Question
Download Solution PDFFor a fixed beam of span 'L', carrying uniformly distributed load 'w' on the entire span, the bending moment at the mid span will be:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanaition:
The moment developed at ends of fixed beam when subjected to UDL are \(\frac{{w{L^2}}}{{12}}\) n magnitude wise.
At the section at mid span, draw FBD of LHS
∑MX = 0
\(\frac{{wL}}{2} \times \frac{L}{4} + M + \frac{{w{L^2}}}{{12}} = \frac{{wL}}{2} \times \frac{L}{2}\)
\(M = \frac{{w{L^2}}}{4} - \frac{{w{L^2}}}{8} - \frac{{w{L^2}}}{{12}}\)
\(M = \frac{{w{L^2}}}{{24}}\)
BM at mid span is \(M = \frac{{w{L^2}}}{{24}}\) (sagging)
(+) → sagging
(-) → Hogging
Last updated on May 28, 2025
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