Question
Download Solution PDFThe Rankine's formula holds good for
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
For short or long columns Rankine’s Formula is used.
\(\frac{1}{P} = \frac{1}{{{P_C}}} + \frac{1}{{{P_E}}}\)
Where P is crippling load by Rankine formula; \({P_C}\;\) is crushing load; \({P_E}\;\) is crippling load by Euler’s formula.
For Short column: \({P_E}\;\) is very large so \(\frac{1}{{{P_E}}}\) will be very small and negligible as compared to \(\frac{1}{{{P_C}}}.\) so
\(\frac{1}{P} = \frac{1}{{{P_C}}} \to P = {P_C}\)
For Long column: \({P_E}\;\) is small so \(\frac{1}{{{P_E}}}\) will be large as compared to. Hence the value of \(\frac{1}{{{P_C}}}\;\) can be neglected. So
\(\frac{1}{P} = \frac{1}{{{P_E}}} \to P = {P_E}\)
Hence the crippling load by Rankine’s formula for a long column is approximately equal to the crippling load by Euler’s formula.Last updated on Jun 24, 2025
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