Develop the flexibility and stiffness matrix for prismatic member AB with reference to the co-ordinates shown in fig.

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M1 : $ΣMa = 0 , Vb = 4/6 = 2/3 = 0.33 \\ ΣV = 0 , Va + Vb = 1 , Va = 4/6 = 0.67$

M2 : $ΣMa = 0 , Vb * 6 =1 , Vb = 1/6 = 0.167 \\ ΣV = 0 , Va + Vb = 0 , Va = -1/6 = -0.167$

Membes	Origin	Length	M1	M2	M
AC	A	0-2	2/3 X = 0.67X	1/6 = 1.67X	0
BC	B	0-4	2/3 X – 1/3 X = 0.33X	-1/6 X = - 1.67X	0

δ11 $= ∫ M1 M1 / EI dx = 1/ EI * [ ∫0 -2 (0.67X)² dx + ∫ 0-4 (0.33X)² dx] \\ = 1/ EI * [1.197 + 2.32] = 3.517 /EI$

δ12 $= ∫ M1 M2 / EI dx = 1/ EI * [ ∫0-2 [( 0.67X) * (-0.67X )] + ∫ 0-4 [(0.33X) * (0.167X)] dx] \\ = -0.298 + 1.176 = 0.878/EI$

δ22 $= ∫ M2 M2 / EI dx = 1/ EI * [ ∫0-2 ( 0.167X)² + ∫ 0-4 (-0.167X )² dx] \\ = 0.0743 + 0.595 = 0.67/EI$

$1/EI*$ $[0.3.517 -0.878 ] \\ [-0.878 0.67 ] ……………..\text{This is the Flexibility Matrix Form.} \\ \text{Stiffness Matrix is the inverse of Flexibility Matrix.}$

EI $[ 0.284 1.138 ] \\ [1.138 1.49 ] ………………………..\text{This is the Stiffness Matrix Form.}$

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