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Breakdown Mechanism
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When a fracture occurs, it is characterized by an area that can no longer bear load and an edge marked by tension that grows inversely to the radius of the fracture tip. This means that the section is diminished, requiring the remaining section to bear a greater load, exacerbating the situation at the fracture tip and facilitating its propagation. Thus, a catastrophic situation ensues where each increase in the fracture adds to the load to be borne, leading to further fracture growth.
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Tensions in the Vecinity of the Break Tip
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The fracture propagates because its tip has an extremely small radius, which implies very high tension, as tension is proportional to the inverse of the square root of the radius.
The advancement of the fracture can be halted if, at some point, the radius increases, reducing the tension at its tip. This is achieved, for example, through material porosity or the insertion of inhomogeneities that act as stress concentration points.
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Model
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Parameters
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Calculations
Calculations
Variable 
Given Calculate 
Target : Equation To be used 
Equations
$ K_I =\sqrt{\displaystyle\frac{ F E }{ l }}$
K_I =sqrt( F * E / l )
$ \sigma_y =\displaystyle\frac{ K_i }{\sqrt{2 \pi r_p }}$
s_y = K_i /sqrt(2* pi * r_p )
$\sigma_x(r,\theta)=\displaystyle\frac{K_i}{\sqrt{2pi r}}\cos\displaystyle\frac{\theta}{2}\left(1-\sin\displaystyle\frac{\theta}{2}\sin\displaystyle\frac{3\theta}{2} \right)$
s_x(r,theta)=(K_i/sqrt(2 pi r))cos(theta/2)(1-sin(theta/2)sin(3theta/2))
$ \sigma_y =\displaystyle\frac{ K_i }{\sqrt{2 \pi r }}\cos\displaystyle\frac{\theta}{2}\left(1-\sin\displaystyle\frac{ \theta }{2}\sin\displaystyle\frac{3 \theta }{2}\right)$
s_y(r,theta)=(K_i/sqrt{2 pi r))cos(theta/2)(1 sin(theta/2)sin(3theta/2))
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Intensity Factor
Equation
The breaking stress is proportional to the intensity factor ($K_I$), which is in turn proportional to the square root of the force ($F$), the modulus of Elasticity ($E$), and the break length ($l$):
| $ K_I =\sqrt{\displaystyle\frac{ F E }{ l }}$ |
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Tension on the tip of the break
Equation
$\sigma_y(r_p,0)=\displaystyle\frac{K_i}{\sqrt{2\pi r_p}}$
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Stress parallel to the rupture
Equation
$\sigma_x(r,\theta)=\displaystyle\frac{K_i}{\sqrt{2\pi r}}\cos\displaystyle\frac{\theta}{2}\left(1-\sin\displaystyle\frac{\theta}{2}\sin\displaystyle\frac{3\theta}{2}\right)$
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Stress perpendicular to the Rupture
Equation
$\sigma_y(r,\theta)=\displaystyle\frac{K_i}{\sqrt{2\pi r}}\cos\displaystyle\frac{\theta}{2}\left(1+\sin\displaystyle\frac{\theta}{2}\sin\displaystyle\frac{3\theta}{2}\right)$
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