10.5 Stress Transformations, Principal Stresses, and Mohr's Circle

10.5 Stress Transformations, Principal Stresses, and Mohr's Circle

Core Engineering Principle: Stress at a point inside a loaded body varies depending on the orientation of the plane cut through that point. Stress transformation equations allow engineers to find critical principal normal stresses and maximum shear stresses necessary for structural design and failure assessment.

2D Plane Stress Transformation Equations

Consider a differential element in a state of plane stress (where $\sigma_z = 0$, $\tau_{xz} = 0$, $\tau_{yz} = 0$) subjected to normal stresses $\sigma_x$ and $\sigma_y$ and shear stress $\tau_{xy}$.

To determine normal stress $\sigma_{x'}$ and shear stress $\tau_{x'y'}$ acting on an inclined plane rotated counterclockwise by an angle $\theta$ from the $x$-axis, use the 2D Plane Stress Transformation Equations:

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

Stress Invariant

An important property of stress transformation is that the sum of normal stresses on any two perpendicular planes is constant (invariant):

$$\sigma_{x'} + \sigma_{y'} = \sigma_x + \sigma_y = 2 \sigma_{avg}$$

where the average normal stress is:

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

Principal Stresses and Maximum In-Plane Shear Stress

Principal Planes and Principal Stresses

Principal planes are planes of orientation $\theta_p$ on which the shear stress vanishes completely ($\tau_{x'y'} = 0$). Setting $\tau_{x'y'} = 0$ in the transformation equation yields the principal plane orientation formula:

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

This equation yields two roots, $\theta_{p1}$ and $\theta_{p2} = \theta_{p1} + 90^\circ$, defining two mutually perpendicular principal planes.

The corresponding maximum and minimum normal stresses acting on these planes are the Principal Stresses $(\sigma_1, \sigma_2)$:

$$\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2} = \sigma_{avg} \pm R$$

where $\sigma_1$ is the algebraically larger principal stress and $\sigma_2$ is the algebraically smaller principal stress.

Maximum In-Plane Shear Stress

The maximum in-plane shear stress $\tau_{max, in-plane}$ occurs on planes rotated by $\theta_s = \theta_p \pm 45^\circ$ relative to the principal planes:

$$\tan 2\theta_s = - \left( \frac{\sigma_x - \sigma_y}{2 \tau_{xy}} \right)$$

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

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

Important Note: On planes of maximum in-plane shear stress, normal stresses are not zero; a normal stress equal to $\sigma_{avg} = \frac{\sigma_x + \sigma_y}{2}$ acts on both orthogonal faces.

Mohr's Circle for 2D Stress Analysis

Mohr's Circle is a geometric representation of 2D stress transformation equations. Points along the circle correspond to normal and shear stress states $(\sigma, \tau)$ on planes of varying orientation $\theta$.

Construction Steps for Mohr's Circle

1. Establish Axes: Plot normal stress $\sigma$ on the horizontal axis (tension positive to the right) and shear stress $\tau$ on the vertical axis.
2. Locate Center ($C$): The center of the circle lies on the horizontal axis at $C = (\sigma_{avg}, 0) = \left( \frac{\sigma_x + \sigma_y}{2}, 0 \right)$.
3. Plot Reference Point ($X$): Plot point $X$ corresponding to stress on the vertical $x$-face: $X = (\sigma_x, -\tau_{xy})$ (using standard sign convention where shear producing clockwise rotation of the element is plotted above/below based on coordinate definition).
4. Calculate Radius ($R$): The radius $R$ is the distance from $C$ to $X$: $$R = \sqrt{\left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2}$$
5. Draw Circle and Identify Critical Points:
   • Principal Stresses: Intersections with horizontal axis $\sigma_1 = \sigma_{avg} + R$ and $\sigma_2 = \sigma_{avg} - R$.
   • Max In-Plane Shear: Top and bottom peaks of circle $\tau_{max, in-plane} = R$.
   • Angle Mapping: A physical element rotation of angle $\theta$ corresponds to an angular sweep of $2\theta$ in the same rotational direction on Mohr's Circle.

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

In 3D stress analysis, three principal stresses exist: $\sigma_1 \ge \sigma_2 \ge \sigma_3$. For 2D plane stress, the third principal stress perpendicular to the plane is zero ($\sigma_3 = 0$).

The Absolute Maximum Shear Stress $\tau_{abs, max}$ is:

$$\tau_{abs, max} = \frac{\sigma_{max} - \sigma_{min}}{2} = \frac{\sigma_1 - \sigma_3}{2}$$

Worked Engineering Problems

Problem 1: Full 2D Stress Transformation & Principal Stresses

Scenario: A state of plane stress at a point in a structural steel bracket is given by $\sigma_x = 80\text{ MPa}$ (tension), $\sigma_y = -40\text{ MPa}$ (compression), and $\tau_{xy} = 35\text{ MPa}$. Calculate (a) average normal stress $\sigma_{avg}$, (b) Mohr's circle radius $R$, (c) principal stresses $\sigma_1$ and $\sigma_2$, (d) principal plane orientation angle $\theta_p$, and (e) maximum in-plane shear stress $\tau_{max, in-plane}$.

Solution:

1. Calculate average normal stress $\sigma_{avg}$:

$$\sigma_{avg} = \frac{\sigma_x + \sigma_y}{2} = \frac{80 + (-40)}{2} = \frac{40}{2} = 20\text{ MPa}$$

1. Calculate Mohr's circle radius $R$:

$$R = \sqrt{\left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2} = \sqrt{\left( \frac{80 - (-40)}{2} \right)^2 + (35)^2}$$ $$R = \sqrt{(60)^2 + (35)^2} = \sqrt{3600 + 1225} = \sqrt{4825} = 69.46\text{ MPa}$$

1. Calculate principal normal stresses $\sigma_1$ and $\sigma_2$:

$$\sigma_1 = \sigma_{avg} + R = 20 + 69.46 = 89.46\text{ MPa (tension)}$$ $$\sigma_2 = \sigma_{avg} - R = 20 - 69.46 = -49.46\text{ MPa (compression)}$$

1. Calculate principal plane orientation angle $\theta_p$:

$$\tan 2\theta_p = \frac{2 \tau_{xy}}{\sigma_x - \sigma_y} = \frac{2(35)}{80 - (-40)} = \frac{70}{120} = 0.58333$$ $$2\theta_p = \arctan(0.58333) = 30.26^\circ \implies \theta_{p1} = 15.13^\circ$$

1. Maximum in-plane shear stress:

$$\tau_{max, in-plane} = R = 69.46\text{ MPa}$$

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Problem 2: Stresses on an Inclined Plane (Weld Seam Analysis)

Scenario: A cylindrical pressure pipe has a spiral weld inclined at an angle $\theta = 30^\circ$ relative to the transverse cross-sectional plane. The stress state relative to the pipe axes is $\sigma_x = 100\text{ MPa}$ (axial tension), $\sigma_y = 40\text{ MPa}$ (hoop tension), and $\tau_{xy} = 0\text{ MPa}$. Compute the normal stress $\sigma_{x'}$ and shear stress $\tau_{x'y'}$ acting directly across the welded seam.

Solution:

1. Identify input stresses and transformation angle:

$$\sigma_x = 100\text{ MPa}, \quad \sigma_y = 40\text{ MPa}, \quad \tau_{xy} = 0\text{ MPa}, \quad \theta = 30^\circ$$

1. Calculate parameters for transformation equations:

$$\sigma_{avg} = \frac{100 + 40}{2} = 70\text{ MPa}, \quad \frac{\sigma_x - \sigma_y}{2} = \frac{100 - 40}{2} = 30\text{ MPa}$$ $$2\theta = 60^\circ \implies \cos 60^\circ = 0.5, \quad \sin 60^\circ = 0.866025$$

1. Calculate normal stress $\sigma_{x'}$ across the weld:

$$\sigma_{x'} = \sigma_{avg} + \left( \frac{\sigma_x - \sigma_y}{2} \right) \cos 2\theta + \tau_{xy} \sin 2\theta$$ $$\sigma_{x'} = 70 + (30)(0.5) + (0)(0.866025) = 70 + 15 = 85.0\text{ MPa}$$

1. Calculate shear stress $\tau_{x'y'}$ along the weld line:

$$\tau_{x'y'} = - \left( \frac{\sigma_x - \sigma_y}{2} \right) \sin 2\theta + \tau_{xy} \cos 2\theta$$ $$\tau_{x'y'} = - (30)(0.866025) + 0 = -25.98\text{ MPa}$$

Magnitude of shear stress along the seam is $25.98\text{ MPa}$.