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Clevegar (Structural)
17 Jan 04 1:57
Please shed some insight for me on the application of 3.24.3

In 3.24.3.1, "Case A-Main Reinforcement Perpendicular to Traffic". The equation given returns a moment in "foot-pounds per foot-width of slab". Is it to be multipied BY THE SPAN LENGTH to find the moment in the center of the span? ie:

S=10 ft span
P=16,000 lb
M = ((10ft+2)/32)*16,000lb*10ft = 60,000ft-lb per foot-width of slab

Is the E which is not explained 1 ft (unity)?

Or am I totally off base on both counts?

In 3.24.3.2, a formula for E is given. Assuming that the load is applied at the midpoint of this parallel span, is moment per unit width of slab calculated as follows?:

S= span parallel to to traffic
P= wheel load
E= width perpendicular that P is distributed over
M= (P/E)*S/8

????

Many thanks. Maybe this should be obvious, but I need a reality check before driving my truck on my slab.
mike80 (Structural)
19 Jan 04 13:51
Clevegar,

The formula in 3.24.3.1 is for designing a one foot wide, transverse strip of the concrete deck, i.e., the strip runs perpendicular to the centerline of the bridge.  The resulting moment is the maximum moment in that one foot strip.  The maximum moment is not related to the span length, only the distance between longitudinal beams supporting the deck.  "E" is not used in this formula.

The formula in 3.24.3.2 is for designing a one foot wide longitudinal strip of the concrete deck, i.e., the strip runs parallel to the centerline of the bridge. In this case, the moment is related to the span length.

Mike
 
Clevegar (Structural)
19 Jan 04 15:54
Mike, thanks for responding.

Have you used these in practice? Can you verify that I'm applying these correctly if I use them as follows:

for 3.24.3.1:

S=10 ft span
P=16,000 lb
M = ((10ft+2)/32)*16,000lb = 6,000ft-lb per foot-width of slab

For 3.24.3.2:

S= 10ft
P= 16000lb
E= 4+0.06*10 = 4.6ft
M= (P/E)*S/8 = ((16000/4.6)*10)/8 =4348ft-lb per wheel.

 
Thanks again.


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