4.2 Combined Loading, Principal Stresses & Mohr's Circle

4.2 Combined Loading, Principal Stresses & Mohr's Circle

Mechanical components—such as power transmission shafts, pressure vessel walls, turbine rotors, and machine frames—rarely operate under simple uniaxial tension or pure torsion. Instead, external forces, bending moments, torques, and internal pressures combine to produce complex multiaxial states of stress. Determining structural integrity requires transforming these combined stress components onto principal planes to calculate maximum normal stresses, maximum in-plane shear, and absolute maximum shear stress.

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3D Stress Tensor & The Plane Stress Condition

The general three-dimensional stress state acting on an infinitesimal cubic element is represented by the symmetric Cauchy stress tensor $\boldsymbol{\sigma}$:

$$\boldsymbol{\sigma} = \begin{bmatrix} \sigma_x & \tau_{xy} & \tau_{xz} \\ \tau_{yx} & \sigma_y & \tau_{yz} \\ \tau_{zx} & \tau_{zy} & \sigma_z \end{bmatrix}$$

Moment equilibrium about the coordinate axes requires shear stress symmetry: $\tau_{xy} = \tau_{yx}$, $\tau_{xz} = \tau_{zx}$, and $\tau_{yz} = \tau_{zy}$, reducing the tensor to six independent stress components.

In many critical engineering applications—including thin plates, structural shells, and free external surfaces of solid bodies—the out-of-plane normal and shear stresses vanish:

$$\sigma_z = 0, \quad \tau_{xz} = 0, \quad \tau_{yz} = 0$$

This condition is defined as plane stress. The in-plane state is fully characterized by normal stresses $\sigma_x$, $\sigma_y$, and shear stress $\tau_{xy}$.

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Plane Stress Transformation Equations

Consider an inclined plane cutting through the stress element at an angle $\theta$, measured counterclockwise positive from the positive $x$-axis. Projecting force equilibrium along the rotated axis $x'$ (normal to the inclined plane) and $y'$ (tangential to the plane) yields the fundamental analytical transformation equations:

$$\sigma_{x'} = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2} \cos 2\theta + \tau_{xy} \sin 2\theta$$ $$\sigma_{y'} = \frac{\sigma_x + \sigma_y}{2} - \frac{\sigma_x - \sigma_y}{2} \cos 2\theta - \tau_{xy} \sin 2\theta$$ $$\tau_{x'y'} = -\frac{\sigma_x - \sigma_y}{2} \sin 2\theta + \tau_{xy} \cos 2\theta$$

An essential mathematical property of stress transformation is the invariance of the sum of orthogonal normal stresses:

$$I_1 = \sigma_{x'} + \sigma_{y'} = \sigma_x + \sigma_y$$

The first stress invariant $I_1$ remains constant under any arbitrary coordinate rotation.

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Principal Stresses and Principal Planes

The planes on which maximum and minimum normal stresses act are called the principal planes. Differentiating $\sigma_{x'}$ with respect to $\theta$ and setting the derivative to zero (or equivalently setting $\tau_{x'y'} = 0$) gives the principal orientation angle $\theta_p$:

$$\tan 2\theta_p = \frac{2 \tau_{xy}}{\sigma_x - \sigma_y}$$

This relationship defines two mutually perpendicular angles, $\theta_{p1}$ and $\theta_{p2} = \theta_{p1} + 90^\circ$. Substituting these angles back into the normal stress transformation equation yields the principal normal stresses $\sigma_1$ and $\sigma_2$:

$$\sigma_1, \sigma_2 = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}$$

By standard convention, $\sigma_1 \ge \sigma_2$.

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Maximum In-Plane vs. Absolute Maximum Shear Stress

Maximum In-Plane Shear Stress

The maximum in-plane shear stress $\tau_{\max, in}$ acts on planes rotated $45^\circ$ from the principal planes. Differentiating $\tau_{x'y'}$ with respect to $\theta$ and setting it to zero gives:

$$\tan 2\theta_s = -\frac{\sigma_x - \sigma_y}{2 \tau_{xy}} = -\frac{1}{\tan 2\theta_p}$$

Because the slopes of $2\theta_s$ and $2\theta_p$ are negative reciprocals, $2\theta_s = 2\theta_p \pm 90^\circ$, confirming that $\theta_s = \theta_p \pm 45^\circ$.

The magnitude of the maximum in-plane shear stress equals the radius of Mohr's circle:

$$\tau_{\max, in} = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} = \frac{\sigma_1 - \sigma_2}{2}$$

Planes of maximum in-plane shear stress are not normal-stress-free; they carry an average normal stress:

$$\sigma_{\text{avg}} = \frac{\sigma_x + \sigma_y}{2} = \frac{\sigma_1 + \sigma_2}{2}$$

Absolute Maximum Shear Stress ($\tau_{\max, abs}$)

Engineering materials fail in three dimensions. To calculate the true worst-case shear stress, all three principal stresses must be ranked algebraically:

$$\sigma_I \ge \sigma_{II} \ge \sigma_{III}$$

The absolute maximum shear stress is given by:

$$\tau_{\max, abs} = \frac{\sigma_I - \sigma_{III}}{2}$$

In a plane stress state, the third out-of-plane principal stress is $\sigma_z = 0$. The relationship between $\tau_{\max, in}$ and $\tau_{\max, abs}$ depends on the algebraic signs of $\sigma_1$ and $\sigma_2$:

1. Opposite Signs ($\sigma_1 > 0$ and $\sigma_2 < 0$): $\sigma_I = \sigma_1, \quad \sigma_{II} = 0, \quad \sigma_{III} = \sigma_2$ $\tau_{\max, abs} = \frac{\sigma_1 - \sigma_2}{2} = \tau_{\max, in}$

2. Both Positive ($\sigma_1 > 0$ and $\sigma_2 > 0$): $\sigma_I = \sigma_1, \quad \sigma_{II} = \sigma_2, \quad \sigma_{III} = 0$ $\tau_{\max, abs} = \frac{\sigma_1 - 0}{2} = \frac{\sigma_1}{2}$

3. Both Negative ($\sigma_1 < 0$ and $\sigma_2 < 0$): $\sigma_I = 0, \quad \sigma_{II} = \sigma_1, \quad \sigma_{III} = \sigma_2$ $\tau_{\max, abs} = \frac{0 - \sigma_2}{2} = \frac{|\sigma_2|}{2}$

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Mohr's Circle Construction and Properties

Mohr's circle is a geometric representation of stress transformation. Squaring and adding the transformation equations for $\sigma_{x'}$ and $\tau_{x'y'}$ eliminates the transformation angle $\theta$:

$$\left(\sigma_{x'} - \sigma_{\text{avg}}\right)^2 + \tau_{x'y'}^2 = R^2$$

This represents a circle in the $(\sigma, \tau)$ plane with:

• Center Coordinates: $C = (\sigma_{\text{avg}}, 0) = \left(\frac{\sigma_x + \sigma_y}{2}, 0\right)$
• Circle Radius: $R = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}$

Construction Rules:

1. Normal tensile stresses are plotted along the positive horizontal axis ($+\sigma$); compressive stresses along the negative axis ($-\sigma$).
2. Use the clockwise-up convention: a face shear stress that tends to rotate the element clockwise is plotted above the $\sigma$-axis, and one that tends to rotate it counterclockwise is plotted below.
3. Plot point $X$ for the vertical $x$-face at abscissa $\sigma_x$. A positive $\tau_{xy}$ on the $x$-face turns the element counterclockwise, so $X$ plots below the axis at a distance $\tau_{xy}$.
4. Plot point $Y$ for the horizontal $y$-face at abscissa $\sigma_y$. Its shear has the opposite sense, so for positive $\tau_{xy}$ the point $Y$ plots above the axis.
5. Connect points $X$ and $Y$ with a straight line. The intersection with the $\sigma$-axis defines the circle center $C$.
6. An angle of rotation $\theta$ on the physical stress element corresponds to an angle of $2\theta$ in the same rotational direction on Mohr's circle. (Some texts instead plot $X$ at $(\sigma_x, +\tau_{xy})$ with shear positive upward; the circle is then mirrored and rotations run in the opposite sense. Either convention works if it is used consistently.)

In three dimensions, the general stress state produces three mutually tangent Mohr's circles plotted between pairs $(\sigma_1, \sigma_2)$, $(\sigma_2, \sigma_3)$, and $(\sigma_1, \sigma_3)$. Any allowable oblique plane stress state falls within the shaded annular region bounded by the three circles.

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Strain Transformations, Rosettes & Generalized Hooke's Law

Strain Transformations

Because strain forms a second-order symmetric tensor analogous to stress, plane strain transformations follow identical trigonometric relationships, substituting tensorial shear strain $\epsilon_{xy} = \gamma_{xy} / 2$ for $\tau_{xy}$:

$$\epsilon_\theta = \frac{\epsilon_x + \epsilon_y}{2} + \frac{\epsilon_x - \epsilon_y}{2} \cos 2\theta + \frac{\gamma_{xy}}{2} \sin 2\theta$$

Strain Rosette Analysis

Strain gages measure normal surface strains along specific orientations. Multi-axis strain rosettes determine the complete in-plane strain tensor $(\epsilon_x, \epsilon_y, \gamma_{xy})$:

• Rectangular ($45^\circ$) Rosette: Three gages positioned at $\theta_a = 0^\circ$, $\theta_b = 45^\circ$, and $\theta_c = 90^\circ$:

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

• Delta ($60^\circ$) Rosette: Three gages positioned at $\theta_a = 0^\circ$, $\theta_b = 60^\circ$, and $\theta_c = 120^\circ$:

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

Generalized Hooke's Law in 3D

For an isotropic linear elastic material, the three-dimensional stress-strain relations are:

$$\epsilon_x = \frac{1}{E}\left[\sigma_x - \nu(\sigma_y + \sigma_z)\right]$$ $$\epsilon_y = \frac{1}{E}\left[\sigma_y - \nu(\sigma_x + \sigma_z)\right]$$ $$\epsilon_z = \frac{1}{E}\left[\sigma_z - \nu(\sigma_x + \sigma_y)\right]$$ $$\gamma_{xy} = \frac{\tau_{xy}}{G} = \frac{2(1+\nu)}{E} \tau_{xy}$$

For plane stress ($\sigma_z = 0$), inverting Hooke's law yields:

$$\sigma_x = \frac{E}{1 - \nu^2}(\epsilon_x + \nu \epsilon_y), \quad \sigma_y = \frac{E}{1 - \nu^2}(\epsilon_y + \nu \epsilon_x)$$

Note that out-of-plane normal strain is non-zero due to Poisson's effect: $\epsilon_z = -\frac{\nu}{E}(\sigma_x + \sigma_y) = -\frac{\nu}{1-\nu}(\epsilon_x + \epsilon_y)$.

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Complete Worked Numerical Example: Transmission Shaft Under Combined Loading

Problem Statement

A solid steel drive shaft of diameter $d = 50.0\,\text{mm}$ operates under simultaneous axial tension, bending, and torsion. At the critical outer surface fiber, the applied loads are:

• Axial tensile force: $P = 40.0\,\text{kN} = 40{,}000\,\text{N}$
• Bending moment: $M = 1.20\,\text{kN}\cdot\text{m} = 1200\,\text{N}\cdot\text{m}$
• Torsional torque: $T = 1.80\,\text{kN}\cdot\text{m} = 1800\,\text{N}\cdot\text{m}$

Determine:

1. The combined state of stress $(\sigma_x, \sigma_y, \tau_{xy})$ at the top outermost fiber.
2. The principal normal stresses $\sigma_1, \sigma_2$ and principal orientation $\theta_p$.
3. The maximum in-plane shear stress $\tau_{\max, in}$ and absolute maximum shear stress $\tau_{\max, abs}$.

Step 1: Compute Geometric Section Properties

$$A = \frac{\pi d^2}{4} = \frac{\pi (0.050\,\text{m})^2}{4} = 1.9635 \times 10^{-3}\,\text{m}^2$$ $$Z = \frac{\pi d^3}{32} = \frac{\pi (0.050\,\text{m})^3}{32} = 1.2272 \times 10^{-5}\,\text{m}^3$$ $$Z_p = \frac{\pi d^3}{16} = \frac{\pi (0.050\,\text{m})^3}{16} = 2.4544 \times 10^{-5}\,\text{m}^3$$

Step 2: Calculate In-Plane Stress Components

The axial force and bending moment produce normal stresses aligned along the $x$-axis:

$$\sigma_{\text{axial}} = \frac{P}{A} = \frac{40{,}000\,\text{N}}{1.9635 \times 10^{-3}\,\text{m}^2} = 20.37 \times 10^6\,\text{Pa} = 20.37\,\text{MPa}$$ $$\sigma_{\text{bending}} = \frac{M}{Z} = \frac{1200\,\text{N}\cdot\text{m}}{1.2272 \times 10^{-5}\,\text{m}^3} = 97.78 \times 10^6\,\text{Pa} = 97.78\,\text{MPa}$$ $$\sigma_x = \sigma_{\text{axial}} + \sigma_{\text{bending}} = 20.37 + 97.78 = 118.15\,\text{MPa}$$ $$\sigma_y = 0\,\text{MPa}$$

The torsional shear stress is:

$$\tau_{xy} = \frac{T}{Z_p} = \frac{1800\,\text{N}\cdot\text{m}}{2.4544 \times 10^{-5}\,\text{m}^3} = 73.34 \times 10^6\,\text{Pa} = 73.34\,\text{MPa}$$

Step 3: Determine Mohr's Circle Parameters

Center of Mohr's circle:

$$\sigma_{\text{avg}} = \frac{\sigma_x + \sigma_y}{2} = \frac{118.15 + 0}{2} = 59.08\,\text{MPa}$$

Radius of Mohr's circle:

$$R = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} = \sqrt{(59.08)^2 + (73.34)^2} = \sqrt{3490.45 + 5378.76} = \sqrt{8869.21} = 94.18\,\text{MPa}$$

Step 4: Calculate Principal Stresses & Principal Plane Orientation

$$\sigma_1 = \sigma_{\text{avg}} + R = 59.08 + 94.18 = 153.26\,\text{MPa}$$ $$\sigma_2 = \sigma_{\text{avg}} - R = 59.08 - 94.18 = -35.10\,\text{MPa}$$ $$\sigma_3 = 0\,\text{MPa} \quad (\text{free surface plane stress})$$

Orientation of principal plane:

$$\tan 2\theta_p = \frac{2 \tau_{xy}}{\sigma_x - \sigma_y} = \frac{2 \times 73.34}{118.15 - 0} = \frac{146.68}{118.15} = 1.2415$$ $$2\theta_p = \arctan(1.2415) = 51.15^\circ \implies \theta_{p1} = 25.58^\circ$$

Step 5: Evaluate Maximum Shear Stresses

1. Maximum in-plane shear stress:

$$\tau_{\max, in} = R = 94.18\,\text{MPa}$$

1. Absolute maximum shear stress: Arranging all three principal stresses algebraically: $\sigma_I = 153.26\,\text{MPa}$, $\sigma_{II} = 0\,\text{MPa}$, $\sigma_{III} = -35.10\,\text{MPa}$.

$$\tau_{\max, abs} = \frac{\sigma_I - \sigma_{III}}{2} = \frac{153.26 - (-35.10)}{2} = \frac{188.36}{2} = 94.18\,\text{MPa}$$

Because the in-plane principal stresses have opposite algebraic signs ($\sigma_1 > 0$ and $\sigma_2 < 0$), the absolute maximum shear stress coincides with the in-plane maximum shear stress.