0
2.8kviews
Analyse the frame and draw BMD by using Flexibility Method

enter image description here

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

Marks: 8 Marks

Year: Dec 2014

1 Answer
0
55views

Θ = tan -1 [ 5/3 ] = 59°

DC inclined distance = sq.rt 5^2 + 3^2 = 5.83 m

Dsi = r – 3 = 5 – 3 = 2

Select support D as redundant. No moment at D bcoz hinge at D.

Remove all actual loads & apply unit load at D,

Members Origin Length M M1 M2
DC D 5.83 0 1xsin 59° 1*cos 59°
CB C 4 -30*x^2 / 2 1*5 1*x + 3
BA B 5 -24x – 3042 -1*x + 4 1*7

δ11 =M1M1/EIdx=05.83(xsin59)2+04(25)+05(4x)2=134.15

δ12 =M1M2/EIdx=05.83(xsin59°cos59°)+04(5x+15)+05(287x)=1169.16

δ22 =M2M2/EIdx=05.83(cos59°)2+04(x+3)2+05(49)=351.88

Δ1L =MM1/EIdx=0+04(530x2/2)+05(24x3042)(4x)=400

Δ2L =MM2/EIdx=0+04(30x2/2)(x+3)+05(168x1680)=8580

Forces P1 & P2 , are calculated as follows ,

[ 400 ] = [ P1 ] [134.15 1169.16]

[ 8580 ] [ P2 ] [1169.16 351.88]

Forces : P1=7.5KNP2=0.52KN

enter image description here

Please log in to add an answer.