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

Marks: 10 Marks

Year: May 2015

Degree of Static Indeterminancy

( Dsi ) e = R-E = 3-3 = 0

( Dsi ) i = m ( 2j – 3 ) = 8 – ( 2*5 – 3 ) = 1

AC considered as a Redundant.

P analysis :

Apply condition of equilibrium

ΣMA = 0

= 20 * 3 + 20*3 – RE * 6 = 0

RE = 20 KN

Σ FX = 0

HA + 20 = 0

HA = -20 KN

ΣFY = 0

VA – 20 -20 + 20 = 0

VA = 20 KN

Θ = tan -1 ( 3/3 ) = 45°

Consider joint A ,

From fig ,

P AB = 20 KN

P AD = 20 KN

Consider joint B ,

Apply condition of equilibrium

ΣFY = 0

-20 +20 + P BD sin 45 = 0

P BD = 0

P BC = 20 KN

Consider joint C ,

Σ FX = 0

-20 – P CE cos 45 = 0

P CE = 28.28 KN

ΣFY = 0

-20 + P CD + 28.28 cos 45 = 0

P CD = 0

Consider joint E ,

Σ FX = 0

-P DE + 28.28 cos 45 = 0

P DE = 20 KN

K – analysis

Σ MA = 0

cos 45 * 3 - sin 45 * 3 – VE *6 = 0

VE = 0

Σ FY = 0

VA + cos 45 – cos 45 + 0 = 0

VA = 0

Σ FX = 0 , HA = 0

Consider joint A ,

K AB = K AD = cos 45 = 0.707 KN

Consider joint E ,

K DE = K CE = 0

KCD = K CB = 0.707 KN

Forces :

Compression – ve

Tension +ve

Correction factor

X = - { [ Σ ( PKL / AE) ] / [ Σ ( K²L / AE ) ] } = -( 4.242 / 10.42 )

X = - 0.414 KN