Stiffness method: frame with 16 t, 5 t

Assumptions

Column AB is 3 m high, with the 5 t horizontal load at joint B. Beam BC is 4 m with the 16 t load at its mid-point. Units: t and m. Axial deformation is ignored (inextensible), so B and C move horizontally by the same sway $\Delta$ and neither joint moves vertically.

Degrees of freedom

Unknowns: $\theta_B$, $\theta_C$ and the sway $\Delta$ (chord rotation of the column $\psi = \Delta/3$). Kinematic indeterminacy = 3. Clockwise end moments are positive.

Fixed-end moments

BC: $PL/8 = 16\times 4/8 = 8$ t·m, so $FEM_{BC} = -8$, $FEM_{CB} = +8$.

Stiffness (slope-deflection) equations

$$\begin{aligned} M_{AB} &= \tfrac{2EI}{3}\,(\theta_B - 3\psi) = \tfrac{2EI}{3}\theta_B - \tfrac{2EI}{3}\Delta \\ M_{BA} &= \tfrac{2EI}{3}\,(2\theta_B - 3\psi) = \tfrac{4EI}{3}\theta_B - \tfrac{2EI}{3}\Delta \\ M_{BC} &= \tfrac{2(2EI)}{4}(2\theta_B + \theta_C) - 8 = EI(2\theta_B+\theta_C) - 8 \\ M_{CB} &= EI(\theta_B + 2\theta_C) + 8 \end{aligned}$$

Equilibrium equations

1. Joint B: $M_{BA} + M_{BC} = 0$
2. Joint C (roller, no moment): $M_{CB} = 0$
3. Horizontal force: the only horizontal restraint is at A, so the column shear equals the 5 t load: $(M_{AB} + M_{BA})/3 = -5$

Substituting:

$$\begin{aligned} \tfrac{10}{3}EI\,\theta_B + EI\,\theta_C - \tfrac{2}{3}EI\,\Delta &= 8 \\ EI\,\theta_B + 2EI\,\theta_C &= -8 \\ 2EI\,\theta_B - \tfrac{4}{3}EI\,\Delta &= -15 \end{aligned}$$

Solving:

$$EI\,\theta_B = \tfrac{117}{11} = 10.636,\quad EI\,\theta_C = -\tfrac{205}{22} = -9.318,\quad EI\,\Delta = \tfrac{1197}{44} = 27.205$$

Member end moments

Joint B: $-3.955 + 3.955 = 0$ (check).

Reactions and diagrams

• Reaction at C (moments about B for beam BC, with $M_{BC} = 3.955$ clockwise): $R_C = (16\times 2 + 3.955)/4 = 8.989$ t up.
• At A: $V_A = 16 - 8.989 = 7.011$ t up, $H_A = 5$ t to the left, $M_A = 11.045$ t·m anticlockwise.
• BMD: column AB is a straight line from 11.045 at A, passing through zero 2.21 m above A, to 3.955 at B on the opposite face. Beam BC: 3.955 (sagging) at B, 17.977 sagging under the 16 t load, 0 at C.
• SFD: column 5 t constant; beam +7.011 from B up to the load, then -8.989 to C.
• Axial: column AB carries 7.011 t compression; beam BC has no axial force.

Answer: $\theta_B = 10.64/EI$, $\theta_C = -9.32/EI$, $\Delta = 27.20/EI$; $M_{AB}=-11.05$, $M_{BA}=-3.95$, $M_{BC}=3.95$, $M_{CB}=0$ t·m.