Unilateral Tension of a Plate with a Small Circular Hole (Kirsch Problem)

Objective: Determination of the stress state of a plate of considerable width and unit thickness with a small circular hole in polar coordinates subjected to unilateral uniform tension.

Initial data files:

File name	Description
4.27_b_20_9_grad.spr	1 variant of the design model – coarse FE mesh
4.27_b_60_4.5_grad.spr	2 variant of the design model – fine FE mesh

 

Problem formulation: The square plate of considerable width and unit thickness with a small circular hole of radius a is subjected to unilateral uniform tension by stresses σ in the direction of the x₁ axis applied in its center. Determine the stress tensor components in polar coordinates σ_rr, σ_θθ, σ_rθ at different radial distances r from the origin at the angles to the x₁ axis θ = 0º and θ = 90º.

References: S.P. Demidov, Theory of Elasticity. — Moscow: High school, 1979.

Initial data:

 

Finite element model: Two variants of the design model are considered.

Variant 1:

Design model – plane frame, width of the plate 2·b = 20.0 m, plate elements – 1088 eight-node elements of type 30 and 32 six-node elements of type 25. The spacing of the finite element mesh in the radial direction from r = 1.00 m to r = 2.00 m is 0.10 m, from r = 2.00 m to r = 10.00 m is 0.50 m and in the tangential direction the spacing is 9º. The direction of the output of internal forces is radial tangential. Number of nodes in the design model – 3409.

Variant 2:

Design model – plane frame, width of the plate 2·b = 60.0 m, plate elements – 5024 eight-node elements of type 30 and 40 six-node elements of type 25. The spacing of the finite element mesh in the radial direction from r = 1.00 m to r = 3.00 m is 0.10 m,  from r = 3.00 m to r = 5.00 m is 0.20 m, from r = 5.00 m to r = 9.00 m is 0.40 m, from r = 9.00 m to r = 21.00 m is 0.80 m, from r = 21.00 m to r = 29.00 m is 1.60 m, and in the tangential direction the spacing is 4.5º. The direction of the output of internal forces is radial tangential. Number of nodes in the design model – 15312.

 

Results in SCAD

Design model. Variant 1

Design model. Variant 2

 
Values of stresses σ_rr (kN/m²) for the design model according to variant 1

Stress diagram σ_rr (kN/m²) at the angle to the Ox₁ axis θ = 0º for the design model according to variant 1

Stress diagram σ_rr (kN/m²) at the angle to the Ox1 axis θ = 90º for the design model according to variant 1

Values of stresses σ_rr (kN/m²) for the design model according to variant 2

Stress diagram σ_rr (kN/m²) at the angle to the Ox₁ axis θ = 0º for the design model according to variant 2

Stress diagram σ_rr (kN/m²) at the angle to the Ox₁ axis θ = 90º for the design model according to variant 2

Values of stresses σ_θθ (kN/m²) for the design model according to variant 1

Stress diagram σ_θθ (kN/m²) at the angle to the Ox₁ axis θ = 0º for the design model according to variant 1

Stress diagram σ_θθ (kN/m²) at the angle to the Ox₁ axis θ = 90º for the design model according to variant 1

Values of stresses σ_θθ (kN/m²) for the design model according to variant 2

Stress diagram σ_θθ (kN/m²) at the angle to the Ox₁ axis θ = 0º for the design model according to variant 2

Stress diagram σ_θθ (kN/m²) at the angle to the Ox1 axis θ = 90º for the design model according to variant 2

Values of stresses σ_rθ (kN/m²) for the design model according to variant 1

Stress diagram σ_rθ (kN/m²) at the angle to the Ox1 axis θ = 0º for the design model according to variant 1

Stress diagram σ_rθ (kN/m²) at the angle to the Ox₁ axis θ = 90º for the design model according to variant 1

Values of stresses σ_rθ (kN/m2) for the design model according to variant 2

Stress diagram σ_rθ (kN/m²) at the angle to the Ox₁ axis θ = 0º for the design model according to variant 2

Stress diagram σ_rθ (kN/m²) at the angle to the Ox₁ axis θ = 90º for the design model according to variant 2

 

Comparison of solutions:

Stress tensor components in polar coordinates σ_rr, σ_θθ, σ_rθ.

Solution	Stresses σrr (kN/m2)					Stresses σθθ (kN/m2)
	θ = 0º			θ = 90º		θ = 0º			θ = 90º
	r  = 1.000 m	r  =  (√6/5)·a = 1.095 m	r  =  (√3/2)·a = 1.225 m	r  = 1.000 m	r  = (√2)·a = 1.414 m	r  = 1.000 m	r  = (√3)·a = 1.732 m	r  = (√6)·a = 2.449 m	r  = 1.000 m
Theory	0.00	-σ/24 = -4.17	0.00	0.00	3·σ/8 = 37.50	-σ = -100.00	0.00	σ/24 = 4.17	3·σ = 300.00
SCAD, DM var.1	-1.32	-5.65	-1.26	2.77	39.43	-100.63	-1.18	3.56	307.46
Deviations, %	─	─	─	─	5.15	0.63	─	─	2.49
SCAD, DM var.2	-0.76	-4.78	-0.36	1.31	37.94	-100.05	-0.04	4.16	299.85
Deviations, %	─	─	─	─	1.17	0.05	─	─	0.05

 

Notes: In the analytical solution the stresses σ_rr, σ_θθ, σ_rθ  in the plate with a small circular hole subjected to unilateral uniform tension are determined according to the following formulas (S.P. Demidov, Theory of Elasticity. — Moscow: High school, 1979, p. 302):

\[ \sigma_{rr} =\frac{\sigma }{2}\cdot \left( {1-\frac{a^{2}}{r^{2}}} \right)+\frac{\sigma }{2}\cdot \left( {1-4\cdot \frac{a^{2}}{r^{2}}+3\cdot \frac{a^{4}}{r^{4}}} \right)\cdot \cos \left( {2\cdot \theta } \right); \] \[ \sigma_{\theta \theta } =\frac{\sigma }{2}\cdot \left( {1+\frac{a^{2}}{r^{2}}} \right)-\frac{\sigma }{2}\cdot \left( {1+3\cdot \frac{a^{4}}{r^{4}}} \right)\cdot \cos \left( {2\cdot \theta } \right); \] \[ \sigma_{r\theta } =-\frac{\sigma }{2}\cdot \left( {1+2\cdot \frac{a^{2}}{r^{2}}-3\cdot \frac{a^{4}}{r^{4}}} \right)\cdot \sin \left( {2\cdot \theta } \right). \]