2.2 Static Failure Theories for Ductile and Brittle Materials

Static Failure Theories for Ductile and Brittle Materials

Predicting whether a machine element will survive multiaxial static loading requires relating complex stress tensors $(\sigma_x, \sigma_y, \sigma_z, \tau_{xy}, \tau_{yz}, \tau_{zx})$ to standard uniaxial tensile test properties: yield strength ($S_y$), ultimate tensile strength ($S_{ut}$), and ultimate compressive strength ($S_{uc}$). Because materials exhibit fundamentally different failure physics depending on ductility, selecting the correct failure theory is vital.

---

1. Material Ductility Classification Criterion

Materials are classified based on their true plastic strain at fracture $\epsilon_f$ (or percent elongation in a standard 2-inch tensile gauge length):

+-----------------------------------------------------------------------------+
|                   DUCTILE VS. BRITTLE CLASSIFICATION                        |
|                                                                             |
|   [DUCTILE MATERIALS]   ---> \epsilon_f \ge 0.05 (Elongation \ge 5\%)         |
|                               - Examples: Structural steels, aluminum,     |
|                                 copper, brass, titanium alloys             |
|                               - Failure Mode: Shear slip / plastic yielding|
|                               - Theories: Tresca (MSS) & von Mises (DE)    |
|                                                                             |
|   [BRITTLE MATERIALS]   ---> \epsilon_f < 0.05 (Elongation < 5\%)          |
|                               - Examples: Gray cast iron, hardened tool    |
|                                 steels, ceramics, structural glass         |
|                               - Failure Mode: Tensile cleavage fracture    |
|                               - Theories: Rankine (MNS), BCM, Mod-Mohr     |
+-----------------------------------------------------------------------------+

---

2. Ductile Yield Failure Theories: Tresca (MSS) vs. von Mises (DE)

1. Maximum Shear Stress Theory (MSS / Tresca Criterion)

Formulated by Henri Tresca, this theory assumes yielding begins when the absolute maximum shear stress in a multiaxial element reaches the maximum shear stress at yielding in a standard uniaxial tensile specimen ($S_y / 2$).

$$\tau_{\text{max, abs}} = \frac{\sigma_1 - \sigma_3}{2} = \frac{S_y}{2 n} \implies \sigma_1 - \sigma_3 = \frac{S_y}{n}$$

In 2D Plane Stress (with $\sigma_A, \sigma_B$ and out-of-plane $\sigma_z = 0$):

• Quadrant 1 ($\sigma_A, \sigma_B > 0$): $\sigma_1 = \sigma_A, \sigma_3 = 0 \implies n = \frac{S_y}{\sigma_A}$
• Quadrant 3 ($\sigma_A, \sigma_B < 0$): $\sigma_1 = 0, \sigma_3 = \sigma_B \implies n = \frac{S_y}{|\sigma_B|}$
• Quadrants 2 & 4 (Opposite signs, $\sigma_A > 0 > \sigma_B$): $\sigma_1 = \sigma_A, \sigma_3 = \sigma_B \implies n = \frac{S_y}{\sigma_A - \sigma_B}$

2. Distortion Energy Theory (DE / von Mises / Hencky-Huber)

Proposed by Richard von Mises, this theory recognizes that total strain energy density ($u$) consists of volumetric strain energy ($u_v$, caused by hydrostatic pressure that changes volume without causing shear yield) and distortion strain energy ($u_d$, caused by deviatoric stress that alters shape).

+-----------------------------------------------------------------------------+
|                   DISTORTION ENERGY FORMULATION (von Mises)                 |
|                                                                             |
|   Total Strain Energy Density:    u = u_v + u_d                             |
|                                                                             |
|   Distortion Energy Density:      u_d = \frac{1+\nu}{3E} \left[\sigma'^2\right] |
|                                                                             |
|   von Mises Effective Stress:     \sigma' \le \frac{S_y}{n}                 |
+-----------------------------------------------------------------------------+

3D General Stress Tensor:

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

2D Plane Stress State ($\sigma_z = \tau_{xz} = \tau_{yz} = 0$):

$$\sigma' = \sqrt{\sigma_x^2 - \sigma_x \sigma_y + \sigma_y^2 + 3\tau_{xy}^2} = \sqrt{\sigma_1^2 - \sigma_1 \sigma_2 + \sigma_2^2}$$

Factor of Safety ($n$):

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

Comparison & The Pure Shear Advantage (15.5% Rule)

                     \sigma_2
                        ^
                        |        von Mises Ellipse
                   +----+----+   /---\
                   |    |    |  /     \
           --------+----+----+-[-------]--------> \sigma_1
                   |    |    |  \     /
                   +----+----+   \---/
                        |       Tresca Hexagon

Consider a shaft in pure shear $\tau$ (where principal stresses are $\sigma_1 = +\tau, \sigma_2 = -\tau, \sigma_3 = 0$):

• Tresca (MSS): $$\tau_{\text{max, abs}} = \frac{\tau - (-\tau)}{2} = \tau = \frac{S_y}{2} \implies S_{sy} = 0.500 S_y$$
• von Mises (DE): $$\sigma' = \sqrt{\tau^2 - \tau(-\tau) + (-\tau)^2} = \sqrt{3\tau^2} = \sqrt{3}\tau = S_y \implies S_{sy} = \frac{S_y}{\sqrt{3}} \approx 0.577 S_y$$

$$\text{Capacity Increase} = \frac{0.57735 - 0.50000}{0.50000} = +15.47\% \approx 15.5\%$$

---

3. Brittle Static Failure Theories: Rankine, Coulomb-Mohr & Modified Mohr

Brittle materials (such as ASTM Class 20-60 Gray Cast Irons) possess compressive strengths significantly higher than their tensile strengths ($S_{uc} \approx 3$ to $4 \times S_{ut}$). Therefore, symmetric ductile theories cannot be used.

+-----------------------------------------------------------------------------+
|                        BRITTLE FAILURE THEORIES                             |
|                                                                             |
|   1. Maximum Normal Stress (Rankine):                                       |
|      \sigma_1 \ge \frac{S_{ut}}{n} \quad \text{or} \quad |\sigma_3| \ge \frac{S_{uc}}{n} |
|      * Unsafe in Quadrant 4 (ignores shear-assisted cleavage).              |
|                                                                             |
|   2. Brittle Coulomb-Mohr (BCM):                                            |
|      - Quadrant 1 (\sigma_1 \ge \sigma_2 \ge 0): n = S_{ut} / \sigma_1      |
|      - Quadrant 4 (\sigma_1 \ge 0 \ge \sigma_2): \frac{\sigma_1}{S_{ut}} - \frac{\sigma_2}{S_{uc}} = \frac{1}{n} |
|      - Quadrant 3 (0 \ge \sigma_1 \ge \sigma_2): n = S_{uc} / |\sigma_2|    |
|      * Conservative straight-line boundary.                                 |
|                                                                             |
|   3. Modified Mohr (MM) Theory (Recommended for PE Exam):                   |
|      - Quadrant 1: n = S_{ut} / \sigma_1                                    |
|      - Quadrant 4 (When |\sigma_2 / \sigma_1| \le 1): n = S_{ut} / \sigma_1 |
|      - Quadrant 4 (When |\sigma_2 / \sigma_1| > 1):                         |
|        \frac{(S_{uc} - S_{ut})\sigma_1}{S_{uc} S_{ut}} - \frac{\sigma_2}{S_{uc}} = \frac{1}{n} |
|      - Quadrant 3: n = S_{uc} / |\sigma_2|                                  |
+-----------------------------------------------------------------------------+

                                \sigma_2
                                   ^
                                   |    Rankine Square (Unsafe in Q4)
                 +-----------------+-----------------+
                 |                 |                 |
                 |                 |     Quadrant 1  |
                 |                 |                 |
    -------------+-----------------+-----------------+-------------> \sigma_1
                 |                 |     Modified Mohr (MM)
                 |   Quadrant 4    |     Coulomb-Mohr (BCM)
                 |                 |
                 +-----------------+
                                   |

---

4. Failure Theory Selection Decision Matrix

---

5. Step-by-Step Worked Engineering Problem

Problem Statement

A critical machine bracket is evaluated under plane stress. Finite element analysis establishes the surface stresses: $\sigma_x = 120\text{ MPa}, \sigma_y = -50\text{ MPa}, \tau_{xy} = 45\text{ MPa}$.

• Part A: Calculate the safety factor $n$ if manufactured from ductile AISI 1040 cold-drawn steel ($S_y = 490\text{ MPa}$) using both (1) Tresca and (2) von Mises.
• Part B: Calculate the safety factor $n$ if cast from brittle ASTM Class 30 gray iron ($S_{ut} = 210\text{ MPa}, S_{uc} = 750\text{ MPa}$) using (1) Coulomb-Mohr and (2) Modified Mohr.

Step 1: Principal Stress Determination

$$\sigma_{\text{avg}} = \frac{120 + (-50)}{2} = 35.0\text{ MPa}$$ $$R = \sqrt{\left(\frac{120 - (-50)}{2}\right)^2 + 45^2} = \sqrt{(85)^2 + 45^2} = \sqrt{7225 + 2025} = 96.177\text{ MPa}$$ $$\sigma_1 = 35.0 + 96.177 = 131.18\text{ MPa}$$ $$\sigma_2 = 35.0 - 96.177 = -61.18\text{ MPa}$$ $$\sigma_3 = 0\text{ MPa}$$

Part A: Ductile Steel ($S_y = 490\text{ MPa}$)

1. Tresca (MSS): Since $\sigma_1 > 0$ and $\sigma_2 < 0$, $\sigma_3 = \sigma_2 = -61.18\text{ MPa}$:

   $$\tau_{\text{max, abs}} = \frac{\sigma_1 - \sigma_2}{2} = \frac{131.18 - (-61.18)}{2} = 96.18\text{ MPa}$$ $$n_{\text{MSS}} = \frac{S_y}{\sigma_1 - \sigma_2} = \frac{490}{192.36} = 2.55$$

2. von Mises (DE):

   $$\sigma' = \sqrt{\sigma_1^2 - \sigma_1\sigma_2 + \sigma_2^2} = \sqrt{(131.18)^2 - (131.18)(-61.18) + (-61.18)^2}$$ $$\sigma' = \sqrt{17208.19 + 8025.59 + 3742.99} = \sqrt{28976.77} = 170.23\text{ MPa}$$ $$n_{\text{DE}} = \frac{S_y}{\sigma'} = \frac{490}{170.23} = 2.88$$

Part B: Brittle Cast Iron ($S_{ut} = 210\text{ MPa}, S_{uc} = 750\text{ MPa}$)

Here $\sigma_1 = 131.18\text{ MPa} > 0$ and $\sigma_2 = -61.18\text{ MPa} < 0$ (Quadrant 4).

1. Brittle Coulomb-Mohr (BCM):

   $$\frac{1}{n} = \frac{\sigma_1}{S_{ut}} - \frac{\sigma_2}{S_{uc}} = \frac{131.18}{210} - \frac{-61.18}{750} = 0.6247 + 0.0816 = 0.7063 \implies n_{\text{BCM}} = \frac{1}{0.7063} = 1.42$$

2. Modified Mohr (MM): Check slope ratio: $\left|\frac{\sigma_2}{\sigma_1}\right| = \frac{61.18}{131.18} = 0.466 \le 1.0$. Because $|\sigma_2 / \sigma_1| \le 1$, failure is governed purely by the tensile strength line:

   $$n_{\text{MM}} = \frac{S_{ut}}{\sigma_1} = \frac{210}{131.18} = 1.60$$

---

6. Common Exam Traps & PE Pro-Tips

• Trap 1 — Using 2D von Mises in 3D Stress Fields: In a pressurized thick pipe where radial stress $\sigma_r \neq 0$, you cannot drop $\sigma_3$. You must use the full 3D von Mises equation.
• Trap 2 — Applying von Mises to Cast Iron: Never use ductile yield theories on gray cast iron or ceramics. Brittle materials do not yield; they shatter under tensile cleavage.
• Trap 3 — Forgetting Quadrant 4 Check in Modified Mohr: If $|\sigma_2 / \sigma_1| > 1$, you must switch to the second Quadrant 4 equation containing $(S_{uc} - S_{ut})$.