created at February 21, 2022

Crossbow Limb strength calculation

We gonna use this example as a basis for the current calculation. There we are evaluating the Limb stiffness and calculating Limb's tip displacement during deformation under applied force (string tension force). So, we have the Limb with a rectangle cross-section of varying size.

Here we gonna calculate Limb strength by given sizes and material specs. Sometimes we happen to know the crossbow parameters before we design the Limb, so we know how much the Limb will deform when the weapon is loaded. Therefore the distance Δ of the Limb's tip between the undeformed state (with NO string installed) and fully loaded state is given.

The Limb is bending, it is considered as a console beam with the force applied to the free end. The bending torque occurs in the Limb, which has the following diagram:

It's known from the Strength of Materials science that under bending the normal stress occurs in the cross-section plane, which diagram is linear. Stress has a direction along the beam center line. There exists a neutral line (center line) where the stress is zero. The stress at specific cross-section point linearly increases while moving away from the neutral line, it's positive on the side where the material fibers are stretching, and negative on the other side, where the fibers are compressing (the image below). The maximal stress will occur on the beam surface, which is the farthest point from the neutral line. The material will start do break at that point first. The bending stress diagram is linear for any cross-section shape. The neutral line of the beam goes through the mass centers of the beam cross-sections.

In our case we have a rectangle cross-section shape, having the mass center at the intersection point of the rectangle diagonals. The maximal positive stress (stretching) will occur on the outer side of the Limb (relative to the crossbow), and the maximal negative stress (compressing) will occur on the opposite side of the limb.

Stress diagramm at the Limb cross-section under bending

It's also known, that the maximal normal stress at specific cross-section is calculated as:

σmax=MymaxIx(1)\sigma_{max} = \frac{M\cdot y_{max}}{I_x}\quad (1)

where M - bending torque in that section, ymax - the distance from the neutral line to the section farthest point (ymax = h/2 in our case), Ix - the section's Moment of Inertia.

In our case the moment is changing along the Limb, the Moment of Inertia Ix does the same. The displacement of the Limb's tip under the applied force F can be calculated with Mohr's integral:

Δ=0LM(z)M1(z)EIxdz(2)\Delta = \int_{0}^{L}\frac{M(z)\cdot M1(z)}{E\cdot I_x}dz\quad (2)

However, now we have an opposite task - to calculate the force F by given displacement Δ. Knowing that:

M(z)=FzFL(3)M(z) = F\cdot z - F\cdot L\quad (3)
M1(z)=zLM1(z) = z - L
Ix(z)=(I2xI1x)Lz+I1x(4)I_x(z) = \frac{(I2_x - I1_x)}{L}\cdot z + I1_x\quad (4)
I1x=b1h1312;I2x=b2h2312I1_x = \frac{b1\cdot h1^3}{12};\quad I2_x = \frac{b2\cdot h2^3}{12}

We will calculate the integral (2) and then express F through Δ. Let's introduce the following designation for simplicity:

k1=Ix2Ix1L;k2=Ix1k1 = \frac{Ix_2 - Ix_1}{L};\quad k2 = Ix_1


F=Ek1Δ1.5L2k2k1L+(k2k1(2L+k2k1)L2)ln(k1k2L+1)F = \frac{E\cdot k1\cdot \Delta}{-1.5\cdot L^2 - \frac{k2}{k1}L + \left (\frac{k2}{k1} \left (2\cdot L + \frac{k2}{k1} \right) - L^2 \right )\cdot ln \left (\frac{k1}{k2}L + 1 \right )}

Now when we know the force F we also know the bending torque dependency (3) on z coordinate (along the Limb). Dependency for the Moment of Inertia is given by (4). Dependency for the Limb height is given by equation:

h(z)=(h2h1)Lz+h1(5)h(z) = \frac{(h2 - h1)}{L}\cdot z + h1\quad (5)

Now we've got everything we need to express the normal stress σz as a function of z coordinate according to equation (1), plot the graph along the Limb and get the maximal stress σzmax.

Then we need to choose some limit value for the Limb material. Let's also introduce some safety factor k as we don't want our Limb to work on the edge. Let's use k = 1.5, as it's a common value used in the aerospace stress calculations. The maximal allowed stress depends on how material behaves under load. If it has no plastic deformation before the total failure (brittle material) - we gonna use the ultimate stress. Otherwise, for the ductile material we gonna use the yield stress as we don't want the deformation remains on the Limb after the shot. For the majority of materials used for limbs I would assume to use yield strength, with except for the composite materials as they usually behave as brittle materials.

After we've chosen the maximal allowed stress for the given material, say yield stress σy, the limit stress is calculated as:

σlim=σyk\sigma_{lim} = \frac{\sigma_y}{k}

If the maximal stress occurring in the Limb is lower than σlim - we can accept the case, otherwise the limb parameters should be revisited to make it more safe. In order to decrease the maximal stress one can make the cross-section thinner or the Limb longer.