Equalization (proof)

A mathematical analysis of static load-sharing (also called load-distributing) 2-point Anchor Systems.

Derivation

Consider the node, where The Two anchor legs join with the main line. At this node, the sum of all forces in the x-direction must equal zero since the system is in mechanical equilibrium.
Fx1 = Fx2 
F1Sin(α) = F2Sin(β) 

The net force in the y-direction must also sum to zero.
Fy1 + Fy2 = Fload 

Substitute F1 from equation () into equation () and factor out F2.
$$F_2\frac{Sin(\beta )}{Sin(\alpha )}Cos(\alpha )+F_2Cos(\beta )=F_{load} \,$$
$$F_2\left [\frac{Sin(\beta )}{Sin(\alpha )}Cos(\alpha )+Cos(\beta )\right ]=F_{load} \,$$
Solve for F2 and simplify.

$$F_2=\frac{F_{load}}{\left [\frac{Sin(\beta )}{Sin(\alpha )}Cos(\alpha )+Cos(\beta )\right ]} \,$$
$$F_2=\frac{F_{load}}{\left [\frac{Sin(\beta )Cos(\alpha )}{Sin(\alpha )}+\frac{Cos(\beta )Sin(\alpha)}{Sin(\alpha)}\right ]} \,$$
$$F_2=\frac{F_{load}}{\left [\frac{Cos(\alpha )Sin(\beta )+Sin(\alpha)Cos(\beta )}{Sin(\alpha )}\right ]} \,$$
$$F_2=F_{load}\frac{Sin(\alpha )}{{Cos(\alpha )Sin(\beta )+Sin(\alpha)Cos(\beta )}} \,$$
Use a trigonometric identity to simplify more and arrive at our final solution for F2.

Then use F2 from equation () and substitute into () to solve for F1

$$F_{1}=F_{load}\frac{Sin(\alpha )}{Sin(\alpha+\beta)}\frac{Sin(\beta )}{Sin(\alpha )} \,$$

Symmetrical Anchor - Special Case

This diagram shows the specific case when the anchor is symmetrical. Notice 2α = θ.

Let us now analyze a specific case in which the two anchors are "symmetrical" along the y-axis.

Start by noticing α and β are the same. Let us start from equation () and substitute β for α and simplify.

$$F_{eachAnchor}=F_{load}\frac{Sin(\alpha )}{Sin(\alpha+\alpha)} \,$$
$$F_{eachAnchor}=F_{load}\frac{Sin(\alpha )}{Sin(2\alpha)} \,$$
And using another trigonometric identity we can simplify the denominator.

$$F_{eachAnchor}=F_{load}\frac{Sin(\alpha )}{2Sin(\alpha)Cos(\alpha)} \,$$
$$F_{eachAnchor}=\frac{F_{load}}{2Cos(\alpha)} \,$$
Remember that α is HALF the angle between the two anchor points. To use the entire angle θ, you must substitute α = θ/2 we can edit our equation thus.

$$F_{eachAnchor}=\frac{F_{load}}{2Cos(\frac{\theta}{2})} \,$$

References