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Calculating bending stresses using transformed area method

bending stress can be calculated for an applied moment on reinforced concrete beam less than the cracking moment by assuming the section properties of the reinforced concrete beam as concrete only(one material) this because concrete is uncracked and resisting tension. also, the amount of reinforcing steel is minimal compared to concrete. area of steel cross-section is 2.00% or less of concrete cross-section area





figure 1

when concrete starts to crack. concrete at this stage won't be able to resist tension anymore. tension will be resisted by the reinforcing bar. usually, the moment caused by service load is more than the cracking moment. therefore, the beam will be cracked and tensile stresses will be resisted by reinforcing bar. the concrete below the neutral axis will not be active and transformed area method will be used to calculate bending stress. RC beam cracks at the extreme fiber of the tensile zone do not mean that the beam will fail. because the reinforcing bar will resist tensile stress. steel will develop an excellent bond with concrete. therefore the strain will be equal. the equal strain does not mean equal stress because the modulus of elasticity is different. the stress will be proportional to the modulus ratio  

Ƞ=Es/Ec
Ƞ=(σs/εs)/(σc/εc), εs=εc 
Ƞ=σs/σc

if Ƞ is 10 then if the stress in concrete equal 1 mpa then the stress in steel equal 10 mpa. area of reinforcing bar can be transformed into an equvilent area of concrete as shown in figure 2. fictitious area (equvilent area)=Ƞ.As. in this method we transfer the area of the reinforced concrete beam into concrete only. therefore elastic beam theory will be applicable. a numerical example below will illustrate this method

Figure 2

Example: calculate bending stresses for the beam shown in figure number 3. knowing that moment equal 150 kn.m and Ƞ=9.

calculating of x can be achieved by calculating the moment of the area around the neutral axis from concrete in compression and transformed area

45*X*X/2-15*15*(X-7.5)=297*(60-X)
22.5X^2-225X+1,687.5=17,820-297X
22.5X^2+72X-16,132.5=0
X=25.22cm



Figure 3

next step is to calculate the moment of inertia using parallel axis theorem
I=I0+A.d^2

Where 
I is the moment of inertia for the composite section
I is the moment of inertia for part of the section
d is the distance from center of the part under investigation to composite section

I=2*(15*25.22^3)/12+2*15*25.22*12.62^2+(15*10.22^3)/12+15*10.22*5.11^2+297*1/12+297*34.78^2=505,450.75cm4=0.005m4

stress at extreme compression=M*Yc/I=150*0.2522/.005=7,566kn/m2=7.566 Mpa

stress at extreme tension=n*M*Ys/I=9*150*0.3478/.005=93,906kn/m2=93.706 Mpa

More examples:









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Comments

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