VM-WB-MECH-072

VM-WB-MECH-072
Thermal Stresses in a Long Cylinder

Overview

Reference:

Timoshenko, S. (1956). Strength of Material, Part II: Elementary Theory and Problems (3rd ed., p. 234, problem 1). New York, NY: D. Van Nostrand Co, Inc.

Solver(s):Ansys Mechanical
Analysis Type(s):Thermal Stress Analysis
Element Type(s):2-D Thermal Solid Elements

Test Case

A long thick-walled cylinder is maintained at a temperature Ti on the inner surface and To on the outer surface. Determine the temperature distribution through the wall thickness. Also determine the axial stress σa and the tangential (hoop) stress σt at the inner and outer surfaces Edge sizing is used for all edges and edge behavior is defined as hard.

Figure 96: Schematic

Schematic

Material PropertiesGeometric PropertiesLoading
E = 30 x 106 psi
α = 1.435 x 10-5/°F
ν = 0.3
k = 8.333 x 10-4 Btu/s-in-°F
a = 0.1875 in
b = 0.625 in
Ti = -1 °F
To = 0 °F

Analysis Assumptions and Modeling Notes for Ansys Mechanical

Because of the symmetry in loading conditions and in the geometry, this problem is solved as an axisymmetric problem in Ansys Mechanical. The axial length is arbitrary and it is taken has 0.1 in. Nodal coupling is used in the static stress analysis. Model is used for the thermal and stress solutions.

Figure 97: Schematic in Ansys Mechanical

Schematic in Ansys Mechanical

Results Comparison for Ansys Mechanical

Thermal AnalysisTargetMechanicalError (%)
T,°F (at X = 0.1875 in)-1-10
T,°F (at X = 0.2788 in)-0.67037-0.670570.0298
T,°F (at X = 0.625 in)000
Static AnalysisTargetMechanicalError (%)
Stressa, psi (at X = 0.1875 in)420.42416.05 -1.039
Stresst, psi (at X = 0.1875 in)420.42405.43-3.565
Stressa, psi (at X = 0.625 in)-194.58-195.090.262
Stresst, psi (at X = 0.625 in)-194.58-195.070.252