How to compute the effective voltage across a DC motor driven by a pulse train?
The effective voltage for producing torque is the battery voltage times the duty cycle.  As you note, this is not the same thing as RMS voltage.

The reason for the discrepancy is that the formula that uses RMS is for finding the power delivered to a resistor.  Power doesn't cause motor torque, current does.  Also, a motor isn't a resistor.  The windings have a significant inductive component in addition to DC resistance.  If your switching frequency is fast enough, then the current changes little over a pulse cycle due to the inductance.  This is in contrast to a resistor where current is always proportional to the applied voltage.

If you switch slowly compared to the resistive/inductive time constant of the windings, then the resistance dominates.  In that case, the RMS formula (after subtracting the back EMF due to motor motion from the applied voltage) tells you how much power you are delivering to the motor, but the torque is still proportional to the average current, which is proportional to the voltage times the duty cycle.

The above statements may sound contradictory, but they're not, and are trying to tell you something important about driving motors with PWM.  Let's keep it simple and say the motor is stalled (so we can ignore back EMF) and we're trying to have it produce a certain torque.  Let's say you get that torque with 50% PWM.  That will be the case whether switching fast or slowly.  However, switching slowly will deliver more power to the motor according to the previous paragraph.  How to reconcile that?

Let's start with the fast switching case.  Each pulse is a small part of the time constant of the windings.  That means the current goes up a little during the on time and down a little during the off time, but this ripple is small compared to the average.  The motor is bascially being driven with a fixed current, which is also a fixed voltage related by the resistance of the windings.  All is well and makes senses.

Now for the slowly switching case.  The pulses are long compared to the time constant of the windings.  Basically the current is full on for half the time and full off for half the time.  The average is still 1/2, so the torque is the same.  However, the power is not.  Since the windings look resistive, the power is 4x the low-ripple average case when full on, and 0 when full off, for a average of 2x the power of the fast switching case.  Where does this extra power go?  To heating the windings.  This is exactly why you want to PWM motors fast enough so that the current stays relatively constant over the whole PWM cycle.

Still not convinced?  Another way to look at this is to decompose the current into its AC and DC parts.  In the first case, the current is steady, so the DC part is 1/2 (relative to what you get with the full applied voltage) and the AC part is 0.  In the second case, the DC part is still 1/2 but the AC part is definitely non-zero.  Now consider that only the DC component goes into providing torque and moving the motor.  The AC part at the PWM frequency just sloshes back and forth thru the windings causing heat due to their resistance.