How Much Load Can a Bolt and Nut Withstand? Strength and Calculation Methods

From industrial design to civil engineering, bolts and nuts play a critical role in ensuring the safety and durability of structures. The load-carrying capacity of a bolt and nut depends on the material type, dimensions, and the design of the joint. In this article, we’ll explore how to calculate the load-carrying capacities of bolts and nuts and the factors to consider during the process.

Load-Carrying Capacities of Bolts and Nuts

The load-carrying capacities of bolts and nuts are determined by parameters such as tensile strength, shear strength, and tightening torque. These parameters depend on the bolt’s dimensions, material, and manufacturing quality.

1. Tensile Strength

Tensile strength refers to a bolt's resistance to axial pulling forces. It is calculated as the product of the bolt's cross-sectional area and the material’s ultimate tensile strength (UTS).

Tensile strength formula:

$F_t = A \times UTS$

Ft: Tensile strength (N)
A: Bolt's cross-sectional area (m²)
UTS: Material's ultimate tensile strength (Pa)

Example: For an M12 grade 8.8 bolt:

UTS for grade 8.8 is 800 MPa.
Cross-sectional area is approximately 84.3 mm².

$F_t = 84.3 \times 10^{-6} \times 800 \times 10^6$
$F_t = 67,440 \, N$

This means the bolt has a tensile strength of approximately 67.4 kN.

2. Shear Strength

Shear strength refers to the resistance of a bolt to forces applied perpendicular to its axis. It is generally considered as half the tensile strength.

Shear strength formula:

$F_s = 0.5 \times F_t$

Using the previous example:
$F_s = 0.5 \times 67,440$
$F_s = 33,720 \, N$

This means the bolt has a shear strength of approximately 33.7 kN.

3. Tightening Torque

Tightening torque ensures proper fastening and maximizes the load-carrying capacity of a bolt and nut. It depends on the bolt’s size and material.

Tightening torque formula:

$T = K \times F \times d$

T: Tightening torque (Nm)
K: Torque coefficient (typically 0.2–0.3)
F: Applied force (N)
d: Nominal diameter of the bolt (m)

Example: For an M12 grade 8.8 bolt:

Torque coefficient: 0.2,
Applied force: 67,440 N,
Nominal diameter: 12 mm (0.012 m).

$T = 0.2 \times 67,440 \times 0.012$
$T = 161.86 \, Nm$

This means the bolt requires a tightening torque of approximately 162 Nm.

What Does Bolt Strength Class Mean? How Is It Read?

Bolt strength is typically represented by a class such as 8.8 or 10.9. These numbers provide critical information:

The first digit (8, 10, etc.) represents the ultimate tensile strength in N/mm². For example, 8.8 means 800 N/mm².
The second digit (0.8, 0.9, etc.) indicates the yield strength ratio. For example, for 8.8, the yield strength is 800 × 0.8 = 640 N/mm².

This classification ensures consistent load-carrying calculations.

Tensile Strength for M4 to M20 Bolts

Below is a table showing the tensile strengths for grade 8.8 bolts of different sizes:

Bolt Size	Cross-Sectional Area (A) [mm²]	Tensile Strength (Ft) [kN]
M4	8.78	7.02
M6	20.1	16.08
M8	36.6	29.28
M10	58	46.4
M12	84.3	67.44
M16	157	125.6
M20	245	196

Note: These values represent tensile strength only. Shear strength and tightening torque can be calculated using the above formulas.

Load-Carrying Capacities of Nuts

Nuts are designed to match the strength of bolts and are influenced by:

Thread profile: Wider or finer threads can enhance capacity.
Material quality: Nuts should be of the same grade as bolts.

Conclusion

The load-carrying capacity of a bolt and nut depends on their material, dimensions, and design. Proper calculations and the selection of suitable materials ensure the safety and durability of structures. For critical applications, adhering to engineering standards is essential.

Design Engineers

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