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

Marks: 8 Marks

Year: May 2015

Degree of Kinematic Indeterminancy ( Dki ) = 2

Assign Co-ordinates , D1 = ΘB , D2 = ΘC

Forces developed in each co-ordinate direction ,

ADL1 = - FE MBA + FE MBC

= (- 20 * 3 * 2² ) / 5² + (20 * 4² ) / 12

= 17.06 KN-m

ADL2 = - FE MCB = (20 * 4² ) / 12 = -26.67 KN-m

Stiffness co-ordinates :

S11 = [ 4EI / L ] AB + [ 4 EI / L ] BC

= 4 EI / 5 + 4 EI / 4

= 1.8 EI

S12 = S21 = [ 2 EI / L ] BC = 2 EI / 4 = 0.5 EI

S22 = [ 4 EI / L ] = 4 EI / 4 = EI

Actual actions at displacement co-ordinates

F1 = 0 , F2 = 0

Apply stiffness equations ,

[ F ] = [ ADL ] + [ S ] [ D ]

[ 0 ] = [ 17.06 ] + EI [ 1.8 0.5 ] [ D1 ]

[ 0 ] = [ -26.67 ] + EI [ 0.5 1 ] [ D2 ]

1.8 D1 + 0.5 D2 = - 17.06

0.5 D1 + D2 = 26.67

We get ,

D1 = ΘB = - 19.60 / EI , D2 = ΘC = 36.47 / EI

End Moments :

MAB = 20 * 2 * 3² / 5² + 2 EI / 5 [ 0 + (- 19.60 / EI ) ]

= 6.56 KN-m

MBA = 20 * 2² * 3 / 5² + 2 EI / 5 [ -2 * (19.60 / EI ) ]

= - 25.28 KN -m

MBC = 20 * 4² / 12 + 2 EI / 4 * [ - 2 * ( 19.60 / EI ) + ( 36.47 / EI ) ]

= - 26.67 - 19.60 + 18.23 = 25.30 KN-m

MCB = - 20 * 4² / 12 + 2 EI / 4 * [ - 2 * (- 19.60 / EI ) + ( 2 * 36.47 / EI ) ]

= - 26.67 – 9.8 + 36.47 = 0