Question
Download Solution PDFIn WSM, if the critical depth of neutral axis is equal to k × d where d is effective depth of beam, then what is the value of k for steel 500 and concrete M25?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
For the given condition:
Given
Grade of Concrete = M20 , Grade of Steel = Fe500
We know
Critical depth of Neutral axis is given by:
xc = k × d
\({\rm{x_c}} = \frac{1}{{1 + \frac{{{{\rm{σ }}_{{\rm{st}}}}}}{{{\rm{m}} × {{\rm{σ }}_{{\rm{cbc}}}}}}}} × {\rm{d}} \)
Where σ st - Permissible stresses in Steel
σ cbc - Permissible stresses in concrete
For Grade M25 → σ cbc = 8.5 N/mm2
For Fe500 → σst = 0.55 × fy = 0.55 × 500 = 275 N/mm2
Equating in the formula
\({\rm{m}} = \frac{{280}}{{3 × {{\rm{σ }}_{{\rm{bc}}}}}} = \frac{{280}}{{3 × 8.5}} = 10.98\)
\(\therefore {{\rm{x}}_{\rm{c}}} = \frac{1}{{1 + \frac{{275}}{{10.95 × 8.5}}}} × d\)
∴ xc = 0.2528 × d ≈ 0.253 × d
Comparing it with xc = k × d ⇒ k values come out to be 0.253
Important PointsFor LSM Method,
As per IS 456:2000, cl.38.1
Grade of steel |
Fe250 |
Fe415 |
Fe500 |
Depth factor |
0.53 |
0.48 |
0.46 |
Last updated on May 19, 2025
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