Ie = (Mcr/Mmax)3Ig + [1 - (Mcr/Mmax) 3]Icr

Where:

f r =2.5%f'm + 2.5%1350 lb/in2 = 91.8 lb/in2

And;

Ig = (7. 62)(23.62)3/12 = 8368 in4

Yt = 23.62/2 = 11.81 in

Icr = The moment of inertia of the cracked section, in4

Icr = 2,393 in4 (From table C-9)

Mmax = The maximum applied moment, ft-lb. (Use the average of the maximum negative moment,

Msi, and the maximum positive moment, Msi, in computing Ie.

Ie = 2899in4 < Ig == 8368 in4; Therefore use Ie in the deflection equations.

dead load and one span is loaded with the uniform live load and is determined as follows:

)mt = )ms + (3)()ml)

Since the long term dead load deflection, )ml, is increased by a factor of 3, the dead and live load deflections

will be determined separately.

mid-span deflection for a uniformly loaded span with unequal end moments. The shear and moment diagrams

for the load case that produces maximum dead load deflection are shown in figure 8-8.

Where:

Msm =The positive moment at mid-span, ft-lb

And;

Mmax = The maximum positive moment in the span, ft-lb

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