Stability of a Cantilever Beam of a Square Cross-Section Subjected to a Load Uniformly Distributed along Its Longitudinal Axis

Objective: Determination of the critical value of the load uniformly distributed along the longitudinal axis of a cantilever beam of a square cross-section corresponding to the moment of its buckling.

Initial data files:

File name

Description

Stability_Bar_8_Bar.SPR

Bar model

Stability_Bar_8_Shell.SPR

Shell element model

Stability_Bar_8_Solid.SPR

Solid element model

 

Problem formulation: The cantilever beam of a square cross-section is subjected to the action of the load q, uniformly distributed along its longitudinal axis. Determine the critical value of the uniformly distributed load qcr, corresponding to the moment of buckling of the cantilever beam.

References: A.S. Volmir. Stability of Deformable Systems, Moscow, Nauka, 1967, p.217;

Initial data:

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

 

Finite element model: Design model – general type system. Three 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 node of the clamped end of the beam in the directions of the degrees of freedom X, Y, Z, UX, UY, UZ. The action with the initial value of the 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. Boundary conditions are provided by imposing constraints on the nodes of the clamped end of the beam in the directions of the degrees of freedom X, Y, Z, UX, UY, UZ. The action with the initial value of the load q uniformly distributed along the line is specified on the upper sides of all beam elements located under the longitudinal axis of the beam. Number of nodes in the design model – 8033.

Solid element model (S), 5120 twenty-node elements of type 37, the spacing of the finite element mesh along the longitudinal axis, width and height of the beam is 0.125 m. Boundary conditions are provided by imposing constraints on the nodes of the clamped end of the beam in the directions of the degrees of freedom X, Y, Z, UX, UY, UZ. The action with the initial value of the load uniformly distributed over the face qA = q/ b is specified on the upper faces of all beam elements located under the longitudinal axis of the beam. Number of nodes in the design model – 24705.

 

Results in SCAD


Design model. Bar model

 


Design model. Reissner-Mindlin shell element model

 


Design model. Solid element model

 


1-st buckling mode. Bar model

 


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

 


1-st buckling mode. Solid element model

 

 

Comparison of solutions:

Critical value of the load qcr, uniformly distributed along the longitudinal axis of the cantilever beam

Design model

Theory

SCAD

Deviation, %

Bar

26933

0,268111∙100000=26811

0,45

Reissner-Mindlin

shell element

26933

0,260448∙100000=26045

3,30

Solid element

26933

0,253906∙100000=25391

5,73

 

 

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

\[ q=\frac{12,85\cdot \sqrt {E\cdot I_{z} \cdot G\cdot I_{x} } }{L^{3}} \quad G=\frac{E}{2\cdot \left( {1+\nu } \right)} \]

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

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

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