--- title: "Beam 6 m span, 46 kN/m, Fe 410" subject: "Design of Steel and Timber Structure" chapter: 7 key: "semester_6/design_of_steel_and_timber_structure:beam-udl-46knm-6m" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/beam-udl-46knm-6m site: BCT Engineering --- # Beam 6 m span, 46 kN/m, Fe 410 Design of Steel and Timber Structure · Chapter 7: Flexure Members - Asked in: 2070 Magh, 14 marks - Video: [#Design of simple supported beam#Design of beams#Design of steel structures#ISMB300#](https://www.youtube.com/watch?v=MmDCXVVc7pY) — CIVIL ENGINEERING COURSE Theory & Practical - 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-udl-46knm-6m - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-7 ## Question A simply supported steel beam of 6 m effective span carries a total uniformly distributed load of 46 KN/m (inclusive of self-weight). Design the beam (Fe 410 steel) if the compression flange is restrained throughout the span against lateral bending. Apply all the necessary checks. ## Short answer ISMB 450 designed for 6 m span with 46 kN/m UDL (incl. self-weight). Factored Mu=310.5 kNm, Vu=207 kN. Md=348.7 kNm > Mu; Vd=555 kN > Vu. Deflection 12.77 mm < L/300 (20 mm). Web buckling and bearing safe with 200 mm stiff bearing. ## 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 $= 200$ mm. #### Loads and design forces The UDL of 46 kN/m (inclusive of self-weight) is the service load; factored $w_u = 1.5\times46 = 69$ kN/m. $L = 6$ m: $M_u = 69\times36/8 = 310.5$ kN·m, $V_u = 207$ kN. The support width is not given, so a stiff bearing of **200 mm is assumed** for the web checks. For deflection the whole 46 kN/m is conservatively taken as imposed. Moment without self-weight $M = 310.5$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 310.5\times10^6\times1.1/250 = 1366$ cm³. #### 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. $$ \begin{aligned} M_u &= \frac{wL^2}{8} = \frac{69.00\times6.00^2}{8} = 310.5\ \text{kN·m} \\ V_u &= \frac{wL}{2} = 207.0\ \text{kN} \end{aligned} $$ #### 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 throughout the span (given), so lateral-torsional buckling is prevented and $M_d$ is the plastic moment capacity. $$ \begin{aligned} M_d &= \frac{\beta_b Z_p f_y}{\gamma_{m0}} = \frac{1534\times10^3\times250}{1.1\times10^6} = 348.7\ \text{kN·m} \\ 1.2 Z_e f_y/\gamma_{m0} &= 1.2\times1350\times10^3\times250/(1.1\times10^6) = 368.2\ \text{kN·m} \end{aligned} $$ The smaller value governs: $M_d = 348.7$ kN·m. $$ M_d = 348.7\ \text{kN·m} > M_u = 310.5\ \text{kN·m}\quad \text{Safe} $$ #### Shear capacity (cl. 8.4) $$ \begin{aligned} A_v &= D t_w = 450\times9.4 = 4230\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{4230\times250}{\sqrt3\times1.1\times10^3} = 555.0\ \text{kN} \end{aligned} $$ $V_u = 207.0$ 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): $$ \delta = \frac{5 w L^4}{384 E I} = \frac{5\times46.00\times(6000)^4}{384\times2\times10^5\times30400\times10^4} = 12.77\ \text{mm} $$ $\delta = 12.77$ mm $< L/300 = 20.00$ mm. **OK.** #### Web buckling and bearing at support (cl. 8.7.3 and 8.7.4) Stiff bearing length $b_1 = 200$ mm. Clear web depth $d = 385.2$ mm; dispersion of load at 45° to mid-depth at an end support $n_1 = d/2 = 192.6$ mm. Effective length $0.7d$ (cl. 8.7.1.5), $\lambda = 0.7d/(t_w/\sqrt{12}) = 99.4$; curve c gives $f_{cd} = 107.9$ N/mm². $$ \begin{aligned} F_{cdw} &= (b_1 + n_1) t_w f_{cd} = (200 + 192.6)\times9.4\times107.9/10^3 = 398.1\ \text{kN} \\ n_2 &= 2.5(t_f + R_1) = 81.0\ \text{mm} \\ F_w &= \frac{(b_1 + n_2) t_w f_y}{\gamma_{m0}} = \frac{(200 + 81.0)\times9.4\times250}{1.1\times10^3} = 600.3\ \text{kN} \end{aligned} $$ Support reaction $R = 207.0$ kN $< F_{cdw}$ and $< F_w$. No web stiffener is needed. #### Final design Provide **ISMB 450** (72.4 kg/m): $M_d = 348.7$ kN·m $> M_u = 310.5$ kN·m, $V_d = 555$ kN $> V_u = 207.0$ kN, $\delta = 12.8$ mm $< 20.0$ mm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The 46 kN/m UDL (including self-weight) is a service load. What are Mu and Vu over L = 6 m? A. Mu = 207 kN.m, Vu = 310.5 kN B. Mu = 310.5 kN.m, Vu = 207 kN C. Mu = 207 kN.m, Vu = 138 kN D. Mu = 276 kN.m, Vu = 138 kN Answer: B. w_u = 1.5 x 46 = 69 kN/m; Mu = wL^2/8 = 310.5 kN.m and Vu = wL/2 = 207 kN. (Key point: Factored load and forces) 2. Because the compression flange is restrained throughout, Md for ISMB 450 is: A. Governed by Mcr of LTB B. Reduced by chi_LT to about 250 kN.m C. The smaller of beta_b Zp fy/gamma_m0 (348.7) and 1.2 Ze fy/gamma_m0 (368.2), i.e. 348.7 kN.m D. The larger of the two, 368.2 kN.m Answer: C. No LTB occurs, so the plastic capacity governs, limited by 1.2 Ze fy/gamma_m0; the smaller value 348.7 kN.m exceeds Mu = 310.5. (Key point: Moment capacity with restrained flange) 3. What is the deflection limit and the result for this floor beam? A. L/300 = 20 mm; actual 12.77 mm is OK B. L/250 = 24 mm; actual 12.77 mm is OK C. L/300 = 20 mm; actual 12.77 mm fails D. L/180 = 33 mm; actual 20 mm is OK Answer: A. Table 6 gives span/300 for floor members; delta = 5wL^4/(384EI) with unfactored 46 kN/m gives 12.77 mm < 20 mm. (Key point: Deflection check) 4. At the support with b1 = 200 mm, what do the web checks show for R = 207 kN? A. R > Fcdw = 398.1 kN, stiffener required B. R < Fcdw = 398.1 kN and R < Fw = 600.3 kN, so no stiffener is needed C. R < Fw but R > Fcdw D. Only Fw is checked, not Fcdw Answer: B. Both web buckling (Fcdw) and web bearing (Fw) capacities exceed the 207 kN reaction. 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