According to AASHTO 3.6.4
Braking Force (BR) can be calculated as greater of the following:
• 25% of the axle weight if the Design Truck or Design Tandem
• 5% of (Design Truck/Design Tandem + Lane Load)
• So,
 25% of Design Truck 0.25*72 18 kips 25% of Design Tandem 0.25*50 12.5 kips 5% of (Design Truck + Lane Load) 0.05*(72+412*0.64) 16.8 kips 5% of (Design Tandem + Lane Load) 0.05*(50+412*0.64) 15.7 kips

So, 25% of Design Truck governs: 18kips
So the number of lanes in a model be: n
Multiple presence factor for the lanes be: m
Then the Braking Force will be maximum of:
 Number of Lanes Multiple Presence Factor Calculation Braking Force 1 Lane 1.2 18 x 1.2 x 1 21.6 kips 2 Lanes 1 18 x 1 x 2 36.0 kips 3 Lanes 0.85 18 x 0.85 x 3 45.9 kips 4 Lanes 0.65 18 x 0.65 x 4 46.8 kips

So, the maximum force should be applied horizontally at the deck level in the longitudinal direction which in turn will generate forces on the columns and other substructures.

EXAMPLE:
The bridge shown here is a 3 span PSC Box Girder bridge having 2 lanes.
Total Span: 410 ft, so can easily fit complete HL-93 Truck and Tandem Load.

The current bridge has 2 lanes:
So, for 2 Lanes Max Braking Load: 36 kips
We have a single line element representing the bridge:

Total Span 410 ft, so a uniformly distributed load of 36/410 = 0.9 or 1 kips/ft has to applied.