Question
Download Solution PDFThe limiting depth of neutral axis for a beam having effective depth of 400 mm with Fe 250 grade steel is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The ratio of limiting depth of neutral axis to the effective depth of the beam is given by
\(\frac{{{x_u}}}{d} = \frac{{{\rm{Max\;strain\;concrete}}}}{{\frac{{{\rm{Grade\;of\;steel}}}}{{{\rm{Modulus\;of\;Elasticity\;of\;steel}} \times {\rm{F}}.{\rm{O}}.{\rm{S}}}} + 0.002 + {\rm{Max\;strain\;concrete}}}}\)
Note:
Without getting into the tedious calculation,
We know the standard values of the ratio of limiting depth of neutral axis to the effective depth (k) of the beam for different steel sections as following:
|
Grade of Steel |
Fe 500 |
Fe 415 |
Fe 250 |
|
‘k’ value |
0.46 |
0.48 |
0.53 |
Calculation:
∴ For Fe 250, xu ≈ 0.53 × 400 = 212 mm.
Last updated on May 28, 2025
-> SSC JE notification 2025 for Civil Engineering will be released on June 30.
-> Candidates can fill the SSC JE CE application from June 30 to July 21.
-> The selection process of the candidates for the SSC Junior Engineer post consists of Paper I, Paper II, Document Verification, and Medical Examination.
-> Candidates who will get selected will get a salary range between Rs. 35,400/- to Rs. 1,12,400/-.
-> Candidates must refer to the SSC JE Previous Year Papers and SSC JE Civil Mock Test, SSC JE Electrical Mock Test, and SSC JE Mechanical Mock Test to understand the type of questions coming in the examination.
-> The Staff Selection Commission conducts the SSC JE exam to recruit Junior Engineers in different disciplines under various departments of the Central Government.