--- title: "Simply supported beam 6 m, 60 kN/m" subject: "Design of Steel and Timber Structure" chapter: 7 key: "semester_6/design_of_steel_and_timber_structure:beam-6m-60knm-udl" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/beam-6m-60knm-udl site: BCT Engineering --- # Simply supported beam 6 m, 60 kN/m Design of Steel and Timber Structure · Chapter 7: Flexure Members - Asked in: 2074 Bhadra, 14 marks - Video: [How To Design a Steel Beam For Beginners: Hand Calculation & Software](https://www.youtube.com/watch?v=ySjLY_idYCo) — BEng Hielscher - 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-60knm-udl - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-7 ## Question Design a simply supported beam with an effective span of 6m for bending, shear and lateral stability. Beam carries a uniformly distributed load of 60 KN/m inclusive of self-weight. The beam is laterally supported. ## Short answer Factored UDL 90 kN/m yields design moment 405 kN·m and shear 270 kN. Provide ISMB 500; its plastic moment capacity 465.4 kN·m, shear capacity 669.2 kN. Deflection 11.2 mm < span/300 (20 mm). Lateral support ensures stability. ## 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 The UDL of 60 kN/m (inclusive of self-weight) is taken as the **service load**; factored load $w_u = 1.5\times60 = 90$ kN/m. Span $L = 6$ m, so $M_u = 90\times6^2/8 = 405$ kN·m and $V_u = 90\times6/2 = 270$ kN. The beam is laterally supported, so bending, shear and deflection are checked; bearing at the supports is not asked and is not checked (the support width is not given). For deflection, the whole service load is conservatively treated as imposed. Self-weight is already included, so $M_u = 405.0$ kN·m and $Z_{p,req} \approx M_u\gamma_{m0}/f_y = 405.0\times10^6\times1.1/250 = 1782$ cm³. #### Trial section Try **ISMB 500** (IS 808): $D = 500$ mm, $b_f = 180$ mm, $t_w = 10.2$ mm, $t_f = 17.2$ mm, $R_1 = 17$ mm; $I_{zz} = 45200$ cm⁴; $Z_e = 1810$ cm³; $Z_p \approx B t_f (D - t_f) + t_w (D - 2t_f)^2/4 = 2048$ cm³; mass $86.9$ kg/m. $$ \begin{aligned} M_u &= \frac{wL^2}{8} = \frac{90.00\times6.00^2}{8} = 405.0\ \text{kN·m} \\ V_u &= \frac{wL}{2} = 270.0\ \text{kN} \end{aligned} $$ #### Section classification (Table 2) $\varepsilon = \sqrt{250/f_y} = 1.00$. Flange: $b/t_f = (180/2)/17.2 = 5.23 < 9.4\varepsilon = 9.4$. Web: $d/t_w = 42.3 < 84\varepsilon = 84$ (with $d = D - 2(t_f + R_1) = 431.6$ mm). The section is **plastic**, so $\beta_b = 1.0$. #### Moment capacity (cl. 8.2.1.2) The beam is laterally supported (given), so the compression flange cannot buckle sideways and lateral stability is satisfied; $M_d$ is the plastic moment capacity. $$ \begin{aligned} M_d &= \frac{\beta_b Z_p f_y}{\gamma_{m0}} = \frac{2048\times10^3\times250}{1.1\times10^6} = 465.4\ \text{kN·m} \\ 1.2 Z_e f_y/\gamma_{m0} &= 1.2\times1810\times10^3\times250/(1.1\times10^6) = 493.6\ \text{kN·m} \end{aligned} $$ The smaller value governs: $M_d = 465.4$ kN·m. $$ M_d = 465.4\ \text{kN·m} > M_u = 405.0\ \text{kN·m}\quad \text{Safe} $$ #### Shear capacity (cl. 8.4) $$ \begin{aligned} A_v &= D t_w = 500\times10.2 = 5100\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{5100\times250}{\sqrt3\times1.1\times10^3} = 669.2\ \text{kN} \end{aligned} $$ $V_u = 270.0$ kN $< V_d$; also $V_u < 0.6V_d = 401.5$ kN, so the shear is low and the moment capacity is not reduced (cl. 9.2.1). Web shear buckling does not govern since $d/t_w = 42.3 < 67\varepsilon = 67$ (cl. 8.4.2.1). #### Deflection (Table 6) Deflection under unfactored service (imposed) loads, limit span/300 (floor and roof members, Table 6): $$ \delta = \frac{5 w L^4}{384 E I} = \frac{5\times60.00\times(6000)^4}{384\times2\times10^5\times45200\times10^4} = 11.20\ \text{mm} $$ $\delta = 11.20$ mm $< L/300 = 20.00$ mm. **OK.** Bearing and web buckling at the supports are not checked because the support width is not given. #### Final design Provide **ISMB 500** (86.9 kg/m): $M_d = 465.4$ kN·m $> M_u = 405.0$ kN·m, $V_d = 669$ kN $> V_u = 270.0$ kN, $\delta = 11.2$ mm $< 20.0$ mm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The 60 kN/m includes self-weight and is taken as a service load. What are wu and Mu? A. 60 kN/m and 270 kN·m B. 90 kN/m and 405 kN·m C. 90 kN/m and 270 kN·m D. 66 kN/m and 297 kN·m Answer: B. wu = 1.5x60 = 90 kN/m; Mu = 90x6^2/8 = 405 kN·m; Vu = 270 kN. (Key point: Design loads) 2. How is lateral stability satisfied? A. By limiting deflection to L/300 B. By using a stiff web C. By checking web bearing D. The beam is laterally supported, so LTB cannot occur and Md is the plastic capacity Answer: D. A laterally supported compression flange cannot buckle sideways, so Md = Zp fy/gamma_m0 = 465.4 kN·m > 405 kN·m. (Key point: Lateral stability) 3. Vu = 270 kN, Vd = 669.2 kN. Conclusion? A. Vu < 0.6Vd = 401.5 kN, so shear is low and Md is not reduced B. Md must be reduced because Vu > 0.4Vd C. Shear fails D. Stiffeners are required Answer: A. Vu is below 0.6Vd, so the shear is low (cl. 9.2.1). (Key point: Shear check) 4. How is deflection checked here? A. Using factored load 90 kN/m B. Using only the part of 60 kN/m that is live C. Using unfactored 60 kN/m (conservatively all imposed): 11.2 mm < L/300 = 20 mm D. Using no check since the beam is laterally supported Answer: C. delta = 5wL^4/(384EI) with w = 60 kN/m gives 11.2 mm, within L/300 = 20 mm. 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