Force method: frame with 60 kN/m, BMD

Reading of the figure: column AB is 7 m high ($2EI$) and fixed at A; the beam BC is 10 m long ($EI$) with 60 kN/m; C is a roller (vertical support); a horizontal load of 80 kN (towards the right) acts on the column at 3 m above A.

$r = 3 + 1 = 4$, so $D_s = 1$. Take the vertical reaction $R_C$ at the roller as the redundant.

 B =================== C  (roller)
 |    60 kN/m
 |
 | 80 kN ->
 |
 A (fixed)

Primary structure (cantilever fixed at A)

($M_0$ at B $= -60\times10\times5 = -3000$; at A it is $-3000 - 80\times3 = -3240$.)

Flexibility coefficient and load term

$$\begin{aligned} f_{11} &= \frac{1}{2EI}\left[10^2(7)\right] + \frac{1}{EI}\left[\frac{10^2\times10}{3}\right] = \frac{350 + 333.33}{EI} = \frac{683.33}{EI} \\ \Delta_{10} &= \frac{1}{2EI}\left[-10\,\frac{3(3240+3000)}{2} - 10(3000)(4)\right] + \frac{1}{EI}\left[-30\int_0^{10}(10-x)^3dx\right] \\ &= \frac{-46800 - 60000 - 75000}{EI} = \frac{-181800}{EI} \end{aligned}$$

Compatibility ($\delta_C = 0$)

$$f_{11}R_C + \Delta_{10} = 0 \ \Rightarrow\ R_C = \frac{181800}{683.33} = 266.05\ \text{kN}\ (\uparrow)$$

Final bending moments $M = M_0 + R_C\,m$ (kNm)

Reactions

• C: $R_C = 266.05$ kN ($\uparrow$)
• A: $V_A = 333.95$ kN ($\uparrow$), $H_A = 80$ kN ($\leftarrow$), $M_A = 579.51$ kNm (anticlockwise)

Check: $V_A + R_C = 600.00$ kN $= 60\times10 = 600$ kN.

Answer: $R_C = 266.05$ kN; $M_A = 579.51$ kNm; $M_B = -339.51$ kNm (hogging); maximum sagging moment $589.85$ kNm in the beam.