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

Marks: 12 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 * 4 ) / 8 + (10 * 3² ) / 12

= 17.5 KN-m

ADL2 = FE MCB + FE MCE + FE MCD

= - (10 *3²) / 12 + ( 20 * 2 * 3 ) / 5² + (20 * 4) / 8

= 16.9 KN-m

Stiffness co-ordinates :

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

= 4 EI / 4 + 4 EI / 3

= 2.33 EI

S12 = S21 = [ 2 EI / L ] = 2 EI / 3 = 0.67 EI

S22 = [ 4 EI / L ] BC + [ 4EI / L ] CD + [ 4 EI / L ] CE = 4 EI / 3 + 4 EI / 4 + 4 EI / 5

= 1.33 EI + EI + 0.8 EI

= 3.13 EI

Actual actions at displacement co-ordinates

F1 = 0 , F2 = 0

Apply stiffness equations ,

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

[ 0 ] = [ 17.5 ] + EI [ 2.33 0.67 ] [ ΘB ]

[ 0 ] = [ 16.9 ] + EI [ 0.67 3.13 ] [ ΘC ]

2.33 EI ΘB + 0.67 EI ΘC = 17.5

0.67 EI ΘB + 3.13 EI ΘC = - 16.9

We get ,

ΘB = - 6.34 / EI , ΘC = - 4.04 / EI

End Moments:

MAB = - 20 * 4 / 8 + 2 EI / 4 * [ 0 + (-6.34 / EI) + 0 ]

= -10 – 3.17 = -13.17 KN-m

MBA = 20 * 4 / 8 + 2 EI / 4 * [ 0 + 2 EI ( -6.34 / EI ) ]

= 10 – 6.34 = 3.66 KN-m

MBC = 10 * 3² / 12 + 2 EI / 3 * [ 2 EI * ( - 6.34 / EI ) + ( -4.04 / EI ) ]

= 7.5 – 8.45 – 2.70 = - 3.65 KN-m

MCB = -10 * 3² / 12 + 2 EI / 3 * [ - 6.34 / EI - 2 * 4.04 / EI ]

= - 7.5 – 9.67 = - 17.16 KN –m

MCD = 20 *4 / 8 + 2 EI / 4 [ 2 EI * (- 4.04 / EI ) + 0 ]

= 10 – 2.02 = 7.98 KN-m

MCE = 20 * 2 * 3² / 5² + 2 EI / 5 [ 2 EI * (- 4.04 / EI ) + 0 ]

= 14.4 – 3.23 = 11.17 KN –m

MEC = 20 * 2 * 3² / 5² + 2 EI / 5 [ EI * (- 4.04 / EI ) + 0 ]

= 7.98 KN –m

MDC = - 20 * 4 / 8 + 2 EI / 4 * [ (-4.04 / EI) + 0 ]

= -10 - 2.02 = - 12.02 KN -m