Universal motor with non-sinusoidal wave

As everyone knows, the line voltage coming out of your wall is in the form of a sine wave. Universal motors are designed to be capable of running on both AC and DC since they are self-commutating. For example, would you burn up a circular saw's motor by running it on a 60 Hz AC square or sawtooth wave instead of a sine wave due to the harmonics?

For example, would you burn up a circular saw's motor by running it on a 60 Hz AC square or sawtooth wave instead of a sine wave due to the harmonics?

That's just example questions? Then what's your actual question? I'll just assume they are your actual questions, if they're not then it your loss.

The Root Mean Square (RMS) value of a DC signal is 1 which is the same as the RMS value for a square wave. (if DC = 1V and square wave oscillates between ½V and -½V).

This means that it will get exactly the same power input if it was an ideal resistor, now since the motor is inductive there will be some extra impedance (imaginary), so the power the motor will receive will be slightly less than if you'd just use a DC signal. So yes, it will work with a square wave.

The RMS for a sawtooth is $\frac{1}{\sqrt3}$, that is less than 1, if DC works then sawtooth will also work.

There's no way that you can break the saw's motor with some crazy waveform. The only way you can burn it is if you have too high power input, one way of doing that is reducing the resistance, which you can do by cooling it down... but then it's cold so that won't work. The other way that you successfully can burn the motor would be to increase the voltage input above what it's rated for. If you're not going to do that then everything will be fine.

However, the RMS for a sinusoid is $\frac{1}{\sqrt2}$, so if your motor is made for only sinusoids with an amplitude of $100V$, then the RMS of that sine is $\frac{100V}{\sqrt2}\approx70.7V$ This means that you can replace the sinusoid with $70.7V$ DC (or $35.35V$ amplitude square wave). Not $100V$. (As I said earlier, you can burn it with too high voltage). The amplitude of the sawtooth would be $70.7V×\sqrt3=122.4V$