SSC JE Civil Engineering 23-Sep-2019 Shift 1 Solved Paper

As we known the that fixed both ends beam-UDL${\mathit{M}}_{\max }$ (at ends) ................................. = $\frac{\mathit{w}{\mathit{L}}^2}{12}$
M${}_1$ (at centre) .................................. = $\frac{\mathit{w}{\mathit{L}}^2}{24}$
M${}_{\mathit{x}}$................................ = $\frac{\mathit{w}}{12}\left(6\mathit{L}\mathit{x}-{\mathit{L}}^2-6{\mathit{x}}^2\right)$
R = V ......................................= $\frac{\mathit{w}\mathit{L}}{2}$
D${}_{\max }$ (at centre) .......................... $\frac{\mathit{w}{\mathit{L}}^4}{384\mathit{E}\mathit{I}}$
D${}_{\mathit{x}}$................................ $\frac{\mathit{w}{\mathit{x}}^2}{24\mathit{E}\mathit{I}}{\left(\mathit{L}-\mathit{x}\right)}^2$
x (points of contraflexure) ...........$\left(\sqrt{3}-3\right)\mathit{L}$
Hence the fixed end moment for the beam carrying udl of 'w' kN/m${}^2$ and length of 'L' m:
${\mathit{M}}_{\mathit{A}}={\mathit{M}}_{\mathit{B}}=\pm \frac{\mathit{W}{\mathit{L}}^2}{12}$