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Intermediate beam design for BM and deflection

Short answer

T-beam with effective flange width 1883 mm. Neutral axis in flange (M_u = 159.90 kN·m). Required A_st = 1009 mm² (4-20 mm bars); deflection safe with L/d = 11.1 << 32.

A floor consists of 125 mm thick RC slab, integrally connected with the beam as shown in figure. Design an intermediate beam for BM and deflection if the floor is subjected to live load of 4 kN/m2^2 and floor finishes of 0.7 kN/m2^2. [Figure: floor plan with one-brick-thick wall at the edge; slab spans between parallel beams (dashed lines) with 4 x 4 m panels; beam span 5 m]

Assumptions: M20 and Fe415; beam simply supported with effective span L=5L=5 m; slab panels 4 m x 4 m, so an intermediate beam carries slab load from 2 m on each side (width 4 m); beam web bw=300b_w=300 mm, overall depth D=500D=500 mm, cover 50 mm, so d=450d=450 mm. The slab and beam act together as a T-beam.

Step 1: Loads per metre run of beam

Slab dead load=0.125×25=3.125 kN/m2Floor finish=0.7 kN/m2,Live load=4 kN/m2Total on slab=7.825 kN/m2 ⇒ 7.825×4=31.30 kN/mRib self weight=0.30×(0.50−0.125)×25=2.812 kN/mw=34.11 kN/m,wu=1.5w=51.17 kN/m\begin{aligned} \text{Slab dead load}&=0.125\times25=3.125\ \text{kN/m}^2\\ \text{Floor finish}&=0.7\ \text{kN/m}^2,\quad\text{Live load}=4\ \text{kN/m}^2\\ \text{Total on slab}&=7.825\ \text{kN/m}^2\ \Rightarrow\ 7.825\times4=31.30\ \text{kN/m}\\ \text{Rib self weight}&=0.30\times(0.50-0.125)\times25=2.812\ \text{kN/m}\\ w&=34.11\ \text{kN/m},\qquad w_u=1.5w=51.17\ \text{kN/m} \end{aligned}

Step 2: Bending moment

Mu=wuL28=51.17×528=159.90 kN⋅mM_u=\frac{w_uL^2}{8}=\frac{51.17\times5^2}{8}=159.90\ \text{kN·m}

Step 3: Effective flange width (IS 456 cl. 23.1.2, T-beam)

bf=l06+bw+6Df=50006+300+6×125=1883 mm (≤4000 mm spacing)b_f=\frac{l_0}{6}+b_w+6D_f=\frac{5000}{6}+300+6\times125=1883\ \text{mm}\ (\le4000\ \text{mm spacing})

Step 4: Position of neutral axis. Moment capacity when xu=Dfx_u=D_f:

Mu,f=0.36fckbfDf(d−0.42Df)=673.76 kN⋅m>MuM_{u,f}=0.36f_{ck}b_fD_f(d-0.42D_f)=673.76\ \text{kN·m}>M_u

So xu<Dfx_u<D_f; design as a rectangular beam of width bfb_f.

Step 5: Tension steel

Mu=0.87fyAstd(1−Astfybfdfck) ⇒ Ast=1009 mm2M_u=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{b_fdf_{ck}}\right)\ \Rightarrow\ A_{st}=1009\ \text{mm}^2

Minimum steel 0.85bwd/fy=2770.85b_wd/f_y=277 mm2^2 is satisfied. Provide 4-20 mm dia (12571257 mm2^2).

Step 6: Deflection check (IS 456 cl. 23.2). bw/bf=0.16<0.3b_w/b_f=0.16<0.3, so the flange factor (Fig. 6) is 0.8. For flanged beams, the steel percentage for Fig. 4 is based on bfdb_fd:

pt=100Astbfd=100×12571883×450=0.15%,fs=0.58fyAst,reqAst,prov=193.2 N/mm2,kt=2.0 (maximum)p_t=\frac{100A_{st}}{b_fd}=\frac{100\times1257}{1883\times450}=0.15\%,\quad f_s=0.58f_y\frac{A_{st,req}}{A_{st,prov}}=193.2\ \text{N/mm}^2,\quad k_t=2.0\ \text{(maximum)} (Ld)allow=20×2.0×0.8=32.0,(Ld)actual=5000450=11.11\left(\frac{L}{d}\right)_{allow}=20\times2.0\times0.8=32.0,\qquad \left(\frac{L}{d}\right)_{actual}=\frac{5000}{450}=11.11

The beam is safe in deflection. (Shear: Vu=wuL/2=127.9V_u=w_uL/2=127.9 kN; stirrups are designed separately.)

Answer: T-beam, bf=1883b_f=1883 mm, bw=300b_w=300 mm, D=500D=500 mm; Ast=1009A_{st}=1009 mm2^2 (4-20 mm bars); safe in deflection.

✓ Answer checked by Claude Opus 5.5 (high effort)

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