What if we have a non-symmetrical cross-section? What happens? Can we still apply the bending formula directly?

The answer is no, because the bending formula only works with principal moments of inertia. Non-symmetrical cross sections have what we call the product of inertia I_yz (analogous to τ_xy), and we need to transform the moment of inertia such that I_y'z' = 0. From this, we then get our principal moments of inertia I₁, I₂ (analogous to σ₁, σ₂).

And yes, you’ve guessed it, we use the Mohr’s circle to do the transformation =) But first, let’s look at the product of inertia.

Product of inertia, I_yz

The definition of I_yz is:

You probably won’t need to calculate this; suffice for you to know that as long as there is an axis of symmetry, I_yz = 0.

We also have the parallel-axis theorem for I_yz:

Consider the following cross-section. We're going to attempt to calculate its product of inertia:

Since the individual shapes 1, 2, 3 all have a local axis of symmetry, their local products of inertia are zero (I_yz1 = I_yz2 = I_yz3 = 0).

Therefore the product of inertia for the shape is:

Principal moment of inertia

The construction process is exactly the same as the standard Mohr’s Circle in Solid Mechanics I, except that:

• initial coordinates are Y (I_y, I_yz) and Z (I_z, -I_yz).
• the I_yz-axis is +ve upwards

These principal moments of inertia I_y' and I_z' can then be used in the bending stress formula. However due to the rotation of the axes (y−z to y'−z' at θ = θ_p1), the coordinates of the point of interest with respect to the new axes y'−z' will also change.

Calculating new coordinates y' and z'

Let’s look at a sample cross-section to derive the formula for y' and z':

Look’s complex? Let’s try to break it down:

Complex? Don’t worry, just remember:

If θ_p1 is clockwise (CW), just plug it into the equation as -θ_p1.

Let’s look at an example now.