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Example 11: flexural stress for doubly reinforced beam using transform area method(SI units)

determine the flexural stresses for the beam shown in figure 1 using the transformed-area method.


Figure 1









Transform area for bottom steel reinforcment=n*As=8*2600=20,800mm2

Transform area for top steel reinforcement=(2*n-1)*As=(2*8-1)*1,018=15,270mm2 
(2*n-1) we subtract 1 to account for holes of reinforcement bar in concrete 

To calculate the neutral axis location we need to equal moments around the neutral axis as shown in figure 2
400*(X)*(X/2)+15,270*(X-70)=20,800*(630-X)
200*X^2+15,270X-1,068,900=13,104,000-20,800X
200X^2+36,070X-14172900=0
X=190.90mm



now we can calculate the moment of inertia using the parallel axis theorem

I=Ic+A*(d^2)



I=(400*(190.9^3))/12+400*190.9*(95.45^2)+15,270*(120.9^2)+20,800*(439.1^2)
I=5161212927mm2
bending stress at extreme compression fiber, y=190.90mm
fc'=(M*y/I)
fc'=(275000000*190.9/5161212927)
fc'=10.20Mpa

bending stress at the center of the bottom reinforcing layer of steel, y=439.1mm

fs=(n*M*y/I)
fs=(8*275000000*439.1/5161212927)
fs=187.70Mpa

bending stress at the center of the top reinforcing layer of steel, y=120.90mm

fs'=(2*n*M*y/I)
fs'=(2*8*275000000*120.90/5161212927)
fs'=103.1Mpa

2*n used for the top reinforcing bar because it will resist shrinkage, creep and deflection of reinforcing beam, therefore with time stress of reinforcing steel bar will increase







Figure 2


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