Critical Stress Intensity Factor Fracture Criterion

Abstract

When a solid is fractured, work is performed to create new material surfaces in a thermodynamically irreversible manner. In Griffith’s theory of ideally brittle materials, the work of fracture is spent in the rupture of cohesive bonds. The fracture surface energy $\gamma$, which represents the energy required to form a unit of new material surface, corresponds to a normal separation of atomic planes. For the fracture of polycrystals, however, the work required for the creation of new surfaces should also include: dissipation associated with nonhomogeneous slip within and between the grains; plastic and viscous deformation; and possible phase changes at the crack surfaces. The energy required for the rupture of atomic bonds is only a small portion of the dissipated energy in the fracture process. There are situations where the irreversible work associated with fracture is confined to a small process zone adjacent to the crack surfaces, while the remaining material is deformed elastically. In such a case the various work terms associated with fracture may be lumped together in a macroscopic term R (resistance to fracture) which represents the work required for the creation of a unit of new material surface. R may be considered as a material parameter. The plastic zone accompanying the crack tip is very small and the state of affairs around the crack tip can be described by the stress intensity factor .

Appendices

Examples

Example 5.1

A three-point bend specimen was tested according to the ASTM E399 procedure . The 0.2% offset yield stress of the material is $\sigma_{Y} = 1200\,{\text{MPa}}$ and the modulus of elasticity is $E = 210\,{\text{GPa}}$. The specimen was tested at a loading rate of $100\,{\text{kN}}/\hbox{min}$ and a type-I load-displacement record was obtained. A chevron starter notch was machined and the specimen was subjected to 30,000 cycles at P_max = 45 kN and P_min = 0. The final stage of fatigue crack growth was conducted for 50,000 cycles at P_max = 30 kN and P_min = 0. The specimen dimensions were measured as

$$ \begin{aligned} & S = 30\,{\text{cm}},W = 8\,{\text{cm}},B = 4\,{\text{cm}} \\ & a_{1} = 3.996\,{\text{cm}} \\ & a_{2} = 4.007\,{\text{cm}} \\ & a_{3} = 3.997\,{\text{cm}} \\ & a({\text{surface}}) = 3.915\,{\text{cm}} \\ & a({\text{surface}}) = 3.952\,{\text{cm}}. \\ \end{aligned} $$

The maximum load and the secant load of the test record were measured as P_max = 86 kN and P_Q = 80 kN.

Determine $K_{\text{Ic}}$.

Solution

We first determine a conditional $K_{\text{Ic}} (K_{Q} )$. We have:

$$ \begin{aligned} & a = \frac{{a_{1} + a_{2} + a_{3} }}{3} = \frac{3.996 + 4.007 + 3.997}{3} = 4.0\,{\text{cm}} \\ & \frac{{a - a({\text{surface}})}}{a} = \frac{4.0 - 3.915}{4.0} = 0.02 < 0.1 \\ & \frac{{a - a({\text{surface}})}}{a} = \frac{4.0 - 3.952}{4.0} = 0.012 < 0.1 \\ & \frac{{P_{\hbox{max} } }}{{P_{Q} }} = \frac{{86\,{\text{kN}}}}{{80\,{\text{kN}}}} = 1.075 < 1.10 \\ \end{aligned} $$

K_Q is determined from Eq. (5.9). This equation can be put in the form

$$ K_{Q} = \frac{PS}{{BW^{3/2} }}f(a/W) $$

(1)

$$ f(a/W) = \frac{{3\left( {\frac{a}{W}} \right)^{1/2} \left[ {1.99 - \frac{a}{W}\left( {1 - \frac{a}{W}} \right)\left( {2.15 - 3.93\frac{a}{W} + 2.7\frac{{a^{2} }}{{W^{2} }}} \right)} \right]}}{{2\left( {1 + 2\frac{a}{W}} \right)\left( {1 - \frac{a}{W}} \right)^{3/2} }}. $$

(2)

To facilitate calculation of K_Q, values of f(a/W) are tabulated for specific values of a/W, according to ASTM

a/W	f (a/W)	a/W	f (a/W)
0.450	2.29	0.500	2.66
0.455	2.32	0.505	2.70
0.460	2.35	0.510	2.75
0.465	2.39	0.515	2.79
0.470	2.43	0.520	2.84
0.475	2.46	0.525	2.89
0.480	2.50	0.530	2.94
0.485	2.54	0.535	2.99
0.490	2.58	0.540	3.04
0.495	2.62	0.545	3.09
–	–	0.550	3.14

For our case (a/W) = 0.5. Thus, f(a/W) = 2.66. The specimen is loaded at a K_I rate, $\Delta K_{\text{I}} /\Delta t$, as

$$ \frac{{\Delta K_{\text{I}} }}{\Delta t} = \frac{(\Delta P)S}{{BW^{3/2} }}f(a/W) $$

or

$$ \frac{{\Delta K_{\text{I}} }}{\Delta t} = \frac{{(100\,{\text{kN}}/60{\text{s}}) \times (0.3\,{\text{m}})}}{{(0.04\,{\text{m}}) \times (0.08\,{\text{m}})^{3/2} }} \times 2.66 = 1.47\,{\text{MPa}}\sqrt {\text{m}} /{\text{s}}. $$

We have

$$ 0.55 < \frac{{\Delta K_{\text{I}} }}{\Delta t} < 2.75\,{\text{MPa}}\sqrt {\text{m}} /{\text{s}}. $$

K_Q is determined as

$$ K_{Q} = \frac{{(80\,{\text{kN}}) \times (0.3\,{\text{m}})}}{{(0.04\,{\text{m}}) \times (0.08\,{\text{m}})^{3/2} }} \times 2.66 = 70.5\,{\text{MPa}}\sqrt {\text{m}} . $$

We have

$$ B,a,W > 2.5\left( {\frac{{K_{Q} }}{{\sigma_{Y} }}} \right)^{2} = 2.5\left( {\frac{{70.5\,{\text{MPa}}\sqrt {\text{m}} }}{{1200\,{\text{MPa}}}}} \right)^{2} = 0.86 \times 10^{ - 2} \,{\text{m}}. $$

The maximum stress intensity factor during the first stage of fatigue growth of precrack is

$$ K_{f(\hbox{max} )} = \frac{{(45\,{\text{kN}}) \times (0.3\,{\text{m}})}}{{(0.04\,{\text{m}}) \times (0.08\,{\text{m}})^{3/2} }} \times 2.66 = 39.6\,{\text{MPa}}\sqrt {\text{m}} $$

so that

$$ K_{f(\hbox{max} )} < 0.6K_{Q} . $$

During the final stage of fatigue growth of precrack K_I is

$$ K_{1} = \frac{{(30\,{\text{kN}}) \times (0.3\,{\text{m}})}}{{(0.04\,{\text{m}}) \times (0.08\,{\text{m}})^{3/2} }} \times 2.66 = 26.4\,{\text{MPa}}\sqrt {\text{m}} $$

and

$$ \frac{{K_{\text{I}} }}{E} = \frac{26.4}{210}\frac{{{\text{MPa}}\sqrt {\text{m}} }}{\text{GPa}} = 0.13 \times 10^{ - 3} \sqrt {\text{m}} < 0.32 \times 10^{ - 3} \sqrt {\text{m}} . $$

The above results indicate that all requirements of a valid $K_{\text{Ic}}$ test according to ASTM E399 standard are satisfied and the conditional value K_Q is equal to $K_{\text{Ic}}$, that is

$$ K_{\text{Ic}} = K_{Q} = 70.5\,{\text{MPa}}\sqrt {\text{m}} . $$

Example 5.2

Figure 5.15 shows the load-displacement record of a compact tension specimen tested according to ASTM E399 procedure to determine $K_{\text{Ic}}$. The 0.2% offset yield stress of the material is 800 MPa. The specimen dimensions were measured as: W = 12 cm, B = 5 cm, a = 6 cm. Determine $K_{\text{Ic}}$.

Fig. 5.15

Determination of P_Q on the load-displacement record of the compact tension specimen of Example 5.2

Solution

A conditional $K_{\text{Ic}} (K_{Q} )$ is first determined from Eq. (5.10). This equation can be put in the form

$$ K_{\text{I}} = \frac{{P_{Q} }}{{BW^{1/2} }}f(a/W) $$

(1)

$$ f(a/W) = \frac{{\left( {2 + \frac{a}{W}} \right)\left[ {0.886 + 0.64\frac{a}{W} - 13.32\left( {\frac{a}{W}} \right)^{2} + 14.72\left( {\frac{a}{W}} \right)^{3} - 5.6\left( {\frac{a}{W}} \right)^{4} } \right]}}{{\left( {1 - \frac{a}{W}} \right)^{3/2} }} $$

(2)

To facilitate calculation of K_Q, values of f(a/W) are tabulated for specific values of a/W, according to ASTM

a/W	f (a/W)	a/W	f (a/W)
0.450	8.34	0.500	9.66
0.455	8.46	0.505	9.81
0.460	8.58	0.510	9.96
0.465	8.70	0.515	10.12
0.470	8.83	0.520	10.29
0.475	8.96	0.525	10.45
0.480	9.09	0.530	10.63
0.485	9.23	0.535	10.80
0.490	9.37	0.540	10.98
0.495	9.51	0.545	11.17
–	–	0.550	11.36

We have a/W = 0.5. Thus, f(a/W)= 9.66. To determine P_Q in Eq. (1) we draw a secant line through the origin of the load-displacement record of Fig. 5.15 with slope equal to 0.95 of the initial slope of the record. We obtain

$$ P_{Q} = 156\,{\text{kN}} $$

and

$$ \frac{{P_{\hbox{max} } }}{{P_{Q} }} = \frac{{169\,{\text{kN}}}}{{156\,{\text{kN}}}} = 1.08 < 1.10 $$

K_Q is determined as

$$ K_{Q} = \frac{{(156\,{\text{kN}})}}{{(0.05\,{\text{m}})(0.12\,{\text{m}})^{1/2} }} \times 9.66 = 87\,{\text{MPa}}\sqrt {\text{m}} . $$

We have

$$ B,a,W > 2.5\left( {\frac{{K_{Q} }}{{\sigma_{Y} }}} \right)^{2} = 2.5\left( {\frac{{87\,{\text{MPa}}\sqrt {\text{m}} }}{{800\,{\text{MPa}}}}} \right)^{2} = 2.96 \times 10^{ - 2} \,{\text{m}} $$

and therefore

$$ K_{\text{Ic}} = K_{Q} = 87\,{\text{MPa}}\sqrt {\text{m}} . $$

Example 5.3

The stress intensity factor of a crack of length 2a in a plate subjected to a tension field is given by

$$ K_{1} = \sigma (a)f(a/b),\quad f(a/b) > 0. $$

(1)

The crack growth resistance curve of the plate is expressed by

$$ K_{R} = m\left[ {a_{0} + \left( {a - a_{0} } \right)^{1/2} } \right] $$

(2)

where a₀ is the initial crack length and m is a constant.

Determine the critical stress $\sigma_{c}$ and the critical length 2a_c at the point of instability.

Solution

At the point of unstable crack growth we have from Eq. (5.11a) and the third Eq. (4.40)

$$ K_{\text{I}} = K_{R} , $$

(3)

$$ \frac{{\partial K_{\text{I}} }}{\partial a} = \frac{{\partial K_{R} }}{\partial a}. $$

(4)

We have from Eqs. (1) and (2)

$$ \frac{{\partial K_{\text{I}} }}{\partial a} = \frac{\partial \sigma }{\partial a}f + \sigma \frac{\partial f}{\partial a},\sigma = \sigma (a), \, f = f(a/b) $$

(5)

$$ \frac{{\partial K_{R} }}{\partial a} = \frac{m}{{2\left( {a - a_{0} } \right)^{1/2} }}. $$

(6)

Equation (4) renders

$$ a - a_{0} = \left( {\frac{m}{2}} \right)^{2} \left( {\frac{\partial \sigma }{\partial a}f + \sigma \frac{\partial f}{\partial a}} \right)^{2} . $$

(7)

Equation (3) becomes

$$ \sigma f = m\left[ {a + \left( {a - a_{0} } \right)^{1/2} } \right]. $$

(8)

Equations (7) and (8) give the critical stress $\sigma_{c}$ and the critical crack length 2a_c at instability.

Example 5.4

Determine the stability condition for a double cantilever beam (DCB) (Fig. 4.14) subjected to an end load in a soft (load-controlled) or a hard (displacement-controlled) testing machine.

Solution

The compliance of the DCB is (Example 4.2)

$$ C = \frac{u}{P} = \frac{{2\eta a^{3} }}{3EI},\quad I = \frac{{Bh^{3} }}{12}. $$

(1)

From Eq. (1) we obtain

$$ \frac{{{\text{d}}C}}{{{\text{d}}A}} = \frac{{3\eta a^{2} }}{{8EB^{2} h^{3} }},\quad \frac{{{\text{d}}^{2} C}}{{\partial A^{2} }} = \frac{3\eta a}{{4EB^{3} h^{3} }}. $$

(2)

For stability in a soft (load-controlled) testing machine Eq. (5.18) becomes

$$ \frac{1}{R}\frac{{{\text{d}}R}}{{{\text{d}}A}} \ge \frac{2}{A}. $$

(3)

From Eq. (1) we obtain

$$ \frac{{{\text{d}}C^{ - 1} }}{{{\text{d}}A}} = - \frac{{24Eh^{3} }}{{\eta a^{4} }},\quad \frac{{{\text{d}}^{2} C^{ - 1} }}{{{\text{d}}A^{2} }} = \frac{{96Eh^{3} }}{{\eta a^{5} B}}. $$

(4)

For stability in a hard (displacement-controlled) testing machine Eq. (5.19) becomes

$$ \frac{1}{R}\frac{{{\text{d}}R}}{{{\text{d}}A}} \ge - \frac{4}{A}. $$

(5)

Equations (3) and (5) express the stability conditions in a soft or a hard testing machine. Note that Eq. (5) is always satisfied for constant R. Equations (3) and (5) show that stability is achieved more easily with a hard than with a soft testing machine.

Example 5.5

A cylindrical pressure vessel with closed ends has a radius R = 1 m and thickness t = 40 mm and is subjected to internal pressure p. The vessel must be designed safely against failure by yielding (according to the von Mises yield criterion) and fracture. Three steels with the following values of yield stress $\sigma_{Y}$ and fracture toughness $K_{\text{Ic}}$ are available for constructing the vessel.

Steel	$\sigma_{Y} ({\text{MPa}})$	$K_{\text{Ic}} ({\text{MPa}}\,\sqrt {\text{m}} )$
A: 4340	860	100
B: 4335	1300	70
C: 350 Maraging	1550	55

Fracture of the vessel is caused by a long axial surface crack of depth a. The vessel should be designed with a factor of safety S = 2 against yielding and fracture.

For each steel:

• (a) Plot the maximum permissible pressure p_c versus crack depth a_c;

• (b) Calculate the maximum permissible crack depth a_c for an operating pressure p = 12 MPa;

• (c) Calculate the failure pressure p_c for a minimum detectable crack depth a = 1 mm.

Solution

Just as in Example 2.5, a material element of the vessel is subjected to a hoop $\sigma_{\theta }$ and a longitudinal $\sigma_{z}$ stress given by

$$ \sigma_{\theta } = \frac{pR}{t},\sigma_{z} = \frac{pR}{2t}. $$

(1)

The von Mises yield criterion expressed by Eq. (1.3) takes the form

$$ \sigma_{\theta }^{2} - \sigma_{\theta } \sigma_{z} + \sigma_{z}^{2} = \left( {\sigma_{Y} /S} \right)^{2} ,\quad S = 2. $$

(2)

From Eqs. (1) and (2) we obtain

$$ p_{c} = \frac{{\sigma_{Y} t}}{\sqrt 3 R} $$

(3)

or

$$ p_{c} = 23.2 \times 10^{ - 3} \sigma_{Y} . $$

(4)

Equation (3) gives the maximum pressure the vessel can withstand without failure by yielding. For the three steels available we obtain

pc (MPa)
A	B	C
19.9	30.0	35.8

For a long axial surface crack of depth a the stress intensity factor is (Fig. 5.16, Case 2 of Appendix 2.1).

Fig. 5.16

A material element for the determination of stress intensity factor for a long axial surface crack of depth a

$$ K_{\text{I}} = 1.12\sigma_{\theta } \sqrt {\pi a} . $$

(5)

The fracture condition is expressed by

$$ K_{\text{I}} = K_{\text{Ic}} /S,\quad S = 2. $$

(6)

From Eqs. (1) to (3) we obtain

$$ p = \frac{{tK_{\text{Ic}} }}{2.24\sqrt \pi R\sqrt a } $$

(7)

or

$$ p = \frac{{10.08 \times 10^{ - 3} K_{\text{Ic}} }}{\sqrt a } $$

(8)

• (a) Based on the results of the previous table and Eq. (8) the maximum pressure versus crack depth curves for the three steels are shown in Fig. 5.17.

  Fig. 5.17

  

  Maximum pressure p_max versus crack depth a for the three steels of Example 5.5

• (b) For p = 12 MPa the maximum permisible crack depths are calculated by Eq. (8) for the three steels as

ac(mm)
A	B	C
7.04	3.64	2.12

Note that at p = 12 MPa material A failed by yielding.

• (c) The failure pressure for a minimum detectable crack depth of a = 1 mm is calculated from Eq. (8) for the three steels as

pc (MPa)
A	B	C
19.90 (31.90)	22.96	17.60

Note that material A although withstands a pressure P_cr = 31.90 MPa for a = 1 mm, it fails at p_c = 19.90 MPa by yielding.

Example 5.6

A cylindrical pipe with inner radius b = 10 cm and outer radius c = 20 cm is thermally stressed due to a temperature difference $\Delta T$ across the wall. Positive $\Delta T$ indicates that the outside wall temperature is higher than the inside. The pipe contains an initial crack of length $a = 1$ mm emanating from its inner radius. The material of the pipe has yield stress $\sigma_{Y} = 1000\,{\text{MPa}}$, Poisson’s ratio v = 0.3, modulus of elasticity $E = 210\,{\text{GPa}}$, coefficient of thermal expansion $\alpha = 6.6 \times 10^{ - 6} \,^\circ {\text{F}}$ and fracture toughness $K_{\text{Ic}} = 100\,{\text{MPa}}\sqrt {\text{m}}$. Determine the maximum temperature difference $(\Delta T)_{c}$ the pipe can withstand without failure, with a factor of safety S = 2 against yielding and S = 3 against fracture.

Solution

The maximum stress at the rim of the inner radius of the pipe is given by (Theory and Elasticity, by S.P. Timoshenko and J.N. Goodier, Third Edition, p. 449)

$$ \sigma_{\hbox{max} } = \frac{\alpha E\Delta T}{2(1 - \nu )}\left[ {\frac{2}{{1 - (b/c)^{2} }} - \frac{1}{\log (c/b)}} \right]. $$

(1)

The condition of failure by yielding is expressed by

$$ \sigma_{\hbox{max} } = \sigma_{Y} /S,\quad S = 2. $$

(2)

From Eqs. (1) and (2) we obtain

$$ \frac{1000}{2} = \frac{{\left( {6.6 \times 10^{ - 6} } \right) \times 210 \times 10^{3} (\Delta T)_{c} }}{2(1 - 0.3)}\left[ {\frac{2}{{1 - 0.5^{2} }} - \frac{1}{\log 2}} \right]. $$

(3)

Thus,

$$ (\Delta T)_{c} = 413\,^\circ {\text{F}} $$

(4)

Equation (4) gives the maximum temperature the pipe can withstand without failure by yielding.

The stress intensity factor $K_{\text{I}}$ at the crack tip is (Case 2 of Appendix 2.1)

$$ K_{\text{I}} = 1.12\sigma_{\hbox{max} } \sqrt {\pi a} $$

(5)

or

$$ K_{\text{I}} = 1.12 \times \left( {1.21(\Delta T)_{c} } \right)\sqrt {\pi \times 10^{ - 3} } $$

or

$$ K_{\text{I}} = 0.076(\Delta T)_{c} . $$

(6)

The condition of failure by fracture is expressed by

$$ K_{\text{I}} = K_{\text{Ic}} /S,\quad S = 3. $$

(7)

From Eqs. (6) and (7) we obtain

$$ (\Delta T)_{c} = 438\,^\circ {\text{F}} . $$

(8)

Equation (8) gives the maximum temperature the pipe can withstand without failure by fracture.

Comparing the values of $(\Delta T)_{c}$ from Eqs. (4) and (8) we see that there is little difference between the prediction based on the maximum stress criterion and that obtained by fracture mechanics.

Example 5.7

A center cracked large plate of steel with thickness 10 mm is subjected to a uniform tension $\sigma = 300\,{\text{MPa}}$ perpendicular to the crack plane. Calculate the maximum crack length the plate can withstand without failure. $\sigma_{Y} = 860\,{\text{MPa}}$, $K_{\text{Ic}} = 100\,{\text{MPa}}\sqrt {\text{m}}$.

Solution

The applied stress $\sigma = 300\,{\text{MPa}}$ is smaller than the yield stress $\sigma_{Y} = 860\,{\text{MPa}}$ of the material. Thus, the plate does not fail by yielding.

The plate thickness B = 10 mm is smaller than the minimum thickness required for plane strain $B_{c} = 2.5\left( {K_{\text{Ic}} /\sigma_{Y} } \right)^{2} = 34\,{\text{mm}}$. Thus, the critical stress intensity factor K_c of the plate is not the plane strain stress intensity factor $K_{\text{Ic}}$. It is calculated from Eq. (5.7) as

$$ K_{c} = 100\sqrt {1 + \frac{1.4}{{\left( {10 \times 10^{ - 3} } \right)^{2} }}\left( {\frac{100}{860}} \right)^{4} } = 189\,{\text{MPa}}\sqrt {\text{m}} . $$

(1)

The stress intensity factor of a crack of length 2a in the plate is calculated as

$$ K_{\text{I}} = \sigma \sqrt {\pi a} = 532\sqrt a . $$

(2)

The fracture condition is expressed by

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

(3)

Equations (1) to (3) show that the maximum crack length the plate can withstand is

$$ 2a = 25.2{\text{cm}} . $$

(4)

Example 5.8

The critical strain energy release rate G_R of the double cantilever beam of Fig. 4.14 subjected to opposite point loads P at its end is given by

$$ G_{R} = g_{0} + g_{1} \varDelta a $$

(1)

where g₀ and g₁ are material constants and Δa = a – a₀ is the crack growth increment with a the current and a₀ the initial crack length.

The beam has a height of 2 h = 20 mm, unit thickness and initial crack length a₀ = 100 mm. The material constants have the following values: E = 100 GPa, ν = 0.3 and plane stress condition prevail.

Determine the critical load P_c for unstable crack growth and the length Δa of stable crack growth for the following cases:

• (a) g₀ = 0.002 and g₁ = 0.05 MN/m

• (b) g₀ = 0.001 and g₁ = 0.015 MN/m

• (c) g₀ = 0.0005 and g₁ = 0.02 MN/m

Solution

The critical load P_c and length Δa of stable crack growth are determined from Eqs. (5.11a) and (4.40) as

$$ G = G_{c} $$

(2)

$$ \frac{\partial G}{\partial a} = \frac{{\partial G_{c} }}{\partial a} $$

(3)

For the double cantilever beam subjected to opposite concentrated loads P at its end for conditions of generalized plane stress we have (Example 4.2)

$$ G = \frac{{12P^{2} a^{2} }}{{Eh^{3} }} $$

(4)

Equations (2) and (3) in conjunction with Eq. (1) take the form

$$ G \, = \frac{{12P^{2} a^{2} }}{{Eh^{3} }} = g_{0} + g_{1} \left( {a{-}a_{0} } \right) $$

(5)

$$ \frac{\partial G}{\partial a} = \frac{{ 24P^{2} a}}{{Eh^{3} }} = \frac{{\partial G_{c} }}{\partial a} = g_{1} $$

(6)

By eliminating P in Eqs. (5) and (6), we obtain for the length Δa of stable crack growth

$$ \varDelta a = a_{0} - \frac{{2g_{0} }}{{g_{1} }} $$

(7)

and for the critical load P_c for unstable crack growth

$$ P_{c} = \sqrt {\frac{{\left( {g_{0} + g_{1} \varDelta a} \right)Eh^{3} }}{{12\left( {1 - \nu^{2} } \right)(a_{0} + \varDelta a)^{2} }}} $$

(8)

From Eqs. (7) and (8), we obtain following numerical results for the above three cases:

• (a)
  $$ \varDelta a = 20\,{\text mm},P_{c} = 43.7\,{\text kN} $$

  (9)
• (b)
  $$ \varDelta a = - 33\,{\text mm}, thus\,\varDelta a = 0,P_{c} = 30.3\,{\text kN} $$

  (10)
• (c)
  $$ \varDelta a = 50\,{\text mm},P_{c} = 24.7\,{\text kN} $$

  (11)

Problems

• 5.1. A three-point bend specimen was tested according to the ASTM E399 procedure. The 0.2 percent offset yield stress of the material is $\sigma_{Y} = 1\,{\text{GPa}}$ and the modulus of elasticity is $E = 210\,{\text{GPa}}$. The specimen was tested at a loading rate of 60 kN/min. A chevron starter notch was machined and the specimen was subjected to 20,000 cycles at P_max = 20 kN and P_min = 0. The final stage of fatigue crack growth was conducted for 40,000 cycles at P_max = 15 kN and P_min = 0. The specimen dimensions were measured as

  $$ \begin{aligned} & S = 40\,{\text{cm}},W = 10\,{\text{cm}},B = 5\,{\text{cm}} \\ & a_{1} = 4.993\,{\text{cm}} \\ & a_{2} = 5.008\,{\text{cm}} \\ & a_{3} = 5.999\,{\text{cm}} \\ & a({\text{surface}}) = 4.925\,{\text{cm}} \\ & a({\text{surface}}) = 4.916\,{\text{cm}}. \\ \end{aligned} $$

  The maximum load and the secant load of the test record were measured as P_max = 108 kN, P_Q = 100 kN. Calculate K_Q and comment on the validity of the test.

• 5.2. A compact tension specimen was tested according to the ASTM E399 procedure. The 0.2% offset yield stress of the material is $\sigma_{Y} = 900\,{\text{MPa}}$ and the modulus of elasticity is $E = 210\,{\text{GPa}}$. The specimen dimensions were measured as W = 10 cm, B = 5 cm, while the crack lengths at equal locations across the crack front were measured as 4.90, 4.93, 5.05, 4.95, 4.85 cm. The maximum load and the secant load of the test record were measured as P_max = 70 kN and P_Q= 65 kN. Calculate K_Q and comment on the validity of the test.

• 5.3. The load-displacement test record of the compact tension specimen of Problem 5.2 is shown in Fig. 5.18. Calculate K_Q and comment on the validity of the test.

  Fig. 5.18

  

  Load-displacement $(P - u)$ record of the compact tension specimen of Problem 5.3

• 5.4. A three-point bend specimen with S = 40 cm, W = 10 cm, a = 5 cm and B = 5 cm was tested according to the ASTM E399 procedure. The 0.2 offset yield stress of the material is 400 MPa. The secant load is P_Q = 60 kN and the maximum load is P_max = 65 kN. Calculate $K_{\text{Ic}}$ and comment on its validity.

• 5.5. A compact tension specimen of a steel with W = 20 cm, a = 9 cm and B = 8 cm was tested according to the ASTM E399 procedure. The 0.2 offset yield stress of the material is $\sigma_{Y} = 900\,{\text{MPa}}$ and the secant load P_Q was measured to be 300 kN. Calculate $K_{\text{Ic}}$ and comment on its validity.

• 5.6. Determine the maximum $K_{\text{Ic}}$ value that may be determined according to ASTM standards on a 30 mm thick plate with 0.2 offset yield stress (a) 400 MPa, (b) 2000 MPa.

• 5.7. Assume that the extension $\Delta a$ of a crack of initial length ao is equal to the plane strain plastic zone radius calculated according to the Irwin model. For plane strain show that

  $$ \frac{\Delta a}{{a_{0} }} \le 0.0212 \simeq 0.02. $$

• 5.8. Use the result of the previous problem to justify the determination of P_Q using a 5% secant offset line on the load-displacement record according to ASTM E399 procedure. For this reason use the result that when the load-displacement $(P - u)$ record takes the form

  $$ \frac{uEB}{P} = f(a/W) $$

  where B and W are the specimen thickness and width, and E the modulus of elasticity, the quantity

  $$ H = \frac{{a_{0} }}{W}\frac{1}{f}\frac{{{\text{d}}f}}{{{\text{d}}\left( {a_{0} /W} \right)}},\quad f = f(a/W) $$

  takes an average value of 2.5 for the recommended range of values a₀/W lying between 0.45 and 0.55.

• 5.9. The crack growth resistance curve of a certain material at a thickness 2 mm is expressed by

  $$ R = \frac{{K_{\text{lc}}^{2} }}{E} + \frac{1}{2}(\Delta a)^{0.5} \,{\text{MJ}}/{\text{m}}^{2} ,K_{\text{Ic}} = 95\,{\text{MPa}}\sqrt {\text{m}} ,E = 210 \times 10^{3} \,{\text{MPa}} . $$

  Consider a center cracked plate of width 10 cm and thickness 2 mm with a crack of length 1 cm. Calculate the length of stable crack growth, the critical crack length, and the critical stress at instability.

• 5.10. As in Problem 5.9 for a center cracked plate of width 5 cm.

• 5.11. Consider a double cantilever beam (DCB, Fig. 4.14) with dimensions: B = 2 cm, a = 50 mm, h = 10 mm. For the material of Problem 5.9 calculate the length of stable crack growth, the critical length, and the critical load at instability.

• 5.12. Determine the stability condition for the double cantilever beam (DCB) of Problem 4.7 tested in a soft (load-controlled) or a hard (displacement-controlled) testing machine.

• 5.13. As in Problem 5.12 for the strip of Problem 4.10.

• 5.14. As in Problem 5.12 for the DCB of Problem 4.24.

• 5.15. As in Problem 5.12 for a center cracked plate.

• 5.16. Calculate the load required for crack growth of a double cantilever beam (Fig. 4.14) with a = 20 cm, h = 4 cm, B = 1 cm, E  = 210 GPa and G_c= 200 kJ/m².

• 5.17. Calculate the critical stress for growth of a crack of length 2 mm in a large thin plate of steel, loaded by a uniform stress perpendicular to the crack plane. Take E = 210 GPa, γ = 2 Jm⁻², γ_p= 2 × 104 Jm⁻².

• 5.18. A sheet of glass with width 50 cm and thickness 1 mm is subjected to a tensile stress of 10 MPa. Determine the minimum crack length that would lead to fracture. Take the following properties of glass: E = 60 GPa, v = 0.25, γ = 0.5 J/m², σ_f = 150 MPa.

• 5.19. A center cracked plate of steel with width 10 cm is subjected to a uniform tension 200 MPa perpendicular to the crack plane. Calculate the maximum crack length the plate can withstand without failure. $K_{\text{Ic}} = 55\,{\text{MPa}}\sqrt {\text{m}}$.

• 5.20. Consider an edge cracked plate of steel with width 5 cm subjected to a uniform tension 200 MPa perpendicular to the crack plane. Calculate the maximum crack length the plate can withstand without failure. $K_{\text{Ic}} = 55\,{\text{MPa}}\sqrt {\text{m}}$.

• 5.21. A crack of length 2 mm emanates from a circular hole of radius 50 mm in a large plate of steel subjected to a uniform stress perpendicular to the crack plane. Calculate the maximum stress the plate can withstand without failure. $\sigma_{Y} = 860\,{\text{MPa}}$, $K_{\text{Ic}} = 100\,{\text{MPa}}\sqrt {\text{m}}$.

• 5.22. A spherical vessel of steel with radius 1 m and thickness 40 mm contains a through crack of length 1 mm. Calculate the maximum internal pressure the vessel can withstand without failure. $\sigma_{Y} = 860\,{\text{MPa}}$, $K_{\text{Ic}} = 100\,{\text{MPa}}\sqrt {\text{m}}$.

• 5.23. A crack of length 2 mm emanates from an elliptical hole with axes 100 and 50 mm in a large plate of steel subjected to a uniform stress perpendicular to the crack plane. Calculate the maximum stress the plate can withstand without failure. $\sigma_{Y} = 860\,{\text{MPa}}$, $K_{\text{Ic}} = 100\,{\text{MPa}}\sqrt {\text{m}}$.

• 5.24. A center cracked plate of width 10 cm contains a crack of length 4 cm. The plate is subjected to a uniform stress perpendicular to the crack plane. Calculate the maximum stress the plate can withstand without failure. $\sigma_{Y} = 450\,{\text{MPa}}$ and $K_{\text{Ic}} = 25\,{\text{MPa}}\sqrt {\text{m}}$.

• 5.25. Calculate the maximum load the three-point bend specimen of Problem 2.32 (Fig. 2.30) can withstand. $\sigma_{Y} = 1500\,{\text{MPa}}$, $K_{\text{Ic}} = 60\,{\text{MPa}}\sqrt {\text{m}}$. The factor of safety for failure by yielding is 3 and by fracture is 2.

• 5.26. Calculate the maximum radial interference the shaft of Problem 2.36 (Fig. 2.33) can withstand. c = 5 cm, b = 2 cm, a =1 mm; E = 210 GPa, v = 0.3, $\sigma_{Y} = 900\,{\text{MPa}}$, $K_{\text{Ic}} = 100\,{\text{MPa}}\sqrt {\text{m}}$. The factor of safety for failure by yielding and fracture is 2.

• 5.27. Calculate the maximum crack length the cylindrical pipe of Problem 2.37 can withstand when it is subjected to a temperature difference $\Delta T = 200\,^\circ {\text{F}}$, across the wall. c = 15 cm, b = 10 cm; E = 210 GPa, v = 0.3, $\alpha = 6.6 \times 10^{ - 6} \,^\circ {\text{F}}$, $\sigma_{Y} = 900\,{\text{MPa}}$, $K_{\text{Ic}} = 100\,{\text{MPa}}\sqrt {\text{m}}$. The factor of safety for failure by yielding and fracture is 2.

• 5.28. Plot the allowable stress versus crack length curves for a large plate with a through crack subjected to a uniform stress perpendicular to the crack plane for the three steels of Example 5.5.

• 5.29. Consider a circular disk of radius c = 20 cm with a hole of radius b = 10 cm rotating at an angular velocity $\omega = 2\pi N/60$, where N is the number of revolutions per minute. The disk contains an initial crack of length a = 1 cm emanating from its hole. The material of the disk has mass density $\rho = 7800\,{\text{kg}}/{\text{m}}^{3}$, yield stress $\sigma_{Y} = 1000\,{\text{MPa}}$, Poisson’s ratio v = 0.3, and fracture toughness $K_{\text{Ic}} = 100\,{\text{MPa}}\sqrt {\text{m}}$. Determine the maximum number of revolutions per minute N_max that the disk can rotate without failure, when the factor of safety is S = 2 and S = 3 against failure by yielding and fracture, respectively.

• 5.30. The circular disk of Problem 5.29 may be constructed from the three steels of Example 5.5. For each steel:

  • (a) Plot the maximum permissible number of revolutions per minute N_max versus crack length;

  • (b) Calculate the maximum permissible crack length for N = 5000 rpm;

  • (c) Calculate N_max for a minimum detectable crack a = 0.05 mm.

• 5.31. The spherical vessel of Problem 5.22 may be constructed from the three steels of Example 5.5. The factor of safety against failure by yielding is 3 and by fracture is 2. For each steel:

  • (a) Plot the maximum permissible pressure p_rnax versus crack length a;

  • (b) Calculate the maximum permissible crack length a_cr for an operating pressure p_c = 10 MPa;

  • (c) Calculate p_max for a minimum detectable crack a = 0.05 mm.

• 5.32. The cylindrical pipe of Example 5.6 may be constructed from the three steels of Example 5.5. For each steel:

  • (a) Plot the maximum permissible temperature T_cr versus crack length;

  • (b) Calculate the maximum permissible crack length a_cr for an operating temperature difference $\Delta T = 200\,^\circ {\text{F}}$;

  • (c) Calculate the critical temperature difference $(\Delta T)_{cr}$ for a minimum detectable crack length a = 0.05 mm.

• 5.33. A rectangular panel of width 1 m is subjected to a tensile load of 100 MN perpendicular to the width. The smallest crack size that can be detected is 1 mm. Two steels with the following values of yield stress and fracture toughness are available for constructing the panel: A. $\sigma_{Y} = 860\,{\text{MPa}}$, $K_{\text{Ic}} = 100\,{\text{MPa}}\sqrt {\text{m}}$ and B. $\sigma_{Y} = 1550\,{\text{MPa}}$, $K_{\text{Ic}} = 55\,{\text{MPa}}\sqrt {\text{m}}$. Find which material gives the smallest thickness of the panel.

• 5.34. Calculate the maximum stress the plate of Problem 5.19 can withstand without failure when it has a thickness of 5 mm.

• 5.35. Calculate the maximum stress the plate of Problem 5.24 can withstand without failure when it has a thickness of 20 mm.

• 5.36. A long steel plate of width 20 cm and thickness 2 cm is subjected to a load 1.2 MN. Calculate the maximum crack length of a central or edge crack the plate can withstand without failure. $\sigma_{Y} = 860\,{\text{MPa}},K_{\text{Ic}} = 100\,{\text{MPa}}\sqrt {\text{m}}$.

Appendix 5.1

Values of critical stress intensity factor K_Ic and 0.2% offset yield stress $\sigma_{Y}$ at room temperature for various alloys.

Material	$\sigma_{Y}$		KIc
	MPa	ksi	MPa $\sqrt {\text{m}}$	ksi $\sqrt {\text{m}}$
300 Maraging Steel	1669	242	93.4	85
350 Maraging Steel	2241	325	38.5	35
D6AC Steel	1496	217	66.0	60
AISI 4340 Steel	1827	265	47.3	43
A533B Reactor Steel	345	50	197.8	180
Carbon Steel	241	35	219.8	200
Al 2014–T4	448	65	28.6	26
Al 2024–T3	393	57	34.1	31
Al 7075–T651	545	79	29.7	27
Al 7079–T651	469	68	33.0	30
Ti 6Al–4 V	1103	160	38.5	35
Ti 6Al-6 V–2Sn	1083	157	37.4	34
Ti 4Al-4Mo–2Sn–0.5Si	945	137	70.3	64