In flight, the fuselage hangs off of the wing. For dimensioning purposes, consider:
- the intersection points to exert no moments and to behave as hinges;
- weight W of the fuselage to be concentrated in the Centre of Gravity;
- lift of each wing to be concentrated in its Centre of Lift.
If we dimension the construction in this way, we over-dimension which is never a bad idea with primary construction bolts. In reality the following factors alleviate the loads:
- The bolts are modelled as hinges which cannot exert a moment, but actually they do exert torque.
- The wing lift is distributed loading, with most of it near the wing root.
Bolts 1, 2 and 3 experience lift and gravity forces from the construction, and exert equal and opposite forces in order for everything to stay in one piece. Fuselage weight is transferred to bar 2-3 which distributes the load evenly over bolts 2 and 3. Remove the bolts and the fuselage falls from the assembly.
Force equilibrium in point 1 from this answer:
- $F_{13} \cdot sinψ = ½L => F_{13} = \frac{L}{2sinψ}$
- $F_{12} = F_{13} \cdot cosψ = \frac{L}{2tanψ}$
In point 2, bolt reaction force in green:
* $F_{V} = ¼W$ $\tag{Vertical}$
* $F_{H} = \frac{L}{2tanψ}$ $\tag{Horizontal}$
In point 3, bolt reaction force in green:
* $F_{V} = ¼W - F_{13} \cdot sinψ = ¼W - ½L$ $\tag{Vertical}$
* $F_{H} = \frac{L}{2tanψ}$ $\tag{Horizontal}$