6.2 Normal, Shear & Triaxial Stress-Strain Relationships

6.2 Normal, Shear & Triaxial Stress-Strain Relationships

1. Uniaxial Stress, Strain, and Hooke's Law

When a structural member is subjected to axial tension or compression forces $P$, internal resistive forces are distributed over the cross-sectional area $A$.

1.1 Axial Normal Stress and Strain

• Average Normal Stress ($\sigma$): $$\sigma = \frac{P}{A}$$ where tensile stress is designated positive ($+$) and compressive stress negative ($-$). Units in SI are Pascals ($1\text{ Pa} = 1\text{ N/m}^2$) or Megapascals ($1\text{ MPa} = 1\text{ N/mm}^2$).
• Axial Normal Strain ($\epsilon$): $$\epsilon = \frac{\delta}{L}$$ where $\delta = L_{\text{final}} - L_0$ is total axial elongation and $L$ is initial gauge length. Strain is dimensionless (often expressed as $\text{mm/mm}$ or $\mu\epsilon$).

1.2 Hooke's Law and Young's Modulus ($E$)

Within the linear elastic regime of a material, stress is directly proportional to strain, as expressed by Hooke's Law:

$$\sigma = E \epsilon$$

where $E$ is Young's Modulus (Modulus of Elasticity), representing the stiffness of the material ($E \approx 200\text{ GPa}$ for structural steel, $70\text{ GPa}$ for aluminum). Substituting $\sigma = P/A$ and $\epsilon = \delta/L$ into Hooke's Law yields the fundamental axial deformation equation:

$$\delta = \frac{P L}{A E}$$

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2. Shear Stress, Shear Modulus, and Elastic Constants

2.1 Shear Stress ($\tau$) and Shear Strain ($\gamma$)

Shear stress acts parallel to the plane of the cross-section, caused by transverse shearing force $V$:

$$\tau = \frac{V}{A}$$

Shear strain $\gamma$ represents the angular change (in radians) between two originally perpendicular lines. Within the elastic limit, shear stress and shear strain obey Hooke's Law for shear:

$$\tau = G \gamma$$

where $G$ is the Shear Modulus (Modulus of Rigidity).

2.2 Poisson's Ratio ($\nu$) and Elastic Relationships

When a bar is pulled in uniaxial tension, it elongates axially while contracting laterally. Poisson's Ratio ($\nu$) is the absolute ratio of lateral strain to axial strain:

$$\nu = - \frac{\epsilon_{\text{lateral}}}{\epsilon_{\text{axial}}}$$

For isotropic materials, the three primary elastic constants—Young's Modulus ($E$), Shear Modulus ($G$), and Bulk Modulus ($K$)—are fundamentally inter-related:

$$G = \frac{E}{2(1 + \nu)}$$ $$K = \frac{E}{3(1 - 2\nu)}$$

For typical metals with $\nu \approx 0.30$, $G \approx 0.385 E$.

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3. Thermal Stress and Statically Indeterminate Axial Members

3.1 Unconstrained Thermal Expansion

A homogeneous isotropic member subjected to a temperature change $\Delta T$ undergoes unconstrained thermal deformation $\delta_{\text{th}}$:

$$\delta_{\text{th}} = \alpha L \Delta T$$

where $\alpha$ is the coefficient of thermal expansion (units: $1/^\circ\text{C}$ or $1/\text{K}$). Unconstrained thermal expansion induces zero internal stress because no external boundary restrains the motion.

3.2 Constrained Thermal Stress

If the ends of a bar are rigidly fixed against two unyielding supports, thermal elongation is completely prevented ($\delta_{\text{net}} = \delta_{\text{th}} - \delta_{\text{reaction}} = 0$). This constraint generates a compressive thermal stress $\sigma_{\text{th}}$:

$$\alpha L \Delta T - \frac{P L}{A E} = 0 \implies \sigma_{\text{th}} = \frac{P}{A} = E \alpha \Delta T$$

Notice that constrained thermal stress depends only on material properties ($E, \alpha$) and temperature rise ($\Delta T$), independent of bar length or cross-sectional area.

3.3 Statically Indeterminate Axially Loaded Members

When internal forces cannot be determined by static equilibrium equations alone (number of unknown forces > number of equilibrium equations), the system is statically indeterminate. Solution requires combining:

1. Equilibrium Equations: e.g., $\sum F = 0 \implies P_1 + P_2 = P$.
2. Compatibility Equations: Kinematic constraints relating deformations (e.g., $\delta_1 = \delta_2$).
3. Load-Displacement Relations: $\delta_i = \frac{P_i L_i}{A_i E_i}$.

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4. Generalized Hooke's Law for Triaxial Stress States

Under multi-axial stress states where normal stresses $\sigma_x, \sigma_y, \sigma_z$ act simultaneously on a differential element, superposition of Poisson contractions yields the Generalized Hooke's Law:

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

Volumetric Strain ($e$): The fractional change in volume under triaxial stress is:

$$\ne = \frac{\Delta V}{V_0} = \epsilon_x + \epsilon_y + \epsilon_z = \frac{1 - 2\nu}{E} (\sigma_x + \sigma_y + \sigma_z)$$

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5. Mohr's Circle for Plane Stress Analysis

Plane stress occurs when stresses perpendicular to one plane are zero ($\sigma_z = 0, \tau_{xz} = 0, \tau_{yz} = 0$). Given stress components $(\sigma_x, \sigma_y, \tau_{xy})$ on an element, transformation equations determine normal and shear stresses on a plane inclined at angle $\theta$:

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

5.1 Mohr's Circle Construction Parameters

Mohr's Circle graphically represents these equations in the $\sigma - \tau$ coordinate plane:

• Center of Circle ($\sigma_{\text{avg}}$): $$\sigma_{\text{avg}} = \frac{\sigma_x + \sigma_y}{2}$$
• Radius of Circle ($R$): $$R = \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 }$$
• Principal Stresses ($\sigma_1, \sigma_2$): The maximum and minimum normal stresses occurring on planes where shear stress is zero: $$\sigma_1, \sigma_2 = \sigma_{\text{avg}} \pm R = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 }$$
• Maximum In-Plane Shear Stress ($\tau_{\text{max}}$): $$\tau_{\text{max}} = R = \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 }$$
• Principal Planes Orientation ($\theta_p$): $$\tan 2\theta_p = \frac{2 \tau_{xy}}{\sigma_x - \sigma_y}$$

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6. Worked Numerical Examples

Step-by-Step Example 1: Thermal Stress in Constrained Bar

Problem: A structural steel bar of cross-sectional area $A = 400\text{ mm}^2$ and length $L = 1.2\text{ m}$ is rigidly held between two unyielding walls. Properties: $E = 200\text{ GPa}$, $\alpha = 12 \times 10^{-6}/^\circ\text{C}$. Calculate the compressive normal stress $\sigma_{\text{th}}$ and reaction force $P$ when temperature rises by $\Delta T = 50^\circ\text{C}$.

Solution:

1. Calculate unconstrained thermal strain $\epsilon_{\text{th}}$: $$\epsilon_{\text{th}} = \alpha \Delta T = (12 \times 10^{-6}/^\circ\text{C}) \times 50^\circ\text{C} = 600 \times 10^{-6} = 0.000600$$
2. Since walls prevent elongation, thermal stress is: $$\sigma_{\text{th}} = E \alpha \Delta T = (200 \times 10^9\text{ Pa}) \times (600 \times 10^{-6}) = 120,000,000\text{ Pa} = 120.0\text{ MPa}\text{ (Compressive)}$$
3. Calculate compressive reaction force $P$: $$P = \sigma_{\text{th}} A = 120.0\text{ N/mm}^2 \times 400\text{ mm}^2 = 48,000\text{ N} = 48.0\text{ kN}$$

Step-by-Step Example 2: Shear Modulus and Bulk Modulus Derivation

Problem: A structural alloy has $E = 205\text{ GPa}$ and Poisson's ratio $\nu = 0.28$. Determine the Shear Modulus $G$ and Bulk Modulus $K$.

Solution:

1. Apply $G = \frac{E}{2(1 + \nu)}$: $$G = \frac{205\text{ GPa}}{2(1 + 0.28)} = \frac{205}{2.56} = 80.078\text{ GPa}$$
2. Apply $K = \frac{E}{3(1 - 2\nu)}$: $$K = \frac{205\text{ GPa}}{3(1 - 2 \times 0.28)} = \frac{205}{3(1 - 0.56)} = \frac{205}{3 \times 0.44} = \frac{205}{1.32} = 155.303\text{ GPa}$$

Step-by-Step Example 3: Mohr's Circle Plane Stress Analysis

Problem: An element in plane stress is subjected to $\sigma_x = 100\text{ MPa}$, $\sigma_y = -40\text{ MPa}$ (compressive), and $\tau_{xy} = 48\text{ MPa}$. Determine $\sigma_{\text{avg}}$, radius $R$, principal stresses $\sigma_1$ and $\sigma_2$, and maximum in-plane shear stress $\tau_{\text{max}}$.

Solution:

1. Calculate average normal stress $\sigma_{\text{avg}}$: $$\sigma_{\text{avg}} = \frac{\sigma_x + \sigma_y}{2} = \frac{100 + (-40)}{2} = \frac{60}{2} = 30.0\text{ MPa}$$
2. Calculate radius $R$: $$\frac{\sigma_x - \sigma_y}{2} = \frac{100 - (-40)}{2} = \frac{140}{2} = 70.0\text{ MPa}$$ $$R = \sqrt{(70.0)^2 + (48.0)^2} = \sqrt{4900 + 2304} = \sqrt{7204} = 84.876\text{ MPa}$$
3. Calculate principal stresses $\sigma_1$ and $\sigma_2$: $$\sigma_1 = \sigma_{\text{avg}} + R = 30.0 + 84.876 = 114.88\text{ MPa}$$ $$\sigma_2 = \sigma_{\text{avg}} - R = 30.0 - 84.876 = -54.88\text{ MPa}$$
4. Maximum in-plane shear stress $\tau_{\text{max}}$: $$\tau_{\text{max}} = R = 84.88\text{ MPa}$$