--- title: "UDL capacity of MB 300 with flange plates" subject: "Design of Steel and Timber Structure" chapter: 7 key: "semester_6/design_of_steel_and_timber_structure:builtup-beam-mb300-plates" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/builtup-beam-mb300-plates site: BCT Engineering --- # UDL capacity of MB 300 with flange plates Design of Steel and Timber Structure · Chapter 7: Flexure Members - Asked in: 2079 Chaitra, 8 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/builtup-beam-mb300-plates - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-7 ## Question Determine uniformly distributed load carrying capacity of a laterally restrained built-up beam having MB 300 and two plates 200 mm×12 mm. Effective length of beam is 3.5 m. ## Short answer Plastic design per IS 800:2007. Shear capacity of the ISMB 300 web governs; moment capacity under high shear is adequate. The factored uniformly distributed load capacity is 173 kN/m (including self-weight). ## Answer **Approach.** Plastic design strength of the built-up section (IS 800:2007 cl. 8.2.1), then reduction of moment capacity under high shear (cl. 9.2), and the load found from the governing check. $f_y = 250$ N/mm², $\gamma_{m0} = 1.10$. The beam is laterally restrained, so lateral-torsional buckling is ignored. Plates are assumed fully connected to the flanges (continuous welds or adequate bolts), one 200×12 plate on each flange. Effective span $L = 3.5$ m. #### Section properties ISMB 300 (IS 808): $D = 300$ mm, $b_f = 140$ mm, $t_w = 7.7$ mm, $t_f = 13.1$ mm, $I_{zz} = 8990$ cm⁴, mass 46.0 kg/m. Plastic modulus of ISMB 300 from its dimensions: $Z_{p,I} = 670$ cm³. $$ \begin{aligned} \text{Overall depth} &= 300 + 2\times12 = 324\ \text{mm} \\ y_c &= 150 + 6 = 156\ \text{mm (plate centroid from N.A.)} \\ I_{zz} &= 8990\times10^4 + 2\left[\frac{200\times12^3}{12} + 200\times12\times156^2\right] = 20677\ \text{cm}^4 \\ Z_e &= \frac{I_{zz}}{162} = 1276\ \text{cm}^3 \\ Z_p &= Z_{p,I} + 2(200\times12)(156) = 670 + 749 = 1419\ \text{cm}^3 \end{aligned} $$ Classification: plate outstand from the web face $= (200 - 7.7)/2 = 96.2$ mm, $b/t = 96.2/12 = 8.0 < 9.4\varepsilon = 9.4$ (plastic); ISMB web $d/t_w = 31.9 < 84$. So $\beta_b = 1$. #### Moment capacity $$ \begin{aligned} M_d &= \frac{Z_pf_y}{\gamma_{m0}} = \frac{1419\times10^3\times250}{1.1\times10^6} = 322.6\ \text{kN·m} \\ 1.2Z_ef_y/\gamma_{m0} &= 348.1\ \text{kN·m} \end{aligned} $$ $M_d = 322.6$ kN·m. Load limited by moment: $w = 8M_d/L^2 = 8\times322.6/3.5^2 = 210.7$ kN/m. #### Shear capacity Shear is resisted by the web of the rolled section: $A_v = D t_w = 300\times7.7 = 2310$ mm² (plates ignored, conservative). $$ V_d = \frac{A_vf_y}{\sqrt3\gamma_{m0}} = \frac{2310\times250}{\sqrt3\times1.1\times10^3} = 303.1\ \text{kN} $$ Load limited by shear: $w = 2V_d/L = 2\times303.1/3.5 = 173.2$ kN/m. Shear governs (173.2 < 210.7). #### Moment capacity under high shear (cl. 9.2.2) At $w = 173.2$ kN/m: $V = V_d = 303.1$ kN $> 0.6V_d$, so the moment capacity is reduced. $M_{fd}$ is the plastic strength of the section without the shear area: $Z_{fd} = Z_p - t_wD^2/4 = 1419 - 173 = 1246$ cm³, $M_{fd} = 283.2$ kN·m. $$ \begin{aligned} \beta &= \left(\frac{2V}{V_d} - 1\right)^2 = 1.00 \\ M_{dv} &= M_d - \beta(M_d - M_{fd}) = 322.6 - 1.00(322.6 - 283.2) = 283.2\ \text{kN·m} \end{aligned} $$ Moment at this load $= wL^2/8 = 173.2\times3.5^2/8 = 265.2$ kN·m $< M_{dv} = 283.2$ kN·m. **OK.** #### Result Self-weight $= (46.0 + 2\times0.2\times0.012\times7850)\times9.81/1000 = 0.82$ kN/m. **Answer: the safe factored uniformly distributed load is $\approx 173$ kN/m** (including self-weight; governed by shear), i.e. a service load of $\approx 115$ kN/m, or about 172 kN/m factored superimposed load after deducting the factored self-weight of 1.23 kN/m. Total factored reaction $= 303$ kN. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is Z_p of the built-up section obtained? A. Z_p of ISMB 300 plus the plate area times its centroid distance, 2(200x12)(156) B. Z_e multiplied by 1.2 C. Only the plates' Z_p D. I_zz divided by the depth Answer: A. The plastic moduli add: 670 cm^3 for the rolled section plus 749 cm^3 from the two plates gives 1419 cm^3. (Key point: Plastic modulus of built-up section) 2. What formula gives the load limited by moment? A. w = 8 M_d / L^2 with M_d = Z_p f_y / gamma_m0 B. w = 2 M_d / L C. w = M_d / L^2 D. w = V_d x L Answer: A. M_d = 322.6 kN·m and w = 8 x 322.6 / 3.5^2 = 210.7 kN/m for the moment check. (Key point: Moment-governed load) 3. What area is used for the shear capacity and what is the shear-governed load? A. Web area D x t_w = 2310 mm^2 giving w = 2V_d/L = 173.2 kN/m B. Total area including plates C. Flange area only D. Plate area 200 x 12 giving 210 kN/m Answer: A. Shear is resisted by the web of the rolled section; V_d = 303.1 kN and w = 2V_d/L = 173.2 kN/m, lower than the moment-limited load. (Key point: Shear area) 4. At w = 173.2 kN/m, V equals V_d which is above 0.6 V_d. What does this require? A. Reducing moment capacity to M_dv (283.2 kN·m) and checking the moment is below it B. Ignoring bending entirely C. Doubling the moment capacity D. Rejecting the section outright Answer: A. Above 0.6 V_d, cl. 9.2.2 reduces moment capacity; the applied 265.2 kN·m is less than M_dv = 283.2 kN·m, so the load is OK. 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