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Doubly reinforced rectangular beam analysis

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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+Mn2
Mn=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 elasticity
fs'=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')


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My name is Mohammad Hamdan, and I am a Civil/bridge Engineer with extensive professional background and experience in the field of construction. I have 10+ years of experience in the construction field. During my professional career, my emphasis has been placed on structural projects, mainly bridges, infrastructure and roads. Reading and practicing the design of structures is my passion.i was lucky to be a part of a big team that executed major highway interchanges. I believe in spreading the knowledge, thus, i compose these courses to share my experience and knowledge. My course focuses on the practical side of construction field. On other word, i am sharing some of what new engineers need to know about construction field. Beside working on construction field, i love writing some articles on my website ( curious civil engineer) Moreover, i wrote some engineering papers, you can check them  by searching my name on Google (Mohammad mamon Hamdan)