Stability of a Cantilever Beam of a Square Cross-Section Subjected to a Load Uniformly Distributed along the Longitudinal Axis of Its Upper Face
Objective: Determination of the critical value of the load uniformly distributed along the longitudinal axis of the upper face of a cantilever beam of a square cross-section corresponding to the moment of its buckling.
Initial data files:
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File name |
Description |
|---|---|
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Bar model |
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Shell element model |
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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 the longitudinal axis of its upper face. Determine the critical value of the uniformly distributed load qcr, corresponding to the moment of buckling of the cantilever beam.
References: S. P. Timoshenko, Stability of Bars, Plates and Shells, Moscow, Nauka, 1971, p.303
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 upper face 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. 11 vertical upward two-node elements of type 100 (3D rigid body) with the length h/2 are adjacent to the nodes of the beam. The action with the initial value of the uniformly distributed load q is specified in the free nodes of the elements of the rigid body (elevated application point) as concentrated forces P = q·b·1.0 = 105 kN (0.5·105 kN for the end nodes). Number of nodes in the design model – 22;
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 upper face 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 upper face 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 upper face of the cantilever beam
|
Design model |
Theory |
SCAD |
Deviation, % |
|---|---|---|---|
|
Bar |
23737 |
0,236895∙100000=23690 |
0,20 |
|
Reissner-Mindlin shell element |
23737 |
0,233316∙100000=23332 |
1,71 |
|
Solid element |
23737 |
0,246094∙100000=24609 |
3,67 |
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:
\[ q=\frac{12,85\cdot \sqrt {E\cdot I_{z} \cdot G\cdot I_{x} } }{L^{3}}\cdot k_{h} \quad k_{h} =f\left( {\frac{h}{2\cdot L}\cdot \sqrt {\frac{E\cdot I_{z} }{G\cdot I_{x} }} } \right) \quad G=\frac{E}{2\cdot \left( {1+\nu } \right)} \]
\( I_{z} =\frac{h\cdot b^{3}}{12} \) – minimum bending inertia moment (out of the moment plane);
\( I_{x} =k_{f} \cdot h\cdot b^{3} \) – free torsional inertia moment, where:
\[ k_{f} =\frac{1}{3}\cdot \left\{ {1-\frac{192}{\pi^{5}}\cdot \frac{b}{h}\cdot \sum\limits_{n=1}^\infty {\left[ {\sin^{2}\left( {\frac{n\cdot \pi }{2}} \right)\cdot \frac{1}{n^{5}}\cdot th\left( {\frac{n\cdot \pi \cdot h}{2\cdot b}} \right)} \right]} } \right\} \]