Question
Download Solution PDFThe minimum eccentricity to be considered for an axially loaded RCC column of size 400 mm × 400 mm with unsupported length of 5 m is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
As per IS 456: 2000, clause 25.4,
Minimum Eccentricity
All columns shall be designed for minimum eccentricity, equal to the addition of the unsupported length of column divided by 500 and lateral dimensions divided by 30, subject to a minimum of 20 mm.
Where biaxial bending is considered, it is sufficient to ensure that eccentricity exceeds the minimum about one axis at a time.
Calculation:
Unsupported length = 5000 mm
Size of the column = 400 mm
Minimum eccentricity = \(\frac{L}{{500}} + \frac{B}{{30}} \)
\(e_{min}= \;\frac{{5000}}{{500}} + \frac{{400}}{{30}} = 23.33\;mm\;\)
But, in no case, the minimum eccentricity should be less than 20 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.