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Derive moment capacity of doubly reinforced WSM

Short answer

Using 1.5m for compression steel, neutral axis is bx^2/2+(1.5m-1)Asc(x-d')=mAst(d-x). Moment of resistance is Mr=0.5σcbc bx(d-x/3)+(1.5m-1)Asc σcbc[(x-d')/x](d-d'). Assumptions include plane sections, no tension in concrete, and perfect bond.

Derive equations for bending moment carrying capacity of a double reinforced rectangular section using assumptions for working stress method given by IS 456:2000.

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 1.5m1.5m (to allow for creep).

Notation: bb = width, dd = effective depth, d′d' = effective cover to compression steel, xx = depth of neutral axis, AstA_{st} = tension steel, AscA_{sc} = compression steel, mm = 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:

σc′=σcbc x−d′x\sigma_c' = \sigma_{cbc}\,\frac{x-d'}{x}

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

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

bx22+(1.5m−1)Asc(x−d′)=mAst(d−x)\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 Cc=12σcbc b xC_c = \tfrac12\sigma_{cbc}\,b\,x acting at x/3x/3 from the top:

Mr=12σcbc b x(d−x3)+(1.5m−1)Asc σc′ (d−d′)M_r = \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) + (1.5m-1)A_{sc}\,\sigma_{c}'\,(d-d')

Substituting σc′\sigma_c':

Mr=12σcbc b x(d−x3)+(1.5m−1)Asc σcbc x−d′x (d−d′)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 M1=Q b d2M_1 = Q\,b\,d^2 when x=xcx = x_c, and the second is the additional moment M2M_2 from the compression steel. The tension steel then follows from Ast=M1σstjd+M2σst(d−d′)A_{st}=\dfrac{M_1}{\sigma_{st}jd}+\dfrac{M_2}{\sigma_{st}(d-d')}.

Checks: σc≤σcbc\sigma_{c}\le\sigma_{cbc}, σst≤σst,perm\sigma_{st}\le\sigma_{st,perm} and σsc=1.5mσc′≤σsc,perm\sigma_{sc}=1.5m\sigma_c'\le\sigma_{sc,perm} (190 N/mm2^2 for Fe415).

✓ Answer checked by Claude Opus 5.5 (high effort)

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