8.2 Principal Stresses, Strain Energy & Mohr's Circle

1. 2D Stress Transformation & Stress Invariants

In complex mining structures—such as continuous miner cutterhead drives, shaft hoist drums, and hydraulic powered roof support canopy beams—material elements are simultaneously subjected to combined axial, bending, and torsional loads, creating multiaxial stress fields.

Stress Transformation Equations

Consider an infinitesimal plane stress element defined by normal stresses $\sigma_x, \sigma_y$ and shear stress $\tau_{xy}$. An oblique plane cut at an angle $\theta$ (measured counterclockwise from the vertical plane on which $\sigma_x$ acts) experiences normal stress $\sigma_{\theta}$ and shear stress $\tau_{\theta}$:

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

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

          σ_y
           ^ 
           |   τ_xy
     +-----|----+----->
     |     |    |    
σ_x <|-----+----+----|> σ_x
     |    /|    |    
     +---/-|----+----->
        /  |   τ_xy
       /   v
      /θ  σ_y
     / (Oblique Plane cut at angle θ)

Stress Invariants in 2D

The transformation equations reveal quantities that remain perfectly invariant under any coordinate rotation $\theta$:

1. First Stress Invariant ($I_1$): $I_1 = \sigma_x + \sigma_y = \sigma_{\theta} + \sigma_{\theta + 90^{\circ}} = \sigma_1 + \sigma_2 = \text{constant}$ Physical meaning: The sum of normal stresses on any two mutually perpendicular planes is constant.
2. Second Stress Invariant ($I_2$): $I_2 = \sigma_x \sigma_y - \tau_{xy}^2 = \sigma_1 \sigma_2 = \text{constant}$

2. Principal Planes, Principal Stresses & Maximum Shear Stress

Principal Planes & Stresses

Principal planes are defined as planes where the shear stress is identically zero ($\tau_{\theta} = 0$). Setting $\tau_{\theta} = 0$ yields the orientation of principal planes:

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

This equation gives two orthogonal solutions for $\theta_p$, spaced $90^{\circ}$ apart ($\theta_{p1}$ and $\theta_{p2} = \theta_{p1} + 90^{\circ}$). Substituting back into $\sigma_{\theta}$ yields the Major ($\sigma_1$) and Minor ($\sigma_2$) principal stresses:

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

Maximum In-Plane Shear Stress

To find the planes of maximum shear stress, we maximize $\tau_{\theta}$ with respect to $\theta$ ($\frac{d\tau_{\theta}}{d\theta} = 0$):

$\tan 2\theta_s = -\frac{\sigma_x - \sigma_y}{2\tau_{xy}} = -\cot 2\theta_p \implies 2\theta_s = 2\theta_p \pm 90^{\circ} \implies \mathbf{\theta_s = \theta_p \pm 45^{\circ}}$

• Planes of maximum shear stress are oriented at $45^{\circ}$ to the principal planes.
• The magnitude of the maximum in-plane shear stress is: $\mathbf{\tau_{\text{in-plane},\max} = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} = \frac{\sigma_1 - \sigma_2}{2}}$
• The normal stress acting on the maximum shear stress plane is non-zero and equals the average normal stress: $\sigma_{\text{avg}} = \frac{\sigma_x + \sigma_y}{2} = \frac{\sigma_1 + \sigma_2}{2}$

Absolute Maximum Shear Stress in 3D ($\tau_{\text{abs},\max}$)

In three dimensions, with ordered principal stresses $\sigma_1 \ge \sigma_2 \ge \sigma_3$, the absolute maximum shear stress is:

$\mathbf{\tau_{\text{abs},\max} = \frac{\sigma_1 - \sigma_3}{2} = \max\left(\left|\frac{\sigma_1 - \sigma_2}{2}\right|, \left|\frac{\sigma_2 - \sigma_3}{2}\right|, \left|\frac{\sigma_3 - \sigma_1}{2}\right|\right)}$

CRITICAL EXAM TRAP FOR PLANE STRESS ($\sigma_3 = 0$):

1. Opposite Signs ($\sigma_1 > 0 > \sigma_2$): Here $\sigma_1 \ge \sigma_3 = 0 \ge \sigma_2$. Thus $\tau_{\text{abs},\max} = \frac{\sigma_1 - \sigma_2}{2} = \tau_{\text{in-plane},\max}$.
2. Like Signs - Both Tensile ($\sigma_1 \ge \sigma_2 > 0$): Here the ordered principal stresses are $\sigma_1 \ge \sigma_2 \ge \sigma_3 = 0$. Thus: $\mathbf{\tau_{\text{abs},\max} = \frac{\sigma_1 - 0}{2} = \frac{\sigma_1}{2}}$ (Occurs on an out-of-plane plane inclined at $45^{\circ}$ to the 1-3 axes, NOT in the x-y plane!).
3. Like Signs - Both Compressive ($0 > \sigma_1 \ge \sigma_2$): Ordered as $\sigma_3 = 0 \ge \sigma_1 \ge \sigma_2$. Thus: $\mathbf{\tau_{\text{abs},\max} = \frac{0 - \sigma_2}{2} = \frac{|\sigma_2|}{2}}$

3. Mohr's Circle of Stress

Mohr's circle is the graphical representation of 2D stress transformation equations, where the horizontal axis represents normal stress $\sigma$ and the vertical axis represents shear stress $\tau$.

                  τ (Shear Stress)
                  ^
                  |          Radius R = τ_in-plane,max
                  |           .---* (σ_y, -τ_xy)
                  |         /     |
                  |        /      |
                  |       |   C   |
        ----------+-------+---*---+------------------> σ (Normal Stress)
                  |        \ (σ_avg, 0)  σ_1
                  |         \     |
                  |          '---* (σ_x, τ_xy)
                  v

Construction Rules

1. Center Coordinates ($C$): $C = \left(\frac{\sigma_x + \sigma_y}{2}, 0\right) = \left(\frac{\sigma_1 + \sigma_2}{2}, 0\right)$. Center always lies on the normal stress axis.
2. Radius ($R$): $R = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} = \frac{\sigma_1 - \sigma_2}{2} = \tau_{\text{in-plane},\max}$.
3. Angle Doubling: An angle rotation of $\theta$ in the physical stress element corresponds to an angle rotation of $2\theta$ in the same direction on Mohr's circle.

Canonical Stress States on Mohr's Circle

4. Thin Pressure Vessels: Cylinders and Spheres

A pressure vessel is classified as thin-walled when the wall thickness $t$ is less than $\frac{1}{20}$ to $\frac{1}{10}$ of its internal diameter $d$ ($t/d \le 0.05 - 0.10$). Radial stress $\sigma_r$ across the thickness is negligible compared to membrane stresses.

1. Thin Cylindrical Shells

Subjected to internal gauge pressure $p$, inside diameter $d$, and thickness $t$:

        +-----------------------------------+  ^ 
  <==   |                                   |  | Longitudinal stress σ_l
  σ_l   |      Internal Pressure p          |  d
  <==   |                                   |  v
        +-----------------------------------+ ----
           Hoop Stress σ_h (Circumferential)   t

1. Hoop / Circumferential Stress ($\sigma_h = \sigma_1$): Acts tangentially along the circumference to resist bursting across longitudinal diametral plane: $\mathbf{\sigma_h = \frac{p d}{2 t}}$
2. Longitudinal / Axial Stress ($\sigma_l = \sigma_2$): Acts parallel to the longitudinal axis to resist bursting across transverse plane: $\mathbf{\sigma_l = \frac{p d}{4 t} = \frac{1}{2}\sigma_h}$
3. Strains: $\epsilon_h = \frac{\sigma_h}{E} - \mu\frac{\sigma_l}{E} = \frac{pd}{4tE}(2 - \mu)$ $\epsilon_l = \frac{\sigma_l}{E} - \mu\frac{\sigma_h}{E} = \frac{pd}{4tE}(1 - 2\mu)$ $\mathbf{\frac{\epsilon_h}{\epsilon_l} = \frac{2 - \mu}{1 - 2\mu}}$
4. Volumetric Strain ($\epsilon_v = \frac{\Delta V}{V}$): $\epsilon_v = 2\epsilon_h + \epsilon_l = \mathbf{\frac{pd}{4tE}(5 - 4\mu)}$
5. Shear Stresses:
   • Maximum In-Plane Shear Stress: $\tau_{\text{in-plane},\max} = \frac{\sigma_h - \sigma_l}{2} = \frac{pd/2t - pd/4t}{2} = \mathbf{\frac{pd}{8t}}$.
   • Absolute Maximum Shear Stress (with $\sigma_3 \approx 0$): $\tau_{\text{abs},\max} = \frac{\sigma_h - 0}{2} = \mathbf{\frac{pd}{4t}}$.

2. Thin Spherical Shells

Due to spherical symmetry, the membrane stress in all tangential directions is identical:

1. Membrane Stress: $\sigma_1 = \sigma_2 = \mathbf{\frac{pd}{4t}}$.
2. Strains: $\epsilon_1 = \epsilon_2 = \frac{\sigma_1}{E}(1 - \mu) = \mathbf{\frac{pd}{4tE}(1 - \mu)}$.
3. Volumetric Strain: $\epsilon_v = 3\epsilon = \mathbf{\frac{3pd}{4tE}(1 - \mu)}$.
4. Shear Stresses:
   • Maximum In-Plane Shear Stress: $\tau_{\text{in-plane},\max} = \frac{\sigma_1 - \sigma_2}{2} = \mathbf{0}$.
   • Absolute Maximum Shear Stress: $\tau_{\text{abs},\max} = \frac{\sigma_1 - 0}{2} = \mathbf{\frac{pd}{8t}}$.

5. Elastic Strain Energy, Resilience & Castigliano's Second Theorem

Elastic Strain Energy Formulations ($U$)

Strain energy is the internal potential energy stored in an elastic body during deformation:

Resilience Terminology

• Resilience: The capacity of a material to absorb energy when deformed elastically and release that energy upon unloading.
• Proof Resilience: The maximum strain energy stored in a body up to the elastic limit without undergoing permanent plastic deformation: $U_{\text{proof}} = \frac{\sigma_y^2}{2E} \times V$
• Modulus of Resilience ($u_r$): The strain energy stored per unit volume at the elastic limit (area under the elastic region of the stress-strain curve): $\mathbf{u_r = \frac{\sigma_y^2}{2E}}$
• Modulus of Toughness: The total work done (energy absorbed) per unit volume up to complete fracture (total area under the entire stress-strain curve).

Castigliano's Second Theorem (Deflection)

In any linearly elastic structure at constant temperature, the first partial derivative of the total internal strain energy $U$ with respect to an applied concentrated force $P_i$ equals the displacement $\delta_i$ of the point of application of $P_i$ in the direction of $P_i$:

$\mathbf{\delta_i = \frac{\partial U}{\partial P_i}}$

Similarly, for an applied moment $M_i$, the angular rotation $\theta_i$ is:

$\mathbf{\theta_i = \frac{\partial U}{\partial M_i}}$

6. Worked Step-by-Step Engineering Calculations

Example 1: Biaxial State and Absolute Maximum Shear Stress

Problem: A structural steel plate element in a mine shaft headframe is subjected to normal stresses $\sigma_x = 120\text{ MPa}$ (tensile), $\sigma_y = 40\text{ MPa}$ (tensile), and shear stress $\tau_{xy} = 30\text{ MPa}$. Find:

1. Principal stresses $\sigma_1$ and $\sigma_2$
2. Orientation of major principal plane $\theta_{p1}$
3. Maximum in-plane shear stress $\tau_{\text{in-plane},\max}$
4. Absolute maximum shear stress in 3D $\tau_{\text{abs},\max}$

Solution:

• 1. Principal Stresses: $\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} = \frac{120 + 40}{2} \pm \sqrt{\left(\frac{120 - 40}{2}\right)^2 + 30^2}$ $\sigma_{1,2} = 80 \pm \sqrt{40^2 + 30^2} = 80 \pm \sqrt{1600 + 900} = 80 \pm 50$ $\sigma_1 = 80 + 50 = \mathbf{130\text{ MPa}}, \qquad \sigma_2 = 80 - 50 = \mathbf{30\text{ MPa}}$
• 2. Major Principal Plane Orientation: $\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x - \sigma_y} = \frac{2(30)}{120 - 40} = \frac{60}{80} = 0.75$ $2\theta_p = \arctan(0.75) = 36.87^{\circ} \implies \mathbf{\theta_{p1} = 18.43^{\circ}}$
• 3. Maximum In-Plane Shear Stress: $\tau_{\text{in-plane},\max} = \frac{\sigma_1 - \sigma_2}{2} = \frac{130 - 30}{2} = \mathbf{50\text{ MPa}}$
• 4. Absolute Maximum Shear Stress in 3D: Since $\sigma_1 = 130\text{ MPa} > 0$ and $\sigma_2 = 30\text{ MPa} > 0$ (both tensile, plane stress $\sigma_3 = 0$): Ordered principal stresses: $\sigma_1 = 130, \sigma_2 = 30, \sigma_3 = 0$. $\tau_{\text{abs},\max} = \frac{\sigma_1 - \sigma_3}{2} = \frac{130 - 0}{2} = \mathbf{65\text{ MPa}}$ (Note: $\tau_{\text{abs},\max} = 65\text{ MPa}$ is greater than $\tau_{\text{in-plane},\max} = 50\text{ MPa}$!).

Example 2: Cantilever Deflection via Castigliano's Theorem

Problem: Use Castigliano's Theorem to find the vertical tip deflection $\delta$ of a cantilever beam of span $L$, flexural rigidity $EI$, carrying a point load $W$ at its free tip.

Solution:

• Coordinate $x$ measured from free end ($0 \le x \le L$):
• Bending moment at section $x$: $M(x) = -W x$.
• Total strain energy in bending: $U = \int_0^L \frac{M^2(x)}{2EI} dx = \int_0^L \frac{(-Wx)^2}{2EI} dx = \frac{W^2}{2EI} \int_0^L x^2 dx = \frac{W^2 L^3}{6EI}$
• Applying Castigliano's theorem with respect to load $W$: $\delta = \frac{\partial U}{\partial W} = \frac{\partial}{\partial W}\left(\frac{W^2 L^3}{6EI}\right) = \mathbf{\frac{W L^3}{3EI}}$