V ?
Designing A Uniformly Loaded Arch Or Cable
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here or at the top of any page.
When you are done with this lesson, click on the
Contents button here or at the top of any page to
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This is the Max Eyth Bridge
in Stuttgart, Germany,
designed by Schlaich &
Bergermann, Engineers.
In this lesson, you will learn
how to shape an arch or cable
to carry a uniform load while
undergoing only axial forces.
V ?
Designing A Uniformly Loaded Arch Or Cable
W = 10,800 lb
36 ft
180 ft
Funicular Polygon
40 ft
The Problem: We are
designing a concrete vault for
the roof of a warehouse. The
span is 180 feet, and we
would like the rise of the vault
to be 36 feet. The total uniform
load on a one-foot-wide strip
of the vault is estimated to be
10,800 lb.
We must find the form of the
vault, and the forces in a onefoot-wide strip.
V ?
Designing A Uniformly Loaded Arch Or Cable
10 loads @ 1080 lb each
P
P
P
P
P
P
P
P
P
P
Loading Diagram
36 ft
180 ft
Funicular Polygon
40 ft
Step 1: Construct a Loading
Diagram and apply interval
notation.
We divide the uniform load
into a convenient number of
increments. We decide to use
ten increments in this case,
although nine or twelve or
eighteen would serve as well.
The load P allotted to each
increment is found by dividing
the total load by the number
of increments:
P=
10,800 lb
= 1080 lb
10
We diagram these loads on a
Loading Diagram.
V ?
Designing A Uniformly Loaded Arch Or Cable
The load on each increment
is represented on the Loading
Diagram by a concentrated
force, P, of 1080 lb., that acts
at the center of the increment.
We extend vertical
construction lines down to a
base line from each force P,
and from the ends of the span.
The base line is located below
the diagram of the vault to
allow adequate space for the
construction of the Funicular
Polygon at a later stage.
10 loads @ 1080 lb each
Loading Diagram
Base Line
Funicular Polygon
40 ft
V ?
Designing A Uniformly Loaded Arch Or Cable
We add interval notation,
completing the Loading
Diagram.
Notice that we have not yet
determined the reactions for
this structure. They will be
determined at a later stage,
in the course of constructing
the Force Polygon.
10 loads @ 1080 lb each
A
B
C
D
E
F
G
H
J
K
L
Loading Diagram
Base Line
Funicular Polygon
40 ft
V ?
Designing A Uniformly Loaded Arch Or Cable
Step 2: Construct a Load
Line to a convenient scale.
Working from left to right, the
forces on the Loading
Diagram are plotted tip-to-tail
onto the Load Line at the
selected scale.
First we plot force ab.
10 loads @ 1080 lb each
A
B
C
D
E
F
G
H
J
K
L
Loading Diagram
a
b
Base Line
1080 lb
Funicular Polygon
40 ft
Load Line
4000 lb
V ?
Designing A Uniformly Loaded Arch Or Cable
The Load Line is complete
when all of the forces have
been plotted.
Because the load increments
are all vertical, the Load Line
is vertical as well.
10 loads @ 1080 lb each
A
B
C
D
E
F
G
H
J
K
L
Loading Diagram
Base Line
10,800 lb
Funicular Polygon
40 ft
Load Line
4000 lb
a
b
c
d
e
f
g
h
j
k
l
V ?
Designing A Uniformly Loaded Arch Or Cable
Step 3: Construct the Force
Polygon and Funicular
Polygon.
In order to find the form and
forces in the vault, we will next
construct a Force Polygon
from the Load Line.
10 loads @ 1080 lb each
A
B
C
D
E
F
G
H
J
K
L
Loading Diagram
Base Line
Funicular Polygon
40 ft
Force Polygon
4000 lb
a
b
c
d
e
f
g
h
j
k
l
V ?
Designing A Uniformly Loaded Arch Or Cable
10 loads @ 1080 lb each
A
B
C
D
E
F
G
H
J
K
L
Loading Diagram
Base Line
Funicular Polygon
40 ft
Tangent to end of
the parabola
2s
Parabola
s
To locate the pole of the Force
Polygon, we use a shortcut that
is based on two facts:
1. The funicular line for a load
that is distributed uniformly
over a horizontal projection
of the span is a parabola.
2. The tangents to the ends
of any parabola intersect
on the vertical centerline
at a distance from the base
line that is twice the height
of the parabola. See the
diagram below left.
a
b
c
d
e
f
g
h
j
k
Force Polygon
l
4000 lb
V ?
Designing A Uniformly Loaded Arch Or Cable
10 loads @ 1080 lb each
A
B
C
D
E
F
G
H
J
K
Loading Diagram
L
On the centerline of our
Funicular Polygon, we make
two marks, one 36 feet above
the base line, the intended
rise of the vault, and one at
twice this height, which is
72 feet.
The axis of the vault will pass
through the 36 foot mark. The
tangents to the ends of the
vault will pass through the
72 ft 72 foot mark.
36 ft
Base Line
Funicular Polygon
40 ft
2s
s
Force Polygon
4000 lb
a
b
c
d
e
f
g
h
j
k
l
V ?
Designing A Uniformly Loaded Arch Or Cable
On the Funicular Polygon we
draw the tangents to the ends
of the parabola, oa and ol .
Parallel to them, we draw
rays oa and ol on the Force
Polygon. These rays intersect
at the final pole, o .
10 loads @ 1080 lb each
A
B
C
D
E
F
G
H
J
K
L
Loading Diagram
72 ft
ol 36 ft
oa
Base Line
Funicular Polygon
40 ft
oa
Pole
o
ol
2s
s
Force Polygon
4000 lb
a
b
c
d
e
f
g
h
j
k
l
Designing A Uniformly Loaded Arch Or Cable
V ?
As we construct each ray, we
scale its length on the Force
Polygon to determine the
magnitude of the force in the
corresponding segment of the
Funicular Polygon.
10 loads @ 1080 lb each
A
B
C
D
E
F
G
H
J
K
L
Loading Diagram
ol
oa
84
4
lb 4
44
84 lb
Funicular Polygon
40 ft
00
80
00
60
Base Line
00
40
oa
00
20
0l
b
o
ol
2s
s
Force Polygon
4000 lb
00
0
,
10
,00
12
a
b
c
d
e
f
g
h
j
k
l
V ?
Designing A Uniformly Loaded Arch Or Cable
Knowing the forces in the
outermost segments of the
vault also allows us to find the
reaction forces at the ends of
the vault.
Although finding these forces
is not necessary to completing
the solution to our problem,
they are easily determined as
being equal and opposite to
the forces in these segments.
10 loads @ 1080 lb each
A
B
C
D
E
F
G
H
J
K
L
Loading Diagram
ol
oa
b
l
44
84
84
4
lb 4
44
84 lb
Funicular Polygon
40 ft
Base Line
84
44
oa
lb
o
ol
2s
s
Force Polygon
4000 lb
a
b
c
d
e
f
g
h
j
k
l
Designing A Uniformly Loaded Arch Or Cable
V ?
We now construct the
remaining rays, beginning with
ob.
Parallel to ob on the Force
Polygon, we draw the
corresponding segment of the
Funicular Polygon, also
named ob. ob lies below
interval B of the Loading
Diagram.
10 loads @ 1080 lb each
A
B
C
D
E
F
G
H
J
K
L
Loading Diagram
oa
44
84 lb
ob
ol
84
4
lb 4
Base Line
oa b
o
Funicular Polygon
40 ft
o
ol
Force Polygon
4000 lb
a
b
c
d
e
f
g
h
j
k
l
Designing A Uniformly Loaded Arch Or Cable
V ?
We scale the length of ray ob
and record the magnitude of
the force on the Funicular
Polygon.
10 loads @ 1080 lb each
A
B
C
D
E
F
G
H
J
K
L
Loading Diagram
oa
44
84 lb
ob
8
777lb
Funicular Polygon
40 ft
ol
84
4
lb 4
00
80
Base Line
00
40
0 lb
00
20
00
60 a
o b
o
o
ol
Force Polygon
4000 lb
0
,00
0
1
a
b
c
d
e
f
g
h
j
k
l
1
Designing A Uniformly Loaded Arch Or Cable
V ?
We next construct and scale
ray oc .
10 loads @ 1080 lb each
A
B
C
D
E
F
G
H
J
K
L
Loading Diagram
oa
44
84 lb
ob
oc
9
756b
l
8
777lb
ol
84
4
lb 4
Base Line
oa b
oc
o
Funicular Polygon
40 ft
o
ol
Force Polygon
4000 lb
a
b
c
d
e
f
g
h
j
k
l
Designing A Uniformly Loaded Arch Or Cable
V ?
We continue working from left
to right in this manner across
the Funicular Polygon.
10 loads @ 1080 lb each
A
B
C
D
E
F
G
H
J
K
L
Loading Diagram
oa
44
84 lb
oc
od
9
688b
9
l
6
75 b
l
8
777lb
ob
ol
84
4
lb 4
Base Line
oa b
oc
o
od
Funicular Polygon
40 ft
o
ol
Force Polygon
4000 lb
a
b
c
d
e
f
g
h
j
k
l
Designing A Uniformly Loaded Arch Or Cable
V ?
Notice how the forces in the
arch lessen in magnitude
toward the apex.
10 loads @ 1080 lb each
A
B
C
D
E
od
oe
F
G
H
J
K
L
Loading Diagram
oc
oa
44
84 lb
9
688b
b
o 7569 l
lb
8
7
7
7 lb
6667
lb
ol
84
4
lb 4
Base Line
oa b
oc
o
od
oe
Funicular Polygon
40 ft
o
ol
Force Polygon
4000 lb
a
b
c
d
e
f
g
h
j
k
l
Designing A Uniformly Loaded Arch Or Cable
V ?
The segment at the apex
passes through the 36 foot
high mark established earlier,
verifying the accuracy of our
construction.
10 loads @ 1080 lb each
A
B
C
D
E
F
od
oe
of
G
H
J
K
L
Loading Diagram
oc
oa
44
84 lb
9
688b
b
o 7569 l
lb
8
7
7
7 lb
6667 6444
lb
lb
ol
84
4
lb 4
Base Line
oa b
oc
o
od
oe
of
Funicular Polygon
40 ft
o
ol
Force Polygon
4000 lb
a
b
c
d
e
f
g
h
j
k
l
Designing A Uniformly Loaded Arch Or Cable
V ?
Segment og has the same
magnitude of force as
segment oe.
Because the arch and the
loads on it are symmetrical,
the distribution of internal
forces is symmetrical as well.
10 loads @ 1080 lb each
A
B
C
D
E
F
G
od
oe
of
og
H
J
K
L
Loading Diagram
oc
oa
44
84 lb
9
688b
b
o 7569 l
lb
8
7
7
7 lb
6667 6444 6667
lb
lb
lb
ol
84
4
lb 4
Base Line
Funicular Polygon
40 ft
o
Force Polygon
4000 lb
a
b
oa b
oc
c
o
od d
oe e
of f
og
g
h
ol
j
k
l
Designing A Uniformly Loaded Arch Or Cable
V ?
We complete the remaining
rays oh through ok.
10 loads @ 1080 lb each
A
B
C
D
E
F
G
od
oe
of
og
H
J
K
L
Loading Diagram
oc
oa
44
84 lb
9
688b
b
o 7569 l
lb
8
7
7
7 lb
oh
6667 6444 6667 68
lb
89
lb
lb
lb
ol
84
4
lb 4
Base Line
Funicular Polygon
40 ft
o
Force Polygon
4000 lb
a
b
oa b
oc
c
o
od d
oe e
of f
og
oh g
h
ol
j
k
l
Designing A Uniformly Loaded Arch Or Cable
V ?
10 loads @ 1080 lb each
A
B
C
D
E
F
G
od
oe
of
og
H
J
K
L
Loading Diagram
oj
6667 6444 6667 68
9
lb
89
lb
lb
688
758
lb
ob 7569 lb
9
l
b
b
l
8
777lb
oc
oa
44
84 lb
oh
ol
84
4
lb 4
Base Line
Funicular Polygon
40 ft
o
Force Polygon
4000 lb
a
b
oa b
oc
c
o
od d
oe e
of f
og
oh g
oj
h
ol
j
k
l
Designing A Uniformly Loaded Arch Or Cable
V ?
The last segment to be
constructed is ok . This
segment closes the Funicular
Polygon, once again verifying
the accuracy of our
construction.
The Force Polygon and
Funicular Polygon are now
complete.
10 loads @ 1080 lb each
A
B
C
D
E
F
G
od
oe
of
og
H
J
K
L
Loading Diagram
oj
6667 6444 6667 68
9
lb
89
lb
lb
688
ok
758
lb
ob 7569 lb
9
lb
lb
777 ol
8
7
77 lb
lb 8
84
4
lb 4
oc
oa
44
84 lb
oh
Base Line
Funicular Polygon
40 ft
o
Force Polygon
4000 lb
a
b
oa b
oc
c
o
od d
oe e
of f
og
oh g
oj
h
ok
ol
j
k
l
V ?
Designing A Uniformly Loaded Arch Or Cable
The actual, smooth parabolic
curve of the vault will pass
through the two end points
that we have plotted, and will
be tangent to the center of
each straight-line segment of
the Funicular Polygon.
W = 10,800 lb
oa
ob
oc
od
oe
of
og
oh
oj
ok
ol
Funicular Polygon
40 ft
o
Force Polygon
4000 lb
a
b
oa b
oc
c
o
od d
oe e
of f
og
oh g
oj
h
ok
ol
j
k
l
V ?
Designing A Uniformly Loaded Arch Or Cable
W
Loading Diagram
By inverting this construction,
we can apply it to a hanging
cable that supports a load that
is uniformly distributed over
its horizontal projection, like
the cables of the Max Eyth
Bridge shown at the beginning
of this lesson.
Funicular Polygon
o
Force Polygon
V ?
Designing A Uniformly Loaded Arch Or Cable
The tangents to the end of the
cable intersect at a distance
from the base line that is twice
the sag of the cable.
Notice also that in this case,
the pole of the Force Polygon
lies to the right of the Load
Line.
W
Loading Diagram
s
Funicular Polygon
2s
o
Force Polygon
V ?
Designing A Uniformly Loaded Arch Or Cable
L/2
This construction may be
applied also to an arch or
cable whose ends do not lie
on the same level, as long as
the load is uniformly
distributed over the horizontal
projection.
Notice how in this case, of
course, the pole of the Force
Polygon is no longer centered
vertically on the Load Line.
L/2
W
Loading Diagram
s
2s
o
Funicular Polygon
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Force Polygon