Truss forces: member AC 10 mm too long

Reading of the figure: A (0, 0) hinged, D (4, 0) roller at the base, B (0, 3) and C (4, 3) at the top, with both diagonals AC and BD. Bars AB and CD are 3 m, BC and AD are 4 m, and AC and BD are 5 m. The truss is indeterminate to the 1st degree (internal); the supports are determinate. Take the force in AC as the redundant $X$ (tension +). Lack of fit $\lambda = +10$ mm $= 0.01$ m (member too long).

No external load acts, so the primary structure carries no force and the fabrication error alone produces the member forces.

Unit force system ($X = 1$ in AC)

$$f_{11} = \sum \frac{n^2 L}{AE} = \frac{17.28}{AE}$$

Compatibility

The member is too long, so after it is forced into place it is compressed. Compatibility requires the net relative displacement at the cut in AC, due to $X$ plus the lack of fit, to be zero:

$$f_{11}X + \lambda = 0 \ \Rightarrow\ X = -\frac{0.01\,AE}{17.28} = -0.0005787\,AE$$

Member forces $S = n X$ (AE in kN)

Example: for $AE = 2\times10^5$ kN, AC = BD $= -115.74$ kN, AB = CD $= 69.44$ kN, BC = AD $= 92.59$ kN.

Check at joint A (horizontal): $S_{AD} + 0.8\,S_{AC} = (0.4630 - 0.8\times0.5787)\times10^{-3}AE = 0$. There is no external load, so all reactions are zero.

Answer: AC = BD $= -5.787\times10^{-4}\,AE$ (compression); AB = CD $= +3.472\times10^{-4}\,AE$; BC = AD $= +4.630\times10^{-4}\,AE$ (tension).