Force method: frame with hinged support

Reading of the figure: column AB is 6 m high (stiffness $2I$) and fixed at A; the beam BC is 10 m long ($1.5I$) with 30 kN/m; the beam end C is on a hinged support at the level of the beam; a 50 kN horizontal load (towards the right) acts on the column 2 m below the beam, at E (4 m above A). Here $I$ stands for $EI$ units: the member stiffnesses are $2EI$ and $1.5EI$.

$r = 3 + 2 = 5$, so $D_s = 2$. Release the hinge at C: $X_1$ = vertical reaction at C ($\uparrow$), $X_2$ = horizontal reaction at C ($\rightarrow$). The primary structure is a cantilever fixed at A.

 B ================= C (hinge)
 |    30 kN/m          X1 up, X2 horizontal
 |
 E ->  50 kN
 |
 A (fixed)

Moments in the primary structure (+ tension on the inner face)

• $M_0$: at A, $-(50\times4 + 30\times10\times5) = -1700$; at E and B, $-1500$ (the column between E and B carries only the beam load); falling parabolically to 0 at C.
• $m_1$ (unit vertical at C): $10$ in the column (from B down to A) and falling linearly from 10 at B to 0 at C.
• $m_2$ (unit horizontal at C): zero in the beam; in the column it grows linearly from 0 at B to $-6$ at A ($-2$ at E).

Flexibility coefficients (in $1/EI$ units)

$$\begin{aligned} f_{11} &= \frac{522.222}{EI} \\ f_{12} &= \frac{-90.000}{EI} \\ f_{22} &= \frac{36.000}{EI} \\ \Delta_{10} &= \frac{-72000.0}{EI} \\ \Delta_{20} &= \frac{14433.3}{EI} \end{aligned}$$

Compatibility

$$\begin{aligned} 522.222\,X_1 - 90.000\,X_2 &= 72000.0 \\ -90.000\,X_1 + 36.000\,X_2 &= -14433.3 \end{aligned}$$ $$X_1 = V_C = 120.84\ \text{kN},\qquad X_2 = H_C = -98.82\ \text{kN}$$

Final bending moments (kNm)

Maximum sagging moment in the beam: 243.38 kNm at 5.97 m from B.

Reactions

• C (hinge): $V_C = 120.84$ kN ($\uparrow$), $H_C = 98.82$ kN (towards the left)
• A (fixed): $V_A = 179.16$ kN ($\uparrow$), $H_A = 48.82$ kN (towards the right), $M_A = 101.35$ kNm (clockwise; the moment at A puts the inner face of the column in tension)

Check: $V_A + V_C = 300.00$ kN $= 30\times10 = 300$ kN; $\Sigma H$: $48.82 + 50 - 98.82 = 0$. Moments about A: the loads give $50(4) + 300(5) = 1700$ kNm clockwise, while $V_C$ ($120.84\times10 = 1208.4$) and $H_C$ ($98.82\times6 = 592.9$) give $1801.3$ kNm anticlockwise, so the support couple is $101.35$ kNm clockwise.

Answer: at the hinged support C, $V_C = 120.84$ kN upward and $H_C = 98.82$ kN towards the left.