Mumbai University > Civil Engineering > Sem 5 > Structural Analysis – II

Marks: 8 Marks

Year: Dec 2014

Area Calculations :

For AB , a = ½ * 2 * 20 = 20

For BC , a = 2/3 * 4 * 20 = 53.33

For CD , a = ½ * [ 4+2] * 20 = 60

Apply 3 – Moment Theorem , to span ABC & BCD ,

At support A & D , MA = MD = 0

ABC , X1 = 1 m from A , X2 = 2 m from C

MA [ L1 ] + 2 MB [ L1 + L2 ] + MC [ L2 ] = - 6a1 x1 / L1 - 6 a2 x2 / L2

2MB [ 2+4 ] + MC *4 = - 6 *20 * 1 / 2 - 6 *53.33 / 4

12 MB + 4 MC = - 140 ………………………………. (1)

BCD , x1 = 2 m from B , x2 = 2 m from D

MB * L1 + 2 MC [ L1 + L2 ] + MD *L2 = - 6a1 x1 / L1 - 6 a2 x2 / L2

4 MB + 2 MC 8 = - - 6 *53.33 * 2 / 4 - 6 60 * 2 / 4

4 MB + 16 MC = - 340 …………………….………… (2)

Solve (1) & (2) ,

MB = - 5 KN-m

MC = -20 KN-m