- 2074 Chaitra · 15 marks
Beam steel at midspan, limit state
Short answer
Mu=451.34 kNm. Doubly reinforced: tension steel Ast=2907 mm² (6-25 mm), compression steel Asc=1543 mm² (5-20 mm).
A beam of rectangular section is 300 mm wide and 500 mm deep to the centre of tensile reinforcement. It has to carry a dead load of 45 kN/m excluding its self weight. Find the steel reinforcement required for the mid span section. The beam has a span of 7 m. Use M20 concrete and Fe 415 steel. Effective cover to compression steel = 40 mm. Use limit state method.
Given: mm, mm, span m, dead load 45 kN/m (excluding self weight), M20, Fe415, effective cover to compression steel mm. Limit state of collapse in flexure.
Assumption: overall depth mm ( plus 50 mm cover); the effective span is taken as 7 m.
Step 1: Loads
Step 2: Design moment at mid-span
Step 3: Limiting moment of resistance (Fe415, )
Since and the size is fixed, the section is doubly reinforced.
Doubly reinforced design
Limiting depth mm. Strain in compression steel (similar triangles):
From the design stress-strain curve of Fe415 (IS 456 Fig. 23): N/mm. Stress in concrete displaced by the steel: N/mm.
Provide:
- Tension steel: 6-25 mm dia ( mm), arranged in two layers.
- Compression steel: 5-20 mm dia ( mm) at mm.
- Stirrups (8 mm, 2-legged) at spacing not more than , 300 mm and to hold the compression bars (shear design separately).
Checks: minimum steel mm OK; and OK.
Answer: Mid-span mm (6-25 mm) and mm (5-20 mm).
✓ Answer checked by Claude Opus 5.5 (high effort)
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