Stiffness method: bent frame member end moments

Assumptions and set-up

Column AC is 4 m high ($3EI$) with A hinged; the 30 kN acts at mid-height. Beam CB is 5 m ($5EI$) with 20 kN/m, and B is fixed. The members are inextensible, and B is fixed, so joint C cannot move: there is no sway. The only unknown displacement is the rotation $\theta_C$ (kinematic indeterminacy = 1). Clockwise moments positive.

Fixed-end moments

• Column AC (30 kN at mid-height): $PL/8 = 30\times4/8 = 15$ kN·m, so $FEM_{CA} = +15$, $FEM_{AC} = -15$. Because A is hinged, use the modified FEM and stiffness $3EI/L$: $FEM'_{CA} = 15 + \tfrac12\times 15 = 22.5$ kN·m.
• Beam CB: $wL^2/12 = 20\times25/12 = 41.667$ kN·m, so $FEM_{CB} = -41.667$, $FEM_{BC} = +41.667$.

Stiffness equations

$$\begin{aligned} M_{CA} &= \tfrac{3(3EI)}{4}\theta_C + 22.5 = 2.25\,EI\theta_C + 22.5 \\ M_{CB} &= \tfrac{2(5EI)}{5}(2\theta_C) - 41.667 = 4\,EI\theta_C - 41.667 \\ M_{BC} &= \tfrac{2(5EI)}{5}(\theta_C) + 41.667 = 2\,EI\theta_C + 41.667 \\ M_{AC} &= 0 \end{aligned}$$

Joint equilibrium at C

$$\begin{aligned} M_{CA} + M_{CB} &= 0 \\ 6.25\,EI\theta_C - 19.167 &= 0 \\ EI\theta_C &= 3.0667 \end{aligned}$$

Member end moments (kN·m)

Joint C: $29.40 - 29.40 = 0$.

Reactions and diagrams

• Beam end shears: 46.32 kN at C and 53.68 kN at B. Maximum sagging moment $= -29.40 + 46.32x - 10x^2$ at $x = 46.32/20 = 2.316$ m from C, giving $+24.23$ kN·m. Hogging moments 29.40 at C and 47.80 at B.
• Column: moment 0 at A, 15.30 kN·m under the 30 kN load, then reversing to 29.40 kN·m at C (outer face tension at C).
• Reactions: A: $H_A = 7.65$ kN (opposing the load), $V_A = 46.32$ kN up. B: $H_B = 22.35$ kN, $V_B = 53.68$ kN up, $M_B = 47.80$ kN·m. Check: $7.65 + 22.35 = 30$ kN; $46.32 + 53.68 = 100 = 20\times5$.

Answer: $\theta_C = 3.067/EI$ (clockwise); $M_{CA} = 29.40$, $M_{CB} = -29.40$, $M_{BC} = 47.80$ kN·m, $M_{AC} = 0$.