Matrix mode A.

In the uniaxial transverse tension test and the in-plane shear test, the plane of fracture is normal to the $x_2$-direction so $\alpha= 0$. From eq. ([eq9.3]) the stresses on the fracture plane are $\sigma_\textit{nn} = \sigma_\textit{22}$, $\sigma_\textit{nt} = \sigma_{23}$, and $\sigma_{n1} = \sigma_\textit{21}$.

L113.25pt

In the transverse tension test $\sigma_\textit{22} = Y_T$ at failure, and all other stresses in the $x_i$-system vanish. For the in-plane shear test all stresses in the $x_i$-system vanish except that $\sigma_\textit{21} = S_L$ at failure. The proposed criterion including these test results is quadratic and of the form $$\begin{aligned} \label{eq9.4} \left(1-c_{1}\right)\left(\frac{\sigma_{n n}}{Y_{T}}\right)^{2}+c_{1}\left(\frac{\sigma_{n n}}{Y_{T}}\right)+\left(\frac{\sigma_{n t}}{S_{T}}\right)^{2}+\left(\frac{\sigma_{n 1}}{S_{L}}\right)^{2}=1 \quad \sigma_{n n} \geq 0.\end{aligned}$$

The shear strength transverse to the fibers in the fracture plane is denoted by $S_T$ in eq. ([eq9.4])¹. In the Cartesian coordinates with axes $\sigma_\textit{nn}$, $\sigma_\textit{nt}$, and $\sigma_{n1}$ the surface given by eq. ([eq9.4]) is an ellipsoid if the constant $c_1 < 1$. In the shear stress plane $\sigma_\textit{nt}$-$\sigma_{n1}$ where $\sigma_\textit{nn}$ is equal to zero, the cross section of the ellipsoid is an ellipse shown in figure [fig9.2]. The equation for the ellipse in the plane $\sigma_\textit{nn}$ equal to zero is $$\begin{aligned} \label{eq9.5} \left(\frac{\sigma_{n t}}{S_{T}}\right)^{2}+\left(\frac{\sigma_{n 1}}{S_{L}}\right)^{2}=1 \quad \sigma_{n n}=0.\end{aligned}$$ The resultant of the shear stress components is denoted by $\sigma_{n\psi}$, and the angle between the line of action of the resultant and $\sigma_\textit{nt}$-axis is denoted by $\psi$. The stress components are related to the resultant by $$\begin{aligned} \label{eq9.6} \sigma_{n t}=\sigma_{n\psi} \cos \psi \quad \sigma_{n 1}=\sigma_{n \psi} \sin \psi.\end{aligned}$$ Substitute eq. ([eq9.6]) for the stress components in eq. ([eq9.5]) to get $$\begin{aligned} \label{eq9.7} \sigma_{n \psi}^{2}\left(\frac{\cos ^{2}{\kern-1pt\psi}}{S_{T}^{2}}+\frac{\sin ^{2}{\kern-1pt\psi}}{S_{L}^{2}}\right)=1 \quad \sigma_{n n}=0.\end{aligned}$$ On the failure ellipse $\sigma_{n \psi}=S_{\psi}$ in eq. ([eq9.7]). Hence, strength $S_{\psi}$ is related to strengths $S_T$ and $S_L$ by $$\begin{aligned} \label{eq9.8} S_{\psi}^{2}\left(\frac{\cos ^{2}{\kern-1pt\psi}}{S_{T}^{2}}+\frac{\sin ^{2}{\kern-1pt\psi}}{S_{L}^{2}}\right)=1 \quad \sigma_{n n}=0.\end{aligned}$$

To interpret constant $c_1$ we take the differential of ([eq9.4]) with respect to $\sigma_\textit{nn}$ followed by setting $\sigma_\textit{nn} = 0$ to get $$\begin{aligned} \label{eq9.9} \frac{c_{1}}{Y_{T}}+\frac{2 \sigma_{n t}}{S_{T}^{2}} \frac{d \dot{\sigma}_{n t}}{d \sigma_{n n}}+\frac{2 \sigma_{n 1}}{S_{L}^{2}} \frac{d \sigma_{n 1}}{d \sigma_{n n}}=0 \quad \sigma_{n n}=0.\end{aligned}$$ Along the curve on the ellipsoid defined by angle $\psi$ equal to a constant, substitute the relations ([eq9.6]) with $\sigma_{n \psi}=S_{\psi}$ into eq. ([eq9.9]) to get $$\begin{aligned} \label{eq9.10} \frac{c_{1}}{2 Y_{T}}+S_{\psi}\left(\frac{\cos ^{2}{\kern-1pt\psi}}{S_{T}^{2}}+\frac{\sin ^{2}{\kern-1pt\psi}}{S_{L}^{2}}\right) \frac{d \sigma_{n \psi}}{d \sigma_{n n}}=0 \quad \sigma_{n n}=0.\end{aligned}$$ Use the result in eq. ([eq9.8]) to write eq. ([eq9.10]) as $$\begin{aligned} \label{eq9.11} \frac{c_{1}}{2 Y_{T}}+\frac{1}{S_{\psi}} \frac{d \sigma_{n \psi}}{d \sigma_{n n}}=0 \quad \sigma_{n n}=0.\end{aligned}$$ Along the curve on the ellipsoid defined by angle $\psi$ equal to a constant, let the negative of the slope of the $\sigma_{n \psi}$ with respect to $\sigma_\textit{nn}$ at $\sigma_\textit{nn} = 0$ be denoted by $p_{n\psi}^{(+)}$. That is, $$\begin{aligned} \label{eq9.12} p_{n \psi}^{(+)}=-\left.\frac{d \sigma_{n \psi}}{d \sigma_{n n}}\right|_{\sigma_{n n}=0}.\end{aligned}$$ Puck defines $p_{n \psi}^{(+)}$ as an inclination parameter. Therefore, the constant $c_1$ is determined from $$\begin{aligned} \label{eq9.13} \frac{c_{1}}{2 Y_{T}}=\frac{p_{n \psi}^{(+)}}{S_{\psi}}.\end{aligned}$$ Substitute the constant $c_1$ determined from eq. ([eq9.13]) into eq. ([eq9.4]) to get the failure criterion for mode A as $$\begin{aligned} \label{eq9.14} \left(1-\frac{2 Y_{T}}{S_{\psi}} p_{n \psi}^{(+)}\right)\left(\frac{\sigma_{n n}}{Y_{T}}\right)^{2}+\frac{2 Y_{T}}{S_{\psi}} p_{n \psi}^{(+)}\left(\frac{\sigma_{n n}}{Y_{T}}\right)+\left(\frac{\sigma_{n t}}{S_{T}}\right)^{2}+\left(\frac{\sigma_{n 1}}{S_{L}}\right)^{2}=1 \quad \sigma_{n n} \geq 0.\end{aligned}$$ The following failure index for mode A is given by Puck: $$\begin{aligned} \label{eq9.15} F I_{M}=\sqrt{\left[1-\frac{Y_{T}}{S_{\psi}} p_{n \psi}^{(+)}\right]^{2}\left(\frac{\sigma_{n n}}{Y_{T}}\right)^{2}+\left(\frac{\sigma_{n t}}{S_{T}}\right)^{2}+\left(\frac{\sigma_{n 1}}{S_{L}}\right)^{2}}+p_{n \psi}^{(+)}\left(\frac{\sigma_{n n}}{S_{\psi}}\right) \quad \sigma_{n n} \geq 0.\end{aligned}$$ To show eqs. ([eq9.14]) and ([eq9.15]) are equivalent: Set $F I_{M}=1$ in ([eq9.15]) and subtract $\frac{Y_{T}}{S_{\psi}} p_{n \psi}^{(+)}\left(\frac{\sigma_{n n}}{Y_{T}}\right)$ from each side. Then square the result to arrive at eq. ([eq9.14]) after some algebraic manipulations.

The inclination parameter $p_{n \psi}^{(+)}$ is related to the inclination parameters defined for the $\psi= 0$ and $\psi=\pi/2$ failure loci on the ellipsoid. The locus of failure initiation for $\psi=0$ is a curve in the $\sigma_\textit{nn}$-$\sigma_\textit{nt}$ plane. At the point on this curve where $\left(\sigma_{n n}, \sigma_{n 1}\right)=(0,0)$ failure initiates when $\sigma_{n t}=S_{T}=S_{\psi}$. The gradient condition at this point from eq. ([eq9.9]) is $$\begin{aligned} \label{eq9.16} \frac{c_{1}}{2 Y_{T}}+\frac{1}{S_{T}} \frac{d \dot{\sigma}_{n t}}{d \sigma_{n n}}=0.\end{aligned}$$ The locus of failure initiation for $\psi=\pi/2$ is a curve in the $\sigma_\textit{nn}$-$\sigma_{n1}$ plane. At the point on this curve where $\left(\sigma_{n n}, \sigma_{n t}\right)=(0,0)$ failure initiates when $\sigma_{n 1}=S_{L}=S_{\psi}$. The gradient condition at this point from eq. ([eq9.9]) is $$\begin{aligned} \label{eq9.17} \frac{c_{1}}{2 Y_{T}}+\frac{1}{S_{L}} \frac{d \sigma_{n 1}}{d \sigma_{n n}}=0.\end{aligned}$$ Define the inclination parameter on the $\psi=0$ curve as $p_{n t}^{(+)}=-\frac{d \dot{\sigma}_{n t}}{d \sigma_{n n}}$, and on the $\psi=\pi/2$ curve as $p_{n 1}^{(+)}=-\frac{d \sigma_{n 1}}{d \sigma_{n n}}$. Combine eqs. ([eq9.13]), ([eq9.16]), and ([eq9.17]) to find $$\begin{aligned} \label{eq9.18} \frac{p_{n \psi}^{(+)}}{S_{\psi}}=\frac{p_{n t}^{(+)}}{S_{T}}, \textrm{ and } \frac{p_{n \psi}^{(+)}}{S_{\psi}}=\frac{p_{n 1}^{(+)}}{S_{L}}.\end{aligned}$$ Multiply the first expression in eq. ([eq9.18]) by $\cos ^{2}{\kern-1pt\psi}$, and add it to the second expression in eq. ([eq9.18]) multiplied by $\sin ^{2}{\kern-1pt\psi}$, to get relationship between the inclination parameters on the tension side of the ellipse in figure [fig9.2] as $$\begin{aligned} \label{eq9.19} \frac{p_{n \psi}^{(+)}}{S_{\psi}}=\frac{p_{n t}^{(+)}}{S_{T}} \cos ^{2}{\kern-1pt\psi}+\frac{p_{n 1}^{(+)}}{S_{L}} \sin ^{2}{\kern-1pt\psi}.\end{aligned}$$

Matrix modes B and C.

These modes are defined for a compressive normal stress, $\sigma_\textit{nn} < 0$, acting on the fracture plane. The motivation of Puck’s criterion for modes B and C is the Coulomb-Mohr (C-M) criterion (Dowling, 1993, pp. 255–261) for failure of brittle materials. In the C-M criterion a compressive normal stress resists fracture caused by the shear stresses $\sigma_\textit{nt}$ and $\sigma_{n1}$. The C-M criterion can be considered to be a shear stress criterion in which the limiting shear stress increases for larger amounts of compression. Consider the case where $\sigma_{\textrm{n1}} = 0$, so on the fracture plane $\sigma_{\textrm{nn}} < 0$ and $\sigma_{n t}\neq 0$. Then the C-M criterion can be written in the form $\left|\sigma_{n t}\right|+\mu \sigma_{n n}=S_{T}$, where $\mu$ is a friction coefficient and $S_T$ is the shear strength transverse to the fibers in the fracture plane. The friction effect, $\mu \sigma_{n n}$, can be used to increase the strength or to decrease the applied shear stress in a C-M criterion. Puck and Schümann (1998) proposed the following criterion $$\begin{aligned} \label{eq9.20} \left(\frac{\sigma_{n t}}{S_{T}-p_{n t}^{(-)} \sigma_{n n}}\right)^{2}+\left(\frac{\sigma_{n 1}}{S_{L}-p_{n 1}^{(-)} \sigma_{n n}}\right)^{2}=1 \quad \sigma_{n n} \leq 0,\end{aligned}$$ in which the strengths in the denominators are increased by the compressive normal stress, and $\left(p_{n t}^{(-)}, p_{n 1}^{(-)}\right)$ are the inclination parameters in compression. Set $\sigma_{n1} = 0$ in eq. ([eq9.20]) to get $\sigma_{n t}=S_{T}-p_{n t}^{(-)} \sigma_{nn}$, and from this expression the inclination parameter is interpreted as the negative slope of $\sigma_\textit{nt}$ with respect to $\sigma_\textit{nn}$, or $$\begin{aligned} \label{eq9.21} p_{n t}^{(-)}=-\left.\left(\frac{d \sigma_{n t}}{d \sigma_{n n}}\right)\right|_{\sigma_{n 1}=0}.\end{aligned}$$ Set $\sigma_\textit{nt} = 0$ in eq. ([eq9.20]) to get $\sigma_{n 1}=S_{L}-p_{n}^{(-)} \sigma_{n n}$, and from this expression the inclination parameter is interpreted as the negative slope of $\sigma_{n1}$ with respect to $\sigma_\textit{nn}$, or $$\begin{aligned} \label{eq9.22} p_{n 1}^{(-)}=-\left.\left(\frac{d \sigma_{n 1}}{d \sigma_{n n}}\right)\right|_{\sigma_{n t}=0}.\end{aligned}$$

Citing better agreement with experimental results, the denominators of the shear stresses in eq. ([eq9.20]) are expanded and the quadratic terms in the normal stress $\sigma_\textit{nn}$ are neglected with respect to the linear terms in $\sigma_\textit{nn}$, so the criterion reduces to $$\begin{aligned} \label{eq9.23} \frac{\sigma_{n t}^{2}}{S_{T}^{2}-2 p_{n t}^{(-)} S_{T} \sigma_{n n}}+\frac{\sigma_{n 1}^{2}}{S_{L}^{2}-2 p_{n 1}^{(-)} S_{L} \sigma_{n n}}=1 \quad \sigma_{n n} \leq 0.\end{aligned}$$ For mathematical simplification Puck and Shümann assume that the inclination parameters are related in a similar way to eq. ([eq9.18]) by $$\begin{aligned} \label{eq9.24} \frac{p_{n t}^{(-)}}{S_{T}}=\frac{p_{n 1}^{(-)}}{S_{L}}=\frac{p}{R}.\end{aligned}$$ With this assumption eq. ([eq9.23]) reduces to the simpler form $$\begin{aligned} \label{eq9.25} \left(\frac{\sigma_{n t}}{S_{T}}\right)^{2}+\left(\frac{\sigma_{n 1}}{S_{L}}\right)^{2}+2\left(\frac{p}{R}\right) \sigma_{n n}=1 \quad \sigma_{n n} \leq 0.\end{aligned}$$ In the Cartesian coordinates with axes $\sigma_\textit{nn}$, $\sigma_\textit{nt}$, and $\sigma_{n1}$ the surface given by eq. ([eq9.25]) is an elliptic paraboloid. Note that the failure surface does not intersect the negative $\sigma_\textit{nn}$-axis according to the hypothesis that a compressive normal stress impedes a shear fracture (i.e., the shear resistance to fracture means the contour lines in the failure surface increase with increasing compression). In the shear stress plane $\sigma_\textit{nt}$-$\sigma_{n1}$ where $\sigma_\textit{nn}$ is equal to zero, the cross section of the ellipsoid is an ellipse shown in figure [fig9.2]. Substitute the relations given by eq. ([eq9.6]) into eq. ([eq9.25]) to get $$\begin{aligned} \label{eq9.26} \sigma_{n \psi}^{2}\left(\frac{\cos ^{2}{\kern-1pt\psi}}{S_{T}^{2}}+\frac{\sin ^{2}{\kern-1pt\psi}}{S_{L}^{2}}\right)+2\left(\frac{p}{R}\right) \sigma_{n n}=1 \quad \sigma_{n n} \leq 0.\end{aligned}$$ Differentiate eq. ([eq9.26]) with respect to $\sigma_\textit{nn}$ to get $$\begin{aligned} \label{eq9.27} \sigma_{n \psi}\left(\frac{\cos ^{2}{\kern-1pt\psi}}{S_{T}^{2}}+\frac{\sin ^{2}{\kern-1pt\psi}}{S_{L}^{2}}\right) \frac{d \sigma_{n \psi}}{d \sigma_{n n}}+\frac{p}{R}=0.\end{aligned}$$ Consider the $\sigma_\textit{nt}$-$\sigma_{n1}$ plane at $\sigma_\textit{nn}=0$. On the failure ellipse $\sigma_{n\psi}=S_{\psi}$ and ([eq9.26]) is $$\begin{aligned} \label{eq9.28} S_{\psi}^{2}\left(\frac{\cos ^{2}{\kern-1pt\psi}}{S_{T}^{2}}+\frac{\sin ^{2}{\kern-1pt\psi}}{S_{L}^{2}}\right)=1.\end{aligned}$$ Evaluate eq. ([eq9.27]) at $\sigma_{n \psi}=S_{\psi}$, followed by the substitution of eq. ([eq9.28]). The result is $$\begin{aligned} \label{eq9.29} \frac{1}{S_{\psi}} \frac{d \sigma_{n \psi}}{d \sigma_{n n}}+\frac{p}{R}=0 \quad \sigma_{n n}=0.\end{aligned}$$ Define the inclination parameter for the curve $\psi$ equal to a constant by $p_{n\psi}^{(-)}=-\frac{d \sigma_{n \psi}}{d \sigma_{n n}}$. Hence, $$\begin{aligned} \label{eq9.30} \frac{p}{R}=\frac{p_{n \psi}^{(-)}}{S_{\psi}}.\end{aligned}$$ Substitute the result ([eq9.30]) into the condition of failure initiation ([eq9.25]) to find $$\begin{aligned} \label{eq9.31} \left(\frac{\sigma_{n t}}{S_{T}}\right)^{2}+\left(\frac{\sigma_{n 1}}{S_{L}}\right)^{2}+2\left(\frac{p_{n \psi}^{(-)}}{S_{\psi}}\right) \sigma_{n n}=1 \quad \sigma_{n n} \leq 0.\end{aligned}$$ The following failure index for $\sigma_{n n} \leq 0$ is given by Puck. $$\begin{aligned} \label{eq9.32} F I_{M}=\sqrt{\left(\frac{\sigma_{n t}}{S_{T}}\right)^{2}+\left(\frac{\sigma_{n 1}}{S_{L}}\right)^{2}+\left(\frac{p_{n \psi}^{-}}{S_{\psi}} \sigma_{n n}\right)^{2}}+\left(\frac{p_{n \psi}^{(-)}}{S_{\psi}}\right) \sigma_{n n} \quad \sigma_{n n} \leq 0.\end{aligned}$$ One can show eq. ([eq9.32]) is equivalent to eq. ([eq9.31]) if we set $F I_{M}=1$ in ([eq9.32]).

Combining eqs. ([eq9.24]) and ([eq9.30]) we get $$\begin{aligned} \label{eq9.33} \frac{p_{n \psi}^{(-)}}{S_{\psi}}=\frac{p_{n t}^{(-)}}{S_{T}}, \textrm{ and } \frac{p_{n \psi}^{(-)}}{S_{\psi}}=\frac{p_{n}^{(-)}}{S_{L}}\end{aligned}$$ Similar to the manipulations to get eq. ([eq9.19]), the expressions in eq. ([eq9.33]) lead to the relationship between the inclination parameters on the compression side of the ellipse of figure [fig9.2] as $$\begin{aligned} \label{eq9.34} \frac{p_{n \psi}^{(-)}}{S_{\psi}}=\frac{p_{n t}^{(-)}}{S_{T}} \cos ^{2}{\kern-1pt\psi}+\frac{p_{n 1}^{(-)}}{S_{L}} \sin ^{2}{\kern-1pt\psi}.\end{aligned}$$

For given values of the stress components $\sigma_{22}$, $\sigma_{33}$, $\sigma_{23}$, $\sigma_{21}$, and $\sigma_{31}$ for which $\sigma_{n n} \leq 0$, the failure index is a function of the angle $\alpha$ of the fracture plane. The condition to find $\alpha$ is to make the failure index a maximum. The necessary condition for a maximum is $\frac{\partial F I_{M}}{\partial \alpha}=0$. To find $\alpha$ that satisfies the necessary condition requires a numerical search.

=-1The section of the failure surface in the $\sigma_\textit{nn}$-$\sigma_\textit{nt}$ plane where $\sigma_{n1} = 0$ is shown in figure [fig9.3](a), and the section of the failure surface in the $\sigma_\textit{nn}$-$\sigma_{n1}$ plane where $\sigma_\textit{nt} = 0$ is shown in figure [fig9.3](b). In addition to the five basic strength data for an FRP composite ply listed in table [tab9.1], Puck’s criterion introduces four new dimensionless parameters: $p_{n1}^{(+)}$, $p_{n t}^{(+)}$, $p_{n t}^{(-)}$, and $p_{n 1}^{(-)}$. The inclination parameters $p_{n t}^{(-)}$ and $p_{n t}^{(+)}$ are the slopes of the failure locus at the $\sigma_{nt}$-axis in figure [fig9.3](a). Inclination parameters $p_{n1}^{(-)}$ and $p_{n 1}^{(+)}$ are the slopes of the failure locus at the $\sigma_{n1}$-axis shown in figure [fig9.3](b). Puck, et al. (2002) recommend that $p_{n t}^{(-)}=p_{n t}^{(+)}$, which makes the slope of the $\sigma_\textit{nn}$-$\sigma_\textit{nt}$ curve continuous at the $\sigma_\textit{nt}$-axis. The inclination parameters $p_{n 1}^{(-)}=0.25$ and $p_{n 1}^{(+)}=0.30$ with $p_{n t}^{(-)}$ computed from eq. ([eq9.24]) were used in the WWFE. Recommended ranges of inclination parameters are listed table [tab9.2].

Fiber modes.

A simple fiber mode criterion that does not interact with the longitudinal shear stresses $\sigma_{21}$ and $\sigma_{31}$ is the maximum stress criterion along the fibers. The fiber failure index $F I_{F}$ is defined by $$\begin{aligned} \label{eq9.35} F I_{F}= \begin{cases}\frac{-\sigma_{11}}{X_{c}} & \sigma_{11}<0 \\[5pt] \frac{\sigma_{11}}{X_{T}} & \sigma_{11}>0\end{cases},\end{aligned}$$ where $0 \leq F I_{F}<1$ for no failure of the fiber, and $F I_{F}=1$ at failure.

Modes B and C for a plane stress state.

Substitute the stress transformation equations ([eq9.36]) into eq. ([eq9.25]) to get $$\begin{aligned} \label{eq9.38} F I_{M}=\left(\frac{\sigma_{22}}{S_{T}}\right)^{2} \sin ^{2} \alpha \cos ^{2} \alpha+\left(\frac{\sigma_{21}}{S_{L}}\right)^{2} \cos ^{2} \alpha+2\left(\frac{p}{R}\right) \sigma_{22} \cos ^{2} \alpha \quad \sigma_{22}<0.\end{aligned}$$ The angle of the fracture plane is determined when index $\textit{FI}_M$ is a maximum value with respect to $\alpha$. Substitute $\sin ^{2} \alpha=1-\cos ^{2} \alpha$ in ([eq9.38]) to express index $F I_{M}$ as a function of $\cos ^{2} \alpha$. Then the necessary condition for a maximum can be written as $$\begin{aligned} \label{eq9.39} \frac{d F I_{M}}{d \alpha}=\frac{d F I_{M}}{d\left(\cos ^{2} \alpha\right)}(-2 \cos \alpha \sin \alpha)=0.\end{aligned}$$ One solution of eq. ([eq9.39]) is $\alpha=0$, which is the mode B fracture where the fracture plane is normal to the $x_2$-direction. $$1=\left(\frac{\sigma_{21}}{S_{L}}\right)^{2}+2 p_{n}^{(-)}\left(\frac{\sigma_{22}}{S_{L}}\right) \quad \sigma_{22}<0 \quad \text {mode B} \label{eq9.40}$$ Now take the derivative of the failure index ([eq9.38]) with respect to $\cos^{2} \alpha$ and set it equal to zero. Solve the resulting expression for $\cos^{2} \alpha$ to find $$\begin{aligned} \label{eq9.41} \cos ^{2} \alpha=\frac{1}{2}\left[1+\left(\frac{S_{T}}{S_{L}}\right)^{2}\left(\frac{\sigma_{21}}{\sigma_{22}}\right)^{2}\right]+p_{n t}^{(-)} \frac{S_{T}}{\sigma_{22}}.\end{aligned}$$ Equation ([eq9.41]) is used to eliminate the trigonometric functions in the failure index ([eq9.38]) to get $$\begin{aligned} \label{eq9.42} F I_{M}=\left\{\frac{1}{2}\left(\frac{S_{T}}{\sigma_{22}}\right)\left[\left(\frac{\sigma_{22}}{S_{T}}\right)^{2}+\left(\frac{\sigma_{21}}{S_{L}}\right)^{2}\right]+p_{n t}^{(-t)}\right\}^{2}.\end{aligned}$$ Note that $-1 \leq \sqrt{F I_{M}} \leq 1$, so that the square root of eq. ([eq9.42]) is $$\begin{aligned} \label{eq9.43} -1 \leq \frac{1}{2}\left(\frac{S_{T}}{\sigma_{22}}\right)\left[\left(\frac{\sigma_{22}}{S_{T}}\right)^{2}+\left(\frac{\sigma_{21}}{S_{L}}\right)^{2}+2 p_{n t}^{(-)}\left(\frac{\sigma_{22}}{S_{T}}\right)\right] \leq 1.\end{aligned}$$ Take the left-hand inequality of eq. ([eq9.43]) and multiply by $-1$ to get the form $$\begin{aligned} \label{eq9.44} 1 \geq \frac{1}{2}\left(\frac{S_{T}}{-\sigma_{22}}\right)\left[\left(\frac{\sigma_{22}}{S_{T}}\right)^{2}+\left(\frac{\sigma_{21}}{S_{L}}\right)^{2}\right]-p_{n t}^{(-)}.\end{aligned}$$ Finally, add $p_{nt}^{(-)}$ to each side of eq. ([eq9.44]) followed by division by $1+p_{n t}^{(-)}$ to get $$\begin{aligned} \label{eq9.45} \frac{1}{2\left(1+p_{n t}^{(-)}\right)}\left[\left(\frac{\sigma_{22}}{S_{T}}\right)^{2}+\left(\frac{\sigma_{21}}{S_{L}}\right)^{2}\right]\left(\frac{S_{T}}{-\sigma_{22}}\right)=F I_{M} \quad \sigma_{22} \leq 0 \quad \text { mode C },\end{aligned}$$ where $\textit{FI}_M = 1$ at failure in mode C. Equation ([eq9.41]) is written in the equivalent form as $$\begin{aligned} \label{eq9.46} \cos ^{2} \alpha=\frac{1}{2}\left(\frac{S_{T}}{\sigma_{22}}\right)^{2}\left[\left(\frac{\sigma_{22}}{S_{T}}\right)^{2}+\left(\frac{\sigma_{21}}{S_{L}}\right)^{2}+2 p_{n t}^{(-)}\left(\frac{\sigma_{22}}{S_{T}}\right)\right].\end{aligned}$$ At failure eq. ([eq9.45]) is solved in the form $$\begin{aligned} \label{eq9.47} \left(\frac{\sigma_{22}}{S_{T}}\right)^{2}+\left(\frac{\sigma_{21}}{S_{L}}\right)^{2}=2\left(1+p_{n t}^{(-t)}\right)\left(\frac{-\sigma_{22}}{S_{T}}\right).\end{aligned}$$ Substitute eq. ([eq9.47]) into eq. ([eq9.46]) and perform some algebra to get final result for the angle of fracture plane for mode C: $$\begin{aligned} \label{eq9.48} \cos ^{2} \alpha=\frac{S_{T}}{-\sigma_{22}} \quad \sigma_{22} \leq-S_{T} \quad \text { mode C }.\end{aligned}$$

Transverse shear strength.

The shear strength transverse to the fibers in the fracture plane $S_T$ cannot be determined from simple tests. Instead $S_T$ is derived from the uniaxial transverse compression test in which $\sigma_\textit{22} = -Y_C$ at failure and all other stresses in the $x_i$-system vanish. In eq. ([eq9.45]) set $\textit{FI}_M = 1$, $\sigma_{\textit{21}} = 0$, and $\sigma_{\textit{22}} = -Y_C$ to evaluate the transverse shear strength $S_T$ at the pure transverse compression condition. The result is $$\begin{aligned} \label{eq9.49} S_{T}=\frac{Y_{C}}{2\left(1+p_{n t}^{(-)}\right)}.\end{aligned}$$ To find the transition values of stresses $\sigma_{\textit{21}}$ and $\sigma_{\textit{22}}$ between modes B and C, solve the eq. ([eq9.40]) for $\sigma_{\textit{21}}$ and substitute this result for $\sigma_{\textit{21}}$ in eq. ([eq9.45]) with $FI_M = 1$. The results are $$\begin{aligned} \label{eq9.50} \sigma_{22}=-S_{T} \quad \sigma_{21}=S_{L} \sqrt{1+2 p_{n}^{(-)}}.\end{aligned}$$ Thus, for plane stress the matrix failure indices are $$\begin{gathered} F I_{M}=\sqrt{\left[1-p_{n 1}^{(+)} \frac{Y_{T}}{S_{L}}\right]^{2}\left(\frac{\sigma_{22}}{Y_{T}}\right)^{2}+\left(\frac{\sigma_{21}}{S_{L}}\right)^{2}}+\frac{p_{n 1}^{(+)} \sigma_{22}}{S_{L}} \quad \sigma_{22} \geq 0 \quad \text { mode A } \label{eq9.51} \\ F I_{M}=\left(\frac{\sigma_{21}}{S_{L}}\right)^{2}+2 p_{n 1}^{(-)}\left(\frac{\sigma_{22}}{S_{L}}\right) \quad-S_{T} \leq \sigma_{22}<0 \quad S_{L}<\left|\sigma_{21}\right| \leq S_{L} \sqrt{1+2 p_{n t}^{(-)}} \quad \text { mode B} \label{eq9.52} \\ F I_{M}=\frac{1}{2\left(1+p_{n t}^{(-)}\right)}\left[\left(\frac{\sigma_{22}}{S_{T}}\right)^{2}+\left(\frac{\sigma_{21}}{S_{L}}\right)^{2}\right]\left(\frac{S_{T}}{-\sigma_{22}}\right) \quad-Y_{C} \leq \sigma_{22} \leq-S_{T} \quad \text { mode C }. \label{eq9.53}\end{gathered}$$

The matrix failure locus is plotted in $\left(\sigma_{n n}, \sigma_{n t}, \sigma_{n 1}\right)$ stress space in figure [fig9.4] for the lamina subject to plane stress, The stress components at selected points are listed in table [tab9.3].

The matrix failure locus shown in figure [fig9.4] is plotted in the $\sigma_{22}$-$\sigma_{21}$ stress plane in figure [fig9.5].

In multidirectional laminates the intralaminar failure predictions are made by the analysis of strains and/or stresses in each lamina, with failure criteria evaluated in each lamina. A failure initiated in one lamina predicts the onset of damage, or first ply failure (FPF), that is usually not the ultimate failure of the laminate. It is insufficient to predict ultimate failure with the failure initiation criteria alone if the composite structure can accumulate damage before ultimate failure.