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Length of Roller Pin of Forked End of Rocker Arm Considering Double Shear Failure of Pin Calculator

Formula Used:

\[ Length of Roller Pin = 1.25 \times \sqrt{\frac{2 \times Force on Roller Pin}{\pi \times Shear Stress in Roller Pin}} \] \[ l2 = 1.25 \times \sqrt{\frac{2 \times Pc}{\pi \times \tau r}} \]

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1. What is the Length of Roller Pin Formula?

The Length of Roller Pin formula calculates the required length of a roller pin in a forked end rocker arm considering double shear failure. This ensures the pin can withstand the applied forces without shearing failure.

2. How Does the Calculator Work?

The calculator uses the formula:

\[ l2 = 1.25 \times \sqrt{\frac{2 \times Pc}{\pi \times \tau r}} \]

Where:

Explanation: The formula accounts for double shear conditions and includes a safety factor of 1.25 to ensure reliable performance under load.

3. Importance of Roller Pin Length Calculation

Details: Proper roller pin length calculation is crucial for mechanical system reliability. It prevents shear failure, ensures load distribution, and maintains structural integrity in rocker arm assemblies.

4. Using the Calculator

Tips: Enter force on roller pin in Newtons and shear stress in Pascals. Both values must be positive numbers. The calculator will compute the required pin length with safety factor included.

5. Frequently Asked Questions (FAQ)

Q1: Why is a safety factor of 1.25 used?
A: The safety factor accounts for material variations, dynamic loads, and unexpected stress concentrations to ensure reliable performance.

Q2: What is double shear failure?
A: Double shear occurs when a pin is subjected to shear forces at two cross-sections simultaneously, typically in forked connections.

Q3: What materials are typically used for roller pins?
A: Roller pins are commonly made from hardened steel, alloy steel, or other high-strength materials with good shear resistance.

Q4: How does pin length affect performance?
A: Proper pin length ensures adequate shear area distribution and prevents excessive stress concentrations that could lead to failure.

Q5: Can this formula be used for single shear applications?
A: No, this specific formula is designed for double shear conditions. Single shear applications require different calculations.

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