Transverse Shear Stress: Jourawski's Formula
Calculates the internal transverse and horizontal shear stress distribution across the depth of a flexural member subjected to shear force V.
\[ \tau = \frac{V \cdot Q}{I \cdot t} \quad \text{where } Q = \int_{y_1}^{y_{top}} y \, dA = A' \cdot \bar{y}' \]
Plain Text Formula:
tau = (V * Q) / (I * t) where Q = A' * y_bar'
📐 Variable Definitions & Parameters
| Symbol | Variable Name | Standard Unit | Physical Definition |
|---|---|---|---|
| \[ \tau \] | Transverse Shear Stress | MPa or Pa | Internal shear stress at height y1 |
| \[ V \] | Total Shear Force | Newtons (N) | Internal vertical shear at the cross-section |
| \[ Q \] | First Moment of Area | m³ or mm³ | Statical moment of the area above the cut plane about neutral axis |
| \[ I \] | Moment of Inertia | m⁴ | Second moment of area of whole cross-section |
| \[ t \] | Web Thickness | m or mm | Width of cross-section at the evaluation cut plane |
💡 Worked Engineering Example
Problem Statement:
Why does maximum shear stress in a rectangular beam occur at the neutral axis (y=0)?
Step-by-Step Solution:
- Q is maximized at the neutral axis where the entire upper half of the cross-section contributes
- At extreme top and bottom fibers, area above cut A'=0, so Q=0 and shear stress tau=0
- At neutral axis of rectangular section, tau_max = 1.5 * (V / A)
🎯 Result: Shear stress follows a parabolic curve, peaking at 1.5 times the average shear stress at the neutral axis.