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Example 8:flexural stress for double reinforced beam using transform area method

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








Transform area for steel in tension area=n.As
As=4*1.0=4in2
Transform area for steel in tension area=4*9=36in2          

Transform area for steel in compression area=(2*n-1)*.As
As=4*0.79=3.16in2
T.A for steel in compression area=(2*9-1)*3.16=53.72in2  

reinforcing steel in compression area will enhance the performance of beam and reduce the size of the beam significantly. also, it will help in positioning stirrups or shear reinforcement. compression steel will contribute to resisting concrete creep and long term deflection. therefore it will be stressed highly with time. transform area will be doubled 2*n*As'. in this example, we used (2*n-1) to deduct the holes of reinforcing bars in the compression area.
     



neutral axis location by equalling moments of the area around the neutral axis as shown in figure 2

53.72*(x-2.5)+18*x*(x/2)=36*(29-x)
53.72x-134.3+9x^2-1044+36x=0
9x^2+89.72x-1178.3=0
x=7.5in

Figure 2

moment of inertia using parallel axis theorem
I=Ic+Ad^2
=(18*7.5^3)/12+5.4*18*3.75^2+53.72*5^2+36*21.5^2
=19983.68in4=0.93ft4


flexural stress at extreme compression fiber where y=7.5in=0.625ft
fc=(M*y)/I
=(320*0.625)/0.93=215k/ft2=1493psi

flexural stress at center of compression reinforcing steel where y=5in=0.417ft
fs'=(2*n*M*y)/I
=(2*9*320*0.417)/0.93=2580.65k/ft2=17921psi

flexural stress at the center of tension steel where y=21.5in=1.79ft

fs=(n*M*y)/I
=(9*320*1.79)/0.93=5548.39k/ft2=38530psi


Comments

  1. thank you for providing us these example. it was very beneficial to me.
    can i understand , why you doubled the area in compression steel. it should be (n-1|) As' , it was nAs'-As' = As' (n-1) that already consider As' area that deducted.

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