Bolt Stretching

Overview

Engineering reference data for Bolt Stretching in material science and properties.

Key Formulas

Stress

$\sigma = \frac{F}{A}$

Force per unit area.

Strain

$\varepsilon = \frac{\Delta L}{L_0}$

Change in length per original length.

Hooke's Law

$\sigma = E \varepsilon$

Stress proportional to strain in elastic region.

Thermal Expansion

$\Delta L = \alpha L_0 \Delta T$

Length change due to temperature.

Variables

References

• Original Source

Example - Imperial Units

This example demonstrates how to calculate bolt stretching, tensile stress area, and tensile stress using imperial units.

Given Parameters

• Stud diameter: 7/8 inches
• Thread pitch (p): 9 threads per inch (n = 9)
• Young's Modulus (E) for steel: 30 × 10⁶ psi
• Designed bolt load (F): 10,000 lb
• Effective length (L_eff): 5 inches

Step-by-Step Calculation

1. Tensile Stress Area (A) Using the formula for the tensile stress area: $A = \frac{\pi}{4} \left(d - \frac{0.9743}{n}\right)^2$ $A = \frac{\pi}{4} \left(\frac{7}{8} \text{ in} - \frac{0.9743}{9}\right)^2 = 0.46 \text{ in}^2$

2. Bolt Elongation (ΔL) Applying Hooke's Law for elongation: $\Delta L = \frac{F \cdot L}{E \cdot A}$ $\Delta L = \frac{10000 \text{ lb} \times 5 \text{ in}}{30 \times 10^6 \text{ psi} \times 0.46 \text{ in}^2} = 0.0036 \text{ in}$

3. Tensile Stress (σ) Calculating the stress in the bolt: $\sigma = \frac{F}{A}$ $\sigma = \frac{10000 \text{ lb}}{0.46 \text{ in}^2} = 21,740 \text{ psi}$

Summary of Results

• Tensile Stress Area: 0.46 in²
• Bolt Elongation: 0.0036 inches
• Resulting Tensile Stress: 21,740 psi

Formulas for Bolt Stretching

1. Elongation (Hook's Law)

$\Delta l = \frac{F \cdot L_{\text{eff}}}{E \cdot A}$

Where:

• $\Delta l$ = change in length of bolt (inches, m)
• $F$ = applied tensile load (lb, N)
• $L_{\text{eff}}$ = effective length of bolt where tensile strength is applied (inches, m)
• $E$ = Young's Modulus of Elasticity (psi, N/m²)
• $A$ = tensile stress area of the bolt (square inches, m²)

2. Tensile Stress Area Calculation

$A = \frac{\pi}{4} \left( d - \frac{0.9743}{n} \right)^2$

Where:

• $A$ = tensile stress area (square inches)
• $d$ = nominal diameter of bolt (inches)
• $n$ = number of threads per inch
• $p$ = pitch = $1/n$ (length per thread, inches)

3. Tensile Stress

$\sigma = \frac{F}{A}$

Where:

• $\sigma$ = tensile stress (psi, N/m²)
• $F$ = applied tensile load (lb, N)
• $A$ = tensile stress area (square inches, m²)

Key Concepts

Effective Bolt Length

The effective length ($L_{\text{eff}}$) is the portion of the bolt that actually experiences tensile stress and stretching. This is typically the distance between the nut and the bolt head, or between two nuts in a double-nut application.

Stress Area vs. Cross-Sectional Area

The tensile stress area ($A$) is not the simple geometric cross-section of the bolt. It accounts for thread relief and represents the minimum area through which tensile load is transmitted. For standard threads, this area is smaller than the nominal diameter cross-section.

Elastic Limit Considerations

The elongation formula (Hook's Law) is valid only within the elastic region of the bolt material. If the tensile stress exceeds the material's yield strength, permanent deformation (plastic elongation) will occur.

Thread Pitch Relationship

The thread pitch ($p$) and threads per inch ($n$) are reciprocally related: $n = \frac{1}{p}$

This relationship is crucial for calculating the tensile stress area using formula 2.

Example Parameters

The following table summarizes the input values used in the imperial unit example calculation for bolt elongation.

Key Calculation Results

Based on the example parameters, the derived engineering values are:

• Tensile Stress Area (A): 0.46 in²
• Elongation (dl): 0.0036 inches
• Tensile Stress (σ): 21,740 psi