Question
Download Solution PDFDeflection of a simply supported beam at center of span, carrying a uniform load of w per unit run over entire span with uniform flexural rigidity is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFThe deflections of the beam with different support conditions are given below,
It can be seen that the deflection for simply supported beam of span ‘l’ carrying distributed load of ‘w’ per unit run over the whole span is given by,
Deflection = \(\frac{{5{\rm{w}}{{\rm{l}}^4}}}{{384{\rm{EI}}}}\)
Last updated on Jun 4, 2025
-> OSSC JE Preliminary Fiinal Answer Key 2025 has been released. Applicants can raise objections in OSSC JE answer key till 06 June 2025.
-> OSSC JE prelim Exam 2025 was held on 18th May 2025.(Advt. No. 1233/OSSC ).
-> The OSSC JE 2024 Notification was released for 759 vacancies through OSSC CTSRE 2024.
-> The selection is based on the written test. This is an excellent opportunity for candidates who want to get a job in the Engineering sector in the state of Odisha.
-> Candidates must refer to the OSSC JE Previous Year Papers to understand the type of questions in the examination.