RD-E: 1000 Bending

Pure bending test with different 3- and 4-nodes shell formulations.

The bending of a straight cantilever beam is studied. The example used is a famous bending test for shell elements. The analytical solution enables the comparison with the quality of the numerical results. Carefully watch the influence from the shell formulation. In addition, the results for the different time step scale factors are compared.

Options and Keywords Used

• Q4 and T3 meshes
• QEPH, Belytshcko & Tsay, BATOZ, and DKT shells
• Mesh, hourglass, quasi-static analysis, and bending test
• Imposed velocity (/IMPVEL)
• Rigid bodies (/RBODY)

At one extremity of the beam, all DOF are blocked. A rotational velocity is imposed on the main node of the rigid body placed on the other side.

This velocity follows a linear function: Y=1

Input Files

Before you begin, copy the file(s) used in this example to your working directory.

Model Description

The purpose of this example is to study a pure bending problem. A cantilever beam with an end moment is studied. The moment variation is modeled by introducing a constant imposed velocity on the free end.

The following system is used: mm, ms, g, N, MPa

Several kinds of element formulation are used.

The material used follows a linear elastic law (/MAT/LAW1) with the following characteristics:
Material Properties
Value
Initial density
0.01 $\left[\frac{g}{m{m}^{3}}\right]$
Reference density
.01 $\left[\frac{g}{m{m}^{3}}\right]$
Young's modulus
1000 $\left[\mathrm{MPa}\right]$
Poisson ratio
0

Model Method

The four models are integrated into one input file. The shell element formulations are:
• Q4 mesh with the Belytshcko & Tsay formulation (Ishell =1, hourglass control TYPE1, TYPE2, and TYPE3)
• Q4 mesh with the QEPH formulation (Ishell =24)
• Q4 mesh with the QBAT formulation (Ishell =12)
• T3 mesh with the DKT18 formulation (Ishell =12)

Results

Numerical Results and Analytical Solution Comparison

As shown in Figure 3, rotation around X and displacement with regard to Y of the free end are studied.

The analytical solution of the Timoshenko beam subjected to a tip moment reads:

$\theta \left(z\right)=\frac{Mz}{EI}$

Which yields the end moment for a complete loop rotation $2\pi$ :

$M=\frac{2\pi EI}{L}=\frac{2\pi \left(1000\right)\left(\frac{48*{20}^{3}}{12}\right)}{500}=4.021×{10}^{5}\text{​}\text{\hspace{0.17em}}\text{KN-mm}$

The following tables summarize the results obtained for the different formulations. From an analytical point of view, the beam deformed under pure bending must satisfy the conditions of the constant curvature which implies that for $\theta$ = $2\pi$ , the beam should form a closed ring. However, depending on the finite element used, a small error can be observed, as shown in the following tables. This is mainly due to beam vibration during deformation as it is highly flexible. Good results are obtained by the QBAT, QEPH and DKT18 elements, respectively. This is mainly due to the good estimation of the curvature in the formulation of these elements. The BT family of under-integrated shell elements is less accurate. With the TYE3 hourglass formulation, the model remains stable until $\theta$ = 6rad. However, the moment-rotation curves do not correspond to the expected response.

To reduce the overall computation error, smaller explicit time steps are used by reducing the scale factor in /DT. The results reported in the end table show that a reduction in the time step enables to reduce the error accumulation, even though the divergence problems for BT elements cannot be avoided.

The following parameters are chosen for drawing curves and displaying animations:
BATOZ QEPH BT DKT
Scale factor 0.6 0.9 0.9 0.2
The following curves show the evolution previously shown (rotation and nodal displacement by moment):
For $\theta =2\pi \ge {M}_{Analytical}=-4.021×{10}^{5}$
BATOZ QEPH BT DKT
Sf=0.9 Sf =0.8 Sf =0.6 Sf =0.9 Sf =0.8 TYPE1 TYPE3 TYPE4 Sf =0.3 Sf =0.2 Sf =0.1
Sf =0.9 Sf =0.1 Sf =0.9 Sf =0.1 Sf =0.9 Sf =0.1
CPU

(normalized)

# cycles

2.18

97600

2.43

109800

3.14

146400

1.23

95800

1.34

107800

42.64

59100

7.07

552600

2.62

182300

108.60

--

1.03

59100

7.17

552600

5.44

364100

8.21

621600

16.21

1243200

Error

$\theta$ = 2 $\pi$

(%)

0% 0% 0% 0% 0% 55.3% 99% 0% 0% 55.9% 99.9% 3.4% 28.88% 3.7%
$\theta$ err =20%

degree

6.91

396°

6.89

395°

-- -- -- 4.36

250°

4.53

260°

6.06

347°

5.98

343°

4.38

251°

4.51

258°

6.37

365°

-- --
Dz $\theta$ = 2 $\pi$

(mm)

-500.5 -500.5 -500.5 -500.5 -500.5 -491.2 -525.8 -518.333 -506.0 -529.8 -433.8 -476.5 -496.5 -499.4
Mx $\theta$ = 2 $\pi$

(x10+5kN-mm)

-4.04 -4.05 -4.06 -4.01 -4.01 -0.21 -0.11 -3.13 -2.38 -0.07 -0.02 -3.09 -3.02

-3.08