TMM4175 Polymer Composites

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Plane stress failure theories¶

The Maximum stress failure criterion for plane stress case is derived by simply eliminating the stress-terms not being relevant in the 3D expression, see Maximum stress criterion: \begin{equation} f_E = max\Big[ -\frac{\sigma_1}{X_C}, \frac{\sigma_1}{X_T}, -\frac{\sigma_2}{Y_C}, \frac{\sigma_2}{Y_T}, -\frac{\sigma_3}{Z_C}, \frac{\sigma_3}{Z_T}, \frac{|\tau_{12}|}{S_{12}}, \frac{|\tau_{13}|}{S_{13}}, \frac{|\tau_{23}|}{S_{23}}\Big] \tag{1} \end{equation}

The 2D criterion is now simplified to

\begin{equation} f_E = max\Big[ -\frac{\sigma_1}{X_C}, \frac{\sigma_1}{X_T}, -\frac{\sigma_2}{Y_C}, \frac{\sigma_2}{Y_T}, \frac{|\tau_{12}|}{S_{12}}\Big] \tag{2} \end{equation}

Implementing a function for the stress exposure factor:

def fE2DMS(s,m):
    s1,s2,s12=s[0],s[1],s[2]
    XT,YT,XC,YC,S12 = m['XT'], m['YT'], m['XC'], m['YC'], m['S12']
    f=max(s1/XT,-s1/XC,s2/YT,-s2/YC,abs(s12/S12))
    return f

The Maximum strain criterion reduced to the 2D case neglects the out-of plane components of strains:

\begin{equation} f_E = max\Big[ -\frac{\epsilon_1}{X_C/E_1}, \frac{\epsilon_1}{X_T/E_1}, -\frac{\epsilon_2}{Y_C/E_2}, \frac{\epsilon_2}{Y_T/E_2}, \frac{|\gamma_{12}|}{S_{12}/G_{12}}\Big] \tag{3} \end{equation}

A function for the stress exposure factor:

def fE2DME(s,m):
    s1,s2,s12=s[0],s[1],s[2]
    XT,YT,XC,YC,S12 = m['XT'],m['YT'],m['XC'],m['YC'],m['S12']
    E1,E2,v12,G12=m['E1'],m['E2'],m['v12'],m['G12']
    e1=   (1/E1)*s1 + (-v12/E1)*s2
    e2=(-v12/E1)*s1 +    (1/E2)*s2
    e12 = s12/G12
    
    f=max( e1/(XT/E1),-e1/(XC/E1),e2/(YT/E2),-e2/(YC/E2),abs(e12/(S12/G12)) )
    return f

Tsai-Wu for plane stress was implemented for the 3D case in Tsai-Wu criterion.

For the plane stress case,

\begin{equation} f = F_1\sigma_1+F_2\sigma_2+F_{11}\sigma_1^2+F_{22}\sigma_2^2+F_{66}\tau_{12}^2+2F_{12}\sigma_1\sigma_2 \tag{4} \end{equation}

where failure is predicted when $f\geq1$

The coefficients are expressed by the basic strength parameters: \begin{equation} \begin{aligned} &F_1=\frac{1}{X_T}-\frac{1}{X_C} && F_2=\frac{1}{Y_T}-\frac{1}{Y_C} \\ &F_{11}=\frac{1}{X_TX_C} && F_{22}=\frac{1}{Y_TY_C} \\ &F_{66}=\frac{1}{S_{12}^2} && F_{12}=f_{12}\sqrt{F_{11}F_{22}} \end{aligned} \tag{5} \end{equation}

Solving the quadratic equation:

\begin{equation} (F_1\sigma_1+F_2\sigma_2+F_3\sigma_3)R+(F_{11}\sigma_1^2+F_{22}\sigma_2^2+F_{66}\tau_{12}^2+2 F_{12}\sigma_1\sigma_2)R^2=1 \tag{6} \end{equation}

Such that

\begin{equation} \begin{aligned} &a=F_{11}\sigma_1^2+F_{22}\sigma_2^2+F_{66}\tau_{12}^2+2 F_{12}\sigma_1\sigma_2 \\ &b= F_1\sigma_1+F_2\sigma_2 \\ &c=-1 \end{aligned} \tag{7} \end{equation}

and

\begin{equation} R=\frac{-b+\sqrt{b^2 -4ac}}{2a} \tag{8} \end{equation}

Observe that the paramtere $a$ can be zero. In such a case the $R$ will be undefined. In those cases we just return zero as written in the code:

def fE2DTW(s,m):
    s1,s2,s12=s[0],s[1],s[2]
    XT,YT,XC,YC,S12,f12 = m['XT'], m['YT'], m['XC'], m['YC'], m['S12'], m['f12']
    F1,  F2  = (1/XT)-(1/XC) , (1/YT)-(1/YC)
    F11, F22 =  1/(XT*XC) , 1/(YT*YC)
    F66 = 1/(S12**2)
    F12 = f12*(F11*F22)**0.5
    a=F11*s1**2 + F22*s2**2 + 2*F12*s1*s2 + F66*s12**2
    b=F1*s1 + F2*s2
    c=-1
    if a==0:
        return 0.0
    R=(-b+(b**2-4*a*c)**0.5)/(2*a)
    fE=1/R
    return fE

Example:

import matlib
m1=matlib.get('Carbon/Epoxy(a)')

stress=(440,36,10)

print(fE2DMS(stress,m1))
print(fE2DME(stress,m1))
print(fE2DTW(stress,m1))

0.9
0.6630769230769231
0.7853539166568826

Hashin:

Left as an exercise.

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