Subject: Structure Analysis 2

Topic: Slope Diflection method

Difficulty: Hard

FEM's (Fixed End Moments)

$AB = M_{ab} = \frac{-wl}{8} = \frac{-100*5}{8} = -62.50KN.m$

$AB = M_{ba} = \frac{wl}{8} = \frac{100*5}{8} = 62.50KN.m$

$BC = M_{bc} = \frac{-wl^2}{12}=\frac{-24*6^2}{12} = -72KN.m$

$BC = M_{cb} = \frac{wl^2}{12}=\frac{24*6^2}{12} = 72KN.m$

2) Slop deflection equation:

Span AB:

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

$=-62.50 + \frac{2EI}{5} (0+Q_b)$

$=-62.50 + \frac{2}{5}EIQ_b$---------------(i)

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

$=62.50 + \frac{2EI}{5} (0+2Q_b)$

$=62.50 + \frac{4}{5}EIQ_b$-----------------(ii)

Span BC:

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

$=-72 - \frac{72}{2} + \frac{3E(3I)}{6} Q_b$

$M_{bc} = -108 + \frac{3}{2}EIQ_b$------------(iii)

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

3) Apply Equilibrium Condition for joints:

Joint B:

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

$(62.50 + \frac{4}{5} Q_b) - (108 + \frac{3}{2}EIQ_b) = 0$

$2.3EIQ_b = 45.5$

$EIQ_B = 19.7826$

4) To find moments at ends:

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

$M_{ab} = -62.50 + \frac{2}{5}(19.7826) = -54.59 KN.m$

$M_{ba} = +62.50 + \frac{4}{5} (19.7826) = 78.33 KN.m$

$M_{bc} = -108 +\frac{3}{2}(19.7826) = -78.33 KN.m$

$M_{cb} =0$