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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. 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

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. Abutment Reactions Loads are transmitted from the arch to abutments at the ends of the arch. The applied gravity loads on the arch are acting downward; therefore the

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abutments need to resist these forces with upward reactions. The vertical reaction, V (lbs or k) at each abutment is:

2* LqV =

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. 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:

dLqH*8* 2

=

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.

Figure 2: Loaded Arch with Reactions

Example: Reactions at the Abutments of the Salginatobel

Figure 3: Elevation of the Salginatobel Bridge

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. Solution: Step 1 -- Calculate the vertical reaction at each abutment:

kftftkLqV 8402295*/7.5

2*

===

Step 2 -- Calculate the horizontal reaction at each abutment:

kftftftk

dLqH 455,1

6.42*8)295(*/7.5

*8* 22

===

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.

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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.

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.

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

Figure 7: Ideal Arch Form Under a Uniformly

Distributed Load

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. 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:

AHfc =

Figure 8: Arch Section Under Compression

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Example: Stress in the Arch of the Salginatobel Bridge Under Dead Load Determine: The stress at the quarterpoints of the Salginatobel Bridge under dead load only. Given: Each quarterpoint has a cross-sectional area, A= 4,291 in2. Recall that the horizontal reaction, H = 1,455k Solution: Calculate the compressive stress in the arch at the quarterpoints:

psiksiink

AHfc 34034.0

291,4455,1

2 ====

Since fc is a compressive stress, fc = -340 psi 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. 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.

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. The horizontal reaction, H (lbs or k) at each abutment due to this loading is:

dLQH*4*

=

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). Example: Compressive Stress in the Arch of the Salginatobel Bridge Under Live Load

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.

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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. Solution: Step 1 -- Calculate the vertical reaction at each abutment:

kQV 55== Step 2 -- Calculate the horizontal reaction at each abutment:

kftftk

dLQH 95

6.42*4295*55

*4*

===

Step 3 -- Calculate the compressive stress in the arch at the quarterpoints:

psiksiink

AHfc 2002.0

291,495

2 ====

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

arch.) From quarterpoint to quarterpoint

the moment is a maximum: 4* LQM =

Therefore the maximum bending moment for the Salginatobel due to the vertical loads is:

ftkftkM −== 050,44295*55

See Figure 10 for the bending moment diagram.

Figure 10: Bending Moment Diagram Due to

Vertical Loads and Reactions 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 dHM *−= and occurs at the center of the arch. At each quarterpoint the bending moment due to the horizontal reactions is

4**3 dHM −= . Therefore for the

Salginatobel the maximum moment is ftkftkM −−=−= 050,46.42*95 and

the bending moment at each quarterpoint

ftkftkM −−=−= 040,34

6.42*95*3 .

See Figure 11 for the bending moment diagram.

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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 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. 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.

Bending stress is still:

flangeb A

Tf =+)(flange

b ACf =−)(

Where Aflange is the area of the flange (in2). Example: Bending Stress in the Salginatobel at the Quarterpoints

Determine: The bending stress at the quarterpoints. 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

Solution: Step 1 -- Calculate the force couple:

kftftk

aMT 79

8.12010,1

=−

==

kTC 79−=−=

Step 2 -- Calculate the bending stress at the quarterpoints:

psiksiink

ATfflange

b 7007.0113,179

2 ====

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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). Total Stress The sum of the stresses applied to an object is the total stress, ftotal. The equation for total stress is:

bctotal fff += Example: Total Stress in the Salginatobel at the Quarterpoints Determine: The minimum and maximum total stresses at the quarterpoints. vi

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.

bLLcDLctotal,min ffff ++= ,,

psipsipsipsif total,min 2907020340 −=+−−=

Step 2 – Calculate the maximum total stress. This occurs on the top face of the arch.

bLLcDLctotal,max ffff ++= ,,

psipsipsipsiftotal,max 4307020340 −=−−−=

The Form of an Arch Bridge Just like for suspension bridges,

dLqH*8* 2

= is an important equation in

determining the form for an arch bridge. Additionally, bending influences the form. The bending moment in a three-hinged 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. Summary of Terms -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]

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Summary of Equations Vertical Reaction at Abutment Due to Dead Load:

2* LqV =

Vertical Reaction at Abutment Due to Live Load:

QV = Horizontal Reaction at Abutment due to Dead Load:

dLqH*8* 2

=

Horizontal Reaction at Abutment due to Live Load:

dLQH*4*

=

Compressive Stress:

AHfc =

Maximum Bending Moment due to Vertical Loads and Reactions:

4* LQM =

Maximum Bending Moment due to Horizontal Reactions:

dHM *−=

Bending Moment at the Quarterpoints due to Horizontal Reactions:

4**3 dHM −=

Force Couple:

aMTC ==−

Bending Stress:

flange

b ATf =+)(

flangeb A

Cf =−)(

Total stress:

bctotal fff += Minimum total stress (stress on the bottom face of the arch):

tensionbctotal,min fff ,+= Maximum total stress (stress on the top face of the arch side):

ncompressiobctotal,max fff ,+=

Notes i Robert Maillart’s Bridges; The Art of Engineering, by David P. Billington, Princeton University Press, Princeton, NJ, 1979, p. 81. 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

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sponsored by the National Endowment for the Humanities

iii Ibid iv The actual compression force, N (lbs or k) at any location along the arch can be found with the following equation:

αcosHN = , where α is the angle

between the main axis of the arch and the horizontal. H and the resulting compressive stress, fc, are adequate approximations for the arch.

v “Official Documents for the Arch Analysis,” in Background Papers for the Second National Conference on Civil Engineering: History, Heritage, and the Humanities II, vi Performing the calculations at the quarterpoints (α = 16.1°) with N instead of H, you get the following stress values: fc,DL = -350 psi fc,LL = -20 psi fb remains unchanged and = ± 70 psi Therefore:

bLLcDLctotal,min ffff ++= ,,

psipsipsipsif total,min 3007020350 −=+−−=

and

bLLcDLctotal,max ffff ++= ,,

psipsipsipsiftotal,max 4407020350 −=−−−=