--- title: "Beam 6 m span with 100 kN secondary beam load" subject: "Design of Steel and Timber Structure" chapter: 7 key: "semester_6/design_of_steel_and_timber_structure:beam-6m-20knm-100kn-midspan" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/beam-6m-20knm-100kn-midspan site: BCT Engineering --- # Beam 6 m span with 100 kN secondary beam load Design of Steel and Timber Structure · Chapter 7: Flexure Members - Asked in: 2076 Bhadra, 10 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-20knm-100kn-midspan - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-7 ## Question Design a simply supported beam having unsupported span of 6m, support width 250 mm. Beam is subjected to imposed load (including self-weight) 20 kN/m and 100 kN from secondary beam at mid-span of beam. Ends and mid of beam is laterally restrained. ## Short answer ISMB 500. Factored moment 360 kN·m < lateral-torsional buckling capacity 375.3 kN·m (unrestrained length 3 m); shear 165 kN < 669.2 kN; deflection 8.71 mm < L/300 = 20 mm. ## 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 $= 250$ mm. #### Loads and design forces The imposed load of $20$ kN/m (including self-weight) and the 100 kN load from the secondary beam are taken as service loads. Factored: $w_u = 1.5\times20 = 30$ kN/m, $W_u = 1.5\times100 = 150$ kN. Self-weight is already included, so it is not added again. Span $L = 6$ m, support width 250 mm. The beam is laterally restrained at both ends and at mid-span, so the unrestrained length is $L_{LT} = 3.0$ m (simply supported segment, torsionally restrained, warping free, $K = 1.0$, Table 15). The two segments are identical, so one is checked using the maximum (mid-span) moment with no allowance for the moment gradient, which is conservative. Design moment (self-weight included in the given load) $M = 360.0$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 360.0\times10^6\times1.1/250 = 1584$ 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. $$ \begin{aligned} M_u &= \frac{wL^2}{8} + \frac{WL}{4} = \frac{30.00\times6.00^2}{8} + \frac{150.0\times6.00}{4} = 360.0\ \text{kN·m} \\ V_u &= \frac{wL}{2} + \frac{W}{2} = 165.0\ \text{kN} \end{aligned} $$ #### 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. 8.2.2) The compression flange is not continuously restrained, so lateral-torsional buckling governs (cl. 8.2.2). Effective length for LTB $L_{LT} = 3.00$ m. $I_y = 1370$ cm⁴; $I_t = (2b_ft_f^3 + d t_w^3)/3 = 77.5$ cm⁴; $I_w = I_y h_f^2/4 = 80\times10^{10}$ mm⁶ ($h_f = D - t_f = 482.8$ mm). $$ \begin{aligned} M_{cr} &= \sqrt{\frac{\pi^2EI_y}{L_{LT}^2}\left(GI_t + \frac{\pi^2EI_w}{L_{LT}^2}\right)} = 839.8\ \text{kN·m} \\ \lambda_{LT} &= \sqrt{\frac{\beta_b Z_p f_y}{M_{cr}}} = 0.781 \\ \phi_{LT} &= 0.5[1 + 0.21(\lambda_{LT} - 0.2) + \lambda_{LT}^2] = 0.866 \\ \chi_{LT} &= \frac{1}{\phi_{LT} + \sqrt{\phi_{LT}^2 - \lambda_{LT}^2}} = 0.807 \\ f_{bd} &= \chi_{LT} f_y/\gamma_{m0} = 183.3\ \text{N/mm}^2 \\ M_d &= \beta_b Z_p f_{bd} = 375.3\ \text{kN·m} \end{aligned} $$ $$ M_d = 375.3\ \text{kN·m} > M_u = 360.0\ \text{kN·m}\quad \text{Safe} $$ #### Shear capacity (cl. 8.4) $$ \begin{aligned} A_v &= D t_w = 500\times10.2 = 5100\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{5100\times250}{\sqrt3\times1.1\times10^3} = 669.2\ \text{kN} \end{aligned} $$ $V_u = 165.0$ 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 the unfactored service loads, limit span/300 (floor and roof members, Table 6): $$ \delta = \frac{5 w L^4}{384 E I} + \frac{W L^3}{48 E I} = \frac{5\times20.00\times(6000)^4}{384\times2\times10^5\times45200\times10^4} + \frac{100000\times(6000)^3}{48\times2\times10^5\times45200\times10^4} = 8.71\ \text{mm} $$ $\delta = 8.71$ mm $< L/300 = 20.00$ mm. **OK.** #### Web buckling and bearing at support (cl. 8.7.3 and 8.7.4) Stiff bearing length $b_1 = 250$ mm. Clear web depth $d = 431.6$ mm; dispersion of load at 45° to mid-depth at an end support $n_1 = d/2 = 215.8$ mm. Effective length $0.7d$ (cl. 8.7.1.5), $\lambda = 0.7d/(t_w/\sqrt{12}) = 102.6$; curve c gives $f_{cd} = 103.6$ N/mm². $$ \begin{aligned} F_{cdw} &= (b_1 + n_1) t_w f_{cd} = (250 + 215.8)\times10.2\times103.6/10^3 = 492.3\ \text{kN} \\ n_2 &= 2.5(t_f + R_1) = 85.5\ \text{mm} \\ F_w &= \frac{(b_1 + n_2) t_w f_y}{\gamma_{m0}} = \frac{(250 + 85.5)\times10.2\times250}{1.1\times10^3} = 777.7\ \text{kN} \end{aligned} $$ Support reaction $R = 165.0$ kN $< F_{cdw}$ and $< F_w$. No web stiffener is needed. #### Final design Provide **ISMB 500** (86.9 kg/m): $M_d = 375.3$ kN·m $> M_u = 360.0$ kN·m, $V_d = 669$ kN $> V_u = 165.0$ kN, $\delta = 8.7$ mm $< 20.0$ mm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The 20 kN/m (including self-weight) and 100 kN are service loads. How are the design loads obtained? A. Multiply by 1.5 and add the beam self-weight again B. Use as given, they are already factored C. Multiply by 1.5 and do not add self-weight again D. Multiply by 1.1 (gamma_m0) Answer: C. wu = 1.5x20 = 30 kN/m and Wu = 1.5x100 = 150 kN. Self-weight is already included in the 20 kN/m. (Key point: Treatment of given loads) 2. What is the design moment Mu for the 6 m simply supported beam? A. 135 kN·m B. 225 kN·m C. 495 kN·m D. 360 kN·m Answer: D. Mu = wL^2/8 + WL/4 = 30x36/8 + 150x6/4 = 135 + 225 = 360 kN·m. (Key point: Design moment) 3. Why is L_LT taken as 3.0 m rather than 6.0 m? A. Because the beam is laterally restrained at mid-span as well as at the ends B. Because the span is halved by the point load C. Because the support width is 250 mm D. Because the section is plastic Answer: A. Lateral restraint at mid-span splits the beam into two 3 m segments, so the unrestrained length is 3 m. (Key point: Effective length for LTB) 4. How is Md found once lambda_LT = 0.781 is known? A. Md = Ze fy/gamma_m0 with no reduction B. chi_LT is found from phi_LT, then fbd = chi_LT fy/gamma_m0 and Md = betab Zp fbd C. Md = Mcr directly D. Md = Zp fy x chi_LT without gamma_m0 and Zp Answer: B. phi_LT gives chi_LT = 0.807, fbd = 0.807x250/1.1 = 183.3 N/mm2 and Md = Zp fbd = 375.3 kN·m, which exceeds 360 kN·m. 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