Non-Load Sharing Anchors

Non-Load Sharing Anchors

As compared to load sharing anchors, where θ1= θ2 and thus FA1= FA2, 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:

  1. If we know the angles, or if the anchors are not at the same level, we solve with trigonometry
  2. 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 FA2 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 FA1 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 FA1 and combining the fractions, we get:

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

Since d1+d2=d, solving for FA1 gives:

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