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Beam design with deflection, 5.5 m span

Short answer

Factored moment 301.53 kN·m exceeds limiting 158.66 kN·m, so doubly reinforced. Required A_st = 1980 mm² (5-25 mm), A_sc = 926 mm² (5-16 mm); deflection check (L/d = 11 < 23.2) is safe.

A simply supported RCC beam of effective span 5.5 meter and overall dimensions 230 mm x 550 mm is subjected to superimposed load of 50 kN/m excluding its self weight. Design the beam for limit state of collapse in flexure. Also check whether the beam is safe in deflection or not. Adopt mild exposure condition and use Fe 415 steel. Take effective cover to re-bars as 50 mm.

Given: L=5.5L=5.5 m, b=230b=230 mm, D=550D=550 mm, superimposed load 50 kN/m (excluding self weight), Fe415, mild exposure, effective cover 50 mm. The grade of concrete is not stated, so M20 (the minimum for mild exposure, IS 456 Table 5) is adopted. d=550−50=500d=550-50=500 mm.

Step 1: Loads and moment

Self weight=0.23×0.55×25=3.16 kN/m,w=53.16 kN/mwu=1.5w=79.74 kN/mMu=wuL28=79.74×5.528=301.53 kN⋅m\begin{aligned} \text{Self weight}&=0.23\times0.55\times25=3.16\ \text{kN/m},\quad w=53.16\ \text{kN/m}\\ w_u&=1.5w=79.74\ \text{kN/m}\\ M_u&=\frac{w_uL^2}{8}=\frac{79.74\times5.5^2}{8}=301.53\ \text{kN·m} \end{aligned}

Step 2: Limiting moment

Mu,lim=0.138fckbd2=0.138×20×230×5002=158.66 kN⋅m<MuM_{u,lim}=0.138f_{ck}bd^2=0.138\times20\times230\times500^2=158.66\ \text{kN·m}<M_u

The section is doubly reinforced (the size is fixed).

Doubly reinforced design

Mu2=Mu−Mu,lim=301.53−158.66=142.87 kN⋅mM_{u2}=M_u-M_{u,lim}=301.53-158.66=142.87\ \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(1−50240.0)=0.00277\varepsilon_{sc}=0.0035\left(1-\frac{d'}{x_{u,max}}\right)=0.0035\left(1-\frac{50}{240.0}\right)=0.00277

From the design stress-strain curve of Fe415 (IS 456 Fig. 23): fsc=351.9f_{sc}=351.9 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′)=142.87×106(351.9−8.92)(500−50)=926 mm2A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')}=\frac{142.87\times10^6}{(351.9-8.92)(500-50)}=926\ \text{mm}^2 Ast1=0.36fck b xu,max0.87fy=1101 mm2,Ast2=Asc(fsc−fcc)0.87fy=879 mm2A_{st1}=\frac{0.36f_{ck}\,b\,x_{u,max}}{0.87f_y}=1101\ \text{mm}^2,\qquad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y}=879\ \text{mm}^2 Ast=Ast1+Ast2=1980 mm2A_{st}=A_{st1}+A_{st2}=1980\ \text{mm}^2

Provide: tension 5-25 mm dia (24542454 mm2^2), compression 5-16 mm dia (10051005 mm2^2), with 8 mm stirrups.

Step 3: Deflection check (IS 456 cl. 23.2)

pt=100Astbd=2.13%,fs=0.58fyAst,reqAst,prov=194.2 N/mm2,kt≈0.95 (Fig. 4)p_t=\frac{100A_{st}}{bd}=2.13\%,\quad f_s=0.58f_y\frac{A_{st,req}}{A_{st,prov}}=194.2\ \text{N/mm}^2,\quad k_t\approx0.95\ \text{(Fig. 4)} pc=0.87%,kc=1.6pcpc+0.275=1.22p_c=0.87\%,\quad k_c=\frac{1.6p_c}{p_c+0.275}=1.22 (Ld)allow=20×kt×kc=20×0.95×1.22=23.2,(Ld)actual=5500500=11.00\left(\frac{L}{d}\right)_{allow}=20\times k_t\times k_c=20\times0.95\times1.22=23.2,\qquad \left(\frac{L}{d}\right)_{actual}=\frac{5500}{500}=11.00

The beam is safe in deflection.

Answer: Mu=301.53M_u=301.53 kN·m, Ast=1980A_{st}=1980 mm2^2 (5-25 mm), Asc=926A_{sc}=926 mm2^2 (5-16 mm); safe in deflection.

✓ Answer checked by Claude Opus 5.5 (high effort)

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