Discussion
A single-storey frame has a lumped mass of 40,000 kg and a lateral stiffness of 16,000 kN/m. Its natural period is nearest to
Answer: A
With \( \omega = \sqrt{k/m} = \sqrt{16{,}000{,}000/40{,}000} = 20 \) rad/s, the period is \( T = 2\pi/\omega = 0.314 \) s. Note that the stiffness must be converted from kN/m to N/m before the root is taken, which is where most slips occur.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development