A 40 m beam under its design load carries 4000 kN·m of bending. Bend that same beam into a parabola, hold its feet from spreading, and the bending drops to zero. Not "small" — zero. This chapter is about why.
A beam in bending is a structure at war with itself. The top fibres squash, the bottom fibres stretch, and the middle does almost nothing — you are paying for material that is barely working. An arch is the attempt to make every fibre carry the same compression, all the way through.
A beam presses straight down on its supports. An arch also shoves them apart, with a horizontal force called the thrust. Every gain in this chapter is bought with that thrust, and if the abutments cannot resist it, the arch flattens and fails.
Get the shape right for the load and the bending moment vanishes over the whole span. The section can then be slim, because it only has to resist axial compression — no lever arm required.
Before computing anything, there is one physical fact that gives you the entire chapter for free.
A chain cannot resist bending — it has no stiffness at all. So when you hang it between two points, it has no choice but to settle into the one shape in which the load is carried by pure tension. That shape is not a guess. It is the only shape that works.
A deck of constant weight per horizontal metre. The chain hangs as a parabola. This is the case that dominates suspension bridges and this course.
The cable's own weight, with nothing hanging from it. The shape is a catenary (a cosh curve). For shallow sags it is within a percent of the parabola, so we use the parabola.
The chain becomes a chain of straight segments with a kink at each load. Flip it and you get the polygonal arch — still no bending, just a different funicular polygon.
Take a cable of span L carrying a uniform load w per horizontal metre, sagging a distance d at midspan. Cut it at the lowest point and look at half of it.
The developed length of a shallow parabolic cable, which you need to order the material and to compute stretch:
For d/L = 0.1 this gives S = 1.0260 L — the cable is only 2.6% longer than the straight line. Sag is cheap in material and enormously valuable in force.
Drag the sag-ratio slider and watch Tmax: at d/L = 1/2 the cable is nearly slack and the force is small; at d/L = 1/30 it is almost straight and the force explodes. The load never changed.
Now invert the cable. Tensions become compressions, the downward sag becomes an upward rise h, and the horizontal pull H becomes the horizontal thrust pushing outwards on the abutments.
Two pinned supports give four reaction components (VA, HA, VB, HB) against three equations — one redundant. Put a hinge at the crown and you gain one equation of condition (\(\Sigma M = 0\) about the hinge, for one side only). Four unknowns, four equations: determinate, and solvable with statics alone.
A two-hinged or fixed arch develops large forces if an abutment settles or the temperature changes, because it is indeterminate. The three-hinged arch simply adjusts its shape and carries on. That robustness is why it is the standard starting point.
Move 1 — vertical reactions: take moments about A and B for the whole arch. These come out exactly as for a simply supported beam of the same span, because H acts through both supports at deck level and contributes nothing.
Move 2 — the thrust: take moments about the crown hinge C, using only the left half. The internal moment there is zero by definition of a hinge:
where \(M_{C,\text{beam}}\) is the bending moment that a simply supported beam of the same span would have at midspan under the same load. For a full UDL that is \(wL^2/8\), giving the formula you will use constantly:
The top panel is the structure with its reactions; the bottom panel is the bending moment along the span, drawn to the same scale for all three shapes so you can see them collapse. Start with the parabolic arch under UDL only, then switch the shape to "straight beam" and watch the moment diagram appear from nothing.
Set P = 0. The moment diagram is a flat line at zero for the whole span. The arch is carrying 800 kN of load with no bending at all.
Same span, same rise, same H — but the axis no longer matches the parabola, so a moment appears. It peaks near the quarter points, at about 4% of the beam value.
Set P = 100 kN at a = 10 m. A kink appears under the load. The UDL is still carried funicularly — only the mismatched part of the load causes bending.
An arch section carries three things. Cut the arch at horizontal position x, keep the left part, and resolve everything that acts on it into components along and across the arch axis — not vertical and horizontal, because the axis is sloped.
Compression. This is the force the section actually lives on. Largest at the springings, where the axis is steepest.
Zero everywhere for a true funicular arch — the resultant lies exactly along the axis, so it has no across-component.
V here is the net vertical force to the left of the cut — the same "beam shear" you already know how to compute.
Real arches always bend a little, and there are only three reasons why.
The arch was shaped for its dead load. Traffic on half the span, snow drifting to one side, or a single heavy vehicle is a different load with a different funicular shape. Asymmetric live load is usually the governing case.
Circular arches are easier to build than parabolic ones, so they get built. The 4% residual moment in the simulator is the price of that convenience — usually acceptable, and always worth knowing about.
If the abutments spread, H drops, and the term \(H\,y\) can no longer cancel the beam moment. The arch starts behaving like a curved beam — and curved beams are bad beams. This is how old masonry bridges fail.
Everything in this chapter is one equation seen from different angles:
Match them and the arch is pure compression. Mismatch them and the difference shows up as bending. The funicular shape, the thrust formula, the sag–force trade and the thrust line are all just readings of this one balance.
Problem: A three-hinged parabolic arch spans L = 40 m with a crown rise h = 8 m and carries a UDL of w = 20 kN/m over the full span. Find the reactions, the maximum bending moment, and the thrust at the springing.
Problem: Remove the UDL and apply a single P = 100 kN at a = 10 m from A. Find H and the bending moment under the load.
Problem: A cable spans L = 30 m with a midspan sag of 3 m and carries w = 5 kN/m along the horizontal. Find H, the maximum and minimum tension, and the cable length.
1. A three-hinged parabolic arch carries a full-span UDL. What is the bending moment at the quarter point?
2. You halve the sag of a cable while keeping the span and load the same. The maximum tension:
3. Why is a hinge placed at the crown of an arch?
4. The abutments of a masonry arch spread slightly outwards over time. What happens?
5. Hooke's principle "as hangs the flexible line, so but inverted will stand the rigid arch" means: