Slope deflection: beam with rotations and settlements

Assumptions

• $EI$ is not given. Take $EI = 10{,}000$ kN·m² (the moments caused by the support rotation and settlement scale with $EI$; load moments do not).
• "Support b rotates" is read as a typing slip for the other fixed support d (b is a free interior joint), so the prescribed rotations are $\theta_a = 0.001$ rad clockwise and $\theta_d = 0.001$ rad anticlockwise. The joints $b$ and $c$ rotate freely and settle 10 mm.
• Spans: ab = 4 m ($3EI$), bc = 5 m ($4EI$), cd = 4 m ($3EI$). Clockwise end moments are positive.

Data

• Chord rotations (clockwise +): $\psi_{ab} = 0.01/4 = +0.0025$, $\psi_{bc} = 0$, $\psi_{cd} = -0.01/4 = -0.0025$.
• FEM: ab: $\pm PL/8 = \pm 10$; bc: $\pm wL^2/12 = \pm 2\times 25/12 = \pm 4.1667$; cd: $\pm 10$.
• $\theta_a = +0.001$, $\theta_d = -0.001$; unknowns $\theta_b$, $\theta_c$.

Slope-deflection equations

$M_{near} = FEM + \dfrac{2EI}{L}(2\theta_{near} + \theta_{far} - 3\psi)$

$$\begin{aligned} M_{ab} &= -10 + 15000(2\times0.001 + \theta_b - 0.0075) = -92.5 + 15000\,\theta_b \\ M_{ba} &= +10 + 15000(2\theta_b + 0.001 - 0.0075) = -87.5 + 30000\,\theta_b \\ M_{bc} &= -4.1667 + 16000(2\theta_b + \theta_c) \\ M_{cb} &= +4.1667 + 16000(2\theta_c + \theta_b) \\ M_{cd} &= -10 + 15000(2\theta_c - 0.001 + 0.0075) = 87.5 + 30000\,\theta_c \\ M_{dc} &= +10 + 15000(-0.002 + \theta_c + 0.0075) = 92.5 + 15000\,\theta_c \end{aligned}$$

Joint equilibrium

$$\begin{aligned} M_{ba} + M_{bc} = 0 &: \; 62000\,\theta_b + 16000\,\theta_c = 91.667 \\ M_{cb} + M_{cd} = 0 &: \; 16000\,\theta_b + 62000\,\theta_c = -91.667 \end{aligned}$$

Solving: $\theta_b = +0.0019928$ rad (clockwise), $\theta_c = -0.0019928$ rad.

Final end moments (kN·m)

The settlement of b and c turns the support moments into sagging moments. The result is symmetrical about the mid-point of bc, as the loading, rotations and settlements are.

Free body diagrams: shears and reactions

Span-wise (shear at left end, taking moments of each span):

• ab: $V_{a} = 32.58$ kN up; after the 20 kN load $V = 12.58$ kN; $V_b$(left) = 12.58 kN.
• bc: $V_b$(right) = +5.00 kN, falling at 2 kN/m to $-5.00$ kN at c.
• cd: $V_c$ = -12.58 kN; after the 20 kN load -32.58 kN at d.

Reactions: $R_a = 32.58$ kN up with $M_a = 62.61$ kN·m anticlockwise; $R_b = 7.58$ kN down; $R_c = 7.58$ kN down; $R_d = 32.58$ kN up with $M_d = 62.61$ kN·m clockwise. Check: $32.58 + 32.58 - 7.58 - 7.58 = 50 = 20+20+2\times5$.

BMD and SFD values

BMD is linear in ab and cd between the loads, parabolic in bc ($M = 27.72 + 5x - x^2$, $x$ from b); SFD is stepped in ab and cd and a straight line through zero at mid-bc.

Answer: $\theta_b = 0.001993$ rad, $\theta_c = -0.001993$ rad; $M_{ab} = -62.61$, $M_{ba} = -27.72$, $M_{bc} = 27.72$, $M_{cb} = -27.72$, $M_{cd} = 27.72$, $M_{dc} = 62.61$ kN·m.