Truss joint E: temperature and lack of fit

Given data and assumed geometry

Wall supports at C (0,0) and D (0,8), both pinned. Bottom chord C-B-A with $CB = BA = 8$ m (B at (8,0), A at (16,0)); E is the top middle joint at (8,4) on the sloping top chord D-E-A. Members: CB, BA, DE, EA, EC, EB. Loads: 8 kN at B and 4 kN at A (downward). $A = 1500$ mm$^2 = 1.5\times10^{-3}$ m$^2$, $E = 200$ kN/mm$^2 = 2\times10^8$ kN/m$^2$, $\alpha = 12\times10^{-6}$/$^\circ$C. A unit vertical load (down) is applied at E.

Lengths: $DE = EA = EC = \sqrt{8^2+4^2} = 8.944$ m, $EB = 4$ m.

Member forces (tension +)

(i) Loading

$$\Delta_1 = \frac{\sum kFL}{AE} = \frac{268.328}{1.5\times10^{-3}\times2\times10^8} = 0.894\ \text{mm}\ (\downarrow)$$

(ii) Temperature rise of 30$^\circ$C in the top chord (DE and EA)

$$\Delta_2 = \sum k\alpha\Delta T L = (1.118)(12\times10^{-6})(30)(8944) = 3.600\ \text{mm}\ (\downarrow)$$

(EA has $k = 0$, so it contributes nothing.)

(iii) AE and EC fabricated 5 mm too long

$$\Delta_3 = \sum k\,e = (0)(5) + (-1.118)(5) = -5.590\ \text{mm}\ (\text{up})$$

Total

$$\Delta_E = 0.894 + 3.600 - 5.590 = -1.096\ \text{mm}$$

Answer: vertical deflection of E $= 1.10$ mm upward (net). Individual effects: load 0.89 mm down, temperature 3.60 mm down, lack of fit 5.59 mm up.