written 8.2 years ago by | modified 3.2 years ago by |
Mumbai University > Civil Engineering > Sem 5 > Structural Analysis – II
Marks: 8 Marks
Year: Dec 2014
written 8.2 years ago by | modified 3.2 years ago by |
Mumbai University > Civil Engineering > Sem 5 > Structural Analysis – II
Marks: 8 Marks
Year: Dec 2014
written 8.2 years ago by |
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 =∫M1∗M1/EIdx=∫0−5.83(xsin59)2+∫0−4(25)+∫0−5(4−x)2=134.15
δ12 =∫M1∗M2/EIdx=∫0−5.83(xsin59°cos59°)+∫0−4(5x+15)+∫0−5(28–7x)=1169.16
δ22 =∫M2∗M2/EIdx=∫0−5.83(cos59°)2+∫0−4(x+3)2+∫0−5(49)=351.88
Δ1L =∫M∗M1/EIdx=0+∫0−4(5∗30x2/2)+∫0−5(−24x−30∗4∗2)∗(4−x)=−400
Δ2L =∫M∗M2/EIdx=0+∫0−4(30x2/2)∗(x+3)+∫0−5(−168x–1680)=−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