5.1 Static Failure Theories for Ductile Materials: von Mises & Tresca

5.1 Static Failure Theories for Ductile Materials: von Mises & Tresca

Quick Summary: In mechanical machine design, ductile materials sustain substantial plastic deformation before rupture ($\%EL \ge 5\%$) and fail predominantly through shear-induced crystallographic slip. Because uniaxial tensile test data ($S_y$) must predict yielding under complex three-dimensional multiaxial stress states, static failure theories map multiaxial stress tensors to a single equivalent scalar stress. This section details the theoretical foundations, governing equations, pure shear yield predictions, and planar yield loci for the Maximum Shear Stress (Tresca) and Distortion Energy (von Mises) theories, culminating in a comprehensive worked shaft design problem under combined bending and torsion.

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1. Fundamentals of Static Failure in Ductile Materials

Under quasi-static loading at ambient temperatures, engineering materials exhibit either ductile or brittle behavior:

• Ductile Classification: A material is classified as ductile if its percent elongation at fracture satisfies $\%EL \ge 5\%$. Ductile materials (such as low- and medium-carbon steels, structural aluminum alloys, austenitic stainless steels, copper, and titanium) undergo significant macroscopic plastic deformation before final rupture.
• Brittle Classification: A material is classified as brittle if $\%EL < 5\%$. Brittle materials (such as gray cast iron, hardened tool steels, and structural ceramics) fracture abruptly with negligible plastic strain.

The Microscopic Mechanism of Ductile Yielding

Plastic deformation in crystalline metals occurs by shear-induced slip—the sliding of atomic planes over one another via the motion of line defects known as dislocations. In a single crystal, slip initiates when the shear stress resolved onto a close-packed crystallographic plane in a close-packed direction reaches the critical resolved shear stress (Schmid's Law):

$\tau_{\text{crss}} = \sigma \cos \phi \cos \lambda$

where $\phi$ is the angle between the loading axis and the slip plane normal, and $\lambda$ is the angle between the loading axis and the slip direction. In randomly oriented polycrystalline aggregates, macroscopic yielding represents the statistical percolation of dislocation glide across millions of grains. Because slip is governed strictly by shearing forces, normal tensile stresses do not drive yielding in ductile metals; they merely pull atoms apart elastically until shear stresses displace them permanently.

Strength Symmetry in Isotropic Metals

For isotropic polycrystalline metals, the resistance to dislocation glide is identical whether the macroscopic normal load is tensile or compressive. Consequently, ductile failure theories assume equal yield strengths in tension and compression:

$S_{yt} \approx S_{yc} = S_y$

Hydrostatic vs. Deviatoric Stress State

To understand why normal stresses alone cannot cause yielding, consider the decomposition of any Cauchy stress tensor $\boldsymbol{\sigma}$ into two mutually orthogonal components: a spherical (hydrostatic) tensor $\boldsymbol{\sigma}_h$ and a deviatoric stress tensor $\mathbf{s}$:

$\sigma_{\text{avg}} = \frac{\sigma_x + \sigma_y + \sigma_z}{3} = \frac{\sigma_1 + \sigma_2 + \sigma_3}{3}$

$\boldsymbol{\sigma} = \boldsymbol{\sigma}_h + \mathbf{s} = \begin{bmatrix} \sigma_{\text{avg}} & 0 & 0 \\ 0 & \sigma_{\text{avg}} & 0 \\ 0 & 0 & \sigma_{\text{avg}} \end{bmatrix} + \begin{bmatrix} \sigma_x - \sigma_{\text{avg}} & \tau_{xy} & \tau_{zx} \\ \tau_{xy} & \sigma_y - \sigma_{\text{avg}} & \tau_{yz} \\ \tau_{zx} & \tau_{yz} & \sigma_z - \sigma_{\text{avg}} \end{bmatrix}$

1. Hydrostatic Stress ($\boldsymbol{\sigma}_h$): Produces uniform triaxial volumetric dilation or compression ($V - V_0$). In landmark high-pressure experiments, Nobel laureate P.W. Bridgman subjected ductile metals to hydrostatic pressures exceeding $2\text{--}3\text{ GPa}$ ($300\text{--}450\text{ ksi}$) without observing any macroscopic plastic deformation or change in yield strength. Pure hydrostatic stress cannot create shear stress on any plane; hence, hydrostatic stress cannot cause plastic yielding.
2. Deviatoric Stress ($\mathbf{s}$): Produces distortion (angular shape changes) at constant volume. The deviatoric stress components generate shear stresses on oblique planes, directly activating dislocation motion and governing plastic flow.

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2. Maximum Shear Stress (MSS / Tresca) Theory

Proposed by French engineer Henri Tresca in 1864, the Maximum Shear Stress (MSS) theory represents the earliest multiaxial static failure criterion for ductile materials.

The Tresca Postulate

Plastic yielding begins when the maximum shear stress in a complex multiaxial stress state reaches or exceeds the maximum shear stress present in a standard uniaxial tensile specimen at the onset of yielding.

Uniaxial Tensile Benchmark

In a standard uniaxial tensile test loaded to the yield strength $S_y$, the principal stresses are:

$\sigma_1 = S_y, \quad \sigma_2 = 0, \quad \sigma_3 = 0$

Constructing Mohr's circle for this state reveals that the maximum shear stress acting on planes oriented at 45° to the tensile axis is:

$\tau_{\max,\text{yield}} = \frac{\sigma_1 - \sigma_3}{2} = \frac{S_y - 0}{2} = \frac{S_y}{2} = 0.50 S_y$

Multiaxial Yield Criterion

For a general three-dimensional stress state, let the principal stresses be ordered algebraically such that:

$\sigma_1 \ge \sigma_2 \ge \sigma_3$

The absolute maximum shear stress acting within the component is:

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

Applying the Tresca postulate, yielding occurs when $\tau_{\max} \ge \tau_{\max,\text{yield}}$:

$\frac{\sigma_1 - \sigma_3}{2} \ge \frac{S_y}{2} \implies \sigma_1 - \sigma_3 \ge S_y$

Factor of Safety Formulation

The design factor of safety ($n$) under the Maximum Shear Stress theory is defined as:

$n = \frac{S_y}{\sigma_1 - \sigma_3} = \frac{S_{sy}}{\tau_{\max}}$

where the Tresca-predicted shear yield strength is:

$S_{sy} = 0.50 S_y$

Critical Algebraic Ordering Trap in 2D Plane Stress

In plane stress, the stress tensor has out-of-plane components $\sigma_z = \tau_{yz} = \tau_{zx} = 0$. Consequently, one of the three principal stresses is always identically zero ($\sigma_z = 0$).

When evaluating in-plane principal stresses $\sigma_A$ and $\sigma_B$ (with $\sigma_A \ge \sigma_B$):

1. Case 1: Both in-plane principal stresses are tensile ($\sigma_A \ge \sigma_B > 0$): Algebraic ordering yields $\sigma_1 = \sigma_A$, $\sigma_2 = \sigma_B$, $\sigma_3 = 0$. $\tau_{\max} = \frac{\sigma_1 - \sigma_3}{2} = \frac{\sigma_A - 0}{2} = \frac{\sigma_A}{2} \implies n = \frac{S_y}{\sigma_A}$
2. Case 2: Both in-plane principal stresses are compressive ($0 > \sigma_A \ge \sigma_B$): Algebraic ordering yields $\sigma_1 = 0$, $\sigma_2 = \sigma_A$, $\sigma_3 = \sigma_B$. $\tau_{\max} = \frac{\sigma_1 - \sigma_3}{2} = \frac{0 - \sigma_B}{2} = \frac{|\sigma_B|}{2} \implies n = \frac{S_y}{|\sigma_B|}$
3. Case 3: In-plane principal stresses have opposite signs ($\sigma_A > 0 > \sigma_B$): Algebraic ordering yields $\sigma_1 = \sigma_A$, $\sigma_2 = 0$, $\sigma_3 = \sigma_B$. $\tau_{\max} = \frac{\sigma_1 - \sigma_3}{2} = \frac{\sigma_A - \sigma_B}{2} \implies n = \frac{S_y}{\sigma_A - \sigma_B}$

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3. Distortion Energy (DE / von Mises) Theory

The Distortion Energy (DE) theory—also known as the von Mises criterion, the Huber-von Mises-Hencky theory, or the Maximum Octahedral Shear Stress theory—is the most physically rigorous and experimentally accurate failure criterion for ductile metals.

Physical Strain Energy Decomposition

For an isotropic linear elastic solid subjected to principal stresses $\sigma_1, \sigma_2, \sigma_3$, the total elastic strain energy per unit volume ($U$) stored within the material is:

$U = \frac{1}{2}(\sigma_1 \epsilon_1 + \sigma_2 \epsilon_2 + \sigma_3 \epsilon_3)$

Applying generalized Hooke's Law for three-dimensional isotropic elasticity:

$\epsilon_1 = \frac{1}{E}\left[\sigma_1 - \nu(\sigma_2 + \sigma_3)\right], \quad \epsilon_2 = \frac{1}{E}\left[\sigma_2 - \nu(\sigma_3 + \sigma_1)\right], \quad \epsilon_3 = \frac{1}{E}\left[\sigma_3 - \nu(\sigma_1 + \sigma_2)\right]$

Substituting these strains into the energy expression yields:

$U = \frac{1}{2E}\left[\sigma_1^2 + \sigma_2^2 + \sigma_3^2 - 2\nu(\sigma_1 \sigma_2 + \sigma_2 \sigma_3 + \sigma_3 \sigma_1)\right]$

This total strain energy is partitioned into two independent components:

$U = U_v + U_d$

1. Volumetric Strain Energy ($U_v$): The energy associated with uniform volumetric dilation without shape change, governed by the mean hydrostatic stress $\sigma_{\text{avg}} = (\sigma_1 + \sigma_2 + \sigma_3)/3$: $U_v = \frac{3(1 - 2\nu)}{2E} \sigma_{\text{avg}}^2 = \frac{1 - 2\nu}{6E}(\sigma_1 + \sigma_2 + \sigma_3)^2$
2. Distortion Strain Energy ($U_d$): The shear strain energy associated strictly with angular distortion at constant volume. Subtracting $U_v$ from $U$: $U_d = U - U_v = \frac{1+\nu}{6E}\left[(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2\right]$ Recalling that the shear modulus is $G = \frac{E}{2(1+\nu)}$, this can be written compactly as: $U_d = \frac{1}{12G}\left[(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2\right]$

The Distortion Energy Postulate

Plastic yielding begins when the distortion strain energy per unit volume ($U_d$) in a multiaxial stress state reaches or exceeds the distortion strain energy absorbed by a uniaxial tensile test specimen at the yield strength $S_y$.

In a uniaxial tension test at yielding ($\sigma_1 = S_y, \sigma_2 = 0, \sigma_3 = 0$):

$U_{d,\text{yield}} = \frac{1+\nu}{6E}\left[(S_y - 0)^2 + (0 - 0)^2 + (0 - S_y)^2\right] = \frac{1+\nu}{3E} S_y^2$

Equating $U_d \ge U_{d,\text{yield}}$:

$\frac{1+\nu}{6E}\left[(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2\right] \ge \frac{1+\nu}{3E} S_y^2$

Dividing both sides by $\frac{1+\nu}{3E}$ defines the von Mises Effective Stress (equivalent stress, $\sigma'$):

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

Yield criterion:

$\sigma' \ge S_y$

Factor of safety:

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

General 3D Cartesian Stress Tensor

When stresses are defined relative to arbitrary $xyz$ Cartesian coordinate axes, $\sigma'$ is computed directly without calculating principal stresses:

$\sigma' = \frac{1}{\sqrt{2}}\sqrt{(\sigma_x - \sigma_y)^2 + (\sigma_y - \sigma_z)^2 + (\sigma_z - \sigma_x)^2 + 6(\tau_{xy}^2 + \tau_{yz}^2 + \tau_{zx}^2)}$

Two-Dimensional Plane Stress State

For plane stress ($\sigma_z = \tau_{yz} = \tau_{zx} = 0$), the von Mises effective stress simplifies into two widely used forms:

1. In terms of in-plane principal stresses ($\sigma_1, \sigma_2$): $\sigma' = \sqrt{\sigma_1^2 - \sigma_1 \sigma_2 + \sigma_2^2}$
2. In terms of general planar stress components ($\sigma_x, \sigma_y, \tau_{xy}$): $\sigma' = \sqrt{\sigma_x^2 - \sigma_x \sigma_y + \sigma_y^2 + 3\tau_{xy}^2}$

Pure Shear Yield Strength Prediction ($S_{sy}$)

Consider a machine component loaded in pure torsional shear ($\sigma_x = 0, \sigma_y = 0, \tau_{xy} = \tau$). The von Mises effective stress is:

$\sigma' = \sqrt{0^2 - (0)(0) + 0^2 + 3\tau^2} = \sqrt{3\tau^2} = \sqrt{3}\tau$

Yielding occurs when $\sigma' = S_y \implies \sqrt{3}S_{sy} = S_y$:

$S_{sy} = \frac{S_y}{\sqrt{3}} = \frac{\sqrt{3}}{3} S_y \approx 0.577 S_y$

Comparing this prediction to the Maximum Shear Stress prediction ($S_{sy} = 0.500 S_y$):

$\frac{S_{sy,\text{DE}} - S_{sy,\text{MSS}}}{S_{sy,\text{MSS}}} = \frac{0.577 S_y - 0.500 S_y}{0.500 S_y} = +15.5\%$

Extensive empirical testing across hundreds of structural steels, aluminum alloys, and brasses confirms that the experimental shear yield strength consistently falls between $0.55 S_y$ and $0.60 S_y$. The von Mises prediction of $0.577 S_y$ matches experimental data with remarkable precision, whereas Tresca underestimates shear capacity by approximately $15\%$.

Octahedral Shear Stress Theory Equivalence

An octahedral plane is an oblique plane whose unit normal vector makes equal angles with the three principal stress axes ($\cos \theta_1 = \cos \theta_2 = \cos \theta_3 = 1/\sqrt{3}$). The shear stress acting on these eight symmetric planes is:

$\tau_{\text{oct}} = \frac{1}{3}\sqrt{(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2} = \frac{\sqrt{2}}{3}\sigma'$

In a uniaxial tensile test at yield ($\sigma_1 = S_y, \sigma_2 = \sigma_3 = 0$):

$\tau_{\text{oct},\text{yield}} = \frac{\sqrt{2}}{3}S_y$

Equating $\tau_{\text{oct}} \ge \tau_{\text{oct},\text{yield}}$ yields an identical criterion to $\sigma' \ge S_y$. Thus, the Octahedral Shear Stress theory and the Distortion Energy theory are mathematically and physically identical.

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4. Planar Yield Loci Comparison on the $\sigma_1 - \sigma_2$ Plane

For 2D plane stress ($\sigma_3 = 0$), plotting the boundary where yielding begins ($n = 1.0$) on coordinate axes of $\sigma_1$ and $\sigma_2$ produces the classic static yield envelopes:

                          σ2
                           ^
                           |   von Mises Ellipse
                       Sy  + - - - - - - - .
                      /    |    Tresca      \
                     /     |    Hexagon      \
                    /      |                  \
                   /       |       (Sy, Sy)    \
                  /        |        *           \
                 /         |       / \           \
       -Sy      /          |      /   \           \       Sy
     <---------+-----------+-----+-----+-----------+-------> σ1
               \           |    /     /           /
                \          |   /     /           /
                 \         |  *     /           /
                  \  (-Sy, -Sy)    /           /
                   \              /           /
                    \            /           /
                     \          /           /
                      \        /           /
                  -Sy  . - - - - - - - - -

Geometric Anatomy of the Envelopes

1. The von Mises Locus: An ellipse governed by $\sigma_1^2 - \sigma_1 \sigma_2 + \sigma_2^2 = S_y^2$. The major axis of the ellipse lies along the 45° line of equal biaxial tension ($\sigma_1 = \sigma_2$) with semi-major axis length $\sqrt{2}S_y$. The minor axis lies along the line $\sigma_1 = -\sigma_2$ with semi-minor axis length $\sqrt{2/3}S_y \approx 0.816 S_y$.
2. The Tresca Locus: An irregular hexagon centered at the origin, governed by the piecewise boundaries: $\sigma_1 = \pm S_y$, $\sigma_2 = \pm S_y$, and $\sigma_1 - \sigma_2 = \pm S_y$.

Mathematical Boundary Contacts

The Tresca hexagon is inscribed entirely within the von Mises ellipse, contacting the ellipse at exactly six points:

• Uniaxial tension and compression along $\sigma_1$: $(\pm S_y, 0)$
• Uniaxial tension and compression along $\sigma_2$: $(0, \pm S_y)$
• Equal biaxial tension and compression: $(S_y, S_y)$ and $(-S_y, -S_y)$

Regions of Divergence

At all points other than the six contact points, the Tresca boundary lies strictly inside the von Mises ellipse:

• Pure Shear State ($\sigma_1 = -\sigma_2 = \tau$): In Quadrants 2 and 4, the Tresca hexagon intersects the pure shear axis at $\tau = 0.500 S_y$, whereas the von Mises ellipse intersects it at $\tau = 0.577 S_y$—a $15.5\%$ expansion in allowable shear capacity.
• Biaxial Stress with $\sigma_1 = 2\sigma_2$ or $\sigma_2 = 2\sigma_1$: In Quadrant 1, Tresca limits $\sigma_1$ to $S_y$, whereas von Mises allows $\sigma_1 = \frac{2}{\sqrt{3}}S_y \approx 1.155 S_y$—again a $15.5\%$ expansion.

Design Philosophy & Code Selection

Because the Tresca envelope never expands beyond the von Mises ellipse:

$n_{\text{MSS}} \le n_{\text{DE}}$

• The Maximum Shear Stress theory is always conservative or equal to experimental data. ASME BPVC Section III (nuclear components) still evaluates stresses as Tresca "stress intensity" ($S = 2\tau_{\max} = \sigma_1 - \sigma_3$). Section VIII Division 2 used stress intensity before its 2007 rewrite, which switched its design-by-analysis rules to the von Mises equivalent stress.
• The Distortion Energy theory represents the true physical yielding threshold. It is the universal standard for weight-critical aerospace and automotive structural design, and serves as the default yield formulation in all modern commercial Finite Element Analysis (FEA) solvers (ANSYS, Abaqus, NASTRAN, SolidWorks Simulation).

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5. Summary Comparison: Maximum Shear Stress vs. Distortion Energy

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6. Worked Numerical Engineering Scenario: Power Transmission Shaft under Combined Bending and Torsion

Problem Statement

A solid drive shaft machined from AISI 1045 carbon steel operates in an industrial speed reducer. The mill certificate for this bar reports a yield strength of $S_y = 450\text{ MPa}$ ($65.3\text{ ksi}$) and an ultimate tensile strength of $S_{ut} = 565\text{ MPa}$. The shaft has a constant outside diameter of $d = 50.0\text{ mm}$ ($0.050\text{ m}$).

At a critical bearing shoulder, static equilibrium analysis indicates that the shaft must sustain:

• A steady bending moment: $M = 1600\text{ N}\cdot\text{m}$
• A steady transmission torque: $T = 2200\text{ N}\cdot\text{m}$
• Axial thrust load: $P = 0\text{ N}$

Calculate the design factor of safety against static yielding at the critical surface element using:

1. The Maximum Shear Stress (MSS / Tresca) theory.
2. The Distortion Energy (DE / von Mises) theory.
3. Quantify the percentage difference between the two predictions and justify which theory should be selected for production.

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Step-by-Step Solution

Step 1: Geometric Properties of the Solid Shaft Cross Section

The cross-sectional area moment of inertia ($I$) and polar moment of inertia ($J$) are:

$I = \frac{\pi d^4}{64} = \frac{\pi (0.050\text{ m})^4}{64} = \frac{\pi (6.25 \times 10^{-6})}{64} = 3.06796 \times 10^{-7}\text{ m}^4$

$J = 2I = \frac{\pi d^4}{32} = 6.13592 \times 10^{-7}\text{ m}^4$

The outer radius is $c = d/2 = 0.025\text{ m}$. The elastic section modulus ($Z$) and polar section modulus ($Z_p$) are:

$Z = \frac{I}{c} = \frac{\pi d^3}{32} = \frac{\pi (0.050\text{ m})^3}{32} = 1.22718 \times 10^{-5}\text{ m}^3$

$Z_p = \frac{J}{c} = \frac{\pi d^3}{16} = 2.45437 \times 10^{-5}\text{ m}^3$

Step 2: Surface Stress Calculation

The critical stress element is located on the extreme outer surface of the shaft where bending tensile stress reaches its maximum value:

$\sigma_x = \frac{M}{Z} = \frac{1600\text{ N}\cdot\text{m}}{1.22718 \times 10^{-5}\text{ m}^3} = 130.380 \times 10^6\text{ Pa} = 130.38\text{ MPa}$

$\sigma_y = 0\text{ MPa}$

The torsional shear stress at the outer surface is:

$\tau_{xy} = \frac{T}{Z_p} = \frac{2200\text{ N}\cdot\text{m}}{2.45437 \times 10^{-5}\text{ m}^3} = 89.636 \times 10^6\text{ Pa} = 89.64\text{ MPa}$

Step 3: Principal Stress Determination

The planar stress state is $\sigma_x = 130.38\text{ MPa}$, $\sigma_y = 0$, $\tau_{xy} = 89.64\text{ MPa}$. Using Mohr's circle relations:

$\sigma_{\text{avg}} = \frac{\sigma_x + \sigma_y}{2} = \frac{130.38 + 0}{2} = 65.19\text{ MPa}$

$R = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} = \sqrt{(65.19)^2 + (89.64)^2} = \sqrt{4249.74 + 8035.33} = \sqrt{12285.07} = 110.84\text{ MPa}$

The in-plane principal stresses are:

$\sigma_A = \sigma_{\text{avg}} + R = 65.19 + 110.84 = 176.03\text{ MPa}$

$\sigma_B = \sigma_{\text{avg}} - R = 65.19 - 110.84 = -45.65\text{ MPa}$

Now establish the three-dimensional algebraic ordering including $\sigma_z = 0$:

$\sigma_1 = +176.03\text{ MPa}, \quad \sigma_2 = 0\text{ MPa}, \quad \sigma_3 = -45.65\text{ MPa}$

Step 4: Maximum Shear Stress (MSS / Tresca) Factor of Safety

The true maximum shear stress acting on the element is:

$\tau_{\max} = \frac{\sigma_1 - \sigma_3}{2} = \frac{176.03 - (-45.65)}{2} = \frac{221.68}{2} = 110.84\text{ MPa}$

The stress difference is $\sigma_1 - \sigma_3 = 221.68\text{ MPa}$.

The factor of safety under Tresca is:

$n_{\text{MSS}} = \frac{S_y}{\sigma_1 - \sigma_3} = \frac{450\text{ MPa}}{221.68\text{ MPa}} = 2.030 \approx 2.03$

Step 5: Distortion Energy (DE / von Mises) Factor of Safety

Using the general 2D plane stress formula in terms of $\sigma_x$ and $\tau_{xy}$:

$\sigma' = \sqrt{\sigma_x^2 - \sigma_x \sigma_y + \sigma_y^2 + 3\tau_{xy}^2} = \sqrt{(130.38)^2 + 3(89.64)^2}$

$\sigma' = \sqrt{16999.0 + 3(8035.3)} = \sqrt{16999.0 + 24105.9} = \sqrt{41104.9} = 202.74\text{ MPa}$

Verifying with 3D principal stresses:

$\sigma' = \sqrt{\frac{(176.03 - 0)^2 + (0 - (-45.65))^2 + (-45.65 - 176.03)^2}{2}}$

$\sigma' = \sqrt{\frac{(176.03)^2 + (45.65)^2 + (-221.68)^2}{2}} = \sqrt{\frac{30986.6 + 2083.9 + 49142.0}{2}} = \sqrt{\frac{82212.5}{2}} = 202.75\text{ MPa}$

The factor of safety under von Mises is:

$n_{\text{DE}} = \frac{S_y}{\sigma'} = \frac{450\text{ MPa}}{202.75\text{ MPa}} = 2.219 \approx 2.22$

Step 6: Percentage Comparison & Engineering Evaluation

Comparing the two calculated factors of safety:

$\Delta n = \frac{n_{\text{DE}} - n_{\text{MSS}}}{n_{\text{MSS}}} \times 100\% = \frac{2.219 - 2.030}{2.030} \times 100\% = +9.31\%$

• Analysis: The Distortion Energy theory predicts a factor of safety that is $9.3\%$ higher than Tresca ($2.22$ vs. $2.03$). This occurs because the stress state contains significant shear ($\tau_{xy} = 89.64\text{ MPa}$), moving the stress point into the region where the von Mises ellipse expands beyond the inscribed Tresca hexagon.
• Engineering Decision: If the governing code or corporate standard specifies Tresca stress intensity (as ASME Section III does) or a conservative maximum-shear check, $n = 2.03$ (Tresca) must be reported as the governing rating. If designing for an aerospace or competitive commercial gearbox where mass and shaft inertia must be minimized, $n = 2.22$ (von Mises) represents the accurate physical margin against yielding.