TMM4175 Polymer Composites
CASE STUDY: Thin-walled pipes¶
Cylindrical pressure vessels and pressurized tubes, or pipes, are important industrial components. Composite vessels and pipes manufactured by filament winding are found in a wide range of applications, from low-pressure water pipeline system to high-pressure storage of natural gas and hydrogen. Laminated tubes with closed ends subjected to partial pressure can be analyzed using laminate theory when a few idealized assumptions are fulfilled:
- Thin-walled refers in general to a vessel having a relatively large radius to thickness ratio, typically 10 and above. However, what we may consider as thin-wall is rather a matter of accuracy, where a greater ratio implies greater accuracy.
- Edge effects are neglected, which is generally a reasonable assumption far from the ends. This assumption implies that the tube wall is uniformly loaded.
Consider a tube with closed ends having a radius R and subjected to a partial pressure P (the difference between inner and outer pressure). A coordinate system x,y,z defines the longitudinal direction (x), the hoop direction (y) and the normal to the cylindrical shell (z).
The section forces $N_x$ and $N_y$ are obtained by considering equilibrium in the two directions:
\begin{equation} \sum F_x = 0: \quad (N_x)(2\pi R) = (P)(\pi R^2) \Rightarrow N_x = \frac{PR}{2} \tag{1} \end{equation}\begin{equation} \sum F_y = 0: \quad (N_y)(2L) = (P)(2RL) \Rightarrow N_y = PR \tag{2} \end{equation}while the shear load $N_{xy}$ is zero in the absence of any torque loading on the tube. Note however that the shear strain can be different from zero according to relation (4) for a unbalanced laminate.
The curvatures are all zero due to the particular symmetry, which also implies that there will be internal moments for non-symmetrical layups. Hence the complete set of equations in the laminate theory framework is:
\begin{equation} \begin{bmatrix} N_x \\ N_y \\ 0 \\ M_x \\ M_y \\ M_{xy} \end{bmatrix} = \begin{bmatrix} A_{xx} & A_{xy} & A_{xs} & B_{xx} & B_{xy} & B_{xs} \\ A_{xy} & A_{yy} & A_{ys} & B_{xy} & B_{yy} & B_{ys} \\ A_{xs} & A_{ys} & A_{ss} & B_{xs} & B_{ys} & B_{ss} \\ B_{xx} & B_{xy} & B_{xs} & D_{xx} & D_{xy} & D_{xs} \\ B_{xy} & B_{yy} & B_{ys} & D_{xy} & D_{yy} & D_{ys} \\ B_{xs} & B_{ys} & B_{ss} & D_{xs} & D_{ys} & D_{ss} \end{bmatrix} \begin{bmatrix} \varepsilon_x^0 \\ \varepsilon_y^0 \\ \varepsilon_{xy}^0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \tag{3} \end{equation}We can now split the set of equations into
\begin{equation} \begin{bmatrix} N_x \\ N_y \\ 0 \end{bmatrix} = \begin{bmatrix} A_{xx} & A_{xy} & A_{xs} \\ A_{xy} & A_{yy} & A_{ys} \\ A_{xs} & A_{ys} & A_{ss} \end{bmatrix} \begin{bmatrix} \varepsilon_x^0 \\ \varepsilon_y^0 \\ \gamma_{xy}^0 \end{bmatrix} \tag{4} \end{equation}and
\begin{equation} \begin{bmatrix} M_x \\ M_y \\ M_{xy} \end{bmatrix} = \begin{bmatrix} B_{xx} & B_{xy} & B_{xs} \\ B_{xy} & B_{yy} & B_{ys} \\ B_{xs} & B_{ys} & B_{ss} \end{bmatrix} \begin{bmatrix} \varepsilon_x^0 \\ \varepsilon_y^0 \\ \gamma_{xy}^0 \end{bmatrix} \tag{5} \end{equation}The latter equation is of no particular interest since the solution of (4) provides all we need for computing the layer results.
Computational example¶
A material:
import laminatelib
import matlib
m1=matlib.get('E-glass/Epoxy')
Layup:
layup1=[{'mat':m1, 'ori':-45, 'thi':1.0},
{'mat':m1, 'ori':+45, 'thi':1.0}]
The submatrix A from the laminate stiffness matrix:
A=laminatelib.laminateStiffnessMatrix(layup1)[0:3,0:3]
print(A)
Radius, pressure and section forces:
R, P = 100, 2
Nx=P*R/2
Ny=P*R
Nxy=0
The laminate mid-plane strains:
import numpy as np
loads=[Nx,Ny,Nxy]
strains=np.linalg.solve(A,loads)
print(strains)
In order to use the function laminatelib.layerResults() we need the deformation as a 6x1 array:
deformations=np.concatenate((strains, [0,0,0]))
print(deformations)
res = laminatelib.layerResults(layup1,deformations)
For example, the stresses in the material coordinate system for the two layers:
res[0]['stress']['123']
res[1]['stress']['123']
Exposure factors for the layers:
res[0]['fail']
res[1]['fail']
Observe that the exposure factors are identical for both layers and position (top/bottom).
What is the optimum angle for a [-$\theta$/+$\theta$] laminate when the objective is to minimize the exposure factor given by the maximum stress criterion?
Solution:
def tubeResults(angle,R,P):
layup =[{'mat':m1, 'ori':-angle, 'thi':1.0},
{'mat':m1, 'ori':+angle, 'thi':1.0}]
A=laminatelib.laminateStiffnessMatrix(layup)[0:3,0:3]
loads=[P*R/2, P*R, 0]
strains=np.linalg.solve(A,loads)
deformations=np.concatenate((strains, [0,0,0]))
res = laminatelib.layerResults(layup,deformations)
return res
import matplotlib.pyplot as plt
%matplotlib inline
angles=np.linspace(0,90)
fE_MS = [tubeResults(angle,R=100,P=2)[0]['fail']['MS']['bot'] for angle in angles]
plt.plot(angles,fE_MS)
plt.grid(True)
plt.show()
Somewhere between 50 and 60. Limit the range:
angles=np.linspace(50,60)
fE_MS = [tubeResults(angle,R=100,P=2)[0]['fail']['MS']['bot'] for angle in angles]
plt.plot(angles,fE_MS)
plt.grid(True)
plt.show()
The failure pressure at the optimum angle is the applied pressure divided by the exposure factor:
fE_MS = tubeResults(53.5,R=100,P=2)[0]['fail']['MS']['bot']
print('Failure pressure for a 53.5 winding angle is',2/fE_MS, 'MPa')
FEA model benchmarking¶
Briefly, a shell mode of a 1000 mm long tube with two layers oriented -53.5 and 53.5 relatively to the longitudinal axis. The ends are considered as rigid and the radius R=100 mm is to the mid-plane of the shell.
tubeResults(53.5,R=100,P=2)[0]['stress']['123']
The results do not compare well at regions close to the ends (top left figure), while the results far from the ends are, for all practial purposes, identical.
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