[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

  • Creation date: 9/4/2016 10:23 AM ()      Updated: 9/5/2017 9:43 AM ()
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