Experimental Methods

Abstract

Experimental methods have extensively been used in fracture mechanics problems. A number of nondestructive testing methods for the detection, location and sizing of defects have been developed.

Examples

Example 15.1

A strain gage of length 2 mm is placed at a distance 8 mm from the crack tip. The reading of the gage is ε_g = 600 με. Determine the angles α and θ and the stress intensity factor K_I.

$$ E = 207\,\text{GPa},\nu = 0.3. $$

Solution

We have from Eqs. (15.4) and (15.5)

$$ { \cos }\,2\alpha = - k = - \frac{1 - \nu }{1 + \nu } = - \frac{1 - 0.3}{1 + 0.3} = - 0.54 $$

(1)

and

$$ \alpha = 61.29^{ \circ } $$

(2)

Then we obtain from Eq. (15.5)

$$ \text{tan}\frac{\theta }{2} = - { \cot }\,2\upalpha = - { \cot }\,\left( {2 \times 61.29^{ \circ } } \right) = 0.64 $$

(3)

and

$$ \theta = 65.16^{ \circ } $$

(4)

With θ = 65.16° and α = 61.29°, we obtain from Eq. (15.6)

$$ \begin{aligned} 2\mu \varepsilon_{{x^{{\prime }} y^{{\prime }} }} & = \frac{{K_{\text{I}} }}{{\sqrt {2\pi r} }}(k\,{ \cos }\frac{\theta }{2} - \frac{1}{2}{ \sin }\,\theta \,{ \sin }\frac{3\theta }{2}{ \cos }\,2\alpha + \frac{1}{2}{ \sin }\,\theta \,{ \cos }\frac{3\theta }{2}{ \sin }\,2\alpha ) \\ & = \frac{{K_{\text{I}} }}{{\sqrt {2\pi r} }}(\left( {0.54} \right){ \cos }\frac{{65.16^{ \circ } }}{2} - \frac{1}{2}{ \sin }\,65.16^{ \circ } { \sin }(\frac{{3 \times 65.16^{ \circ } }}{2}){ \cos }\left( {2 \times 61.29^{ \circ } } \right) \\ &\quad + \frac{1}{2}{ \sin }\,65.16^{ \circ } { \cos } \left( {\frac{{3 \times 65.16^{ \circ } }}{2}} \right){ \sin }\left( {2 \times 61.29^{ \circ } } \right)) \\ & = \frac{{0.65K_{\text{I}} }}{{\sqrt {2\pi r} }} \\ \end{aligned} $$

(5)

Then, we obtain from Eq. (15.5) with 2μ = E/(1+ ν), E = 207 GPa, ν = 0.3, r = 0.008 m

$$\begin{aligned} K_{\text{I}} &= 2.97E\sqrt r \varepsilon_{{x^{{\prime }} y^{{\prime }} }} = 2.97\,E\sqrt r \varepsilon_{g} \\ &= 2.97 \times (207 \times 10^{9} )\sqrt {0.008} \,(600 \times 10^{ - 6} ) = 33\,\text{MPa}\,\sqrt {\text{m}} \end{aligned}$$

(6)

Equations (2) and (4) give the angles α and θ and Eq. (6) gives the stress intensity factor K_I.

Example 15.2

Determine the distance between the center of a gage and the true strain location for the following cases:

• (a) A strain gage of length 1 mm placed at a distance 4 mm from the crack tip.

• (b) A strain gage of length 2 mm placed at a distance 4 mm from the crack tip.

• (c) A strain gage of length 2 mm placed at a distance 8 mm from the crack tip.

Solution

We have from Eq. (15.16)

• (a)
  $$ \begin{aligned} & \frac{\varDelta r}{{r_{c} }} = \frac{1}{2}\left\{ 1 \, {-}\left[ {1{-}\left( {\frac{L}{{2r_{c} }}} \right)^{2} } \right]^{1/2} \right\} = \frac{1}{2}\left\{ 1 \, {-} \, \left[1 \, {-}\left( { \, \frac{1}{2 \times 4}} \right)^{2} \right]^{1/2} \right\} = 0.0039 \\ & \varDelta r = 0.0039\,\times \,4 {\text{mm}} = 0.0157\,{\text{mm}} \\ \end{aligned} $$

  (1)
• (b)
  $$ \begin{aligned} & \frac{\varDelta r}{{r_{c} }} = \frac{1}{2}\left\{ 1 \, {-} \, \left[1{-}\left( {\frac{L}{{2r_{c} }}} \right)^{2} \right]^{1/2} \right\} = \frac{1}{2} \left\{ 1{-}\left[1 \, {-}\left( {\frac{2}{2 \times 4}} \right)^{2} \right]^{1/2} \right\} = 0.016 \\ & \varDelta r = 0.016\,\times\,4\,{\text{mm}} = 0.064\,{\text{mm}} \\ \end{aligned} $$

  (2)
• (c)
  $$ \begin{aligned} & \frac{\varDelta r}{{r_{c} }} = \frac{1}{2}\left\{ 1{-}\left[1{-}\left( {\frac{L}{{2r_{c} }}} \right)^{2} \right]^{1/2} \right\} = \frac{1}{2}\left\{ 1{-}\left[1{-}\left( {\frac{2}{2 \times 8}} \right)^{2} \right]^{1/2} \right\} = 0.0039 \\ & \varDelta r = 0.0039\,\times\,8\,{\text{mm}} = 0.0314\,{\text{mm}} \\ \end{aligned} $$

  (3)

Equations (1), (2) and (3) give the distance between the center of a gage and the true strain location for the above three cases.

Example 15.3

For a material with Poisson’s ratio ν = 1/3 under mode-I and mode-II loading show that by placing the electrical resistance strain gages at positions α = θ = +60° and −60° the sum of the strains along these positions is independent of K_II and the difference of the stresses is independent of K_I.

Solution

From Eq. (15.17) we first observe that when Eq. (15.5) is satisfied, then the coefficients of the terms A₁, B₀ and D₁ vanish. Introducing the values of α = θ =+60° into Eq. (15.17), we obtain

$$ {\text{E}}{\upvarepsilon_{{\text{x}}^{{\prime }} {\text{y}}^{{\prime }} }}^{ + } = \frac{\sqrt 3 }{2}A_{0 } \,r^{ - 1/2} - 2{B}_{1} r + \frac{1}{2}C_{0 } r^{ - 1/2} + C_{1 } r^{1/2} $$

(1)

Similarly for α = θ = −60°, we obtain

$$ {\text{E}}{\upvarepsilon_{{\text{x}}^{{\prime }} {\text{y}}^{{\prime }} }}^{ - } = \frac{\sqrt 3 }{2}A_{0 } r^{ - 1/2} - 2{B}_{1} r - \frac{1}{2}C_{0 } r^{ - 1/2} - C_{1 } r^{1/2} $$

(2)

By adding Eqs. (1) and (2) we obtain

$$ {\text{E}}({\upvarepsilon_{\text{x}^{\prime } \text{y}^{\prime}}}^{ + } +{\upvarepsilon_{{\text{x}}^{{\prime }} {\text{y}}^{{\prime }} }}^{ - } ) = \sqrt {3 } A_{0 } r^{ - 1/2} - 4{B}_{1} r $$

(3)

and, by subtracting Eqs. (1) and (2) we obtain

$$ {\text{E}}({\upvarepsilon_{{{\text{x}}^{{\prime }} {\text{y}}^{{\prime }} }}}^{ + } - {\upvarepsilon_{{{\text{x}}^{{\prime }} {\text{y}}^{{\prime }} }}}^{ - } ) = - C_{0 } r^{ - 1/2} + 2C_{1 } r^{1/2} $$

(4)

A₀ and C₀ are related to the stress intensity factors K_I and K_II by

$$ K_{\text{I}} = \sqrt {2\pi } \,A_{0} , K_{\text{II}} = \sqrt {2\pi }\, C_{0} $$

(5)

From Eq. (3) we observe that the sum (ε_x′y′⁺ + ε_x′y′⁻) measured at the same distance r from the crack tip along the radial lines +60° and −60° is independent of C₀, and therefore, of the shearing mode stress intensity factor K_II.

From Eq. (4) we observe that the difference (ε_x′y′⁺ − ε_x′y′⁻) measured at the same distance r from the crack tip along the radial lines +60° and −60° is independent of A₀, and therefore, of the opening mode stress intensity factor K_I.

Example 15.4

The following data were obtained from the isochromatic fringe pattern of a large plate of thickness d = 4 mm with an edge crack of length a = 10 cm subjected to opening-mode loading:

$$ \begin{aligned} N_{1} = 3, r_{m1} = 0.6\,{\text{cm}},\theta_{m1} = 76^{ \circ } \hfill \\ N_{2} = 5, r_{m2} = 0.2\,{\text{cm}},\theta_{m2} = 80^{ \circ } \hfill \\ \end{aligned} $$

where N₁, N₂ are the isochromatic fringe loop orders, and, r_m1, θ_m1 and r_m2, θ_m2 are the polar coordinates of the farthest points of the loops 1, 2 at the crack tip.

The photoelastic constant of the material of the plate is f_c = 20 kN/m.

Determine the stress intensity factor K_I using the Irwin and the Bradley and Kobayashi methods.

Solution

The stress intensity factor K_I is calculated according to the Irwin method for point 1 from Eq. (15.25) as

$$ K_{\text{I}} = \frac{{Nf_{c} \sqrt {2\pi r_{m} } }}{{d\,{ \sin }\,\theta_{m} }}\left[ {1 + \left( {\frac{2}{{3\,{ \tan }\,\theta_{m} }}} \right)^{2} } \right]^{ - 1/2} \left( {1 + \frac{{2\,{ \tan }(3\theta_{m} /2)}}{{3\,{ \tan }\,\theta_{m} }}} \right) $$

(1)

Substituting the values of N₁ = 3, r_m1 = 0.6 cm, θ_m1 = 76°, d = 4 mm and f_c = 20 kN/m for point 1 in Eq. (1), we obtain

$$ K_{\text{I}} = \frac{{3 \times 20 \times 10^{3} \sqrt {2\pi \times 6 \times 10^{ - 3} } }}{{4 \times 10^{ - 3} { \sin }\,76^{\circ} }}\left[ {1 + \left( {\frac{2}{{3\,{ \tan }\,76^{\circ} }}} \right)^{2} } \right]^{ - 1/2} \left( {1 + \frac{{2\,{ \tan }114^{\circ} }}{{3\,{ \tan }\,76^{\circ} }}} \right) = 1.87\,{\text{MPa}} \sqrt {\text{m}} $$

(2)

Substituting the values of N₂ = 5, r_m2 = 0.2 cm, θ_m2 = 80°, d = 4 mm and f_c = 20 kN/m for point 2 in Eq. (1), we obtain

$$ K_{\text{I}} = \frac{{5 \times 20 \times 10^{3} \sqrt {2\pi \times 2 \times 10^{ - 3} } }}{{4 \times 10^{ - 3} { \sin }\,80^{\circ} }}\left[ {1 + \left( {\frac{2}{{3\,{ \tan }\,80^{\circ} }}} \right)^{2} } \right]^{ - 1/2} \left( {1 + \frac{{2\,{ \tan }\,120^{\circ} }}{{3\,{ \tan }\,80^{\circ} }}} \right) = 2.26\,{\text{MPa}} \sqrt {\text{m}} $$

(3)

For the Bradley and Kobayashi method, K_I is determined from (Eq. (15.29)) :

$$ K_{\text{I}} = \frac{{f_{c} }}{d}\sqrt {2\pi } \frac{{\sqrt {r_{1} r_{2} } (N_{2} - N_{1)} }}{{f_{2} \sqrt {r_{1} } + f_{1} \sqrt {r_{2} } }} $$

(4)

with

$$ f = [{ \sin }^{2} \theta + 2\delta \sqrt {2r/a}\,{ \sin }\,\theta \,{ \sin }(3\theta /2) + 2r\delta^{2} /a]^{1/2} $$

(5)

We have for the value of f_1,2 for the points 1 and 2 (putting δ = 1)

$$ f_{1} = \left[ {{ \sin }^{2}\,76^{ \circ } + \, 2\left( {2 \times 0.6/10} \right)^{0.5} { \sin }\,76^{ \circ } { \sin }\,114^{ \circ } + 2 \times 0.6/10} \right]^{0.5} = 1.29 $$

(6)

$$ f_{2} = \left[ {{ \sin }^{2}\,80^{ \circ } + 2\left( {2 \times 0.2/10} \right)^{0.5} { \sin }\,80^{ \circ } { \sin }\,120^{0} + 2 \times 0.2/10} \right]^{0.5} = 1.16 $$

(7)

K_I is determined from Eq. (4) as

$$ K_{\text{I}} = \frac{{20 \times 10^{3} \times (2\pi )^{0.5} (\left( {2 \times 10^{ - 3} } \right) \times (6 \times 10^{ - 3} ))^{0.5 } \left( {5 - 3} \right)}}{{\left( {4 \times 10^{ - 3} } \right)((1.16 \times (6 \times 10^{ - 3} )^{0.5} - 1.29 \times \left( {2 \times 10^{ - 3} )^{0.5} } \right)}} = \, 2.70 \times 10^{6} \,{\text{MPa}}\,\sqrt {\text{m}} $$

(8)

Example 15.5

For mixed-mode loading, consider the photoelastic data obtained along the lines θ = π and θ = π/2. Develop a method for the determination of stress intensity factors K_I and K_II (it is assumed that the photoelastic fringes intersect the above two lines).

Solution

We obtain from Eq. (15.21) with θ = π

$$ \tau_{ \text{max} }^{2} = \frac{1}{2\pi r}{K}_{\text{II}}^{2} + \frac{{\sigma_{0x} }}{{\sqrt {2\pi r} }}K_{\text{II}} + \frac{{\sigma_{0x}^{2} }}{4} $$

(1)

and

$$ \frac{{Nf_{\sigma } }}{d} = 2\uptau_{ \text{max} } = \pm \left( {\frac{{2K_{II} }}{{\sqrt {2\pi r} }} + \sigma_{ox} } \right) $$

(2)

Consider two fringes N₁ and N₂ (N₁> N₂) which intersect the upper edge of the crack (θ = π) at distances r₁ and r₂ from the crack tip. The isochromatic fringes intersect the line θ = π when K_II < 0, and intersect the line θ = −π when K_II > 0.

Applying Eq. (2) at positions r₁ and r₂ and taking the negative sign in Eq. (2), we obtain

$$ \frac{{N_{1} f_{\sigma } }}{d} = - \left( {\frac{{2K_{\text{II}} }}{{\sqrt {2\pi r_{1} } }} + \sigma_{ox} } \right) $$

(3)

$$ \frac{{N_{2} f_{\sigma } }}{d} = - \left( {\frac{{2K_{\text{II}} }}{{\sqrt {2\pi r_{2} } }} + \sigma_{ox} } \right) $$

(4)

From Eqs. (3) and (4), we obtain

$$ K_{\text{II}} = \frac{{f_{\sigma } }}{d}\sqrt {\frac{\pi }{2}} \left( {\frac{{\sqrt {r_{1} r_{2} } }}{{\sqrt {r_{1} } - \sqrt {r_{2} } }}} \right)(N_{1} {-} \, N_{2} ) $$

(5)

$$ \sigma_{ox} = - \frac{{N_{1} f_{\sigma } }}{d} - \frac{{2K_{\text{II}} }}{{\sqrt {2\pi r_{1} } }} $$

(6)

Equations (5) and (6) give the stress intensity factor K_II and the constant stress $\sigma_{ox} .$

We obtain from Eq. (15.21) with θ = π/2 and r = r₃

$$ \left( {\frac{{N_{3} f_{\sigma } }}{d}} \right)^{2} = \frac{1}{{2\pi r_{3} }}(K_{\text{I}}^{2} + K_{\text{II}}^{2} ) + \frac{{\sigma_{0x} }}{{\sqrt {\pi r_{3} } }} \left( {K_{\text{I}} + K_{\text{II}} } \right) + \sigma_{0x}^{2} $$

(7)

By solving the above quadratic equation the value of stress intensity factor K_I is obtained (using the values of K_II and $\sigma_{ox}$ determined from Eqs. (5) and (6)).

Example 15.6

Isopachic fringe patterns obtained from interferometric methods represent the contours of the out-of-plane displacement for conditions of generalized plane stress or the contours of the sum of the principal stresses. Consider only two terms of the strain field around the crack tip (Eq. (15.1)) and select data along the line θ = 0. Find an equation for the stress intensity factor K_I from the isopachic fringe orders N₁ and N₂ at positions r₁ and r₂ along the crack axis. This method for determining K_I was developed by Dudderar and Gorman [32].

Solution

The stress-optical law for isopachic patterns is given by

$$ \sigma_{xx} + \sigma_{yy} = \frac{{Nf_{p} }}{d} $$

(1)

where N is the fringe order, f_p is the optical constant and d is the thickness of the plate.

From Eq. (1) in combination with the stress–strain equation under condition of plane stress (σ_z = 0), we obtain for θ = 0

$$ \frac{{\sigma_{xx} + \sigma_{yy} }}{2} = A_{0} r^{ - 1/2} + B_{0} = \frac{{Nf_{p} }}{2d} $$

(2)

Applying Eq. (2) at positions r₁ and r₂ with fringe orders N₁ and N₂, respectively, we obtain

$$ A_{0} r_{1}^{ - 1/2} + B_{0} = \frac{{N_{1} f_{p} }}{2d} $$

(3)

$$ A_{0} r_{2}^{ - 1/2} + B_{0} = \frac{{N_{2} f_{p} }}{2d} $$

(4)

Subtracting Eqs. (3) and (4), we obtain

$$ A_{0} \left( {r_{1}^{ - 1/2} - r_{2}^{ - 1/2} } \right) = \frac{{\left( {N_{1} - N_{2} } \right)f_{p} }}{2d} $$

(5)

or

$$ A_{0} = \frac{{\sqrt {r_{1} r_{2} } }}{{\sqrt {r_{2} } - \sqrt {r_{1} } }}\frac{{\left( {N_{1} - N_{2} } \right)f_{p} }}{2d} $$

(6)

K_I is determined as

$$ K_{\text{I}} = \sqrt {2\pi } A_{0} = \sqrt {\frac{\pi }{2}} \frac{{\sqrt {r_{1} r_{2} } }}{{\sqrt {r_{2} } - \sqrt {r_{1} } }}\frac{{\left( {N_{1} - N_{2} } \right)f_{p} }}{d} $$

(7)

Example 15.7

A PMMA (Plexiglas) plate of thickness d = 2 mm with a crack is illuminated by a divergent light emanating from a point source placed at a distance z_i = 0.4 m from the plate. A caustic of transverse diameter D_t = 20 mm is formed at a screen placed at a distance z₀ = 1 m behind the plate. The optical constant of PMMA is c = 108 × 10⁻¹² m²/N.

• (a) Determine the K_I stress intensity factor and the radius of the initial curve of the caustic r₀.

• (b) The screen is then placed at a distance z₀ = 2 m from the plate. Determine the diameter of the caustic D_t and the radius r_0.

Solution

• (a) The magnification factor of the optical arrangement is

$$ m = \frac{{z_{0} + z_{i} }}{{z_{i} }} = \frac{1 + 0.4}{0.4 } = 3.50 $$

(1)

We obtain from Eq. (15.65) for the stress intensity factor K_I

$$ K_{\text{I}} = \frac{0.0939}{{{\text{z}}_{0} {\text{c}}dm^{3/2} }}D_{t}^{5/2} = \frac{0.0939}{{1 \times \left( {108 \times 10^{ - 12} } \right)\left( {0.002} \right)(3.50)^{3/2} }}\,0.02^{5/2} = \, 3.76\,{\text{MPa}}\sqrt {\text{m}} $$

(2)

The radius of the initial curve r₀ is obtained from Eq. (15.67) as

$$ r_{0} = 0.316\,D_{t} /m = 0.316\, \times \,20/3.5 = 1.81\,{\text{mm}} $$

(3)

• (b) When the screen is placed at a distance z₀ = 2 m from the plate, the magnification factor m is

$$ m = \frac{{z_{0} + z_{i} }}{{z_{i} }} = \frac{200 + 40}{40 } = 6 $$

(4)

The diameter D_t of the caustic is given by Eq. (2) as

$$ D_{t} = \left( {\frac{{K_{I} cz_{0} dm^{3/2} }}{0.0939}} \right)^{2/5} = \left( {\frac{{\left( {3.76 \times 10^{6} } \right) \times \left( {108 \times 10^{ - 12} } \right) \times 2 \times \left( {0.002} \right) \times \left( {6^{3/2} } \right)}}{0.0939}} \right)^{2/5} = 36.5\,{\text{mm}} $$

(5)

and the radius of the initial curve r₀ by

$$ r_{0} = 0.316\,D_{t} /m = 0.316\, \times \,36.5/6 = 1.92\,{\text{mm}} $$

(6)