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Example 1: determination of reduction factor and nominal moment for rectangular beam

Determine εt, ϕ  and ϕ *Mn for the beam shown in figure 1.
Figure 1







C=T
As*fy=0.85*fc'*a*b

a=(As*fy)/(0.85*fc'*b)
a=(4*1*60,000)/(0.85*4,000*12)
a=5.88 in

C=(a/β1)
β1=0.85-(fc'-4000)*0.05/(1000)≥0.65
β1=0.85
C=a/β11
C=5.88/.85
C=6.92in

0.003/C=(εt+0.003)/d
0.003/6.92=(εt+0.003)/24
εt=0.0074>0.005
from figure 3,  ϕ =0.90

 ϕ *Mn= ϕ *As*fy*(d-a/2)
 ϕ *Mn=0.9*4*60,000*(24-5.88/2)
 ϕ *Mn=4,548,960 lb-in
 ϕ *Mn=379.1K-ft
Figure 2

Figure 3









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