TEST CASE 6.3 CIRCULAR PLATE ON A SINGLESIDED ELASTIC FOUNDATION
Reference:
П. Панагиотопулос, Неравенства в механике и их приложения, Москва: «Мир», 1989, стр. 384.
Problem description:
Circular linearelastic plate was motionless and supported by singlesided elastic foundation. Then this plate was loaded by uniformly areadistributed load q and by edgedistributed bending moment M. Determine rotational angle of plate edge ψ and domain with contact of plate and foundation at coefficient а value 1; 2; 3; 4; 5; 5.1.
Problem sketch:
Type of created problem:
The spatial structure (X, Y, Z, UX, UY, UZ).
Geometric characteristics:
Plate radius: R = 6 m;
Plate thickness: t = 0.06 m.
Material properties:
Elastic modulus: E = 2.1·10^{11} Pa;
Poisson’s ratio: μ = 0.
Boundary conditions:
All plate nodes: X = Y = UX = UZ = 0;
Plate central assembly: UY = 0;
Plate boundary nodes: Z = 0.
Loads:
q = 600 N/m^{2};
M = a·q·R^{2}/32.
Model description:
The system is modeled by 95 finite elements of thin shell (FE type is 44). Because of symmetry only a sector of circular plate is considered. For nonlinear solution iterative process is used.
Analytical solution:
Rotational angle of plate edge is an angle between the perpendiculars of deformed and nondeformed plate surface. It is defined via expression ψ = dw/dn  derivative of plate displacement along normal to the boundary (along radius).
Calculation results:
а 
Target value  Number of iterations  Analytical solution  Numerical solution  LIRA 10  Deviation, % 
1 
ψ·10^{5}, rad 
5700 
7.50 
7.29 
7.119 
5.35 
2 
7100 
23.00 
22.236 
21.872 
5.16 

3 
12000 
43.90 
43.80 
43.407 
1.14 

4 
14600 
71.50 
72.379 
71.503 
0.004 

5 
16400 
107.14 
107.176 
106.52 
0.58 

5.1 
26000 
126.50 
126.117 
121.40 
4.03 

1 
Contact domain, m 
5700 
3.70 / 3.78 
3.70 
3.70 
0.00 
2 
7100 
2.58 / 2.64 
2.58 
2.58 
0.00 

3 
12000 
1.56 / 1.62 
1.56 
1.56 
0.00 

4 
14600 
0.54 / 0.60 
0.54 
0.60 
0.00 

5 
16400 
0.00 
0.00 
0.00 
0.00 

5.1 
26000 
No contact 
No contact 
No contact 
0.00 