A cracked beam is subjected to a pair of forces at the center of the crack (see Figure).

a) Find the energy release release G.

b) Find the minimum P that can split the beam. Assume E =
70 GPa and G_{c} = 200 N.m/m^{2}.

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strain energyrelease rate.G=\pi \times \sigma^{2} \times \frac{a}{E}a \doteq kalf crank leugth=10 \mathrm{~cm}=0.1 \mathrm{~m}\begin{array}{l}E=70 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2} \\G=200 \\\sigma=\sqrt{4.456 \times 10^{13}} \Rightarrow 6675581.178 \mathrm{~N} / \mathrm{m}^{2}\end{array}To find the load P\begin{array}{l}\sigma=\frac{P}{A} \\P=6675581.178 \times\left(0.2 \times 0.5 \times 10^{-2}\right) \\P=6675.58 \mathrm{~N} \\P=6.675 \mathrm{kN}\end{array} Solution(a)Draw Free Body DiagramFind roment of Inertia for specified dimensiousI=\frac{t h^{3}}{12}here t=0.5 \mathrm{~cm}=0.005 \mathrm{~m}\begin{array}{l}\quad h=1 \mathrm{~cm}=0.01 \mathrm{~m} \\I=\frac{0.005(0.01)^{3}}{12}\end{array}I=4.166666667 \times 10^{-10} \mathrm{~m}^{4}Let beam (1) is a cantilevar beam with tip load P / 2Slope at tip \left(Q_{P}\right)=(P / 2) \frac{a^{2}}{2 E I}SLope due to tip moment (\theta \mathrm{m})\begin{array}{c}\theta_{m}=-\frac{m a}{E I} \\\theta p+\theta_{m}=0 \\\frac{\left(\frac{p}{2}\right) a^{2}}{2 E I}-\frac{m a}{E I}=0 \\m=\frac{p}{4} a\end{array}The moment distribution along LengthM(x)=\frac{p}{2} x-\frac{p}{4} aStram energies\begin{array}{l}v_{1}=v_{2}=\int_{0}^{a} \frac{M\left(x^{2}\right) d x}{2 E I} \\v_{1}=U_{2}=\frac{1}{2 E I} \int_{0}^{a}\left[\frac{p x}{2}-\frac{p a}{u}\right]^{2} d x \\=\frac{1}{2 E I} \int_{0}^{a}\left[\frac{p^{2} x^{2}}{4}+\frac{p^{2} a^{2}}{16}-\frac{p^{2}}{\frac{x a}{4}}\right] d x \\\end{array}Integrate above equation\begin{array}{l}U_{1}=U_{2}=\frac{1}{2} E I\left[\frac{p^{2} x^{3}}{12}+\frac{p^{2} a^{2} x}{16}-\frac{p^{2} x^{2} a}{8}\right]_{0}^{a} \\U_{1}=v_{2}=\frac{1}{2} E I\left[\frac{p^{2} a^{3}}{12}+\frac{p^{2} a^{3}}{16}-\frac{p^{2} q^{3}}{8}\right]\end{array}\begin{array}{l}U_{1}=U_{2}=\frac{1}{2 E I}\left[\frac{8 p^{2} a^{3}+6 p^{2} a^{3}-8 p^{2} a^{3}}{96}\right] \\U_{1}=U_{2}=\frac{1}{2 E I}\left[\frac{2 p^{2} a^{3}}{96}\right] \\U_{1}=U_{2}=\frac{p^{2} a^{3}}{96 E I}\end{array}Total Strain energles (V)=v_{1}+v_{2}+v_{3}\begin{array}{l}U=\frac{p^{2} a^{3}}{96 E I}+\frac{p^{2} a^{3}}{96 E I}+0 \\U=\frac{p^{2} a^{3}}{48 E I} \rightarrow \begin{array}{l}\text { Total } \\\text { Struin } \\\text { energy }\end{array}\end{array}Strain energy recease rate\begin{array}{l}G=\frac{1}{t} \frac{d u}{d a}=\frac{1}{t} \frac{d}{d a}\left[\frac{p^{2} a^{3}}{48 E I}\right] \\G=\frac{1}{t} \frac{p^{2}\left(3 a^{2}\right)}{48 E I} \\G=\frac{p^{2} a^{2}}{16 E I t} \rightarrow \rightarrow \begin{array}{c}\text { Strain ehergy } \\\text { release rate }\end{array}\end{array}Crack will propagate if G \geq G G\begin{aligned}G=G_{c} & =\frac{P_{\min }^{2} a^{2}}{16 E I t} \\P_{\min } & =\sqrt{\frac{16 E I t G_{c}}{a^{2}}} \\\text { Here } E & =70 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2} \\G_{c} & =200 \mathrm{~N} \cdot \mathrm{m} / \mathrm{m}^{2}\end{aligned}\begin{array}{l}t=0.5 \mathrm{~cm}=0.005 \mathrm{~m} \\ I=4.16666667 \times 10^{-10} \mathrm{~m}^{4} \\ a=10 \mathrm{~cm}=0.1 \mathrm{~m} \\ P_{\text {min }}=\sqrt{\frac{16 \times 70 \times 10^{9} \times 4.167 \times 10^{-10} \times 0.005 \times 200}{(0.1)^{2}}} \\ P_{\text {min }}=\sqrt[46666.66666]{P_{\text {min }}=216.02 \mathrm{~N}} \\ P_{\text {min }} \approx 216 \mathrm{~N} \rightarrow \text { Result }\end{array} ...