Question

3a fr オ Given: A steel railroad track which was installed at 30° F. Let E = 29 x 10.6 psi & α = 6.5 x 10.6 /。F Find: If the temperature reaches 125° F, a) The normal stress in the rails which are welded together to form a continuous track. b) The normal stress in the rails assuming the rails are 39 ft. long with 1/4" gap between them.

Given a steel railroad track which was installed at 30 F. Let E = 29 x 10^6 psi and α = 6.5 x 10^-6 / F

If the temperature reaches 125 F,

a) Find the normal stress in the rails which are welded together to form a continuous track.

b) Find the normal stress in the rails assuming the rails are 39 ft. long with 1/4" gap between them

Answer 1

"Given L = 39 ft standard temp, t_0 = 30 degree F E = 29 times 10^6 psi alpha = 6.5 times 10^-6/degree F T = 125 degree F when the steel is welded together then normal stress = alpha (T - T_0) E = 6.5 times 10^-6 times (125 - 30) times 29 times 10^6 = 17907.5 psi when the rails are 39 ft long with 1/4"" gap between them Delta T = 125 - 30 = 95 degree F Free expansion = L alpha Delta T = 39' times 6.5 times 10^-6 times 95 = 0.024' = 0.2889 inch actual expansion = 0.25"" expansion resisted = 0.2889"" = 0.25"" = 0.0389"" strain resisted = 0.0389""/39 times 12"" = 8.33119 times 10^-5 stress developed = strain times E = 8.33119 times 10^-5 times 29 times 10^6 psi Therefore Normal stress = 2416.047 psi"