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

Marks: 6 Marks

Year: May 2016

For AB , $A = ( 2/3 ) * 31.25 * 5 = 104.68 m² \\ X = 5/2 = 2.5 m$

For BC , $A = 26.67 * 6-2 * ½ * 26.67 * 2 = 106.68 m² \\ X = 6/2 = 3.5 m$

Use 3 – Moment Theorem

Assume imaginary span A’A on left side of fixed support

For span A’AB ,

MA’ * L’ + 2 MA (L’ + L1 ) + MB * L1 = 6 A’X’ / L’ + 6 A1 X1 / L1

0 + 2 MB ( 0 + 5 ) + 5 * MB = 0 + 6 * 104.68 * 2.5 / 5

10 MA + 5 MB = 314.04 ……………………………….. (1)

For span ABC ,

MA * L1 + 2 MB (L1 + L2 ) + MC * L2 = 6 A1 L1 / L1 + 6 A2 X2 / L2

5 MA + 2 MB ( 5 + 6 ) + 0 * 6 = 6 * 104.68 * 2.5 / 5 + 6 * 106.68 * 3.5 / 6

5 MA + 22 MB = 687.42 …………………………....… (2)

By solving (1) & (2) ,

MA = 17.80 KN-m

MB = 27.2 KN-m