Forces in members of three-bar system

The three bars are hinged to the ceiling at A, B, C (on one level) and meet at D, so there are 3 unknown forces but only 2 equilibrium equations. One compatibility equation (from the displacement of D) is needed.

Data: $A = 600\ \text{mm}^2$, $E = 2\times10^5\ \text{N/mm}^2$, $AE = 1.2\times10^8\ \text{N}$, $P = 30\ \text{kN}$. Bar BD is vertical: $L_{BD} = 3000$ mm. AD is at $50^\circ$ and CD at $40^\circ$ to the vertical. Assume A and C lie on opposite sides of B.

$$L_{AD} = \frac{3000}{\cos 50^\circ} = 4667.2\ \text{mm}, \qquad L_{CD} = \frac{3000}{\cos 40^\circ} = 3916.2\ \text{mm}$$

Equilibrium of joint D

$$\begin{aligned} \Sigma H = 0:&\quad F_{AD}\sin 50^\circ = F_{CD}\sin 40^\circ \\ \Sigma V = 0:&\quad F_{BD} + F_{AD}\cos 50^\circ + F_{CD}\cos 40^\circ = 30\,000 \end{aligned}$$

Compatibility

Let D move down by $v$ and sideways (away from A) by $u$. Small-displacement elongations are:

$$\delta_{BD} = v, \qquad \delta_{AD} = v\cos 50^\circ + u\sin 50^\circ, \qquad \delta_{CD} = v\cos 40^\circ - u\sin 40^\circ$$

and each force is $F = \dfrac{AE}{L}\,\delta$. Substituting into the two equilibrium equations gives two equations in $u$ and $v$:

$$u = 0.0384\ \text{mm}, \qquad v = 0.4386\ \text{mm}$$

Forces (all tensile)

Check: $\Sigma H = 6.13 - 6.13 = 0$ and $\Sigma V = 17.55 + 5.15 + 7.31 = 30.00\ \text{kN}$.

Answer: $F_{AD} = 8.01\ \text{kN}$, $F_{BD} = 17.55\ \text{kN}$, $F_{CD} = 9.54\ \text{kN}$ (all tension).