Geometric Properties of a Semicircle

Aim: To check the accuracy of the geometric properties calculation for a rod cross-section in the form of a semicircle.

Name of a file with the initial data: Disk2.cns

Formulation: Check the accuracy of the torsional geometric properties calculation for a rod cross-section in the form of a semicircle.

References: Young W.C., Budynas R.G., Roark's Formulas for Stress and Strain, New York , McGraw-Hill,  New York, 2002.

Weisstein E.W., Torsional Rigidity, From MathWorld — A Wolfram Web Resource. http://mathworld.wolfram.com/TorsionalRigidity.html

Cowper G.R., The Shear Coefficient in Timoshenko's Beam, ASME Journal of Applied Mechanics, 1966, 33, 335-341.

Novojilov V.V., Theory of Elasticity, Moscow, State Union Publishing House of the Shipbuilding Industry, 1958, (§ VI.21).

Marinetti A., On the Accuracy of Shear Factors for Elastic Uniform Beams: Evaluation Using the Boundary Element Method, In "Materiali e Metodi Innovativi nell'Ingegneria Strutturale" Aracne Editrice, ISBN 978-88-548-2451-5, (2009).

Initial data:

 ν = 0.1 - Poisson’s ratio; d = 10 cm - diameter of a circle.

Design model: The design model is created by triangulation (the number of triangles ≈ 3000) on the basis of a model of the external contour. The external contour is a polygon approximating a semicircle. The number of vertices of the contour in a model is 33.

Results Obtained in Consul

Design model, coordinate and principal axes, center of mass, ellipse of inertia, core of the section

Comparison of results:

Parameter

Theory

CONSUL

Deviation, %

Cross-sectional area, A cm2

39,269

39,207

0.16

Conventional shear area along the principal U-axis, Av,y cm2

30,239

29,071

3,862

Conventional shear area along the principal V-axis, Av,z cm2

34,691

33,29

4,03

Torsional moment of inertia, It cm4

182,25

183,05

0,438

Y-coordinate of the shear center, yb cm

5

5

0

Z-coordinate of the shear center Z, zb cm

2,55497

2,555

0,00117

Notes: Geometric properties can be determined analytically by the following formulas:

$A_{v,y} \approx A\frac{1+\nu }{1,305+1,273\nu };$ $A_{v,z} \approx A\frac{(1+\nu )^{2}}{\frac{7}{6}+\left( {2+\frac{64}{45\pi ^{2}}} \right)\cdot \nu +\left[ {\frac{2}{3}+\frac{64}{15\pi ^{2}}-\sum\limits_{т=1}^\infty {\frac{1}{(2n-1)^{2}(2n+3)^{2}}} } \right]\cdot \nu^{2}};$ $I_{t} =\left( {\frac{\pi }{2}-\frac{4}{\pi }} \right)\left( {\frac{d}{2}} \right)^{4};$ $y_{b} =d/2;$ $z_{b} \approx \left( {\frac{4}{15\pi }\cdot \frac{3+3,05\nu }{1+\nu }} \right)d.$

Differences in the position of the shear center are partially explained by the difference in the definition of the shear center (see Definitions of Geometric Properties).