SSC JE Civil Engineering 30-Oct-2020 Shift 2 Paper

Mohr’s Theorem I:
The angle between the two tangents drawn on the elastic line is equal to the area of the Bending Moment Diagram between those two points divided by flexural rigidity.
$\mathrm{\theta }=\frac{\left[\mathrm{Area}\mathrm{of}\mathrm{bending}\mathrm{moment}\mathrm{diagram}\right]}{\mathrm{EI}}$
Mohr’s Theorem II:
The deviation of a point away from the tangent drawn from the other point is given by the moment of area of bending moment diagram about the first point divided by flexural rigidity.
$\mathrm{\delta }={\mathrm{BB}}^{}=\frac{\left[\mathrm{Area}\mathrm{of}\mathrm{bending}\mathrm{moment}\mathrm{diagram}\right]\times \overline{\mathrm{x}}}{\mathrm{EI}}$