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Beam steel at midspan, limit state

Short answer

Mu=451.34 kNm. Doubly reinforced: tension steel Ast=2907 mm² (6-25 mm), compression steel Asc=1543 mm² (5-20 mm).

A beam of rectangular section is 300 mm wide and 500 mm deep to the centre of tensile reinforcement. It has to carry a dead load of 45 kN/m excluding its self weight. Find the steel reinforcement required for the mid span section. The beam has a span of 7 m. Use M20 concrete and Fe 415 steel. Effective cover to compression steel = 40 mm. Use limit state method.

Given: b=300b=300 mm, d=500d=500 mm, span L=7L=7 m, dead load 45 kN/m (excluding self weight), M20, Fe415, effective cover to compression steel d′=40d'=40 mm. Limit state of collapse in flexure.

Assumption: overall depth D=550D=550 mm (dd plus 50 mm cover); the effective span is taken as 7 m.

Step 1: Loads

Self weight=0.30×0.55×25=4.12 kN/mw=45+4.12=49.12 kN/mwu=1.5×49.12=73.69 kN/m\begin{aligned} \text{Self weight}&=0.30\times0.55\times25=4.12\ \text{kN/m}\\ w&=45+4.12=49.12\ \text{kN/m}\\ w_u&=1.5\times49.12=73.69\ \text{kN/m} \end{aligned}

Step 2: Design moment at mid-span

Mu=wuL28=73.69×728=451.34 kN⋅mM_u=\frac{w_uL^2}{8}=\frac{73.69\times7^2}{8}=451.34\ \text{kN·m}

Step 3: Limiting moment of resistance (Fe415, xu,max=0.48dx_{u,max}=0.48d)

Mu,lim=0.138fckbd2=0.138×20×300×5002=206.95 kN⋅mM_{u,lim}=0.138f_{ck}bd^2=0.138\times20\times300\times500^2=206.95\ \text{kN·m}

Since Mu>Mu,limM_u>M_{u,lim} and the size is fixed, the section is doubly reinforced.

Doubly reinforced design

Mu2=Mu−Mu,lim=451.34−206.95=244.39 kN⋅mM_{u2}=M_u-M_{u,lim}=451.34-206.95=244.39\ \text{kN·m}

Limiting depth xu,max=0.48d=240.0x_{u,max}=0.48d=240.0 mm. Strain in compression steel (similar triangles):

εsc=0.0035(1−d′xu,max)=0.0035(1−40240.0)=0.00292\varepsilon_{sc}=0.0035\left(1-\frac{d'}{x_{u,max}}\right)=0.0035\left(1-\frac{40}{240.0}\right)=0.00292

From the design stress-strain curve of Fe415 (IS 456 Fig. 23): fsc=353.2f_{sc}=353.2 N/mm2^2. Stress in concrete displaced by the steel: fcc=0.446fck=8.92f_{cc}=0.446f_{ck}=8.92 N/mm2^2.

Asc=Mu2(fsc−fcc)(d−d′)=244.39×106(353.2−8.92)(500−40)=1543 mm2A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')}=\frac{244.39\times10^6}{(353.2-8.92)(500-40)}=1543\ \text{mm}^2 Ast1=0.36fck b xu,max0.87fy=1436 mm2,Ast2=Asc(fsc−fcc)0.87fy=1471 mm2A_{st1}=\frac{0.36f_{ck}\,b\,x_{u,max}}{0.87f_y}=1436\ \text{mm}^2,\qquad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y}=1471\ \text{mm}^2 Ast=Ast1+Ast2=2907 mm2A_{st}=A_{st1}+A_{st2}=2907\ \text{mm}^2

Provide:

  • Tension steel: 6-25 mm dia (29452945 mm2^2), arranged in two layers.
  • Compression steel: 5-20 mm dia (15711571 mm2^2) at d′=40d'=40 mm.
  • Stirrups (8 mm, 2-legged) at spacing not more than 16ϕc16\phi_c, 300 mm and bb to hold the compression bars (shear design separately).

Checks: minimum steel 0.85bd/fy=3070.85bd/f_y=307 mm2^2 OK; pt=1.96%<4%p_t=1.96\%<4\% and pc=1.05%p_c=1.05\% OK.

Answer: Mid-span Ast=2907A_{st}=2907 mm2^2 (6-25 mm) and Asc=1543A_{sc}=1543 mm2^2 (5-20 mm).

✓ Answer checked by Claude Opus 5.5 (high effort)

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