Stability of a Simply Supported Rectangular Plate Uniformly Compressed in One Direction

Objective: Determination of the critical value of the compressive forces uniformly distributed along two opposite sides of a simply supported rectangular plate corresponding to the moment of its buckling.

Initial data files:

File name

Description

6.6_a_4_n_4.spr

Design model with the ratios of the sides of the plate a/b = 0.5 from four-node shell elements of type 44

6.6_a_4_n_8.spr

Design model with the ratios of the sides of the plate a/b = 0.5 from eight- node shell elements of type 50

6.6_a_8_n_4.spr

Design model with the ratios of the sides of the plate a/b = 1.0 from four-node shell elements of type 44

6.6_a_8_n_8.spr

Design model with the ratios of the sides of the plate a/b = 1.0 from eight- node shell elements of type 50

6.6_a_12_n_4.spr

Design model with the ratios of the sides of the plate a/b = 1.5 from four-node shell elements of type 44

6.6_a_12_n_8.spr

Design model with the ratios of the sides of the plate a/b = 1.5 from eight- node shell elements of type 50

 

Problem formulation: The simply supported rectangular plate is subjected to the action of compressive forces σ, uniformly distributed along two opposite sides. Determine the critical value of the compressive forces σcr, corresponding to the moment of buckling of the rectangular plate.

References:

S. P. Timoshenko, Stability of Bars, Plates and Shells. — Moscow. Nauka. — 1971. — p. 621.

A.S. Volmir. Stability of Deformable Systems. — Moscow. — Nauka. — 1967. — p. 328.

Initial data:

a = 4.0; 8.0; 12.0 m - side of the rectangular plate free from forces (along the X axis of the global coordinate system);
b = 8.0 m - side of the rectangular plate subjected to the compressive forces (along the Y axis of the global coordinate system);
h = 0.08 m - thickness of the rectangular plate;
E = 1.0·107 kN/m2 - elastic modulus of the rectangular plate material;
ν = 1/3 - Poisson’s ratio;
σ = 1.25·103 kN/m2 - initial value of the compressive forces.

 

Finite element model: Design model – general type system. Two design models with four-node shell elements of type 44 and eight-node shell elements of type 50 are considered for three cases with the ratios of the sides of the plate a/b = 0.5; 1.0; 1.5. The spacing of the finite element mesh along the sides of the plate (along the X and Y axes of the global coordinate system) is 1.0 m. Number of elements in the models – 32; 64; 96. Boundary conditions are provided by imposing constraints on the nodes of the support contour of the plate in the direction of the degree of freedom Z. A load uniformly distributed along the line with the initial value p = σ∙h = 100 kN/m is specified on one of the two opposite sides of the plate subjected to the compressive forces, and the constraints in the respective direction (along the X axis of the global coordinate system) are imposed on the nodes of the other one. The dimensional stability of the design model is provided by imposing constraints in the normal direction (along the Y axis of the global coordinate system) on the nodes of one of the two opposite sides of the plate free from forces, and by imposing constraints in the UZ direction of the global coordinate system on the node of one of the corners of the plate. Number of nodes in the models – 45 (121); 81 (225); 117 (329).

 

Results in SCAD


Design models with the ratio of the sides of the plate a/b = 0.5

 


Design models with the ratio of the sides of the plate a/b = 1.0

 


Design models with the ratio of the sides of the plate a/b = 1.5

 


Buckling modes for the design models with the ratio of the sides of the plate a/b = 0.5

 


Buckling modes for the design models with the ratio of the sides of the plate a/b = 1.0

 


Buckling modes for the design models with the ratio of the sides of the plate a/b = 1.5

 

Comparison of solutions:

Critical value of the compressive forces σcr, kN/m2

Plate sides ratio

Design model

Theory

SCAD

Deviation, %

a/b = 0.5

Member type 44

n = 4 nodes

5783

4.716991∙100/0.08 =

= 5896

1.95

Member type 50

n = 8 nodes

4.626558∙100/0.08 =

= 5783

0.00

a/b = 1.0

Member type 44

n = 4 nodes

3701

2.998497∙100/0.08 =

= 3748

1.27

Member type 50

n = 8 nodes

2.960899∙100/0.08 =

= 3701

0.00

a/b = 1.5

Member type 44

n = 4 nodes

4016

3.264680∙100/0.08 =

= 4081

1.62

Member type 50

n = 8 nodes

3.212803∙100/0.08 =

= 4016

0.00

 

 

Notes: In the analytical solution the critical value of the compressive forces σcr, corresponding to the moment of buckling of the rectangular plate can be determined according to the following formula:

\[ \sigma_{cr} =k\cdot \frac{\pi^{2}\cdot D}{b^{2}\cdot h}, \quad where:\quad \] \[ D=\frac{E\cdot h^{3}}{12\cdot \left( {1-\nu^{2}} \right)}, \quad k=\left( {\frac{m\cdot b}{a}+\frac{a}{m\cdot b}} \right)^{2}, \]

m = 1, 2, 3 … – number of half waves of the buckling mode in the direction of the compression of the plate; its minimum value is determined from the following expression:

\[ \frac{a}{b}\le \sqrt {m\cdot \left( {m+1} \right)} . \]