Steel requirement and doubly reinforced moment

Requirement of steel as reinforcement

Concrete is strong in compression but weak in tension (tensile strength is about 10% of compressive strength), so it cracks at low loads. Steel is therefore provided as reinforcement for the following reasons.

• Tension: steel carries the tensile force in beams, slabs, footings and ties, with concrete protecting it from corrosion and fire.
• Shear and torsion: stirrups and bent-up bars resist diagonal tension.
• Compression: longitudinal bars share the load in columns and raise capacity of beams (doubly reinforced section).
• Control of cracks and shrinkage: distribution and temperature steel limit crack width.
• Bond: steel has a similar coefficient of thermal expansion to concrete ($\approx 1.2\times10^{-5}$ per $^\circ$C), and ribs/deformations give good bond.
• Ductility: steel gives warning before failure.

Moment of resistance of doubly reinforced section

A beam is doubly reinforced when the applied moment exceeds the balanced moment $Q\,b\,d^2$ of the singly reinforced section, or when depth is restricted, or for reversal of moment. Compression steel $A_{sc}$ is added at effective cover $d'$ from the compression face. The moment of resistance is

$$M_r = M_1 + M_2$$

where $M_1 = Q\,b\,d^2$ is the balanced-section moment and $M_2$ the additional moment taken by the compression steel and the additional tension steel.

Derivation

Assumptions (IS 456:2000, Annex B-1.2)

• Plane sections remain plane after bending.
• Concrete is elastic in compression and carries no tension.
• Perfect bond between steel and concrete; stress in steel $=$ modular ratio $\times$ stress in surrounding concrete.
• For compression steel, effective modular ratio is $1.5m$ (to allow for creep).

Notation: $b$ = width, $d$ = effective depth, $d'$ = effective cover to compression steel, $x$ = depth of neutral axis, $A_{st}$ = tension steel, $A_{sc}$ = compression steel, $m$ = modular ratio.

   Section    Strain      Stresses/forces
   +------+   |\         Cc  (concrete)
 d'| Asc  |   | \ x      Cs  (comp. steel)
   |      |  -+--- NA
   |      |   | /
   | Ast  |   |/         T   (tension steel)
   +------+

Stress in concrete at the level of compression steel

By similar triangles of the strain diagram:

$$\sigma_c' = \sigma_{cbc}\,\frac{x-d'}{x}$$

Stress in compression steel $\sigma_{sc} = 1.5\,m\,\sigma_c'$. The concrete displaced by the steel is allowed for by using $(1.5m-1)A_{sc}$ as the equivalent concrete area.

Neutral axis (first moment of the equivalent area about the neutral axis is zero)

$$\frac{b x^2}{2} + (1.5m-1)A_{sc}(x-d') = m A_{st}(d-x)$$

Moment of resistance

Taking moments about the tension steel, with concrete compression $C_c = \tfrac12\sigma_{cbc}\,b\,x$ acting at $x/3$ from the top:

$$M_r = \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) + (1.5m-1)A_{sc}\,\sigma_{c}'\,(d-d')$$

Substituting $\sigma_c'$:

$$M_r = \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) + (1.5m-1)A_{sc}\,\sigma_{cbc}\,\frac{x-d'}{x}\,(d-d')$$

The first term is the moment of the singly reinforced (balanced) part $M_1 = Q\,b\,d^2$ when $x = x_c$, and the second is the additional moment $M_2$ from the compression steel. The tension steel then follows from $A_{st}=\dfrac{M_1}{\sigma_{st}jd}+\dfrac{M_2}{\sigma_{st}(d-d')}$.

Checks: $\sigma_{c}\le\sigma_{cbc}$, $\sigma_{st}\le\sigma_{st,perm}$ and $\sigma_{sc}=1.5m\sigma_c'\le\sigma_{sc,perm}$ (190 N/mm$^2$ for Fe415).