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Example 2: 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
0.85*fc'*a*b=As*fy
a=(As*fy)/(0.85*fc'*b)
a=(7*1.56*80,000)/(0.85*6,000*20)
a=8.56in

C=(a/β1)

β1=0.85-(fc'-4000)*0.05/(1000)≥0.65
β1=0.85-(6,000-4,000)*0.05/1000
β1=0.75
C=(8.56/0.75)
C=11.41in

0.003/C=(0.003+εt)/d
0.003/11.41=(0.003+εt)/27
εt=0.004

from figure 3 ϕ can be calculated as follows

ϕ=0.65+((εt-εy)*0.25)/(0.005-εy)
εy=fy/Es
εy=80,000/29,000,000
εy=0.0027
ϕ=0.79

ϕ*Mn=ϕ*As*fy(d-a/2)
ϕ*Mn=0.79*10.92*80,000*(27-8.56/2)
ϕ*Mn=1,308.83k-ft

Figure 2

Figure 3





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