Subject: Structure Analysis 2

Topic: Slope Diflection method

Difficulty: Hard

FEM's (Fixed End Moments)

$AB = M_{ab} = \frac{-wl}{8} = \frac{-32*4}{8} = -16KN.m$

$AB = M_{ba} = \frac{wl}{8} = \frac{32*4}{8} = 16KN.m$

$BC = M_{bc} = \frac{-wl}{8} = \frac{-48*4}{8} = -18KN.m$

$BC = M_{cb} = \frac{wl}{8} = \frac{48*4}{8} = 18KN.m$

$CD = M_{cd} = 0$

$CD = M_{dc} = 0$

2) Slop deflection equation:

Member AB:

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

$=-16 + \frac{2E(2I)}{4} (0+Q_b)$

$=-16 + EIQ_b$-------------(i)

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

$=16 + \frac{2E(2I)}{4} (0+2Q_b)$

$=16 + 2EIQ_b$-----------(ii)

Member BC:

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

$=-18 + \frac{2EI}{3} (2Q_b + Q_c)$

$=-18 + \frac{4}{3}EIQ_b + \frac{2}{3}Q_c$ ------------------(iii)

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

$=18 + \frac{2EI}{3} (2Q_c + Q_b)$

$=18 + \frac{4}{3}EIQ_c + \frac{2}{3}Q_b$

Member CD:

$M_{cd} = M_{cd} + \frac{2EI}{l}(2Q_c + Q_d)$

$=0 + \frac{2EI(2I)}{4} (2Q_c + 0)$

$=2EIQ_c$ ------------------(iv)

$M_{dc} = M_{dc} + \frac{2EI}{l}(2Q_d + Q_c)$

$=0 + \frac{2EI(2I)}{4} (0+ Q_c)$

$=EIQ_c$ ---------------------(v)

3) Apply Equilibrium Condition for joints:

Joint B:

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

$16 + 2EIQ_b - 18 + \frac{4}{3}EIQ_b + \frac{2}{3} EIQ_c = 0$

$-5EIQ_b + EIQ_c = 3$ -----------(I)

Joint C:

$M_{cb} + M_{cd} = 0$

$18 + \frac{2}{3}EIQ_b + \frac{4}{3}EIQ_c + \2EIQ_c = 0$

$EIQ_b + 5EIQ_c = -27$ -----------(II)

Solving (I) & (II) we get:

$EIQ_b = 1.75$

$EIQ_c = -5.75$

4) To find moments at ends:

Substituting value of $EIQ_b$ & $EIQ_c$ in final moments are obtained

$M_{ab} = -16 + 1.75 = -14.25 KN.m$

$M_{ba} = 16 + 2(1.75) = 19.50 KN.m$

$M_{bc} = -18 + 1.75(\frac{4}{3}) + \frac{2}{3}(-5.75) = -19.50 KN.m$

$M_{cb} =18 + 1.75(\frac{2}{3}) + \frac{4}{3}(-5.75) = 11.50 KN.m$

$M_{cd} = 2(-5.75) = -11.50KN.m$

$M_{dc} = -5.75 KN.m$

5) BMD