Mechanical Engineering Design · Chapter 5
Mechanical Engineering Design · Part 2 — Failure Prevention

Failures Resulting from Static Loading

How to predict static failure from limited material data: the yield criteria for ductile metals (maximum shear stress, distortion energy, Coulomb-Mohr), the fracture criteria for brittle materials (maximum normal stress, Mohr modifications), criterion selection, and an introduction to linear elastic fracture mechanics.

Learning objectives

  • Recognize whether yielding, brittle fracture, or crack instability controls a static-load decision.
  • Convert a free-body diagram and nominal loads into a signed local stress state.
  • Select and apply MSS, distortion-energy, Coulomb-Mohr, or LEFM equations.
  • Check factors of safety against assumptions, dimensions, manufacturing details, and data provenance.
  • Iterate geometry or material while keeping strength distinct from stress and stiffness.

Physical meaning

A static load does not imply a single stress. Equilibrium supplies loads, geometry converts them to local stresses, and a failure criterion compares that multiaxial state with independently measured material strength.

Problem-solving strategy

  1. Draw the free-body diagram and identify the section, surface, or crack tip most likely to control.
  2. Declare tensile stress positive, calculate nominal stress, and apply a justified local concentration model.
  3. Classify service behavior as ductile yielding, brittle fracture, or crack-controlled instability before selecting an equation.
  4. Order all three principal stresses, including the zero principal in plane stress, then compute the governing utilization.
  5. Compare the achieved factor with the design target and iterate dimensions, radii, material, inspection, or process.

5–1Static Strength

Ideal vs. Available Data

Ideally a designer would have strength tests on specimens of the exact material, heat treatment, surface finish, and size, loaded exactly as the part will be in service. Such data are justified when failure endangers life or when production volume is huge (appliances achieve high reliability because per-part test cost is tiny). In reality, design usually proceeds from published yield strength, ultimate strength, percent reduction in area, and percent elongation — so the methods of this chapter exist to stretch that meager data across static and dynamic loads, 2-D and 3-D stress states, and a wide range of temperatures and sizes.

Four Design Categories

(1) Failure endangers life, or the part is made in enormous quantity — an elaborate test program is justified. (2) Quantity supports a moderate test series. (3) Quantity (or schedule) is too small for testing. (4) The part is already designed, built, and found unsatisfactory — analysis is needed to learn why and how to improve it.

5–2Stress Concentration

When to Set Kt = 1 (Ductile, Static)

Stress concentration is highly localized. For a ductile material under static load, local yielding at a notch strain-strengthens that tiny region and raises its yield strength, so the part carries the load with no general yielding — hence designers set the geometric factor \(K_t = 1\). The argument: in an idealized elastic-perfectly-plastic material with \(K_t = 2\) and \(S_y = 40\) kpsi, a 20-kip load gives \(K = 40/20 = 2\), a 30-kip load \(K = 40/30 = 1.33\), and a 40-kip load \(K = 40/40 = 1\) — the notch stress is capped at \(S_y\) while the bulk approaches general yield.

When Kt Must Be Applied

Apply \(K_t\) (or \(K_{ts}\)) to brittle materials, where there is no plastic range — the raised stress can initiate fracture and catastrophic failure. Exception: a brittle material whose own microstructure already contains worse stress raisers (gray cast iron's graphite flakes are essentially built-in cracks; its reported tensile strength already includes them), where added \(K_t\) need not be applied. Always confirm a "ductile" material is not liable to brittle fracture in its service environment (low temperature, cyclic or rapid loading, defects).

Stress-Concentration Factors (from Ch. 3)
\[ \sigma_{\max} = K_t\,\sigma_{nom} \qquad\qquad \tau_{\max} = K_{ts}\,\tau_{nom} \]Eq. (3–48)

5–3Failure Theories

No Universal Theory

Distortion, permanent set, cracking, and rupture are all "failure." There is no single universal failure theory, so several tested hypotheses are accepted as working "theories." Metals are classed as ductile (\(\epsilon_f \ge 0.05\), with an identifiable yield strength usually equal in tension and compression, \(S_{yt} = S_{yc} = S_y\)) or brittle (\(\epsilon_f < 0.05\), no clear yield, characterized by \(S_{ut}\) and \(S_{uc}\)).

The Accepted Criteria

Ductile (yield criteria): maximum shear stress (MSS), distortion energy (DE), ductile Coulomb-Mohr (DCM). Brittle (fracture criteria): maximum normal stress (MNS), brittle Coulomb-Mohr (BCM), modified Mohr (MM). The following sections describe the basis of each and apply them.

5–4Maximum-Shear-Stress Theory for Ductile Materials

Maximum-Shear-Stress (MSS / Tresca) Theory
Yielding begins when the maximum shear stress in any element equals or exceeds the maximum shear stress in a tension specimen of the same material at yield. Since the tension specimen yields at \(\tau_{\max} = S_y/2\), for a general state with ordered principal stresses \(\sigma_1 \ge \sigma_2 \ge \sigma_3\), yield is predicted when \(\tau_{\max} = (\sigma_1 - \sigma_3)/2 \ge S_y/2\).

Physical Basis

Ductile tension specimens form slip (Lüders) lines and fracture lines at roughly 45° to the axis — where shear is maximum — suggesting shear drives yield. MSS is an acceptable but conservative predictor, so it is widely used. It implies a shear yield strength \(S_{sy} = 0.5 S_y\), about 15% low.

MSS Yield Condition
\[ \tau_{\max} = \frac{\sigma_1 - \sigma_3}{2} \ge \frac{S_y}{2} \quad\Leftrightarrow\quad \sigma_1 - \sigma_3 \ge S_y \]Eq. (5–1)
Shear Yield Strength (MSS)
\[ S_{sy} = 0.5\,S_y \]Eq. (5–2)
MSS Design Equation
\[ \tau_{\max} = \frac{S_y}{2n} \quad\Leftrightarrow\quad \sigma_1 - \sigma_3 = \frac{S_y}{n} \]Eq. (5–3)

Plane Stress Is Three-Dimensional

Plane stress always has a third principal stress of zero out of plane, so the in-plane principals from Eq. (3–13) — labelled \(\sigma_A \ge \sigma_B\) — must be ordered together with the zero before applying Eq. (5–1). Three cases result (assuming \(\sigma_A \ge \sigma_B\)): Case 1 (\(\sigma_A \ge \sigma_B \ge 0\)): \(\sigma_A \ge S_y\); Case 2 (\(\sigma_A \ge 0 \ge \sigma_B\)): \(\sigma_A - \sigma_B \ge S_y\); Case 3 (\(0 \ge \sigma_A \ge \sigma_B\)): \(\sigma_B \le -S_y\). These form the hexagonal MSS yield envelope. Shaft problems (bending or axial normal stress plus torsional shear, with \(\sigma_y = 0\)) are Case 2.

5–5Distortion-Energy Theory for Ductile Materials

Distortion-Energy (DE / von Mises) Theory
Yielding occurs when the distortion strain energy per unit volume reaches the distortion strain energy per unit volume at yield in simple tension. Equivalently, yield occurs when the von Mises stress \(\sigma'\) — a single effective stress capturing the whole 3-D state — reaches \(S_y\).

Origin: Distortion, Not Volume Change

Ductile materials under hydrostatic (equal) principal stresses show yield strengths far above the tension-test value — so yield must relate to angular distortion, not volume change. Splitting the stress state into a hydrostatic part \(\sigma_{av} = (\sigma_1+\sigma_2+\sigma_3)/3\) (pure volume change) and a remainder (pure distortion), and subtracting the volumetric strain energy from the total, gives the distortion energy — which is zero when \(\sigma_1 = \sigma_2 = \sigma_3\). DE predicts no yield under pure hydrostatic stress and fits ductile data best, so it is the recommended default.

Distortion Energy
\[ u_d = \frac{1+\nu}{3E}\left[\frac{(\sigma_1-\sigma_2)^2 + (\sigma_2-\sigma_3)^2 + (\sigma_3-\sigma_1)^2}{2}\right] \]Eq. (5–8)
Von Mises Stress — Principal Form
\[ \sigma' = \left[\frac{(\sigma_1-\sigma_2)^2 + (\sigma_2-\sigma_3)^2 + (\sigma_3-\sigma_1)^2}{2}\right]^{1/2} \]Eq. (5–12)
Von Mises Stress — Plane Stress (principal)
\[ \sigma' = \left(\sigma_A^2 - \sigma_A\sigma_B + \sigma_B^2\right)^{1/2} \]Eq. (5–13)
Von Mises Stress — 3-D Components
\[ \sigma' = \frac{1}{\sqrt{2}}\left[(\sigma_x-\sigma_y)^2 + (\sigma_y-\sigma_z)^2 + (\sigma_z-\sigma_x)^2 + 6(\tau_{xy}^2 + \tau_{yz}^2 + \tau_{zx}^2)\right]^{1/2} \]Eq. (5–14)
Von Mises Stress — Plane Stress (components)
\[ \sigma' = \left(\sigma_x^2 - \sigma_x\sigma_y + \sigma_y^2 + 3\tau_{xy}^2\right)^{1/2} \]Eq. (5–15)
DE Design Equation
\[ \sigma' = \frac{S_y}{n} \]Eq. (5–19)
Shear Yield Strength (DE)
\[ S_{sy} = 0.577\,S_y \]Eq. (5–21)

Octahedral-Shear-Stress View & Why MSS Is Conservative

DE is identical to the octahedral-shear-stress theory: failure when the octahedral shear stress reaches its tension-test value \(\sqrt{2}\,S_y/3\). The DE shear yield strength \(S_{sy} = 0.577 S_y\) is about 15% higher than the MSS value \(0.5 S_y\). The difference is that MSS ignores the normal stresses on the 45° planes (which are \(P/2A\), not the hydrostatic \(P/3A\)) — making the MSS hexagon sit inside the DE ellipse, hence MSS always gives an equal or lower factor of safety.

σ_A σ_B DE MSS S_y S_y
FIG. 5.1Plane-stress yield envelopes: the DE ellipse and the inscribed MSS hexagon. MSS is the more conservative criterion. (Original diagram, redrawn.)

5–6Coulomb-Mohr Theory for Ductile Materials

Unequal Tension and Compression Strengths

Some materials are not equally strong in tension and compression (magnesium's compressive yield can be half its tensile; gray iron's compressive ultimate is 3–4× its tensile). Mohr's 1900 theory builds a failure envelope tangent to three test circles (tension, compression, shear); the Coulomb-Mohr (internal-friction) variant assumes that envelope is a straight line, needing only the tension and compression strengths. With ordered principals \(\sigma_1 \ge \sigma_2 \ge \sigma_3\), similar triangles give the failure relation below (yield or ultimate strengths may be used).

Coulomb-Mohr Failure Relation
\[ \frac{\sigma_1}{S_t} - \frac{\sigma_3}{S_c} = 1 \]Eq. (5–22)
Coulomb-Mohr Design Equation
\[ \frac{\sigma_1}{S_t} - \frac{\sigma_3}{S_c} = \frac{1}{n} \]Eq. (5–26)
Shear Yield Strength (Coulomb-Mohr)
\[ S_{sy} = \frac{S_{yt}\,S_{yc}}{S_{yt} + S_{yc}} \]Eq. (5–27)

Plane-Stress Cases

For \(\sigma_A \ge \sigma_B\): Case 1 (\(\sigma_A \ge \sigma_B \ge 0\)): \(\sigma_A \ge S_t\); Case 2 (\(\sigma_A \ge 0 \ge \sigma_B\)): \(\sigma_A/S_t - \sigma_B/S_c \ge 1\); Case 3 (\(0 \ge \sigma_A \ge \sigma_B\)): \(\sigma_B \le -S_c\).

5–7Failure of Ductile Materials — Summary

Which Ductile Theory to Use

Experimental data (Marin and others) show both MSS and DE are acceptable for ductile materials. For design, MSS is quick, easy, and conservative. To learn why a part failed, DE passes closer to the center of the data and predicts better — but a center-of-data curve has only ~50% reliability, so use a larger safety factor with it. When \(S_{yt} \ne S_{yc}\), the full Mohr theory is best, but ductile Coulomb-Mohr (tension and compression strengths only) is the practical equation-form alternative.

5–8Maximum-Normal-Stress Theory for Brittle Materials

Maximum-Normal-Stress (MNS) Theory
Failure occurs whenever one of the three principal stresses equals or exceeds the corresponding strength: with \(\sigma_1 \ge \sigma_2 \ge \sigma_3\), failure when \(\sigma_1 \ge S_{ut}\) or \(\sigma_3 \le -S_{uc}\) (\(S_{ut}\), \(S_{uc}\) positive).
MNS Design Equations (plane stress, \(\sigma_A \ge \sigma_B\))
\[ \sigma_A = \frac{S_{ut}}{n} \qquad\text{or}\qquad \sigma_B = -\frac{S_{uc}}{n} \]Eq. (5–30)

Why MNS Is Not Recommended

MNS predicts poorly in the fourth quadrant of the \(\sigma_A,\sigma_B\) plane (one tensile, one compressive principal), so it is retained mainly for historical reasons; the Mohr modifications below represent brittle data better.

5–9Modifications of the Mohr Theory for Brittle Materials

Brittle Coulomb-Mohr and Modified Mohr

Two plane-stress design forms (with safety factor \(n\)) improve on MNS in the fourth quadrant. Brittle Coulomb-Mohr (BCM) applies the straight Coulomb-Mohr line. Modified Mohr (MM) expands the fourth quadrant to match observed gray-iron data far better, switching form at the load-line slope \(|\sigma_B/\sigma_A| = 1\). A parabolic refinement exists but adds nonlinearity for a minor gain and is omitted.

Brittle Coulomb-Mohr (BCM)
\[ \sigma_A = \frac{S_{ut}}{n}\ \ (\sigma_A \ge \sigma_B \ge 0) \quad\;\; \frac{\sigma_A}{S_{ut}} - \frac{\sigma_B}{S_{uc}} = \frac{1}{n}\ \ (\sigma_A \ge 0 \ge \sigma_B) \quad\;\; \sigma_B = -\frac{S_{uc}}{n}\ \ (0 \ge \sigma_A \ge \sigma_B) \]Eq. (5–31)
Modified Mohr (MM) — fourth-quadrant form (|σ_B/σ_A| > 1)
\[ \frac{(S_{uc} - S_{ut})\,\sigma_A}{S_{uc}\,S_{ut}} - \frac{\sigma_B}{S_{uc}} = \frac{1}{n} \]Eq. (5–32b)

5–10Failure of Brittle Materials — Summary

What the Cast-Iron Data Show

For nominal grade-30 cast iron under biaxial stress: in the first quadrant MNS, Coulomb-Mohr, and modified Mohr coincide and fit the data well; in the fourth quadrant modified Mohr fits best and MNS does not; in the third quadrant there are too few points to fix a locus. Be alert that normally ductile materials can fracture in a brittle manner below the transition temperature.

5–11Selection of Failure Criteria

Decision Guide

Ductile behavior (\(\epsilon_f \ge 0.05\)): use distortion energy (DE) by preference, or maximum shear stress (MSS) for simplicity and conservatism; if \(S_{yt} \ne S_{yc}\), use ductile Coulomb-Mohr (DCM). Brittle behavior (\(\epsilon_f < 0.05\)): the curved original Mohr hypothesis is best, but for hand work choose brittle Coulomb-Mohr (BCM, conservative) or modified Mohr (MM). MNS is excluded — the others represent the data better.

ε_f < 0.05 ? brittle vs. ductile yes (brittle) Conservative? BCM Eq. (5–31) Mod. Mohr (MM) Eq. (5–32) no (ductile) S_yt ≈ S_yc ? DCM Eq. (5–26) Conservative? DE / MSS DE Eqs. (5–15),(5–19) MSS Eq.(5–3)
FIG. 5.2Failure-criterion selection: brittle vs. ductile by εf, then by equal strengths and the conservatism preference. (Original diagram, redrawn.)

5–12Introduction to Fracture Mechanics

Damage-Tolerant Design

Cracks exist before service and grow during it. Damage-tolerant design follows crack growth until it becomes critical, then retires the part — analyzed by linear elastic fracture mechanics (LEFM), with mandated periodic crack inspection where safety is at stake. Stress-concentration factors fail here: as a crack-tip radius approaches zero, \(K_t \to \infty\), and local plasticity further invalidates the elastic factor. Even normally ductile materials (hard steels, strong aluminum alloys, low-carbon steel below the transition temperature, glass) can fracture in a "relatively brittle" way — without yielding across the section.

The Elliptical Flaw and Crack Modes

For an elliptical flaw (semi-axes \(a\) across the load, \(b\) along it) in an infinite plate under stress \(\sigma\), the tip stress is \((\sigma_y)_{\max} = (1 + 2a/b)\sigma\) — giving the familiar factor of 3 for a circular hole (\(a = b\)) and \(\to \infty\) as \(b/a \to 0\) (an idealization plasticity prevents). Griffith showed crack growth occurs when the energy release rate exceeds the rate needed to create new surface. Three crack modes exist: Mode I (opening, tensile — most common and the focus here), Mode II (sliding, in-plane shear), Mode III (tearing, out-of-plane shear).

Elliptical-Flaw Tip Stress (Inglis)
\[ (\sigma_y)_{\max} = \left(1 + 2\frac{a}{b}\right)\sigma \]Eq. (5–33)
Mode I (opening) Mode II (sliding) Mode III (tearing)
FIG. 5.3The three crack-propagation modes; Mode I (tensile opening) is the most common in practice. (Original diagram, redrawn.)

Stress Intensity Factor and Fracture Toughness

Near a Mode I crack tip the stress field scales with the stress intensity factor \(K_I = \sigma\sqrt{\pi a}\) (units MPa·√m or kpsi·√in) — not to be confused with \(K_t\). Geometry enters through a modification factor \(\beta\): \(K_I = \beta\,\sigma\sqrt{\pi a}\). When \(K_I\) reaches the critical value \(K_{Ic}\) — the fracture toughness, a material property (plane-strain, Mode I) — the crack propagates. Typical metals: \(20 \le K_{Ic} \le 200\) MPa·√m; polymers/ceramics \(1\text{–}5\) MPa·√m. The strength-to-stress ratio gives the safety factor \(n = K_{Ic}/K_I\). A high \(S_y/S_u\) ratio and low-temperature service both signal brittle-fracture risk.

Mode I Stress Intensity Factor
\[ K_I = \sigma\sqrt{\pi a} \qquad\qquad K_I = \beta\,\sigma\sqrt{\pi a} \]Eqs. (5–37, 5–37 general)
Fracture-Mechanics Factor of Safety
\[ n = \frac{K_{Ic}}{K_I} \]Eq. (5–38)

5–13Important Design Equations

Quick Reference

The governing relations of this chapter, with \(\sigma_A,\sigma_B\) the plane-stress principals from Eq. (3–13): MSS \(\tau_{\max} = (\sigma_1-\sigma_3)/2 = S_y/(2n)\); DE \(\sigma' = S_y/n\) with \(\sigma'\) from Eq. (5–14)/(5–15); ductile Coulomb-Mohr \(\sigma_1/S_t - \sigma_3/S_c = 1/n\); brittle Coulomb-Mohr and modified Mohr per Eqs. (5–31)/(5–32); fracture \(K_I = \beta\sigma\sqrt{\pi a}\), \(n = K_{Ic}/K_I\).

EXAMPLE 1Auditing a biaxial bracket hotspot
At a ductile steel bracket hotspot, finite-element stresses are \(\sigma_x=80\ \text{MPa}\), \(\sigma_y=-20\ \text{MPa}\), and \(\tau_{xy}=30\ \text{MPa}\). With certified room-temperature yield strength \(S_y=250\ \text{MPa}\), find static factors of safety by distortion energy and MSS.

Recognition strategy

The supplied state is plane stress in a ductile metal, so yielding—not fracture—is the immediate family. DE is the usual correlation and MSS supplies a conservative comparison.

Given information

  • Local signed stress components, \(S_y=250\ \text{MPa}\), static proportional loading, and the requested factors \(n_{DE}\) and \(n_{MSS}\).

Model and interpretation

Use tensile normal stress as positive and positive shear as reported. The out-of-plane principal is zero; no additional notch factor is applied because the values are already local stresses.

Step-by-step solution
1. Plan. Separate demand from capacity: the three reported stresses describe load response, whereas \(S_y\) is material strength from a specified certificate.
2. Formulate. For plane stress, write \(\sigma'=\sqrt{\sigma_x^2-\sigma_x\sigma_y+\sigma_y^2+3\tau_{xy}^2}\) before inserting values.
3. Calculate. Substitution gives \(\sigma'=\sqrt{80^2-80(-20)+(-20)^2+3(30)^2}=105.4\ \text{MPa}\), hence \(n_{DE}=250/105.4=2.37\).
4. Resolve. The in-plane principals are \(30\pm\sqrt{50^2+30^2}=88.3,-28.3\ \text{MPa}\). Including zero, MSS gives \(n_{MSS}=250/[88.3-(-28.3)]=2.14\).

Verification

Units: Every term under the square root has units of stress squared, so the equivalent stress and principal-stress range are both in MPa.

Signs and direction: Compression remains negative, and ordering \(88.3\ge 0\ge -28.3\) ensures MSS uses the true largest principal separation.

Plausibility: MSS should not predict a larger factor than DE for this ductile state; \(2.14<2.37\), and both exceed unity without implying an approved design margin.

Final result: \(\boxed{n_{DE}=2.37,\qquad n_{MSS}=2.14}\).

Engineering interpretation

MSS is about ten percent more conservative here. Acceptance still depends on the specified design factor, stress convergence, temperature, residual stress, and confirmation that the quoted strength applies to the delivered material.

EXAMPLE 2Brittle housing under mixed principal stress
A brittle pump housing has local principal stresses \(\sigma_1=45\ \text{MPa}\), \(\sigma_2=0\), and \(\sigma_3=-90\ \text{MPa}\). Supplier-certified strengths are \(S_{ut}=150\ \text{MPa}\) and \(S_{uc}=600\ \text{MPa}\). Determine the brittle Coulomb-Mohr factor of safety.

Recognition strategy

Opposite-signed principal stresses and unequal brittle tensile and compressive strengths identify the fourth-quadrant Coulomb-Mohr case. A ductile von Mises comparison would use the wrong failure evidence.

Given information

  • Ordered local principals, positive magnitudes \(S_{ut}=150\ \text{MPa}\) and \(S_{uc}=600\ \text{MPa}\), static loading, room-temperature supplier data, and unknown \(n\).

Model and interpretation

Treat the housing as linear elastic to fracture, tensile stress positive. The stated stresses already include geometry effects; the strength magnitudes remain positive in the denominator.

Step-by-step solution
1. Plan. Confirm the ordering \(45\ge0\ge-90\ \text{MPa}\), which selects the mixed tension-compression branch rather than either single-sign branch.
2. Formulate. Write the utilization as \(1/n=\sigma_1/S_{ut}-\sigma_3/S_{uc}\); the minus sign converts the negative compressive principal into positive utilization.
3. Calculate. Insert the certified strengths to obtain \(1/n=45/150-(-90)/600=0.300+0.150=0.450\).
4. Resolve. Invert the dimensionless utilization: \(n=1/0.450=2.22\). The tensile contribution is twice the compressive contribution despite the larger compression magnitude.

Verification

Units: Each stress-to-strength ratio is dimensionless, so their sum and its reciprocal factor of safety are dimensionless.

Signs and direction: The compressive principal is negative in the numerator but \(S_{uc}\) is entered as a positive magnitude; this makes both failure contributions positive.

Plausibility: If compression strength tended to infinity, the result would approach \(150/45=3.33\); finite compressive sensitivity correctly lowers it to 2.22.

Final result: \(\boxed{n_{BCM}=2.22}\) for the supplied local stress state and certified strengths.

Engineering interpretation

Casting porosity, wall-thickness variation, low temperature, and proof-test history can dominate brittle reliability. Drawing radii and machining allowances must preserve the analyzed geometry, and production acceptance requires the governing code plus lot-specific material evidence.

EXAMPLE 3Screening a detectable edge crack
A tension link carries a local membrane stress of \(120\ \text{MPa}\). Inspection may leave an edge crack of depth \(a=2.00\ \text{mm}\); a validated geometry solution gives \(\beta=1.12\). With \(K_{Ic}=45\ \text{MPa}\sqrt{\text{m}}\) and \(S_y=360\ \text{MPa}\), calculate fracture and yield safety factors and the critical crack depth.

Recognition strategy

A known crack invalidates ordinary finite-notch stress concentration at its tip. LEFM provides crack demand, while a separate nominal yield check identifies the competing static failure mode.

Given information

  • \(\sigma=120\ \text{MPa}\), \(a=0.00200\ \text{m}\), validated \(\beta=1.12\), certified \(K_{Ic}=45\ \text{MPa}\sqrt{\text{m}}\), \(S_y=360\ \text{MPa}\).

Model and interpretation

Assume Mode I, linear elasticity, plane-strain toughness applicability, tensile stress positive, and geometry represented by the supplied beta. Residual stress and plastic-zone corrections are excluded.

Step-by-step solution
1. Plan. Use \(K_I=\beta\sigma\sqrt{\pi a}\) for fracture demand and \(n_y=S_y/\sigma\) for general yielding; these are independent limit states.
2. Formulate. The current crack gives \(K_I=1.12(120)\sqrt{\pi(0.00200)}=10.65\ \text{MPa}\sqrt{\text{m}}\).
3. Calculate. Therefore \(n_K=K_{Ic}/K_I=45/10.65=4.23\), while the separate nominal yield factor is \(n_y=360/120=3.00\).
4. Resolve. At fixed stress, solve \(a_c=[K_{Ic}/(\beta\sigma)]^2/\pi\), giving \(a_c=0.0355\ \text{m}=35.5\ \text{mm}\).

Verification

Units: Using crack depth in metres makes stress intensity units \(\text{MPa}\sqrt{\text{m}}\); the safety factors are dimensionless and the recovered crack size is in metres.

Signs and direction: Positive membrane tension opens the edge crack. Compression would close an ideal Mode I crack, so this equation must not be reused unchanged for a negative stress.

Plausibility: The computed critical depth exceeds the current 2 mm depth, and the smaller factor is the yield factor of 3.00; both trends are physically consistent for the supplied toughness.

Final result: \(\boxed{K_I=10.65\ \text{MPa}\sqrt{\text{m}},\ n_K=4.23,\ n_y=3.00,\ a_c=35.5\ \text{mm}}\).

Engineering interpretation

Yield governs this simplified snapshot, but crack growth, residual tension, corrosion, and inspection probability can reverse that conclusion. A professional disposition must use code-qualified geometry factors, valid toughness orientation and temperature, nondestructive-examination capability, and fatigue crack-growth data.

Chapter recap

  • Start with equilibrium and local stress; strength comes from independent material evidence.
  • Use DE or MSS for comparable ductile yield strengths, Mohr forms for asymmetric brittle behavior, and LEFM for explicit cracks.
  • Check every candidate against all credible static limit states and the required reliability process.

Before moving on

You should be able to order principal stresses, select a defensible criterion, calculate and audit a factor of safety, identify missing data, and explain how geometry or material would be iterated.

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