What happens when dc shunt generator winding turn off during work?

If the field is opened ( a dangerous operation in itself,, because of high voltage due to Lf dif/dt) .. the armature Votage ( back emf will collapse) and armature current will rise as given by Vapp = Ka dia/dt +Ra*ia +vback(due to remanent magnetism ) with ia=ia0 at no load. The motor decelerates due to friction and windage .

The torque due to remanent magnetism will not be significant.

I think that the Output voltage will lose and the generator will overheated