Subject: Structure Analysis 2

Topic: Slope Diflection method

Difficulty: Hard

FEM's (Fixed End Moments)

$AB = M_{ab} = \frac{-wab^2}{l^2} = \frac{-30*3*2^2}{5^2} = -14.4KN.m$

$AB = M_{ba} = \frac{wa^2b}{l^2} = \frac{30*3^2*2}{5^2} = 21.6KN.m$

$BC = M_{bc} =M_{cb} =0 **2) Slop deflection equation:** *Member AB:* $M_{ab} = M_{ab} + \frac{2EI}{l}(2Q_a + Q_b)$ $=-14.4 + \frac{2E(2I)}{5} (0+Q_b)$ $M_{ab} =-14.4 + 0.8EIQ_b$ $M_{ba} = M_{ba} + \frac{2EI}{l}(2Q_b + Q_a)$ $=21.6 + \frac{2E(2I)}{5} (0+2Q_b)$ $M_{ba}=21.6 + 1.6EIQ_b$ *Member BC[Note, the end c is hinged]:* $M_{bc} = M_{bc} - \frac{M_{cb}}{2} + \frac{3EIQ_b}{l}$ $=0 - 0 + \frac{3EI}{2.5} Q_b$ $M_{bc} =1.2EIQ_b$ **3) Apply Equilibrium Condition for joints:** *Joint B:* $M_{ba} + M_{bc} = 0$ $21.6 +1.6EIQ_b + 1.2 EIQ_b = 0$ $2.8EIQ_b = -21.6$ $EIQ_B = \frac{-54}{7}$ **4) To find moments at ends:** Substituting value of $Q_b$ in moment equation $M_{ab} = -14.4 + (0.8)\frac{-54}{7} = -20.57 KN.m$ $M_{ba} = +21.6 + (1.6)\frac{-54}{7} = 9.26 KN.m$ $M_{bc} = 1.2(\frac{-54}{7}) = -9.26 KN.m$ $M_{cb} =0$

5) BMD