When the bending moment to be borne by a beam becomes greater than the balanced moment of resistance, the section becomes over-reinforced section and IS 456 does not recommend an over-reinforced section due to its brittle nature of the failure.
If there is no restriction on the size of the beam, we can increase its size so that the beam becomes under-reinforced. But if the size is restricted due to some reason then either we can increase the concrete mix to increase the capacity of the section or we can provide compression reinforcement in compression zone to give additional strength to the concrete in compression and such beams are called doubly reinforced beam.
Advantages of compression reinforcement:
where,
${A}_{sc}=\text{Area of Compression Steel}$
${A}_{st}=\text{Area of Tension Steel}$
$m=\text{Modular Ratio of tension steel}$
$m'=1.5m=\text{Modular Ratio of compression steel}$
${f}_{sc}=\text{Stress in compression steel} $
${f}_{st}=\text{Stress in tension stell}$
The value of the modular ratio of compression steel $(m')$ is higher than that of tension steel because of the long-term plastic deformation is known as creep. The creep deformation of concrete produces additional strain in compression steel and gradually raises the level of stress. To account for this increase in stress, the modular ratio of compression steel is increased.
Depth of Neutral Axis,
Stresses in Concrete and steel,
Moment of Resistance of Doubly Reinforced Section
If neutral axis coefficient ${n}_{0}$ for singly reinforced balanced section is greater than actual neutral axis coefficient $n$
${n}_{0}> n$ then, ${f}_{st}={\sigma}_{st}$
${f}_{cbc}=\frac{{\sigma}_{st}}{m}(\frac{x}{d-x})$
$\text{MOR}={C}_{1}(d-\frac{x}{3})+{C}_{2}(d-d')$
${C}_{1}=\frac{1}{2}\times{{f}_{cbc}\times{xb}}$
${C}_{2}=(m'-1){A}_{sc}\times{{f}_{sc}}=(m'-1){A}_{sc}\times {f}_{cbc}\frac{x-d'}{x}$
If neutral axis coefficient ${n}_{0}$ for singly reinforced balanced section is smaller than actual neutral axis coefficient $n$
${n}_{0}< n$ then, ${f}_{cbc}={\sigma}_{cbc}$
$\text{MOR}={C}_{1}(d-\frac{x}{3})+{C}_{2}(d-d')$
${C}_{1}=\frac{1}{2}\times{{\sigma}_{cbc}\times{xb}}$
${C}_{2}=(m'-1){A}_{sc}\times {\sigma}_{cbc}\frac{x-d'}{x}$
Area of steel in tension and compression zone
To calculate or rather to design a doubly reinforced beam, the beam is divided in to two parts. First part will resemble a balanced singly reinforced section while the other will show only tension and compression reinforcement.
Area of Steel, ${A}_{st}={A}_{st1}+{A}_{st2}$
${A}_{st}=\frac{{R}_{w}bd^2}{{\sigma}_{st}({J}_{0}d)}+\frac{M-{R}_{w}bd^2}{{\sigma}_{st}(d-d')}$
$M-{{R}_{w}}b{{d}^{2}}=(m'-1){{A}_{sc}}\left( {{\sigma }_{cbc}}\frac{x-d'}{x} \right)(d-d')$
We can canculate the compression steel from above expression.
If there is no restriction on the size of the beam, we can increase its size so that the beam becomes under-reinforced. But if the size is restricted due to some reason then either we can increase the concrete mix to increase the capacity of the section or we can provide compression reinforcement in compression zone to give additional strength to the concrete in compression and such beams are called doubly reinforced beam.
Advantages of compression reinforcement:
- It permits smaller size beams which look aesthetic.
- It reduces the long-term deflection and increases the ductility of the beam.
- It can be used as anchor bars for positioning the shear reinforcement.
- As the compression reinforcement increases the ductility of the beam, they are provided in the seismic zone to withstand repeated reversals produces.
where,
${A}_{sc}=\text{Area of Compression Steel}$
${A}_{st}=\text{Area of Tension Steel}$
$m=\text{Modular Ratio of tension steel}$
$m'=1.5m=\text{Modular Ratio of compression steel}$
${f}_{sc}=\text{Stress in compression steel} $
${f}_{st}=\text{Stress in tension stell}$
The value of the modular ratio of compression steel $(m')$ is higher than that of tension steel because of the long-term plastic deformation is known as creep. The creep deformation of concrete produces additional strain in compression steel and gradually raises the level of stress. To account for this increase in stress, the modular ratio of compression steel is increased.
Depth of Neutral Axis,
$\frac{b{{x}^{2}}}{2}+(m'-1){{A}_{sc}}(x-d')=m{{A}_{st}}(d-x)$
Stresses in Concrete and steel,
-
By flexure formula,
Moment of Inertia about NA, $I=\frac{b{{x}^{3}}}{3}+(m'-1){{A}_{sc}}{{(x-d')}^{2}}+m{{A}_{st}}{{(d-x)}^{2}}$
${{f}_{cbc}}=\frac{Mx}{I}$
$\frac{{{f}_{sc}}}{m'}=\frac{M(x-d')}{I}$
$\frac{{{f}_{st}}}{m}=\frac{M(d-x)}{I}$ -
By Internal Couple Method,
${C}_{1}=\frac{1}{2}\times{{f}_{cbc}\times{xb}}$
${C}_{2}=(m'-1){A}_{sc}\times{{f}_{sc}}$
${f}_{sc}={f}_{cbc}\frac{x-d'}{x}$
$M={C}_{1}(d-\frac{x}{3})+{C}_{2}(d-d')$
Using above equations ${f}_{cbc}$ can be found out.
$\frac{{f}_{sc}}{m'}={f}_{cbc}\frac{x-d'}{x}$
$\frac{{f}_{st}}{m}={f}_{cbc}\frac{d-x}{x}$
${f}_{sc}$,${f}_{st}$ can be found out using above equations
${C}_{1}=\text{Compression carried by concrete}$
${C}_{2}=\text{Additional force carried by compression steel}$
Moment of Resistance of Doubly Reinforced Section
If neutral axis coefficient ${n}_{0}$ for singly reinforced balanced section is greater than actual neutral axis coefficient $n$
${f}_{cbc}=\frac{{\sigma}_{st}}{m}(\frac{x}{d-x})$
$\text{MOR}={C}_{1}(d-\frac{x}{3})+{C}_{2}(d-d')$
${C}_{1}=\frac{1}{2}\times{{f}_{cbc}\times{xb}}$
${C}_{2}=(m'-1){A}_{sc}\times{{f}_{sc}}=(m'-1){A}_{sc}\times {f}_{cbc}\frac{x-d'}{x}$
If neutral axis coefficient ${n}_{0}$ for singly reinforced balanced section is smaller than actual neutral axis coefficient $n$
$\text{MOR}={C}_{1}(d-\frac{x}{3})+{C}_{2}(d-d')$
${C}_{1}=\frac{1}{2}\times{{\sigma}_{cbc}\times{xb}}$
${C}_{2}=(m'-1){A}_{sc}\times {\sigma}_{cbc}\frac{x-d'}{x}$
Area of steel in tension and compression zone
To calculate or rather to design a doubly reinforced beam, the beam is divided in to two parts. First part will resemble a balanced singly reinforced section while the other will show only tension and compression reinforcement.
${A}_{st}=\frac{{R}_{w}bd^2}{{\sigma}_{st}({J}_{0}d)}+\frac{M-{R}_{w}bd^2}{{\sigma}_{st}(d-d')}$
$M-{{R}_{w}}b{{d}^{2}}=(m'-1){{A}_{sc}}\left( {{\sigma }_{cbc}}\frac{x-d'}{x} \right)(d-d')$
We can canculate the compression steel from above expression.
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