9.1 Failure Theories, Static & Fluctuating Stress Design

9.1 Failure Theories, Static & Fluctuating Stress Design

Machine components in mechanical systems are subjected to static, dynamic, and fluctuating multiaxial loads. Mechanical engineering design requires predicting when a component will cease to fulfill its intended function due to excessive elastic deformation, permanent plastic deformation (yielding), or macroscopic fracture. The PRC Mechanical Engineering Licensure Examination (MELE) heavily tests static failure theories for both ductile and brittle materials, stress concentration factor calculations, and cyclic fatigue design criteria.

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1. Static Failure Theories for Ductile Materials

Ductile materials are defined as those exhibiting a tensile elongation at fracture of $\epsilon_f \ge 0.05$ (5%) and possessing yield strengths ($S_y$) nearly equal in tension and compression ($S_{yt} \approx S_{yc}$). Failure in ductile materials is characterized by yield-initiated plastic slip along maximum shear planes.

Maximum Shear Stress (MSS) Theory (Tresca Criterion)

The Maximum Shear Stress theory states that yielding begins in a multiaxial stress state whenever the maximum shear stress $\tau_{max}$ in any element equals or exceeds the maximum shear stress present in a uniaxial tensile test specimen at the onset of yield ($S_y / 2$).

For a three-dimensional principal stress state where $\sigma_1 \ge \sigma_2 \ge \sigma_3$:

$\tau_{max} = \frac{\sigma_1 - \sigma_3}{2} \ge \frac{S_y}{2}$

Expressing this in terms of the design Factor of Safety ($N$):

$N = \frac{S_y}{\sigma_1 - \sigma_3} = \frac{S_y}{2 \tau_{max}}$

For a two-dimensional (plane stress) state $(\sigma_z = 0, \tau_{xz} = \tau_{yz} = 0)$, the principal stresses are calculated as:

$\sigma_{A,B} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}$

• Case 1 (Same Signs): $\sigma_A, \sigma_B > 0$ or $\sigma_A, \sigma_B < 0$. The third principal stress $\sigma_3 = 0$ lies outside or inside, making $\tau_{max} = \max(|\sigma_A|, |\sigma_B|) / 2$.
• Case 2 (Opposite Signs): $\sigma_A > 0 > \sigma_B$. Then $\sigma_1 = \sigma_A$ and $\sigma_3 = \sigma_B$, so $\tau_{max} = (\sigma_A - \sigma_B) / 2$.

Note: The MSS theory is conservative (underpredicts strength by up to 15%), making it safe and easy to apply in general machine design.

Distortion Energy (DE) Theory (von Mises / Octahedral Shear Stress)

The Distortion Energy theory states that yielding occurs when the distortion strain energy per unit volume in a stress state equals or exceeds the distortion strain energy per unit volume at the yield point in uniaxial tension or compression.

The equivalent von Mises stress $\sigma'$ in a general 3D principal stress state is given by:

$\sigma' = \sqrt{\frac{(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2}{2}}$

For 2D plane stress states with normal stresses $\sigma_x, \sigma_y$ and shear stress $\tau_{xy}$:

$\sigma' = \sqrt{\sigma_x^2 - \sigma_x \sigma_y + \sigma_y^2 + 3 \tau_{xy}^2} \le S_y$

The factor of safety under the Distortion Energy theory is:

$N = \frac{S_y}{\sigma'}$

Pure shear yield strength predicted by von Mises theory is $S_{sy} = 0.577 S_y$ (compared to $S_{sy} = 0.50 S_y$ under MSS). DE theory matches experimental yield data for ductile metals (steels, aluminum, titanium) with high precision.

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2. Static Failure Theories for Brittle Materials

Brittle materials (fracture elongation $\epsilon_f < 0.05$, such as gray cast iron, hardened steel, and ceramics) fail by catastrophic fracture without significant plastic yield. Compressive ultimate strength ($S_{uc}$) is typically much higher than tensile ultimate strength ($S_{ut}$) ($S_{uc} \gg S_{ut}$).

Maximum Normal Stress (MNS) Theory (Rankine Criterion)

States that failure occurs when the maximum principal tensile stress $\sigma_1$ reaches the ultimate tensile strength $S_{ut}$, or when the minimum principal compressive stress $\sigma_3$ reaches the ultimate compressive strength $S_{uc}$:

$N = \frac{S_{ut}}{\sigma_1} \quad \text{or} \quad N = \frac{S_{uc}}{|\sigma_3|}$

Limitation: MNS theory works well in Quadrant I (both principal stresses tensile), but overpredicts strength in Quadrant IV where shear components are high.

Coulomb-Mohr & Modified Mohr Theories

Coulomb-Mohr accounts for unequal tensile and compressive properties by constructing envelope tangents to compression and tension stress circles:

$\frac{\sigma_1}{S_{ut}} - \frac{\sigma_3}{S_{uc}} = \frac{1}{N} \quad (\text{for } \sigma_1 \ge 0 \ge \sigma_3)$

Modified Mohr Theory adjusts the failure locus in Quadrant IV $(\sigma_1 > 0 > \sigma_3)$ to better fit experimental brittle fracture data:

• If $\sigma_1 > 0 > \sigma_3$ and $|\sigma_3 / \sigma_1| \le 1$: Failure occurs when $\sigma_1 = S_{ut}$.
• If $\sigma_1 > 0 > \sigma_3$ and $|\sigma_3 / \sigma_1| > 1$: $\frac{(S_{uc} - S_{ut}) \sigma_1}{S_{uc} S_{ut}} - \frac{\sigma_3}{S_{uc}} = \frac{1}{N}$.

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3. Factor of Safety ($N$) Determination

The Factor of Safety $N$ is defined as:

$N = \frac{\text{Material Strength (Yield or Ultimate)}}{\text{Allowable Design Stress}}$

Selection Guidelines in ME Practice

• $N = 1.25 - 1.50$: Highly controlled conditions, well-known loads, lightweight aerospace applications.
• $N = 1.50 - 2.00$: Standard mechanical engineering design with known materials and moderate environmental shock.
• $N = 2.00 - 3.00$: Applications with loading uncertainty, dynamic impact, or brittle materials.
• $N = 3.00 - 4.00+$: Hazardous environments, human safety critical structures (elevators, pressure vessels).

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4. Fluctuating Stress & Fatigue Failure Design

Machine elements like crankshafts, turbine blades, and gears experience cyclic loading where fracture occurs at stresses far below static yield strength due to microcrack nucleation and propagation.

Stress Components in Cyclic Loading

For a stress waveform varying between maximum stress $\sigma_{max}$ and minimum stress $\sigma_{min}$:

1. Mean Stress ($\sigma_m$): $\sigma_m = \frac{\sigma_{max} + \sigma_{min}}{2}$
2. Alternating Stress ($\sigma_a$): $\sigma_a = \frac{|\sigma_{max} - \sigma_{min}|}{2}$
3. Stress Ratio ($R$): $R = \frac{\sigma_{min}}{\sigma_{max}}$
4. Amplitude Ratio ($A$): $A = \frac{\sigma_a}{\sigma_m}$

Modified Endurance Limit ($S_e$)

The rotary-beam specimen endurance limit $S_e'$ is modified for real components using Marin factors:

$S_e = k_a k_b k_c k_d k_e k_f S_e'$

Where $S_e' \approx 0.50 S_{ut}$ for steels with $S_{ut} \le 1400 \text{ MPa}$ ($200 \text{ kpsi}$).

• $k_a$: Surface factor (ground, machined, cold-drawn, forged)
• $k_b$: Size factor
• $k_c$: Reliability factor
• $k_d$: Temperature factor
• $k_e$: Duty cycle / operating stress factor
• $k_f$: Miscellaneous effects factor

Stress Concentration & Fatigue Notch Factor

Geometric discontinuities (fillets, keyways, holes) increase local stresses. The geometric stress concentration factor $K_t$ is converted to the fatigue notch factor ($K_f$) using notch sensitivity $q$ ($0 \le q \le 1$):

$K_f = 1 + q(K_t - 1)$

$\sigma_{a,actual} = K_f \sigma_{a,nominal}$

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5. Fatigue Failure Criteria (Fluctuating Stress with Non-Zero Mean Stress)

When a machine part experiences both alternating stress $\sigma_a$ and tensile mean stress $\sigma_m$, safe fatigue life boundaries are represented on an S-N mean stress diagram.

  Alternating Stress (σa)
    ^
 S_e| \  (Soderberg Line: connects S_e & S_y)
    |  \  (Goodman Line: connects S_e & S_ut)
    |   \  (ASME Elliptic: curve between S_e & S_y)
    |    \ 
    +-----+-------+-------------> Mean Stress (σm)
    0    S_y     S_ut

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6. Worked Machine Design Example: Fluctuating Bending Calculation

Problem Statement

A machine shaft made of AISI 1045 cold-drawn steel has an ultimate tensile strength $S_{ut} = 650 \text{ MPa}$ and yield strength $S_y = 400 \text{ MPa}$. The modified endurance limit of the shaft at a critical fillet section is $S_e = 200 \text{ MPa}$. The nominal bending stress at the fillet fluctuates dynamically between $\sigma_{min} = -50 \text{ MPa}$ and $\sigma_{max} = +250 \text{ MPa}$. Theoretical stress concentration factor at the fillet is $K_t = 1.65$, and the material notch sensitivity is $q = 0.80$.

Calculate:

1. The fatigue notch factor $K_f$.
2. The actual nominal mean stress $\sigma_m$ and alternating stress $\sigma_a$.
3. The design factor of safety $N$ using the Goodman criterion.
4. The design factor of safety $N$ using the Soderberg criterion.

Step-by-Step Solution

Step 1: Compute Fatigue Notch Factor $K_f$ $K_f = 1 + q(K_t - 1) = 1 + 0.80(1.65 - 1) = 1 + 0.80(0.65) = 1 + 0.52 = 1.52$

Step 2: Calculate Mean Stress $\sigma_m$ and Alternating Stress $\sigma_a$ $\sigma_m = \frac{\sigma_{max} + \sigma_{min}}{2} = \frac{250 + (-50)}{2} = \frac{200}{2} = 100 \text{ MPa}$

$\sigma_{a,nominal} = \frac{|\sigma_{max} - \sigma_{min}|}{2} = \frac{250 - (-50)}{2} = \frac{300}{2} = 150 \text{ MPa}$

Applying $K_f$ to alternating stress (mean stress is unnotched for ductile materials under fatigue): $\sigma_a = K_f \cdot \sigma_{a,nominal} = 1.52 \times 150 \text{ MPa} = 228 \text{ MPa}$

Step 3: Factor of Safety using Modified Goodman Line $\frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_{ut}} = \frac{1}{N}$

$\frac{228}{200} + \frac{100}{650} = \frac{1}{N}$

$1.1400 + 0.1538 = 1.2938 = \frac{1}{N}$

$N = \frac{1}{1.2938} = 0.773$

Interpretation: Because $N < 1.0$, the component will fail in fatigue prior to infinite life under the Goodman criterion.

Step 4: Factor of Safety using Soderberg Line $\frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_y} = \frac{1}{N}$

$\frac{228}{200} + \frac{100}{400} = \frac{1}{N}$

$1.1400 + 0.2500 = 1.3900 = \frac{1}{N}$

$N = \frac{1}{1.3900} = 0.719$

Conclusion: Soderberg criterion yields $N = 0.719$, which is even more conservative than Goodman ($N = 0.773$), confirming finite life.