Truss: loads, temperature rise, lack of fit

Reading of the figure: A (0, 0) hinged and D (4, 0) roller at the base, B (0, 3) and C (4, 3) at the top, members AB, CD (verticals, 3 m), BC, AD (horizontals, 4 m) and diagonals AC, BD (5 m). Loads: 30 kN horizontal at B (towards C) and 50 kN downward at C. The truss has $D_s = 6 + 3 - 8 = 1$ (internal). Take the force in AC as the redundant $X$ (tension +).

Effects to be included

• Diagonals AC and BD: temperature rise 20$^\circ$C, so $e = \alpha\,\Delta t\,L = 2.06\times10^{-6}\times20\times5 = 2.06\times10^{-4}$ m (elongation).
• Verticals AB and CD: 5 mm too long, so $e = +0.005$ m.
• $AE = 5000$ kN for all members.

Reactions (statics)

$\Sigma M_A = 0$: $V_D(4) = 50(4) + 30(3)$, so $V_D = 72.50$ kN ($\uparrow$). $\Sigma F_y$: $V_A = 50 - 72.50 = -22.50$ kN, i.e. 22.50 kN downward. $\Sigma F_x$: $H_A = 30$ kN acting towards the left.

Primary structure (AC removed) and unit force system

$S_0$ = force due to loads; $n$ = force due to $X = 1$ in AC. Tension is positive.

Compatibility

$$\frac{\sum S_0 nL}{AE} + X\frac{\sum n^2L}{AE} + \sum n\,e = 0$$ $$\frac{-234.000}{5000} + X\,\frac{17.280}{5000} - 0.005588 = 0$$ $$X = \frac{-(-234.000 + 5000 \times -0.005588)}{17.280} = 15.159\ \text{kN}$$

(The thermal and lack-of-fit terms enter only through $\sum n\,e$, which is $-0.005588$ m.)

Final forces $S = S_0 + nX$ (kN)

Check at joint C (vertical): $-S_{CD} - 0.6\,S_{AC} = 50.00$ kN, equal to the 50 kN load.

Answer: AB $= +13.40$ kN, BC $= -12.13$ kN, CD $= -59.10$ kN, AD $= +17.87$ kN, AC $= +15.16$ kN, BD $= -22.34$ kN (positive = tension). Reactions: $H_A = 30$ kN (left), $V_A = 22.50$ kN (downward), $V_D = 72.50$ kN (upward).