- 2068 Magh (old course) · 14 marks
Beam with slab, 20 kN/m, 350 mm wall
Short answer
Use ISMB 300 (46 kg/m). Factored moment 135 kN·m, moment capacity 152.4 kN·m; shear capacity 303 kN > 90 kN; deflection 18.8 mm < L/300 = 20 mm. Web crippling and buckling checks are satisfied without stiffeners.
Design a beam of 6m span carrying UDL of 20KN/m including self wt. The beam is laterally restrained by a concrete slab and is simply supported at the ends on wall of 350mm width. Check for shear, deflection, web crippling and buckling.
Data
- Steel Fe 410 (E250): N/mm², N/mm², N/mm², (IS 800:2007, Table 5). Limit state design.
- Effective span m, bearing length of support mm.
Loads and design forces
The UDL of 20 kN/m (inclusive of self-weight) is the service load; factored kN/m. m, wall (support) width 350 mm so mm: kN·m, kN.
Moment without self-weight kN·m, so cm³.
Trial section
Try ISMB 300 (IS 808): mm, mm, mm, mm, mm; cm⁴; cm³; cm³; mass kg/m.
Section classification (Table 2)
. Flange: . Web: (with mm). The section is plastic, so .
Moment capacity (cl. 8.2.1.2)
The concrete slab restrains the compression flange along the full length, so lateral-torsional buckling is prevented.
The smaller value governs: kN·m.
Shear capacity (cl. 8.4)
kN ; also kN, so the shear is low and the moment capacity is not reduced (cl. 9.2.1). Web shear buckling does not govern since (cl. 8.4.2.1).
Deflection (Table 6)
Deflection under unfactored service (imposed) loads, limit span/300 (floor and roof members, Table 6):
mm mm. OK.
Web buckling and bearing at support (cl. 8.7.3 and 8.7.4)
Stiff bearing length mm. Clear web depth mm; dispersion of load at 45° to mid-depth at an end support mm. Effective length (cl. 8.7.1.5), ; curve c gives N/mm².
Support reaction kN and . No web stiffener is needed.
Final design
Provide ISMB 300 (46.0 kg/m): kN·m kN·m, kN kN, mm mm.
✓ Answer checked by Claude Opus 5.5 (high effort)
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