Doubly reinforced rectangular beam refers to a beam with reinforcement in compression and tension zone. In some cases, the dimension of the beam is limited, and the tensile reinforcement is not sufficient to resist the factored expected bending moment. Therefore compression reinforcement is added. The figure no:1 showing a doubly reinforced beam.

*Figure 1*

The compression reinforcement will contribute to increasing the nominal moment of the beam. However, the contribution is relatively small. Compression reinforcement will ease the process of fixing the shear stirrups, shear stirrups will be tied to both tensile and compression reinforcement. Adding compression reinforcement will result in increasing the value of **εt, **which means increasing the ductility of the beam.

Increasing the ductility or **εt **is directly related to strength reduction factor ϕ, if **εt **value is larger than 0.005, the value of ϕ=0.90, for a beam with lesser ductility or **εt, **the value of reduction factor will decrease, and this means the reducing of beam strength or nominal moment. Compression reinforcement will also reduce the long term deflection of beams.

*Figure 2*

*Figure 3*

The nominal moment of the doubly reinforced beam can be calculated for two cases:

case 1, when the compression steel yield. The nominal moment can be calculated using the following equations.

when the compression steel yield fs’=fy

from the equilibrium of forces

C+Cs’=T As’*fy+0.85fc’*b*a=As*fy

a=((As-As’)fy/(0.85fc’*b)

The total nominal moment is equal to the sum of Mn1 and Mn2. Mn1 is the moment created by the force of compression steel As’ and an equal area of reinforcement in the tensile zone.

Mn1=As’*fy(d-d’)

Mn2 is the moment created by the rest of tensile reinforcement (As-As’) and the compression force.

Mn2=(As-As’)*fy(d-a/2)

Mn=Mn1+Mn2Mn=As’*fy(d-d’)+(As-As’)*fy(d-a/2)

Case 2, when the compression steel not yielding (fs'<fy). Therefore the value of fs’ is unknown. We need to determine the value of fs’ to determine the depth of the concrete compression block. From the strain diagram.

(C/0.003)=((c-d’)/εs’)

εs’=(0.003*(c-d’)/c)εs’=fs’/Es, Es is the steel modulus of elasticityfs’=0.003*Es(1-(d’/c))

The neutral axis depth can be obtained by the equilibrium.

C+Cs’=T

As’*fs’+0.85fc’*b*a=As*fy

As’*0.003*Es(1-(d’/c))+0.85fc’*b*a=As*fy

a=β1C

As’*0.003*Es(1-(d’/c))+0.85fc’*b*β1C=As*fy

multiplying the whole equation by C

As’*0.003*Es(C-(d’))+0.85fc’*b*β1C^2=As*fy*C

rearranging the equation

(0.85fc’*b*β1C^2)-As*fy*C+As’*0.003*Es*C-As’*0.003*Es*d’=0

(0.85fc’*b*β1C^2)+C(0.003*As’*Es-As*fy)-As’*0.003*Es*d’=0

this quadratic equation, C, can be calculated using the general equation for a quadratic equation. by calculating C, we will be able to determine fs’ from fs’=0.003*Es(1-(d’/c)).

the nominal moment Mn when the steel doesn’t yield in compression zone can be calculated by the following equation

Mn=0.85fc’*a*b(d-a/2)+As’*fs'(d-d’)