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

Calculation:
Given, Bending moment = 590.63 KN-m
Partial safety factor = 1.10 and Yield strength of steel ( ${\mathrm{f}}_{\mathrm{y}}$ ) = 250 N/ ${\mathrm{mm}}^2$
Plastic moment( ${\mathrm{M}}_{\mathrm{p}}$ ),
${\mathrm{M}}_{\mathrm{p}}$ = Bending moment × F.O.S
$=$ 590.63 × 1.1
$=$ 649.693 KN-m
${\mathrm{M}}_{\mathrm{p}}$ = 649.693 × ${10}^6$ N-mm
Relation between Plastic moment and plastic section modulus of a beam,
${\mathrm{M}}_{\mathrm{p}}={\mathrm{f}}_{\mathrm{y}}\times {\mathrm{Z}}_{\mathrm{p}}$
$649.693\times {10}^6=250\times {\mathrm{Z}}_{\mathrm{p}}$
${\mathrm{Z}}_{\mathrm{p}}=2.59877\times {10}^6{\mathrm{mm}}^3$
${\mathrm{Z}}_{\mathrm{p}}=2598.77{\mathrm{cm}}^{3.}$