Do you have engineering data on tempered glass?
Wagner does not supply glass. As such, we do not have engineering data on the glass. However, the following information is based on data provided by The California Glass Association publication "The Source".
Building code requirements vary across the country. Generally, guard load requirements will either be expressed as a "uniform load" or a "concentrated load". These loads are not to be applied concurrently. Most codes will require one or the other or the more conservative of the two.
The most common load requirement is 200 lbs. concentrated or 50 lb/ft uniform.
The other issue to keep in mind is that the glass industry recommends a safety factor of four for tempered glass. The ICC presently also requires a safety factor of four.
The basic formula for determining bending moment and stresses is:
f = M/S Note: Rupture modulus of annealed glass in tension is 6,000 psi therefore the modulus of rupture of tempered glass is 4 x 6,000 or 24,000 psi. 

Requirements for Balustrade Assembly using the example to the left
200 lb Concentrated Load (recalculate as necessary for 300 lb load)
f = M/S
f = Modulus of Rupture (mean)/Safety Factor
f = 24,000/4 = 6000 psi (maximum allowable stress)
M = h x P = 38 in x 200 lb = 7,600 inlb
Section Modulus for 1/2" glass (0.469" is the minimum ASTM standard for nominal 1/2" glass; assuming a 48" wide panel)::
S = (w x t^{2})/6 =(48" x .469^{2})/6 = 1.759 in^{3}
f = M/S
f = 7,600 inlb/1.759 in^{3} = 4,320 psi
4,320 psi < maximum allowable stress of 6,000 psi therefore 1/2" tempered glass is acceptable
Requirements for Top Rail Load using the example to the left
50 lb/ft (as applied to a 12" section)
f = M/S
S = (w x t^{2})/6 = (12 x .469^{2})/6 = .44 in^{3}
M = h x 50 lbs = 38" x 50 lbs = 1,900 inlb
f = M/S = 1,900 inlb/.44 in^{3 }= 4318 psi
4,318 < 6,000 psi therefore 1/2" tempered glass is acceptable