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Beam with slab, 20 kN/m, 350 mm wall

Short answer

Use ISMB 300 (46 kg/m). Factored moment 135 kN·m, moment capacity 152.4 kN·m; shear capacity 303 kN > 90 kN; deflection 18.8 mm < L/300 = 20 mm. Web crippling and buckling checks are satisfied without stiffeners.

Design a beam of 6m span carrying UDL of 20KN/m including self wt. The beam is laterally restrained by a concrete slab and is simply supported at the ends on wall of 350mm width. Check for shear, deflection, web crippling and buckling.

Data

  • Steel Fe 410 (E250): fy=250f_y = 250 N/mm², fu=410f_u = 410 N/mm², E=2×105E = 2\times10^5 N/mm², γm0=1.10\gamma_{m0} = 1.10 (IS 800:2007, Table 5). Limit state design.
  • Effective span L=6.00L = 6.00 m, bearing length of support =350= 350 mm.

Loads and design forces

The UDL of 20 kN/m (inclusive of self-weight) is the service load; factored wu=1.5×20=30w_u = 1.5\times20 = 30 kN/m. L=6L = 6 m, wall (support) width 350 mm so b1=350b_1 = 350 mm: Mu=30×36/8=135M_u = 30\times36/8 = 135 kN·m, Vu=90V_u = 90 kN.

Moment without self-weight M=135.0M = 135.0 kN·m, so Zp,req≈Mγm0/fy=135.0×106×1.1/250=594Z_{p,req} \approx M\gamma_{m0}/f_y = 135.0\times10^6\times1.1/250 = 594 cm³.

Trial section

Try ISMB 300 (IS 808): D=300D = 300 mm, bf=140b_f = 140 mm, tw=7.7t_w = 7.7 mm, tf=13.1t_f = 13.1 mm, R1=14R_1 = 14 mm; Izz=8990I_{zz} = 8990 cm⁴; Ze=599Z_e = 599 cm³; Zp≈Btf(D−tf)+tw(D−2tf)2/4=670Z_p \approx B t_f (D - t_f) + t_w (D - 2t_f)^2/4 = 670 cm³; mass 46.046.0 kg/m.

Mu=wL28=30.00×6.0028=135.0 kN⋅mVu=wL2=90.0 kN\begin{aligned} M_u &= \frac{wL^2}{8} = \frac{30.00\times6.00^2}{8} = 135.0\ \text{kN·m} \\ V_u &= \frac{wL}{2} = 90.0\ \text{kN} \end{aligned}

Section classification (Table 2)

ε=250/fy=1.00\varepsilon = \sqrt{250/f_y} = 1.00. Flange: b/tf=(140/2)/13.1=5.34<9.4ε=9.4b/t_f = (140/2)/13.1 = 5.34 < 9.4\varepsilon = 9.4. Web: d/tw=31.9<84ε=84d/t_w = 31.9 < 84\varepsilon = 84 (with d=D−2(tf+R1)=245.8d = D - 2(t_f + R_1) = 245.8 mm). The section is plastic, so βb=1.0\beta_b = 1.0.

Moment capacity (cl. 8.2.1.2)

The concrete slab restrains the compression flange along the full length, so lateral-torsional buckling is prevented.

Md=βbZpfyγm0=670×103×2501.1×106=152.4 kN⋅m1.2Zefy/γm0=1.2×599×103×250/(1.1×106)=163.4 kN⋅m\begin{aligned} M_d &= \frac{\beta_b Z_p f_y}{\gamma_{m0}} = \frac{670\times10^3\times250}{1.1\times10^6} = 152.4\ \text{kN·m} \\ 1.2 Z_e f_y/\gamma_{m0} &= 1.2\times599\times10^3\times250/(1.1\times10^6) = 163.4\ \text{kN·m} \end{aligned}

The smaller value governs: Md=152.4M_d = 152.4 kN·m.

Md=152.4 kN⋅m>Mu=135.0 kN⋅mSafeM_d = 152.4\ \text{kN·m} > M_u = 135.0\ \text{kN·m}\quad \text{Safe}

Shear capacity (cl. 8.4)

Av=Dtw=300×7.7=2310 mm2Vd=Avfy3 γm0=2310×2503×1.1×103=303.1 kN\begin{aligned} A_v &= D t_w = 300\times7.7 = 2310\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{2310\times250}{\sqrt3\times1.1\times10^3} = 303.1\ \text{kN} \end{aligned}

Vu=90.0V_u = 90.0 kN <Vd< V_d; also Vu<0.6Vd=181.9V_u < 0.6V_d = 181.9 kN, so the shear is low and the moment capacity is not reduced (cl. 9.2.1). Web shear buckling does not govern since d/tw=31.9<67ε=67d/t_w = 31.9 < 67\varepsilon = 67 (cl. 8.4.2.1).

Deflection (Table 6)

Deflection under unfactored service (imposed) loads, limit span/300 (floor and roof members, Table 6):

δ=5wL4384EI=5×20.00×(6000)4384×2×105×8990×104=18.77 mm\delta = \frac{5 w L^4}{384 E I} = \frac{5\times20.00\times(6000)^4}{384\times2\times10^5\times8990\times10^4} = 18.77\ \text{mm}

δ=18.77\delta = 18.77 mm <L/300=20.00< L/300 = 20.00 mm. OK.

Web buckling and bearing at support (cl. 8.7.3 and 8.7.4)

Stiff bearing length b1=350b_1 = 350 mm. Clear web depth d=245.8d = 245.8 mm; dispersion of load at 45° to mid-depth at an end support n1=d/2=122.9n_1 = d/2 = 122.9 mm. Effective length 0.7d0.7d (cl. 8.7.1.5), λ=0.7d/(tw/12)=77.4\lambda = 0.7d/(t_w/\sqrt{12}) = 77.4; curve c gives fcd=140.4f_{cd} = 140.4 N/mm².

Fcdw=(b1+n1)twfcd=(350+122.9)×7.7×140.4/103=511.2 kNn2=2.5(tf+R1)=67.8 mmFw=(b1+n2)twfyγm0=(350+67.8)×7.7×2501.1×103=731.1 kN\begin{aligned} F_{cdw} &= (b_1 + n_1) t_w f_{cd} = (350 + 122.9)\times7.7\times140.4/10^3 = 511.2\ \text{kN} \\ n_2 &= 2.5(t_f + R_1) = 67.8\ \text{mm} \\ F_w &= \frac{(b_1 + n_2) t_w f_y}{\gamma_{m0}} = \frac{(350 + 67.8)\times7.7\times250}{1.1\times10^3} = 731.1\ \text{kN} \end{aligned}

Support reaction R=90.0R = 90.0 kN <Fcdw< F_{cdw} and <Fw< F_w. No web stiffener is needed.

Final design

Provide ISMB 300 (46.0 kg/m): Md=152.4M_d = 152.4 kN·m >Mu=135.0> M_u = 135.0 kN·m, Vd=303V_d = 303 kN >Vu=90.0> V_u = 90.0 kN, δ=18.8\delta = 18.8 mm <20.0< 20.0 mm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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