Truss: force method, 100 kN and 50 kN

Reading of the figure: A (0, 0) hinged, B (4, 0) roller, C (4, 3) and D (0, 3) at the top; members AB, CD (4 m), BC, DA (3 m) and diagonals AC, BD (5 m). Loads: 100 kN downward at C and 50 kN horizontal at D (towards the right). $AE$ is constant. $D_s = 6 + 3 - 8 = 1$. Take the force in AC as the redundant $X$.

Reactions

• $\Sigma M_A = 0$: $V_B(4) = 100(4) + 50(3) \Rightarrow V_B = 137.50$ kN ($\uparrow$)
• $V_A = 100 - 137.50 = -37.50$ kN, i.e. $37.50$ kN ($\downarrow$)
• $H_A = 50$ kN ($\leftarrow$)

Force tables (tension +)

$$f_{11} = \frac{\sum n^2L}{AE} = \frac{17.280}{AE},\qquad \Delta_{10} = \frac{\sum S_0nL}{AE} = \frac{-360.000}{AE}$$

Compatibility

$$f_{11}X + \Delta_{10} = 0 \ \Rightarrow\ X = -\frac{-360.000}{17.280} = 20.83\ \text{kN}$$

Final forces $S = S_0 + nX$

Answer: AB $= +33.33$, BC $= -112.50$, CD $= -16.67$, DA $= +25.00$, AC $= +20.83$, BD $= -41.67$ kN (positive = tension).