--- title: "Beam 4 m span, 45 kN/m and 100 kN midspan" subject: "Design of Steel and Timber Structure" chapter: 7 key: "semester_6/design_of_steel_and_timber_structure:beam-4m-45knm-100kn" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/beam-4m-45knm-100kn site: BCT Engineering --- # Beam 4 m span, 45 kN/m and 100 kN midspan Design of Steel and Timber Structure · Chapter 7: Flexure Members - Asked in: 2078 Chaitra, 12 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-4m-45knm-100kn - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-7 ## Question Design a laterally supported beam of 4 m span subjected to design imposed load of 45 kN/m and a concentrated load of 100 kN at mid span. The load is transferred through stiff bearings of 300 mm width at the supports. The depth of beam is limited to 350 mm. Also check for deflection. ## Short answer Trial section ISMB 350: plastic moment capacity 199.3 kN·m > 191.5 kN·m, shear capacity 372 kN > 141.5 kN, deflection 6.94 mm < 13.33 mm. Web buckling and bearing are safe. ISMB 350 is adequate. ## 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 $L = 4.00$ m, bearing length of support $= 300$ mm. #### Loads and design forces The imposed load of 45 kN/m and the concentrated load of 100 kN at mid-span are taken as the **design (factored) loads**; self-weight is added with $\gamma_f = 1.5$. Span $= 4.0$ m, depth limited to 350 mm, so ISMB sections up to ISMB 350 are considered. Stiff bearing at supports: 300 mm. For the deflection check the unfactored imposed loads are $45/1.5 = 30.0$ kN/m and $100/1.5 = 66.7$ kN. Moment without self-weight $M = 190.0$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 190.0\times10^6\times1.1/250 = 836$ cm³ (self-weight adds a little). #### Trial section Try **ISMB 350** (IS 808): $D = 350$ mm, $b_f = 140$ mm, $t_w = 8.1$ mm, $t_f = 14.2$ mm, $R_1 = 14$ mm; $I_{zz} = 13600$ cm⁴; $Z_e = 779$ cm³; $Z_p \approx B t_f (D - t_f) + t_w (D - 2t_f)^2/4 = 877$ cm³; mass $52.4$ kg/m. Self-weight $= 52.4\times9.81/1000 = 0.514$ kN/m, factored $= 1.5\times0.514 = 0.771$ kN/m. Total factored UDL $w = 45.00 + 0.77 = 45.77$ kN/m. $$ \begin{aligned} M_u &= \frac{wL^2}{8} + \frac{WL}{4} = \frac{45.77\times4.00^2}{8} + \frac{100.0\times4.00}{4} = 191.5\ \text{kN·m} \\ V_u &= \frac{wL}{2} + \frac{W}{2} = 141.5\ \text{kN} \end{aligned} $$ #### Section classification (Table 2) $\varepsilon = \sqrt{250/f_y} = 1.00$. Flange: $b/t_f = (140/2)/14.2 = 4.93 < 9.4\varepsilon = 9.4$. Web: $d/t_w = 36.2 < 84\varepsilon = 84$ (with $d = D - 2(t_f + R_1) = 293.6$ mm). The section is **plastic**, so $\beta_b = 1.0$. #### Moment capacity (cl. 8.2.1.2) The beam is laterally supported, so lateral-torsional buckling does not occur and $M_d$ is the plastic moment capacity. $$ \begin{aligned} M_d &= \frac{\beta_b Z_p f_y}{\gamma_{m0}} = \frac{877\times10^3\times250}{1.1\times10^6} = 199.3\ \text{kN·m} \\ 1.2 Z_e f_y/\gamma_{m0} &= 1.2\times779\times10^3\times250/(1.1\times10^6) = 212.5\ \text{kN·m} \end{aligned} $$ The smaller value governs: $M_d = 199.3$ kN·m. $$ M_d = 199.3\ \text{kN·m} > M_u = 191.5\ \text{kN·m}\quad \text{Safe} $$ #### Shear capacity (cl. 8.4) $$ \begin{aligned} A_v &= D t_w = 350\times8.1 = 2835\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{2835\times250}{\sqrt3\times1.1\times10^3} = 372.0\ \text{kN} \end{aligned} $$ $V_u = 141.5$ kN $< V_d$; also $V_u < 0.6V_d = 223.2$ kN, so the shear is low and the moment capacity is not reduced (cl. 9.2.1). Web shear buckling does not govern since $d/t_w = 36.2 < 67\varepsilon = 67$ (cl. 8.4.2.1). #### Deflection (Table 6) Deflection under unfactored service (imposed) loads, limit span/300 (floor and roof members, Table 6): $$ \delta = \frac{5 w L^4}{384 E I} + \frac{W L^3}{48 E I} = \frac{5\times30.00\times(4000)^4}{384\times2\times10^5\times13600\times10^4} + \frac{66667\times(4000)^3}{48\times2\times10^5\times13600\times10^4} = 6.94\ \text{mm} $$ $\delta = 6.94$ mm $< L/300 = 13.33$ mm. **OK.** #### Web buckling and bearing at support (cl. 8.7.3 and 8.7.4) Stiff bearing length $b_1 = 300$ mm. Clear web depth $d = 293.6$ mm; dispersion of load at 45° to mid-depth at an end support $n_1 = d/2 = 146.8$ mm. Effective length $0.7d$ (cl. 8.7.1.5), $\lambda = 0.7d/(t_w/\sqrt{12}) = 87.9$; curve c gives $f_{cd} = 124.1$ N/mm². $$ \begin{aligned} F_{cdw} &= (b_1 + n_1) t_w f_{cd} = (300 + 146.8)\times8.1\times124.1/10^3 = 449.3\ \text{kN} \\ n_2 &= 2.5(t_f + R_1) = 70.5\ \text{mm} \\ F_w &= \frac{(b_1 + n_2) t_w f_y}{\gamma_{m0}} = \frac{(300 + 70.5)\times8.1\times250}{1.1\times10^3} = 682.1\ \text{kN} \end{aligned} $$ Support reaction $R = 141.5$ kN $< F_{cdw}$ and $< F_w$. No web stiffener is needed. #### Final design Provide **ISMB 350** (52.4 kg/m): $M_d = 199.3$ kN·m $> M_u = 191.5$ kN·m, $V_d = 372$ kN $> V_u = 141.5$ kN, $\delta = 6.9$ mm $< 13.3$ mm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are the factored moment and shear computed for the UDL plus central point load? A. M = wL^2/8 + WL/4 and V = wL/2 + W/2 B. M = wL^2/8 + WL/2 and V = wL + W C. M = wL^2/2 + WL and V = wL/2 D. M = wL^2/8 only Answer: A. Superposing simply supported results gives M_u = 45.77x16/8 + 100x4/4 = 191.5 kN·m and V_u = 141.5 kN. (Key point: Design forces) 2. Why does ISMB 350 satisfy bending? A. M_d = 199.3 kN·m exceeds M_u = 191.5 kN·m, with the beam laterally supported so there is no LTB reduction B. M_d = 100 kN·m is above the point load C. M_d is reduced by LTB but still exceeds M_u D. M_d = 212.5 kN·m, because 1.2 Z_e governs Answer: A. Lateral support means M_d = beta_b Z_p f_y / gamma_m0 = 199.3 kN·m (smaller than the 212.5 limit), above 191.5. (Key point: Moment capacity check) 3. In the deflection check, what loads are used? A. Unfactored loads: 30 kN/m and 66.7 kN, giving 6.94 mm < L/300 B. Factored loads 45 kN/m and 100 kN, giving 10 mm C. Dead load only D. Self-weight only Answer: A. Since the given loads are factored, they are divided by 1.5 to get service loads for deflection, which total 6.94 mm, under the 13.33 mm limit. (Key point: Deflection load) 4. What is checked at the 300 mm stiff bearings at the supports? A. Web buckling and bearing strength against the reaction (R = 141.5 kN) B. Flange local buckling only C. Torsional rigidity D. Shear lag Answer: A. F_cdw = 449.3 kN and F_w = 682.1 kN both exceed the reaction, so no web stiffener is needed. 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