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L-beam reinforcement for given moment

Short answer

Limiting moment 495.04 kN·m < 615 kN·m, so doubly reinforced. Provide tension steel 2943 mm² (6-25 mm) and compression 592 mm² (3-16 mm); 8 mm two-legged stirrups at 200 mm c/c satisfy shear and minimum requirements.

A L-beam has a flange of effective width 900 mm and depth of 100 mm. The web below is 250 mm x 500 mm. Determine the amount of reinforcement required for the cross-section if it has to carry a factored bending moment of 615 kN-m and SF of 50 kN. Adopt M20 concrete mix and Fe 500 grade steel.

Given: bf=900b_f=900 mm, Df=100D_f=100 mm, bw=250b_w=250 mm, web depth below flange 500 mm; Mu=615M_u=615 kN·m, Vu=50V_u=50 kN (factored); M20, Fe500.

Assumptions: overall depth D=500+100=600D=500+100=600 mm; effective cover 50 mm, so d=550d=550 mm and d′=50d'=50 mm.

Step 1: Limiting moment of the L-section. Df/d=100/550=0.182≤0.2D_f/d=100/550=0.182\le0.2, so the flange thickness DfD_f is used directly. For Fe500, xu,max=0.46d=253.0x_{u,max}=0.46d=253.0 mm >Df>D_f (neutral axis in the web).

Mu,lim=0.36fckbwxu,max(d−0.416xu,max)+0.45fck(bf−bw)Df(d−Df2)=202.54+292.50=495.04 kN⋅m\begin{aligned} M_{u,lim}&=0.36f_{ck}b_wx_{u,max}(d-0.416x_{u,max})+0.45f_{ck}(b_f-b_w)D_f\left(d-\frac{D_f}{2}\right)\\ &=202.54+292.50=495.04\ \text{kN·m} \end{aligned}

Since Mu=615>Mu,lim=495.04M_u=615>M_{u,lim}=495.04 kN·m, compression steel is required (doubly reinforced L-beam).

Step 2: Compression steel

Mu2=615−495.04=119.96 kN⋅mM_{u2}=615-495.04=119.96\ \text{kN·m} εsc=0.0035(1−50253.0)=0.00281 ⇒ fsc=414.2 N/mm2 (Fe500 curve),fcc=0.446×20=8.92\varepsilon_{sc}=0.0035\left(1-\frac{50}{253.0}\right)=0.00281\ \Rightarrow\ f_{sc}=414.2\ \text{N/mm}^2\ (\text{Fe500 curve}),\quad f_{cc}=0.446\times20=8.92 Asc=Mu2(fsc−fcc)(d−d′)=119.96×106(414.2−8.92)(550−50)=592 mm2A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')}=\frac{119.96\times10^6}{(414.2-8.92)(550-50)}=592\ \text{mm}^2

Step 3: Tension steel

Ast1=0.36fckbwxu,max+0.45fck(bf−bw)Df0.87fy=2392 mm2,Ast2=Asc(fsc−fcc)0.87fy=552 mm2A_{st1}=\frac{0.36f_{ck}b_wx_{u,max}+0.45f_{ck}(b_f-b_w)D_f}{0.87f_y}=2392\ \text{mm}^2,\qquad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y}=552\ \text{mm}^2 Ast=Ast1+Ast2=2943 mm2A_{st}=A_{st1}+A_{st2}=2943\ \text{mm}^2

Provide 6-25 mm dia (29452945 mm2^2, in layers) and 3-16 mm dia compression bars (603603 mm2^2).

Step 4: Shear (Vu_u = 50 kN)

τv=Vubwd=50×103250×550=0.364 N/mm2\tau_v=\frac{V_u}{b_wd}=\frac{50\times10^3}{250\times550}=0.364\ \text{N/mm}^2

pt=100Ast/(bwd)=2.14%p_t=100A_{st}/(b_wd)=2.14\%, τc=0.80\tau_c=0.80 N/mm2^2 (IS 456 Table 19, M20). Since τv<τc\tau_v<\tau_c, only minimum shear reinforcement is needed: Asvb sv≥0.40.87fy\dfrac{A_{sv}}{b\,s_v}\ge\dfrac{0.4}{0.87f_y}.

For 8 mm 2-legged stirrups (Asv=100.5A_{sv}=100.5 mm2^2): sv≤0.87×500×100.50.4×250=437s_v\le\dfrac{0.87\times500\times100.5}{0.4\times250}=437 mm; also ≤0.75d=412\le0.75d=412 mm and 300 mm. Provide 8 mm 2-legged stirrups at 200 mm c/c (closer than 300 mm to support the compression bars).

Answer: Ast=2943A_{st}=2943 mm2^2 (6-25 mm), Asc=592A_{sc}=592 mm2^2 (3-16 mm); shear: 8 mm 2-legged stirrups at 200 mm c/c.

✓ Answer checked by Claude Opus 5.5 (high effort)

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