Subject: Structure Analysis 2

Topic: Slope Diflection method

Difficulty: Hard

1) Fixed End moments(FEM's)

AB => $M_{ab} = \frac{-wl^2}{12} = \frac{-4*3^2}{12} = -3KN.m$

$M_{ba} = \frac{wl^2}{12} = \frac{4*3^2}{12} = 3KN.m$

BC => $M_{bc} = \frac{-wl^2}{12} = \frac{-5*4^2}{12} = -6.67KN.m$

$M_{cb} = \frac{wl^2}{12} = \frac{5*4^2}{12} = 6.67KN.m$

2) blop deflection equation:

we know that slope at A=$Q_a = 0$

and slope at B=$Q_b =0$

Span AB:

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

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

$=-3 + \frac{2}{3} EIQ_B = -3 +0.67EIQ_B$--------------------(i)

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

$= 3 + \frac{2EI}{3} (2Q_a + 0)$

$=3+ \frac{4}{3} Q_b$

$=3+1.33EIQ_b$ ----------(ii)

Span BC:

$M_{bc} = M_{bc} + \frac{2EI}{l} (2Q_a + Q_c)$

$= -6.67 + \frac{2EI}{4} (2Q_a + 0)$

$= -6.67 + \frac{4}{4} Q_b$

$= -6.67 + EIQ_b$ ----------------(iii)

$M_{cb} = M_{cb} + \frac{2EI}{l} (2Q_c + Q_b)$

$= 6.67 + \frac{2EI}{4} (0+Q_b)$

$= 6.67 + 0.5EIQ_b$ ----------------(iv)

3) Apply Equilibrium Condition for joint:

Joint B:

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

$3 + 1.33EIQ_b - 6.67 + EIQ_b = 0$

$Q_b = \frac{1.57286}{EI}$

4) End Moments of each member:

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

$M_{ab} = -3 + \frac{2}{3}(1.57286) = -1.95KN.m$

$M_{ba} = +3 + \frac{4}{3}(1.57286) = 5.10KN.m$

$M_{bc} = -6.76 + (1.57286) = -5.10KN.m$

$M_{cb} = 6.76 + \frac{1}{2}(1.57286) = 7.46KN.m$

5) B.M.D: