Crossbow limb's stiffness
Let's estimate the stiffness of the crossbow limb. Will consider the limb as a bending beam, which is clamped on the one side and free on the other (console beam). The force F is applied orthogonal to the free end.
The stiffness of the spring in general is a ratio of applied force to deformation. Say, the spring was loaded with 100 Newton force and it got deformation 1 cm (0.01 meters), then the stiffness is:
In the same way we gonna estimate the stiffness of the limb as a ratio of the limbs free end point displacement to the force applied to the end. The applied force generates a torque within the limb, which diagram looks like this:
The maximal torque will act at the clamp (the crossbow riser), thus that place is the most vulnerable to cracking. The curvature of the limb directly depends on the torque, so the curvature will be greater as closer to the riser. In order to flatten the curvature along the limb, one can use a variable cross-section limb with a thicker cross-section on the riser side and the thinner section on the string side. Let's consider we gonna have two given sections S1 and S2 at the ends, and the intermediate sections will change smoothly between S1 and S2.
In order to find the stiffness, we should calculate the displacement of the free end (where the string force is applied). According to the strength of materials, we should apply the unit force to the point of interest, so that the unit force direction is coincident with the desired displacement direction. So, the unit force is applied in the same way as the force F, but it has a magnitude of 1. Then we construct the diagram of the torque, caused by the unit force. Finally, the displacement can be calculated with Mohr's integral:
where E - elastic modulus of the limb's material, Ix - the moment of Inertia of the limb cross-section in the bending plane, M(z) - the torque caused by the force F, M1(z) - the torque caused by the unit force. The last two analytical functions for M and M1 could be obtained from the torque diagrams. The unit force torque diagram looks similar to the force F torque diagram, but the maximal value will be equal to L.
If the cross-section is constant along the limb, one can take out 1/E·Ix from the integral sign, then after integration of (1) we obtain:
Let's consider the cross-section to be rectangle. The rectangle's Moment of Inertia is:
where b - the width, h - the height in the bending plane.
In our case the cross-section is changing linearly from S1 to S2, so the Moment of Inertia will be the function from z:
Now we can put all obtained dependencies into (1) and calculate the integral, which is preferable to do in Dysolve app. The link to the calculation document is below. The stiffness can be obtained as F/Δ.
Let's calculate some additional parameters, which could be useful - Limb's mass and center of gravity position. The mass can be calculated as:
where ρ - material density, A1 and A2 - riser and free end limb's cross-section areas. In general, the resulting center of gravity coordinate of the system containing N bodies can be calculated as:
where M - the total mass of all bodies, xi and mi - the center of gravity coordinates for the bodies. If we consider splitting the limb into small pieces, the sum from the equation above will turn into the integral:
where A - the cross section as a function of coordinate z. In the same way as it was done for the Moment of Inertia, one can express the Area as:
So, the relative position of the center of gravity is:
One should keep in mind that the mass and center of gravity are being calculated for the working part of the limb, i.e. the part which is bending. Also, L is the length of the working part. The total Limb's length will be greater as there is a part fixed to the riser, that part is fixed (clamped) and is not bending.