Question 2.8: A steel railroad track (E = 200 GPa, α = 11.7 × 10^-6/°C) wa...
A steel railroad track (E = 200 GPa, α = 11.7 × 10^{-6}/°C) was laid out at a temperature of 0°C. Determine the normal stress in a rail when the temperature reaches 50°C, assuming that the rails are (a) welded to form a continuous track, or (b) 12 m long with 6-mm gaps between them.
Given: Geometry of problem, material properties, imposed temperature change.
Find: Normal stress (a) when continuous or (b) when gaps are left.
Assume: Hooke’s law applies.
Based on our understanding of thermal stresses, we expect the stress calculated in part (b) to be lower than that in part (a): We have learned that thermal stresses are induced only when a part is prevented from experiencing its natural thermal deformation, so the space left to accommodate thermal expansion in part (b) should help relieve the induced stress. We will see whether this expectation is met.
A schematic helps to illustrate the problem (Figure 2.41)
(a) The total deformation of a steel track segment is δ_{_{T}} + δ_{_{P}} = 0 , as the welding allows no net change to the length of the segments. Hence, we add the deformations due to thermal effects and compressive forces:
0=\underbrace{\alpha (\Delta T)L}_{tends to stretch}+\underbrace{\frac{-PL}{AE}}_{tends to squash} ,
so
\alpha \Delta T= \frac{P}{AE}= \frac{\sigma }{E}\sigma= \alpha \Delta T E=(11.7\times 10^{-6}(°\textrm{C})^{-1})(50-0°\textrm{C})(200\times 10^9 \textrm{Pa})
σ = 117 MPa (compressive)
when welded.
(b) If a gap of 6 mm is left between rails, we allow each segment a net stretch of 6 mm:
so
\sigma= \frac{\alpha \Delta TL-0.006 \textrm{m}}{L} E =\frac{(11.7\times 10^{-6}(°\textrm{C})^{-1}(50°\textrm{C})(12 \textrm{m})-0.006 \textrm{m}}{12 \textrm{m}} )(200\times 10^9 \textrm{Pa})σ = 17 MPa (compressive)
when a gap is left.
