Dynamic Fracture

Abstract

The analysis of crack systems considered so far concerned only quasi-static situations in which the kinetic energy is relatively insignificant compared with the other energy terms and can be omitted. The crack was assumed either to be stationary or to grow in a controlled stable manner, and the applied loads varied quite slowly. The present chapter is devoted entirely to dynamically loaded stationary or growing cracks. In such cases rapid motions are generated in the medium and inertia effects become important.

Appendices

Examples

Example 8.1

The singular elastodynamic stress and displacement fields of a crack subjected to tearing-mode of deformation are expressed by

$$ \sigma_{13} = \frac{{K_{\text{III}} }}{{\sqrt {2\pi r} }}F\left( {\beta_{2} } \right)h\left( {\beta_{2} } \right),\sigma_{23} = \frac{{K_{\text{III}} }}{{\sqrt {2\pi r} }}\beta_{2} F\left( {\beta_{2} } \right)g\left( {\beta_{2} } \right) $$

(1a)

$$ u_{3} = - \frac{{K_{\text{III}} }}{\mu }\sqrt {\frac{2r}{\pi }} \left( {\cos^{2} \theta + \beta_{2}^{2} \sin^{2} \theta } \right)^{1/2} F\left( {\beta_{2} } \right)h\left( {\beta_{2} } \right) $$

(1b)

where the functions $g\left( {\beta_{2} } \right)$ and $h\left( {\beta_{2} } \right)$ are given by Eq. (8.43).

For an infinite plate with a central crack extending on both ends at constant speed the function $F\left( {\beta_{2} } \right)$ is given by

$$ F\left( {\beta_{2} } \right) = \frac{2}{\pi }K\left( {\frac{V}{{C_{2} }}} \right) $$

(2)

where K denotes the complete elliptic integral of the first kind with argument $V/C_{2}$.

Calculate the strain energy release rate G_III and plot its variation with $V/C_{2}$.

Solution

The strain energy release rate $G_{\text{III}}$ is calculated from Eq. (8.38) by choosing the integration path C as a circle of radius R centered at the crack tip, as

$$ G_{\text{III}} = R\int\limits_{ - \pi }^{\pi } {\left[ {\omega \cos \theta + \frac{1}{2}\rho V^{2} \left( {\frac{{\partial u_{3} }}{\partial x}} \right)^{2} \cos \theta - \left( {\sigma_{13} \cos \theta + \sigma_{23} \sin \theta } \right)\frac{{\partial u_{3} }}{\partial x}} \right]} {\text{d}}\theta . $$

(3)

The strain energy density function $\omega$ is given by

$$ \omega = \frac{1}{2\mu }\left( {\sigma_{13}^{2} + \sigma_{23}^{2} } \right). $$

(4)

Substituting the values of stresses $\sigma_{13}$ and $\sigma_{23}$ and displacement $u_{3}$ from Eq. (1) into Eqs. (3) and (4) and performing the integration in Eq. (3) we obtain

$$ \frac{{G_{\text{III}} }}{{G_{\text{III}}^{*} }} = \beta_{2} F^{2} \left( {\beta_{2} } \right) $$

(5)

where $G_{\text{III}}^{*}$ is the static value of G and can be obtained by putting V = 0 as (Eq. 4.27)

$$ G_{\text{III}}^{*} = G_{{{\text{III}}, V=0}} = \frac{{K_{\text{III}}^{2} }}{2\mu }. $$

(6)

The variation of $G_{\text{III}} /G_{\text{III}}^{*}$ versus $V/C_{2}$ is shown in Fig. 8.6. Note that the influence of crack speed on $G_{\text{III}}$ is negligible for small values of $V/C_{2}$, while at very high values of $V/C_{2}$, $G_\text{III}$ diminishes rapidly and becomes zero at $V=C_{2}$.

Fig. 8.6

Normalized strain energy release rate versus crack speed of an expanding crack under anti-plane shear

Example 8.2

A double cantilever beam (DCB) of height 2h with a crack of length a (Fig. 4.14) is made of a nonlinear material whose stress-strain relation is described by

$$ \sigma = \alpha \varepsilon (\varepsilon )^{(\beta - 1)/2} ,\quad 0 < \beta \le 1 $$

(1)

where $\alpha$ measures the stiffness of the material and is equal to the modulus of elasticity E for $\beta = 1$ (linear material). Equation (1) is shown in Fig. 8.7. The DCB is subjected to an end load P that remains constant during rapid crack propagation. Let a_o denote the initial crack length and P_c the load at crack propagation.

Fig. 8.7

Nonlinear stress-strain curves

Calculate the speed V and acceleration a_c of the crack and plot their variation versus $a_{0} /a$ for various values of $\beta$ [11].

Solution

The energy balance equation during crack growth (Eq. 4.1) takes the form

$$ P_{c} \left( {u - u_{c} } \right) = U(a) - U\left( {a_{0} } \right) + K + \gamma \left( {a - a_{0} } \right). $$

(2)

The left-hand side of Eq. (2) represents the work supplied to the system during growth of the crack from its initial length a₀ to the length a. u_c and u represent the load-point displacements at crack lengths a₀ and a, respectively. The right-hand side of Eq. (2) is composed of the term U(a)−U(a₀) that represents the change of strain energy, the term K that represents the kinetic energy and the term $\gamma (a - a_{0} )$ that represents the change of the surface energy.

From beam theory we have for the stress $\sigma$ and strain $\varepsilon$ at position x of the DCB

$$ \sigma = \frac{Pxy}{I}\left( {y^{2} } \right)^{(\beta - 1)/2} $$

(3)

$$ \varepsilon = y\left( {\frac{aI}{Px}} \right)^{ - 1/\beta } $$

(4)

where

$$ I = 2\int\limits_{0}^{h/2} {y^{\beta + 1} } {\text{d}}y. $$

(5)

The deflection y(x) of each beam of the DCB at position x during crack growth is calculated from beam theory as

$$ y(x) = \frac{\beta }{\beta + 1}\left( {\frac{{P_{c} }}{aI}} \right)^{1/\beta } \left[ {\frac{\beta }{2\beta + 1}x^{(2\beta + 1)/\beta } - a^{(\beta + 1)/\beta } x + \frac{\beta + 1}{2\beta + 1}a^{(2\beta + 1)/\beta } } \right] $$

(6)

and the deflection of each beam at the point of application of the load is

$$ u = y(0) = \frac{\beta }{2\beta + 1}\left( {\frac{{P_{c} }}{aI}} \right)^{1/\beta } a^{(2\beta + 1)/\beta } . $$

(7)

The strain energy of each beam is

$$ U(a) = \int\limits_{V} {\left( {\int\limits_{0}^{\varepsilon } {\sigma {\text{d}}\varepsilon } } \right)} {\text{d}}V. $$

(8)

Substituting the values of σ and ɛ from Eqs. (3) and (4) we obtain

$$ U(a) = \frac{\beta }{(\beta + 1)(2\beta + 1)}P_{c}^{(\beta + 1)/\beta } a^{(2\beta + 1)/\beta } (aI)^{ - 1/\beta } . $$

(9)

Equation (9) can be put in the form

$$ U(a) = \frac{{P_{c} u}}{\beta + 1}. $$

(10)

The kinetic energy K due to the motion of the beams along the y-direction is

$$ K = \frac{1}{2}\int\limits_{0}^{a} {\rho h\left( {\frac{{{\text{d}}y}}{{{\text{d}}t}}} \right)^{2} {\text{d}}x} . $$

(11)

Substituting the value of y = y(x) from Eq. (6) into Eq. (11) we obtain

$$ K = \frac{1}{6}\rho h\left( {\frac{{P_{c} }}{aI}} \right)^{2/\beta } a^{(2 + 3\beta )/\beta } V^{2} $$

(12)

where $V = {\text{d}}a/{\text{d}}t$ is the crack velocity.

Substituting the values of u, $u_{c} = u\left( {a = a_{0} } \right)$, U(a), U(a₀) and K from Eqs. (7), (9) and (12) into the energy balance Eq. (2) we obtain for the crack speed during crack growth

$$ \begin{aligned} & V^{2} = \frac{{6\beta^{2} }}{(2\beta + 1)(\beta + 1)}\frac{{(aI)^{1/\beta } }}{\rho h}P_{c}^{(\beta - 1)/\beta } a^{ - (\beta + 1)/\beta } \\ & \left[ {1 - \left( {\frac{{a_{0} }}{a}} \right)^{(2\beta + 1)/\beta } - n\left( {1 - \frac{{a_{0} }}{a}} \right)\left( {\frac{{a_{0} }}{a}} \right)^{(\beta + 1)/\beta } } \right]. \\ \end{aligned} $$

(13)

Here

$$ na_{0}^{(\beta + 1)/\beta } = \frac{(\beta + 1)(2\beta + 1)}{{\beta^{2} }}\gamma (aI)^{1/\beta } \left( {P_{c} } \right)^{ - (1 + \beta )/\beta } . $$

(14)

By differentiation from Eq. (13) with respect to time we obtain the crack acceleration $a_{c} = {\text{d}}V/{\text{d}}t$

$$ \begin{aligned} & a_{c} = \frac{{3\beta^{2} }}{(2\beta + 1)(\beta + 1)}\frac{{(aI)^{1/\beta } }}{\rho h}P_{c}^{(\beta - 1)/\beta } a^{ - (2\beta + 1)/\beta } \\ & \left[ {(1 - n)\left( {\frac{3\beta + 2}{\beta }} \right)\left( {\frac{{a_{0} }}{a}} \right)^{(2\beta + 1)/\beta } + \frac{\beta + 1}{\beta }\left[ {2n\left( {\frac{{a_{0} }}{a}} \right)^{(\beta + 1)/\beta } - 1} \right]} \right]. \\ \end{aligned} $$

(15)

As the initial acceleration of the crack $a_{c} = a_{c} \left( {a = a_{0} } \right)$ must be positive we obtain from Eq. (15)

$$ (1 - n)\left( {\frac{3\beta + 2}{\beta }} \right) + (2n - 1)\left( {\frac{\beta + 1}{\beta }} \right) > 0 $$

(16)

so that

$$ n < \frac{2\beta + 1}{\beta }. $$

Numerical results were obtained for $n = 0.99[(2\beta + 1)/\beta ]$ and various values of $\beta$. Figure 8.8 presents the variation of normalized crack speed V versus $a_{0} /a$ for $\beta = 0.2$, 0.4, 0.6, 0.8 and 1.0. The value $\beta = 1.0$ corresponds to a linear material. Note that crack speed increases during crack growth, from zero value at $a_{0} /a = 1$ reaches a maximum, and then decreases and becomes zero at $a_{0} /a \to 0$.

Fig. 8.8

Normalized crack velocity versus crack length at constant force

Figure 8.9 shows the variation of normalized crack acceleration $a_{c}$ versus $a_{0} /a$ for various values of $\beta$. The crack first accelerates, and then decelerates before coming to a complete stop at $a_{0} /a \to 0$. From Figs. 8.8 and 8.9 we observe that the crack travels more slowly as $\beta$ decreases. This should be expected as the material becomes stiffer with decreasing $\beta$.

Fig. 8.9

Nonnalized crack acceleration versus crack length at constant force

Example 8.3

An infinite strip of height 2 h with a semi-infinite crack is rigidly clamped along its upper and lower faces at y = ±h (Fig. 6.16). The upper and lower faces are moved in the positive and negative y-direction over distances $u_{0}$, respectively. Determine the dynamic stress intensity factor $K(t)$ during steady state crack propagation.

Solution

The dynamic stress intensity factor $K(t)$ is computed from Eq. (8.40) as

$$ K(t) = \left[ {\frac{EG}{1 + \nu }\frac{{4\beta_{1} \beta_{2} - \left( {1 + \beta_{2}^{2} } \right)^{2} }}{{\beta_{1} \left( {1 - \beta_{2}^{2} } \right)}}} \right]^{1/2} . $$

(1)

The strain energy release rate G is computed from Eq. (8.38) by taking the same integration path as in Example 6.1. Observing that $\partial u_{i} /\partial x = 0$ along the line BC we find that G for the dynamic problem is equal to its value for the static problem. Substituting the value of $G = J$ from Eq. (9) of Example 6.1 into Eq. (1) we obtain for $K(t)$

$$ K(t) = \left[ {\frac{{(1 - \nu )\left[ {4\beta_{1} \beta_{2} - \left( {1 + \beta_{2}^{2} } \right)^{2} } \right]}}{{(1 + \nu )^{2} (1 - 2\nu )\beta_{1} \left( {1 - \beta_{2}^{2} } \right)}}} \right]^{1/2} \frac{{Eu_{0} }}{{h^{1/2} }} $$

(2)

under conditions of plane strain, and

$$ K(t) = \left[ {\frac{{\left[ {4\beta_{1} \beta_{2} - \left( {1 + \beta_{2}^{2} } \right)^{2} } \right]}}{{(1 + \nu )^{2} (1 - \nu )\beta_{1} \left( {1 - \beta_{2}^{2} } \right)}}} \right]^{1/2} \frac{{Eu_{0} }}{{h^{1/2} }} $$

(3)

under conditions of generalized plane stress.

Example 8.4

A crack of length 20 mm propagates in a large steel plate under a constant stress of 400 MPa. The dynamic toughness of the material $K_{ID}$ can be expressed by the following empirical equation

$$ K_{{{\text{I}}D}} = \frac{{K_{{{\text{I}}A}} }}{{1 - \left( {\frac{V}{{V_{l} }}} \right)^{m} }} $$

(1)

where $K_{{{\text{I}}A}}$ is the arrest toughness of the material, $V_{l}$ is the limiting crack speed and m is an empirical parameter. Using Rose’s approximation for the dynamic stress intensity factor determine the speed of crack during propagation. Take: C₁ = 5940 m/s, C₂ = 3220 m/s, C_R = 2980 m/s; K_IA = 100 MPa $\sqrt {\mathfrak{m}}$, m = 2, $V_{l} = 1600\,\,{\text{m/}}\text{s}$.

Solution

Crack propagation is governed by the following Eq. (8.41)

$$ K_{1} (t) = K_{1D} $$

(2)

of the material.

Using Rose’s approximation for $K_{\text{I}} (t)$ we have (Eq. 8.26)

$$ K_{\text{I}} (t) = \sigma \sqrt {\pi a}\, k(V) $$

(3)

or

$$ K_{\text{I}} (t) = 800\,{\text{MPa}}\sqrt {\pi \times \left( {10 \times 10^{ - 3} } \right){\text{m}}} k(V) = 141.8\,k(V)\,{\text{MPa}}\sqrt {\text{m}} . $$

(4)

$K(V)$ is computed from Eq. (8.28), where h is given by Eq. (8.29).

We have

$$ h = \frac{2}{5940}\left( {\frac{3220}{2980}} \right)^{2} \left( {1 - \frac{3220}{5940}} \right)^{2} = 0.0824 \times 10^{ - 3} {\text{s}}/{\text{m}} $$

(5)

and

$$ k(V) = \left( {1 - \frac{V}{2980}} \right)\left( {1 - 0.0824 \times 10^{ - 3} V} \right)^{ - 1/2} . $$

(6)

$K_{{{\text{I}}D}}$ is computed as

$$ K_{{{\text{I}}D}} = \frac{{100\,{\text{MPa}}\sqrt {\text{m}} }}{{1 - \left( {\frac{V}{1600}} \right)^{2} }}. $$

(7)

Substituting the values of $K_{\text{I}} (t)$ and $K_{{{\text{I}}D}}$ from Eqs. (3) and (7) into Eq. (2) we obtain an equation involving the unknown crack speed. From a numerical solution of this equation we obtain

$$ V = 571.3\,\,{\text{m}}/{\text{s}} . $$

(8)

Problems

• 8.1 Show that the strain energy release rate G defined by Eq. (8.38) is path independent.

• 8.2 The double cantilever beam of Example 8.2 is subjected to an end displacement u_c which remains constant during rapid crack propagation. Calculate the speed and acceleration of the crack and plot their variation versus $a_{0} /a$ for various values of $\beta$. Take $n = 0.6(2\beta + 1)/\beta$.

• 8.3 The double cantilever beam of Example 8.2 is made of a linear material. Calculate the speed of the crack and plot its variation versus $a/a_{0}$ when the applied load P is kept constant during crack propagation. Take various values of the ratio $n = P/P_{0}$, where $P_{0}$ is the load at crack initiation.

• 8.4 The double cantilever beam of Example 8.2 is made of a linear material. Calculate the speed of the crack and plot its variation versus $a/a_{0}$ when the applied displacement u is kept constant during crack propagation. Take various values of the ratio $n = u/u_{0}$, where $P_{0}$ is the displacement at crack initiation.

• 8.5 The Yoffé crack model considers a crack of fixed length propagating with constant speed in an infinite plate under a uniform tensile stress normal to the crack line. It is assumed that the crack retains its original length during propagation by resealing itself at the trailing end. For this problem the strain energy release rate G is given by

  $$ \frac{G}{{G^{*} }} = \frac{4}{\kappa + 1}\beta_{1} \left( {1 - \beta_{2}^{2} } \right)\left[ {4\beta_{1} \beta_{2} - \left( {1 + \beta_{2}^{2} } \right)^{2} } \right]F^{2} \left( {\beta_{1} ,\beta_{2} } \right) $$

  where G* is the strain energy release rate at zero crack speed. G* and F are given by

  $$ \begin{aligned} & G^{*} = \frac{\kappa + 1}{8\mu }K^{2} \\ & F = \left[ {4\beta_{1} \beta_{2} - \left( {1 + \beta_{2}^{2} } \right)^{2} } \right]^{ - 1} . \\ \end{aligned} $$

  Plot the variation of $G/G^{*}$ versus crack speed $V/C_{2}$ under conditions of plane strain for various values of Poisson’s ratio v.

• 8.6 For Problem 8.5 show that G becomes infinite at a crack speed V computed from the following equation

  $$ (\kappa + 1)\left( {\frac{V}{{C_{2} }}} \right)^{4} \left[ {\left( {\frac{V}{{C_{2} }}} \right)^{2} - 8} \right] + 8(\kappa + 5)\left( {\frac{V}{{C_{2} }}} \right)^{2} - 32 = 0. $$

  This equation gives the Rayleigh wave speed $C_{R} \left( {C_{R} < C_{2} < C_{1} } \right)$.

• 8.7 Plot the variation of normalized Rayleigh wave speed $C_{R} /C_{2}$ versus Poisson’s ratio $\nu$ under conditions of plane stress and plane strain according to Equation of Problem 8.6. Also plot the variation of $C_{1} /C_{2}$ versus v.

• 8.8 Calculate the dilatational, C₁, the shear, C₁, and the Rayleigh, C_R, wave speeds for (a) steel with E = 210 GPa, ρ = 7800 kg/m³, v = 0.3 and (b) copper with E = 130 GPa, ρ = 8900 kg/m³, v = 0.34.

• 8.9 Broberg considered a crack in an infinite plate extending on both ends with constant speed. For this case the strain energy release rate G is computed from Equation of Problem 8.5 where the function $F\left( {\beta_{1} ,\beta_{2} } \right)$ is given by

  $$ \begin{aligned} F\left( {\beta_{1} ,\beta_{2} } \right) & = \beta_{1} \left[ {\left[ {\left( {1 + \beta_{2}^{2} } \right)^{2} - 4\beta_{1}^{2} \beta_{2}^{2} } \right]K\left( {\beta_{1} } \right) - 4\beta_{1}^{2} \left( {1 - \beta_{2}^{2} } \right)K\left( {\beta_{2} } \right)} \right. \\ & \left. { - \left[ {4\beta_{1}^{2} + \left( {1 + \beta_{2}^{2} } \right)^{2} } \right]E\left( {\beta_{1} } \right) + 8\beta_{1}^{2} E\left( {\beta_{2} } \right)} \right]^{ - 1} \\ \end{aligned} $$

  and K and E are complete elliptic integrals of the first and second kind respectively.

  They are given by

  $$ K(k) = \frac{\pi }{2}\left[ {1 + \left( {\frac{1}{2}} \right)^{2} k^{2} + \left( {\frac{3}{2.4}} \right)^{2} k^{4} + \left( {\frac{3.5}{2.4.6}} \right)^{2} k^{6} + \ldots } \right] $$
  $$ E(k) = \frac{\pi }{2}\left[ {1 - \left( {\frac{1}{{2^{2} }}} \right)^{2} k^{2} - \left( {\frac{{3^{2} }}{{2^{2} \cdot 4^{2} }}} \right)^{2} \frac{{k^{4} }}{3} - \left( {\frac{{3^{2} \cdot 5^{2} }}{{2^{2} \cdot 4^{2} \cdot 6^{2} }}} \right)^{2} \frac{{k^{6} }}{5} - \cdots } \right]. $$

  Plot the variation of $G/G^{*}$ versus crack speed $V/C_{2}$ under conditions of plane strain for various values of Poisson’s ratio v. Note that G becomes zero as V tends to the Rayleigh wave speed C_R.

• 8.10 An infinite strip of height 2h with a semi-infinite crack is rigidly clamped along its upper and lower faces at y = ± h (Fig. 6.16). The upper and lower faces are moved in the positive and negative z-direction over distances ω₀, respectively. Determine the dynamic stress intensity factor $K_{\text{III}} (t)$ during steady state crack propagation. Note that $K_{\text{III}} (t)$ is related to the strain energy release rate G_III by

  $$ G_{\text{III}} = \frac{{K_{\text{III}}^{2} (t)}}{{2\mu \beta_{2} }}. $$

• 8.11 Show that the strain energy release rate $G_{\text{III}}$ for mode-l dynamic crack propagation can be determined by the following equation

  $$ G_{1} = \mathop {\lim }\limits_{\delta \to 0} \frac{1}{\delta }\int\limits_{0}^{\delta } {\sigma_{22} } \left( {x_{1} ,0} \right)u_{2} \left( {\delta - x_{1} ,\pi } \right){\text{d}}x_{1} $$

  where $\delta$ is a segment of crack extension along the x₁-axis. Note that this expression of $G_{\text{I}}$ is analogous to Eq. (4.19) of the static crack.

• 8.12 Establish a relationship between the dynamic, G, and static, G(0), energy release rates using Rose’s approximation to k(V). Plot the variation of $G/G(0)$ versus $V/C_{R}$ for a thick steel plate with: C₁ = 5940 m/s, C₂ = 3220 m/s and C_R = 2980 m/s.

• 8.13 A crack of length 2a propagates in a large steel plate under a constant stress of 300 MPa. The dynamic toughness of the material $K_{{{\text{I}}D}}$ can be expressed by the following empirical equation

  $$ K_{{{\text{I}}D}} = \frac{{K_{{{\text{I}}A}} }}{{1 - \left( {\frac{V}{{V_{l} }}} \right)^{m} }} $$

  where $K_{{{\text{I}}A}}$ is the arrest toughness of the material, $V_{l}$ is the limiting crack speed and m is an empirical parameter. Plot the variation of crack speed V versus initial crack length for 10 mm < 2a < 100 mm, using Rose’s approximation for the dynamic stress intensity factor. Take C₁ = 5940 m/s, C₂ = 3220 m/s, C_R = 2980 m/s; $K_{\text{IA}} = 100\,{\text{MPa}}\sqrt {\text{m}}$, $m = 2$, $V_{l} = 1600\,{\text{m/s}}$.