Truss joint E: temperature and lack of fit

Given data

A(0,0) hinge, E(4,0), B(8,0) roller; top joints C(2, 3.464) and D(6, 3.464); all members 4 m (60$^\circ$). 20 kN down at C. $E = 200$ kN/mm$^2 = 2\times10^8$ kN/m$^2$, $A = 1000$ mm$^2 = 10^{-3}$ m$^2$, $\alpha = 12\times10^{-6}$/$^\circ$C. Virtual load: a unit downward load at E.

Reactions (real): $\sum M_B = 0$: $A_y\times8 = 20\times6 \Rightarrow A_y = 15$ kN, $B_y = 5$ kN.

Member forces (tension +)

(i) Effect of the 20 kN load shown in the figure

$$\Delta_1 = \frac{\sum kFL}{AE} = \frac{80.000}{10^{-3}\times2\times10^8} = 0.400\ \text{mm}\ (\downarrow)$$

(ii) Temperature rise of 20$^\circ$C in CD (top chord)

$$\Delta L_{CD} = 12\times10^{-6}\times20\times4000 = 0.96\ \text{mm},\qquad \Delta_2 = k_{CD}\,\Delta L = (-0.577)(0.96) = -0.554\ \text{mm}\ (\uparrow)$$

(iii) CE 5 mm too long

$$\Delta_3 = k_{CE}\,e = (0.577)(5) = 2.887\ \text{mm}\ (\downarrow)$$

Total

$$\Delta_E = 0.400 - 0.554 + 2.887 = 2.732\ \text{mm}$$

Answer: vertical deflection of E = 2.73 mm downward (including the 20 kN load; the temperature change and lack of fit alone give 2.33 mm downward).