Truss: rectangular, diagonal 2 cm too short

Setup: joints A(0,0) and B(12,0) at the base (both hinges), C(12,9) and D(0,9) at the top. Members: verticals AD, BC; top chord CD; diagonals AC and BD (bottom chord AB is not a member between the two hinges, or if present it carries no force because A and B cannot move). 100 kN acts horizontally to the right at D (the loaded joint). The diagonal BD joining the loaded joint is 20 mm short. $E=2\times10^5$ MPa, $A=1000$ mm², so $EA=2\times10^5$ kN and lengths: AD = BC = 9 m, CD = 12 m, AC = BD = 15 m.

Degree of indeterminacy

$m=5$, $r=4$, $j=4$: $D_s=5+4-8=1$. Take the horizontal reaction at B, $X=H_B$ (to the right +), as the redundant. In the primary truss B is a roller on the base line.

Forces in the primary truss

$S_0$ = bar forces due to the 100 kN load (the lack of fit is treated separately). $S_1$ = bar forces due to a unit force at B acting to the right.

Lack of fit term

The bar BD is shorter by 20 mm, i.e. its free elongation is $e_0=-0.02$ m. It contributes $S_1e_0=1.2500\times(-0.02)=-0.02500$ m to the displacement at B.

Compatibility ($H_B$ gives no horizontal movement at B)

$$\frac{\sum S_0S_1L}{EA}+\sum S_1e_0+X\frac{\sum S_1^2L}{EA}=0$$ $$\frac{4050.000}{2\times10^5}+(-0.02500)+X\frac{69.0000}{2\times10^5}=0\;\Rightarrow\; X=H_B=13.77\ \text{kN}$$

(positive: $H_B$ acts to the right, in the same direction as the 100 kN load.)

Bar forces ($S=S_0+XS_1$, + tension, - compression)

BC = -85.33, CD = -113.77, DA = -10.33, AC = 142.21, BD = 17.21 kN.

Reactions: $H_B=13.77$ kN (right), $H_A=-113.77$ kN, i.e. 113.77 kN to the left; vertical reactions $V_A=-75.00$ kN and $V_B=75.00$ kN (couple of the 100 kN load). The result was checked with a full stiffness analysis.

Answer: BC = -85.33, CD = -113.77, DA = -10.33, AC = 142.21, BD = 17.21 kN (+ tension, - compression); AB carries zero.