2.1 Stress, Strain, Principal Stresses & Combined Loading

Stress, Strain, Principal Stresses & Combined Loading

Mechanical components in industrial machinery, aerospace structures, and pressure containment systems rarely experience isolated single-axis loads. Real-world shafts, pressure vessels, structural beams, and connecting rods simultaneously endure axial tension or compression, multi-plane bending, torque transmission, and transverse shear. Evaluating component safety and structural integrity requires constructing the 3D stress tensor at critical points, resolving principal normal and shear stresses via Mohr's circle transformations, and applying generalized Hooke's law.

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1. Stress Components and Superposition under Combined Loading

When multiple external forces and moments act simultaneously on a structural element, the total state of stress at any point $(x, y, z)$ is obtained by superimposing the individual normal stresses ($\sigma$) and shear stresses ($\tau$) acting on orthogonal planes.

+-----------------------------------------------------------------------------+
|                   INDIVIDUAL STRESS COMPONENT FORMULAS                      |
|                                                                             |
|   [AXIAL STRESS]          ---> \sigma_{\text{axial}} = \pm \frac{P}{A}      |
|                                                                             |
|   [BENDING STRESS]        ---> \sigma_b = \pm \frac{M y}{I}                 |
|                                                                             |
|   [TORSIONAL SHEAR]       ---> \tau_t = \frac{T r}{J}                       |
|                                                                             |
|   [TRANSVERSE SHEAR]      ---> \tau_v = \frac{V Q}{I b}                     |
+-----------------------------------------------------------------------------+

Formulations & Cross-Sectional Geometry

1. Axial Normal Stress: $\sigma_{\text{axial}} = \frac{P}{A}$ Uniformly distributed across cross-sectional area $A$ for concentric loading (tensile positive, compressive negative).

2. Bending Normal Stress (Flexure Formula): $\sigma_b = -\frac{M y}{I}$ Where $M$ is the internal bending moment, $y$ is the perpendicular distance from the neutral axis (NA), and $I$ is the area moment of inertia. Stress varies linearly from zero at the neutral axis to maximum tensile/compressive values at the outermost fibers ($y = \pm c$).

3. Torsional Shear Stress: $\tau_t = \frac{T r}{J}$ Where $T$ is the applied torque, $r$ is the radial distance from the center, and $J$ is the polar moment of inertia. For a solid circular cross-section of diameter $d$: $J = \frac{\pi d^4}{32}, \quad Z_p = \frac{J}{c} = \frac{\pi d^3}{16} \implies \tau_{\text{max}} = \frac{16 T}{\pi d^3}$ For a hollow circular cross-section with outer diameter $d_o$ and inner diameter $d_i$: $J = \frac{\pi (d_o^4 - d_i^4)}{32}$

4. Transverse Shear Stress: $\tau_v = \frac{V Q}{I b}$ Where $V$ is internal transverse shear force, $Q = \int_{y_1}^c y , dA = \bar{y}' A'$ is the first moment of area above the level where shear is computed, $I$ is moment of inertia, and $b$ is width of the section at the cut.

   • Rectangular Beam ($b \times h$): $\tau_{\text{max}} = \frac{3 V}{2 A}$ (at the neutral axis, parabolic distribution).
   • Solid Circular Cross-Section: $\tau_{\text{max}} = \frac{4 V}{3 A}$ (at the neutral axis).
   • Thin-Walled Hollow Circular Tube: $\tau_{\text{max}} = \frac{2 V}{A}$ (at the neutral axis).

Locating Critical Stress Elements

Consider a solid circular cantilevered shaft carrying a vertical transverse end load $F_y$, an axial tension $P$, and a torque $T$:

               y                                                    
               ^   F_y                                              
               |   |                                                
               |   v                                                
     +---------+---------+=============================+    
     | Element A (Top)   |                             |    
     |                   |       Torque T              |    
   --+-------------------+-----------(=======>---------+---> x
     | Element B (Side)  |                             |    
     |                   |       Axial P --->          |    
     +---------+---------+=============================+    
              /                                                     
             z (Neutral Axis for vertical bending)                  

• Element A (Top Outer Fiber, $y = +c, z = 0$):
  • Bending stress is maximum tensile: $\sigma_b = +\frac{M c}{I}$.
  • Axial stress adds directly: $\sigma_{\text{axial}} = +\frac{P}{A}$.
  • Total normal stress: $\sigma_x = \frac{P}{A} + \frac{M c}{I}$.
  • Torsional shear is maximum: $\tau_{xy} = \frac{T c}{J}$.
  • Transverse shear is zero ($Q = 0$ at outer boundary).
• Element B (Side Outer Fiber on Neutral Axis, $y = 0, z = +c$):
  • Bending stress is zero ($y = 0$).
  • Axial stress is active: $\sigma_x = \frac{P}{A}$.
  • Torsional shear is maximum: $\tau_{xz} = \frac{T c}{J}$.
  • Transverse shear is maximum: $\tau_v = \frac{4 V}{3 A}$, which adds vectorially to torsional shear.

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2. 2D & 3D Mohr's Circle and Principal Stress Analysis

2D Plane Stress Transformation Equations

Given normal stresses $\sigma_x, \sigma_y$ and shear stress $\tau_{xy}$ acting on an element, the transformed stresses $\sigma_{x'}, \tau_{x'y'}$ on a plane inclined at angle $\theta$ (counterclockwise positive) are:

$\sigma_{x'} = \frac{\sigma_x + \sigma_y}{2} + \left(\frac{\sigma_x - \sigma_y}{2}\right) \cos 2\theta + \tau_{xy} \sin 2\theta$

$\tau_{x'y'} = -\left(\frac{\sigma_x - \sigma_y}{2}\right) \sin 2\theta + \tau_{xy} \cos 2\theta$

+-----------------------------------------------------------------------------+
|                        2D MOHR'S CIRCLE PARAMETERS                          |
|                                                                             |
|   Center Coordinate:   C = (\sigma_{\text{avg}}, 0) = (\frac{\sigma_x + \sigma_y}{2}, 0) |
|                                                                             |
|   Circle Radius:       R = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}    |
|                                                                             |
|   Principal Stresses:  \sigma_{1,2} = \sigma_{\text{avg}} \pm R            |
|                                                                             |
|   Principal Angle:     \tan 2\theta_p = \frac{2 \tau_{xy}}{\sigma_x - \sigma_y}           |
|                                                                             |
|   Max In-Plane Shear:  \tau_{\text{max, in-plane}} = R \quad (\text{at } \theta_s = \theta_p \pm 45^\circ) |
+-----------------------------------------------------------------------------+

                     \tau (Shear Stress)
                       ^
                       |          * (\sigma_2, \tau=0)
                       |         /|\
                       |       /  |  \
                       |      /   |   \
          -------------+-----[C]--+----+----* (\sigma_1, \tau=0)----> \sigma
                       |      \   |   /
                       |       \  |  /
                       |         \|/
                       |          * (\sigma_{\text{avg}}, -R)
                       |

3D Mohr's Circle and Absolute Maximum Shear Stress

In three dimensions, any state of stress exhibits three mutually perpendicular principal axes with normal stresses ordered algebraically as: $\sigma_1 \ge \sigma_2 \ge \sigma_3$

For a state of plane stress ($\sigma_z = \tau_{xz} = \tau_{yz} = 0$), the out-of-plane principal stress is identically zero ($0$). The three 3D principal stresses are the two in-plane principal stresses $\sigma_A, \sigma_B$ and $0$.

+-----------------------------------------------------------------------------+
|               ABSOLUTE MAXIMUM SHEAR STRESS IN PLANE STRESS                 |
|                                                                             |
|   Formula:  \tau_{\text{max, abs}} = \frac{\sigma_1 - \sigma_3}{2}           |
|                                                                             |
|   Case I:   \sigma_A > 0 \text{ and } \sigma_B < 0 \implies \sigma_1=\sigma_A, \sigma_2=0, \sigma_3=\sigma_B |
|             \tau_{\text{max, abs}} = \frac{\sigma_A - \sigma_B}{2} = R (\text{In-plane governs})    |
|                                                                             |
|   Case II:  \sigma_A > \sigma_B > 0 \implies \sigma_1=\sigma_A, \sigma_2=\sigma_B, \sigma_3=0         |
|             \tau_{\text{max, abs}} = \frac{\sigma_A - 0}{2} = \frac{\sigma_A}{2} > R (Out-of-plane) |
|                                                                             |
|   Case III: 0 > \sigma_A > \sigma_B \implies \sigma_1=0, \sigma_2=\sigma_A, \sigma_3=\sigma_B         |
|             \tau_{\text{max, abs}} = \frac{0 - \sigma_B}{2} = \frac{|\sigma_B|}{2} > R (Out-of-plane)|
+-----------------------------------------------------------------------------+

[!WARNING] Critical Exam Trap — Absolute vs. In-Plane Shear Stress: If both non-zero principal stresses have the same sign (both tensile or both compressive), the in-plane Mohr's circle radius $R$ is not the absolute maximum shear stress. The absolute maximum shear stress occurs on a plane oriented at $45^\circ$ into the thickness of the material and equals $\sigma_1 / 2$ or $|\sigma_3| / 2$.

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3. Generalized Hooke's Law, Plane Stress vs. Plane Strain & Elastic Moduli

For an isotropic, linear-elastic material, the triaxial strains are related to triaxial stresses through Young's modulus $E$ and Poisson's ratio $\nu$:

$\epsilon_x = \frac{1}{E}\left[\sigma_x - \nu(\sigma_y + \sigma_z)\right], \quad \gamma_{xy} = \frac{\tau_{xy}}{G}$ $\epsilon_y = \frac{1}{E}\left[\sigma_y - \nu(\sigma_x + \sigma_z)\right], \quad \gamma_{yz} = \frac{\tau_{yz}}{G}$ $\epsilon_z = \frac{1}{E}\left[\sigma_z - \nu(\sigma_x + \sigma_y)\right], \quad \gamma_{zx} = \frac{\tau_{zx}}{G}$

Elastic Moduli Relationships

Plane Stress vs. Plane Strain Comparison

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4. Strain Gauge Rosette Transformation

Electrical resistance strain gauges measure normal strains only along the axis of their grid. Because shear strain cannot be measured directly by a single gauge, a strain gauge rosette consisting of three independent gauges is mounted to the free surface (plane stress state) to resolve the full 2D surface strain tensor $(\epsilon_x, \epsilon_y, \gamma_{xy})$.

+-----------------------------------------------------------------------------+
|                      STRAIN GAUGE ROSETTE GEOMETRIES                        |
|                                                                             |
|   [RECTANGULAR ROSETTE (0°-45°-90°)]       [DELTA ROSETTE (0°-60°-120°)]    |
|                  y                                      y                   |
|                  ^                                      ^                   |
|                  | Gauge c (90°)                        |                   |
|                  |   / Gauge b (45°)            Gauge b | Gauge c           |
|                  |  /                            (60°) \ / (120°)           |
|                  | /                                    |                   |
|                  +--------> x                           +--------> x        |
|                    Gauge a (0°)                           Gauge a (0°)      |
+-----------------------------------------------------------------------------+

General Transformation Equation for any Gauge at Angle $\theta$:

$\epsilon_\theta = \epsilon_x \cos^2 \theta + \epsilon_y \sin^2 \theta + \gamma_{xy} \sin \theta \cos \theta$ $\epsilon_\theta = \frac{\epsilon_x + \epsilon_y}{2} + \left(\frac{\epsilon_x - \epsilon_y}{2}\right)\cos 2\theta + \frac{\gamma_{xy}}{2}\sin 2\theta$

1. Rectangular Rosette ($\theta_a = 0^\circ, \theta_b = 45^\circ, \theta_c = 90^\circ$):

• $\epsilon_x = \epsilon_a$
• $\epsilon_y = \epsilon_c$
• $\gamma_{xy} = 2\epsilon_b - (\epsilon_a + \epsilon_c)$

Principal Strains: $\epsilon_{1,2} = \frac{\epsilon_a + \epsilon_c}{2} \pm \sqrt{\left(\frac{\epsilon_a - \epsilon_c}{2}\right)^2 + \left(\epsilon_b - \frac{\epsilon_a + \epsilon_c}{2}\right)^2}$

2. Delta Rosette ($\theta_a = 0^\circ, \theta_b = 60^\circ, \theta_c = 120^\circ$):

• $\epsilon_x = \epsilon_a$
• $\epsilon_y = \frac{2(\epsilon_b + \epsilon_c) - \epsilon_a}{3}$
• $\gamma_{xy} = \frac{2(\epsilon_b - \epsilon_c)}{\sqrt{3}}$

Principal Strains: $\epsilon_{\text{avg}} = \frac{\epsilon_a + \epsilon_b + \epsilon_c}{3}, \quad R_\epsilon = \sqrt{\left(\epsilon_a - \epsilon_{\text{avg}}\right)^2 + \frac{(\epsilon_b - \epsilon_c)^2}{3}}, \quad \epsilon_{1,2} = \epsilon_{\text{avg}} \pm R_\epsilon$

Calculating Principal Stresses from Principal Strains (Plane Stress):

$\sigma_1 = \frac{E}{1-\nu^2}(\epsilon_1 + \nu\epsilon_2)$ $\sigma_2 = \frac{E}{1-\nu^2}(\epsilon_2 + \nu\epsilon_1)$

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5. Stress Concentration Factors ($K_t$) and Notch Sensitivity ($q$)

Discontinuities such as geometric notches, shoulder fillets, keyways, and transverse cross-holes disrupt nominal stress trajectories, causing steep localized stress gradients.

+-----------------------------------------------------------------------------+
|                     STRESS CONCENTRATION & SENSITIVITY                      |
|                                                                             |
|   Theoretical / Geometric Factor:   K_t = \frac{\sigma_{\text{max}}}{\sigma_{\text{nom}}} |
|                                                                             |
|   Fatigue / Effective Factor:       K_f = 1 + q(K_t - 1)                    |
|                                                                             |
|   Notch Sensitivity Index:          q = \frac{K_f - 1}{K_t - 1}             |
+-----------------------------------------------------------------------------+

• Static Loading Behavior:
  • Ductile Materials ($\epsilon_f \ge 0.05$): Localized yielding at the notch root causes plastic redistribution without macroscopic failure. Consequently, $K_t$ is typically neglected for static ductile design ($K_f \approx 1.0$).
  • Brittle Materials ($\epsilon_f < 0.05$): No plastic yielding occurs; the full theoretical stress concentration factor $K_t$ must be applied directly to nominal stresses ($K_f = K_t$).
• Dynamic / Fatigue Loading: Materials exhibit notch sensitivity $0 \le q \le 1$. If notch root radius $r \to 0$, $q \to 0$ (microscopic scratches in soft metals have negligible fatigue penalty); for large radii or hard steels, $q \to 1.0$.

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6. Step-by-Step Worked Engineering Problem

Problem Statement

A solid cylindrical drive shaft made of alloy steel ($E = 205\text{ GPa}, \nu = 0.29$) with diameter $d = 50\text{ mm}$ is subjected to simultaneous combined loading:

• Axial tensile load: $P = 40\text{ kN}$
• Bending moment: $M = 1.25\text{ kN}\cdot\text{m}$
• Torsional torque: $T = 1.80\text{ kN}\cdot\text{m}$

Determine: (a) the state of stress on the top outer fiber, (b) the principal stresses $\sigma_1, \sigma_2, \sigma_3$, and (c) the absolute maximum shear stress $\tau_{\text{max, abs}}$.

                      M = 1.25 kN·m (Bending)
                              ^
                              |
            P = 40 kN =====[ SHAFT ]=====> P = 40 kN (Tension)
                              |
                              v
                      T = 1.80 kN·m (Torque)
                      Diameter d = 50 mm

Step 1: Geometric Properties of Cross-Section

$A = \frac{\pi d^2}{4} = \frac{\pi (0.050)^2}{4} = 1.9635 \times 10^{-3}\text{ m}^2$ $I = \frac{\pi d^4}{64} = \frac{\pi (0.050)^4}{64} = 3.0680 \times 10^{-7}\text{ m}^4$ $J = 2 I = 6.1359 \times 10^{-7}\text{ m}^4$ $c = \frac{d}{2} = 0.025\text{ m}$

Step 2: Stress Components at Top Fiber ($y = +c$)

$\sigma_{\text{axial}} = \frac{P}{A} = \frac{40 \times 10^3}{1.9635 \times 10^{-3}} = 20.372\text{ MPa}$ $\sigma_{\text{bending}} = \frac{M c}{I} = \frac{1250 \times 0.025}{3.0680 \times 10^{-7}} = 101.860\text{ MPa}$ $\sigma_x = \sigma_{\text{axial}} + \sigma_{\text{bending}} = 20.372 + 101.860 = 122.232\text{ MPa}$ $\sigma_y = 0$ $\tau_{xy} = \frac{T c}{J} = \frac{1800 \times 0.025}{6.1359 \times 10^{-7}} = 73.340\text{ MPa}$

Step 3: Mohr's Circle Center and Radius

$\sigma_{\text{avg}} = \frac{\sigma_x + \sigma_y}{2} = \frac{122.232 + 0}{2} = 61.116\text{ MPa}$ $R = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} = \sqrt{(61.116)^2 + (73.340)^2} = \sqrt{3735.17 + 5378.76} = 95.467\text{ MPa}$

Step 4: Principal Normal Stresses

$\sigma_A = \sigma_{\text{avg}} + R = 61.116 + 95.467 = 156.58\text{ MPa}$ $\sigma_B = \sigma_{\text{avg}} - R = 61.116 - 95.467 = -34.35\text{ MPa}$

Ordering the 3D principal stresses (with free surface $\sigma_z = 0$): $\sigma_1 = 156.58\text{ MPa}, \quad \sigma_2 = 0\text{ MPa}, \quad \sigma_3 = -34.35\text{ MPa}$

Step 5: Absolute Maximum Shear Stress

Because $\sigma_A > 0$ and $\sigma_B < 0$, $\sigma_3 = \sigma_B$: $\tau_{\text{max, abs}} = \frac{\sigma_1 - \sigma_3}{2} = \frac{156.58 - (-34.35)}{2} = \frac{190.93}{2} = 95.47\text{ MPa}$ (Note: Here in-plane maximum shear equals absolute maximum shear because principal stresses have opposite signs).

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7. Common Exam Traps & PE Pro-Tips

• Trap 1 — Direct Stress Addition: Never add bending stress ($\sigma$) and torsional shear stress ($\tau$) directly together! Normal and shear stresses act on different tensor orientations; they must be combined using Mohr's circle or failure theory effective stresses.
• Trap 2 — Sign Convention in Rosettes: In rosette analysis, ensure counterclockwise angles from Gauge A are positive. Mixing up signs on $\gamma_{xy}$ shifts the principal angle $\theta_p$ by $90^\circ$.
• Trap 3 — Transverse Shear vs. Torsional Shear on Shaft Surfaces: Transverse shear stress is zero at the top and bottom outer fibers ($y = \pm c$) and maximum at the neutral axis ($y = 0$). Torsional shear is maximum everywhere on the outer perimeter ($r = c$). Therefore, at the top fiber, only torsional shear exists.