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

As we known the that fixed both ends beam-UDLMmax (at ends) ................................. =

wL2

12

M‌1 (at centre) .................................. =

wL2

24

M‌x................................ =

w

12

(6Lx−L2−6x2)
R = V ......................................=

wL

2

D‌max (at centre) ..........................

wL4

384EI

D‌x................................

wx2

24EI

(L−x)2
x (points of contraflexure) ...........(√3−3)L
Hence the fixed end moment for the beam carrying udl of 'w' kN/m‌2 and length of 'L' m:
MA=MB=±

WL2

12