Stability of a Beam of a Square Cross-Section Simply Supported in and out of the Bending Plane Subjected to a Transverse Load Uniformly Distributed along Its Longitudinal Axis

Stability Bar 1-5

Objective: Determination of the critical value of the transverse load uniformly distributed along the longitudinal axis of a beam of a square cross-section simply supported in and out of the bending plane corresponding to the moment of its buckling.

Initial data files:

File name

Description

Stability_Bar_13_Bar.SPR

Bar model

Stability_Bar_13_Shell.SPR

Shell element model

 

Problem formulation: The beam of a square cross-section simply supported in and out of the bending plane is subjected to the action of the transverse load q, uniformly distributed along its longitudinal axis. Determine the critical value of the transverse uniformly distributed load qcr, corresponding to the moment of buckling of the simply supported beam.

References: A.S. Volmir. Stability of Deformable Systems, Moscow, Nauka, 1967, p.220

Initial data:

L = 10.0 m - length of the simply supported beam;
h = b = 1.0 m - side of the square cross-section of the simply supported beam;
E = 3.0·107 kN/m2 - elastic modulus of the simply supported beam material;
ν = 0.2 - Poisson’s ratio;
q = 105 kN/m - initial value of the transverse load uniformly distributed along the longitudinal axis of the beam.

 

Finite element model: Design model – general type system. Two design models are considered:

Bar model (B), 10 elements of type 5, the spacing of the finite element mesh along the longitudinal axis is 1.0 m. Boundary conditions are provided by imposing constraints on the nodes of the simply supported ends of the beam in the directions of the degrees of freedom X, Y, Z, UX. The action with the initial value of the transverse uniformly distributed load q is specified on all elements of the beam. Number of nodes in the design model – 11;

Reissner-Mindlin shell element model (P), 2560 eight-node elements of type 150, the spacing of the finite element mesh along the longitudinal axis and along the height of the beam is 0.0625 m. The shell is supported by vertical high-rigidity bars (h = b = 1.0 m; E = 3.0·109 kN/m2; ν = 0.2), 64 elements of type 5. Boundary conditions are provided by imposing constraints on the nodes of the ends of the beam lying on its longitudinal axis in the directions of the degrees of freedom X, Y, Z and on all other nodes of the ends of the beam in the direction of the degree of freedom Y. The action with the initial value of the transverse load q uniformly distributed along the line is specified on the lower sides of all beam elements located above its longitudinal axis. Number of nodes in the design model – 8033.

 

Results in SCAD

ScreenShot715
Design model. Bar model

 

ScreenShot717
Design model. Reissner-Mindlin shell element model

 

ScreenShot718
1-st buckling mode. Bar model

 

ScreenShot719
1- st buckling mode. Reissner-Mindlin shell element model

 

 

Comparison of solutions:

Critical value of the transverse load qcr (kN/m), uniformly distributed along the longitudinal axis of the beam simply supported in and out of the bending plane

Design model

Theory

SCAD

Deviation, %

Bar

59337

0,590213∙100000=59021

0,53

Reissner-Mindlin

shell element

59337

0,578880∙100000=57888

2,44

 

Notes: In the analytical solution the critical value of the transverse uniformly distributed load qcr, corresponding to the moment of buckling of the simply supported beam can be determined according to the following formula:

\[ q=\frac{28,31\cdot \sqrt {E\cdot I_{z} \cdot G\cdot I_{x} } }{L^{3}} \quad G=\frac{E}{2\cdot \left( {1+\nu } \right)} \]

\( I_{z} =\frac{h\cdot b^{3}}{12} \) – minimum bending inertia moment (out of the moment plane);

\( I_{y} =\frac{b\cdot h^{3}}{12} \) – maximum bending inertia moment (in the moment plane);

\( I_{x} =k_{f} \cdot h\cdot b^{3} \) – free torsional inertia moment, where:

\[ k_{f} =\frac{1}{3}\cdot \left\{ {1-\frac{192}{\pi^{5}}\cdot \frac{b}{h}\cdot \sum\limits_{n=1}^\infty {\left[ {\sin^{2}\left( {\frac{n\cdot \pi }{2}} \right)\cdot \frac{1}{n^{5}}\cdot th\left( {\frac{n\cdot \pi \cdot h}{2\cdot b}} \right)} \right]} } \right\} \]