Average Shear Stress

While normal stress acts perpendicular to a material's surface to pull it apart or push it together, shear stress acts parallel to the surface. It is the stress component that tends to slide or shear one part of a material past another.

What is Average Shear Stress?

When a force is applied parallel to a cross-sectional area, it creates an internal shear force $V$. The intensity of this force distributed over the area $A$ is called the average shear stress ($\tau$, tau).

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

where:

• $\tau$ = average shear stress [Pa or N/m²]
• $V$ = internal shear force [N]
• $A$ = shear area (area parallel to the applied force) [m²]

Engineering Note: We call this "average" shear stress because, in reality, shear stress is not distributed perfectly uniformly across a section. However, for simple connections like pins and bolts, assuming an average uniform distribution is a standard and safe engineering practice.

Single Shear vs. Double Shear

One of the most common applications of average shear stress is in analyzing connections (e.g., bolts, pins, or rivets). These connections typically fail in shear.

Single Shear

In a single shear connection, the applied force $F$ is resisted by a single cross-sectional area of the pin. The internal shear force $V$ equals the applied force $F$.

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

Double Shear

In a double shear connection, the pin connects three plates, and the force is resisted by two cross-sectional areas. The internal shear force $V$ at each cross-section is half of the applied force $F$.

$V = \frac{F}{2}$ $\tau = \frac{V}{A} = \frac{F}{2A}$

Engineering Design Principle: Using a double-shear connection halves the shear stress on the pin compared to a single-shear connection of the same diameter, allowing the connection to safely carry twice the load.

Worked Example

Problem: A steel pin is used to connect two plates in a single shear configuration. The pin has a diameter of 12 mm. If the plates are pulled apart with a force of 15 kN, what is the average shear stress in the pin?

Solution:

Step 1 — Calculate the shear area: The shear area is the cross-sectional area of the pin. $A = \frac{\pi d^2}{4} = \frac{\pi (0.012)^2}{4} = 1.131 \times 10^{-4} \text{ m}^2$

Step 2 — Determine the internal shear force: Since this is a single shear connection, the internal shear force equals the applied load. $V = F = 15{,}000 \text{ N}$

Step 3 — Calculate the average shear stress: $\tau = \frac{V}{A} = \frac{15{,}000}{1.131 \times 10^{-4}} = 132.6 \text{ MPa}$

Engineering Check: If the allowable shear stress for this grade of steel pin is 100 MPa, this design would fail (132.6 > 100). The engineer would need to increase the pin diameter or switch to a double-shear configuration.

Common Mistakes

1. Confusing Normal Area with Shear Area: In normal stress ($\sigma = F/A$), $A$ is perpendicular to the force. In shear stress ($\tau = V/A$), $A$ is parallel to the force.
2. Forgetting to check for double shear: Always draw a free-body diagram of the pin to determine if one or two planes are resisting the load. Missing a double shear condition will result in calculating twice the actual stress.