Castigliano: square truss member forces

Reading of the figure: A (0, 0) hinged, B (4, 0) roller, C (4, 4) and D (0, 4); members AB, BC, CD, DA (4 m) and diagonals AC, BD (5.657 m); 120 kN downward and 90 kN horizontal (towards the left) at D. $AE$ is constant. $D_s = 6 + 3 - 8 = 1$. Take the force in AC as the redundant $X$.

Reactions

$\Sigma M_A = 0$: $V_B(4) + 90(4) = 0$, hence $V_B = -90.00$ kN (downward 90 kN). $V_A = 120 + 90 = 210.00$ kN ($\uparrow$), $H_A = 90$ kN ($\rightarrow$).

Member forces in terms of $X$ ($S = S_0 + nX$)

Castigliano's theorem

The strain energy of the truss is $U = \sum \dfrac{S^2L}{2AE}$. The relative displacement of the cut ends of AC must be zero, so

$$\frac{\partial U}{\partial X} = \sum \frac{S\,n\,L}{AE} = \frac{1}{AE}\left[\sum S_0nL + X\sum n^2L\right] = 0$$ $$\sum S_0nL = 1568.528,\qquad \sum n^2L = 19.314$$ $$X = S_{AC} = -\frac{1568.528}{19.314} = -81.21\ \text{kN}$$

Forces in all members

Answer: AB $= -32.57$, BC $= +57.43$, CD $= +57.43$, DA $= -152.57$, AC $= -81.21$, BD $= +46.07$ kN (positive = tension).