Question 2.7: Each bar in the truss shown in Figure 2.38 has a 2 in.² cros...
Each bar in the truss shown in Figure 2.38 has a 2 in.² cross-sectional area, modulus of elasticity E = 14 × 10^6 psi, and coefficient of thermal expansion α = 11 × 10^{-6} (°F)^{-1} . If their temperature is increased by 40°F from their initial temperature T, what is the resulting displacement of point A? What upward force must be applied to prevent this displacement?
Given: Dimensions and properties of truss; imposed temperature change.
Find: Displacement of point A; force necessary to prevent this displacement.
Assume: Hooke’s law applies. Neglect weight of bars.
The geometry of the problem allows us to find the original length of bars AB and AC (Figure 2.39)
L = 36 in./sin (60 ̊) = 41.56 in.
The change in the length of each bar due to the change in temperature ΔT is then
\delta _{_{T}}=L\alpha \Delta T=(41.56 \textrm{in.})(11\times 10^{-6}(°\textrm{F})^{-1})(40°\textrm{F})=0.018 \textrm{in.}So, the new vertical distance from the fixed surface to point A is
( 41.56+0.018) ⋅ sin ( 60°) = 36.008 in.
The horizontal displacements of AB and AC will be equal and opposite, so the net displacement of point A is only vertical and is 0.008 in. The upward force applied to prevent this must “undo” the thermal expansion of the two bars; it must induce a compressive axial load P in both bars such that, by Hooke’s law,
\frac{PL}{AE}= -0.018 \textrm{in.} , so P=\frac{(-0.018 \textrm{in.})(14\times 10^6 \textrm{psi})(2 \textrm{in.}^2)}{41.56 \textrm{in.}} = -12,127 lb (compressive!) .
We construct an FBD and use equilibrium to find the force F necessary to induce this compressive load P in both bars (Figure 2.40)
ΣF_y = 0 = F – 2P sin(60°), or F = 2P sin(60°)
F = 21,327 lb.
