How does curvature of an arch bridge affect its strength?

AI Thread Summary
The curvature of an arch bridge significantly influences its strength and load-bearing capacity. As the curvature increases, the angles in the force vector diagram change, leading to shorter force vectors and reduced stress on the bridge. This relationship allows for the derivation of equations to quantify stress based on curvature. Calculating the maximum load a bridge can carry involves understanding material properties and potential failure modes, such as block crushing or foundation displacement. Overall, optimizing the design for cost and strength requires a detailed analysis of these factors.
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Hey, I've been looking into some civil/structural engineering for a school project, and came across bridge design. I've decided to try some integrating and optimising to do with an arch bridge (optimal cost/strength proportions). The math isn't too hard, but what I'm struggling with so far is finding any relationship between the curvature of the arch and the bridge's maximum load. I'm sure there must be one, there were some indicators of it in my research, but these were always very superficial. If anyone could point me in the right direction on this, I would be very grateful.
 
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Here's one way to understand it. Draw a diagram of three consecutive blocks in the arch, and idealise them as point masses each with a fixed weight W, connected by stiff, massless rods.

The lower rod pushes up diagonally on the central point mass and the upper rod pushes down diagonally at a slightly shallower angle than that of the lower rod. The force in the lower rod will be larger than in the upper rod (the lower members bear the weight of all members above them).

Draw the vectors of the forces in the two rods end to end, in such a way that the head (arrow end) of the second vector (point A) is exactly above the tail of the first one (point B) (It doesn't matter which vector you draw first - the net result is the same). For the bridge to be stationary, the vector from A to B must be the weight of the central mass.

That vector AB is fixed, regardless of the curvature of the bridge - it's the weight of a standard block. Now, as you change the curvature, the two other sides of the vector triangle change. If you increase (decrease) the curvature they shorten (lengthen) because the angle opposite the AB vector increases (decreases). So you can see that increasing the curvature increases that angle and thus shortens the two force vectors, which means less stress in the bridge.

Given that framework, it's pretty easy to write some equations to quantify that.
 
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andrewkirk said:
Here's one way to understand it. Draw a diagram of three consecutive blocks in the arch, and idealise them as point masses each with a fixed weight W, connected by stiff, massless rods.

The lower rod pushes up diagonally on the central point mass and the upper rod pushes down diagonally at a slightly shallower angle than that of the lower rod. The force in the lower rod will be larger than in the upper rod (the lower members bear the weight of all members above them).

Draw the vectors of the forces in the two rods end to end, in such a way that the head (arrow end) of the second vector (point A) is exactly above the tail of the first one (point B) (It doesn't matter which vector you draw first - the net result is the same). For the bridge to be stationary, the vector from A to B must be the weight of the central mass.

That vector AB is fixed, regardless of the curvature of the bridge - it's the weight of a standard block. Now, as you change the curvature, the two other sides of the vector triangle change. If you increase (decrease) the curvature they shorten (lengthen) because the angle opposite the AB vector increases (decreases). So you can see that increasing the curvature increases that angle and thus shortens the two force vectors, which means less stress in the bridge.

Given that framework, it's pretty easy to write some equations to quantify that.

Thank you! That's very helpful, exactly what I was hoping for I'm not completely sure of how your diagram would work, if I'm not wrong there should be two rods connecting the three point masses, so what rods are you referring to with the lower and upper rods? And why would the upper rod push down at a slightly shallower angle than the lower one pushes up? I tried visualising it the way you suggested, but I feel like I'm not understanding it correctly as of yet.
 
maxf11 said:
And why would the upper rod push down at a slightly shallower angle than the lower one pushes up?
Unfortunately I haven't taught myself to do diagrams on the web yet, so I'll use coordinates in the number plane instead.

Imagine the three point masses are at points U=(-2,1), V=(0,0) and W=(1,-2). The upper rod is UV and the lower is VW. The rods can only push in a direction that aligns with their length, so the lines of force of UV have gradient 0.5 and 2.0 respectively.
 
andrewkirk said:
Unfortunately I haven't taught myself to do diagrams on the web yet, so I'll use coordinates in the number plane instead.

Imagine the three point masses are at points U=(-2,1), V=(0,0) and W=(1,-2). The upper rod is UV and the lower is VW. The rods can only push in a direction that aligns with their length, so the lines of force of UV have gradient 0.5 and 2.0 respectively.

OK, I see where my thinking was off. Just to be sure I'm getting it right now, I've attached a screenshot of how I understand it: http://imgur.com/KHjEH4v. I also see how changing the curvature will affect the internal angle of the vector diagram, and the lengths of the vectors. Could you please explain how longer vectors mean more stress on the bridge? And how (if possible) I can calculate the amount of weight the bridge can carry based on this stress?
 
maxf11 said:
Could you please explain how longer vectors mean more stress on the bridge
The stress is just another name for the magnitude of the force transmitted through its members - which in this case is the blocks. The length of a force vector is its magnitude. So the longer the force vector, the bigger the stress.

Calculating the weight it can carry will depend on what component would be the first to fail. Engineers would be the best placed to suggest that. Possibilities I can see are

(1) a block being crushed under the stress and disintegrating, thereby creating a gap and allowing the arch to collapse
(2) the foundation at either end being displaced by the stress, so that the arch flattens and collapses
(3) the stresses on either side of a block squeezing the block out of position

Each of these will depend on complex properties of the materials like compressibility and coefficients of friction, as well as the design of the foundations. Calculating the stress at various points in a bridge is much easier than estimating the load at which it will fail.
 
andrewkirk said:
(1) a block being crushed under the stress and disintegrating, thereby creating a gap and allowing the arch to collapse
Suppose I focus on only one of them, and assume the construction in itself is perfect, that would essentially leave me only with the quoted option to investigate, right? If I choose a specific material, do you think I can calculate when this material will collapse under the stress?
Some more context: I'm doing all this in a mathematics-subject based investigation, so I'm fine with doing lots of calculations. In the end, what I intend to do is optimise cost (basically just amount of material used in the bridge) and 'strength' of the bridge, so this research is focussed mainly on finding equations for the stress in a bridge.
 
Could anyone else just briefly let me know if they have any insights whatsoever regarding this topic? It would be greatly appreciated, thanks.
 
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  • #10
There are basically two classes of arch bridge :

(1) The simple free standing inverted U shape bridge .

(2) The embedded arch bridge where the basic U shape is built into a larger masonry structure .

Stress analysis of a real arch bridge is quite complicated .

http://www.ikbrunel.org.uk/maidenhead-bridge
 
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  • #12
Thank you for all the responses, you have all been very helpful!
 
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