Consistent deformation: frame with 12 kN, SFD and BMD

Reading of the figure: the horizontal member AB is 6 m long ($2I$) with a 12 kN load at mid-span D (3 m from A), A hinged; the vertical member BC is 4 m long ($I$) with C hinged at its foot. $E$ is constant.

$r = 4$, so $D_s = 1$. Take the horizontal reaction $H_A$ at A as the redundant. Release A horizontally (roller): the primary structure is a simply supported frame.

        12 kN
          v
  A o-----D-----B
  H_A->         |
                | 4 m
                o C
  |<-- 3 -->|<-- 3 -->|

Primary structure (A on a roller)

$V_A = V_C = 6$ kN. A unit horizontal force at A is balanced by a horizontal reaction at C and a vertical couple: $V = 4/6 = 0.667$ kN.

(With $m$ negative at D and B because a unit force at A produces hogging in AB: $m_D = -0.667\times3 = -2$, $m_B = -4$.)

Deformations

$$\begin{aligned} f_{11} &= \frac{1}{2EI}\left[\frac{3}{3}(4) + \frac{3}{3}(4 + 8 + 16)\right] + \frac{1}{EI}\left[\frac{4}{3}(16)\right] = \frac{16 + 21.333}{EI} = \frac{37.33}{EI} \\ \Delta_{10} &= \frac{1}{2EI}\left[\frac{3}{3}(18)(-2) \cdot 1 + \frac{3}{6}\left\{2(18)(-2) + 18(-4)\right\}\right] = \frac{-54.00}{EI} \end{aligned}$$

Consistent deformation (A does not move horizontally)

$$\Delta_{10} + f_{11}H_A = 0 \ \Rightarrow\ H_A = \frac{54.00}{37.33} = 1.446\ \text{kN}\ (\rightarrow)$$

Reactions

• A: $H_A = 1.446$ kN ($\rightarrow$), $V_A = 5.036$ kN ($\uparrow$)
• C: $H_C = 1.446$ kN ($\leftarrow$), $V_C = 6.964$ kN ($\uparrow$)

Bending moment $M = M_0 + H_A\,m$ (kNm)

Shear force and axial force (kN)

Answer: $H_A = H_C = 1.446$ kN; $V_A = 5.036$ kN and $V_C = 6.964$ kN; $M_D = 15.11$ kNm (sagging), $M_B = -5.79$ kNm (hogging).