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

Marks: 10 Marks

Year: Dec 2014

Degree of static indeterminancy, Dsi = 5 – 3 = 2

Assume Dsi = Number of Redundants

ΔL1 $= ∫ Mxx * M1 dx/EI \\ = ∫0-1 (0) + ∫0-1 (-20X)* (X+1) dx / EI + ∫0-4 [ (-20X * (X+1)- 6X²) * X+2 ] dx/EI \\ = -16.67 – 1066.67 – 640 = -1723.34 / EI$

ΔL2 $= ∫ Mxx * M2 dx/EI \\ = ∫ 0-4 [ -20 (X+1) – 6X² ] * X dx / EI = - 970.67 / EI$$

δ11 $= ∫ M1 M1 / EI dx = ∫0 -1 ( X )² dx/EI + ∫0-1 ( X+1 )² dx / EI + ∫0-4 ( X+2 )² dx/EI \\ =0.33 + 2.33 + 69.33 = 71.99/EI$

δ12 $= ∫ M1 M2 / EI dx = ∫0-4 ( X+2 ) * X dx/EI = 37.33 / EI$

δ22 $= ∫ M2 M2 / EI dx = ∫0-4 (X * X) dx/EI = 21.33/EI$

Apply Flexibility Equation ,

-1 723.34/EI + 71.99/EI 37.33/EI RA = 0

-970.67/EI 37.33/EI 21.33/EI RB = 0

By solving above matrix ,

RA = 3.69 KN , RB = 39.05 KN

Calculations : SFD & BMD

ΣMC = 0 , 3.69 * 6 + 39.05 * 4 - 20 * 5 - 12 * 4²/2 + MC = 0

MC = 17.66 KN

HC = 0

ΣFY = 0 , 3.69 – 20 + 39.05 – 12 * 4 + VC = 0

VC = 25.26 KN