--- title: "Safe UDL on ISLB 600 beam, restrained vs unrestrained" subject: "Design of Steel and Timber Structure" chapter: 7 key: "semester_6/design_of_steel_and_timber_structure:beam-moment-islb600" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/beam-moment-islb600 site: BCT Engineering --- # Safe UDL on ISLB 600 beam, restrained vs unrestrained Design of Steel and Timber Structure · Chapter 7: Flexure Members - Asked in: 2080 Chaitra, 3+5 marks - Video: [Numerical on design of Laterally Supported beam|Design of beams|Numerical](https://www.youtube.com/watch?v=3aByGMbfwRQ) — ISHA MISHRA - 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-moment-islb600 - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-7 ## Question An ISLB 600 is used as a simply supported beam over 7 m span. Determine the safe uniform load that the beam can carry in flexure when a) the compression flange of beam is restrained against lateral buckling. b) the compression flange of beam is not restrained against lateral buckling with beam fully restrained in torsion but both the flanges are free to warp at the ends. ## Short answer With compression flange restrained, M_d = 626 kN·m and safe factored UDL = 102.1 kN/m. With flange unrestrained (torsion restrained, warping free), lateral-torsional buckling reduces M_d to 227 kN·m, safe factored UDL = 37.0 kN/m. ## Answer **Approach.** IS 800:2007 cl. 8.2: design bending strength $M_d = \beta_b Z_p f_y/\gamma_{m0}$ for a laterally supported beam, and $M_d = \beta_b Z_p f_{bd}$ with $f_{bd} = \chi_{LT}f_y/\gamma_{m0}$ for an unrestrained beam. $f_y = 250$ N/mm² (Fe 410), $\gamma_{m0} = 1.10$. Safe load found from $M_d = wL^2/8$. #### Section data: ISLB 600 @ 99.5 kg/m (IS 808) $D = 600$ mm, $b_f = 210$ mm, $t_w = 10.5$ mm, $t_f = 15.5$ mm, $R_1 = 20$ mm, $I_{zz} = 72800$ cm⁴, $I_{yy} = 1820$ cm⁴, $Z_e = 2430$ cm³. Plastic modulus (from dimensions): $Z_p = b_ft_f(D - t_f) + t_w(D - 2t_f)^2/4 = 2752$ cm³. Classification: flange $b/t_f = 105/15.5 = 6.77 < 9.4$; web $d/t_w = 50.4 < 84$. The section is plastic ($\beta_b = 1$). #### (a) Compression flange restrained against lateral buckling $$ \begin{aligned} M_d &= \frac{Z_pf_y}{\gamma_{m0}} = \frac{2752\times10^3\times250}{1.1\times10^6} = 625.6\ \text{kN·m} \\ \text{Limit } 1.2Z_ef_y/\gamma_{m0} &= 662.7\ \text{kN·m} \\ M_d &= 625.6\ \text{kN·m} \end{aligned} $$ Safe factored UDL (including self-weight): $w = 8M_d/L^2 = 8\times625.6/7^2 = 102.1$ kN/m. Shear check: $V = wL/2 = 357.5$ kN $< V_d = D t_wf_y/(\sqrt3\gamma_{m0}) = 827$ kN, and below $0.6V_d$, so bending is not reduced. Service load $= 102.1/1.5 = 68.1$ kN/m. Deducting the self-weight ($99.5\times9.81/1000 = 0.98$ kN/m) the superimposed service load is $\approx 67.1$ kN/m. **Answer (a): $M_d = 626$ kN·m; safe factored load $\approx 102.1$ kN/m ($\approx 68.1$ kN/m unfactored, including self-weight).** #### (b) Not restrained against lateral buckling (torsion restrained, flanges free to warp) For simply supported ends with torsion restrained and warping free, $L_{LT} = 1.0L = 7000$ mm (Table 15). $$ \begin{aligned} I_t &= \frac{2b_ft_f^3 + (D-2t_f)t_w^3}{3} = 74.1\ \text{cm}^4 \\ h_f &= D - t_f = 584.5\ \text{mm},\quad I_w = \frac{I_yh_f^2}{4} = 155.4\times10^{10}\ \text{mm}^6 \\ M_{cr} &= \sqrt{\frac{\pi^2EI_y}{L_{LT}^2}\left(GI_t + \frac{\pi^2EI_w}{L_{LT}^2}\right)} = 296.1\ \text{kN·m} \end{aligned} $$ (with $G = 0.769\times10^5$ N/mm²). Rolled section, so $\alpha_{LT} = 0.21$ (Table 13): $$ \begin{aligned} \lambda_{LT} &= \sqrt{Z_pf_y/M_{cr}} = 1.524 \\ \phi_{LT} &= 0.5[1 + 0.21(\lambda_{LT} - 0.2) + \lambda_{LT}^2] = 1.801 \\ \chi_{LT} &= \frac{1}{\phi_{LT} + \sqrt{\phi_{LT}^2 - \lambda_{LT}^2}} = 0.362 \\ f_{bd} &= \chi_{LT}f_y/\gamma_{m0} = 82.3\ \text{N/mm}^2 \\ M_d &= Z_pf_{bd} = 226.7\ \text{kN·m} \end{aligned} $$ Safe factored UDL: $w = 8M_d/L^2 = 8\times226.7/49 = 37.0$ kN/m (service $\approx 24.7$ kN/m). **Answer (b): $M_d = 227$ kN·m; safe factored load $\approx 37.0$ kN/m ($\approx 24.7$ kN/m unfactored, including self-weight).** | Case | $M_d$ (kN·m) | Safe factored UDL (kN/m) | |---|---|---| | (a) Restrained | 625.6 | 102.1 | | (b) Unrestrained | 226.7 | 37.0 | Lack of lateral restraint reduces the strength to about 36% of the restrained value. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For case (a), the compression flange restrained, what is the design moment capacity formula for the plastic section? A. M_d = beta_b Z_p f_y / gamma_m0 B. M_d = Z_e f_y x gamma_m0 C. M_d = chi_LT Z_p f_y D. M_d = M_cr Answer: A. With full lateral restraint there is no LTB reduction, so M_d is the plastic moment divided by gamma_m0 (625.6 kN·m). (Key point: Restrained beam capacity) 2. How is the safe uniform load found from M_d for the simply supported beam? A. From M_d = wL^2/8, so w = 8M_d/L^2 B. From M_d = wL^2/2 C. From M_d = wL/4 D. From M_d = wL^2/12 Answer: A. The maximum moment of a simply supported beam under UDL is wL^2/8, so w = 8 x 625.6 / 49 = 102.1 kN/m. (Key point: Safe load from moment) 3. In case (b), what does the elastic critical moment M_cr depend on? A. Unbraced length, EI_y, torsional constant GI_t and warping constant EI_w B. Only the yield stress C. Only the web thickness D. Only the span and the load Answer: A. M_cr combines lateral bending, St Venant torsion and warping resistance over the effective length L_LT. (Key point: Elastic critical moment) 4. How does the unrestrained moment compare to the restrained one, and why? A. It is much lower (227 vs 626 kN·m) since chi_LT reduces the bending stress to f_bd B. It is the same, because the section is plastic C. It is higher, because the flange is free to warp D. It is slightly lower, about 95% of the restrained value Answer: A. Lateral-torsional buckling gives chi_LT = 0.362, so f_bd falls to 82.3 N/mm^2 and capacity drops to about 36% of the restrained case. 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