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Lever Rule For Computation Of Live Load Distribution Factors In CONSPAN [TN]

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Lever Rule For Computation Of Live Load Distribution Factors In CONSPAN [TN]

Document Information

Document Type: TechNote 

Product(s): LEAP CONSPAN

Version(s): All

Original Author: Bentley Technical Support Group

EXAMPLE 1: COMPUTATION OF LEVER RULE FOR EXTERIOR BEAM

 

The distribution factor is the Reaction, R, about the Hinge.

Case (A). If Only One Lane is Loaded
The first axle is placed 2 feet from the face of the curb. The reaction is computed by summing moments about one support to find the reaction at the other support by assuming that the supported component is hinged at interior supports.

If P is lane (axle) load, then
   R × (16) = 0.5 × P × (20) + 0.5 × P × (14)
   R = 0.5 × P × (34) / (16)
   R = 0.5 × P × (2.125)

Multiple Presence Factor for single lane loaded = 1.20
   R = 0.5 × P × (2.125) × 1.20
   R = 1.275 P

Case (B). If Two Lanes Are Loaded
If P is lane (axle) load, then
   R × (16) = 0.5 × P × (20) + 0.5 × P × (14) + 0.5 × P × (8) + 0.5 × P × (2)
   R = 0.5 × P × (44) / (16)
   R = 0.5 × P × (2.75)
Multiple Presence Factor for two lanes loaded = 1.00
   R = 0.5 × P × (2.75) × 1.00
   R = 1.375 P

The distribution factor using Lever Rule is taken as the larger of these two cases, which
in this case, is 1.375.


EXAMPLE 2: COMPUTATION OF LEVER RULE FOR INTERIOR BEAM

The distribution factor is the Reaction, R, is independently computed about the Hinge on
both the right and the left sides.

Case (A). If Only One Lane is Loaded
The first wheel line is placed directly over the interior girder of interest to generate the largest reaction. Since the girder spacing is only 16 feet, it cannot accommodate an additional two axles from a second truck on the right hand side. The reaction is computed by summing moments about the right hinge.

If P is lane (axle) load, then
   R × (16) = 0.5 × P × (10) + 0.5 × P × (16)
   R = 0.5 × P × (26) / (16)
   R = 0.5 × P × (1.625)
Multiple Presence Factor for single lane loaded = 1.20
   R = 0.5 × P × (1.625) × 1.20
   R = 0.975 P

Case (B). If Two Lanes are Loaded
In addition to the axles placed to the right of the interior beam, two other wheel lines can be placed to the left of the interior beam being studied. The reaction is computed by summing moments about the hinges independently on the right side and then the left side
as shown below.
If P is lane (axle) load, then
   R × (16) = 0.5 × P × (10) + 0.5 × P × (16) + 0.5 × P × (6) + 0.5 × P × (12)
   R = 0.5 × P × (44) / (16)
   R = 0.5 × P × (2.75)
Multiple Presence Factor for two lanes loaded = 1.00
   R = 0.5 × P × (2.75) × 1.00
   R = 1.375 P

Case (C). If Three Lanes are Loaded
In addition to the axles placed for the two-lane loading, two more axles are placed, if they
fit, to the right. The reaction is computed by summing moments about the hinges independently on the right side and then the left side as shown below.

If P is lane (axle) load, then
   R × (16) = 0.5 × P × (10) + 0.5 × P × (16) + 0.5 × P × (6) + 0.5 × P × (12) +0.5x P × (4)
   R = 0.5 × P × (48) / (16)
   R = 0.5 × P × (3.0)
Multiple Presence Factor for two lanes loaded = 0.85
   R = 0.5 × P × (3.0) × 0.85
   R = 1.275 P
The distribution factor using Lever Rule is taken as the largest of these three cases, which in this example is 1.375.

See Also

Product TechNotes and FAQs

 

External Links

Bentley Technical Support KnowledgeBase

Bentley LEARN Server

 

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Comments
  • Example 2, Case C for 3 lanes loaded:  I believe the moment arm for the third axle should be 4' vice 6' .  

    This makes R = .5*P*46/16 = 1.44 and multiplied by mpf of .85 R = 1.22*P (Case B still controls.)