If a column is fixed at one end and free at the other end, the effective length is ______ the original length.

This question was previously asked in
NHPC JE Mechanical 6 April 2022 (Shift 2) Official Paper
View all NHPC JE Papers >
  1. thrice
  2. four times
  3. half
  4. twice

Answer (Detailed Solution Below)

Option 4 : twice
Free
NHPC & THDC JE Civil Full Test 1
5.3 K Users
200 Questions 200 Marks 180 Mins

Detailed Solution

Download Solution PDF

Explanation:

Buckling load:
The load at which column buckle is termed as buckling load. Buckling load is given by:

\({P_b} = \frac{{{\pi ^2}EI}}{{L_e^2}}\)
where E = Young's modulus of elasticity, Imin = Minimum moment of inertia, and Le = Effective length

End conditions Le Buckling load
Both ends hinged Le = L \({P_b} = \frac{{{\pi ^2}EI}}{{L_e^2}}\)
Both ends fixed Le = L/2 \({P_b} = \frac{{{4\pi ^2}EI}}{{L_e^2}}\)
One end fixed and another end is free Le = 2L \({P_b} = \frac{{{\pi ^2}EI}}{{4L_e^2}}\)
One end fixed and another end is hinged \({L_e} = \frac{L}{{\sqrt 2 }}\) \({P_b} = \frac{{{\pi ^2}EI}}{{2L_e^2}}\)
Latest NHPC JE Updates

Last updated on May 12, 2025

-> The exam authorities has released the NHPC JE tender notice under supervisor posts through CBT.

->NHPC JE recruitment 2025 notification will be released soon at the official website. 

-> NHPC JE vacancies 2025 will be released for Mechanical, Electrical, Civil and Electronics & Communication disciplines.

-> NHPC JE selection process comprises online computer based test only.

-> Candidates looking for job opportunities as Junior Engineers are advised to refer to the NHPC JE previous year question papers for their preparations. 

-> Applicants can also go through the NHPC JE syllabus and exam pattern for their preparations. 

Get Free Access Now
Hot Links: teen patti lotus teen patti download apk teen patti master apk