Subject: Structure Analysis 2

Topic: Slope Diflection method

Difficulty: Hard

1) FEM's (Fixed End Moments)

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

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

$BC = M_{bc} = \frac{-wl^2}{12} = \frac{-25*4^2}{12} = -33.33KN.m$

$BC = M_{cb} = \frac{wl^2}{12} = \frac{25*4^2}{12} = 33.33KN.m$

$CD = M_{cd} = \frac{-wab^2}{l^2} = \frac{-50*1.5*3.5^2}{5^2} = -36.75KN.m$

$CD = M_{dc} = \frac{-wa^2b}{l^2} = \frac{-50*1.5^2*3.5}{5^2} = 15.75KN.m$

The Given frame is Sway Frame:

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

2) Slop deflection equation:

Member AB:

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

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

$\frac{1}{2} EIQ_b - \frac{3}{8} EI\gamma$ ---------(i)

$M_{ba} = M_{ba} + \frac{2EI}{l}(2Q_b + Q_a - \frac{3\gamma}{4})$

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

$=EIQ_b - \frac{3EI \gamma}{8}$-----------(ii)

Member BC:

$M_{bc} = -33.33 + \frac{2E(2I)}{4}(2Q_b + Q_c)$

$=-33.33 + 2EIQ_b + EIQ_c$ ------------------(iii)

$M_{cb} = 33.33 + \frac{2E(2I)}{4}(2Q_c + Q_b)$

$=33.33 + EIQ_b + 2EIQ_c$ -----------(iv)

Member CD:

$M_{cd} = -36.75 + \frac{2E(1.25I)}{5}(2Q_c + 0 - \frac{3\gamma}{5})$

$=-36.75 + EIQ_c - \frac{3}{10} EI \gamma$ -------------------(v)

$M_{dc} = 15.75 + \frac{2E(125I)}{5}(0 + Q_c - \frac{3 \gamma}{5})$

$=15.75 + \frac{1}{2}EIQ_c - \frac{3}{10} EI \gamma$ ---------------------(vi)

3) Apply Equilibrium Condition for joints:

Joint B:

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

$EIQ_b - \frac{3}{8}EI\gamma - 33.33 + 2EIQ_b + EIQ_c = 0$

$24EIQ_b + 8EIQ_c - 3EI\gamma = 266.64$ -----------(I)

Joint C:

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

$33.33 + EIQ_b + 2EIQ_c - 36.75 + EIQ_c - \frac{3}{10} EI\gamma = 0$

$10EIQ_b + 30EIQ_c - 3EI\gamma= 34.2$ -----------(II)

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

$Let us create 1 more structural cond^n$

$ΣM_B =0$

$M_{ab} + M_{ba} - H_a*4 = 0$

$H_a = \frac{M_{ab} + M_{ba}}{4}$ ---> (1)

$ΣM_C =0$

$M_{cd} + M_{dc} + 50*1.5 - H_d*5 = 0$

$H_d = \frac{M_{cd} + M_{dc} + 50*1.5}{5}$ ---> (2)

$ΣF_x = 0$

$H_a + H_d - 50$ ---> (3)

$\frac{M_{ab} + M_{ba}}{4} + \frac{M_{cd} + M_{dc} + 50*1.5}{5} - 50 =0$

$150 EIQ_b + 120EIQ_c - 123EI\gamma = -27680$ --------(III)

Solving all three equations we get:

$EIQ_b = -2.6389$

$EIQ_c = -12.2448$

$EI\gamma = -142.6440$

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

$M_{ab} = 52.17KN.m$

$M_{ba} = 50.85KN.m$

$M_{bc} = -50.85KN.m$

$M_{cb} = 6.20KN.m$

$M_{cd} = -6.20KN.m$

$M_{dc} = 52.42 KN.m$

6) BMD