How does a motor perform when it is given a load greater than its rated ability?
If the power supply is able to deliver the extra current that the motor demands, then the motor will draw more current, and deliver more torque. The increased voltage drop in the motor windings will reduce the speed of the motor. 

The increased current in the windings is now dissipating more power than is rated for the motor. The extra heating in the motor will be approximately (1/0.87)^2 = 1.32x. This is OK for a while, while the copper absorbs the extra heat by getting hotter. It may take minutes for the motor to reach its rated temperature under this small extra heating.

If left like this while the motor reaches thermal equilibrium, you would expect the eventual temperature rise above ambient of the motor to be 1.32x higher than rated. This might make the motor smoke, and if it didn't destroy it by damaging the insulation, would certainly reduce the life of the motor.