Fatigue and Environment-Assisted Fracture

Abstract

It was first realized in the middle of the nineteenth century that engineering components and structures often fail when subjected to repeated fluctuating loads whose magnitude is well below the critical load under monotonic loading.

Appendices

Examples

Example 9.1

A large plate contains a crack of length 2a₀ and is subjected to a constant-amplitude tensile cyclic stress normal to the crack which varies between 100 and 200 MPa. The following data were obtained: for 2a₀ = 2 mm it was found that N = 20,000 cycles were required to grow the crack to 2a_f = 2.2 mm, while for 2a = 20 mm it was found that N = 1000 cycles were required to grow the crack to 2a_f = 22 mm. The critical stress intensity factor is $K_{c} = 60\,{\text{MPa}}\sqrt m$. Determine the constants in the Paris (Eq. (9.3)) and Forman (Eq. (9.4)) equations.

Solution

The stress intensity factor range ∆K is calculated as

$$ \Delta K = \Delta \sigma \sqrt {\pi a_{0} } . $$

(1)

For the crack of initial length 2a₀ = 2 mm we have

$$ \Delta K = 100\sqrt {\pi \times \left( {1 \times 10^{ - 3} } \right)} = 5.60\,{\text{MPa}}\,\sqrt {\text{m}} $$

(2)

and for the crack of initial length 2a₀ = 20 mm we have

$$ \Delta K = 100\sqrt {\pi \times \left( {10 \times 10^{ - 3} } \right)} = 17.72\,{\text{MPa}}\,\sqrt {\text{m}} . $$

(3)

The crack growth rate da/dN is calculated for the crack of initial length 2a₀ = 2 mm as

$$ \frac{{{\text{d}}a}}{{{\text{d}}N}} = \frac{{(1.1 - 1.0) \times 10^{ - 3} \,{\text{m}}}}{{20000\,{\text{cycle}}}} = 5 \times 10^{ - 9} \,m/{\text{cycle}} $$

(4)

and for the crack of initial length 2a₀ = 20 mm as

$$ \frac{{{\text{d}}a}}{{{\text{d}}N}} = \frac{{(11 - 10) \times 10^{ - 3} \,{\text{m}}}}{{1000\,{\text{cycle}}}} = 10^{ - 6} \,m/{\text{cycle}} . $$

(5)

• (a) Paris equation (Eq. (9.3)). Taking the logarithm of both sides of Eq. (9.3) we obtain

$$ \log \frac{{{\text{d}}a}}{{{\text{d}}N}} = \log C + m\log \Delta K. $$

(6)

Introducing into Eq. (6) the data of our problem we obtain the following two equations for the determination of the two unknowns C and m

$$ \begin{aligned} & \log \left( {5 \times 10^{ - 9} } \right) = \log C + m\,\log 5.6 \\ & \log \left( {10^{ - 6} } \right) = \log C + m\,\log 17.72 \\ \end{aligned} $$

(7)

or

$$ \begin{aligned} & - 8.30 = \log C + 0.748\,m \\ & - 6.00 = \log C + 1.248\,m. \\ \end{aligned} $$

(8)

From Eqs. (7) and (8) we obtain

$$ C = 1.82 \times 10^{ - 12} MN^{ - 4.6} m^{7.9} /{\text{cycle}},\quad m = 4.6. $$

(9)

• (b) Forman equation (Eq. (9.4)). Taking the logarithm of both sides of Eq. (9.4) we obtain

$$ \log \left[ {\left[ {(1 - R)K_{c} - \Delta K} \right]\frac{{{\text{d}}a}}{{{\text{d}}N}}} \right] = \log C + n\log \Delta K. $$

(10)

Introducing into Eq. (10) the data of our problem with $R = K_{\hbox{min} } /K_{\hbox{max} } = \sigma_{\hbox{min} } /\sigma_{\hbox{max} } = 100/200 = 0.5$ we obtain the following two equations for the determination of the two unknowns C and n

$$ \begin{aligned} & \log \left[ {(0.5 \times 60 - 5.6) \times 5 \times 10^{ - 9} } \right] = \log C + n\log 5.6 \\ & \log \left[ {(0.5 \times 60 - 17.72) \times 10^{ - 6} } \right] = \log C + n\log 17.72 \\ \end{aligned} $$

(11)

or

$$ \begin{aligned} & - 6.914 = \log C + 0.748\,n \\ & - 4.911 = \log C + 1.248\,n. \\ \end{aligned} $$

(12)

From Eq. (12) we obtain

$$ C = 1.229\,MN^{ - 3.3} m^{5.95} /{\text{cycle}},\quad n = 4.006. $$

(13)

Example 9.2

A large thick plate contains a crack of length 2a₀ = 10 mm and is subjected to a constant-amplitude tensile cyclic stress normal to the crack which varies between σ_min = 100 MPa and σ_max = 200 MPa. The critical stress intensity factor is $K_{{{\text{I}}c}} = 60\,{\text{MPa}}\sqrt m$ and fatigue crack growth is governed by the equation

$$ \frac{{{\text{d}}a}}{{{\text{d}}N}} = 0.42 \times 10^{ - 11} (\Delta K)^{3} $$

(1)

where da/dN is expressed in m/cycle and ∆K in ${\text{MPa}}\sqrt m$. Plot a curve of crack growth, a, versus number of cycles, N, up to the point of crack instability.

If a lifetime of 10⁶ cycles is required for the plate, discuss the options the designer has for an improved lifetime.

Solution

The critical crack length a_c at instability is calculated from equation

$$ K_{\text{I}} = K_{{{\text{I}}c}} $$

(2)

which becomes

$$ 200\sqrt {\pi a_{c} } = 60 $$

(3)

and gives

$$ a_{c} = 28.65\,{\text{mm}}. $$

(4)

The curve of crack growth, a, versus number of cycles N_c up to the crack length a_c is calculated by Eq. (9.3) with m = 3, C = 0.42 × 10⁻¹¹MN⁻³m^5.5/cycle and $\Delta K_{0} = 100\sqrt {\pi \times \left( {5 \times 10^{ - 3} } \right)} = 12.53\,{\text{MPa}}\sqrt {\text{m}}$. Thus, for the crack to grow from its initial length a₀ = 5 mm to a length a = 7 mm the number of cycles N required is given by

$$ N = \frac{{2 \times \left( {5 \times 10^{ - 3} } \right)}}{{1 \times \left( {0.42 \times 10^{ - 11} } \right)(12.53)^{3} }}\left[ {1 - \left( {\frac{5}{7}} \right)^{0.5} } \right] = 187{,}412\,{\text{cycles}} . $$

In a similar manner the number of cycles required for a crack of length a₀ = 7 mm to grow to a length a = 9 mm is given by

$$ N = \frac{{2 \times \left( {7 \times 10^{ - 3} } \right)}}{{1 \times \left( {0.42 \times 10^{ - 11} } \right)[100\sqrt {\pi \times \left( {7 \times 10^{ - 3} } \right)} ]^{3} }}\left[ {1 - \left( {\frac{7}{9}} \right)^{0.5} } \right] = 120{,}696\,{\text{cycles}}. $$

The above procedure is repeated for crack growth of steps 2 mm. Results are shown in Table 9.1 and are plotted in Fig. 9.8.

Table 9.1 Fatigue crack growth calculations

Fig. 9.8

Crack length versus number of cycles

The total number of cycles required to propagate a crack from 5 mm to 28.65 mm is calculated as

$$ N_{c} = \frac{{2 \times \left( {5 \times 10^{ - 3} } \right)}}{{1 \times \left( {0.42 \times 10^{ - 11} } \right) \times (12.53)^{3} }}\left[ {1 - \left( {\frac{5}{28.65}} \right)^{0.5} } \right] = 704,697\,{\text{cycles}} $$

which is close to the value 704,308 obtained previously (Table 9.1).

If a lifetime of 10⁶ cycles is required for the plate the designer may make the following changes:

• (a) Employ a different metal with higher K_Ic, so as to increase the critical crack length a_c at instability.

• (b) Reduce the maximum value of the applied stress σ_max.

• (c) Reduce the stress range ∆σ.

• (d) Improve the inspection so as to reduce the assumed initial crack length. If for example the initial crack length was reduced from 2a₀ = 10 mm to 2a₀ = 6 mm, the lifetime of the plate would be increased to 1,056,097 cycles, which is more than the required number of 10⁶ cycles.

Example 9.3

A plate of width 2b = 50 mm contains a central crack of length 2a₀ = 10 mm, and is subjected to a constant-amplitude tensile cyclic stress normal to the crack which varies between σ_min = 100 MPa and σ_max = 200 MPa. The crack growth is dictated by Eq. (1) of Example 9.2. Calculate the number of cycles required for the crack to propagate to a length 2a = 20 mm. The stress intensity factor K_I of the plate is given by

$$ K_{\text{I}} = \sigma (\pi a)^{1/2} \left[ {\frac{2b}{\pi a}\tan \left( {\frac{\pi a}{2b}} \right)} \right]^{1/2} . $$

(1)

Solution

In this case the function $f(a) = [(2b/\pi a)\tan (\pi a/2b)]^{1/2}$ of Eq. (9.10) is not equal to its initial value f(a₀), but depends on the crack length. Thus, Eq. (9.14) cannot be applied for the calculation of the lifetime N of the plate. A step-by-step procedure is adapted where the number of cycles for the crack to grow at intervals of 2 mm is calculated. We have for the stress intensity factor fluctuations ∆K₁₀ and ∆K₁₂ for cracks of length 10 mm and 12 mm under stress $\Delta \sigma = \sigma_{\hbox{max} } - \sigma_{\hbox{min} } = 100\,{\text{MPa}}$:

$$ \begin{aligned} \Delta K_{10} & = 100\left( {\pi \times 5 \times 10^{ - 3} } \right)^{1/2} \left[ {\frac{50}{\pi \times 5}\tan \left( {\frac{\pi \times 5}{50}} \right)} \right]^{1/2} = 12.75\,{\text{MPa}}\sqrt m \\ \Delta K_{12} & = 100\left( {\pi \times 6 \times 10^{ - 3} } \right)^{1/2} \left[ {\frac{50}{\pi \times 6}\tan \left( {\frac{\pi \times 6}{50}} \right)} \right]^{1/2} = 14.07\,{\text{MPa}}\sqrt m . \\ \end{aligned} $$

da/dN for a crack of length 2a = 10 mm and 2a = 12 mm is calculated from Eq. (1) of Example 9.2 as

$$ \begin{aligned} \left( {\frac{{{\text{d}}a}}{{{\text{d}}N}}} \right)_{10} & = 0.42 \times 10^{ - 11} \times (12.75)^{3} = 8.7 \times 10^{ - 9} m/{\text{cycle}} \\ \left( {\frac{{{\text{d}}a}}{{{\text{d}}N}}} \right)_{12} & = 0.42 \times 10^{ - 11} \times (14.07)^{3} = 11.70 \times 10^{ - 9} m/{\text{cycle}}. \\ \end{aligned} $$

The mean value of da/dN is then calculated as

$$ \frac{{{\text{d}}a}}{{{\text{d}}N}} = \frac{{8.7 \times 10^{ - 9} + 11.70 \times 10^{ - 9} }}{2} = 10.2 \times 10^{ - 9} {\text{m}}/{\text{cycle}} $$

and the number of cycles dN to propagate a crack from length 2a₀ = 10 mm to 2a = 12 mm is calculated as

$$ {\text{d}}N = \frac{{1 \times 10^{ - 3} }}{{10.2 \times 10^{ - 9} }} = 98{,}039\,{\text{cycles}}. $$

Crack growth calculations are arranged in Table 9.2.

Table 9.2 Fatigue crack growth calculations

Thus, the number of cycles required to propagate the crack of initial length 2a₀ = 10 mm to a length of 2a_c = 20 mm is N_c = 315,351 cycles.

If the variation of f(a) during crack growth were ignored, N_c would be calculated from Eq. (9.14) as

$$ N_{c} = \frac{{2 \times \left( {5 \times 10^{ - 3} } \right)}}{{1 \times \left( {0.42 \times 10^{ - 11} } \right) \times (12.75)^{3} }}\left[ {1 - \left( {\frac{5}{10}} \right)^{0.5} } \right] = 336{,}457\,{\text{cycles}}. $$

Example 9.4

A double cantilever beam (DCB) (Fig. 4.14) with height 2h = 60 mm, thickness B = 20 mm and an initial crack of length a₀ = 200 mm is subjected to a tensile load which varies between P_min = 5 kN and P_max = 10 kN. The critical stress intensity factor of the beam is $K_{{{\text{I}}C}} = 100\,{\text{MPa}}\sqrt m$ and crack growth is dictated by equation

$$ \frac{{{\text{d}}a}}{{{\text{d}}N}} = 5 \times 10^{ - 15} (\Delta K)^{4} $$

(1)

where da/dN is expressed in m/cycle and ∆K in ${\text{MPa}}\sqrt m$. Calculate the critical crack length a_c for instability and the fatigue lifetime of the beam.

Solution

The critical crack length a_c at instability is calculated from equation

$$ K_{\text{I}} = K_{{{\text{I}}c}} $$

(2)

where the stress intensity factor K_I for the DCB is given by Eq. (7) of Example 4.2.

We have for P_max = 10 kN

$$ K_{\text{I}} = \sqrt {\frac{12}{{\left( {30 \times 10^{ - 3} m} \right)^{3} }}} \frac{{\left( {10 \times 10^{ - 3} MN} \right)a_{c} }}{{20 \times 10^{ - 3} m}} = 333a_{c} \,{\text{MPa}}/\sqrt m . $$

(3)

From Eqs. (2) and (3) with $K_{{{\text{I}}c}} = 100\,{\text{MPa}}\sqrt m$ we obtain

$$ a_{c} = 300\,{\text{mm}}. $$

(4)

The fatigue life N_c of the beam is calculated from Eq. (9.3) where

$$ \Delta K = \sqrt {\frac{12}{{h^{3} }}} \frac{(\Delta P)a}{B} $$

(5)

We have for $\Delta P = 5\,{\text{kN}}$

$$ \Delta K = \sqrt {\frac{12}{{\left( {30 \times 10^{ - 3} } \right)^{3} }}} \frac{{\left( {5 \times 10^{ - 3} } \right)a}}{{20 \times 10^{ - 3} }} = 166.7\,a. $$

(6)

Equation (9.3) with C = 5 × 10⁻¹⁵ and m = 4 becomes

$$ {\text{d}}N = \frac{{{\text{d}}a}}{{\left( {5 \times 10^{ - 15} } \right)(166.7a)^{4} }} $$

(7)

or

$$ {\text{d}}N = \frac{{2.59 \times 10^{5} {\text{d}}a}}{{a^{4} }}. $$

(8)

Equation (8) gives

$$ N_{c} = \frac{{2.59 \times 10^{5} }}{3}\left( {\frac{1}{{a_{0}^{3} }} - \frac{1}{{a_{c}^{3} }}} \right) $$

(9)

or

$$ N_{c} = \frac{{2.59 \times 10^{5} }}{3}\left( {\frac{1}{{0.2^{3} }} - \frac{1}{{0.3^{3} }}} \right) = 76 \times 10^{6} \,{\text{cycles}}. $$

(10)

Example 9.5

A large plate with an initial crack of length 2a₀ is subjected to a series of cyclic stress amplitudes $\Delta \sigma_{i} (i = 1,2, \ldots ,n)$ normal to the crack. Assume that the final crack length at instability 2a_f is the same for all stress amplitudes $\Delta \sigma_{i}$. If fatigue crack growth is governed by equation

$$ \frac{{{\text{d}}a}}{{{\text{d}}N}} = C(\Delta K)^{2} $$

(1)

show that

$$ \sum\limits_{i = 1}^{n} {\frac{{N_{i} }}{{\left( {N_{i} } \right)_{f} }}} = 1 $$

(2)

where N_i is the number of cycles required to grow the crack from 2a_i−1 to 2a_i and (N_i)_f is the total number of cycles required to grow the crack from its initial length 2a₀ to its final length 2_af at instability.

Equation (2) is known as the Miner rule of fatigue crack growth under variable load amplitudes.

Solution

From Eq. (1) we obtain for the number of cycles N₁ required to grow the crack from its initial length 2a₀ to a length 2a₁

$$ {\text{d}}N_{1} = \frac{{{\text{d}}a}}{{C\left( {\Delta \sigma_{1} \sqrt {\pi a} } \right)^{2} }} $$

(3)

and by integration we have

$$ N_{1} = \frac{1}{{\pi C\left( {\Delta \sigma_{1} } \right)^{2} }}\int\limits_{{a_{0} }}^{{a_{1} }} {\frac{{{\text{d}}a}}{a}} = \frac{1}{{\pi C\left( {\Delta \sigma_{1} } \right)^{2} }}\ln \left( {\frac{{a_{1} }}{{a_{0} }}} \right). $$

(4)

In a similar manner we obtain for the number of cycles (N₁)_f required to grow the crack from its initial length 2a₀ to the final length 2a_f at instability

$$ \left( {N_{1} } \right)_{f} = \frac{1}{{\pi C\left( {\Delta \sigma_{1} } \right)^{2} }}\ln \left( {\frac{{a_{f} }}{{a_{0} }}} \right). $$

(5)

From Eqs. (4) and (5) we obtain

$$ \frac{{N_{1} }}{{\left( {N_{1} } \right)_{f} }} = \ln \left( {\frac{{a_{1} }}{{a_{0} }}} \right)/\ln \left( {\frac{{a_{f} }}{{a_{0} }}} \right). $$

(6)

In a similar manner we obtain

$$ \frac{{N_{2} }}{{\left( {N_{2} } \right)_{f} }} = \ln \left( {\frac{{a_{2} }}{{a_{1} }}} \right)/\ln \left( {\frac{{a_{f} }}{{a_{0} }}} \right) $$

(7)

and

$$ \frac{{N_{i} }}{{\left( {N_{i} } \right)_{f} }} = \ln \left( {\frac{{a_{i} }}{{a_{i - 1} }}} \right)/\ln \left( {\frac{{a_{f} }}{{a_{0} }}} \right) $$

(8)

and so on.

From Eqs. (6) to (8) we obtain

$$ \begin{aligned} & \frac{{N_{1} }}{{\left( {N_{1} } \right)_{f} }} + \frac{{N_{2} }}{{\left( {N_{2} } \right)_{f} }} + \cdots + \frac{{N_{i} }}{{\left( {N_{i} } \right)_{f} }} + \cdots \frac{{N_{n} }}{{\left( {N_{n} } \right)_{f} }} \\ & = \left[ {\ln \left( {\frac{{a_{1} }}{{a_{0} }}} \right) + \ln \left( {\frac{{a_{2} }}{{a_{1} }}} \right) + \cdots + \ln \left( {\frac{{a_{i} }}{{a_{i - 1} }}} \right) + \cdots + \ln \left( {\frac{{a_{n} }}{{a_{n - 1} }}} \right)} \right]/\ln \left( {\frac{{a_{f} }}{{a_{0} }}} \right) \\ \end{aligned} $$

(9)

or

$$ \sum\limits_{i = 1}^{n} {\frac{{N_{i} }}{{\left( {N_{i} } \right)_{f} }}} = \ln \left[ {\left( {\frac{{a_{1} }}{{a_{0} }}} \right)\left( {\frac{{a_{2} }}{{a_{1} }}} \right) \ldots \left( {\frac{{a_{n} }}{{a_{n - 1} }}} \right) \ldots \left( {\frac{{a_{n} }}{{a_{n - 1} }}} \right)} \right]/\ln \left( {\frac{{a_{f} }}{{a_{0} }}} \right) $$

(10)

or for a_n = a_f

$$ \sum\limits_{i = 1}^{n} {\frac{{N_{i} }}{{\left( {N_{i} } \right)_{f} }}} = 1 $$

(11)

which is Eq. (2).

Example 9.6

A large thick plate contains a crack of length 2a₀ = 20 mm and is subjected to a series of triangular stress sequences normal to the crack as shown in Fig. 9.9. The stress varies between $\sigma_{\hbox{min} } = 0$ and $\sigma_{\hbox{max} } = 200\,{\text{MPa}}$ and it takes 1000 cycles in the triangle to reach $\sigma_{\hbox{max} }$. The critical stress intensity factor is $K_{{{\text{I}}c}} = 100\,{\text{MPa}}\sqrt m$ and fatigue crack growth is governed by equation

Fig. 9.9

Triangular cyclic stress profile

$$ \frac{{{\text{d}}a}}{{{\text{d}}N}} = 10^{ - 12} (\Delta K)^{3} $$

(1)

where da/dN is expressed in m/cycle and ∆K in ${\text{MPa}}\sqrt m$.

Calculate the number of triangular stress sequences (N_c)₁ required to grow the crack to instability. Compare (N_c)₁ to the number of cycles (N_c)₂ required to grow the crack to instability when the plate is subjected to a constant amplitude stress cycle between the stresses $\sigma_{\hbox{min} } = 0$ and $\sigma_{\hbox{max} } = 200\,{\text{MPa}}$.

Solution

The root-mean-square value of the stress intensity factor range ∆K_rms is calculated from Eq. (9.16). We have for the stress σ at N cycles in the triangular stress sequence

$$ \sigma = \Delta \sigma \frac{N}{{N_{p} }} $$

(2)

so that

$$ \Delta K = \Delta \sigma \sqrt {\pi a} \frac{N}{{N_{p} }}. $$

(3)

Since we have a large number of cycles (N_p = 1000) in each triangular stress sequence ∆K_rms can be calculated by integration as

$$ \left( {\Delta K_{\text{rms}} } \right)^{2} = \frac{1}{{N_{p} }}\int\limits_{0}^{{N_{p} }} {(\Delta K)^{2} } {\text{d}}N $$

(4)

and substituting the value of ∆K from Eq. (3) we obtain

$$ \left( {\Delta K_{\text{rms}} } \right)^{2} = \frac{{(\Delta \sigma \sqrt {\pi a} )^{2} }}{{N_{p} }}\int\limits_{0}^{{N_{p} }} {\frac{{N^{2} }}{{N_{p}^{2} }}} {\text{d}}N $$

(5)

or

$$ \left( {\Delta K_{\text{rms}} } \right)^{2} = \frac{{(\Delta \sigma )^{2} \pi a}}{3}. $$

(6)

The critical crack length a_c at instability is calculated from equation

$$ K_{\text{I}} = K_{{{\text{I}}c}} $$

(7)

which becomes

$$ 200\sqrt {\pi a_{c} } = 100 $$

(8)

and gives

$$ a_{c} = 79.6\,{\text{mm}}. $$

(9)

Equation (1) with ∆K = ∆K_rms gives

$$ \left( {N_{c} } \right)_{1} = \int\limits_{0.01}^{0.0796} {\frac{{{\text{d}}a}}{{10^{ - 12} (200)^{3} (\pi a)^{1.5} (1/3)^{1.5} }}} $$

(10)

or

$$ \left( {N_{c} } \right)_{1} = 116{,}645 \times 2\left( {0.01^{ - 0.5} - 0.0796^{ - 0.5} } \right) = 1.5 \times 10^{6} \,{\text{cycles}} . $$

(11)

The number of cycles (N_c)₂ required to grow the crack to instability when the plate is subjected to a constant amplitude stress cycle between the stress $\sigma_{\hbox{min} } = 0$ and $\sigma_{\hbox{max} } = 200\,{\text{MPa}}$ is calculated from Eq. (1) as

$$ \left( {N_{\text{c}} } \right)_{2} = \int\limits_{0.01}^{0.0796} {\frac{{{\text{d}}a}}{{10^{ - 12} (200)^{3} (\pi a)^{1.5} }}} $$

(12)

or

$$ \left( {N_{c} } \right)_{2} = 0.29 \times 10^{6} \,{\text{cycles}} $$

(13)

which is equal to (N_c)₁ divided by 3^1.5.

Example 9.7

The contoured double cantilever beam (DCB) of Example 4.3 is subjected to a fatigue load P fluctuating between P_min and P_max according to Paris’s law

$$ \frac{\text{da}}{{{\text{d}}N}} = C \Delta K_{I}^{m} $$

(1)

Determine the constants C and m when the DCB is subjected to two steps 1 and 2 with N = 10⁶ cycles in each step, and

Step 1: P_min = 1 kN, P_max = 4 kN

Step 2: P_min = 2 kN, P_max = 6 kN

The crack growth after the tests is 10 mm for step 1 and 20 mm for step 2.

The height of the DCB at the tip of the crack tip at position x₀ = 15 mm is h₀ = 5 mm and the thickness is B = 5 mm.

Solution

The compliance of the contoured DCB is (Eq. (2) of Example 4.3 with n = 2/3)

$$ {\text{C}} = \frac{24\,a}{{E\,B\,h_{0}^{3} }} $$

(2)

The stress intensity factor K_I is determined from Eq. (4.34), for conditions of plane stress, as

$$ K_{I}^{2} = \frac{{EM^{2} }}{2B}\left( {\frac{{{\text{d}}C}}{{{\text{d}}a}}} \right) = \frac{{12M^{2} }}{{B^{2} h_{0}^{3} }} = \frac{{12P^{2} x_{0}^{2} }}{{B^{2} h_{0}^{3} }} $$

(3)

where M = Px₀ is the moment at the crack tip due to the applied load P. We have from Eq. (3)

$$ K_{I} = \frac{{2\sqrt 3 Px_{0} }}{{Bh_{0} \sqrt {h_{0} } }} $$

(4)

We have from Eq. (1) for steps 1 and 2

$$ \frac{{\Delta a_{1} }}{\Delta N} = C\left[ {\frac{{2\sqrt 3 x_{0} }}{{Bh_{0} \sqrt {h_{0} } }} \cdot \left( {P_{\hbox{max} 1} - P_{\hbox{min} 1} } \right)} \right]^{m} $$

(5)

$$ \frac{{\Delta a_{2} }}{\Delta N} = C\left[ {\frac{{2\sqrt 3 x_{0} }}{{Bh_{0} \sqrt {h_{0} } }}\left( {P_{\hbox{max} 2} - P_{\hbox{min} 2} } \right)} \right]^{m} $$

(6)

From Eqs. (5) and (6), we have

$$ \ln \left( {\frac{{\Delta a_{1} }}{{\Delta a_{2} }}} \right) = m\ln \left( {\frac{{P_{\hbox{max} 1} - P_{\hbox{min} 1} }}{{P_{\hbox{max} 2} - P_{\hbox{min} 2} }}} \right) $$

(7)

and

$$ \ln \left( {\frac{10}{20}} \right) = m\ln \left( {\frac{4 - 1}{6 - 2}} \right) $$

(8)

so that

$$ m = 2.41 $$

(9)

Then we obtain from Eq. (5)

$$ \frac{0.01}{{10^{6} }} = C\left[ {\frac{2\sqrt 3 \times (0.015)}{{(0.005) \times (0.005)\sqrt {0.005} }}\left( {4 \times 10^{3} - 1 \times 10^{3} } \right)} \right]^{2.41} $$

(10)

and

$$ C = 7.13 \times 10^{ - 28} $$

(11)

Equations (9) and (11) give the values of the constants m and C of the Paris law.

Example 9.8

A double cantilever beam of height 2h = 200 mm (Fig. 4.14) under plane stress is subjected to a given displacement u₀ = 1 mm at the end of each of its two beams. The stress corrosion crack growth follows the equation

$$ \frac{{{\text{d}}a}}{{{\text{d}}t}} = CK_{I}^{m} \left( {a\,in\,meters,t\,in\,sec,K_{I} \,in\,MPa\,m^{1/2} } \right) $$

(1)

Calculate the time for the crack to grow from a length a₀ = 200 mm to a length a_f = 300 mm.

It is given:

$$ C = 10^{ - 10} ,m = 3,E = 210\,{\text{GPa}} $$

Solution

We have from Eq. (6) of Example 4.2 (with u = 2u₀)

$$ K_{I} = \frac{{\sqrt {3h^{3} } Eu_{0} }}{{2a^{2} }} $$

(2)

Then, Eq. (1) takes the form

$$ \frac{{{\text{d}}a}}{{{\text{d}}t}} = C\left( {\frac{{\sqrt {3h^{3} } Eu_{0} }}{{2a^{2} }}} \right)^{m} $$

(3)

Solving for t, we obtain

$$ t(a) = \frac{1}{c}\left( {\frac{{\sqrt {3h^{3} } Eu_{0} }}{2}} \right)^{ - m} \int\limits_{{a_{0} }}^{a} {a^{2m} } {\text{d}}a = \frac{1}{c}\left( {\frac{{\sqrt {3h^{3} } Eu_{0} }}{2}} \right)^{ - m} \frac{1}{2m + 1}\left( {a^{2m + 1} - a_{0}^{2m + 1} } \right) $$

(4)

From Eq. (4), we obtain (E in MPa)

$$ \begin{aligned} t(a) &= \frac{1}{{10^{ - 10} }}\left( {\frac{{\sqrt {3(0.1)^{3} } \times \left( {210 \times 10^{3} } \right) \times (0.001)}}{2}} \right)^{ - 3} \frac{1}{2 \times 3 + 1}\left( {0.3^{2 \times 3 + 1} - 0.2^{2 \times 3 + 1} } \right)\\ &= 1.55 \times 10^{3} { \sec } \end{aligned} $$

(5)

Equation (5) gives the time for the crack to grow from a length a₀ = 200 mm to a length a_f = 300 mm.

Example 9.9

A semi-circular surface crack of radius a₀ in a semi-infinite body grows under stress corrosion cracking according to

$$ \frac{{{\text{d}}a}}{{{\text{d}}t}} = CK_{I}^{m} ,K_{I} = 0.66\,\sigma \,\sqrt {\pi a} $$

(1)

The crack remains semi-circular during growth. The applied stress increases linearly with time (σ = σ₀t, where σ₀ is a constant).

Determine the relation between the critical time to fracture t_c and the critical stress σ_c at fracture. Fracture takes place at a K_IC critical stress intensity factor.

Solution

We have from Eq. (1)

$$ \frac{{{\text{d}}a}}{{{\text{d}}t}} = C\left( {0.66\,\sigma \,\sqrt {\pi a} } \right)^{m} = C\left( {0.66\,\sigma_{0} t\,\sqrt {\pi a} } \right)^{m} $$

(2)

By separation of variables Eq. (2) becomes

$$ a^{ - m/2} {\text{d}}a = C\left( {0.66\,\sigma_{0} \sqrt \pi } \right)^{m} t^{m} {\text{d}}t $$

(3)

and integrating both parts, we obtain

$$ \frac{1}{1 - m/2}\left( {a_{c}^{1 - m/2} - a_{0}^{1 - m/2} } \right) = \frac{1}{m + 1}C\left( {0.66\,\sigma_{0} \sqrt \pi } \right)^{m} t_{c}^{m + 1} $$

(4)

where t_c is the critical time to fracture.

From Eq. (4), we obtain for t_c

$$ t_{c} = \left[ {\frac{2(m + 1)}{{C\left( {0.66\,\sigma_{0} \sqrt \pi } \right)^{m} (m - 2)}}\left( {\alpha_{0}^{{\left( {1 - \dfrac{m}{2}} \right)}} - \alpha_{c}^{{\left( {1 - \dfrac{m}{2}} \right)}} } \right)} \right]^{1/(m + 1)} $$

(5)

The critical crack length a_c is obtained from

$$ K_{IC} = 0.66\,\sigma_{c} \sqrt {\pi a_{c} } $$

(6)

as

$$ a_{c} = \left( {\frac{{K_{Ic} }}{{0.66\sqrt \pi \sigma_{c} }}} \right)^{2} $$

(7)

Introducing the value of a_c into Eq. (5), we obtain for the critical time t_c to fracture

$$ t_{c} = \left[ {\frac{2(m + 1)}{{C\left( {0.66\,\sigma_{0} \sqrt \pi } \right)^{m} (m - 2)}}\left( {a_{0}^{{\left( {1 - \dfrac{m}{2}} \right)}} - \left( {\frac{{K_{Ic} }}{{0.66\sqrt \pi \sigma_{c} }}} \right)^{(2 - m)} } \right)} \right]^{1/(m + 1)} $$

(8)

Equation (8) gives the relation between the critical time to fracture t_c and the critical stress σ_c at fracture.

Example 9.10

The double cantilever beam of Example 4.2 (Fig. 4.14) with an initial crack of length a₀ is subjected to a cyclic load which varies between 0 and P_max. The growth rate increases with the number of cycles N according to

$$ {\text{d}}a/{\text{d}}N = kN^{m} $$

(1)

The material follows the Paris fatigue law

$$ {\text{d}}a/{\text{d}}N = C\Delta K_{I}^{n} $$

(2)

Determine P_max as a function of the number of cycles, N, and the initial crack length a₀. Conditions of generalized plane stress dominate.

Solution

The stress intensity factor, according to Example 4.2, is

$$ K_{I} = \sqrt {\frac{12}{{h^{3} }}} \frac{Pa}{B} $$

(3)

Equation (2) (with ΔK_I = K_I) becomes

$$ {\text{d}}a/{\text{d}}N = C\left( {\sqrt {\frac{12}{{h^{3} }}} \frac{{P_{\hbox{max} } a}}{B}} \right)^{\text{n}} $$

(4)

By integrating Eq. (1), we obtain

$$ a = a_{0} + \frac{{kN^{m + 1} }}{m + 1} $$

(5)

From Eqs. (1) and (4), we obtain

$$ kN^{m} = C\left( {\sqrt {\frac{12}{{h^{3} }}} \frac{{P_{\hbox{max} } a}}{B}} \right)^{\text{n}} $$

(6)

and

$$ P_{\hbox{max} } = B\sqrt {\frac{{h^{3} }}{12}} \frac{1}{{\left( {a_{0} + \dfrac{{kN^{m + 1} }}{m + 1}} \right)}}\left( {\frac{{kN^{m} }}{C}} \right)^{{1/{\text{n}}}} $$

(7)

Equation (7) gives the value of P_max as a function of the number of cycles, N, and the initial crack length, a₀.

Problems

• 9.1. A large thick plate contains a crack of length 2a₀ and is subjected to a constant amplitude cyclic stress normal to the crack which varies between $\sigma_{\hbox{max} }$ and $\sigma_{\hbox{min} }$.

  Fatigue crack growth is governed by the equation

  $$ \frac{{{\text{d}}a}}{{{\text{d}}N}} = C\left[ {(\Delta K)^{2} - \left( {\Delta K_{\text{th}} } \right)^{2} } \right] $$

  where C is a material parameter and ∆K_th is the threshold value of ∆K.

  Show that the number of cycles N_c required to grow the crack to instability is given by

  $$ N_{c} = \frac{1}{{C\pi (\Delta \sigma )^{2} }}\ln \frac{{\left( {K_{\text{IC}} /\sigma_{\hbox{max} } } \right)^{2} - \left( {\Delta K_{\text{th}} /\Delta \sigma } \right)^{2} }}{{\pi a_{0} - \left( {\Delta K_{\text{th}} /\Delta \sigma } \right)^{2} }} $$

  where K_Ic is the critical stress intensity factor and $\Delta \sigma = \sigma_{\hbox{max} } - \sigma_{\hbox{min} }$.

• 9.2. Show that, if m = 4 in Eq. (9.3), the time it takes for a crack in a large plate to quadruple its length is equal to 1.5 times the time it takes to double its length.

• 9.3. A crack grows at a rate (da/dN)₁ = 8.84 × 10⁻⁷ m/cycle when the stress intensity factor amplitude is $(\Delta K)_{1} = 50\,{\text{MPa}}\sqrt m$ and at a rate (da/dN)₂ = 4.13 × 10⁻⁵ when $(\Delta K)_{2} = 150\,{\text{MPa}}\sqrt m$. Determine the parameters C and m in Paris equation (Eq. (9.3)).

• 9.4. A plate of width 2b = 40 mm contains a center crack of length 2a₀ and is subjected to a constant-amplitude tensile cyclic stress which varies between 100 and 200 MPa. The following data were obtained: For 2a₀ = 2 mm N = 20,000 cycles were required to grow the crack to 2a_f = 2.2 mm, while for 2a₀ = 20 mm N = 1000 cycles were needed to grow the crack to 2a_f = 22 mm. The critical stress intensity factor is $K_{1c} = 60\,{\text{MPa}}\sqrt m$. Determine the constants in the Paris (Eq. (9.3)) and Forman (Eq. (9.4)) equations.

• 9.5. A large thick plate contains a crack and is subjected to a cyclic stress normal to the crack which varies between 0 and 400 MPa. Determine the maximum allowable crack length the plate can withstand when $K_{1c} = 90\,{\text{MPa}}\sqrt m$. If the initial crack had a length 6 mm calculate the number of loading cycles the plate can withstand, when C = 10⁻²⁰ MN⁻³ m^5.5/cycle, m = 3.

• 9.6. A large thick plate contains a crack of length 8 mm and is subjected to a constant amplitude tensile cyclic stress normal to the crack which varies between 200 and 400 MPa. Plot a curve of crack growth versus number of cycles up to the point of crack instability. The material parameters are as follows: $K_{1c} = 90\,{\text{MPa}} \sqrt m$, C = 10⁻¹² MN⁻⁴ m⁷/cycle, m = 4.

• 9.7. A large thick plate contains an edge crack of length 1 mm and is subjected to a constant-amplitude tensile cyclic stress normal to the crack which varies between 100 and 300 MPa. Which of the following three materials gives the longest lifetime for the plate.

Material	$K_{Ic} ({\text{MPa}}\sqrt m )$	C	m
A	55	5 × 10−14 MN−4m7/cycle	4.0
B	70	2 × 10−12 MN−3.5 m6.25/cycle	3.5
C	90	3 × 10−11 MN−3m5.5/cycle	3.0

• 9.8. A large thick plate contains a crack of length 2a₀ and is subjected to a constant amplitude tensile cyclic stress normal to the crack with maximum stress σ_max = 200 MPa and stress range ∆σ = σ_max − σ_min. The fatigue crack growth is governed by equation

  $$ \frac{{{\text{d}}a}}{{{\text{d}}N}} = 3.9 \times 10^{ - 14} (\Delta K)^{3.7} $$

  where da/dN is expressed in m/cycle and ∆K in ${\text{MPa}}\sqrt m$. Determine the fatigue lifetime of the plate for the following conditions:

  • (a) 2a₀ = 2 mm, $K_{1c} = 24\,{\text{MPa}}\sqrt m$, ∆σ = 50 MPa

  • (b) 2a₀ = 2 mm, $K_{1c} = 24\,{\text{MPa}}\sqrt m$, ∆σ = 100 MPa

  • (c) 2a₀ = 2 mm, $K_{1c} = 44\,{\text{MPa}}\sqrt m$, ∆σ = 50 MPa

  • (d) 2a₀ = 1 mm, $K_{1c} = 24\,{\text{MPa}}\sqrt m$, ∆σ = 50 MPa

  • (e) 2a₀ = 1 mm, $K_{1c} = 44\,{\text{MPa}}\sqrt m$, ∆σ = 100 MPa.

  Discuss the influence on fatigue lifetime of the initial crack length 2a₀, the fracture toughness K_Ic and the stress range ∆σ.

• 9.9. A large thick plate contains a crack of length 2a₀ = 2 mm and is subjected to a constant amplitude tensile cyclic stress normal to the crack which varies between 0 and 100 MPa. The fatigue crack growth is governed by equation of Problem 9.6. The material of the plate has been heat treated to the following conditions with the corresponding values of K_Ic.

  $$ \begin{array}{*{20}l} {\text{ Condition }} \hfill & {\text{A}} \hfill & {\text{B}} \hfill & {\text{C}} \hfill \\ {K_{{{\text{I}}c}} \left( {{\text{MPa}}\,\sqrt m } \right)} \hfill & {24} \hfill & {30} \hfill & {44} \hfill \\ \end{array} $$

  For each of the three material conditions, determine the fatigue lifetime N_f of the plate for crack lengths varying between 2a₀ and 2a_f, where the latter length corresponds to crack instability. Plot the variation of N_c versus a (a₀ < a < a_f) for the three material conditions.

• 9.10. A large thick plate has two equal cracks of length a = 1 mm emanating from both sides of a hole with radius R = 10 mm. The plate is subjected to a constant amplitude tensile cyclic stress normal to the crack which varies between 100 and 200 MPa and the cracks grow at equal rates. Calculate the number of loading cycles the plate can withstand when $K_{1c} = 90\,{\text{MPa}}\sqrt m$, and crack growth is governed by equation

  $$ \frac{{{\text{d}}a}}{{{\text{d}}N}} = 0.42 \times 10^{ - 14} (\Delta K)^{3} $$

  where da/dN is expressed in m/cycle and ∆K in ${\text{MPa}}\sqrt m$. The stress intensity factor at the crack tip is given by

  $$ K_{\text{I}} = k\sigma \sqrt {\pi a} $$

  where k depends on the a/R according to the following table.

  a/R	0	0.1	0.2	0.3	0.4	0.5	0.6	0.8	1.0	1.5	2.0	3.0	∞
  k	3.39	2.73	2.41	2.15	1.96	1.83	1.71	1.58	1.45	1.29	1.21	1.14	1.00

• 9.11. A thick plate of width b = 50 mm contains an edge crack of length 10 mm and is subjected to a constant amplitude tensile cyclic stress normal to the crack which varies between 100 and 200 MPa. Fatigue crack growth is dictated by equation of Problem 9.10 and $K_{1c} = 55\,{\text{MPa}}\sqrt m$. Calculate the number of cycles the plate can withstand.

• 9.12. A cylindrical pressure vessel has a radius R = 1 m and thickness t = 40 mm and contains a long axial surface crack of depth a = 2 mm. The vessel is subjected to internal pressure p which varies between 0 and 200 MPa. Calculate the number of loading cycles the vessel can withstand. $K_{1c} = 55\,{\text{MPa}}\sqrt m$ and fatigue crack growth is dictated by equation of Problem 9.10.

• 9.13. An ASTM three-point bend specimen of span S = 30 cm, width W = 8 cm, thickness B = 4 cm with a crack of length a₀ = 3.5 em is subjected to a constant amplitude cyclic load which varies between 30 and 50 kN. Calculate the number of cycles required to grow the crack to a length of a_f = 4.0 cm and plot a curve of crack length versus number of cycles.

• 9.14. An ASTM compact tension specimen of width W = 12 cm, thickness B = 6 cm with a crack of length a₀ = 5 cm is subjected to a constant amplitude cyclic load which varies between 20 kN and 40 kN. Calculate the number of cycles required to grow the crack to a length a_f = 5 cm and plot a curve of crack length versus number of cycles.

• 9.15. A thick plate contains a semicircular surface crack and is subjected to a constant amplitude tensile cyclic stress normal to the crack which varies between 200 and 400 MPa. The fatigue crack growth is governed by Equation of Problem 9.8. Assume that during fatigue the crack grows in a self-similar manner. Calculate the number of cycles required to grow the crack from an initial radius a₀ to a final radius a_f for the following cases (a) a₀ = 2 mm, a_f = 20 mm, (b) a₀ = 2 mm, a_f = 60 mm, (c) a₀ = 4 mm, a_f = 20 mm and (d) a₀ = 4 mm, a_f = 60 mm. The stress intensity factor along the crack is given by

$$ K_{\text{I}} = \frac{2.24}{\pi }\sigma \sqrt {\pi a} . $$

• 9.16. A large thick plate contains a crack of length 2 mm and is subjected to a cyclic stress normal to the crack which varies between 0 and σ. Crack growth is governed by Paris’ equation with C = 5 × 10⁻¹⁴ MN⁻⁴ m⁷/cycle and m = 4. The critical stress intensity factor is $K_{1c} = 100\,{\text{MPa}}\sqrt m$. Calculate the value of σ if it is required that the plate withstand 10⁸ cycles.

• 9.17. A large plate contains a central crack of length 2 mm and is subjected to a cyclic stress normal to the crack which varies between 100 and 300 MPa. Crack growth is governed by Paris’ equation with C = 5 × 10⁻¹⁵ MN⁻⁴ m⁷/cycle and m = 4. Calculate the crack length after 10⁶ cycles.

• 9.18. A large plate contains a central crack of length 2 mm and is subjected to a cyclic wedge load which varies between 1 and 3 MN (Fig. 2.8, Example 2.2). Crack growth is governed by Paris’ equation with C = 5 × 10⁻¹⁵ MN⁻⁴ m⁷/cycle and m = 4. Calculate the crack length after 10³ cycles.

• 9.19. A double cantilever beam (Fig. 4.14) with height 2h = 40 mm, thickness B = 20 mm and a crack of length a = 100 mm is subjected to a tensile cyclic load which varies between 0 and 8 kN. Crack growth is governed by Paris’ equation with C = 5 × 10⁻¹⁵ MN⁻⁴ m⁷/cycle and m = 4. Calculate the crack length after 10⁶ cycles.

• 9.20. A large plate contains an edge crack of length 1 mm and is subjected to a tensile cyclic stress which varies between 0 and σ. Determine the value of σ so that the crack does not grow in fatigue. $\Delta K_{\text{th}} = 7\,{\text{MPa}}\sqrt m$.

• 9.21. In Problem 9.18 the threshold value, $\Delta K = 5\,{\text{MPa}}\sqrt m$. Calculate the number of cycles to crack arrest.

• 9.22. A large plate with a crack of length 10 mm was tested to stress corrosion in salt water. The time to failure for an applied stress normal to the crack σ = 100, 200 and 300 MPa was 2000, 50 and 2 h. Estimate K_ISSC and calculate the amount of crack growth for each applied stress. $K_{1c} = 55\,{\text{MPa}}\sqrt m$.