Subject: Structure Analysis 2

Topic: Slope Diflection method

Difficulty: Hard

1) Fixed end moments (FEM's)

$AB =\gt M_{ab} = \frac{-wab^2}{l^2} = \frac{-45*2*4^2}{6^2} = -40KN.m$

$AB =\gt M_{ba} = \frac{-wa^2b}{l^2} = \frac{45*2^2*4}{6^2} = 20KN.m$

$BC =\gt M_{bc} = \frac{-wab^2}{l^2} = \frac{-48*3*1^2}{4^2} = -9KN.m$

$BC =\gt M_{cb} = \frac{-wa^2b}{l^2} = \frac{48*3^2*1}{4^2} = 27KN.m$

2) Slop deflection equation:

Span AB:

$M_{ab} = M_{ab} + 2\frac{EI}{l} (2Q_a + Q_b)$

$=-40 + \frac{1}{3} EIQ_b $ ---------(i)

$M_{ba} = M_{ba} + 2\frac{EI}{l} (2Q_b + Q_a)$

$=20 + \frac{2}{3} EIQ_b $ ---------(ii)

Span BC:

$M_{bc} = M_{bc} + 2\frac{EI}{l} (2Q_b + Q_c)$

$=-9 + EIQ_b $ ---------(iii)

$M_{cb} = M_{cb} + 2\frac{EI}{l} (2Q_c + Q_b)$

$=27 + \frac{1}{2} EIQ_b $ ---------(iv)

3) Apply equillibrium condition for joints:

Joint B:

$M_{ba} + M_{bc} =0$

$20 + \frac{2}{3} EIQ_b - 9 + EIQ_b=0$

$\frac{5}{3}EIQ_b = -11$

$EIQ_b = -6.60$

4) To find moments atends:

Substituting value of $Q_b$ in moment eg (i) to (iv)

Final moments:

$M_{ab} = -40 + \frac{1}{3}(-6.60) = -42.20KN.m$

$M_{ba} = 20 + \frac{2}{3}(-6.60) = 15.60KN.m$

$M_{bc} = -9 +(-6.60) = -15.60KN.m$

$M_{cb} = 27 + \frac{1}{2}(-6.60) = 23.70KN.m$

BMD:

Reaction:

$B.MatB = V_a *6-47*4 - 42.20 = -15.60 =\gtV_a =34.433 KN $

$B.MatB = V_c*4 - 48*3 - 23.70 = -15.60 =\gt V_c = 38.025KN$