# Non-Load Sharing Anchors

As compared to load sharing anchors, where θ_{1}= θ_{2} and thus F_{A1}= F_{A2}, without those assumptions the calculations get a little more complex.

As before, we start with the equation:

$${F_{Load}} = {F_{A1}}\cos {\theta{ _1}} + {F_{A2}}\cos {\theta _2}$$The trick to solving this equation is in realizing that the horizontal forces on the anchors must be equal and opposite, giving us:

$${F_{A{1_H}}} = {F_{A{2_H}}} = {F_{A1}}\sin {\theta _1} = {F_{A2}}\sin {\theta _2}$$

From here we have two options:

- If we know the angles, or if the anchors are not at the same level, we solve with trigonometry
- If the anchors are at the same level, we can use the distance between the anchors, the length of the anchor legs and the total height of the system.

### Option 1, using trigonometry

Substitution of F_{A2} with equivalencies gives us:

$${F_{Load}} = {F_{A1}}\cos {\theta _1} + {{{F_{A1}}\sin {\theta _1}} \over {\sin {\theta _2}}}\cos {\theta _2}$$

We then remove the F_{A1} to reach:

$${F_{Load}} = {F_{A1}}\left[ {{{\cos {\theta _1}\sin {\theta _2} + \sin \theta \cos {\theta _2}} \over {\sin {\theta _2}}}} \right]$$

From the trigonometric equality of the sum of angles:

$$\sin ({\theta _1} + {\theta _2}) = \sin {\theta _1}\cos {\theta _2} + \cos {\theta _1}\sin {\theta _2}$$

we get:

$${F_{Load}} = {F_{A1}}\left[ {{{\sin ({\theta _1} + {\theta _2})} \over {\sin {\theta _2}}}} \right]$$

and, finally:

$${F_{A1}} = {F_{Load}}{{\sin {\theta _2}} \over {\sin \theta }}$$

### Option 2, using distance, length and height

If all distances are known as shown in the diagram, we can substitute trigonometric relations for sinθ and cosθ:

$${\sin _{{\theta _{1or2}}}} = {{{d_{1or2}}} \over {{L_{1or2}}}} \quad \textrm and \quad {\cos _{{\theta _{1or2}}}} = {{{h_{1or2}}} \over {{L_{1or2}}}}$$

Substituting these identities into the first equation gives us:

$${F_{Load}} = {F_{A1}}{h \over {{L_1}}} + \left( {{F_{A1}}{{{d_1}{L_2}} \over {{d_2}{L_1}}}} \right){h \over {{L_2}}}$$

By removing F_{A1} and combining the fractions, we get:

$${F_{Load}} = {F_{A1}}\left[ {{{h{d_2} + h{d_2}} \over {{d_2}{L_1}}}} \right]$$

Since d_{1}+d_{2}=d, solving for F_{A1} gives:

$${F_{A1}} = {F_{Load}}{{{d_2}{L_1}} \over {hd}}$$