Cantilever Beam Subjected to a Concentrated Shear Force
Objective: Determination of the strain state of a cantilever beam subjected a concentrated shear force.
Initial data files:
| CS06_c_v11_3.spr | bar model |
| CS06_p_v11_3.spr | plane-stress model |
Problem formulation: The cantilever beam of a rectangular cross-section is subjected to a concentrated shear force Р applied at its free end. Determine the displacement z of the free end of the beam taking into account the effect of the transverse shear.
Initial data:
| E = 3.0·107 Pa | - elastic modulus, |
| ν = 0.0 | - Poisson’s ratio, |
| L = 10.0 m | - beam length; |
| t = 0.1 m | - width of the beam cross-section; |
| h = 1.0 m | - height of the beam cross-section; |
| k = 1.2 | - shear coefficient; |
| P = 1.0 N | - value of the concentrated force |
Finite element model: Two design models are considered:
Bar model (B), design model – plane frame, 10 elements of type 10. The spacing of the finite element mesh along the longitudinal axis is 1.0 m. Boundary conditions at the clamped end are provided by imposing constraints in the directions of the degrees of freedom: X, Z, UY. Number of nodes in the design model – 11.
Plane-stress model (P), 10 eight-node elements of type 30. The spacing of the finite element mesh along the longitudinal axis is 1.0 m. Boundary conditions at the clamped end are provided by imposing constraints in the directions of the degrees of freedom: X, Z. Number of nodes in the design model – 53.
Results in SCAD
Design model. Bar model
Deformed model. Bar model
Displacements z (m). Bar model
Design model. Plane-stress model
Deformed model. Plane-stress model
Displacements z (m). Plane-stress model
Comparison of solutions:
|
Model |
Displacements z, m |
Deviations, % |
|---|---|---|
|
Bar (B) |
-1.341∙10-3 |
0.00 |
|
Plane-stress (P) |
-1.340∙10-3 |
0.07 |
|
Theory |
-1.341∙10-3 |
─ |
Notes: In the analytical solution, the displacement z of the free end of the beam taking into account the effect of the transverse shear is determined according to the following formula:
\[ z=\frac{4\cdot P\cdot L^{3}}{E\cdot t\cdot h^{3}}\cdot \left( {1+\frac{k\cdot \left( {1+\nu } \right)\cdot h^{2}}{2\cdot L^{2}}} \right). \].