Truss: temperature decrease in vertical members

Reading of the figure: bottom chord A-B-C-D with three panels of 3 m (A hinged, D roller); top joints F above B and E above C at 3 m height; members AB, BC, CD, FE, AF, ED, the verticals BF and CE, and the two diagonals BE and CF. The truss has $m = 10$, $j = 6$, $r = 3$, so $D_s = m + r - 2j = 1$ (internal). Take the force in CF as the redundant $X$.

Data

• $E = 2\times10^5$ N/mm$^2 = 2\times10^8$ kN/m$^2$, $A = 35$ cm$^2 = 35\times10^{-4}$ m$^2$, so $AE = 700000$ kN.
• Vertical members BF and CE: $\Delta t = -20^\circ$C, so each shortens by $e = \alpha\,\Delta t\,L = 12\times10^{-6}(-20)(3) = -7.20\times10^{-4}$ m $= -0.72$ mm.

There is no external load, so the primary structure carries no force. The temperature change alone causes the forces.

Force $n$ due to $X = 1$ in CF

The outer members AB, CD, AF and ED carry zero force in this system.

Compatibility

$$\begin{aligned} f_{11} &= \sum\frac{n^2 L}{AE} = \frac{4(0.5)(3) + 2(1)(4.243)}{AE} = \frac{14.485}{AE} = 2.0693\times10^{-5}\ \text{m/kN} \\ \Delta_{1t} &= \sum n\,e = 2(-0.7071)(-0.00072) = 1.0182\times10^{-3}\ \text{m} \end{aligned}$$ $$f_{11}X + \Delta_{1t} = 0 \ \Rightarrow\ X = -\frac{1.0182\times10^{-3}}{2.0693\times10^{-5}} = -49.21\ \text{kN}$$

Member forces $S = nX$

Answer: bar force in AB $= 0$; bar force in EF $= 34.79$ kN (tension).