University of Engineering & Technology
Peshawar, Pakistan
CE215: Structure Analysis I
Module 06:
Analysis of Cables and Arches
By:
Prof. Dr. Bashir Alam
Civil Engineering Department
UET , Peshawar
Topics to be Covered
• Cables
• Cables subjected to concentrated loading
• Examples
• Cables subjected to uniformly distributed loading
•
Examples
• Arches
Cables
 Cables:
• Cables are often used in engineering structures for support and to
transmit loads from one member to another.
• Cables used to support suspension roofs, bridges, and trolley wheels,
cables form the main load-carrying element in the structure.
• In the force analysis of such systems, the weight of the cable itself
may be neglected; however, when cables are used as guys for radio
antennas, electrical transmission lines, and derricks, the cable weight
may become important and must be included in the structural
analysis.
Cables
 Cables:
• Cables are usually flexible and carry their loads in tension.
• Cables stretch well and are light, so they are useful in large span
structures.
• Cable is flexible and in-extensible; hence does not resist any
bending moment or shear force; same as truss bar.
• Cable is in-extensible, hence the length is always constant.
Cables
Cables are mostly used in long span bridges.
Suspension Bridge
Cable Stayed Bridge
Cables
Hanger
Tower
Cable
Cable
Anchorage
Road Way
Golden Gate Bridge, San Francisco
Cables
 Cables:
Two cases will be considered in the sections that follow:
• A cable subjected to concentrated loads and
• A cable subjected to a distributed load.
Provided these loadings are coplanar with the cable, the requirements
for equilibrium are formulated in an identical manner.
Cables
 Cables:
• When deriving the necessary relations between the force in the
cable and its slope, we will make the assumption that the cable is
perfectly flexible and inextensible.
• Due to its flexibility, the cable offers no resistance to shear or
bending and, therefore, the force acting in the cable is always
tangent to the cable at points along its length.
Cables
 Cables:
Being inextensible, the cable has a constant length both before and
after the load is applied.
As a result, once the load is applied, the geometry of the cable
remains fixed, and the cable or a segment of it can be treated as a
rigid body.
Cables
1. Cables subjected to concentrated loads:
When a cable of negligible weight supports several concentrated
loads, the cable takes the form of several straight-line segments, each
of which is subjected to a constant tensile force.
A
q
D
yB
yC
B
C
P1
P2
L1
L2
L
L3
Cables
1. Cables subjected to concentrated loads:
• To determine the nine unknowns consisting of the tension in each
of the three segments, the four components of reaction at A and D,
and the sags yB and yC at the two points B and C.
• For the solution we can write two equations of force equilibrium at
each of points A, B, C, and D. This results in a total of eight
equations.
• To complete the solution, it will be necessary to know something
about the geometry of the cable in order to obtain the necessary
ninth equation.
Cables
1. Cables subjected to concentrated loads:
For example, if the cable’s total length is specified, then the
Pythagorean theorem can be used to relate to each of the three
segmental lengths, written in terms of and Unfortunately, this type of
problem cannot be solved easily by hand.
A
q
D
yC
yD
B
C
P1
P2
L1
L2
L
L3
Cables
1. Cables subjected to concentrated loads:
• Another possibility, however, is to specify one of the sags, either or
instead of the cable length.
• By doing this, the equilibrium equations are then sufficient for
obtaining the unknown forces and the remaining sag.
• Once the sag at each point of loading is obtained, force can then be
determined by trigonometry.
A
q
D
yC
yD
B
P1
P2
L1
L2
L
C
L3
Cables
1. Cables subjected to concentrated loads:
When performing an equilibrium analysis for a problem of this type,
the forces in the cable can also be obtained by writing the equations
of equilibrium for the entire cable or any portion thereof.
A
q
D
yC
yD
B
P1
P2
L1
L2
L
C
L3
Cables
1. Cables subjected to concentrated loads:
Example 01: Determine the tension in each segment of the cable
shown in Figure. Also, what is the dimension h?
A
2m
h
D
B
2m
C
3 kN
2m
8 kN
2m
1.5 m
Cables
1. Cables subjected to concentrated loads:
Example 01: Solution:
• 4 unknown external reactions (Ax, Ay, Dx and Dy)
• 3 unknown cable tensions
• 1 geometrical unknown h
A
2m
h
D
B
• 8 unknowns
• 8 equilibrium equations
3 kN
8 kN
(two at each joint)
2m
• So cable is statically determinate
2m
C
2m
1.5 m
Cables
1. Cables subjected to concentrated loads:
Example 01: Solution:
Ay
Ax
A
TCD
5 4
D
3
h
B
2m
C
3 kN
2m
8 kN
2m
1.5 m
+ SMA = 0:
TCD(3/5)(2 m) + TCD(4/5)(5.5 m) - 3kN(2 m) - 8 kN(4 m) = 0
TCD = 6.79 kN
Cables
1. Cables subjected to concentrated loads:
Example 01: Solution:
y
Joint C:
TCB
TBA
TCD = 6.79 kN
C 5
qBC
y
Joint B:
4
3
qBA
x
8 kN
+
B 32.3o
x
TBC = 4.82 kN
3 kN
+ SF = 0: 6.79(3/5) - TCB cos qBC = 0
x
+ SF = 0: - TBA cos qBA + 4.82cos 32.3o = 0
x
SFy = 0: 6.79(4/5) - 8 + TCB sin qBC = 0
+
SFy = 0: TBA sin qBA - 4.82sin 32.3o -3 = 0
Solving both equations simultaneously
Solving both equations simultaneously
qBC = 32.3o
qBA = 53.8o
TCB = 4.82 kN
h = 2tanqBA = 2tan53.8o = 2.74 m
TBA = 6.90 kN
Cables
1. Cables subjected to concentrated loads:
Example 02: Determine the tension in each segment of the cable
shown in Figure.
A
B
C
3’
3’
D
5k
4’
5k
3’
Cables
1. Cables subjected to concentrated loads:
Example 02: Solution:
• 4 unknown external reactions (Ax, Ay, Bx and By)
• 3 unknown cable tensions
• 7 unknowns
A
B
C
• 8 equilibrium equations
(two at each joint)
• So cable is statically determinate
3’
3’
D
5k
4’
5k
3’
Cables
1. Cables subjected to concentrated loads:
Example 02: Solution:
A
B
3’
C
3’
D
5k
4’
5k
3’
MB= 0 ⇒ Ay*10 – 5*7 – 5*3 = 0 ⇒ Ay= 5 k
Similarly
By = 5k
Ax = 5k , Bx = 5k
Cables
1. Cables subjected to concentrated loads:
Example 02: Solution:
y
Joint C:
TCA
x
TCD
45
C
5 kN
F𝑦 = 0
TAC sin45 = 5 ⇒ TAC = 7.07 k
TAC Cos45 – TCD = 0 ⇒ TCD = 5k
Cables
2. Cables subjected to uniform distributed loads:
Cables provide a very effective means of supporting the dead weight
of girders or bridge decks having very long spans.
A suspension bridge is a typical example, in which the deck is
suspended from the cable using a series of close and equally spaced
hangers.
wo = force / horizontal distance
y
B
A
x
L
Cables
2. Cables subjected to uniform distributed loads:
T
y
T
q
q
W
To
To
x
W
x=L
T cos q = To = FH = Constant
T sin q = W
𝑑𝑦
𝑤
= 𝑡𝑎𝑛𝜃 =
𝑑𝑥
𝑇𝑜
Cables
2. Cables subjected to uniform distributed loads:
• Parabolic cable subjected to UDL:
wo = force / horizontal distance
y
B
A
x
L
Cables
2. Cables subjected to uniform distributed loads:
• Parabolic cable subjected to UDL:
𝑥
2
y
wo x
qx
To
T
x
x
T
qx
To
Tmax
wox
For x=L & y=H
To = FH =woL2/2H
woL
qB
To
Cables
2. Cables subjected to uniform distributed loads:
• Derivation:
The x , y axes have their origin located at the lowest point on the
cable such that the slope is zero at this point.
Since the tensile force in the cable changes continuously in both
magnitude & direction along the cable’s length, this load is denoted
by T.
y
wo
h
x
x
L
x
Cables
2. Cables subjected to uniform distributed loads:
• Derivation:
wo(x)
Δ𝑥
2
T + T
O
q
T
+ SF = 0:
x
s
x
q+ q
y
-Tcosq + (T + T) cos (q + q) = 0
SFy = 0:
-Tsinq - wo(x) + (T + T)sin (q + q) = 0
+ SMO = 0:
wo(x)(x/2) - T cos qy + T sinq(x) = 0
+
Cables
2. Cables subjected to uniform distributed loads:
• Derivation:
Dividing each of these equations by x and taking the limit as x
y 0, q 0, and T 0, we obtain
Integrating Eq. 5-1, where T = FH at x = 0, we have:
0, and hence
Cables
2. Cables subjected to uniform distributed loads:
• Derivation:
Integrating Eq. 5-2, where T sin q = 0 at x = 0, gives
Dividing Eq. 5-5 Eq. 5-4 eliminates T. Then using Eq. 5-3, we can obtain the
slope at any point,
Performing a second integration with y = 0 at x = 0 yields
Cables
2. Cables subjected to uniform distributed loads:
• Derivation:
Equation 5-7 is the equation of a parabola. The constant FH may be obtained by
using the boundary condition y = h at x = L. Thus,
Finally, substituting into Eq. 5-7 yields
From Eq. 5-4, the maximum tension in the cable occurs when q is maximum; i.e.,
at x = L. Hence, from Eqs. 5-4 and 5-5,
Cables
2. Cables subjected to uniform distributed loads:
• Derivation:
From Eq. 5-4, the maximum tension in the cable occurs when q is maximum; i.e.,
at x = L. Hence, from Eqs. 5-4 and 5-5,
Or, using Eq. 5-8, we can express Tmax in terms of wo, i.e.,
• We have neglect the weight of the cable which is uniform along the length
• A cable subjected to its own weight will take the form of a catenary curve
• If the sag-to-span ratio is small, this curve closely approximates a parabolic shape.
Cables
2. Cables subjected to uniform distributed loads:
Example 03: The cable shown supports a girder which weighs
12kN/m. Determine the tension in the cable at points A, B, and C.
30 m
A
C
12 m
B
wo = 12 kN/m
6m
Cables
2. Cables subjected to uniform distributed loads:
Example 03: Solution
Here origin or zero slope point is at point B.
TA
qA
30 m
A
y
TC
C
12 m
qC
6m
B
x
wo = 12 kN/m
30 - L´
x2
L´
x1
Cables
2. Cables subjected to uniform distributed loads:
Example 03: Solution: Taking Right side section of point B
TC
y
To
C
qC
6m
B
x
wo = 12 kN/m
12 L´
L´
x1
T
q
To
12x1
Cables
2. Cables subjected to uniform distributed loads:
Example 03: Solution: Taking left side section of point B
TA
qA
y
A
12 m
12 x2
B
To
wo = 12 kN/m
12 (30 - L´)
30 - L´
x2
x
T
q
To
Cables
2. Cables subjected to uniform distributed loads:
Example 03: Solution: Taking left side section of point B
12 x2
T
q
To
Cables
2. Cables subjected to uniform distributed loads:
Example 03: Solution:
From equation 1 & 2 we have
Solving equation 1 & 2 both simultaneously
L´ = 12.43 m, To = 154.50 kN
Also from cable geometry it is clear that
TB = To = 154.50 kN
Cables
2. Cables subjected to uniform distributed loads:
Example 03: Solution:
From right side we have
TC
qC
To
12 L´
From left side we have
TA
12 (30 - L´ )
qA
To
Cables
2. Cables subjected to uniform distributed loads:
Example 04: Determine the maximum and minimum tension in cable
shown below.
Cables
2. Cables subjected to uniform distributed loads:
Example 04: Solution
Arches
Arches
Arches:
The arch achieves its strength in compression, since it has a reverse
curvature to that of the cable.
An arch must also resist bending and shear depending upon how it is
loaded & shaped (an ideal arch will have no secondary effects).
Chinese Moon Bridge
Arches
Arches:
• In particular, if the arch has a parabolic shape and it is subjected to
a uniform horizontally distributed vertical load, then from the
analysis of cables it follows that only compressive forces will be
resisted by the arch.
• Under these conditions the arch shape is called a funicular arch
because no bending or shear forces occur within the arch.
• What is difference between an arch and a curved beam?
Arches
1. Types of arches based on the support conditions or hinges:
indeterminate
indeterminate
indeterminate
determinate
Arches
2. Types of Arches according to their shapes
• Circular or curved or segmental arch
• Parabolic arch
• Elliptical arch
• Polygon arch
3. Types of Arches according to materials
• Steel arches
• Reinforced concrete arches
• Masonry arches (stone or brick)
Arches
Eddy’s Theorem :
Eddy’s theorem states that “ The bending moment at any section of an
arch is proportional to the vertical intercept between the linear arch (or
theoretical arch) and the center line of the actual arch.”
OR
The bending moment at any point on the arch is the difference between
simple span bending moment and product Hy” .
Where H is the horizontal thrust at supports (springings), y is the rise of
arch at a distance 𝑋 from the origin.
Arches
Eddy’s Theorem :
• Shape of simple span bending moment diagram due to applied loads
is also called linear arch.
• Hy may also be termed as equation of centerline of actual arch
multiplied by a constant (H).
• Consider the arch shown on next slide carrying the loads P1, P2 and
P3. The shaded area is the BMD.
Arches
Eddy’s Theorem :
Bending moment at 𝑋 is
𝑀𝑋 = 𝑉𝑎𝑋 − 𝐻𝑦 − 𝑃1(𝑋 − 𝑎)
𝑀𝑋 = µ𝑋 − 𝐻𝑦. (𝐸𝑑𝑑𝑦’ 𝑠 𝑡ℎ𝑒𝑜𝑟𝑒𝑚)
 Where µ𝑋 = 𝑉𝑎 × 𝑋 − 𝑃1(𝑋 − 𝑎) =
Simple span bending moment considering
the arch to be a simple beam.
Arches
Three hinged Arch:
• To provide some insight as to how arches transmit loads, we will
now consider the analysis of a three-hinged arch.
• In this case, the third hinge is located at the crown and the supports
are located at different elevations.
Arches
Three hinged Arch:
In order to determine the reactions at the supports, the arch is
disassembled and the free-body diagram of each member is shown
Arches
Three hinged Arch:
One method of solving this problem is to apply the moment
equilibrium equations about points A and B.
Simultaneous solution will yield the reactions and The support
reactions are then determined from the force equations of equilibrium.
Three hinged Arch:
Arches
• Once the reactions obtained, the internal normal force, shear, and
moment loadings at any point along the arch can be found using the
method of sections.
• The section should be taken perpendicular to the axis of the arch at
the point considered.
• For
example,
the
free-body
diagram
for
segment
AD.
Arches
Example 04:
The three-hinged open-spandrel arch bridge has a parabolic shape
and supports the uniform load. Show that the parabolic arch is
subjected only to axial compression at an intermediate point such as
point D. Assume the load is uniformly transmitted to the arch ribs.
Arches
Example 04: Solution
Applying the equations of equilibrium, we have:
+ SMA = 0:
Cy (40m) – 320 (20 m) = 0
Cy = 160 kN
Arches
Example 04: Solution
Applying the equations of equilibrium to segment BC, we have:
+ SMB = 0:
-160 kN (10m) + 160 (20 m) – Cx(10m) = 0
Cx = 160 kN
+ SF = 0: 𝐵 = 160 kN
𝑥
x
+ SFy = 0: By – 160 + 160 = 0 ⇒ By = 0
Arches
Example 04: Solution
Applying the equations of equilibrium to a section of the arch taken
through point D we have:
𝑥 = 10𝑚
y = -10(10)2/(20)2 = -2.5m
tan 𝜃 =
𝑑𝑦
−20
=
𝑥|
= −0.5
𝑑𝑥
20 2 𝑥 = 10𝑚
+ SF =0
x
160 kN - ND cos 26.6o – VD sin 26.6o = 0 ----- Eq-A
Arches
Example 04: Solution
Applying the equations of equilibrium to a section of the arch taken
through point D we have:
+ S Fy = 0
- 80 kN - ND sin 26.6o – VD cos 26.6o = 0 ---- Eq-B
Solving Eq- A and B simultaneously we have
VD = 0 kN
ND = 178.9 kN
+ S MD = 0
MD+ 80 (5 m) – 160(2.5m) = 0
MD = 0
References
• Structural Analysis by R. C. Hibbeler
• Structure Analysis by Alexander Chajes