Truss joint B: temperature and vertical members long

Given data and assumed geometry

From the figure: top chord joints D(0,6), E(5,6), F(10,6), G(15,6), H(20,6); bottom joints A(5,0), B(10,0), C(15,0). Members: top chord DE, EF, FG, GH; bottom chord AB, BC; verticals EA, FB, GC; diagonals DA, EB, GB, HC (13 members, 8 joints, 3 reactions: determinate). A hinge, C roller. Loads: 50 kN down at F, 30 kN up and 20 kN to the right at H. $A = 1500$ mm$^2 = 1.5\times10^{-3}$ m$^2$, $E = 2\times10^8$ kN/m$^2$, $\alpha = 12\times10^{-6}$/$^\circ$C. Unit vertical load (down) at B.

Member forces (tension +); $F$ real, $k$ unit load at B

(i) Loading

$$\Delta_1 = \frac{\sum kFL}{AE} = \frac{578.073}{1.5\times10^{-3}\times2\times10^8} = 1.927\ \text{mm}\ (\downarrow)$$

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

$$\Delta_2 = \sum k\,\alpha\,\Delta T\,L = (-0.417)(12\times10^{-6})(20)(5000)\times2 = -1.000\ \text{mm}$$

(only EF and FG have non-zero $k$).

(iii) Vertical members 10 mm too long

$$\Delta_3 = \sum k\,e = (-0.5)(10) + (0)(10) + (-0.5)(10) = -10.000\ \text{mm}$$

Total

$$\Delta_B = 1.927 - 1.000 - 10.000 = -9.073\ \text{mm}$$

Answer: vertical deflection of joint B = 9.07 mm upward (net). Loading gives 1.93 mm down, temperature 1.00 mm up and lack of fit 10.00 mm up.