*Disclaimer: This subtopic in bending is slightly advanced and you should learn this only if your lecturer covers this in the course
In all previous chapters we’ve dealt only with cross-sections that are symmetrical, such as:
*Disclaimer: This subtopic in bending is slightly advanced and you should learn this only if your lecturer covers this in the course
In all previous chapters we’ve dealt only with cross-sections that are symmetrical, such as:
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 I_{1}, I_{2} (analogous to σ_{1}, σ_{2}).
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.
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:
The construction process is exactly the same as the standard Mohr’s Circle in Solid Mechanics I, except that:
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.
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.
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 I_{1}, I_{2} (analogous to σ_{1}, σ_{2}).
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.
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:
The construction process is exactly the same as the standard Mohr’s Circle in Solid Mechanics I, except that:
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.
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.