Subject: Structure Analysis 2

Topic: Slope Diflection method

Difficulty: Hard

FEM's (Fixed End Moments)

$AB = M_{ab} = \frac{-wl^2}{12} = \frac{-10*6^2}{12} = -30KN.m$

$AB = M_{ba} = \frac{wl^2}{12} = \frac{10*6^2}{12} = 30KN.m$

$BC = M_{bc} = \frac{-wl^2}{12}=\frac{-20*3^2}{12} = -15KN.m$

$BC = M_{cb} = \frac{wl^2}{12}=\frac{20*3^2}{12} = 15KN.m$

2) Slop deflection equation:

Span AB:

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

$=-30 + \frac{2EI}{6} (0+Q_b)$

$=-30 + \frac{1}{3}EIQ_b$---------------(i)

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

$=30 + \frac{2EI}{6} (0+2Q_b)$

$=30 + \frac{2}{3}EIQ_b$-----------------(ii)

Span BC:

$M_{bc} = M_{bc} - \frac{M_{cb}}{2} + \frac{3EI}{l}(Q_b)$

(Note, c is simply supported)

$=-15 - \frac{15}{2} + \frac{3EI}{3} Q_b$

$M_{bc} = -22.5 + EIQ_b$------------(iii)

$M_{cb} = 0$ (Simply Supported)

3) Apply Equilibrium Condition for joints:

Joint B:

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

$(30 + \frac{2}{3} Q_b) + (-22.5 + EIQ_b) = 0$

$\frac{5}{3} EIQ_b = -7.5$

$EIQ_B = -4.5$

4) To find moments at ends:

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

$M_{ab} = -30 - 1.50 = -31.50 KN.m$

$M_{ba} = +30 - 3 = 27 KN.m$

$M_{bc} = -22.5 +EI(-4.5) = -27 KN.m$

$M_{cb} =0$