--- title: "Laterally supported I-section for 700 kNm" subject: "Design of Steel and Timber Structure" chapter: 7 key: "semester_6/design_of_steel_and_timber_structure:beam-restrained-700knm" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/beam-restrained-700knm site: BCT Engineering --- # Laterally supported I-section for 700 kNm Design of Steel and Timber Structure · Chapter 7: Flexure Members - Asked in: 2071 Magh, 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-restrained-700knm - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-7 ## Question Design a simply supported I section to support a moment of 700 kNm. The beam is laterally supported and grade of steel is Fe410. ## Short answer ISMB 600 provided for laterally supported beam with Mu=700 kNm. Plastic moment capacity Md=775 kNm > 700 kNm; shear capacity Vd=945 kN. Section classified plastic; no LTB check needed. Self-weight included in Mu. ## 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. - Span not given (not needed for bending design). #### Loads and design forces The factored moment is $M_u = 700$ kN·m (taken to include the beam's own weight). The beam is laterally supported, so the plastic moment capacity applies. No span, shear or load is given, so only bending is designed here, and the shear capacity is reported for reference. Moment without self-weight $M = 700.0$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 700.0\times10^6\times1.1/250 = 3080$ cm³. #### Trial section Try **ISMB 600** (IS 808): $D = 600$ mm, $b_f = 210$ mm, $t_w = 12.0$ mm, $t_f = 20.3$ mm, $R_1 = 20$ mm; $I_{zz} = 91800$ cm⁴; $Z_e = 3060$ cm³; $Z_p \approx B t_f (D - t_f) + t_w (D - 2t_f)^2/4 = 3410$ cm³; mass $123.0$ kg/m. #### Section classification (Table 2) $\varepsilon = \sqrt{250/f_y} = 1.00$. Flange: $b/t_f = (210/2)/20.3 = 5.17 < 9.4\varepsilon = 9.4$. Web: $d/t_w = 43.3 < 84\varepsilon = 84$ (with $d = D - 2(t_f + R_1) = 519.4$ 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 plastic moment capacity governs; lateral-torsional buckling is not checked. $$ \begin{aligned} M_d &= \frac{\beta_b Z_p f_y}{\gamma_{m0}} = \frac{3410\times10^3\times250}{1.1\times10^6} = 775.0\ \text{kN·m} \\ 1.2 Z_e f_y/\gamma_{m0} &= 1.2\times3060\times10^3\times250/(1.1\times10^6) = 834.5\ \text{kN·m} \end{aligned} $$ The smaller value governs: $M_d = 775.0$ kN·m. $$ M_d = 775.0\ \text{kN·m} > M_u = 700.0\ \text{kN·m}\quad \text{Safe} $$ #### Shear capacity (cl. 8.4) $$ \begin{aligned} A_v &= D t_w = 600\times12.0 = 7200\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{7200\times250}{\sqrt3\times1.1\times10^3} = 944.8\ \text{kN} \end{aligned} $$ The shear force is not given. The section can resist a shear of $V_d = 945$ kN; for $V \le 0.6V_d = 567$ kN the moment capacity is not reduced (cl. 9.2.1). The web does not buckle in shear since $d/t_w = 43.3 < 67\varepsilon = 67$. Web bearing and deflection depend on the span and support, which are not given. #### Final design Provide **ISMB 600** (123.0 kg/m): $M_d = 775.0$ kN·m $> M_u = 700.0$ kN·m, $V_d = 945$ kN. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the required plastic section modulus found for Mu = 700 kN.m, fy = 250, gamma_m0 = 1.10? A. Zp = M fy / gamma_m0 = 1590 cm^3 B. Zp = M gamma_m0 / fy = 3080 cm^3 C. Zp = M / fy = 2800 cm^3 D. Zp = 1.2 M gamma_m0 / fy = 3696 cm^3 Answer: B. From Md = Zp fy/gamma_m0 >= M, Zp >= 700e6 x 1.1 / 250 = 3080 cm^3. (Key point: Required plastic modulus) 2. For ISMB 600, b/tf = 5.17 and d/tw = 43.3 with epsilon = 1.0. What is the classification and beta_b? A. Plastic, beta_b = 1.0 B. Compact, beta_b = 1.2 C. Semi-compact, beta_b = Ze/Zp D. Slender, beta_b = 0.8 Answer: A. 5.17 < 9.4 and 43.3 < 84, so the section is plastic and beta_b = 1.0. (Key point: Section classification) 3. Plastic capacity is 775.0 kN.m and 1.2 Ze fy/gamma_m0 is 834.5 kN.m. Which is Md? A. 834.5 kN.m, the larger B. The average of the two C. Mu = 700 kN.m D. 775.0 kN.m, the smaller Answer: D. The smaller value governs (it limits the capacity to avoid irreversible deformation at service), so Md = 775.0 kN.m > 700 kN.m. (Key point: Governing moment capacity) 4. Why is lateral-torsional buckling not checked here? A. The beam is laterally supported, so the plastic moment capacity applies B. The span is not given so it cannot be checked C. Because shear governs D. Because the section is slender Answer: A. Lateral support to the compression flange prevents LTB, so Md = beta_b Zp fy/gamma_m0. 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