Analyse the frame and draw BMD by using Flexibility Method

54views

written 7.8 years ago by

teamques10 ★ 68k

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 $= ∫ M1*M1 /EI dx = ∫0-5.83 ( x sin 59 )^2 + ∫0-4 (25 ) + ∫0-5 (4 - x)^2 \\ = 134.15$

δ12 $= ∫ M1*M2 / EI dx = ∫0-5.83 (x sin 59° cos 59° ) + ∫0-4 (5x + 15 ) + ∫0-5 ( 28 – 7x ) \\ = 1169.16$

δ22 $= ∫ M2*M2 / EI dx = ∫0-5.83 ( cos 59° )^2 + ∫0-4 ( x+3 )^2 + ∫0-5 ( 49) \\ = 351.88$

Δ1L $= ∫ M*M1 / EI dx = 0 + ∫0-4 (5* 30x^2 / 2 ) + ∫0-5 ( -24x - 30*4*2 )* (4-x) \\ = -400$

Δ2L $= ∫ M*M2 /EI dx = 0 + ∫0-4 ( 30x^2 / 2 ) * (x+3) + ∫0-5 ( - 168x – 1680 ) \\ = -8580$

Forces P1 & P2 , are calculated as follows ,

[ 400 ] = [ P1 ] [134.15 1169.16]

[ 8580 ] [ P2 ] [1169.16 351.88]

Forces : $P1 = 7.5 KN \\ P2 = -0.52 KN$

enter image description here

ADD COMMENT EDIT