Truss deflection at B: load, temperature, lack of fit

Given data and method

$A = 600$ mm$^2 = 6\times10^{-4}$ m$^2$, $E = 200$ GPa $= 2\times10^8$ kN/m$^2$, $\alpha = 19\times10^{-6}$/$^\circ$C. Joints: A(0,0), B(3,0), C(6,0), D(0,4), E(3,4), F(6,4); members AB, BC, DE, EF, AD, BE, CF, DB, BF. A is a hinge and C a roller; 80 kN acts downward at D and at F. Apply a unit vertical load at B (downward) and use the virtual work method. Tension is positive.

Support reactions

By symmetry, $A_y = C_y = 80$ kN. The 80 kN loads act directly over the supports, so they pass straight down the verticals AD and CF ($F = -80$ kN, compression); all other members have zero force.

Forces in members ($F$ = real loading, $k$ = unit load at B)

(The real force $F$ is non-zero only in AD and CF.)

(i) External loading

$$\Delta_{B1} = \sum\frac{kFL}{AE} = \frac{320.00}{6\times10^{-4}\times2\times10^8} = 2.667\ \text{mm}\ (\downarrow)$$

(ii) Temperature fall of 15$^\circ$C in the inclined members DB and BF

Change in length of each: $\alpha\,\Delta T\,L = 19\times10^{-6}\times(-15)\times5000 = -1.425$ mm.

$$\Delta_{B2} = \sum k\,\alpha\Delta T L = 2\times(0.625)(-1.425) = -1.781\ \text{mm}$$

(Negative: joint B moves up.)

(iii) Lack of fit: vertical members 8 mm too long

$$\Delta_{B3} = \sum k\,e = (-0.5)(8) + (0)(8) + (-0.5)(8) = -8.000\ \text{mm}$$

(BE has $k = 0$.)

Total

$$\Delta_B = 2.667 - 1.781 - 8.000 = -7.115\ \text{mm}$$

Answer: (i) 2.67 mm down, (ii) 1.78 mm up, (iii) 8.00 mm up. The net vertical deflection of B is 7.11 mm upward.