5.2 Static Failure Theories for Brittle Materials & Fracture Mechanics

5.2 Static Failure Theories for Brittle Materials & Fracture Mechanics

Quick Summary: Brittle materials sustain negligible plastic deformation prior to catastrophic fracture ($\%EL < 5\%$) and exhibit markedly asymmetric strength, with ultimate compressive strengths exceeding tensile capacities by factors of three to four. Classical stress concentration formulations break down at sharp flaws where theoretical notch radii approach zero. This section covers the governing brittle failure criteria—Maximum Normal Stress, Brittle Coulomb-Mohr, and the Modified Mohr (Dowling) theory—and introduces Linear Elastic Fracture Mechanics (LEFM), including stress intensity ($K_I$), ASTM E399 plane strain fracture toughness ($K_{Ic}$), valid thickness constraints, NDE crack detection limits, and the Leak-Before-Break (LBB) design criterion.

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1. Mechanical Behavior and Asymmetric Strength of Brittle Materials

Engineering materials exhibiting percent elongation at fracture $\%EL < 5\%$ are classified as brittle. Common engineering brittle materials include gray cast iron, white cast iron, hardened tool steels (quenched without tempering), structural ceramics (silicon nitride, alumina, zirconia), glass, concrete, and powder metallurgy components.

Cleavage Fracture vs. Plastic Slip

Unlike ductile metals that deform via dislocation glide along close-packed slip planes, brittle solids undergo cleavage fracture—the rapid, direct separation of atomic crystallographic planes normal to the maximum tensile stress direction. Because cleavage requires very little energy dissipation compared to plastic shear slip, brittle fracture is sudden, catastrophic, and unaccompanied by gross dimensional warning (no necking or yield plateau).

The Physical Origin of Compressive-to-Tensile Strength Asymmetry

The defining mechanical characteristic of brittle materials is their pronounced strength asymmetry:

$S_{uc} \gg S_{ut}$

For standard grades of gray cast iron (ASTM A48 Class 20 to Class 60), the ultimate compressive strength is typically three to four times greater than the ultimate tensile strength ($S_{uc} \approx 3\text{--}4 S_{ut}$):

This extreme asymmetry arises from internal microstructural defects. Gray cast iron contains an interconnected network of sharp, brittle graphite flakes dispersed throughout a pearlitic or ferritic iron matrix:

1. Under Tensile Stress: The sharp tip radii of graphite flakes act as severe internal stress concentrators. Tensile stresses open micro-cracks at flake tips, which link together and propagate across the cross section at low nominal tensile loads ($S_{ut}$).
2. Under Compressive Stress: Compressive forces push the surfaces of the graphite flakes and micro-cracks into mechanical contact. Flaws are squeezed shut, allowing the solid metallic iron matrix to transfer compressive load across the entire gross cross-sectional area. Failure can only occur when the iron matrix itself crushes or shears along oblique planes at very high stress levels ($S_{uc}$).

Fracture Morphology in Tension vs. Torsion

Because brittle fracture is governed by normal tensile cleavage:

• Uniaxial Tension: Fractures along a flat plane oriented precisely perpendicular (90°) to the tensile loading axis.
• Pure Torsion: Under pure shear stress $\tau$, Mohr's circle indicates that the principal stresses are $\sigma_1 = +\tau$ and $\sigma_3 = -\tau$, acting on planes inclined at 45° to the shaft longitudinal axis. Consequently, a brittle cast iron shaft tested in pure torsion fractures along a characteristic 45° helical cleavage plane.

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2. Maximum Normal Stress (MNS) Theory

Formulated by Scottish engineer William Rankine, the Maximum Normal Stress (MNS) theory is the simplest static failure model for brittle materials.

The MNS Postulate

Brittle fracture occurs when the maximum principal tensile stress in a multiaxial stress state reaches the ultimate tensile strength $S_{ut}$, or when the minimum principal stress reaches the ultimate compressive strength $-S_{uc}$.

Mathematical Formulation

Ordering principal stresses algebraically such that $\sigma_1 \ge \sigma_2 \ge \sigma_3$, the MNS failure criterion states that fracture occurs if:

$\sigma_1 \ge S_{ut} \quad \text{or} \quad \sigma_3 \le -S_{uc}$

The design factor of safety ($n$) is evaluated separately for tension and compression:

$n_t = \frac{S_{ut}}{\sigma_1} \quad (\text{if } \sigma_1 > 0), \qquad n_c = \frac{S_{uc}}{|\sigma_3|} \quad (\text{if } \sigma_3 < 0)$

The governing factor of safety is $n = \min(n_t, n_c)$.

Evaluation and Severe Limitations of MNS

In plane stress ($\sigma_3 = 0$), the MNS criterion plots as a simple rectangle on the $\sigma_1 - \sigma_2$ plane bounded by $\sigma_1 = S_{ut}$, $\sigma_1 = -S_{uc}$, $\sigma_2 = S_{ut}$, and $\sigma_2 = -S_{uc}$:

• Quadrant 1 (Biaxial Tension: $\sigma_1 > 0, \sigma_2 > 0$): MNS exhibits excellent agreement with experimental fracture data. Superimposed transverse tensile stress does not significantly alter the tensile fracture capacity.
• Quadrant 4 (Combined Tension and Compression: $\sigma_1 > 0, \sigma_2 < 0$): Extensive experimental testing (notably by Coffin, Grassi, and Cornet on thin-walled gray iron tubes) reveals that MNS is dangerously non-conservative. When compressive stress is applied simultaneously with tensile stress, the tensile stress required to trigger brittle fracture drops well below $S_{ut}$. Designing with MNS in Quadrants 2 and 4 can lead to unexpected catastrophic rupture.

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3. Coulomb-Mohr and Modified Mohr (MM) Theories

To account for strength asymmetry ($S_{uc} > S_{ut}$) and overcome the non-conservatism of MNS in mixed tension-compression stress states, two Mohr-based failure criteria are used in machine design.

Brittle Coulomb-Mohr (BCM) Theory

The Brittle Coulomb-Mohr (BCM) theory (also known as the internal friction theory) constructs a straight-line failure envelope tangent to the uniaxial tensile test circle (diameter $S_{ut}$) and the uniaxial compressive test circle (diameter $S_{uc}$) on a Mohr's stress plane.

For 2D plane stress with ordered in-plane principal stresses $\sigma_A \ge \sigma_B$:

1. Quadrant 1 ($\sigma_A \ge \sigma_B \ge 0$): $n = \frac{S_{ut}}{\sigma_A}$
2. Quadrants 2 & 4 ($\sigma_A \ge 0 \ge \sigma_B$): $\frac{\sigma_A}{S_{ut}} - \frac{\sigma_B}{S_{uc}} = \frac{1}{n} \implies n = \frac{1}{\frac{\sigma_A}{S_{ut}} - \frac{\sigma_B}{S_{uc}}}$
3. Quadrant 3 ($0 \ge \sigma_A \ge \sigma_B$): $n = -\frac{S_{uc}}{\sigma_B} = \frac{S_{uc}}{|\sigma_B|}$

While BCM prevents the dangerous non-conservatism of MNS in Quadrant 4, extensive testing shows that BCM is excessively conservative when compressive stress is small ($|\sigma_B| \le S_{ut}$).

Modified Mohr (MM) Theory (Dowling Equations)

The Modified Mohr (MM) theory—standardized by N.E. Dowling and adopted in Shigley's Mechanical Engineering Design—modifies the BCM envelope to match experimental biaxial test data with exceptional precision. It recognizes that a minor compressive stress does not degrade the tensile fracture capacity until the compressive stress exceeds $S_{ut}$.

For ordered in-plane principal stresses $\sigma_A \ge \sigma_B$:

                               σB
                                ^
                                |   Quadrant 1
                            Sut + - - - - - - - .
                                |               |
                                |   Modified    |
                                |     Mohr      |
               -Suc             |    Envelope   |       Sut
     <-----------+--------------+---------------+--------+------> σA
                 |              |               |
                 |              |               |
                 |              |   Quadrant 4  |
                 |              * - - - - - - - '
                 |             /   (Sut, -Sut)
                 |            /
                 |           /
                 |          /
            -Suc . - - - - '
                   Quadrant 3

The factor of safety ($n$) is evaluated using Dowling's four distinct piecewise conditions:

Condition 1: Biaxial Tension (Quadrant 1: $\sigma_A \ge \sigma_B \ge 0$)

Both principal stresses are tensile. Failure is governed strictly by tensile cleavage:

$n = \frac{S_{ut}}{\sigma_A}$

Condition 2: Tension-Dominant Mixed Stress (Quadrant 4: $\sigma_A \ge 0 \ge \sigma_B$ with $|\sigma_B / \sigma_A| \le 1$)

Here, the magnitude of the compressive stress is less than or equal to the tensile stress (meaning $\sigma_B \ge -\sigma_A$, which keeps the stress state above the $-45^\circ$ diagonal line). Experimental data confirm that tensile fracture remains completely uninhibited by the transverse compression:

$n = \frac{S_{ut}}{\sigma_A}$

Condition 3: Compression-Dominant Mixed Stress (Quadrant 4: $\sigma_A \ge 0 \ge \sigma_B$ with $|\sigma_B / \sigma_A| > 1$)

When compressive stress exceeds tensile stress ($|\sigma_B| > \sigma_A$), the compressive stress significantly accelerates crack propagation. Dowling's linear interaction equation governs:

$\frac{(S_{uc} - S_{ut})\sigma_A}{S_{uc} S_{ut}} - \frac{\sigma_B}{S_{uc}} = \frac{1}{n}$

Solving explicitly for the factor of safety $n$:

$n = \frac{S_{uc} S_{ut}}{(S_{uc} - S_{ut})\sigma_A - S_{ut} \sigma_B}$

Condition 4: Biaxial Compression (Quadrant 3: $0 \ge \sigma_A \ge \sigma_B$)

Both principal stresses are compressive. Failure is governed strictly by compressive crushing or shear splitting along the minimum principal stress axis:

$n = -\frac{S_{uc}}{\sigma_B} = \frac{S_{uc}}{|\sigma_B|}$

Quadrant-by-Quadrant Theory Comparison

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4. Linear Elastic Fracture Mechanics (LEFM)

The Breakdown of Classical Stress Concentration

Classical elasticity analyzes geometric notches (fillets, holes, keyways) using stress concentration factors ($K_t = \sigma_{\max} / \sigma_{\text{nom}}$). For an elliptical hole with major axis $2a$ and notch root radius $\rho$ oriented perpendicular to a tensile stress, the Inglis solution yields:

$\sigma_{\max} = \sigma\left(1 + 2\sqrt{\frac{a}{\rho}}\right)$

In actual manufactured hardware, physical cracks, fatigue tears, and weld lack-of-fusion defects have atomic-scale root radii approaching zero ($\rho \to 0$). Under classical elasticity, as $\rho \to 0$, $K_t \to \infty$ and $\sigma_{\max} \to \infty$. Infinite stress predictions make classical stress analysis incapable of assessing whether a cracked component can safely carry load. Linear Elastic Fracture Mechanics (LEFM) resolves this mathematical singularity by treating the crack tip not as an infinite point stress, but through energy dissipation and stress field intensity.

Griffith Energy Criterion (1921)

A.A. Griffith established that a crack propagates catastrophically when the rate of elastic strain energy released by infinitesimal crack advance ($dU/da$) equals or exceeds the rate of surface energy required to create two new fracture surfaces:

$G = -\frac{dU}{da} \ge G_c = 2\gamma_s$

For an ideal brittle material, the critical fracture stress is:

$\sigma_f = \sqrt{\frac{2E\gamma_s}{\pi a}}$

In engineering metals, E. Orowan and G.R. Irwin showed that localized dislocation plasticity always occurs in a microscopic zone at the crack tip, absorbing thousands of times more energy than surface tension alone ($\gamma_p \gg \gamma_s$). Irwin modified the criterion to:

$G_c = 2(\gamma_s + \gamma_p) \approx 2\gamma_p \implies \sigma_f = \sqrt{\frac{E G_c}{\pi a}}$

The Three Modes of Crack Surface Displacement

Crack tip stress fields are categorized into three fundamental kinematic modes:

1. Mode I (Opening Mode): Tensile normal stresses act perpendicular to the crack plane, pulling crack faces directly apart ($y$-direction). Mode I is the predominant failure mode in mechanical machine design, pressure vessels, and rotating shafts.
2. Mode II (Sliding / In-Plane Shear Mode): Shear stresses act parallel to the crack plane and perpendicular to the crack front ($x$-direction), causing crack faces to slide over one another.
3. Mode III (Tearing / Anti-Plane Shear Mode): Out-of-plane shear stresses act parallel to the crack front ($z$-direction), causing crack faces to twist and tear apart.

Mode I Stress Intensity Factor ($K_I$)

G.R. Irwin demonstrated that the asymptotic elastic stress field at distance $r$ and angle $\theta$ from a Mode I crack tip is governed by a single scalar parameter:

$\sigma_{ij}(r, \theta) = \frac{K_I}{\sqrt{2\pi r}} f_{ij}(\theta) + \mathcal{O}(r^0)$

The Mode I Stress Intensity Factor ($K_I$) scales the entire local crack-tip stress field and is defined macroscopically as:

$K_I = Y \sigma \sqrt{\pi a}$

where:

• $\sigma$ is the nominal gross far-field tensile stress acting normal to the crack plane.
• $a$ is the characteristic crack dimension ($a = \text{depth of an edge crack}$; $2a = \text{total length of an internal crack}$, so $a = \text{half-crack length}$).
• $Y$ (also denoted $\beta$) is the dimensionless geometry correction factor, which accounts for crack geometry, component thickness, finite width ($a/W$), and boundary constraints.
  • Central through-crack of length $2a$ in an infinitely wide plate: $Y = 1.0$.
  • Single edge crack of depth $a$ in a semi-infinite plate: $Y = 1.12$ (the $12\%$ increase reflects the loss of mechanical constraint at the free edge).
• Units: $\text{MPa}\sqrt{\text{m}}$ in SI; $\text{ksi}\sqrt{\text{in}}$ in USCS. Conversion factor: $1\text{ ksi}\sqrt{\text{in}} = 1.09884\text{ MPa}\sqrt{\text{m}}$

Plane Strain Fracture Toughness ($K_{Ic}$) & Fracture Criterion

When the applied stress intensity factor reaches a critical threshold, crack growth becomes unstable and propagates through the component at high speed (a sizable fraction of the material's sound speed, on the order of $1\text{ km/s}$ in steel). This critical threshold under maximum triaxial constraint is the Plane Strain Fracture Toughness ($K_{Ic}$)—a fundamental material property measured experimentally in accordance with ASTM E399.

The fracture mechanics design criterion is:

$K_I \ge K_{Ic} \implies \text{Unstable Brittle Fracture}$

The factor of safety against brittle fracture is:

$n = \frac{K_{Ic}}{K_I} = \frac{K_{Ic}}{Y \sigma \sqrt{\pi a}}$

ASTM E399 Valid Plane Strain Thickness Requirement

Fracture toughness is strongly dependent on specimen thickness ($B$):

• Thin Plates (Plane Stress): At a free surface, out-of-plane normal stress is zero ($\sigma_z = 0$). The material is free to contract laterally ($\epsilon_z \ne 0$), allowing extensive plastic yielding in a large yield zone ahead of the crack tip. This blunts the crack, dissipates substantial plastic work, and produces high apparent toughness ($K_c$).
• Thick Plates (Plane Strain): Deep inside a thick plate, surrounding elastic material prevents lateral contraction ($\epsilon_z = 0$). This constraint generates a large triaxial tensile stress ($\sigma_z \approx \nu(\sigma_x + \sigma_y)$). Triaxial tension severely suppresses plastic slip, producing a microscopic plastic zone and an invariant, lower-bound material toughness plateau: $K_{Ic}$.

To ensure that experimental measurements or design calculations are valid lower-bound plane strain values, textbooks use the classic ASTM E399 size criterion:

$B \ge 2.5\left(\frac{K_{Ic}}{S_y}\right)^2 \quad \text{and} \quad a \ge 2.5\left(\frac{K_{Ic}}{S_y}\right)^2$

where $S_y$ is the $0.2\%$ offset yield strength of the material. Since the 2012 revision, ASTM E399 makes the uncracked ligament the mandatory size check, $W - a \ge 2.5(K_{Ic}/\sigma_{YS})^2$, together with standard specimen proportions (such as $B \approx W/2$) and a $P_{\max}/P_Q \le 1.10$ limit. These rules qualify a laboratory test specimen; in design, $K_{Ic}$ is used as a conservative lower-bound property for thick, highly constrained parts.

Irwin Plastic Zone Size

Because actual engineering metals yield at stresses exceeding $S_y$, a small zone of plastic deformation develops at the crack tip. Irwin estimated the radius of this plastic zone ($r_y$) as:

$r_y = \frac{1}{2\pi}\left(\frac{K_I}{S_y}\right)^2 \quad (\text{Plane Stress})$

$r_y = \frac{1}{6\pi}\left(\frac{K_I}{S_y}\right)^2 \quad (\text{Plane Strain})$

Under plane strain, triaxial constraint reduces the plastic zone volume by a factor of 3, promoting sudden cleavage.

Non-Destructive Evaluation (NDE) Crack Detection Thresholds

In modern damage-tolerant aerospace and pressure vessel design, structures are assumed to contain initial flaws ($a_0$) equal to the largest flaw that could reliably escape detection during quality inspection (typically established at a $90\%$ Probability of Detection with a $95\%$ confidence limit, 90/95 POD):

Leak-Before-Break (LBB) Design Criterion

In the design of high-pressure piping, gas storage cylinders, and reactor pressure vessels (ASME Section XI), Leak-Before-Break (LBB) is the primary safety criterion against catastrophic explosion.

At a design operating stress $\sigma$, the critical crack size for unstable fast fracture is:

$K_I = K_{Ic} \implies Y \sigma \sqrt{\pi a_c} = K_{Ic} \implies a_c = \frac{1}{\pi}\left(\frac{K_{Ic}}{Y \sigma}\right)^2$

Let $t$ represent the wall thickness of the pressure vessel:

1. Leak-Before-Break Satisfied ($a_c > t$): If the critical crack length exceeds the wall thickness, a growing crack will breach the exterior surface ($a = t$) while remaining stable. The pressurized fluid escapes through the through-wall slit, producing a detectable drop in system pressure or visible fluid venting. Operators can execute an orderly emergency shutdown before catastrophic fracture occurs.
2. Leak-Before-Break Violated ($a_c < t$): If the critical crack depth is less than the wall thickness, the crack reaches unstable fast-fracture velocity while still partially embedded inside the wall. The vessel ruptures catastrophically into multiple flying fragments without prior leakage warning.

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5. Worked Numerical Engineering Scenario: Linear Elastic Fracture Mechanics of a Heavy Pressure Wall

Problem Statement

A thick-walled tension plate forming part of a high-pressure hydraulic accumulator is fabricated from high-strength quenched and tempered alloy steel (AISI 4340). Quality control records document the following mechanical properties:

• Yield strength: $S_y = 1200\text{ MPa}$ ($174\text{ ksi}$)
• Ultimate tensile strength: $S_{ut} = 1400\text{ MPa}$ ($203\text{ ksi}$)
• Plane strain fracture toughness: $K_{Ic} = 65.0\text{ MPa}\sqrt{\text{m}}$ ($59.2\text{ ksi}\sqrt{\text{in}}$)

The plate has a rectangular cross-section with width $W = 200\text{ mm}$ ($0.200\text{ m}$) and thickness $B = t = 25.0\text{ mm}$ ($0.025\text{ m}$). It is subjected to a uniform nominal gross tensile stress of $\sigma = 350\text{ MPa}$ ($50.8\text{ ksi}$).

During a scheduled ultrasonic maintenance inspection (UT), an edge crack of depth $a = 6.0\text{ mm}$ ($0.0060\text{ m}$) is detected normal to the tensile stress axis. For this single-edge notched geometry, the geometry correction factor is governed by:

$Y = 1.12 - 0.231\left(\frac{a}{W}\right) + 10.55\left(\frac{a}{W}\right)^2 - 21.72\left(\frac{a}{W}\right)^3 + 30.39\left(\frac{a}{W}\right)^4$

Perform the following engineering evaluations:

1. Verify whether the component satisfies the ASTM E399 minimum thickness requirement for valid plane strain conditions.
2. Calculate the operating Mode I stress intensity factor ($K_I$) and the resulting factor of safety against catastrophic brittle fracture.
3. Determine the critical crack depth ($a_c$) that would cause sudden brittle fracture under the current operating stress.
4. Determine whether this component satisfies the Leak-Before-Break (LBB) safety criterion.

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Step-by-Step Solution

Step 1: ASTM E399 Valid Plane Strain Thickness Verification

The minimum thickness required for valid plane strain fracture toughness behavior is:

$B_{\min} = 2.5\left(\frac{K_{Ic}}{S_y}\right)^2$

Substitute the material properties:

$B_{\min} = 2.5\left(\frac{65.0\text{ MPa}\sqrt{\text{m}}}{1200\text{ MPa}}\right)^2 = 2.5(0.054167\sqrt{\text{m}})^2 = 2.5(0.002934\text{ m}) = 0.007335\text{ m} = 7.34\text{ mm}$

Comparing with the actual plate dimensions:

• Actual thickness: $B = 25.0\text{ mm}$, more than three times $7.34\text{ mm}$, so the plate is thick enough for high crack-tip constraint.
• Actual crack depth: $a = 6.0\text{ mm}$, slightly less than $7.34\text{ mm}$, so a specimen this shallow would not meet the classic size rule. A shallow crack loses some constraint, which raises the apparent toughness, so using the lower-bound $K_{Ic}$ is conservative.

The size criteria qualify a laboratory test; for this component the conclusion is that $K_{Ic} = 65.0\text{ MPa}\sqrt{\text{m}}$ is an appropriate (conservative) design toughness.

Step 2: Operating Stress Intensity Factor ($K_I$) & Factor of Safety

First, determine the crack-to-width ratio:

$\frac{a}{W} = \frac{6.0\text{ mm}}{200.0\text{ mm}} = 0.030$

Calculate the geometry correction factor $Y$:

$Y = 1.12 - 0.231(0.030) + 10.55(0.030)^2 - 21.72(0.030)^3 + 30.39(0.030)^4$

$Y = 1.12 - 0.00693 + 0.009495 - 0.000586 + 0.000025 = 1.1220 \approx 1.122$

Now calculate the applied Mode I stress intensity factor:

$K_I = Y \sigma \sqrt{\pi a} = (1.122)(350\text{ MPa})\sqrt{\pi (0.0060\text{ m})}$

$\sqrt{\pi (0.0060)} = \sqrt{0.0188496} = 0.137294\sqrt{\text{m}}$

$K_I = (1.122)(350)(0.137294) = 53.915\text{ MPa}\sqrt{\text{m}} \approx 53.92\text{ MPa}\sqrt{\text{m}}$

The factor of safety against static brittle fracture is:

$n = \frac{K_{Ic}}{K_I} = \frac{65.0\text{ MPa}\sqrt{\text{m}}}{53.92\text{ MPa}\sqrt{\text{m}}} = 1.205 \approx 1.21$

Note on Yielding: The nominal stress is $\sigma = 350\text{ MPa}$, giving a static tensile yield factor of safety $n_y = S_y / \sigma = 1200 / 350 = 3.43$. However, the presence of the $6\text{ mm}$ crack reduces the true safety margin to a precarious $1.21$!

Step 3: Critical Crack Depth ($a_c$) Under Operating Stress

At the point of catastrophic fracture, $K_I = K_{Ic} = 65.0\text{ MPa}\sqrt{\text{m}}$. Using $Y \approx 1.12$ for shallow edge cracks:

$K_{Ic} = Y \sigma \sqrt{\pi a_c} \implies a_c = \frac{1}{\pi}\left(\frac{K_{Ic}}{Y \sigma}\right)^2$

$a_c = \frac{1}{\pi}\left(\frac{65.0\text{ MPa}\sqrt{\text{m}}}{(1.12)(350\text{ MPa})}\right)^2 = \frac{1}{\pi}\left(\frac{65.0}{392.0}\right)^2 = \frac{1}{\pi}(0.165816)^2 = \frac{0.027495}{\pi} = 0.008752\text{ m} = 8.75\text{ mm}$

Refining with $a/W = 8.75/200 = 0.0438$ and all five terms of the geometry polynomial gives $Y = 1.128$:

$a_c = \frac{1}{\pi}\left(\frac{65.0}{(1.128)(350)}\right)^2 = \frac{1}{\pi}\left(\frac{65.0}{394.8}\right)^2 = \frac{1}{\pi}(0.16464)^2 = 0.00863\text{ m} = 8.63\text{ mm}$

(A second iteration changes nothing at this precision.) Catastrophic fracture will occur if the crack grows by an additional $8.63 - 6.00 = 2.63\text{ mm}$.

Step 4: Leak-Before-Break (LBB) Evaluation

Treat the $25.0\text{ mm}$ plate thickness as the pressure wall and apply the simplified textbook LBB check, comparing the critical crack size with the wall thickness $t$ (a full LBB assessment uses surface-crack and through-wall-crack solutions for the real vessel geometry):

$a_c = 8.63\text{ mm} < t = 25.0\text{ mm}$

• Engineering Finding: The critical crack size ($8.63\text{ mm}$) is far smaller than the $25\text{ mm}$ wall thickness ($a_c / t = 0.345$, roughly one-third of the wall).
• Safety Conclusion: The component fails the Leak-Before-Break criterion. A growing crack would reach the fast-fracture size long before it could breach the wall and leak, so the wall would rupture without a warning leak.
• Corrective Engineering Action: For the crack to reach the full $25\text{ mm}$ wall before going unstable ($Y \approx 1.22$ at $a/W = 0.125$), the material would need $K_{Ic} \ge (1.22)(350)\sqrt{\pi(0.025)} \approx 120\text{ MPa}\sqrt{\text{m}}$ at the present stress (for example, a tougher grade or a lower-strength temper of the steel). Alternatively, keeping $K_{Ic} = 65\text{ MPa}\sqrt{\text{m}}$, the operating stress would have to drop to about $65/(1.22\sqrt{\pi(0.025)}) \approx 190\text{ MPa}$.