Subject: Structure Analysis 2

Topic: Slope Diflection method

Difficulty: Hard

1) FEM's (Fixed End Moments)

$AB = M_{ab} = 0$ (Because of no load at member AB)

$AB = M_{ba} = 0$ (Because of no load at member AB)

$BC = M_{bc} = \frac{-wl}{8} = \frac{-48*4}{8} = -24KN.m$

$BC = M_{cb} = \frac{wl}{8} = \frac{48*4}{8} = 24KN.m$

$CD = M_{cd} = 0$

$CD = M_{dc} = 0$

[Assumed Sway always at right]

The Given frame is Sway Frame:

Assuming that the frame Sway by an amount '\gamma' towards right

2) Slop deflection equation:

Member AB:

$M_{ab} = 0$ {Frame Kase}

$M_{ba} = M_{ba} - \frac{M_{ab}}{2} +\frac{3EI}{l}(Q_b - \frac{\gamma}{l})$

$=0 - \frac{0}{2} + \frac{3E(2I)}{6} (Q_b - \frac{\gamma}{6})$

$=EIQ_b - \frac{1EI \gamma}{6}$-----------(i)

Member BC:

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

$=-24 + \frac{2E(2I)}{4} (2Q_c + Q_b)$

$=-24 + 2EIQ_b + 2EIQ_c$ ------------------(ii)

$M_{cb} = M_{cb} + \frac{2EI}{l}(2Q_c + Q_d - \frac{3 \gamma}{l})$

$=18 + 2EIQ_c + 2EIQ_c$ -----------(iii)

Member CD:

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

$=0 + \frac{2EI}{4} (2Q_c + 0 - \frac{3 \gamma}{4})$

$=EIQ_c - \frac{3}{8} EI \gamma$ ------------------(iv)

$M_{dc} = M_{dc} + \frac{2EI}{l}(2Q_d + Q_c - \frac{3 \gamma}{l})$

$=0 + \frac{2EI)}{4} (0+ Q_c - \frac{3 \gamma}{4})$

$=\frac{1}{2}EIQ_c - \frac{3}{8} EI \gamma$ ---------------------(v)

3) Apply Equilibrium Condition for joints:

Joint B:

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

$EIQ_b - \frac{1}{6}EI\gamma -24 + 2EIQ_b + EIQ_c = 0$

$18EIQ_b + 6EIQ_c - EI\gamma = 144$ -----------(I)

Joint C:

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

$24 + EIQ_b + 2EIQ_c + EIQ_c - \frac{3}{8} EI\gamma = 0$

$8EIQ_b + 24EIQ_c - 3EI\gamma= -192$ -----------(II)

These are three unknown $Q_b, Q_c$ & $\gamma$, where as equation are only two:

$Let us create 1 more structural cond^n$

$ΣM_B =0$

$M_{ab} + M_{ba} - H_a*6 = 0$

$H_a = \frac{M_{ab} + M_{ba}}{6}$ ---> (1)

$ΣM_C =0$

$M_{cd} + M_{dc} - H_d*4 = 0$

$H_d = \frac{M_{cd} + M_{dc}}{4}$ ---> (2)

$ΣF_x = 0$

$H_a + H_d$ ---> (3)

$\frac{M_{ab} + M_{ba}}{6} + \frac{M_{cd} + M_{dc}}{4} =0$

$24 EIQ_b + 54EIQ_c - 31EI\gamma = 0$ --------(III)

Solving all three equations we get:

$EIQ_b = 11.7671$

$EIQ_c = -13.7852$

$EI\gamma = -14.9030$

5) Substituting value of $EIQ_b, EI\gamma$ & $EIQ_c$ in final moments are determined

$M_{ab} = 0$

$M_{ba} = 11.7671 - \frac{1}{6}(-14.9030) = 14.25 KN.m$

$M_{bc} = -24 + 2(11.7671) -13.7852 = -14.25 KN.m$

$M_{cb} =24 + 11.7671 + 2(-13.7852) = 8.20 KN.m$

$M_{cd} = -13.7852 - \frac{3}{8} (-14.9030) = -8.20KN.m$

$M_{dc} = -1.30 KN.m$

6) BMD