TMM4175 Polymer Composites

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CASE STUDY: Compression testing of UD composites¶

Properties and dimensions for loading along the 0-direction:¶

Material: E-glass/Epoxy
Width $b$ = 10 mm
Thickness $t$ = 2 mm
Tab length = 50 mm
Gauge length $l_g$: to be decided

Specimen

According to Euler buckling of a column:

\begin{equation} F_{crit} = - \frac{\pi^2 E I}{(KL)^2} \end{equation}

Searching for a gauge length that enables compressive stress well above 1000 MPa (or actually below since it is negative...):

from math import pi
E=40000
b=10
h=2
K=0.5
I = (b*h**3)/12
 
for L in (25,24,23,22,21,20):
    Fc = -((pi**2)*E*I)/((K*L)**2)
    print('L=',L,'Fc=',Fc, 's2=',Fc/(2*10))

L= 25 Fc= -16844.12484452584 s2= -842.2062422262919
L= 24 Fc= -18277.045187202515 s2= -913.8522593601258
L= 23 Fc= -19900.903644288563 s2= -995.0451822144281
L= 22 Fc= -21751.194272373243 s2= -1087.559713618662
L= 21 Fc= -23872.059020019613 s2= -1193.6029510009807
L= 20 Fc= -26318.945069571622 s2= -1315.9472534785812

The results suggest that a gauge length of 20 mm is sufficiently small to avoid premature buckling before compressive failure.

Assumptions for a simplified approach¶

We can reasonably assume that the first buckling mode is symmetric about both the y-z plane and the x-z plane (x-axis is along the length and y-axis is along the width). Therefore, only a 1/4 of the model is needed.
The tabs transfer a uniform shear force to the specimen, and the grips imposes a uniform displacement constraint on the specimen. Therefore, we do not need to include the tabs in the model.

Specimen

Material, orientation and mesh¶

Material properties from E-glass/Epoxy:

import matlib
m=matlib.get('E-glass/Epoxy')
print(m)

{'name': 'E-glass/Epoxy', 'units': 'MPa-mm-Mg', 'type': 'UD', 'fiber': 'E-glass', 'Vf': 0.55, 'rho': 2e-09, 'description': 'Typical UD E-glass/Epoxy from TMM4175', 'E1': 40000, 'E2': 10000, 'E3': 10000, 'v12': 0.3, 'v13': 0.3, 'v23': 0.4, 'G12': 3800, 'G13': 3800, 'G23': 3400, 'a1': 7e-06, 'a2': 2.2e-05, 'a3': 2.2e-05, 'XT': 1000, 'YT': 40, 'ZT': 40, 'XC': 700, 'YC': 120, 'ZC': 120, 'S12': 70, 'S13': 70, 'S23': 40, 'f12': -0.5, 'f13': -0.5, 'f23': -0.5}

Element size: 0.5 mm
No. of elements through the thickness: 6
Element type: C3D8R

Specimen

Static step and boundary conditions¶

According to the assumptions:

Symmetrical BC about the y-z plane
Symmetrical BC about the x-z plane
Fixed uz on surfaces connecting to the tabs

Specimen

Loading¶

Compressive normal stress: -1000 MPa
Need to compute the corresponding shear stress (traction) on the tab's area:

s2 = -1000                 # compressive stress (MPa)
force = s2*5*2             # stress multiplied with cross section area
shear = force/(5*50*2)    # shear traction (MPa) on tab's surface (both sides)
print('shear traction to be imposed:',shear)

shear traction to be imposed: -20.0

Specimen

Solution and results¶

Specimen

Interpretation of the results is left as exercise

Specimen

Buckling simulation and results¶

Specimen

The eigenvalue of 0.73287 implies that the nominal longitudinal stress at the point of buckling is

eigenvalue = 0.73287
s1_applied = -1000
s1_atBuckling = eigenvalue*s1_applied

print(s1_atBuckling)

-732.87

Apparantly, the Euler buckling theory is not applicable for this case when the FEA result is jugded more accurate. Hence, the thickness should be increased or the gauge length should be reduced. See the recomendation given in Compression testing based on ASTM 3410.

Further interpretation of results is left as exercise.

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