SSC JE Civil Engineering 29-Oct-2020 Shift 1 Paper

The moment developed at ends of fixed beam when subjected to UDL are $\frac{\mathit{w}{\mathit{L}}^2}{12}$ n magnitude wise. At the section at mid span, draw FBD of LHS $\sum {\mathrm{M}}_{\mathrm{X}}=0$
$\frac{\mathit{w}\mathit{L}}{2}\times \frac{\mathit{L}}{4}+\mathit{M}+\frac{\mathit{w}{\mathit{L}}^2}{12}=\frac{\mathit{w}\mathit{L}}{2}\times \frac{\mathit{L}}{2}$
$\mathit{M}=\frac{\mathit{w}{\mathit{L}}^2}{4}-\frac{\mathit{w}{\mathit{L}}^2}{8}-\frac{\mathit{w}{\mathit{L}}^2}{12}$
$\mathit{M}=\frac{\mathit{w}{\mathit{L}}^2}{24}$
BM at mid span is $\mathit{M}=\frac{\mathit{w}{\mathit{L}}^2}{24}$ (sagging)(+) → sagging
(-) → Hogging