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Example 4: determination of reduction factor and nominal moment for rectangular beam(SI units)

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*804*420)/(0.85*28*460)
a=123.38mm

C=(a/β1)
β1=0.85 for concrete less than 30 Mpa
β1=0.85
C=a/β11
C=123.38/.85
C=145.15 mm

0.003/C=(εt+0.003)/d
0.003/145.15=(εt+0.003)/310
εt=0.00341
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=249.87Kn-m

Figure 2

Figure 3








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