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Example 7:flexural stress for triangle shape beam using transform area method

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








Transform area for each layer=n*As
As=0.79*1=0.79in2
Transform area=9*0.79=7.11in2


we will determine neutral axis location by equalling moment of the area from both parts as shown in figure 2


2*0.50*0.41x(x)(2*x/3)+(20-0.82x)(x)(x/2)=7.11(15-X)+7.11(18-X)

0.273x^3+10x^2-0.41x^3-234.63-14.22x=0

-0.137x^3+10x^2-14.22x-234.63=0

X=5.88in
Figure 2

now we will calculate the moment of inertia using parallel axis theorem
I=Ic+Ad^2
=(15.178*5.88^3)/12+15.178*5.88*2.94^2+2*((2.41*5.88^3)/36+0.5*2.41*5.88*3.92^2)+7.11*9.12^2+7.11*12.12^2=2906.35in4=0.14ft4


flexural stress at extreme compression where y=5.88in=0.49ft

fc=(M*y)/I
=(90*0.49)/0.14=315k/ft2=2187psi

flexural stress at the center of the reinforcing steel layer. y=16.5in=1.375ft


fs=(n*M*y)/I
=(9*90*1.375)/0.14=7955.36k/ft2=55245psi







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