--- title: "Beam with slab, 20 kN/m, 350 mm wall" subject: "Design of Steel and Timber Structure" chapter: 7 key: "semester_6/design_of_steel_and_timber_structure:beam-concrete-slab-20knm-350mm-wall" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/beam-concrete-slab-20knm-350mm-wall site: BCT Engineering --- # Beam with slab, 20 kN/m, 350 mm wall Design of Steel and Timber Structure · Chapter 7: Flexure Members - Asked in: 2068 Magh (old course), 14 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/beam-concrete-slab-20knm-350mm-wall - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-7 ## Question 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. ## 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. ## Answer #### Data - Steel Fe 410 (E250): $f_y = 250$ N/mm², $f_u = 410$ N/mm², $E = 2\times10^5$ N/mm², $\gamma_{m0} = 1.10$ (IS 800:2007, Table 5). Limit state design. - Effective span $L = 6.00$ m, bearing length of support $= 350$ mm. #### Loads and design forces The UDL of 20 kN/m (inclusive of self-weight) is the service load; factored $w_u = 1.5\times20 = 30$ kN/m. $L = 6$ m, wall (support) width 350 mm so $b_1 = 350$ mm: $M_u = 30\times36/8 = 135$ kN·m, $V_u = 90$ kN. Moment without self-weight $M = 135.0$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 135.0\times10^6\times1.1/250 = 594$ cm³. #### Trial section Try **ISMB 300** (IS 808): $D = 300$ mm, $b_f = 140$ mm, $t_w = 7.7$ mm, $t_f = 13.1$ mm, $R_1 = 14$ mm; $I_{zz} = 8990$ cm⁴; $Z_e = 599$ cm³; $Z_p \approx B t_f (D - t_f) + t_w (D - 2t_f)^2/4 = 670$ cm³; mass $46.0$ kg/m. $$ \begin{aligned} M_u &= \frac{wL^2}{8} = \frac{30.00\times6.00^2}{8} = 135.0\ \text{kN·m} \\ V_u &= \frac{wL}{2} = 90.0\ \text{kN} \end{aligned} $$ #### Section classification (Table 2) $\varepsilon = \sqrt{250/f_y} = 1.00$. Flange: $b/t_f = (140/2)/13.1 = 5.34 < 9.4\varepsilon = 9.4$. Web: $d/t_w = 31.9 < 84\varepsilon = 84$ (with $d = D - 2(t_f + R_1) = 245.8$ mm). The section is **plastic**, so $\beta_b = 1.0$. #### Moment capacity (cl. 8.2.1.2) The concrete slab restrains the compression flange along the full length, so lateral-torsional buckling is prevented. $$ \begin{aligned} M_d &= \frac{\beta_b Z_p f_y}{\gamma_{m0}} = \frac{670\times10^3\times250}{1.1\times10^6} = 152.4\ \text{kN·m} \\ 1.2 Z_e f_y/\gamma_{m0} &= 1.2\times599\times10^3\times250/(1.1\times10^6) = 163.4\ \text{kN·m} \end{aligned} $$ The smaller value governs: $M_d = 152.4$ kN·m. $$ M_d = 152.4\ \text{kN·m} > M_u = 135.0\ \text{kN·m}\quad \text{Safe} $$ #### Shear capacity (cl. 8.4) $$ \begin{aligned} A_v &= D t_w = 300\times7.7 = 2310\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{2310\times250}{\sqrt3\times1.1\times10^3} = 303.1\ \text{kN} \end{aligned} $$ $V_u = 90.0$ kN $< V_d$; also $V_u < 0.6V_d = 181.9$ kN, so the shear is low and the moment capacity is not reduced (cl. 9.2.1). Web shear buckling does not govern since $d/t_w = 31.9 < 67\varepsilon = 67$ (cl. 8.4.2.1). #### Deflection (Table 6) Deflection under unfactored service (imposed) loads, limit span/300 (floor and roof members, Table 6): $$ \delta = \frac{5 w L^4}{384 E I} = \frac{5\times20.00\times(6000)^4}{384\times2\times10^5\times8990\times10^4} = 18.77\ \text{mm} $$ $\delta = 18.77$ mm $< L/300 = 20.00$ mm. **OK.** #### Web buckling and bearing at support (cl. 8.7.3 and 8.7.4) Stiff bearing length $b_1 = 350$ mm. Clear web depth $d = 245.8$ mm; dispersion of load at 45° to mid-depth at an end support $n_1 = d/2 = 122.9$ mm. Effective length $0.7d$ (cl. 8.7.1.5), $\lambda = 0.7d/(t_w/\sqrt{12}) = 77.4$; curve c gives $f_{cd} = 140.4$ N/mm². $$ \begin{aligned} F_{cdw} &= (b_1 + n_1) t_w f_{cd} = (350 + 122.9)\times7.7\times140.4/10^3 = 511.2\ \text{kN} \\ n_2 &= 2.5(t_f + R_1) = 67.8\ \text{mm} \\ F_w &= \frac{(b_1 + n_2) t_w f_y}{\gamma_{m0}} = \frac{(350 + 67.8)\times7.7\times250}{1.1\times10^3} = 731.1\ \text{kN} \end{aligned} $$ Support reaction $R = 90.0$ kN $< F_{cdw}$ and $< F_w$. No web stiffener is needed. #### Final design Provide **ISMB 300** (46.0 kg/m): $M_d = 152.4$ kN·m $> M_u = 135.0$ kN·m, $V_d = 303$ kN $> V_u = 90.0$ kN, $\delta = 18.8$ mm $< 20.0$ mm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A 6 m span carries 20 kN/m service load including self-weight. What are the factored Mu and Vu? A. Mu = 90 kN.m, Vu = 135 kN B. Mu = 90 kN.m, Vu = 60 kN C. Mu = 135 kN.m, Vu = 90 kN D. Mu = 135 kN.m, Vu = 60 kN Answer: C. w_u = 1.5 x 20 = 30 kN/m; Mu = 30 x 36/8 = 135 kN.m and Vu = 30 x 6/2 = 90 kN. (Key point: Design forces) 2. How does the concrete slab affect the moment capacity check? A. It adds composite strength to Md B. It restrains the compression flange along its length, so LTB is prevented and Md is the plastic capacity C. It reduces shear on the beam D. It increases the span Answer: B. A flange restrained by the slab cannot buckle laterally, so Md = Zp fy/gamma_m0 = 152.4 kN.m (limited by 1.2 Ze fy/gamma_m0). (Key point: Role of the slab) 3. Deflection is 18.77 mm against a limit of L/300. What is the limit and conclusion? A. 20 mm; OK but close to the limit B. 24 mm; OK with wide margin C. 20 mm; fails D. 12 mm; fails Answer: A. L/300 = 6000/300 = 20 mm, and 18.77 mm is just inside it. (Key point: Deflection governs) 4. How does the 350 mm wall width enter the web checks? A. It is the span of the beam B. It is the stiff bearing length b1 = 350 mm used in Fcdw and Fw C. It is added to the web thickness D. It is ignored Answer: B. The support width gives the stiff bearing b1, so Fcdw = (b1 + n1) tw fcd = 511.2 kN and Fw = 731.1 kN, both above R = 90 kN. 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