--- title: "Beam supporting RCC slab, DL 10 kN/m, IL 30 kN" subject: "Design of Steel and Timber Structure" chapter: 7 key: "semester_6/design_of_steel_and_timber_structure:beam-6m-rcc-slab-10dl-30il" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/beam-6m-rcc-slab-10dl-30il site: BCT Engineering --- # Beam supporting RCC slab, DL 10 kN/m, IL 30 kN Design of Steel and Timber Structure · Chapter 7: Flexure Members - Asked in: 2075 Bhadra, 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-6m-rcc-slab-10dl-30il - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-7 ## Question A simply supported beam of span 6 m supports a reinforcement concrete slab. The compression flange of the beam is restrained due to its connection with the slab. The beam is subjected to a dead load of 10 kN/m and imposed load of 30 kN. Design the beam. Assume the beam is sufficiently stiff against bearing. ## Short answer ISMB 450. Factored moment 274.8 kN·m < capacity 348.7 kN·m; shear 183.2 kN < 555 kN; deflection 8.33 mm < 20 mm. Lateral buckling is prevented by the continuous slab restraint. ## 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. #### Loads and design forces Span $L = 6$ m. The "30 kN" imposed load is read as **30 kN/m** (a UDL, like the dead load of 10 kN/m); dead load is taken to exclude the beam's own weight. Factored load $w_u = 1.5(10 + 30) = 60$ kN/m, plus factored self-weight. The beam is "sufficiently stiff against bearing", so the web buckling and bearing checks are not required. Moment without self-weight $M = 270.0$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 270.0\times10^6\times1.1/250 = 1188$ cm³ (self-weight adds a little). #### Trial section Try **ISMB 450** (IS 808): $D = 450$ mm, $b_f = 150$ mm, $t_w = 9.4$ mm, $t_f = 17.4$ mm, $R_1 = 15$ mm; $I_{zz} = 30400$ cm⁴; $Z_e = 1350$ cm³; $Z_p \approx B t_f (D - t_f) + t_w (D - 2t_f)^2/4 = 1534$ cm³; mass $72.4$ kg/m. Self-weight $= 72.4\times9.81/1000 = 0.710$ kN/m, factored $= 1.5\times0.710 = 1.065$ kN/m. Total factored UDL $w = 60.00 + 1.07 = 61.07$ kN/m. $$ \begin{aligned} M_u &= \frac{wL^2}{8} = \frac{61.07\times6.00^2}{8} = 274.8\ \text{kN·m} \\ V_u &= \frac{wL}{2} = 183.2\ \text{kN} \end{aligned} $$ #### Section classification (Table 2) $\varepsilon = \sqrt{250/f_y} = 1.00$. Flange: $b/t_f = (150/2)/17.4 = 4.31 < 9.4\varepsilon = 9.4$. Web: $d/t_w = 41.0 < 84\varepsilon = 84$ (with $d = D - 2(t_f + R_1) = 385.2$ mm). The section is **plastic**, so $\beta_b = 1.0$. #### Moment capacity (cl. 8.2.1.2) The compression flange is restrained by the RCC slab along its full length, so lateral-torsional buckling is prevented. $$ \begin{aligned} M_d &= \frac{\beta_b Z_p f_y}{\gamma_{m0}} = \frac{1534\times10^3\times250}{1.1\times10^6} = 348.7\ \text{kN·m} \\ 1.2 Z_e f_y/\gamma_{m0} &= 1.2\times1350\times10^3\times250/(1.1\times10^6) = 368.2\ \text{kN·m} \end{aligned} $$ The smaller value governs: $M_d = 348.7$ kN·m. $$ M_d = 348.7\ \text{kN·m} > M_u = 274.8\ \text{kN·m}\quad \text{Safe} $$ #### Shear capacity (cl. 8.4) $$ \begin{aligned} A_v &= D t_w = 450\times9.4 = 4230\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{4230\times250}{\sqrt3\times1.1\times10^3} = 555.0\ \text{kN} \end{aligned} $$ $V_u = 183.2$ kN $< V_d$; also $V_u < 0.6V_d = 333.0$ kN, so the shear is low and the moment capacity is not reduced (cl. 9.2.1). Web shear buckling does not govern since $d/t_w = 41.0 < 67\varepsilon = 67$ (cl. 8.4.2.1). #### Deflection (Table 6) Deflection under unfactored service (imposed) loads, limit span/300 (floor and roof members, Table 6): $$ \delta = \frac{5 w L^4}{384 E I} = \frac{5\times30.00\times(6000)^4}{384\times2\times10^5\times30400\times10^4} = 8.33\ \text{mm} $$ $\delta = 8.33$ mm $< L/300 = 20.00$ mm. **OK.** Web bearing/buckling at supports is not checked because the question states the beam is sufficiently stiff against bearing. #### Final design Provide **ISMB 450** (72.4 kg/m): $M_d = 348.7$ kN·m $> M_u = 274.8$ kN·m, $V_d = 555$ kN $> V_u = 183.2$ kN, $\delta = 8.3$ mm $< 20.0$ mm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With dead load 10 kN/m and the imposed load read as 30 kN/m, what is the factored UDL before self-weight? A. 40 kN/m B. 50 kN/m C. 60 kN/m D. 75 kN/m Answer: C. wu = 1.5(10 + 30) = 60 kN/m; 40 kN/m is the unfactored total. (Key point: Factored load) 2. Mu without self-weight is 270 kN·m. Why is ISMB 450 (Zp = 1534 cm3) a suitable trial? A. Because Zp is above the required about 1188 cm3, leaving room for self-weight B. Because its depth is 450 mm C. Because Ze equals the required Zp D. Because it is the lightest ISMB available Answer: A. Zp,req = M gamma_m0/fy = 270x10^6x1.1/250 = 1188 cm3, and 1534 cm3 covers this plus self-weight. (Key point: Trial section selection) 3. The slab restrains the compression flange. Md is then: A. Reduced by chi_LT B. Equal to Mcr C. Zp fy/gamma_m0 = 348.7 kN·m, limited to 1.2 Ze fy/gamma_m0 D. Ze fy/gamma_m0 only Answer: C. LTB is prevented, so Md = betab Zp fy/gamma_m0 = 348.7 kN·m, which is below 1.2Ze fy/gamma_m0 = 368.2, and exceeds Mu = 274.8. (Key point: Moment capacity with slab restraint) 4. Vu = 183.2 kN and Vd = 555 kN. What follows? A. Moment capacity must be reduced since Vu > 0.5Vd B. Web must be stiffened C. Shear fails because Vu > 0.6Vd D. Shear is low (Vu < 0.6Vd = 333 kN), so Md is not reduced Answer: D. Since Vu is less than 0.6Vd, shear is low and the moment capacity is not reduced (cl. 9.2.1). (Key point: Shear check) ## Close variants set in other exams - [Beam supporting RCC slab, DL 25, IL 20 kN/m](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-rcc-slab-6m-25dl-20il) — Design of Steel and Timber Structure, chapter 7, asked 2× ## Related in Design of Steel and Timber Structure - [Beam 6 m span with 100 kN secondary beam load](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-6m-20knm-100kn-midspan) — Design of Steel and Timber Structure, chapter 7 - [Simply supported beam 6 m, 60 kN/m](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-6m-60knm-udl) — Design of Steel and Timber Structure, chapter 7 ## Related in other subjects - [Simply supported I-beam: stress at flange-web junction](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ai-beam-flange-web-junction-stress) — Strength of Materials, chapter 7 - [Design simply supported beam, 6 m span](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Adesign-simply-supported-beam-6-m) — Design of RCC Structure, chapter 4 - [Simply supported T-beam: max bending stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3At-beam-6m-100kn-10kn) — Strength of Materials, chapter 7 - [Mid-span deflection of beam, 10 kN load](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ass-beam-deflection-mid-span-4m) — Strength of Materials, chapter 7