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Example 3: 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









a=(As*fy)/(0.85*fc'*b)
a=(4*1.27*60,000)/(0.85*4,000*18)
a=4.98in

c=(a/β1)

β1=0.85-(fc'-4000)*0.05/(1000)≥0.65
β1=0.85

C=(a/β1)
C=5.86in

0.003/C=(εt+0.003)/d
0.003/5.86=(εt+0.003)/12
εt=0.00314

ϕ from figure 3

ϕ=0.65+(εt-0.002)*250/3
ϕ=0.745

ϕ*Mn=ϕ*As*fy*(d-a/2)
ϕ*Mn=179.95K-ft

Figure 2


Figure 3




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