Equalization (proof)

The load is pointing straight down, and is being held up by two forces pointing up at various angles $\alpha$ and $\beta$.

Equalization is 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.

$F_{x1}=F_{x2} \,$

$F_{1}Sin(\alpha )=F_2Sin(\beta ) \,$

$F_{1}=F_2\frac{Sin(\beta )}{Sin(\alpha )} \,$

(1)

The net force in the y-direction must also sum to zero.

$F_{y1}+F_{y2}=F_{load} \,$

$F_{1}Cos(\alpha )+F_2Cos(\beta )=F_{load} \,$

(2)

Substitute $F_{1}$ from equation (1) into equation (2) and factor out $F_{2}$.

$F_2\frac{Sin(\beta )}{Sin(\alpha )}Cos(\alpha )+F_2Cos(\beta )=F_{load} \,$

$F_2\left [\frac{Sin(\beta )}{Sin(\alpha )}Cos(\alpha )+Cos(\beta )\right ]=F_{load} \,$

Solve for $F_{2}$ 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 $F_{2}$.

$F_2=F_{load}\frac{Sin(\alpha )}{Sin(\alpha+\beta)} \,$

(3)

Then use $F_2$ from equation (3) and substitute into (1) to solve for $F_1$

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

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

(4)

Symmetrical Anchor - Special Case

This diagram shows the specific case when the anchor is symmetrical. Notice $2\alpha = \theta$.

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

Start by noticing $\alpha$ and $\beta$ are the same. Let us start from equation (4) and substitute $\beta$ for $\alpha$ 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.

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

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

Note that $\alpha$ is half of the angle $\theta$ between the two anchor points. To express the force at each anchor using the entire angle $\theta$, we substitute $\alpha=\theta/2$.

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

References

• "Non-Load Sharing Anchors." Ratref. Web. 11 February 2012. <http://ratref.com/content/non-load-sharing-anchors>.