VM32

VM32
Thermal Stresses in a Long Cylinder

Overview

Reference:S. Timoshenko, Strength of Material, Part II, Elementary Theory and Problems, 3rd Edition, D. Van Nostrand Co., Inc., New York, NY, 1956, pg. 234, problem 1.
Analysis Type(s):Static Analysis (ANTYPE = 0)
Element Type(s):
2D Thermal Solid Elements (PLANE55)
2D Structural Solid Elements (PLANE182)
Input Listing:vm32.dat

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.

Figure 42: Long Cylinder Problem Sketch

Long Cylinder Problem Sketch

Material PropertiesGeometric PropertiesLoading
E = 30 x 106 psi
α = 1.435 x 10-5 in/in-°C
υ = 0.3
k = 3Btu/hr-in-°C
a = 0.1875 inches
b = 0.625 inches
Ti = -1°C
To = 0°C

Analysis Assumptions and Modeling Notes

The axial length is arbitrary. Two element types are defined so that the same model can be used for the thermal and stress solutions. A radial grid with nonuniform spacing ratio (1:2) is used since the largest rate of change of the thermal gradient occurs at the inner surface. Surface stresses are requested on element 1 and 7 to obtain more accurate axial and hoop stresses at the inner and outer radii. Nodal coupling is used in the static stress analysis.

Results Comparison

Thermal AnalysisTargetMechanical APDLRatio
T,°C (at X = 0.1875 in)-1.0000-1.00001.000
T,°C (at X = 0.2788 in)-0.67037-0.670611.000
T,°C (at X = 0.625 in)0.00000.0000-
Static AnalysisTargetMechanical APDLRatio
Stressa, psi (at X = 0.1875 in)420.42432.591.029
Stresst, psi (at X = 0.1875 in)420.42426.491.014
Stressa, psi (at X = 0.625 in)-194.58-190.050.977
Stresst, psi (at X = 0.625 in)-194.58-189.760.975