Draw BMD

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

Marks: 8 Marks

Year: Dec 2015 ,May 2016

A1 $= 2/3 * h * l = 2/3 * 20 * 4 \\ = 106.67 m²$

A2 $= ½ * b* h = ½ * 5 * 48 \\ = 120 m²$

Cg1 = cg of triangle from Right hand side = ( L + b ) / 3 = ( 5 + 3 ) / 3 = 2.67 m

Cg2 = cg of udl = L / 2 = 4 / 2 = 2 m

Loading of triangle = wab / L = 40 * 2 * 3 / 5 = 48

Loading of udl = wL² / 8 = 20 * 4² / 8 = 40

For A’A & AB , Use 3 - moment theorem

MA’ * L1 + 2 MA ( L1 + L2 ) + MB * l2

= -(6 * A1 * cg1 / L1 ) – (6 * A2 * cg2 / L2 )

0 + 2 MA * 5 + MB * 4 = 0 – ( 6 * 106 * 2/4 )

10 MA + 4 MB = - 318 ……………………………………………………………. ( 1 )

For AB & BC , Use 3 - moment theorem

MA *5 + 2 MB * (9) + MC *4 = - (6 * 106 *2 / 4 ) – (6 * 120 * 2.67 / 5 )

5 MA + 18 MB + 0 = - 318 - 384.48

5 MA + 18 MB = - 702. 48 ………………………………………………………………… ( 2 )

By equating equations (1) & (2) ,

MA = -18.21 KN-m

MB = - 34 KN-m