Editing Tensile structure (section)

== Simple mathematics of cables ==

=== Transversely and uniformly loaded cable ===

A uniformly loaded cable spanning between two supports forms a curve intermediate between a [[catenary]] curve and a [[parabola]]. The simplifying assumption can be made that it approximates a circular arc (of radius ''R'').

[[Image:Catenary cable diagram.svg|700px|left]]
{{Clear}}

'''By [[mechanical equilibrium|equilibrium]]:'''

The horizontal and vertical reactions :

:<math>H = \frac{wS^2}{8d}  </math>
:<math>V = \frac{wS}{2}</math>

'''By [[geometry]]:'''

The length of the cable:

:<math>L = 2R\arcsin\frac{S}{2R}</math>

The tension in the cable:

:<math>T = \sqrt{H^2+V^2}</math>

By substitution:

:<math>T = \sqrt{\left(\frac{wS^2}{8d}\right)^2 + \left(\frac{wS}{2}\right)^2}</math>

The tension is also equal to:

:<math>T = wR</math>

The extension of the cable upon being loaded is (from [[Hooke's law]], where the axial stiffness, ''k,'' is equal to <math>k = \frac{{EA}}{{L}}</math>):

:<math>e = \frac{TL}{EA}</math>

where ''E'' is the [[Young's modulus]] of the cable and ''A'' is its cross-sectional [[area]].

If an initial pretension, <math>T_0</math> is added to the cable, the extension becomes:

:<math>e = L - L_0 = \frac{L_0(T-T_0)}{EA}</math>

Combining the above equations gives:

:<math>\frac{L_0(T-T_0)}{EA} + L_0 = \frac{2T\arcsin\left(\frac{wS}{2T}\right)}{w}</math>

By plotting the left hand side of this equation against ''T,'' and plotting the right hand side on the same axes, also against ''T,'' the intersection will give the actual equilibrium tension in the cable for a given loading ''w'' and a given pretension <math>T_0</math>.

=== Cable with central point load ===

[[Image:point-loaded cable.svg|800px|left]]
{{Clear}}
A similar solution to that above can be derived where:

'''By equilibrium:'''

:<math>W = \frac{4Td}{L}</math>

:<math>d = \frac{WL}{4T}</math>

'''By geometry:'''

:<math>L = \sqrt{S^2 + 4d^2} = \sqrt{S^2 + 4\left(\frac{WL}{4T}\right)^2}</math>

This gives the following relationship:

:<math>L_0 + \frac{L_0(T-T_0)}{EA} = \sqrt{S^2 + 4\left(\frac{W(L_0+\frac{L_0(T-T_0)}{EA})}{4T}\right)^2}</math>

As before, plotting the left hand side and right hand side of the equation against the tension, ''T,'' will give the equilibrium tension for a given pretension, <math>T_0</math> and load, ''W''.