How to Calculate Bolt Capacity — AISC 360, EN 1993, AS 4100 Worked Example

Bolt capacity calculations are fundamental to every steel connection design. Whether you are designing a simple shear tab, a moment end plate, or a bracing connection, getting the bolt shear, bearing, and tension capacities right is a non-negotiable step. Yet the formulas differ meaningfully across the three major international codes — AISC 360 (United States), EN 1993-1-8 (Europe), and AS 4100 (Australia).

In this guide: We derive bolt shear, bearing, and tension capacity across all three codes, then work through a complete M20 Grade 8.8 bolt example in a single-shear connection. By the end, you will understand exactly how each code handles bolt limit states and where the differences matter.

PRELIMINARY — NOT FOR CONSTRUCTION. All results discussed are for educational and reference use only. Must be independently verified by a licensed Professional Engineer or Structural Engineer before use in any project.

Bolt Shear Capacity — AISC 360-22 Section J3.6

AISC 360 provides the most commonly used bolt shear formulas in North American practice. The key distinction is whether the threads are included (N) or excluded (X) from the shear plane.

Nominal Shear Strength

Per AISC 360-22 Section J3.6:

$$R_n = F_{nv} \times A_b$$

Where:

For Group A bolts (A325, A325M, F1852):

For Group B bolts (A490, A490M, F2280):

LRFD Design Strength

$$\phi R_n = 0.75 \times R_n$$

ASD Allowable Strength

$$R_n / \Omega = R_n / 2.00$$

Worked Example — M20 Bolt, AISC 360

For an M20 bolt (nominal diameter 20 mm, $A_b$ = 314 mm²):

For a single-shear connection, the bolt capacity is the single-plane value. For a double-shear connection, multiply by 2.

Bolt Shear Capacity — EN 1993-1-8 Table 3.4

The Eurocode approach uses a shear resistance formula with an $\alpha_v$ factor that depends on bolt grade and whether the shear plane passes through the threaded or unthreaded portion.

Design Shear Resistance per Shear Plane

$$F_{v,Rd} = \frac{\alpha_v \times f_{ub} \times A}{\gamma_{M2}}$$

Where:

Worked Example — M20 Grade 8.8, EN 1993-1-8

$$F_{v,Rd} = \frac{0.6 \times 800 \times 245}{1.25} = \frac{117,600}{1.25} = \mathbf{94.1\text{ kN}}$$

Key difference from AISC: The Eurocode uses the tensile stress area $A_s$ (245 mm²) rather than the nominal body area $A_b$ (314 mm²), which is about 22% smaller. This makes the Eurocode more conservative when threads are in the shear plane.

Bolt Shear Capacity — AS 4100 Clause 9.3

AS 4100 uses a formula that accounts for both the number of shear planes with threads intercepted ($n_x$) and those without ($n_n$), allowing more precise modeling of bolt behavior.

Design Shear Capacity

$$V_f = \phi \times 0.62 \times f_{uf} \times (n_n \times A_c + n_x \times A_o)$$

Where:

Alternatively, using the simplified method:

$$V_f = \phi \times 0.62 \times f_{uf} \times A_o$$

Worked Example — M20 Grade 8.8/S, AS 4100

$$V_f = 0.80 \times 0.62 \times 830 \times (0 \times 245 + 1 \times 314)$$ $$V_f = 0.80 \times 0.62 \times 830 \times 314$$ $$V_f = 0.80 \times 161,584 = \mathbf{129.3\text{ kN}}$$

Note: AS 4100 provides a higher shear capacity than both AISC LRFD and EN 1993-1-8 for the same bolt. This is partly due to the higher capacity factor ($\phi = 0.80$ vs 0.75) and the use of the full nominal area.

Cross-Code Comparison — M20 Grade 8.8, Single Shear

Code Design Shear Capacity Basis
AISC 360-22 (LRFD) 87.6 kN $\phi = 0.75$, threads in shear plane (N)
AISC 360-22 (ASD) 58.4 kN $\Omega = 2.00$, threads in shear plane (N)
EN 1993-1-8 94.1 kN $\gamma_{M2} = 1.25$, $A_s$ = 245 mm²
AS 4100 129.3 kN $\phi = 0.80$, $A_o$ = 314 mm²

The variation of 58.4 to 129.3 kN — more than a factor of 2 — underscores why code-specific calculations are essential. You cannot simply convert between codes with a unit change.

Bolt Bearing Capacity — AISC 360-22 Section J3.10

Bearing failure occurs when the connected plate material crushes or tears out at the bolt hole. This is a plate limit state, not a bolt limit state.

Nominal Bearing Strength

Per AISC 360-22 J3.10, for standard, oversized, and short-slotted holes loaded in any direction:

$$R_n = 1.2 \times l_c \times t \times F_u \leq 2.4 \times d \times t \times F_u$$

Where:

For long-slotted holes perpendicular to the load:

$$R_n = 1.0 \times l_c \times t \times F_u \leq 2.0 \times d \times t \times F_u$$

Bearing vs. Tearout

The $1.2 l_c t F_u$ term covers tearout (shear rupture along two parallel lines from the bolt hole to the plate edge). The $2.4 d t F_u$ term is the upper bound for pure bearing (crushing of the plate in front of the bolt). The controlling value is the lesser of the two.

Worked Example — Bearing, 10mm Plate, A36 Steel

For an M20 bolt in a 10 mm thick A36 plate ($F_u = 400$ MPa), with 40 mm edge distance and 70 mm bolt spacing:

Bolt Tension Capacity — AISC 360-22 Section J3.6

Nominal Tension Strength

For bolts in direct tension:

$$R_n = F_{nt} \times A_b$$

Where $F_{nt} = 0.75 \times F_u$ from Table J3.2 (90 ksi / 621 MPa for Group A bolts).

The tension capacity is typically higher than the shear capacity for the same bolt, which is why shear often governs connection design.

Combined Shear and Tension — AISC 360-22 Section J3.7

When a bolt is subjected to both shear and tension simultaneously (common in moment connections with prying), an interaction check is required.

Per AISC 360-22 J3.7:

$$\frac{V_u}{\phi R_{nv}} + \frac{T_u}{\phi R_{nt}} \leq 1.0$$

Where $V_u$ and $T_u$ are the required shear and tension forces, and $R_{nv}$ and $R_{nt}$ are the nominal shear and tension strengths.

The available tensile strength is also reduced when shear is present, per the equation in J3.7. For a bolt with significant shear demand, the available tension drops below the pure-tension value.

Practical Application — Using Our Bolt Calculators

Manual bolt capacity calculations are tedious and error-prone when you need to check multiple bolt diameters, grades, and limit states simultaneously. Our free online bolt calculators automate the process:

Both tools run entirely in your browser with no signup required.

FAQ

How do you calculate the shear capacity of a bolt?

Bolt shear capacity is calculated as the product of the nominal shear strength and the resistance factor. Per AISC 360-22 J3.6, the nominal shear strength per shear plane is $F_{nv} \times A_b$, where $F_{nv}$ is the nominal shear stress (0.563 x Fu for Group A bolts with threads in the shear plane) and $A_b$ is the nominal bolt area. For an M20 8.8 bolt, this gives approximately 94.2 kN per shear plane. Per EN 1993-1-8 Table 3.4, shear resistance is $\alpha_v \times f{ub} \times A / \gamma*{M2}$, and per AS 4100 Clause 9.3.2.1, $V_f = \phi \times 0.62 \times f*{uf} \times (n_n \times A_c + n_x \times A_o)$.

What is the difference between bolt shear and bolt bearing?

Bolt shear failure occurs when the bolt shank itself shears across the shear plane. Bearing failure occurs when the connected plate material crushes or tears out around the bolt hole. Both must be checked independently. In single-shear connections, bearing is often the governing limit state for thinner plates, while shear governs for thicker plates or high-strength plates. Per AISC 360 J3.10, bearing strength depends on bolt diameter, plate thickness, edge distance, and bolt spacing.

How does bolt grade affect capacity?

Bolt grade directly determines the tensile strength $f_{ub}$, which governs both shear and tension capacity. Grade 4.6 bolts have $f_{ub} = 400$ MPa, Grade 8.8 has $f_{ub} = 800$ MPa, and Grade 10.9 has $f_{ub} = 1000$ MPa. A Grade 10.9 bolt provides roughly 2.5x the shear capacity of a Grade 4.6 bolt of the same diameter. Higher grade bolts also permit higher preload values for slip-critical connections per AISC 360 J3.8.

Is this calculator a replacement for professional engineering judgment?

No — this is an educational reference only. All bolt capacity calculations must be independently verified by a licensed Professional Engineer before use in any project. Results are PRELIMINARY — NOT FOR CONSTRUCTION.

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