Subject: Structure Analysis 2

Topic: Slope Diflection method

Difficulty: Hard

FEM's (Fixed End Moments)

$AB = M_{ab} = \frac{-wl^2}{12} = \frac{-40*3^2}{12} = -30KN.m$

$AB = M_{ba} = \frac{wl^2}{12} = \frac{40*3^2}{12} = 30.m$

$BC = M_{bc} = \frac{-wl}{8}=\frac{-100*2}{8} = -25KN.m$

$BC = M_{cb} = \frac{wl}{8}=\frac{100*2}{8} = 25KN.m$

$CD = M_{cd} = \frac{-wl^2}{12} = \frac{-50*3^2}{12} = -37.5KN.m$

$AB = M_{dc} = \frac{wl^2}{12} = \frac{50*3^2}{12} = 37.5.m$

2) Slop deflection equation:

Span AB:

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

$=-30 + \frac{2(7000)}{3} (Q_b - \frac{3*2.5}{3000} )$

$=-41.667 + \frac{14000}{3}Q_b$

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

$=30 + \frac{2(7000)}{2} (2Q_b + 0 - \frac{3*2.5}{3000})$

$=18.333 + \frac{28000}{3}Q_b$

Span BC:

$M_{bc} = M_{bc} + \frac{2EI}{l}[2Q_b + Q_c + \frac{3 \gamma}{l}]$

$=-25 + \frac{2(7000)}{2} [2Q_b + Q_c + \frac{3*2.5}{2000}$

$= 1.25 + 14000Q_b + 7000Q_c$

$M_{cd} = M_{cd} + \frac{2EI}{l}[2Q_c + Q_b + \frac{3 \gamma}{l}]$

$=25 + \frac{2(7000)}{2} [2Q_c + Q_b + \frac{3*2.5}{3000}$

$= 51.25 + 7000Q_b + 14000Q_c$

Span CD:

$M_{cd} = M_{cd} + \frac{2EI}{l}[2Q_c + Q_d]$

$=-37.50 + \frac{2(7000)}{3} [2Q_c]$

$= -37.50 + \frac{28000}{3}Q_c$

$M_{dc} = M_{dc} + \frac{2EI}{l}[2Q_d + Q_c]$

$=37.50 + \frac{2(7000)}{3} [Q_c]$

$= 37.50 + \frac{14000}{3}Q_c$

3) Apply Equilibrium Condition for joints:

Joint B:

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

$(18.3333 + \frac{28000}{3} Q_b) + (1.25 + 14000Q_b + 7000Q_c) = 0$

$\frac{70000}{3}Q_b + 7000Q_c = -19.5833$

$10Q_B + 3Q_c = -0.008392848429$-----------(i)

Joint C:

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

$(51.25 + 7000Q_b + 14000Q_c) - 37.50 + \frac{28000}{3}Q_c) = 0$

$7000Q_b + \frac{7000}{3}Q_c = -13.75$

$3Q_b + 10Q_c = -0.0058928571$-----------(i)

Solving (i) & (ii) we get

$Q_b = 0.000728$

$Q_c = 0.000370879$

Substituting value of $Q_b$ & $Q_c$ in moment equation (i) to (iii)

$M_{ab} = -41.6667 + \frac{14000}{3}(-0.000728) = -45.11 KN.m$

$M_{ba} = 11.54 KN.m$

$M_{bc} = -11.5KN.m$

$M_{cb} = 40.96$

$M_{cd} = -40.96$

$M_{dc} = 35.77$