Question
Download Solution PDFAccording to Euler's column theory, the crippling load for a column of length l with one end fixed and the other end hinged, is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
According to Euler's column theory, the crippling load for a column of length L,
\({P_{cr}}= \;\frac{{{\pi ^2}EI}}{{{{\left( {L_e} \right)}^2}}}\)
Where Leq is the effective length of the column.
|
Support Condition |
Effective Length |
|
Both ends hinged |
\({L_e} = L\) |
|
Both ends fixed |
\({L_e} = \frac{L}{2}\) |
| One end fixed other end free | \({L_e} = 2L\) |
| One end fixed other end hinged | \({L_e} = \frac{L}{\sqrt{2}}\) |
Now,
From the above table
According to Euler's column theory, the crippling load of a column of length (l), with one end is fixed and the other end is hinged is \({P_{cr}}= \;\frac{{{2\pi ^2}EI}}{{{{\left( {L} \right)}}}}\)
Last updated on Jun 24, 2025
-> WBPSC JE recruitment 2025 notification will be released soon.
-> Candidates with a Diploma in the relevant engineering stream are eligible forJunior Engineer post.
-> Candidates appearing in the exam are advised to refer to the WBPSC JE syllabus and exam pattern for their preparations.
-> Practice WBPSC JE previous year question papers to check important topics and chapters asked in the exam.