Leonardo da Vinci’s bridge

If you apply a vertical load W to the top centre bar (C), what is downward force exerted on the horizontal bars?

Keith’s analysis

For load W on pin 3 reactions at each foot of bridge must be W/2;

thinking only of static equilibrium for each beam in vertical direction;

force on pin 2 from A is W/2; force on A from B through pin 1 is W;

force on pin 3 from B must also be W and by symmetry from D also W;

force on pin3 is 2W from Band D and with load on pin 3 of W, load on C from 3 is 3W;

that makes force on pin 2 from C to be 3W/2;

force of pin 2 on B is 3W/2+W/2=2W which is consistent with reactions at the ends of B of W.

Chris’s analysis

The beam C transmits the load to the pins 2 and 4, each taking W/2 vertically.

This vertical load is supported by pins 1 and 3 and pins 4 and 5, each taking W/4 vertically.

Therefore pin3 has an applied load of W/2 vertically as well as the original W, which means that the load at pins 2 and 4 is 3W/4 not W/2.

This tends towards an actual load of 2W on beam C for an applied load of W on pin 3.

David H’s analysis

Let E₅ be the force exerted by pin 5 on beam E.
So –E₅ is the force exerted on pin 5 by beam E, and hence on beam D by pin 5.

Following Keith’s argument, the force exerted by the ground on beam E is ½W vertically upwards.
Considering the moments on beam E about pin 4 we have:
      2 × ½ W cos β = E₅ sin φ
             where β is the angle between beam E and the horizontal and φ is the angle between beam E and the force E₅. Considering the moments on beam D about pin 4 we have:
      E₅ sin φ = D₃
             φ is the angle between beam D and the force D₅. This is so because of the isosceles triangle formed by the two tangents to pin 5.
Furthermore, any notches locating pin3 would make the system over-rigid, so we assume that beam D is smoothly in contect with pin 3. The downward force on beam C is therefore:
      C₃ = W + 2 D₃ cos α
             where α is the angle between beam D and the horizontal.

      C₃ = W ( 1 + 2 cos β cos α )
Using the unstated assumptions that the pins are all the same diameter and that the spacing beteen pins is equal to that between pin 5 and the end of beam E, we can see that: β = 2 α, and hence:
      C₃ = W ( 1 + 2 cos 2α cos α )

David H’s other analysis

Even quicker we can consider moments on the assembly of beams D and E and pins 4 and 5 about the point at which pin 4 acts upon beam C.

As before the anti-clockwise moment is (assume each beam is of length 2):
      2 × ½ W cos β
to be balanced by a clockwise moment:
      D₃

      \ D₃ = 2 × ½ W cos β
and the rest of the analysis carries on as before.

Why did I not think of this first time?