--- title: "Beam with flange plates, 30 kN/m + 16 kN" subject: "Design of Steel and Timber Structure" chapter: 7 key: "semester_6/design_of_steel_and_timber_structure:beam-additional-plates-30knm-16kn" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/beam-additional-plates-30knm-16kn site: BCT Engineering --- # Beam with flange plates, 30 kN/m + 16 kN Design of Steel and Timber Structure · Chapter 7: Flexure Members - Asked in: 2069 Bhadra, 16 marks - Video: [PSAD SITUATION 49: WF Steel Beam Strengthened by Cover Plates | CELE November 2022](https://www.youtube.com/watch?v=Cl5fw92MkU0) — Engr. Harf Miranda (Sir Harf Engineering Tutorial) - 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-additional-plates-30knm-16kn - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-7 ## Question A beam of effective span 6.0m carries a uniformly distributed load of 30KN/m with a concentrated load of 16KN at mid span. The depth of the beam is limited to 300mm. Design the beam with additional plates to the flanges. Assume that the beam is laterally supported throughout. The grade of steel is E250. M16 bolts of property class 4.6 and product grade C may be used for connection. Checks for shear, deflection and lateral stability are necessary. ## Short answer With depth limited to 300 mm, use ISMB 250 with 145×20 mm flange plates. Factored moment 244 kN·m; shear-reduced plastic moment capacity 279.5 kN·m, shear 150.7 kN < 226.3 kN. Deflection 18.4 mm < 20 mm. Lateral support ensures stability. Plates connected by M16 bolts. ## 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 $L = 6.00$ m, maximum overall depth 300 mm. A rolled I-section alone is not enough, so flange plates are added. #### Loads and design forces Working loads: UDL 30 kN/m and a point load of 16 kN at mid-span (taken as imposed loads). Factored: $w_u = 1.5\times30 = 45$ kN/m, $W_u = 1.5\times16 = 24$ kN. Span $L = 6$ m; the beam is laterally supported throughout. The support bearing length is not given, so no bearing check is made. M16 bolts, grade 4.6, product grade C, are used for the plate connection. Without self-weight $M = 238.5$ kN·m; no rolled ISMB within the depth limit can supply $Z_p = M\gamma_{m0}/f_y = 1049$ cm³, so plates are added to both flanges. #### Trial section Try **ISMB 250 with two plates 145 × 20 mm** (one on each flange). ISMB 250: $D = 250$, $b_f = 125$, $t_w = 6.9$, $t_f = 12.5$, $R_1 = 13$ mm, $I_{zz} = 5130$ cm⁴, mass 37.3 kg/m. Overall depth $= 250 + 2\times20 = 290$ mm $\le 300$ mm. $$ \begin{aligned} y_c &= D/2 + t_p/2 = 135\ \text{mm} \\ I_{zz} &= 5130\times10^4 + 2\left[\frac{145\times20^3}{12} + 145\times20\times135^2\right] = 15720\ \text{cm}^4 \\ Z_e &= I_{zz}/(290/2) = 1084\ \text{cm}^3 \\ Z_p &= Z_{p,ISMB} + 2(b_pt_p)y_c = 458 + 783 = 1241\ \text{cm}^3 \end{aligned} $$ Mass of section $= 37.3 + 2\times0.145\times0.020\times7850 = 82.8$ kg/m, self-weight $= 0.813$ kN/m (factored 1.219 kN/m). Total factored UDL $w = 45.00 + 1.22 = 46.22$ kN/m. $$ \begin{aligned} M_u &= \frac{wL^2}{8} + \frac{WL}{4} = \frac{46.22\times6.00^2}{8} + \frac{24.0\times6.00}{4} = 244.0\ \text{kN·m} \\ V_u &= 150.7\ \text{kN} \end{aligned} $$ #### Section classification $\varepsilon = 1$. Plate outstand (from the web face) $b/t = (145 - 6.9)/2/20 = 3.45 < 9.4\varepsilon$ (plastic; plates continuously connected). ISMB flange $b/t_f = 5.00$, web $d/t_w = 28.8 < 84$. Plastic section, $\beta_b = 1$. #### Moment capacity The beam is laterally supported throughout, so lateral-torsional buckling cannot occur (lateral stability is satisfied) and $M_d$ is the plastic capacity. $$ \begin{aligned} M_d &= \frac{Z_pf_y}{\gamma_{m0}} = \frac{1241\times10^3\times250}{1.1\times10^6} = 282.1\ \text{kN·m} \\ 1.2Z_ef_y/\gamma_{m0} &= 295.7\ \text{kN·m} \end{aligned} $$ #### Shear capacity Shear is taken by the web of the rolled section: $V_d = D t_wf_y/(\sqrt3\gamma_{m0}) = 250\times6.9\times250/(\sqrt3\times1.1\times10^3) = 226.3$ kN; $V_u = 150.7$ kN. $V_u > 0.6V_d = 135.8$ kN, so the moment capacity is reduced (cl. 9.2.2): $M_{fd} = 257.6$ kN·m (plastic capacity excluding the shear area $Dt_w$), $\beta = (2V/V_d - 1)^2 = 0.110$, $$ M_{dv} = M_d - \beta(M_d - M_{fd}) = 282.1 - 0.110(282.1 - 257.6) = 279.5\ \text{kN·m} $$ $M_{dv} = 279.5$ kN·m $> M_u = 244.0$ kN·m. **Safe in bending and shear.** #### Deflection (limit span/300, Table 6) $$ \delta = \frac{5wL^4}{384EI} + \frac{WL^3}{48EI} = 18.39\ \text{mm} < \frac{L}{300} = 20.00\ \text{mm} $$ (unfactored imposed loads $w = 30.0$ kN/m, $W = 16.0$ kN, $I_{zz} = 15720$ cm⁴). **OK.** #### Connection of flange plates (M16, grade 4.6 bolts) Horizontal shear per unit length between plate and flange at the support: $q = \dfrac{V A_py_c}{I_{zz}} = \dfrac{150657\times2900\times135}{15720\times10^4} = 375.2$ N/mm. M16 bolt (single shear): $V_{dsb} = 29.0$ kN; bearing on the thinner part ($t = 12.5$ mm, $k_b = 0.556$): $V_{dpb} = 91.1$ kN. Bolt value $= 29.0$ kN. With two bolts per pitch (two lines): $$ p = \frac{2\times29.0\times10^3}{375.2} = 154\ \text{mm} $$ Maximum pitch in a compression member (cl. 10.2.3.2): $12t = 12\times12.5 = 150$ mm or 200 mm, whichever is less, $= 150$ mm. Provide **two lines of M16 bolts at 150 mm pitch** (this pitch is based on the shear flow at the support, so it is safe along the whole span). The plates run the full length of the beam. #### Final design ISMB 250 with a 145 × 20 mm plate on each flange (overall depth 290 mm); plates connected by M16 bolts at 150 mm pitch. $M_{dv} = 279.5$ kN·m $> 244.0$ kN·m, $\delta = 18.4$ mm $< 20$ mm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why are plates added to the flanges of the ISMB 250? A. To increase web thickness B. To avoid lateral buckling C. A rolled section within the 300 mm depth limit cannot supply the required Zp of about 1049 cm3 D. To reduce the bolt size Answer: C. The depth limit rules out a deeper rolled section, so flange plates raise Zp to 1241 cm3 while keeping overall depth 290 mm. (Key point: Need for plates) 2. Vu = 150.7 kN and 0.6Vd = 135.8 kN. What happens? A. Nothing, shear is always ignored B. The section fails in shear C. The moment capacity is reduced to Mdv = 279.5 kN·m, which is still above Mu = 244.0 kN·m D. The plates are removed from the check Answer: C. Vu > 0.6Vd means high shear, so Md is reduced (cl. 9.2.2) with beta = (2V/Vd - 1)^2 = 0.110; the beam is still safe. (Key point: Shear effect on moment) 3. Which part resists the shear in the built-up beam? A. Web of the rolled ISMB 250 (Av = D tw) B. The added plates C. The bolts D. The whole 290 mm depth Answer: A. Vd = 250x6.9x250/(sqrt3 x 1.1) = 226.3 kN comes from the rolled web only. (Key point: Shear resistance) 4. The shear flow gives p = 154 mm but 150 mm is provided. Why? A. Rounded for convenience only B. The maximum pitch for compression plates is 12t = 150 mm, which is less than 154 C. Bolt value is doubled D. M16 bolts need 150 mm edge distance Answer: B. p = 2x29.0x10^3/375.2 = 154 mm, but the maximum pitch (cl. 10.2.4.2) is the lesser of 12t = 150 mm or 200 mm. (Key point: Bolt pitch) ## Related in Design of Steel and Timber Structure - [Beam 6 m span, 46 kN/m, Fe 410](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-udl-46knm-6m) — Design of Steel and Timber Structure, chapter 7 - [Beam 4 m span, 45 kN/m and 100 kN midspan](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-4m-45knm-100kn) — Design of Steel and Timber Structure, chapter 7 - [Beam 6 m span with 100 kN secondary beam load](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-6m-20knm-100kn-midspan) — Design of Steel and Timber Structure, chapter 7 - [UDL capacity of MB 300 with flange plates](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abuiltup-beam-mb300-plates) — Design of Steel and Timber Structure, chapter 7 - [Beam supporting RCC slab, DL 10 kN/m, IL 30 kN](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-6m-rcc-slab-10dl-30il) — Design of Steel and Timber Structure, chapter 7, asked 2× - [Simply supported beam 6 m, 60 kN/m](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-6m-60knm-udl) — Design of Steel and Timber Structure, chapter 7 ## Related in other subjects - [Mid-span deflection of beam, 10 kN load](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ass-beam-deflection-mid-span-4m) — Strength of Materials, chapter 7 - [Beam mid-span deflection with temperature gradient](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Abeam-midspan-defl-temperature) — Theory of Structures I, chapter 3 - [Bent-up bars shear, 550 mm effective depth](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Abent-up-bars-shear-550-mm) — Design of RCC Structure, chapter 5, asked 3× - [Beam reactions and mid-span deflection (2EI)](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Abeam-support-reactions-midspan-k8) — Computational Techniques in Civil Engineering, chapter 4, asked 2×