--- title: "Unsupported beam, M=470 kNm, V=180 kN" subject: "Design of Steel and Timber Structure" chapter: 7 key: "semester_6/design_of_steel_and_timber_structure:beam-unsupported-3-5m-470knm" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/beam-unsupported-3-5m-470knm site: BCT Engineering --- # Unsupported beam, M=470 kNm, V=180 kN Design of Steel and Timber Structure · Chapter 7: Flexure Members - Asked in: 2072 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-unsupported-3-5m-470knm - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-7 ## Question Design a simply supported beam of span 3.5 m subjected to a factored bending moment of 470 KN-m and factored shear of 180 KN. The beam is laterally unsupported. Steel grade of Fe 410. Check for web buckling, web crippling and maximum deflection is required. ## Short answer ISMB 600 selected for 3.5 m laterally unsupported span. Moment capacity with LTB is 627.8 kNm > 470 kNm; shear capacity 944.8 kN > 180 kN. Deflection 2.18 mm < L/300 (11.67 mm). Web buckling and crippling checks satisfied with 100 mm 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 $L = 3.50$ m, bearing length at the support $= 100$ mm (assumed). #### Loads and design forces Given: factored $M_u = 470$ kN·m and factored $V_u = 180$ kN, span $L = 3.5$ m, beam laterally **unsupported** (so $L_{LT} = 3.5$ m, simply supported with torsional restraint and warping free). Self-weight is taken as already included in $M_u$ and $V_u$. The bearing length is not given; **assume** a stiff bearing $b_1 = 100$ mm and reaction $R = V_u = 180$ kN for the web checks. For deflection, the service moment is $470/1.5 = 313$ kN·m, treated as coming from a UDL. Required $Z_{p} \approx M_u\gamma_{m0}/f_y = 470.0\times10^6\times1.1/250 = 2068$ cm³ (before allowing for lateral-torsional buckling). #### 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 compression flange is not laterally restrained, so lateral-torsional buckling governs (cl. 8.2.2). Effective length for LTB $L_{LT} = 3.50$ m. $I_y = 2650$ cm⁴; $I_t = [2b_ft_f^3 + (D - 2t_f) t_w^3]/3 = 149.3$ cm⁴; $I_w = I_y h_f^2/4 = 223\times10^{10}$ mm⁶ ($h_f = D - t_f = 579.7$ mm). $$ \begin{aligned} M_{cr} &= \sqrt{\frac{\pi^2EI_y}{L_{LT}^2}\left(GI_t + \frac{\pi^2EI_w}{L_{LT}^2}\right)} = 1422.1\ \text{kN·m} \\ \lambda_{LT} &= \sqrt{\frac{\beta_b Z_p f_y}{M_{cr}}} = 0.774 \\ \phi_{LT} &= 0.5[1 + 0.21(\lambda_{LT} - 0.2) + \lambda_{LT}^2] = 0.860 \\ \chi_{LT} &= \frac{1}{\phi_{LT} + \sqrt{\phi_{LT}^2 - \lambda_{LT}^2}} = 0.810 \\ f_{bd} &= \chi_{LT} f_y/\gamma_{m0} = 184.1\ \text{N/mm}^2 \\ M_d &= \beta_b Z_p f_{bd} = 627.8\ \text{kN·m} \end{aligned} $$ $$ M_d = 627.8\ \text{kN·m} > M_u = 470.0\ \text{kN·m}\quad \text{Safe} $$ #### Shear capacity (cl. 8.4) $$ \begin{aligned} A_v &= D t_w = 600\times12.0 = 7200\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{7200\times250}{\sqrt3\times1.1\times10^3} = 944.8\ \text{kN} \end{aligned} $$ $V_u = 180.0$ kN $< V_d$; also $V_u < 0.6V_d = 566.9$ kN, so the shear is low and the moment capacity is not reduced (cl. 9.2.1). Web shear buckling does not govern since $d/t_w = 43.3 < 67\varepsilon = 67$ (cl. 8.4.2.1). #### Deflection (Table 6) The load is not given, so the deflection is estimated from the moment, taking the service moment $= M_u/1.5 = 313.3$ kN·m from a UDL: $\delta = \frac{5ML^2}{48EI}$. $$ \delta = \frac{5\times313.3\times10^6\times(3500)^2}{48\times2\times10^5\times91800\times10^4} = 2.18\ \text{mm} $$ $\delta = 2.18$ mm $< L/300 = 11.67$ mm. **OK.** #### Web buckling and web crippling (bearing) at the support (cl. 8.7.3 and 8.7.4) Stiff bearing length $b_1 = 100$ mm. Clear web depth $d = 519.4$ mm; dispersion of load at 45° to mid-depth at an end support $n_1 = d/2 = 259.7$ mm. Effective length $0.7d$ (cl. 8.7.1.5), $\lambda = 0.7d/(t_w/\sqrt{12}) = 105.0$; curve c gives $f_{cd} = 100.6$ N/mm². $$ \begin{aligned} F_{cdw} &= (b_1 + n_1) t_w f_{cd} = (100 + 259.7)\times12.0\times100.6/10^3 = 434.4\ \text{kN} \\ n_2 &= 2.5(t_f + R_1) = 100.8\ \text{mm} \\ F_w &= \frac{(b_1 + n_2) t_w f_y}{\gamma_{m0}} = \frac{(100 + 100.8)\times12.0\times250}{1.1\times10^3} = 547.5\ \text{kN} \end{aligned} $$ Support reaction $R = 180.0$ kN $< F_{cdw}$ and $< F_w$. No web stiffener is needed. #### Final design Provide **ISMB 600** (123.0 kg/m): $M_d = 627.8$ kN·m $> M_u = 470.0$ kN·m, $V_d = 945$ kN $> V_u = 180.0$ kN, $\delta = 2.2$ mm $< 11.7$ mm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The beam is laterally unsupported. What does this mean for the moment capacity? A. Md = Zp fy / gamma_m0 as for a supported beam B. Lateral-torsional buckling governs and Md = beta_b Zp fbd with fbd = chi_LT fy / gamma_m0 C. Only the shear capacity needs to be checked D. Md is taken as 1.5 times the elastic moment Answer: B. An unrestrained compression flange can buckle sideways, so the design bending strength fbd is reduced by chi_LT (cl. 8.2.2). (Key point: LTB governs for unsupported beam) 2. Why is Md = 627.8 kN.m smaller than the plastic capacity Zp fy/gamma_m0 (about 775 kN.m) for ISMB 600? A. Because gamma_m0 is applied twice B. Because the section is semi-compact C. Because chi_LT = 0.810 (< 1) reduces the bending strength for LTB D. Because shear reduces the moment capacity Answer: C. chi_LT (from lambda_LT = 0.774) is less than 1, so fbd = 184.1 N/mm^2 instead of fy/gamma_m0 = 227 N/mm^2. (Key point: Reduction factor chi_LT) 3. No load is given for the deflection check. How is the service moment found? A. Mu x 1.5 = 705 kN.m B. Mu / 1.5 = 313 kN.m, treated as from a UDL C. Mu = 470 kN.m is used directly D. Mu / 2 = 235 kN.m Answer: B. Deflection uses unfactored loads, so the factored moment is divided by the load factor 1.5. (Key point: Deflection from moment) 4. The checks at the support compare the reaction R = 180 kN with Fcdw (web buckling) and Fw (bearing). What is the result? A. R exceeds Fcdw, so a stiffener is needed B. R exceeds Fw, so a stiffener is needed C. R is less than both Fcdw (434.4 kN) and Fw (547.5 kN), so no stiffener is needed D. Both checks are skipped for an unsupported beam Answer: C. The load-carrying capacities 434.4 kN and 547.5 kN both exceed 180 kN, so the web is adequate without stiffeners. 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