Force on Door
Just to put some figures on the forces involved. If the door is 4ft by 3ft, then its area is 4x3x144, = 1728 square inches.
If the differential pressure is 5 psi, then the total force on the door is 8640 pounds, or about 4 tons.
As the door would weigh much less than 1/4 ton, it would experience an acceleration of over 16G., if all the fasteners failed at once.