Truss: lack of fit, temperature, bars BC BF

Assumptions (figure read as follows): bottom joints A(0,0), B(5,0), C(10,0); top joints D(0,3), E(5,3), F(10,3). Members: bottom chord AB, BC; top chord DE, EF; verticals AD, BE, CF; inclined members DB and BF. A is a hinge; B and C are rollers (vertical reactions only). All areas 40 cm², $E=2\times10^5$ N/mm², so $EA=8\times10^5$ kN. The inclined members DB and BF are 2 mm too long; the verticals AD, BE, CF are cooled by 15°C ($\alpha=10.8\times10^{-6}$/°C).

Degree of indeterminacy

$m=9$, $r=4$, $j=6$: $D_s=9+4-12=1$. The truss is internally determinate, so the lack of fit and the temperature change only produce forces because of the extra support. Take the vertical reaction at B as the redundant $X$.

Free (initial) elongations $e_0$

• Inclined bars too long: $e_0=+2$ mm (DB and BF).
• Verticals cooled: $e_0=-\alpha\,\Delta T\,L=-10.8\times10^{-6}\times15\times3000=-0.486$ mm.

Unit load on the primary truss

The primary truss (B roller removed) is loaded by a unit upward force at B, giving $S_1$. Since no external load acts, $S_0=0$ and only the initial strains cause displacement of B.

$$\Delta_{B}=\sum S_1e_0=-3.8873+(-0.4860)=-4.3733\ \text{mm}\quad(\text{negative: B moves 4.373 mm downward in the primary truss})$$ $$f_{11}=\frac{\sum S_1^2L}{EA}=\frac{19.458}{8\times10^5}=0.02432\ \text{mm/kN}$$

Compatibility ($B$ cannot move): $\Delta_B+f_{11}X=0$

$$X=-\frac{-4.3733}{0.02432}=179.80\ \text{kN}$$

Bar forces $S=XS_1$ ($S_0=0$, because the initial strains alone produce no forces in the statically determinate primary truss):

Answer: force in BC = 0.00 kN and force in BF = -174.73 kN (+ tension, - compression). Other bars: AB = 0.00, DE = 149.83, EF = 149.83, AD = 89.90, BE = 0.00, CF = 89.90, DB = -174.73 kN.