine the cable
·t: forces at specific points along its length.
• Generate an efficient funicular shape of an arch in direct compressive stress by utilizing an inverted
cable under its self-weight.

- '
r·······································································································
__1 Introduction

- As we discussed in Section 1.5, cables constructed of high-strength steel

wires are completely flexible and have a tensile strength four or five times
greater than that of structural steel. Because of their great strength-to-weight
compression
ratio, designers use cables to construct long-span structures, including suspen- cable ring
sion bridges and roofs over large arenas and convention halls. To use cable
construction effectively, the designer must deal with two problems:
1. Preventing large displacements and oscillations from developing in
cables that carry live loads whose magnitude or direction changes
with time.
2, Providing an efficient means of anchoring the large tensile force
carried by cables.
To take advantage of the cable's high strength while minimizing its
negative features, designers must use greater inventiveness and imagination
tension vertical
than are required in conventional beam and column structures. For example, ring support
Figure 6.1 shows a schematic drawing of a roof composed of cables
connected to a center tension ring and an outer compression ring. The small Figure 6.1: Cable-supported roof composed
of three elements: cables, a center tension
center ring, loaded symmetrically by the cable reactions, is stressed primarily ring, and an outer compression ring.
in direct tension while the outer ring carries mostly axial compression. By
creating a self-balancing system composed of members in direct stress, the
227
228 Chapter 6 • Cables

l+--------L -------+!

(a) (b)

Figure 6.2: Ve1tically loaded cables: (a)

cable with an inclined chord-the vertical
distance between the chord and the cable, h, is
called the sag; (b) free body of a cable seg-
ment carrying vertical loads; although the re-
sultant cable force T. varies with the slope of
the cable, }:,Fx = 0 requires that H, the hori-
zontal component of T. is constant from sec-
tion to section.

Photo 6.1: Terminal building at Dulles

airport. Roof supported on a net of steel ca-
bles spanning between massive, sloping;re-
inforced concrete pylons.

designer creates an efficient structural form for gravity loads that requires
only vertical supports around its perimeter. A number of sports arenas,
including Madison Square Garden in New York City, are roofed with a cable
system of this type.
In a typical cable analysis the designer establishes the position of the end
supports, the magnitude of the applied loads, and the elevation of one other
point on the cable axis (often the sag at midspan; see Figure 6.2a). Based on
these parameters, the designer applies cable theory to ctlmpute the end reac-
tions, the force in the cable at all other points, and the position of other points
along the cable axis .

.. ;
6.2
Characteristics of Cables
Cables, which are made of a group of high-strength wires twisted together to
2
form a strand, have an ultimate tensile strength of approximately 270 kips/in
(1862 MPa). The twisting operation imparts a spiral pattern to the individual
~
wires.
 6.3 • Variation of Cable Force 229

Photo 6.2: Cable-stayed bridge over

Tampa bay.

While the drawing of wires through dies during the manufacturing

process raises the yield point of the steel, it also reduces its ductility. Wires can
undergo an ultimate elongation of 7 or 8 percent compared to 30 to 40 percent
for structural steel with a moderate yield point, say, 36 kips/in 2 (248 MPa).
Steel cables have a modulus of elasticity of approximately 26,000 kips/in2
(179 GPa) compared to a modulus of 29,000 kips/in 2 (200 GPa) for structural
steel bars. The lower modulus of the cable is due to the uncoiling of the wire's
spiral structure under load.
Since a cable carries only direct stress, the resultant axial force Ton all
sections must act tangentiallY' to the longitudinal axis of the cable (see
Figure 6.2b). Because a cable lacks flexural rigidity, designers must use
great care when designing cable structures to ensure that live loads do not
induce either large deflections or vibrations. In early prototypes, many
cable-supported bridges and roofs developed large wind-induced displace-
ments (flutter) that resulted in failure of the structure. The complete destruc-
tion of the Tacoma Narrows Bridge on November 7, 1940 by wind-induced
oscillations is one of the most spectacular examples of a structural failure
of a large cable-supported structure. The bridge, which spanned 5939 ft
(1810 m) over Puget Sound near the City of Tacoma, Washington, devel-
oped vibrations that reached a maximum amplitude in the vertical direction
of28 ft (8.53 m) before the floor system broke up and dropped into the water
below (see Photo 2.1).

··! , ..................•....................•....................•..........................................
6.3
Variation of Cable Force
If a cable supports vertical load only, the horizontal component Hof the cable
tension Tis constant at all sections along the axis of the cable. Thi~ conclu-
sion can be demonstrated by applying the equilibrium equation 2,F_y = 0 to ,a
 230 Chapter 6 • Cables

segment of cable (see Figure 6.2b ). If the cable tension is expressed in terms
of the horizontal component Hand the cable slope e,

H
T=-- (6.1)
cos (J

At a point where the cable is horizontal (e.g., see point B in Figure 6.2a),
equals zero. Since cos O = 1, Equation 6.1 shows that T = H. The maximum
(J
value of T typically occurs at the support where the cable slope is largest.

'
,·······················································································~···············
.
Analysis of a Cable Supporting Gravity (Vertical) Loads
When a set of concentrated loads is applied to a cable of negligible weight, the
cable deflects into a series of linear segments (Figure 6.3a). The resulting
Ay Dy shape is called the funicular polygon. Figure 6.3b shows the forces acting at
pointB on a cable segment of infinitesimal length. Since the segment is in equi-
H H
librium, the vector diagram consisting of the cable forces and the applied load
forms a closed force polygon (see Figure 6.3c ).
A cable supporting vertical load (e.g., see Figure 6.3a) is a determinate
B member. Four equilibrium equations are available to compute the four reac-
Pz tion components supplied by the supports. These equations include the three
P, equations of static equilibrium applied to the free body of the cable and a con-
(a) dition equation, ""2:,Mz = 0. Since the moment at all sections of the cable is
zero, the condition equation can be written at any section as long as the cable
FAB sag (the vertical distance between the cable chord and the cable) is known.
Typically, the designer sets the maximum sag to ensure both a required clear-
ance and an economical design.
Fsc To illustrate the computations of the support reactions and the forces at
B various points along the cable axis, we will analyze the cable in Figure 6.4a.
The cable sag at the location of the 12-kip load is set at 6 ft. In this analysis
we will assume that the weight of the cable is trivial (compared to the load)
and neglect it.

STEP 1 Compute Dy by surmning moments about supp?rt A.

(b)

r,L>
o+
=o ""2:,MA
(12 kips)(30) + (6 kips)(70) - Dy(lOO) = 0
Dy= 7.8 kips (6.2)

Fnc
STEP 2 Compute Ay.
(c) +
t ""2:,Fy =0
Figure 6.3: Vector diagrams: (a) cable with
two vertical loads; (b) forces acting on an 0 = Ay - 12 - 6 + 7.8
infinitesimal segment of\ cable at B; (c) force ff

polygon for vectors in (b). ' Ay = 10.2 kips (6.3)

1
 6.4 • Analysis of a Cable Supporting Gravity (Vertical) Loads 231

3 Compute H; sum moments about B (Figure 6.4b).

C+ "'2:.Ms = 0
0 = Ay(30) - Hh 8

h8 H = (10.2)(30) (6.4)

Setting h8 = 6 ft yields
H = 51 kips
After H is computed, we can establish the cable sag at C by
1.-- 30'---l-- 40'----f._ 30'__,.j
considering a free body of the cable just to the right of C '. (a)
(Figure 6.4c).
Ay = 10.2 kips

C+ 2-Mc = 0
-Dy(30) + Hhc = 0
30Dy 30(7.8)
h = -- = = 4.6 ft (6.5)
c H 51
(b)

To compute the force in the three cable segments, we establish Dy= 7.8kips
and (} c and then use Equation 6.1.
(}A, (} 8 ,

H=5lkips
Compute TAB·
6
tan (} A = and (}A = 11.31 °
30
1
H 51 (c)
'I';is = -- = -- = 51.98 kips
COS (}A 0.981 12kips 6kips

Compute Tse·
6 - 4.6 A
A !B !c A
D

tan () 8 =

H
40.
= 0.035

51
and
t-
10.2 kips
30' ~ 40' ___..(._ 30' _,f
7.8 kips
Tse= cos Os = 0.999 = 51.03 kips

Compute Tco·
4.6
=
tan Oc
30 = 0.153 and
(d)
H 51
Teo= - - = -- = 51.62 kips Figure 6.4: (a) Cable loaded with vertical
cos Oc 0.988
forces, cable sag atB set at 6 ft; (b) free body
Since the slopes of all cable segments in Figure 6.4a are relatively of cable to left of B; (c) free body of cable to
small, the computations above show that the difference in magnitude right of C; (d) a simply supported beam with
between the horizontal component of cable tension H and the total cable same loads and span as cable (moment dia-
force Tis small. ~ gram below).
232 Chapter 6 • Cables

6.5
........................................................................................................
:

General Cable Theorem

As we carried out the computations for the analysis of the cable in
Figure 6.4a, you may have observed that certain of the computations are
similar to those you would make in analyzing a simply supported beam with a
span equal to that of the cable and carrying the same loads applied to the cable.
For example, in Figure 6.4c we apply the cable loads to a beam whose span
equals that of the cable. If we sum moments about support A to compute the
vertical reaction Dy at the right support, the moment equation is identical to
Equation 6.2 previously written to compute the vertical reaction at the right
support of the cable. In addition, you will notice that the shape of the cable and
the moment curve for the beam in Figure 6.4 are identical. A comparison of the
computations between those for a cable and those for a simply supported beam
that supports the cable loads leads to the following statement of the general
cable theorem:

At any point on a cable supporting vertical loads, the product of the cable
sag h and the horizontal component H of the cable tension equals the
bending moment at the same point in a simply supported beam that car-
ries the same loads in the same position as those on the cable. The span of
the beam is equal to that of the cable.

The relationship above can be stated by the following equation:

(6.6)

where H = horizontal component of cable tension

hz = cable sag at point z where Mz is evaluated
Mz = moment at point z in a simply supported beam carrying the
loads applied to the cable

Since H is constant at all sections, Equation 6.6 shows that the cable sag h is
proportional to the ordinates of the moment curve.
To verify the general cable theorem given\_by Equation 6.6, we will
show that at an arbitrary point z on the cable axis the product of the
horizantal component H of cable thrust and the cable sag hz equals
the moment at the same point in a simply supported beam carrying the cable
loads (see Figure 6.5). We will also assume that the end supports of the
cable are located at different elevations. The vertical distance between
the two supports can be expressed in terms of a, the slope of the cable chord,
and the cable span L as
y=Ltana (6.7)

Directly below the cable we show a simply supported beam to which we apply
the cable loads. The distance between loads is the same in both members. In
both the cable and the beam,~the arbitrary section at which we will evaluate
the terms in Equation 6.6 is located a ?istance x~to the right of the left support.
 I
6.5 • General Cable Theorem 233

chord
0 :TH
\ ------------- y=Ltana
-----

x
I: L

(a)

A 1' j2 T TB;i,
A jz

t t
Rn

x
I: L

(b)

Figure 6.5

We begin by expressing the ve1tical reaction of the cable at support A in terms

of the vertical loads and H (Figure 6.5a).

C+ "'2:,M8 =0
0 = AyL - Lm 8 + H(L tan a) (6.8)
\_

where Lm 8 represents the moment about suppmt B of the ve1tical loads

(P 1 through P 4 ) applied to the cable.
In Equation 6.8 the forces Ay and Hare the unknowns. Considering a free
body to the left of point z, we sum moments about point z to produce a second
equation in terms of the unknown reactions Ay and H.

0 = Ayx + H(x tan a - hz) - Lmz (6.9)

where Lmz represents the moment about z of the loads on a free body of the
cable to the left of point z. Solving Equation 6.8 for Ay gives ~

Lm 8 - H(L tan a)
Ay = L (6.10)
234 Chapter 6 • Cables

Substituting Ay from Equation 6.10 into Equation 6.9 and simplifying,

we find

(6.11)

We next evaluate Mz, the bending moment in the beam at point z (see
Figure 6.5b):
(6.12)

To evaluate RA in Equation 6.12, we sum moments of the forces about the

roller at B. Since the loads on the beam and the cable are identical, as are the
spans of the two structures, the moment of the applied loads (P 1 through P4 )
about B also equals Lm8 •

c+ "2:,MB = o
0 = RAL - L,,nB
"2:,,nB
RA=-- (6.13)
L
Substituting RA from Equation 6.13 into Equation 6.12 gives

"2:,,nB
M
z
= x - - - "2:,m
L z
(6.14)

Since the right sides of Equations 6.11 and 6.14 are identical, we can equate
the left sides, giving, Hhz = Mz, and Equation 6.6 is verified ..

............•...........................•...............................................................
6.6 !
' Establishing the Funicular Shape of an Arch
'11he material required to construct an arch is minimized when all sections
along the axis of the arch are in direct stress. For a particular set of loads the
arch profile in direct stress is called the funicular arch. By imagining that the
loads carried by the arch are applied to a cable, the designer can automatically
Figure 6.6: Establishing the shape of the
funicular arch: (a) loads supported by arch 1 generate a funicular shape for the loads. If the cable shape is turned upside
applied to a cable whose sag h3 at midspan down, the designer produces a funicular arch. Since dead loads are usually
equals the midspan height of the arch; much greater than the live loads, a designer might use t~em to establish the
(b) arch (produced by inverting the cable funicular shape (see Figure 6.6). '
profile) in direct stress.

(a) (b)
~
 6.6 • Establishing the Funicular Shape of an Arch 235

EXAMPLE 6.1
etermine the reactions at the supports produced by the 120-kip load at
··dspan (Figure 6.7) (a) using the equations of static equilibrium and (b)
ing the general cable theorem. Neglect the weight of the cable.

Since supports are not on the same level, we must write two equilibrium
' equations to solve for the unknown reactions at support C. First consider
Figure 6.7a.

c+ ~MA= o
0 = 120(50) + SH - lOOCy (1)

Next consider Figure 6.7b.

0+ ~Mn= o
o = 10.sH - soc)'
50
H = 10.5 CY (2) 120 kips
!:--- 5 0 ' - - - 50' ______.j
bstitute H from Equation 2 into Equation 1.
(a)

c>'
o = 6000 + s(~cy)
10.5
- wocr
IO~H
Cy = 78.757 kips Ans.
I
T 1...------ 50' ______.j
bstituting Cy into Equation 2 yields (b)

50 .
H = ~(78.757) = 375 kips Ans.
10.5
g the general cable theorem, apply Equation 6.6 at midspan where the AA B ~
• e sag h2 = 8 ft and M 2 = 3000 kip· ft (see Figure 6.7c).
} -50'~,..------ 50' ---4
RA = 60 kips Re = 60 kips

H(8) = 3000 M = 3000 kip·ft

~
H = 375 kips Ans.
r His evaluated, sum moments about A in Figure 6.7a to compute
78.757 kips. (c)

Figure 6.7: (a) Cable with a vertical load at

Although the vertical reactions at the supports for the cable in midspan; (b) free body to the right of B;
• 7a and the beam in Figure 6.7c are not the same, the final results are (c) simply supported beam with same length
as cable. Beam supports cable load.
~

~ - - - - - - - - - - - - - - _ , ,_ _ ---=--~~~~-';LE,~,~~~r.G
236 Chapter 6 • Cables

EXAMPLE 6.2
A cable-suppmted roof carries a uniform load w = 0.6 kip/ft (see Figure 6.8a). If
the cable sag at midspan is set at 10 ft, what is the maximum tension in the cable
(a) between points B and D and (b) between points A and B?

Solution
(a) Apply Equation 6.6 at midspan to analyze the cable between points B and D.
Apply the uniform load to a simply supported beam and compute the
w =0.6 kip/ft moment Mz at midspan (see Figure 6.8c). Since the moment curve is a

A=- ~~r,~10:~ 1 I I I I parabola, the cable is also a parabola between points Band D.

wL2
Hh=M = -
z 8

0.6(120)2
H(lO) =
8
1--40 1 --4---60' 60' .1. 40•-J
H = 108 kips
(a)
Tihe maximum cable tension in span BD occurs at the supports where the
slope is maximum. To establish the slope at the supports, we differentiate
the equation of the cable y = 4hx 2/L 2 (see Figure 6.8b).

dy 8hx
tanf)=-=-
dx L2

atx = 60ft, tan()= 8(10)(60)/(120) 2 = t and()= 18.43°:

(b)
cos 6 = 0.949
w =0.6 kip/ft

B£ 1 1 L 1 1 ): Substituting into

H
T=-- (6.1)
1
~60 - - 6 0
1
-{
cos ()
108
wL2 T = - - = 113.8 kips Ans.

~
0.949

(b) If we neglect the weight of the cable between points A and B, the cable can
moment diagram be treated as a straight member. Since the cable slope () is 45°, the cable
(c) tension equals

Figure 6.8 H 108 .

T = - - = - - = 152.76 kips Ans.
cos 6 0.707
 Bixby Creek Bridge in Big Sur, California \
The wide use at elegant open-spandre/, reinforced concrete arch bridges in the first halt
at the 20th century has resulted in many notable historic bridges in the United States,
including the Bixby Creek Bridge. It is 714 ft long with a 320-ft main span and is over
280 ft high. The seismic retrofit, completed in 1996, ensures not only the bridge will
remain stable during intense earthquake shaking but a/so addresses important aesthetic
and environmental concerns during the retrofit construction over the ecologically
sensitive and inaccessible canyon.

j
.1;~·. -.'
'C
I "_-
-
-
--

-
Arches

Chapter Objectives
• Study the characteristics, types, and behavior of arch structures.
• Analyze determinate three-hinged arches and trussed arches.
• Establish a funicular arch such that the forces are in direct compression along the arch, producing
an efficient minimum weight arch utilizing the general cable theorem.

7.1
-,
'.• .......................................................................................................
Introduction
As we discussed in Section 1.5, the arch uses material efficiently because
applied loads create mostly axial compression on all cross sections. In this
chapter we show that for a particular set of loads, the designer can establish
one shape of arch- the funicular shape- in which all sections are in direct
compression (moments are zero).
Typically, dead load constitutes the major load supported by the arch. If
a funicular shape is based on the dead load distribution, moments will be cre-
ated on cross sections by live loads whose distribution differs from that of the
dead load. But normally in most arches, the bending stresses produced by live
load moments are so small compared to the axial stresses that net compression
stresses exist on all sections. Because arches use material efficiently, designers
often use them as the main structural elements in long-span bridges (say, 400
to 1800 ft) or buildings that require large column-free areas, for example, air-
plane hangers, field houses, or convention halls.
In this ·chapter we consider the behavior and analysis of three-hinged
arches. As pa1t of this study, we derive the equation for the shape of a funic-
ular arch that supports a uniformly distributed load, and we apply the general
cable theory (Section 6.5) to produce the funicular arch for an arbitrary set of
concentrated loads. Finally, we apply the concept of structural optimization to
establish the minimum weight of a simple three-hinged arch carrying a con-
centrated load.

243
 -----
244 Chapter 7 • Arches

•·I 1::2~ ~ r·······································································································

I

i .I Types of Arches
Arches are often classified by the number of hinges they contain or by the
The Romans mastered the manner in which their bases are constructed. Figure 7.1 shows the three main
construction of arch structures types: three-hinged, two-hinged, and fixed-ended. The three-hinged arch is stat-
ically determinate; the other two types are indeterminate. The three-hinged arch
thousands of years ago by
is the easiest to analyze and construct. Since it is determinate, temperature
empirical methods of changes, support settlements, and fabrication errors do not create stresses. On
proportioning, as exampled in the other hand, because it contains three hinges, it is more flexible than the
Photo 1.2. However, the other arch types. :
theory and analysis of masonry Fixed-ended arches are often constructed of masonry or concrete when the
arches was formalized much base of an arch bears on rock, massive blocks of masonry, or heavy reinforced
concrete foundations. Indeterminate arches can be analyzed by the flexibility
later In particular, Philippe de
method covered in Chapter 11 or more simply and rapidly by any general-
LaHire (1640-1718) applied purpose computer program. To determine the forces and displacements at
statics to geometrical solution arbitrary points along the axis of the arch using a computer, the designer treats
of funicular polygons (1695), the points as joints that are free to displace.
and found that semicircular In long-span bridges, two main arch ribs are used to support the roadway
arches are unstable and rely beams. The roadway beams can be supported either by tension hangers from
the arch (Figure l .9a) or by columns that bear on the arch (Photo 7.1). Since
on grout bond or friction
the arch rib is mostly in compression, the designer must also consider the pos-
between masonry or stone sibility of its buckling-particularly if it is slender (Figure 7 .2a). If the arch
wedges to prevent sliding. is constructed of steel members, a built-up rib or a box section may be used
Further important develop- to increase the bending stiffness of the cross section and to reduce the likeli-
ment was made by Charles hood of buckling. In many arches, the floor system or wind bracing is used to
Coulomb (1736-1806), in stiffen the arch against lateral buckling. In the case of the trussed arch shown
in Figure 7 .2b, the vertical and diagonal members brace the arch rib against
which he established design
buckling in the vertical plane.
equations for determining the
limiting values of arch thrust in
order to achieve stability.

(a)

(b)

Figure 7 .1 Types of arches: (a) three-

hinged arch, stable and determinate; (b) two-
hinged arch, indeterminate to the first degree;
(c) fixed-end arch, indeterminate to the third
degree. (c')
 7 .2 • Types of Arches 245

1V Figure 7 :2: (a) Buckling of an unsupported

i I I I I I l arch; (b) trussed arch, the vertical and diag-
onal members brace the arch rib against
buckling in the vertical plane; (c) two types
of built-Up steel cross sections used to con-
struct an arch rib.

(a)

(b)

.
steel
L
channel plates

1
cross section welded box section

(c)

Since many people find the arch form aesthetically pleasing, designers often
se low arches to span small rivers or roads in parks and other pubhc places. At
ites where rock sidewalls exist, designers often construct short-span highway
, 'dges using bar.rel arches (see Figure 7.3). Constructed of accurately fitted
,, asonry blocks or reinforced concrete, the barrel arch consists of a wide, shal-
i.w arch that supports a heavy, compacted fill on which the engineer places the
1'adway slab. The large weight of the fill induces sufficient compression in the

el arch to neutralize any tensile bending stresses created by even the heavi-
t vehicles. Although the loads supported by the barrel arch may be large, direct
ses in the arch itself are typically low-on the order of 300 to 500 psi
' cause the cross-sectional area of the arch is large. A study by the senior author
' a number of masonry barrel-arch bridges built in Philadelphia in the mid-
_eteenth century showed that they have the capacity to support vehicles three Photo 7.1: Railroad bridge (1909) over the
• ive times heavier than the standard AASHTO truck (see Figure 2.7), which Landwasser Gorge, near Wiesen, Switzerland.
Masonry construction. The main arch is para-
·:hway bridges are currently designed to support. Moreover, while many steel
bolic, has a span of 55 m and a rise of 33 m.
• reinforced concrete bridges built in the past 100 years are no longer service- The bridge is narrow as the railway is single-
' e because of corrosion produced by salts used to melt snow, many masonry track, The arch ribs are a mere 4.8 m at the
es, constructed of good-quality stone, show no deterioration. crown, tapering to 6 m at the supports.
246 Chapter 7 • Arches

Figure 7.3: (a) Barrel arch resembles a

cuived slab; (b) barrel arch used to support a
compacted fill and roadway slab.

(a)

roadway
slab fill

rock

(b)

7.3
Three-Hinged Arches
To demonstrate certain of the characteristics of arches, we will consider how
the bar forces vary as the slope (} of the bars changes in the pin-jointed arch
in Figure 7.4a. Since the members carry axial load only, this configuration
represents the funicular shape for an arch supporting a single concentrated
load at midspan.
Because of symmetry, the vertical components of the reactions at sup-
ports A and Care identical in magnitude and equal to P/2. Denoting the slope
of bars A'B and CB by angle (}, we can express the bar forces FAB and F CB in
terms of P and the slope angle(} (see Figure 7.4b) as
. P/2 P/2
sm (} = - - = - -
FAB FcB
P/2
FAB = FcB = -.-- (7.1)
sm (}
Equation 7.1 shows that as (} increases from O to 90°, the force in each bar
decreases from infinity to P/2. We can also observe that as the slope angle (}
increases, the length of the bars-and consequently the material required-also
p

P/2
FAR= sinO

Figure 7.4: (a) Three-hinged arch with a

concentrated load; (b) vector diagram of:
forces acting on the hinge at B, forces Fc8
t---
p
L/2 -+i+--L/2 ---------t
p
and FAB are equal because of: symmetry; 2 2
(c) components of: force in bar AB. (a) (b) (c)
 7.4 • Funicular Shape for an Arch That Supports a Uniformly Distributed Load 247

increases. 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 sup-
port the load P in terms of the geometry of the structure and the compressive
strength of the material
V = 2ALs (7.2)
where A is the area of one bar and Ls is the length of a bar.
To express the required area of the bars in terms of load P, we divide the
bar forces given by Equation 7.1 by the allowable compressive stress O'allow:

P/2
A=----- (7.3)
(sin 8)0'anow
We will also express the bar length Ls in terms of 8 and the span length L as

L/2
LB=-- (7.4)
cos 8 '
'
: minimum
' volume
Substituting A and Ls given by Equations 7.3 and 7 .4 into Equation 7.2, simpli- ~at0=45°
fying, and using the trigonometric identity sin 28 = 2 sin 8 cos 8, we calculate '
0 15 30 45 60 75 80
PL
V=----- (7.5) slope () (deg)
20' allow sin 28
Figure 7 .5: Variation of volume of material
If Vin Equation 7.5 is plotted as a function of 8 (see Figure 7.5), we observe with slope of bars in Figure 7 .4a.
that the minimum volume of material is associated with an angle of 8 = 45°.
Figure 7.5 also shows that very shallow arches (8 :S 15°) and very deep arches
(8 ~ 75°) require a large volume of material; on the other hand, the fiat cur-
vature in Figure 7 .5 when 8 varies between 30 and 60° indicates that the
volume of the bars is not sensitive to the slope between these limits. ~here-
fore, the designer can vary the shape of the structure within this range with-
out significantly affecting either its weight or its cost.
In the case of a curved arch carrying a distributed load, the engineer will
also find that the volume of material required in the structure, within a certain
ange, is not sensitive to the depth of the arch. Of course, the cost of a very
shallow or very deep arch will be greater than that of an arch of moderate
depth. Finally, in establishing the shape of an arch, the designer will also con-
ider the profile of the site, the location of solid bearing material for the foun-
dations, and the architectural and functional requirements of the project.

I Funicular Shape for an Arch That Supports

a Uniformly Distributed Load
any arches carry dead loads that have a uniform or nearly uniform distri-
, . tion over the span of the structure. For example, the weight peffiunit length
' · the floor system of a bridge will typically be constant. To establish fpr a
'formly loaded arch the funicular shape-the form required if only direct
248 Chapter 7 • Arches

stress is to develop at all points along the axis of an arch- we will consider
the symmetric three-hinged arch in Figure 7 .6a. The height (or rise) of the
arch is denoted by h. Because of symmetry, the vertical reactions at sup-
ports A and Care equal to wL/2 (one-half the total load supported by the
structure).
The horizontal thrust Hat the base of the arch can be expressed in terms
of the applied load w and the geometry of the arch by considering the free
body to the right of the center hinge in Figure 7 .6b. Sulllllling moments about
the center hinge at B, we find

c+ "'2:.Mn = o
0= ( ~L) i -(~L) f + ~h

wL2
H=- (7.6)
8h

To establish the equation of the axis of the arch, we superimpose a rec-

tangular coordinate system, with an origin o located at B, on the arch. The
positive sense of the vertical y axis is directed downward. We next express the
moment Mat an arbitrary section (point Don the arch's axis) by considering
the free body of the arch between D and the pin at C.

C+ "'2:.Mv = 0

0 = ( f - x)2; - f - x) + ~L ( H (h - y) +M

Solving for M gives

wL2y wx 2
M=---- (7.7)
8h 2
Figure 7 .6: Establishing the funicular shape
for a unifonnly loaded arch.

Ill
!lllll:ll
,___ _ x _ _ L/2-x _.,..>--------',,.---:-;---+i

hinge
H

f.--u2 L/2--t l+----L/2---+I

wL wL
2 2 wL
2
(a) (b) (c)
 7.4 • Funicular Shape for an Arch That Supports a Uniformly Distributed Load 249

the arch axis follows the funicular shape, M = 0 at all sections. Substitut- .~····································
ing this value for Minto Equation 7 .7 and solving for y establishes the fol- Robert Hooke (1635-1703)
\owing mathematical relationship between y and x: established that the shape
1,
4h of a hanging chain, when
y=-x2 (7.8)
L2 inverted, would generate an
Equation 7 .8, of course, represents the equation of a parabola. Even if the par- efficient funicular shape of an
abolic arch in Figure 7 .6 were a fixed-ended arch, a uniformly distributed arch in direct compressive
load-assuming no significant change in geometry from axial shortening- stress. Antonio Gaudi
would still produce direct stress at all sections because the arch conforms to (1852-1926) utilized Hooke's
the funicular shape for a uniform load. theory and:built funicular
From a consideration of equihbrium in the horizontal direction, we can see
arches by scaling P.hysical
that the horizontal thrust at any section of an arch equals H, the horizontal reac-
tion at the support. In the case of a uniformly loaded parabolic arch, the total models that consisted of a
axial thrust Tat any section, a distance x from the origin at B (see Figure 7 .6b), network of strings with
can be expressed in terms of H and the slope at the given section as hanging weights. Gaudi 1/sed
H this method in the
T=-- (7.9)
cos 8 construction of Co/onia GQell
in Santa Coloma, Spain.
To evaluate cos 8, we first differentiate Equation 7 .8 with respect to x to give
Another iconic exanip/~'qf a ,
dy 8hx funicular arch is St Louis
tan 8 =- =- 2 (7.10)
dx L Gateway Arch in St Louis,
1ihe tangent of 8 can be shown graphically by the triangle in Figure 7 .6c. From Missouri, completed in 1967.
.this triangle we can compute the hypotenuse r using r 2 = x 2 + y2:

(7.11)

From the relationship between the sides of the triangle in Figure 7 .6c and the
cosine function, we can write

(7.12)

ubstituting Equation 7.12 into Equation 7.9 gives

(7.13)

uation 7.13 shows that the largest value of thrust occurs at the supports
here x has its maximum value of L/2. If w or the span of the arch is large, the
~esigner may wish to vary (taper) the cross section in direct proportion to the
alue of Tso that the stress on the cross section is constant.
Example 7.1 illustrates the analysis of a three-hinged trussed arch for both
a set of loads that corresponds to the funicular shape of the arch a~ well as for
single concentrated load. Example 7.2 illustrates the use of cabl~ theory to
Bstablish a funicular shape for the set of vertical loads in Example 7 .1. '
250 Chapter 7 • Arches

EXAMPLE 7.1
The geometry of the bottom chord of the arch is the funicular shape for the
loads shown. Analyze the three-hinged trussed arch in Figure 7 .7 a for the
dead loads applied at the top chord. Member Kl, which is detailed so that it
cannot transmit axial force, acts as a simple beam instead of a member of the
truss. Assume joint D acts as a hinge.
30 kips 60 kips 60 kips 60 kips 30 kips 60 kips 60 kips 60 kips

fo.--------- 6@ 30' = 180' _ _ _ _ _ _ _ __,,t

t
180
(a) (b)

Figure 7.7

Solution
Because the arch and its loads are symmetric, the vertical reactions at A and G
are equal to 180 kips (one-half the applied load). Compute the horizontal re-
action at support G.
Consider the free body of the arch to the left of the hinge atD (Figure 7 .7b),
and sum moments about D.
c+ 2-MD = o
0 = - 60(30) - 60(60) - 30(90) + 180(90) - 36H
H = 225 kips
We now analyze the truss by the method of joints starting at support A.
Figure 7.8 Results of the analysis are shown on a sketch of the truss in Figure 7 .7b.

90 kips NOTE. Since the arch rib is the funic-

ular shape for the',loads applied at the
top chord, the only members that carry
load-other than the rib-are the verti-
cal columns, which transmit the load
down to the arch. The diagonals and top
chords will be stressed when a loading
75 pattern that does not conform to the fu-
nicular shape acts. Figure 7 .8 shows the
forces produced in the same truss by a
t
60
t
~o
single concentrated load at joint L.
f 7 .4 • Funicular Shape for an Arch That Supports a Uniform~ Distdbuted Load 251

f EXAMPLE 7.2
Establish the shape of the funicular arch for the set of
loads acting on the trussed arch in Figure 7 .7. The rise of
the arch at midspan is set at 36 ft.

Solution
We imagine that the set of loads is applied to a cable that
spans the same distance as the arch (see Figure 7 .9a). The
sag of the cable is set at 36 ft-the height of the arch at
midspan. Since the 30-kip loads at each end of the span act
directly at the supports, they do not affect the force or the
shape of the cable and may be neglected. Applying the
60 kips
general cable theory, we imagine that the loads supported 60 kips 60 kips
by the cable are applied to an imaginary simply supported 60 kips
beam with a span equal to that of the cable (Figure 7 .9b). (a)
We next construct the shear and moment curves.
According to the general cable theorem at every point,
60 kips 60 kips 60 kips 60 kips 60 kips
M=Hy (6.6)
l l l l l
where M = moment at an arbitrary point in the beam AA ~
H = horizontal component of support reaction
y = cable sag at an arbitrary point t.·-----
.
150 kips
--------i·t 6 @ 30' = 180'
150 kips
Since y = 36 ft at midspan and M = 8100 kip· ft, we
can apply Equation 6.6 at that point to establish H. (b)

M 8100 .
H = - = - - = 225 kips
y 36

With H established we next apply Equation 6.6 at

30 and 60 ft from the supports. Compute y 1 at 30 ft:

M 4500 (c)
Y1 = - = - - = 20 ft
H 225

Compute Y2 at 60 ft: 7200 8100

M 7200
Y2 = - = - - = 32 ft moment
H 225 (kip. ft)

(d)
A cable profile is always a funicular structure because
a cable can only carry direct stress. If the cable profile is Figure 7. 9: Use of cable theory to establish
turned upside down, a funicular arch is produced. When the funicular shape of an arch.
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 cor-
responding sections.