Truss joint D: loading, lack of fit, temperature

Given data and assumptions

A(0,0) hinge, D(4,0), B(8,0) roller; top joints E(2, 3.464) and C(6, 3.464); all members are 4 m long (60$^\circ$). Loads: 15 kN at E and 30 kN at C, downward. $E = 200\times10^3$ N/mm$^2$, $\alpha = 12\times10^{-6}$/$^\circ$C. The member area is not given; it is assumed to be the same for all members, $A = 1000$ mm$^2 = 10^{-3}$ m$^2$ (the loading part scales as $1/A$; the other parts do not depend on $A$). A unit downward load is applied at D.

Reactions (moments about A): $B_y\times8 = 15\times2 + 30\times6 = 210 \Rightarrow B_y = 26.25$ kN, $A_y = 18.75$ kN.

Member forces (tension +)

(i) Loading

$$\Delta_1 = \frac{\sum kFL}{AE} = \frac{180.000}{10^{-3}\times2\times10^8} = 0.900\ \text{mm}\ (\downarrow)$$

(ii) DE and DC each 5 mm too long

$$\Delta_2 = \sum k\,e = (0.577)(5) + (0.577)(5) = 5.774\ \text{mm}\ (\downarrow)$$

(iii) Temperature rise of 20$^\circ$C in CD

$$\Delta L = 12\times10^{-6}\times20\times4000 = 0.96\ \text{mm},\qquad \Delta_3 = k\,\Delta L = 0.577\times0.96 = 0.554\ \text{mm}\ (\downarrow)$$

Total

$$\Delta_D = 0.900 + 5.774 + 0.554 = 7.228\ \text{mm}$$

Answer: vertical deflection of D = 7.23 mm downward (for $A = 1000$ mm$^2$; the load part scales as $1/A$).