--- title: "Built-up beam, unsupported 4 m, 40 kN/m + 100 kN" subject: "Design of Steel and Timber Structure" chapter: 7 key: "semester_6/design_of_steel_and_timber_structure:builtup-beam-unsupported-4m-40knm" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/builtup-beam-unsupported-4m-40knm site: BCT Engineering --- # Built-up beam, unsupported 4 m, 40 kN/m + 100 kN Design of Steel and Timber Structure · Chapter 7: Flexure Members - Asked in: 2073 Bhadra, 15 marks - Video: [Design of Built up Beam](https://www.youtube.com/watch?v=4cLgLVn_viA) — Harihanandh M - 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/builtup-beam-unsupported-4m-40knm - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-7 ## Question Design a built up beam having laterally unsupported span of 4 m, support width 300 mm. Beam is subjected to design imposed load of 40 KN/m and 100 KN at mid span. Depth of beam is limited to 350 mm. ## Short answer Use a built-up section: ISMB 300 with two 180 × 14 mm flange plates (total depth 328 mm). Lateral-torsional buckling capacity M_d = 202.9 kN·m > M_u = 182.5 kN·m; shear and deflection are safe. ## Answer #### Data - Fe 410 (E250): $f_y = 250$ N/mm², $f_u = 410$ N/mm², $E = 2\times10^5$ N/mm², $\gamma_{m0} = 1.10$. Limit state design (IS 800:2007). - Effective span $L = 4.00$ m, maximum overall depth 350 mm. A rolled I-section alone is not enough, so flange plates are added. #### Loads and design forces The imposed UDL of 40 kN/m and the 100 kN load at mid-span are the **design (factored)** loads. Span $L = 4$ m, support width 300 mm (stiff bearing $b_1 = 300$ mm), depth limited to 350 mm. The beam is laterally **unsupported** over the full span, so $L_{LT} = 4.0$ m (simple supports, torsion restrained, warping free). Unfactored loads for deflection: $40/1.5 = 26.7$ kN/m and $100/1.5 = 66.7$ kN. Without self-weight $M = 180.0$ kN·m, so even with full lateral support $Z_p \ge M\gamma_{m0}/f_y = 792$ cm³. The deepest rolled section within the depth limit, ISMB 350 ($Z_p \approx 890$ cm³), would lose much of this capacity to lateral-torsional buckling over the unsupported 4 m span, so plates are added to both flanges to increase both $Z_p$ and the lateral stiffness $I_y$. #### Trial section Try **ISMB 300 with two plates 180 × 14 mm** (one on each flange). ISMB 300: $D = 300$, $b_f = 140$, $t_w = 7.7$, $t_f = 13.1$, $R_1 = 14$ mm, $I_{zz} = 8990$ cm⁴, mass 46.0 kg/m. Overall depth $= 300 + 2\times14 = 328$ mm $\le 350$ mm. $$ \begin{aligned} y_c &= D/2 + t_p/2 = 157\ \text{mm} \\ I_{zz} &= 8990\times10^4 + 2\left[\frac{180\times14^3}{12} + 180\times14\times157^2\right] = 21421\ \text{cm}^4 \\ Z_e &= I_{zz}/(328/2) = 1306\ \text{cm}^3 \\ Z_p &= Z_{p,ISMB} + 2(b_pt_p)y_c = 670 + 791 = 1462\ \text{cm}^3 \end{aligned} $$ Mass of section $= 46.0 + 2\times0.180\times0.014\times7850 = 85.6$ kg/m, self-weight $= 0.839$ kN/m (factored 1.259 kN/m). Total factored UDL $w = 40.00 + 1.26 = 41.26$ kN/m. $$ \begin{aligned} M_u &= \frac{wL^2}{8} + \frac{WL}{4} = \frac{41.26\times4.00^2}{8} + \frac{100.0\times4.00}{4} = 182.5\ \text{kN·m} \\ V_u &= 132.5\ \text{kN} \end{aligned} $$ #### Section classification $\varepsilon = 1$. Plate outstand (from the web face) $b/t = (180 - 7.7)/2/14 = 6.15 < 9.4\varepsilon$ (plastic; plates continuously connected). ISMB flange $b/t_f = 5.34$, web $d/t_w = 31.9 < 84$. Plastic section, $\beta_b = 1$. #### Moment capacity The beam is laterally unsupported, so lateral-torsional buckling governs. $L_{LT} = 4.0$ m. Properties of the built-up section: $I_y = I_{y,ISMB} + 2t_pb_p^3/12 = 1847$ cm⁴; $I_t = 58.1$ cm⁴ ($\Sigma bt^3/3$); distance between flange centroids $h_f = 303$ mm; $I_w = I_yh_f^2/4 = 42\times10^{10}$ mm⁶. Welded built-up section: $\alpha_{LT} = 0.49$ (Table 13). $$ \begin{aligned} M_{cr} &= \sqrt{\frac{\pi^2EI_y}{L_{LT}^2}\left(GI_t + \frac{\pi^2EI_w}{L_{LT}^2}\right)} = 469.7\ \text{kN·m} \\ \lambda_{LT} &= \sqrt{Z_pf_y/M_{cr}} = 0.882,\ \phi_{LT} = 1.056,\ \chi_{LT} = 0.611 \\ f_{bd} &= \chi_{LT}f_y/\gamma_{m0} = 138.8\ \text{N/mm}^2 \\ M_d &= Z_pf_{bd} = 202.9\ \text{kN·m} \end{aligned} $$ #### Shear capacity Shear is taken by the web of the rolled section: $V_d = D t_wf_y/(\sqrt3\gamma_{m0}) = 300\times7.7\times250/(\sqrt3\times1.1\times10^3) = 303.1$ kN; $V_u = 132.5$ kN. $V_u < 0.6V_d = 181.9$ kN, so the shear is low and the moment capacity is not reduced. $M_d = 202.9$ kN·m $> M_u = 182.5$ kN·m. **Safe.** #### Deflection (limit span/300, Table 6) $$ \delta = \frac{5wL^4}{384EI} + \frac{WL^3}{48EI} = 4.15\ \text{mm} < \frac{L}{300} = 13.33\ \text{mm} $$ (unfactored imposed loads $w = 26.7$ kN/m, $W = 66.7$ kN, $I_{zz} = 21421$ cm⁴). **OK.** #### Web buckling and bearing at the support $b_1 = 300$ mm, $d = 245.8$ mm, $n_1 = d/2 = 122.9$ mm, $\lambda = 0.7d/(t_w/\sqrt{12}) = 77.4$, $f_{cd} = 140.4$ N/mm² (curve c): $F_{cdw} = (b_1 + n_1)t_wf_{cd} = 457.1$ kN. Bearing: $n_2 = 2.5(t_f + R_1) = 67.8$ mm, $F_w = (b_1 + n_2)t_wf_y/\gamma_{m0} = 643.6$ kN. Both exceed $R = 132.5$ kN. **OK.** #### Connection of plates to the flanges (fillet welds) Horizontal shear flow at the support: $$ q = \frac{V_uA_py_c}{I_{zz}} = \frac{132.5\times10^3\times2520\times157}{21421\times10^4} = 244.7\ \text{N/mm} $$ Use a 6 mm fillet weld along each edge of each plate: $f_{wd} = f_u/(\sqrt3\gamma_{mw}) = 410/(\sqrt3\times1.25) = 189.4$ N/mm², strength per weld $= 0.7\times6\times189.4 = 795$ N/mm, two welds $= 1591$ N/mm $> 244.7$ N/mm. **OK.** #### Final design ISMB 300 with a 180 × 14 mm plate on each flange (overall depth 328 mm), plates connected by 6 mm fillet welds along both edges. $M_d = 202.9$ kN·m $> 182.5$ kN·m, $\delta = 4.1$ mm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The 40 kN/m and 100 kN are design loads. What follows? A. Multiply by 1.5 for Mu B. Use them as factored for Mu, and divide by 1.5 (26.7 kN/m and 66.7 kN) for deflection C. Use them unfactored for strength D. Add 1.5 x self-weight only Answer: B. Design loads are already factored, so strength uses them directly (plus self-weight) and deflection uses the unfactored values. (Key point: Load interpretation) 2. Why is a plated ISMB 300 used? A. Because lateral buckling needs wider webs B. Because the support width is 300 mm C. Because the load is a point load D. A rolled section within the 350 mm depth limit cannot supply Zp of about 792 cm3 Answer: D. Adding 180x14 plates gives Zp = 1462 cm3 with an overall depth of 328 mm, within 350 mm. (Key point: Reason for built-up section) 3. Why is Md (202.9 kN·m) less than the plastic capacity Zp fy/gamma_m0 (about 332 kN·m)? A. Shear has reduced the capacity B. The beam is unsupported over 4 m so LTB reduces it (chi_LT = 0.611) C. Section is semi-compact D. Gamma_m0 was applied twice Answer: B. Mcr = 469.7 kN·m gives lambda_LT = 0.882 and chi_LT = 0.611, so fbd = 138.8 N/mm2 and Md = Zp fbd. (Key point: LTB governs) 4. At the end support with stiff bearing b1 = 300 mm, the dispersion length n1 is: A. 2.5(tf + R1) B. b1/2 C. d/2 = 122.9 mm D. d = 245.8 mm Answer: C. At an end support the load disperses at 45 degrees to the mid-depth, so n1 = d/2; F_cdw = 457.1 kN exceeds R = 132.5 kN. 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