The critical mode I stress intensity factor is denoted by \(\textbf{\textit{K}}_{\textbf{\textit{Ic}}}\), and it is assumed to be a material parameter.

• \(K_{I}<K_{I c}\). There is no crack growth, and the material resists the crack without brittle fracture.

• \(K_{I}=K_{I c}\). The crack begins to propagate and brittle fracture occurs.

The critical mode I stress intensity factor is also called the fracture toughness. A tough material has a large value of \(K_{I c}\), which means it is effective in resisting crack growth. At crack growth, the remote stress is denoted by \(\sigma_{\mathrm{cr}}\) and is given by \[\begin{aligned} \label{eq13.14} \sigma_{\mathrm{cr}}=\frac{K_{I c}}{F \sqrt{\pi a}}.\end{aligned}\] A representative graph of eq. ([eq13.14]) is shown in figure [fig13.6]. The value of the half crack length when \(\sigma_{c r}=\sigma_{Y}\), is called the transition crack length \(a_{t}\), where \(\sigma_{Y}\) denotes the yield strength of the material. Set \(\sigma_{\mathrm{cr}}=\sigma_{Y}\) in eq. ([eq13.14]) to find

\[\begin{aligned} \label{eq13.15} a_{t}=\frac{1}{\pi}\left(\frac{K_{I c}}{\sigma_{Y}}\right)^{2}.\end{aligned}\] The transition crack length is the approximate length above which strength is limited by fracture. If \(a>a_{t}\) then fracture limits strength. If \(a<a_{t}\) then yielding dominates strength. Materials with

1. high \(K_{I c}\) and low \(\sigma_{Y}\) imply a long \(a_{t}\) and small cracks are no problem, and

2. low \(K_{I c}\) and high \(\sigma_{Y}\) imply a short \(a_{t}\) and small cracks can be a problem.

Typical values of the fracture toughness at room temperature are listed in table [tab13.2].

[ex13.1]The configuration shown in figure [fig13.7] is subject to a tensile load P, has a crack length denoted by \(a\), and a thickness denoted by \(t\). It is the configuration of the ASTM standard compact specimen. The mode I stress intensity factor is given by (Dowling, p. 295)

\[K=F_{P}(\alpha) \frac{P}{t \sqrt{b}}, \label{eq13.1.a}\tag{a}\]

L72pt

where \(\alpha=a/b\). The nondimensional function \(F_p\) is \[F_{p}(\alpha)=\frac{(2+\alpha)}{(1-\alpha)^{3 / 2}}\left[0.866+4.64 \alpha-13.32 \alpha^{2}+14.72 \alpha^{3}-5.6 \alpha^{4}\right] \quad \alpha \geq 0.2. \label{eq13.1.b}\tag{b}\]

For \(P=22 \textrm{~kN}\), \(b= 50\) mm, and \(t= 25\) mm, determine the critical crack length for brittle fracture of 2024-T351 aluminum alloy. The numerical factor multiplying \(F_p\) is \[\frac{P}{t \sqrt{b}}=\frac{22{,}000 \mathrm{~N}}{25 \mathrm{~mm} \sqrt{50 \mathrm{~mm}\left(10^{3} \mathrm{~mm} / \mathrm{m}\right)}}=\frac{22{,}000}{25 \sqrt{50{,}000}}\left(\frac{\mathrm{N}}{\mathrm{mm}^{2}}\right) \sqrt{\mathrm{m}}=3.93548 \textrm{ MPa} \sqrt{\mathrm{m}}. \label{eq13.1.c}\tag{c}\] A graph of the stress intensity factor as a function of the crack length in the range from 15 mm to 35 mm is shown in figure [fig13.8]. The critical mode I stress intensity factor is 34 MPa \(\sqrt{\mathrm{m}}\) from table [tab13.2], and it plots as horizontal line in the graph. Using a root finding procedure, or a trial and error method, the intersection of the two lines in the graph occurs at 23.1 mm. Thus, the critical crack length for brittle fracture is 23.1 mm.

The example that follows illustrates how the methods of strength of materials and fracture mechanics are employed to analyze the strength of a cantilever beam.

[ex13.2]This example is adapted from Kanninen and Popelar (1985, pp. 10–12). Determine the maximum value of the tip load \(Q\) acting on a cantilever beam depicted in figure [fig13.9] by the strength of materials method and the fracture mechanics method. The beam has a length \(L=300 \mathrm{~mm}\), height \(h=30 \mathrm{~mm}\), and rectangular cross section with width \(b=15 \mathrm{~mm}\). Assume a factor of safety (FS) of 1.5. The material is aluminum alloy 2014-T651 with properties listed in table [tab13.2]. Plot the maximum value of Q versus the crack length \(a\).

(a) Strength of materials approach.

The maximum load is determined such that the maximum stress in the beam is less than the yield strength of the material. The maximum tensile normal stress occurs at the bottom ofthe beam at its built-in end. The bending moment at the built-in end is \(M=L Q\), and the flexure formula for the maximum normal stress is \[\sigma_{\max }=\frac{M(h / 2)}{I}, \label{eq13.2.a}\tag{a}\] where the second area moment of the rectangular cross section is \(I=\left(b h^{3}\right) / 12\). Hence, \[\sigma_{\max }=\frac{6 Q L}{b h^{2}}<\frac{\sigma_{Y}}{F S}. \label{eq13.2.b}\tag{b}\] Solve the inequality for the load in eq. ([eq13.2.b]) to get \[Q<\frac{b h^{2}}{6(F S) L} \sigma_{Y}=\frac{(15 \mathrm{~mm})(30 \mathrm{~mm})^{2}}{6(1.5)(300 \mathrm{~mm})}(415 \mathrm{~N} / \mathrm{mm}^{2}). \label{eq13.2.c}\tag{c}\] Hence, \(Q<2075 \mathrm{~N}\).

(b) Fracture mechanics approach.

Consider that the beam, instead of being defect free, contains an edge crack of length \(a\) normal to the free edge. Further suppose that, as shown in figure [fig13.9](b), the crack is located where the maximum tensile stress is anticipated. This geometry and loading configuration is not the center-cracked plate. However, for a relatively small crack, an LEFM-based analysis of the flawed beam shown in figure [fig13.9](b) would give a reasonable approximation in the following expression for the stress intensity factor: \[K_{I}=1.12 \sigma_{\max } \sqrt{\pi a}, \label{eq13.2.d}\tag{d}\] where \(\sigma_{\max }\) is the stress that would occur at the crack location in the absence of the crack. The beam is safe from fracture if \(K_{I} \leq K_{I c}\). Further assurance can be obtained by having \(K_{I}<K_{I c} /(F S)\), where, just as in the strength of materials approach, the number FS is the factor of safety. Using eq. ([eq13.2.b]) to replace \(\sigma_{\max }\) in eq. ([eq13.2.d]) then leads to \[K_{I}=1.12\left(\frac{6 Q L}{b h^{2}}\right) \sqrt{\pi a}<\frac{K_{I c}}{\textit{FS}}. \label{eq13.2.e}\tag{e}\] Solve eq. ([eq13.2.e]) for the load to get \[Q<\left(\frac{b h^{2}}{6(\textit{FS}) L}\right) \frac{K_{I c}}{1.12 \sqrt{\pi a}}. \label{eq13.2.f}\tag{f}\] Substitute numerical values into eq. ([eq13.2.f]) to get \[Q<\left[\frac{(15 \mathrm{~mm})(30 \mathrm{~mm})^{2}}{6(1.5)(300 \mathrm{~mm})}\right] \frac{\left(24 \mathrm{~N} / \mathrm{mm}^{2}\right) \sqrt{\mathrm{m}\left(10^{3} \mathrm{~mm} / \mathrm{m}\right)}}{1.12 \sqrt{\pi a}}. \label{eq13.2.g}\tag{g}\]

Hence, \(Q<\frac{1911.56 \mathrm{~N} \sqrt{\mathrm{mm}}}{\sqrt{a}}\), which is the fracture mechanics estimate of the safe operating load. A graph of the failure load is shown in figure 1.2. From the plots in the graph, yielding governs failure for \(0<a<0.849 \mathrm{~mm}\), whereas fracture governs failure for \(a>0.849 \mathrm{~mm}\). The transition crack length, \(a_{t}=0.849 \mathrm{~mm}\), determines the crack length for which it can be expected that fracture rather than yielding governs the mode of failure. In this example the transition crack length is only 2.8 percent of the height, \(h\), of the beam.

A comparison between eq. ([eq13.2.c]) and eq. ([eq13.2.g]) is instructive. It can be seen that the structural geometry and the factor of safety enter both relations in exactly the same way – that is, through the multiplicative parameter \(\left(b h^{2}\right) /(6(F S) L)\). Also, both relations contain a basic, albeit different, material property. The essential difference is that the fracture mechanics approach explicitly introduces a new physical parameter: the size of a (real or postulated) crack-like flaw. In fracture mechanics the size of the crack is the dominant structural parameter. It is the specification of this parameter that distinguishes fracture mechanics from conventional failure analyses.

The critical mode II stress intensity factor is denoted by \(\textbf{\textit{K}}_{\textbf{\textit{IIc}}}\), and it is assumed to be a material parameter.

• \(K_{I I}<K_{I I c}\). There is no crack growth, and the material resists the crack without brittle fracture.

• \(K_{I I}=K_{I I c}\). The crack begins to propagate and brittle fracture occurs.

The critical mode II stress intensity factor is also called the mode II fracture toughness.

The displacements on the two crack surfaces are antisymmetric with respect to the \(x\)-axis. Let \(u\) denote the displacement component in the \(x\)-direction, and let \(v\) denote the displacement component in the \(y\)-direction. The expression for the upper surface displacement \(u(x)\) as given by Sun (1998, p. 163) is \[\begin{aligned} \label{eq13.21} u=\frac{(\kappa+1)(1+v)}{2 E} \tau \sqrt{a^{2}-x^{2}} \textrm{ and } v(x)=0, |x| \leq a.\end{aligned}\] The origin of the \(x\)-axis is located at the center of the crack as is shown in figure 1.1, and \(\kappa\) depends on Poisson’s ratio, which is given by eq. ([eq13.13]). Under antisymmetric loading the crack surfaces are in sliding contact with each other and do not open as they do in mode I.

Griffith criterion

The rationale for the Griffith energy criterion is the first law of thermodynamics and the theorem of minimum potential energy for an elastic body. The critical condition for equilibrium crack growth is that there is no net change in the total energy (Anderson, 1995, p. 36). Consider a cracked plate of thickness \(t\) with a through crack of length \(a\) as shown in figure [fig13.12]. Then the infinitesimal increase in the crack area is \(d A=t d a\).

Let E denote the total energy, \(U\) the total strain energy contained in the plate, \(W_{e}\) the work performed by the external force, and \(W_{s}\) the work to create new surfaces. Then, \(E=U-W_{e}+W_{s}\) and the critical condition for crack growth in the plate is given by \[\begin{aligned} \label{eq13.22} \frac{d E}{t d a}=\frac{d}{t d a}\left[U-W_{e}+W_{s}\right]=0.\end{aligned}\] Rearrange the previous equation to \[\begin{aligned} \label{eq13.23} \underbrace{\frac{1}{t} \frac{d W_{s}}{d a}}_{R}=\underbrace{\frac{1}{t} \frac{d}{d a}\left[W_{e}-U\right]}_{G}.\end{aligned}\]

In the latter equation \(R\) denotes the material resistance to crack growth, and \(G\) denotes the energy release rate. The energy release rate is a measure of the energy available for an increment of crack extension, and since it is obtained from the derivative of a potential, it is also called the crack extension force or crack driving force. \[\begin{gathered} \boxed{ \begin{array}{ll} \textrm{If $G<R$ no crack growth} \\ \textrm{If $G=R$ crack growth} \end{array}} \label{eq13.24}\end{gathered}\] When \(G=R\) there is sufficient energy in the system to form an additional crack size dA. For the plate under the action of the load P, the load application points undergo a relative displacement \(v\). When the crack length increases by da, the displacement will increase by dv. The work done by the external load is Pdv. Hence, for mode I loading \[\begin{aligned} \label{eq13.25} G=\frac{1}{t} \frac{d}{d a}\left[W_{e}-U\right]=\frac{1}{t}\left[P \frac{d v}{d a}-\frac{d U}{d a}\right].\end{aligned}\] Prior to crack growth \(v=c P\), where \(c\) is the compliance of the plate, and the strain energy \(U=\frac{1}{2} P v=\frac{1}{2} c P^{2}\). as is shown in figure [fig13.13].

Hence, \[\begin{aligned} \label{eq13.26} G=\frac{1}{t}\left[P \frac{d}{d a}(c P)-\frac{d}{d a}\left(\frac{1}{2} c P^{2}\right)\right].\end{aligned}\] Performing the differentiations in eq. ([eq13.26]) we get \[\begin{aligned} \label{eq13.27} G=\frac{1}{t}\left[P^{2} \frac{d c}{d a}+c P \frac{d P}{d a}-\frac{P^{2}}{2} \frac{d c}{d a}-c P \frac{d P}{d a}\right].\\[-4pt] \raisebox{4pt}{$\blacktriangle$}\hspace*{-3.6pt}\rule{1pt}{5pt}\hspace*{-0.5pt}\rule{6pc}{1pt}\hspace*{-0.5pt}\rule{1pt}{5pt}\hspace*{-3.8pt}\raisebox{4pt}{$\blacktriangle$}\hspace*{1.1pc}\nonumber\\[-7pt] \mbox{add to zero}\hspace*{2.5pc}\nonumber\end{aligned}\] Therefore, the strain energy release rate is independent of whether the load changes or not during crack growth, and we get \[\begin{aligned} \label{eq13.28} G=\frac{P^{2}}{2 t} \frac{d c}{d a}.\end{aligned}\] Consider two types of loading as shown in figure [fig13.14].

1. Load \(P= \textrm{constant}\) during crack growth

   The strain energy and its differential with respect to the compliance at a fixed value of the load is \[U=\left.\frac{1}{2} c P^{2} \quad d U\right|_{P}=\frac{P^{2}}{2} d c,\]

   or \[\begin{aligned} \label{eq13.29} \left.\frac{1}{t} \frac{d U}{d a}\right|_{P}=\frac{P^{2}}{2 t} \frac{d c}{d a}.\end{aligned}\] Comparing eq. ([eq13.28]) and eq. ([eq13.29]), we find \[\begin{aligned} \label{eq13.30} G=\left.\frac{1}{t} \frac{d U}{d a}\right|_{P}.\end{aligned}\] That is, \(d U>0\) for \(d a>0\), and for \(P\) fixed in value.

2. Fixed grips with \(v = \textrm{constant}\)

   In this case \(U=\left(c P^{2}\right) / 2=\left(c\left(\frac{v}{c}\right)^{2}\right) / 2=v^{2} /(2 c)\) since \(P=v / c\). Hence, the differential of the strain energy with fixed grips is \[\begin{aligned} \label{eq13.31} \left.d U\right|_{v}=d\left(\frac{v^{2}}{2 c}\right)=\left(\frac{v^{2}}{2}\right)\left(-\frac{d c}{c}\right)=-\frac{1}{2}\left(\frac{v}{c}\right)^{2} d c.\end{aligned}\] Substitute \(v=c P\) for \(v\) in eq. ([eq13.31]) to get \[\begin{aligned} \label{eq13.32} \left.\frac{1}{t} \frac{d U}{d a}\right|_{v}=-\left(\frac{P^{2}}{2 t} \frac{d c}{d a}\right).\end{aligned}\] Comparing eq. ([eq13.28]) and eq. ([eq13.32]), we find \[\begin{aligned} \label{eq13.33} G=-\left.\left(\frac{1}{t} \frac{d U}{d a}\right)\right|_{v}.\end{aligned}\] So \(d U<0\) for \(d a>0\), and for \(v\) fixed in value, and no external work is done by load P.

Equations ([eq13.30]) and ([eq13.33]) show that the strain energy release rate is always equal to the derivative of the strain energy apart from the sign. Crack extension occurs when \(G_{I}=R\) where R is the work of fracture, or the material resistance to crack extension. For an ideally brittle material, like glass, \(R \sim 2 \mathrm{~J} / \mathrm{m}^{2}\), which is twice the surface energy to create two new surfaces. For ductile metals, the work of fracture is much larger than the surface energy to create new free surfaces because of plastic deformation occurring in front of the crack tip. During crack extension energy is expended by deformation of a new plastic zone at the tip of the advancing crack. Note that the Griffith energy criterion is based on linear elastic material behavior, so the formation of a nonlinear effect such as plasticity seems to negate the analysis. However, if the plastic zone ahead of the crack tip is very small compared to the bulk material remaining elastic, then the criterion is applicable. If the increase in energy required for plastic deformation is independent of the increase in crack area, then \(R=\text { constant }\). From fracture tests of a material, the value of G at crack growth, which equals R, is called the critical strain energy release rate, denoted by \(G_{\mathrm{c}}\), and is assumed to be a material parameter. \[\begin{gathered} \textrm{If $G<G_{\mathrm{c}}$ no crack growth} \nonumber\\ \textrm{If $G=G_{\mathrm{c}}$ crack growth initiates.} \label{eq13.34}\end{gathered}\]

Mixed mode fracture

Generally, crack growth occurs under mixed mode loading. Under this type of loading, crack growth might occur before any of the energy release rate components attain their individual critical value. Failure interaction criteria are established from mixed mode fracture test configurations. The following criterion has been shown to fit test data for many materials quite well: \[\begin{aligned} \label{eq13.41} \frac{G_{I}}{G_{I c}}+\frac{G_{I I}}{G_{I I c}}=1,\end{aligned}\] where G\(_\textit{Ic}\) and G\(_\textit{IIc}\) are the single mode critical energy release rates for modes I and II, respectively. From the relations between G and K in eqs. ([eq13.38]) and ([eq13.40]), the previous criterion ([eq13.41]) in terms of the stress intensity factors is \[\begin{aligned} \label{eq13.42} \left(\frac{K_{I}}{K_{\textit{Ic}}}\right)^{2}+\left(\frac{K_{I I}}{K_{\textit{IIc}}}\right)^{2}=1.\end{aligned}\]

Solution.

We use Castigliano’s second theorem to determine the relative displacement \(v\). Complementary strain energy is stored in each arm of the beam for \(0 \leq x \leq a\), and not in the un-stressed section of the beam from \(a<x \leq L\). The complementary strain energy is \[U^{*}=\frac{1}{2} \int_{0}^{a} \frac{M_{1}^{2}}{E I} d x+\frac{1}{2} \int_{0}^{a} \frac{M_{2}^{2}}{E I} d x, \label{eq13.3.a}\tag{a}\] where the bending moment in the upper arm is \(M_{1}\), the bending moment in the lower arm is \(M_{2}\), and the second area moment of each arm is \[I=\frac{t h^{3}}{12}=12.9373 \mathrm{~mm}^{4}. \label{eq13.3.b}\tag{b}\] Note that the modulus of elasticity \(E=E_{1}\). The distribution of the bending moment in each arm is determined from equilibrium. The results are \[M_{1}(x)=-M_{2}(x)=-P x \quad 0 \leq x \leq a. \label{eq13.3.c}\tag{c}\] Hence, \[\nu=\frac{\partial U^{*}}{\partial P}=\frac{1}{E I}\left\{\int_{0}^{a}(-P x)(-x) d x+\int_{0}^{a}(P x)(x) d x\right\}=\frac{2 P}{E I} \int_{0}^{a} x^{2} d x. \label{eq13.3.d}\tag{d}\]

Performing the integration in eq. ([eq13.3.d]) we get \[v=\frac{2}{3} \frac{P a^{3}}{E I}. \label{eq13.3.e}\tag{e}\] From this last equation, the compliance of the beam is given by \(c=(2 a^{3}) /(3 E I)\). The strain energy release rate is determined from eq. ([eq13.28]), which yields \[G_{I}=\frac{P^{2}}{2 t} \frac{d c}{d a}=\frac{P^{2} a^{2}}{t E I}. \label{eq13.3.f}\tag{f}\] At the initiation of crack growth \(G_{I}=G_{I c}\), so \[G_{I c}=\frac{P^{2} a^{2}}{t E I} \text { at crack growth }. \label{eq13.3.g}\tag{g}\] Solve eq. ([eq13.3.g]) for the crack length a to get \[a=\left(t E I G_{I c}\right)^{1 / 2}\left(\frac{1}{P}\right). \label{eq13.3.h}\tag{h}\] Substitute eq. ([eq13.3.h]) for crack length \(a\) into eq. ([eq13.3.e]) to get \[v=\frac{2}{3} \frac{\left(t G_{I c} E I\right)^{3 / 2}}{E I P^{2}} \text { for the propagating crack }. \label{eq13.3.i}\tag{i}\]

Prior to crack growth eq. ([eq13.3.e]) determines the response as \[v=\frac{2}{3} \frac{(50 \mathrm{~mm})^{3} P}{\left(150,000 \mathrm{~N} / \mathrm{mm}^{2}\right)\left(12.9373 \mathrm{~mm}^{4}\right)}=(0.04294 \mathrm{~mm} / \mathrm{N}) P. \label{eq13.3.j}\tag{j}\] For the propagating crack eq. ([eq13.3.h]) evaluates to \[a=\big[(20 \mathrm{~mm})\big(150{,}000 \mathrm{~N} / \mathrm{mm}^{2}\big)\big(12.9373 \mathrm{~mm}^{4}\big)(0.352 \mathrm{~N} / \mathrm{mm})\big]^{1/2}\left(\frac{1}{P}\right)=\frac{3{,}696.19 \textrm{ Nmm}}{P}. \label{eq13.3.k}\tag{k}\] For the initial crack length \(a_{0}=50 \mathrm{~mm}\) eq. ([eq13.3.k]) determines the maximum load, and then either eq. ([eq13.3.e]) or eq. ([eq13.3.j]) determines the corresponding displacement; i.e., \[P=73.9238 \mathrm{~N} \textrm{ and } v=3.1744 \textrm{ mm at the initiation of crack growth.}\] For the propagating crack eq. ([eq13.3.i]) evaluates to \[v=\frac{2}{3} \frac{[20 \mathrm{~mm}(0.352 \mathrm{~N} / \mathrm{mm})\left(150,000 \mathrm{~N} / \mathrm{mm}^{2}\right)\left(12.9373 \mathrm{~mm}^{4}\right)]^{3 / 2}}{\left(150,000 \mathrm{~N} / \mathrm{mm}^{2}\right)\left(12.9373 \mathrm{~mm}^{4}\right)}\left(\frac{1}{P^{2}}\right)=17,347.4 \mathrm{~N}^{2} \mathrm{~mm}\left(\frac{1}{P^{2}}\right). \label{eq13.3.l}\tag{l}\] Equations ([eq13.3.j]) and ([eq13.3.l]) are used to plot the load-displacement response shown in figure [fig13.16].

In load control, where P is specified and increased slowly from zero, a sudden dynamic increase in crack growth occurs at \(P=73.9238 \mathrm{~N}\) since there is no stable adjacent equilibrium state. In displacement control, where \(v\) is specified and is increased slowly from zero to 3.17 mm, the load P increases to 73.92 N. For \(v>3.17 \mathrm{~mm}\) the load decreases as the crack increases in length.\(\blacksquare\)

[ex13.4]The end load split (ELS) configuration shown in figure 1.4 has an initial crack length denoted by \(a=a_{0}\), length by \(L\), height by \(2 h\), and thickness normal to the x-y plane by \(t\). The lower arm at the tip is subject to vertical force \(P\). Both arms below and above the crack are identical and are subject to the same load as shown in the free body diagrams of figure 1.5. Hence, both arms have the same lateral displacement \(v(x)\) and rotation \(-v^{\prime}(x)\). The\(x\)-direction displacement at the crack tip of the upper arm is \(\frac{h}{2}\left[v^{\prime}(a)\right]\), and the \(x\)-direction displacement ofthe lower arm at the crack tip is \(\frac{h}{2}\left[-v^{\prime}(a)\right]\), Thus, the relative axial displacement between the lower surface of the upper arm and the upper surface of the lower arm is \(h\left[v^{\prime}(a)\right]\), which is a mode II displacement loading. Determine the strain energy release rate G\(_\textrm{II}\). Analytical and numerical solutions for this example were originally given by Chen et al. (1999).

Solution.

The strain energy is \[U=\int_{0}^{a} \frac{M_{1}^{2}}{2 E I} d x+\int_{0}^{a} \frac{M_{2}^{2}}{2 E I} d x+\int_{a}^{L} \frac{M_{3}^{2}}{2 E(8 I)} d x, \label{eq13.4.a}\tag{a}\] where \(\textit{M}_1\) denotes the bending moment in the arm below the crack, \(\textit{M}_2\) the bending moment in the arm above the crack, and \(\textit{M}_3\) the bending moment in the segment not containing the crack. The second area moment of each arm is denoted by I, and \(I=\left(t h^{3}\right)/12\). From the free body diagrams shown in figure 1.5, equilibrium determines the bending moments as \[\begin{gathered} M_{1}(x)=M_{2}(x)=-\left(\frac{P}{2}\right) x \quad 0 \leq x \leq a, \textrm{ and} \label{eq13.4.b}\tag{b}\\ M_{3}(x)=-(x-a) P+M_{1}(a)+M_{2}(a)=-P x \quad a \leq x \leq L. \label{eq13.4.c}\tag{c}\end{gathered}\] Substitute eqs. ([eq13.4.b]) and ([eq13.4.c]) for the bending moments in eq. ([eq13.4.a]), and perform the integrations to get \[U=\frac{(3 a^{3}+L^{3})P^{2}}{48 E I}. \label{eq13.4.d}\tag{d}\] The mode II strain energy release rate is \[G_{II}=\frac{\partial U}{t \partial a}=\frac{3 a^{2} P^{2}}{16 t E I}.\label{eq13.4.e}\tag{e}\]

Mixed mode fracture

Consider a cantilever beam of length L containing a through crack of length \(a\) centered at its free end. This configuration subject to load P shown in part (a) of figure [fig13.19] is labeled FRMM, which means fixed ratio mixed mode. By the method of superposition FRMM is equivalent to the DCB test configuration shown in part (b) of the figure plus the ELS test configuration shown in part (c) of the figure. Hence, the FRMM configuration is a mixed mode I and II fracture test.

From eq. ([eq13.3.f]) in example [ex13.3] the mode I strain energy release rate for the configuration in part (b) of figure [fig13.19] is \[\begin{aligned} \label{eq13.43} G_{I}=\left.\frac{P^{2} a^{2}}{t E I}\right|_{P \rightarrow P / 2}=\frac{P^{2} a^{2}}{4 t E I}.\end{aligned}\] From eq. ([eq13.4.e]) in example [ex13.4] the mode II strain energy release rate for the configuration in part (c) of figure [fig13.19] is \[\begin{aligned} \label{eq13.44} G_{I I}=\frac{3 a^{2} P^{2}}{16 t E I}.\end{aligned}\] Therefore, the mixed mode ratio for the FRMM configuration is \(G_{I}/G_{I I}=4/3\).

[ex13.5]Take the height of the arms \(h=1.5 \mathrm{~mm}\), thickness \(t=10 \mathrm{~mm}\), length \(L=100 \mathrm{~mm}\), so that \(I=2.8125 \mathrm{~mm}^{4}\). The initial crack length \(a_{0}=40 \mathrm{~mm}\). From table [tab13.4] where \(E=150{,}000 \mathrm{~N} / \mathrm{mm}^{2}\), \(G I_{c}=0.352 \mathrm{~N} / \mathrm{mm}\) and \(G I I_{c}=1.45 \mathrm{~N} / \mathrm{mm}\). Determine the load-displacement response, and the crack-length-displacement response.