Stress Concentration at Bolt Hole (Mechanical & Civil Engineering) Solution

A structural engineer analyzes stress distribution around a circular bolt fastening hole in a structural steel gusset plate under heavy tensile loading. Geometric discontinuities cause severe localized stress risers that can initiate micro-cracking and brittle fatigue fracture.

From high-resolution Finite Element Analysis (FEA) simulation, the Von Mises mechanical stress $\sigma(x, y)$ (in megapascals, $\text{MPa}$) near the bolt hole boundary is described by:

where $x$ and $y$ represent planar spatial coordinates in millimeters ($\text{mm}$) measured from the center of the fastening aperture.

1. State the Quotient Rule and Reciprocal Power Rule representations used to compute $\frac{\partial \sigma}{\partial x}$ and $\frac{\partial \sigma}{\partial y}$.
2. Determine general algebraic expressions for the spatial stress gradients $\frac{\partial \sigma}{\partial x}$ and $\frac{\partial \sigma}{\partial y}$.
3. Evaluate the instantaneous stress gradients at the critical monitoring coordinate $(x, y) = (1, 1)\text{ mm}$.
4. State the physical units of these derivatives. Geometrically interpret what these values mean if an inspector moves away from the bolt hole along the $+x$ vs $+y$ directions.

Step-by-Step Solution:

1. State the Differentiation Rules:

   • Writing $\sigma(x, y) = 400(x^2 + 2y^2 + 1)^{-1}$, apply the Generalized Power Chain Rule:
     $$\frac{\partial}{\partial x}[c \cdot u^{-1}] = -c \cdot u^{-2} \cdot \frac{\partial u}{\partial x} = -\frac{c}{u^2} \frac{\partial u}{\partial x}$$
     (Defined in Partial Derivatives and Geometric Interpretation of Partial Derivatives.)

2. Compute the General Algebraic Expressions:

   • Let $u(x, y) = x^2 + 2y^2 + 1$.
   • Differentiating with respect to $x$ (holding $y$ constant):
     $$\frac{\partial u}{\partial x} = 2x$$
     $$\frac{\partial \sigma}{\partial x} = -\frac{400}{(x^2 + 2y^2 + 1)^2} \cdot (2x) = -\frac{800x}{(x^2 + 2y^2 + 1)^2}$$
   • Differentiating with respect to $y$ (holding $x$ constant):
     $$\frac{\partial u}{\partial y} = 4y$$
     $$\frac{\partial \sigma}{\partial y} = -\frac{400}{(x^2 + 2y^2 + 1)^2} \cdot (4y) = -\frac{1600y}{(x^2 + 2y^2 + 1)^2}$$

3. Evaluate at the Point $(1, 1)\text{ mm}$:

   • Evaluate denominator $u(1, 1)$:
     $$u(1, 1) = 1^2 + 2(1)^2 + 1 = 1 + 2 + 1 = 4$$
     $$u^2 = 4^2 = 16$$
   • Evaluate $\frac{\partial \sigma}{\partial x}(1, 1)$:
     $$\left.\frac{\partial \sigma}{\partial x}\right|_{(1, 1)} = -\frac{800(1)}{16} = -50\text{ MPa/mm}$$
   • Evaluate $\frac{\partial \sigma}{\partial y}(1, 1)$:
     $$\left.\frac{\partial \sigma}{\partial y}\right|_{(1, 1)} = -\frac{1600(1)}{16} = -100\text{ MPa/mm}$$

4. Physical Units and Geometric Interpretation:

   • Units: Megapascals per millimeter ($\text{MPa/mm}$).
   • Geometric Interpretation: At $(1, 1)$, moving in the positive $x$-direction (holding $y = 1\text{ mm}$) causes localized stress to drop at an instantaneous rate of $50\text{ MPa/mm}$. Moving in the positive $y$-direction (holding $x = 1\text{ mm}$) causes stress to plunge twice as rapidly, at $-100\text{ MPa/mm}$.
   • Engineering Consequence: The stress gradient is significantly steeper along the $y$-axis because of the coefficient $2$ in the ellipse geometry. Crack initiation risk diminishes twice as rapidly as you step along $y$ away from the notch than along $x$.