Arches & Cables

How an arch works

An arch is a curved member supported at its ends so that the supports can provide horizontal reactions (thrust) as well as vertical ones. The horizontal thrust $H$ produces a moment that opposes the beam moment caused by the loads:

where ${\mu }_x$ is the bending moment in a simply supported beam of the same span under the same loads, and $y$ is the height of the arch axis at the section. Because $Hy$ cancels much of ${\mu }_x$, an arch carries load mainly by compression, with small bending. That is why masonry — strong in compression, weak in tension — could span large openings as arches.

Linear arch and Eddy's theorem

For a given loading, the linear arch (line of thrust, funicular shape) is the arch shape in which the bending moment is zero everywhere — the shape of a hanging cable under the same loads, turned upside down.

• For a UDL on the horizontal span, the linear arch is a parabola.
• For concentrated loads, it is a polygon.

Three-hinged arches

Unknowns: $V_A$, $H_A$, $V_B$, $H_B$ (4). Equations: 3 of equilibrium + $M=0$ at the crown hinge = 4. So the arch is determinate.

Parabolic arch geometry

With origin at the left support, span $L$ and rise $h$:

Circular arch geometry

Radius $R$ from span and rise: $R=\frac{L^2}{8h}+\frac{h}{2}$. At a horizontal distance $x^{\prime }$ from the crown, $y=\sqrt{R^2-x^{\prime 2}}-(R-h)$.

UDL over the whole span (parabolic three-hinged arch)

$V_A=V_B=wL/2$. Moment at crown $=0$:

At any section, ${\mu }_x=\frac{wx(L-x)}{2}$ and $Hy=\frac{w{L}^2}{8h}\cdot \frac{4hx(L-x)}{L^2}=\frac{wx(L-x)}{2}$.

Normal thrust and radial shear

At a section where the arch axis makes angle $\theta$ with the horizontal, resolve $H$ and the vertical shear $V$ (beam shear at that section):

Cables as structural members

A cable is a flexible member that can carry only tension — it has no bending or compression resistance. Under any load it takes the shape in which every part is in pure tension (the funicular shape). Cables are used in suspension and cable-stayed bridges, ropeways, transmission lines, guyed masts and cable roofs.

Assumptions in elementary analysis: the cable is perfectly flexible and inextensible (or its elongation is small), self-weight is either neglected or treated as a UDL, and loads are vertical.

Cable under point loads

Between loads the cable is straight. At each load point, equilibrium gives a polygon. Because loads are vertical, the horizontal component of tension $H$ is the same everywhere in the cable.

For supports at the same level, the reactions follow from treating the cable like a beam, and $H$ follows from one known sag.

General cable theorem

It is the arch relationship with $M=0$: a cable is an arch turned upside down that can only be in tension.

Cable under a UDL (parabolic cable)

A UDL $w$ per unit horizontal length (e.g. the deck of a suspension bridge, which is much heavier than the cable) makes the cable a parabola. With span $L$ and central sag $h$ (supports at the same level):

Tension is minimum at the lowest point ($=H$) and maximum at the supports.

A cable hanging under its own weight per unit length of cable forms a catenary; for small sag–span ratios the parabola is an excellent approximation.

Supports at different levels

Let the lowest point be at horizontal distances $x_1$ and $x_2$ from the supports, with sags $h_1$ and $h_2$ below them ($x_1+x_2=L$). Each side behaves like half a parabola:

Vertical reactions: $V_1=w{x}_1$, $V_2=w{x}_2$.

Length of a parabolic cable

For supports at the same level:

Each half of a cable with unequal supports: ${\ell }_i\approx x_i+\frac{2h_i^2}{3x_i}$.

Change of sag with length or temperature

From $\ell =L+8h^2/3L$: $\delta \ell =\frac{16h}{3L}\delta h$ → $\delta h=\frac{3L}{16h}\delta \ell$.

A temperature rise $T$ lengthens the cable by $\delta \ell =\alpha T\ell \approx \alpha TL$, which increases the sag and reduces $H$. Elastic stretch under tension does the same.