VI. Arches: The Salginatobel Bridge

VI. Arches:
The Salginatobel Bridge
Figure 1: The Salginatobel Bridge, Schiers, Switzerland
The Salginatobel Bridge located in
Switzerland crosses the Salgina Valley
connecting the village of Schuders to the
town of Schiers.i Built in 1930 and
designed by Robert Maillart, the bridge
spans 295 ft, has an arch rise of 42.6 ft,
and is approximately 10 ft wide.ii In
creating the form of this three-hinged
concrete arch bridge Maillart borrowed
elements from his previous works. The
Salginatobel stands as one of the finest
20th century concrete structures.
vertical members, the dead load of the
bridge deck can be assumed to be
uniformly distributed, q (lb/ft or k/ft).
The dead load of the arch itself will be
assumed to be uniformly distributed.
Live load from the traffic is also
transferred to the arch through the
vertical members. However, assuming
the live load is uniformly distributed
across the entire span is not conservative
and we’ll discuss this in further detail
later. Gravity loads are the dominant
load on this structure. Wind loads will
not be considered in this study.
What is an Arch Bridge?
An arch bridge is one in which the
bridge deck is supported by an arch.
Attaching the deck to the arch are
vertical members. The vertical members
transfer the loads from the deck to the
arch. As you increase the number of
Abutment Reactions
Loads are transmitted
abutments at the ends
applied gravity loads
acting
downward;
1
from the arch to
of the arch. The
on the arch are
therefore
the
abutments need to resist these forces
with upward reactions. The vertical
reaction, V (lbs or k) at each abutment
is:
q*L
V=
2
Determine: The vertical and horizontal
reactions at each abutment of the
Salginatobel Bridge under dead load
only. See Figure 3.
Given: The dead load of the bridge,
qDL = 5.7 k/ft iii
The span, L = 295 ft.
The depth of the arch, d = 42.6 ft.
where q is the uniformly distributed load
applied to the bridge (lbs/ft or k/ft) and
L (ft) is the length of the span of the
bridge.
Solution:
Step 1 -- Calculate the vertical reaction
at each abutment:
In addition, the loads on the bridge act to
spread the abutments away from each
other; therefore the abutments need to
resist these forces with reactions that act
to keep the abutments in place. The
horizontal reaction, H (lbs or k) at each
abutment is:
q * L2
H=
8* d
V=
q * L 5.7k / ft * 295 ft
=
= 840k
2
2
Step 2 -- Calculate the horizontal
reaction at each abutment:
H=
where d is the depth of the arch (ft) at
midspan. Once again q is the distributed
load and L is length of the span. See
Figure 2 below.
q * L2 5.7k / ft * (295 ft ) 2
=
= 1,455k
8*d
8 * 42.6 ft
Cable vs. Arch
Recall that a cable is a flexible structural
element that resists applied loads in
tension. Because a cable is only able to
take axial tensile forces, it adjusts its
form in accordance with the applied
loads to achieve this behavior.
An arch is rigid and cannot change its
form in accordance with the applied
loads. Arches are most efficient in
compression; therefore the most efficient
form is one in which the arch is entirely
in compression. To determine the ideal
form for an arch under a particular
loading, you can imagine applying that
load to a cable, freezing the shape, and
flipping the cable over.
Figure 2: Loaded Arch with Reactions
Example: Reactions at the Abutments
of the Salginatobel
Figure 3: Elevation of the Salginatobel Bridge
2
For example, the shape of a cable under
point loads, Q applied at each of the two
quarterpoints takes on the shape shown
in Figure 4.
Compressive Stress
Recall from structural studies “I.
Columns: The Washington Monument,”
and “III. Cables: The George
Washington Bridge” that an axial load is
directed along the main axis of the
structural element. For columns the
loads were compressive and for cables
the loads were tensile. For arches the
loads are compressive. The resulting
stresses we’ll refer to as compressive
stresses, fc.
Figure 4: Cable Shape Under Point Loads at the
Quarterpoints
If the cable is frozen and flipped over it
gives the ideal form for an arch (the arch
entirely in compression) under the point
loads applied at the quarterpoints. See
Figure 5.
If an arch takes on a form like that
shown in Figure 7, and the compressive
forces are in line with the axis of the
arch, then the direction of the
compressive
force
is
constantly
changing. The magnitude of the force
increases as the slope of the arch
increases and is therefore greatest at the
abutments. For our purposes we’ll
consider the compressive force at the
midspan as an approximation for the
force in the arch at any location.iv At the
midspan of the bridge the arch is
horizontal and therefore the compressive
force is in the horizontal direction. This
compressive force is equal (in magnitude
and direction) to the horizontal reaction,
H at the abutment. We will use this
compressive force to calculate the
compressive stress, fc in the arch. See
Figure 8:
Figure 5: Ideal Arch Form Under Point Loads at
the Quarterpoints
Likewise, Figure 6 shows the form (a
parabola) that a cable takes on under a
uniformly distributed load, q, while
Figure 7 shows the ideal form (a
parabola) for an arch under a uniformly
distributed load, q.
Figure 6: Cable Shape Under a Uniformly
Distributed Load
fc =
Figure 7: Ideal Arch Form Under a Uniformly
Distributed Load
H
A
Figure 8: Arch Section Under Compression
3
Example: Stress in the Arch of the
Salginatobel Bridge Under
Dead Load
Live Load
The location of the live traffic loads on
the bridge is not constant. While a cable
can alter its form to adapt to this change,
an arch cannot. When designing a bridge
we must position the live load so that its
effect on the bridge is the greatest.
Placing a concentrated live load, Q (lbs
or k) at each quarterpoint results in the
maximum bending. The vertical
reaction, V (lbs or k) at each abutment
due to this loading is simply Q.
Determine:
The
stress
at
the
quarterpoints of the Salginatobel Bridge
under dead load only.
Given: Each quarterpoint has a crosssectional area, A= 4,291 in2.
Recall that the horizontal reaction,
H = 1,455k
Solution:
The horizontal reaction, H (lbs or k) at
each abutment due to this loading is:
Calculate the compressive stress in the
arch at the quarterpoints:
fc =
H=
H
1,455k
=
= 0.34ksi = 340 psi
A 4,291in 2
Q*L
4*d
Where L (ft) is the length of the span of
the bridge, d is the depth of the arch (ft)
at midspan, and Q is still the live load at
the quarterpoints (lbs or k).
Since fc is a compressive stress,
fc = -340 psi
Example: Compressive Stress in the
Arch of the Salginatobel
Bridge Under Live Load
Bending in Arches
Arches are most efficient when the arch
is entirely in compression. As
demonstrated earlier for a particular
loading there is only one corresponding
form that is in pure compression. An
arch is unable to change its form to
accommodate the loads that are placed
on it; therefore, the selected form for an
arch must be able to carry different
loadings. The arch resists these different
loadings through a second type of
structural behavior, bending.
Figure 9: Elevation of the Salginatobel Bridge
Determine: The vertical and horizontal
reactions at each abutment of the
Salginatobel Bridge under concentrated
live loads placed at the quarterpoints.
Also, find the compressive stress at the
quarterpoint of the Salginatobel Bridge
under live load only. See Figure 9.
The Salginatobel Bridge was designed as
a three-hinged arch which means that at
the supports and at the crown the ends
are free to rotate and therefore no
bending occurs at these locations. In
contrast, the other points along the arch
experience bending under certain
loadings.
4
Given: The live load on the bridge, Q =
55k v
The span, L = 295 ft.
The depth of the arch, d = 42.6 ft.
The quarter point has an area, A =
4,291 in2.
arch.) From quarterpoint to quarterpoint
Q*L
the moment is a maximum: M =
4
Therefore the maximum bending
moment for the Salginatobel due to the
vertical loads is:
Solution:
55k * 295 ft
= 4,050k − ft
4
See Figure 10 for the bending moment
diagram.
M =
Step 1 -- Calculate the vertical reaction
at each abutment:
V = Q = 55k
Step 2 -- Calculate the horizontal
reaction at each abutment:
H=
Q * L 55k * 295 ft
=
= 95k
4*d
4 * 42.6 ft
Step 3 -- Calculate the compressive
stress in the arch at the quarterpoints:
fc =
Figure 10: Bending Moment Diagram Due to
Vertical Loads and Reactions
H
95k
=
= 0.02ksi = 20 psi
A 4,291in 2
Step 2: Draw the bending moment
diagram due to the horizontal reactions
at the abutments. (This bending moment
is not present in a beam.) The maximum
bending moment due to the horizontal
reactions is M = − H * d and occurs at
the center of the arch. At each
quarterpoint the bending moment due to
the horizontal reactions is
3* H * d
. Therefore for the
M =−
4
Salginatobel the maximum moment is
M = −95k * 42.6 ft = −4,050k − ft and
the bending moment at each quarterpoint
3 * 95k * 42.6 ft
M =−
= −3,040k − ft .
4
See Figure 11 for the bending moment
diagram.
Since fc is a compressive stress,
fc = -20 psi
Bending Moments
The uniformly distributed dead load
doesn’t cause bending in the arch, but
the concentrated live loads placed at the
quarterspans do. To determine the
bending moments along the arch, we
need to perform the following three
steps:
Step 1: Draw the bending moment
diagram due to the vertical loads. (This
bending moment diagram is the same as
what you would get if the structure was a
simply supported beam instead of an
5
Bending stress is still:
T
C
f b ( +) =
f b ( −) =
A flange
A flange
Where Aflange is the area of the flange
(in2).
Example: Bending Stress in the
Salginatobel at the
Quarterpoints
Figure 11: Bending Moment Diagram Due to
Horizontal Reactions
Step 3: Add the two plots together to get
the total bending moment along the arch.
See Figure 12.
Figure 12: Total Bending Moment Diagram
Determine: The bending stress at the
quarterpoints.
From the bending moment diagram you
can see that the bending moment at
locations where there are hinges (the
supports and at the crown) is equal to
0 k-ft. At the quarterpoints the bending
moment is greatest and is equal to
1,010 k-ft for the Salginatobel.
Given: The bending moment, M = 1,010
k-ft
The distance from the center of the top
flange to the center of the bottom flange,
a = 12.8 ft
The area of the flange, Aflange = 1,113 in2
Force Couple and Bending Stress
As presented in structural studies “II.
Cantilevers: The Eiffel Tower” and “IV.
Beams: The Alexander Road Overpass,”
a bending moment can be represented by
a force couple, a set of equal and
opposite forces separated by a distance,
a. For the I-beam presented in structural
study IV the distance a, was the distance
from the center of the top flange to the
center of the bottom flange. In the case
of the Salginatobel the cross-section is a
hollow box and a, is still the distance
from the center of the top flange to the
center of the bottom flange.
Solution:
Step 1 -- Calculate the force couple:
T=
M 1,010k − ft
=
= 79k
a
12.8 ft
C = −T = −79k
Step 2 -- Calculate the bending stress at
the quarterpoints:
T
79k
fb =
=
= 0.07 ksi = 70 psi
A flange 1,113in 2
6
The bending stress on the bottom face
will be 70si in tension (or fb = +70psi).
The bending stress on the top face will
be 70psi in compression (or
fb = -70psi).
The Form of an Arch Bridge
Just like for suspension bridges,
q * L2
is an important equation in
H=
8* d
determining the form for an arch bridge.
Additionally, bending influences the
form. The bending moment in a threehinged arch is largest at the quarterpoints
and zero at the supports and crown. An
efficient form puts more material in
locations where bending is large and less
material in locations where bending is
small. In other words an efficient form
mimics the bending moment diagram.
Total Stress
The sum of the stresses applied to an
object is the total stress, ftotal. The
equation for total stress is:
f total = f c + f b
Example: Total Stress in the
Salginatobel at the Quarterpoints
Summary of Terms
Determine: The minimum and maximum
total stresses at the quarterpoints. vi
-a: distance between C and T [ft]
-A: total cross-sectional area, [in2]
-Aflange: cross-sectional area of the
flange, [in2]
-C: compressive force, [lbs] or [k]
-d: depth of the arch, [ft]
-fb: bending stress, [psi] or [ksi]
-fc: compressive stress, [psi] or [ksi]
-fc,DL: compressive stress due to DL,
[psi] or [ksi]
-fc,LL: compressive stress due to LL, [psi]
or [ksi]
-ftotal: total stress, [psi] or [ksi]
-ftotal,min: total stress on the bottom face
of the arch, [psi] or [ksi]
-ftotal,max: total stress on the top face of
the arch, [psi] or [ksi]
-H: horizontal reaction at abutment and
compressive force in arch, [lbs] or
[k]
-L: main span length [ft]
-M: moment, [lbs-ft] or [k-ft]
-q: distributed load, [lbs/ft] or [k/ft]
-Q: concentrated load, [lbs] or [k]
-T: tensile force, [lbs] or [k]
-V: vertical reaction at tower, [lbs] or [k]
Given: The compressive stress due to
DL, fc,DL = -340 psi
The compressive stress due to LL, fc,LL =
-20 psi
The bending stress, fb = +/- 71 psi
Solution:
Step 1 – Calculate the minimum total
stress. This occurs on the bottom face of
the arch.
f total,min = f c ,DL + f c ,LL + f b
f total,min = −340 psi − 20 psi + 70 psi = −290 psi
Step 2 – Calculate the maximum total
stress. This occurs on the top face of the
arch.
f total,max = f c ,DL + f c ,LL + f b
f total,max = −340 psi − 20 psi − 70 psi = −430 psi
7
Bending Moment at the Quarterpoints
due to Horizontal Reactions:
3* H * d
M =−
4
Summary of Equations
Vertical Reaction at Abutment Due to
Dead Load:
V=
Force Couple:
q*L
2
−C =T =
Vertical Reaction at Abutment Due to
Live Load:
Bending Stress:
V =Q
f b ( +) =
Horizontal Reaction at Abutment due to
Dead Load:
A flange
f b ( −) =
C
A flange
f total = f c + f b
Minimum total stress (stress on the
bottom face of the arch):
Horizontal Reaction at Abutment due to
Live Load:
f total,min = f c + f b ,tension
Q*L
4*d
Maximum total stress (stress on the top
face of the arch side):
f total,max = f c + f b ,compression
Compressive Stress:
fc =
T
Total stress:
q * L2
H=
8* d
H=
M
a
H
A
Notes
i
Robert Maillart’s Bridges; The Art of
Engineering, by David P. Billington,
Princeton University Press, Princeton,
NJ, 1979, p. 81.
Maximum Bending Moment due to
Vertical Loads and Reactions:
M =
Q*L
4
ii
“Official Documents for the Arch
Analysis,” in Background Papers for the
Second National Conference on Civil
Engineering: History, Heritage, and the
Humanities II, October 4, 5, 6, 1972,
Princeton University, Edited by John F.
Abel, Conference Co-directors: David P.
Billington and Robert Mark, Conference
Maximum Bending Moment due to
Horizontal Reactions:
M = −H * d
8
sponsored by the National Endowment
for the Humanities
iii
f total,max = f c ,DL + f c ,LL + f b
f total,max = −350 psi − 20 psi − 70 psi = −440 psi
Ibid
iv
The actual compression force, N (lbs
or k) at any location along the arch can
be found with the following equation:
H
, where α is the angle
cos α
between the main axis of the arch and
the horizontal. H and the resulting
compressive stress, fc, are adequate
approximations for the arch.
N=
“Official Documents for the Arch
Analysis,” in Background Papers for the
Second National Conference on Civil
Engineering: History, Heritage, and the
Humanities II,
v
Performing the calculations at the
quarterpoints (α = 16.1°) with N instead
of H, you get the following stress values:
vi
fc,DL = -350 psi
fc,LL = -20 psi
fb remains unchanged and = ± 70 psi
Therefore:
f total,min = f c ,DL + f c ,LL + f b
f total,min = −350 psi − 20 psi + 70 psi = −300 psi
and
9