Arch bridges is not very common in our culture but it is one of the most aesthetical bridges that be found in nature (see Figure 1). A young engineer once mentioned that he built an arch bridge with no reinforcement. This claim generated bits of argument among fellow young engineers. How can this be possible? Some young inexperienced engineers may believe that it would be impossible to achieve strength and stability of any structure without reinforcement. A structure without moments may not need reinforcement. One such situation where this could be obtainable is in arches.
There are basically three types of arches namely: hingeless arches, two hinged arches and three hinged arches. Among these three, only three hinged arch is statically determinate. Three hinged arches are further sub-divided into parabolic arch and circular arch.
The distinguishing feature of an arch from a beam can be explained by comparing its response to loading to that of a beam. A concentrated or point load applied to a beam results in vertical reactions and moment in the beam. Thus to resist these moments, reinforcements are required. If the same load is supported by a two hinged arch, the arch resists the load by vertical and horizontal components of reaction. The horizontal components of reaction in effect reduces the moment from that in a simple beam. Thus, the arch supports loading with less moment than simple beam configuration. In this process, stresses which may largely be compressive stresses are generated in the arch and these can conveniently be resisted by the mass of concrete.
Advantages of arches
- Arches have axial forces.
- They are structurally more efficient than beam and can carry loads L-times more than beam and thus, they are economical.
- Arches have no problem of bending and deflection and thus requires little or no reinforcement.
A comparison is made in this article with analysis of simple beam and three hinged arch subjected to the same system of loads to determine the stresses acting on them.
A simple beam and three hinged parabolic arch each of span 20 metres and central rise of 5 metres (arch) carries a point load of 20 kN at 6 metres from the left-hand support as shown in Figures 3 and 4.
- Find the reactions at the supports A and B and
- Draw the bending moment diagram for the arch, and indicate the position of maximum bending moment.
From Figure 1, AB = 20000 mm = 20 m; AC = 6000 mm = 6 m; HC = 5000 mm = 5 m
ΣMB = 0
VA x 20 – 20 x 14 = 0
VA = 280/20 = 14kN
VA + VB = 20; VB = 20 – VA = 20 – 14 = 6kN
Moment at any distance, x along the span = RAx – 20 (x – 6)
Maximum moment at C, x = 6 m = 14 x 6 – 20 = 84 – 20 = 64kNm
See Figure 5 for the bending moment diagram
From Figure 2, Let
VA = vertical reaction at A;
VB = vertical reaction at B;
T= horizontal thrust at A and B
As determined previously, VA = 14kN.
The beam moment at C, MC due to the external loading = VB x 10 = 6 x 10 = 60 kNm or VA x 10 – (20 x 4) = (14 x 10) – 80 = 140 – 80 = 60kNm
Horizontal trust at A and B, T = MC/HC = 60/5 = 12kN
Reaction at A, RA = √VA2 + T2 = √142 + 122 = √340 = 18.44kN
Reaction at B, RB = √VB2 + T2 = √62 + 122 = √180 = 13.42kN
How to draw the bending moment diagram
- Draw the arch ACB with its given span and rise.
- Since the bending moment at A, B and C are zero, Join B and C and extend this line.
- Draw a vertical line through D, meeting the line BC at E.
- Join AE.
- AEB is the required bending moment diagram. The maximum moment occurs at the position of the load.
We can see from the diagram how the positive and negative moments came close to cancelling each other out, thus, small moment effect would be felt. This is unlike a simple beam where the effect of moment is very obvious and thus reinforcement is required for strength and stability.
See Figure 6 for bending moment diagram
How to determine the maximum positive and negative moment in the arch
Maximum positive moment
y = rise of the arch at D (point of maximum bending moment)
Using the relation, y = 4HC/AB2 x R (AB – R) = 4 x 5/202 x 6 (20 – 6) = 4.2m
Maximum positive moment at D = VA x 6 – T x y = 14 x 6 – 12 x 4.2 = 84 – 50.4 = 33.6kNm
Maximum negative moment
Let the maximum negative moment take place at the distance, S from B.
The rise of the arch at the distance, S from B = y = 4HC/AB2 x S (AB – S) = 4 x 5/202 x S (20 – S) = S – (S2/20)
The bending moment at S distance from B = VB x S – T x y = 6 x S – 12 (S – (S2/20)) = 6S – 12S – 3S2/5 = 3S2/5 – 6S.
For maximum bending moment, differentiate the above equation with respect to S and equate to zero
dMS/dS (3S2/5 – 6S) = 0
Thus, 6S/5 – 6 = 0; 6S/5 = 6; S = 5m
This implies that the maximum moment is at the rise of the arch at 5m from B.
Rise of the arch at 5m from B, y = S – (S2/20) = 5 – (25/20) = 5 – 1.25 = 3.75m
Maximum negative moment = 6 x 5 – 12 x 3.75 = 30 – 45 = -15kN/m
In summary, the curvature of the arch reduces the moment from 64 kN/m of the simple beam to 18.6 kN/m.