[LOAD]     Braking Load Application

BRAKING LOAD APPLICATION

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.

Go to Load > Static Load > Static Load Cases:
Name: Braking Load
Type: Braking Load
Click Add

Then Go to Load > Static Loads > Element Beam Loads

Load Case Name: Braking Load

Load Type: Uniform Loads

Direction: Global X

Value: Relative, x1:0, x2:1, w:1

Select All the elements

Click Apply