SCAD: Pathological Tests

Twisted Cantilever Beam with Concentrated Shear Forces at Its Free End

Objective: Check of the obtained values of the transverse displacements of the free end of a twisted cantilever beam subjected to concentrated shear forces.

Initial data files:

File name

Description

Twisted_cantilever_beam_Shell_42.spr

Design model with the elements of type 42

Twisted_cantilever_beam_Shell_142.spr

Design model with the elements of type 142

Twisted_cantilever_beam_Shell_44.spr

Design model with the elements of type 44

Twisted_cantilever_beam_Shell_144.spr

Design model with the elements of type 144

Twisted_cantilever_beam_Shell_45.spr

Design model with the elements of type 45

Twisted_cantilever_beam_Shell_145.spr

Design model with the elements of type 145

Twisted_cantilever_beam_Shell_50.spr

Design model with the elements of type 50

Twisted_cantilever_beam_Shell_150.spr

Design model with the elements of type 150

Twisted_cantilever_beam_Solid_36.spr

Design model with the elements of type 36

Twisted_cantilever_beam_Solid_37.spr

Design model with the elements of type 37

 

Problem formulation: The isotropic cantilever beam of a rectangular cross-section twisted along the longitudinal axis is subjected to the concentrated shear Py, Pz forces (bending in and out of the plane of the beam height at the free end). Check the obtained values of the transverse displacements Y, Z of the free end of the twisted cantilever beam from the respective actions.

References: R. H. Macneal, R. L. Harder, A proposed standard set of problems to test finite element accuracy, North-Holland, Finite elements in analysis and design, 1, 1985, p. 3-20.

Initial data:

E = 2.9·107 kPa - elastic modulus of the beam material;
ν = 0.22 - Poisson’s ratio;
b = 0. 32 m - width of the beam;
h = 1. 10 m - height of the beam;
L = 12.0 m - length of the longitudinal axis of the beam;
α = π/2 rad - twist angle of the longitudinal axis of the beam;
Py = 1.0 kN - value of the shear force acting along the height of the beam at the free end;
Pz = 1.0 kN - value of the shear force acting along the width of the beam at the free end.

 

Finite element model: Design model – general type system. Ten design models with a regular finite element mesh 12x2 are considered:

 

Model 1 - 48 three-node shell elements of type 42. 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 concentrated shear Py, Pz forces are given in the form of two nodal forces (Py = 2∙0.5 kN, Pz = 2∙0.5 kN). Number of nodes in the model – 39.

Model 2 - 48 three-node shell elements allowing for shear of type 142. 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 concentrated shear Py, Pz forces are given in the form of two nodal forces (Py = 2∙0.5 kN, Pz = 2∙0.5 kN). Number of nodes in the model – 39.

Model 3 - 24 four-node shell elements of type 44. 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 concentrated shear Py, Pz forces are given in the form of two nodal forces (Py = 2∙0.5 kN, Pz = 2∙0.5 kN). Number of nodes in the model – 39.

Model 4 - 24 four-node shell elements allowing for shear of type 144. 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 concentrated shear Py, Pz forces are given in the form of two nodal forces (Py = 2∙0.5 kN, Pz = 2∙0.5 kN). Number of nodes in the model – 39.

Model 5 - 48 six-node shell elements of type 45. 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 concentrated shear Py, Pz forces are given in the form of two nodal forces (Py = 2∙0.5 kN, Pz = 2∙0.5 kN). Number of nodes in the model – 125.

Model 6 - 48 six-node shell elements allowing for shear of type 145. 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 concentrated shear Py, Pz forces are given in the form of two nodal forces (Py = 2∙0.5 kN, Pz = 2∙0.5 kN). Number of nodes in the model – 125.

Model 7 - 24 eight-node shell elements of type 50. 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 concentrated shear Py, Pz forces are given in the form of two nodal forces (Py = 2∙0.5 kN, Pz = 2∙0.5 kN). Number of nodes in the model – 101.

Model 8 - 24 eight-node shell elements allowing for shear of type 150. 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 concentrated shear Py, Pz forces are given in the form of two nodal forces (Py = 2∙0.5 kN, Pz = 2∙0.5 kN). Number of nodes in the model – 101.

Model 9 - 24 eight-node isoparametric solid elements of type 36. 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 concentrated shear Py, Pz forces are given in the form of four nodal forces (Py = 4∙0.25 kN, Pz = 4∙0.25 kN). Number of nodes in the model – 78.

Model 10 - 24 twenty-node isoparametric solid elements of type 37. 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 concentrated shear Py, Pz forces are given in the form of four nodal forces (Py = 4∙0.25 kN, Pz = 4∙0.25 kN). Number of nodes in the model – 241.

 

Results in SCAD


Models 1 and 2. Design model


Models 1 and 2. Deformed model

 


Model 1. Values of the transverse displacements Y, Z of the free end of the twisted cantilever beam (m, m)

 


Model 2. Values of the transverse displacements Y, Z of the free end of the twisted cantilever beam (m, m)

 


Models 3 and 4. Design model

 


Models 3 and 4. Deformed model

 


Model 3. Values of the transverse displacements Y, Z of the free end of the twisted cantilever beam (m, m)

 


Model 4. Values of the transverse displacements Y, Z of the free end of the twisted cantilever beam (m, m)

 


Models 5 and 6. Design model

 


Models 5 and 6. Deformed model

 


Model 5. Values of the transverse displacements Y, Z of the free end of the twisted cantilever beam (m, m)

 


Model 6. Values of the transverse displacements Y, Z of the free end of the twisted cantilever beam (m, m)

 


Models 7 and 8. Design model

 


Models 7 and 8. Deformed model

 


Model 7. Values of the transverse displacements Y, Z of the free end of the twisted cantilever beam (m, m)

 


Model 8. Values of the transverse displacements Y, Z of the free end of the twisted cantilever beam (m, m)

 


Model 9. Design model

 


Model 9. Deformed model

 


Model 9. Values of the transverse displacements Y, Z of the free end of the twisted cantilever beam (m, m)

 


Model 10. Design model

 


Model 10. Deformed model

 


Model 10. Values of the transverse displacements Y, Z of the free end of the twisted cantilever beam (m, m)

 

Comparison of solutions:

Model

Parameter

Theory

SCAD

Deviation, %

1

(Member type 42)

Transverse displacement Y

of the free end

of the cantilever beam, m

0.005426

0.005314

2.06

Transverse displacement Z

of the free end

of the cantilever beam, m

0.001746

0.001463

16.21

2

(Member type 142)

Transverse displacement Y

of the free end

of the cantilever beam, m

0.005426

0.006220

14.63

Transverse displacement Z

of the free end

of the cantilever beam, m

0.001746

0.001710

2.06

3

(Member type 44)

Transverse displacement Y

of the free end

of the cantilever beam, m

0.005426

0.001899

65.00

Transverse displacement Z

of the free end

of the cantilever beam, m

0.001746

0.000906

48.11

4

(Member type 144)

Transverse displacement Y

of the free end

of the cantilever beam, m

0.005426

0.002158

60.23

Transverse displacement Z

of the free end

of the cantilever beam, m

0.001746

0.000961

44.96

5

(Member type 45)

Transverse displacement Y

of the free end

of the cantilever beam, m

0.005426

0.005388

0.70

Transverse displacement Z

of the free end

of the cantilever beam, m

0.001746

0.001750

0.23

6

(Member type 145)

Transverse displacement Y

of the free end

of the cantilever beam, m

0.005426

0.006588

21.42

Transverse displacement Z

of the free end

of the cantilever beam, m

0.001746

0.002178

24.74

7

(Member type 50)

Transverse displacement Y

of the free end

of the cantilever beam, m

0.005426

0.003948

27.24

Transverse displacement Z

of the free end

of the cantilever beam, m

0.001746

0.001481

15.18

8

(Member type 150)

Transverse displacement Y

of the free end

of the cantilever beam, m

0.005426

0.004094

24.55

Transverse displacement Z

of the free end

of the cantilever beam, m

0.001746

0.001505

13.80

9

(Member type 36)

Transverse displacement Y

of the free end

of the cantilever beam, m

0.005426

0.001119

79.38

Transverse displacement Z

of the free end

of the cantilever beam, m

0.001746

0.000583

66.61

10

(Member type 37)

Transverse displacement Y

of the free end

of the cantilever beam, m

0.005426

0.005403

0.42

Transverse displacement Z

of the free end

of the cantilever beam, m

0.001746

0.001757

0.63
 

Notes: In the analytical solution the values of the transverse displacements Y, Z of the free end of the twisted cantilever beam from the respective actions are determined according to the following formulas:

\[ Y=\frac{12\cdot P_{y} \cdot L^{3}}{E\cdot b^{3}\cdot h^{3}}\cdot \left[ {\left( {\frac{1}{6}-\frac{1}{\pi^{2}}} \right)\cdot h^{2}+\left( {\frac{1}{6}+\frac{1}{\pi^{2}}} \right)\cdot b^{2}} \right]; \quad Z=\frac{12\cdot P_{z} \cdot L^{3}}{E\cdot b^{3}\cdot h^{3}}\cdot \left[ {\left( {\frac{1}{6}+\frac{1}{\pi^{2}}} \right)\cdot h^{2}+\left( {\frac{1}{6}-\frac{1}{\pi^{2}}} \right)\cdot b^{2}} \right]. \]