Force method: frame reactions at hinged support

Reading of the figure: A (0, 0) fixed; B (0, 5) with column AB 5 m of stiffness $1.5EI$; beam BC of span 10 m with stiffness $2EI$ carrying 50 kN/m; column CD 3 m of stiffness $EI$ with D hinged. As the columns are of different lengths, D is taken 2 m above the level of A. Here $EI$ stands for $E\times I$ of the lightest member.

The frame has $r = 3 + 2 = 5$, so $D_s = 2$. Release the hinge at D and take the reactions at D as redundants: $X_1$ = horizontal ($\rightarrow$ positive) and $X_2$ = vertical ($\uparrow$ positive). The primary structure is a cantilever fixed at A.

  B ================ C   50 kN/m
  |                  |
  | 5 m              | 3 m
  |                  o D (hinge)
 A///

Moment diagrams (sign: + tension on the inner face of the frame)

• $M_0$ (50 kN/m on the cantilever): $M_B = M_{A} = -50(10)^2/2 = -2500$ kNm, falling parabolically to 0 at C.
• $m_1$ (unit horizontal force at D): CD varies 0 to 3, beam constant 3, column from 3 at B down to $-2$ at A.
• $m_2$ (unit vertical force at D): beam varies 10 to 0 from B to C, column constant 10.

Flexibility coefficients and load terms

$$\begin{aligned} f_{11} &= \frac{61.778}{EI} \\ f_{12} &= \frac{91.667}{EI} \\ f_{22} &= \frac{500.000}{EI} \\ \Delta_{10} &= \frac{-16666.7}{EI} \\ \Delta_{20} &= \frac{-114583.3}{EI} \end{aligned}$$

Compatibility ($\Delta_{D,x} = \Delta_{D,y} = 0$)

$$\begin{aligned} 61.778\,X_1 + 91.667\,X_2 &= 16666.7 \\ 91.667\,X_1 + 500.000\,X_2 &= 114583.3 \end{aligned}$$

Solving:

$$X_1 = -96.51\ \text{kN},\qquad X_2 = 246.86\ \text{kN}$$

Final bending moments $M = M_0 + X_1 m_1 + X_2 m_2$ (kNm)

The maximum sagging moment in BC is 319.87 kNm at 5.07 m from B.

Reactions

• Hinge D: $H_D = 96.51$ kN ($\leftarrow$), $V_D = 246.86$ kN ($\uparrow$)
• Fixed support A: $H_A = 96.51$ kN ($\rightarrow$), $V_A = 253.14$ kN ($\uparrow$), $M_A = 161.62$ kNm (clockwise; the moment at A puts the inner face of the column in tension)

Check: $\Sigma V = 253.14 + 246.86 = 500.00$ kN $= 50 \times 10 = 500$ kN; $\Sigma H = 96.51 - 96.51 = 0$. Moments about A: the UDL gives $2500$ kNm clockwise, while $V_D$ ($246.86\times10 = 2468.6$) and $H_D$ ($96.51\times2 = 193.0$) give $2661.6$ kNm anticlockwise, so the support couple must be $161.62$ kNm clockwise.

Answer: at the hinged support D, $H_D = 96.51$ kN (towards the left) and $V_D = 246.86$ kN (upward).