Anchor bolt in tension
Four mechanisms compete and the weakest governs: the steel itself, the concrete cone that pulls out with it, the head bearing against the concrete, and — near an edge — the side-face blowout. The breakout cone is usually the one that surprises people, because it scales with hef^1.5 rather than with the bolt area.
| Mode | Capacity | Basis |
|---|---|---|
| Steel tension | 18.8 kip | 0.75 Ase futa |
| Concrete breakout | 24.0 kip | 0.70 × 24 λ √f’c hef^1.5 × area and edge factors |
| Pullout | 33.6 kip | 0.70 × 8 Abrg f’c |
| Governing | 18.8 kip | Steel |
| Applied load | 10.0 kip | 53% of capacity |
| Breakout area ratio | 1.00 | ANc / ANco |
| Edge factor ψed,N | 1.00 | 0.7 + 0.3 ca1 / 1.5hef |
| Effective embedment | 8.0 in | hef |
Steel governs at 18.8 kip per bolt. The breakout cone is full, so the capacity is set by hef^1.5 rather than by the bolt area.
φ Nsa = 0.75 Ase futa · Bolt capacity.
0.70 × 24 λ √f’c hef^1.5 · Cast-in anchor, single.
0.7 + 0.3 ca1 / 1.5hef ≤ 1 · Truncates near a free edge.
0.70 × 8 Abrg f’c · Head bearing.
A clear path from inputs to results.
Four mechanisms compete and the weakest governs: the steel itself, the concrete cone that pulls out with it, the head bearing against the concrete, and — near an edge — the side-face blowout. The breakout cone is usually the one that surprises people, because it scales with hef^1.5 rather than with the bolt area.
How to use
- Choose the bolt diameter, grade and embedment depth.
- Set the edge distance and the concrete strength.
- Enter the head bearing area and the applied tension.
- Read the governing mode and the utilisation.
The formula
Reference: ACI 318-19 Chapter 17, §17.6.1 (steel), §17.6.2 (breakout) and §17.6.3 (pullout).
Worked example
A 3/4 inch bolt embedded 8 inches
- 3/4 in F1554 Gr 55 bolt, hef = 8 in.
- 12 in edge distance, 4,000 psi normal-weight concrete.
- 1.5 in² head bearing area, 10 kip applied tension.
- 1
The tensile area is π/4 (0.75 − 0.9743/10)² = 0.3345 in².
- 2
Steel capacity is 0.75 × 0.3345 × 75,000 = 18.8 kip.
- 3
The breakout cone is full at this edge distance, so Ncb = 0.70 × 24 × √4000 × 8^1.5 = 24.0 kip.
- 4
Pullout is 0.70 × 8 × 1.5 × 4,000 = 33.6 kip.
- 5
Steel governs at 18.8 kip, and the 10 kip load uses 53% of it.
18.8 kip · steel governs · 53% utilised.
Interpreting the result
The governing mode changes with the detail rather than with the load: a shallow embedment is breakout-controlled, a deep one is steel-controlled, and a close edge is breakout-controlled again however deep the bolt. That is why moving an anchor is often cheaper than lengthening it.
Assumptions & limitations
- A single cast-in headed anchor in tension.
- φ = 0.75 for steel and 0.70 for the concrete modes.
- Normal-weight concrete unless lightweight is selected.
- No supplementary reinforcement is counted.
Frequently asked questions
Why does embedment matter more than diameter?
Because breakout capacity goes as hef^1.5 while steel capacity goes as the area, which is d². For a shallow anchor the concrete always governs.
Can I count the rebar as reinforcement?
Yes — §17.5.2.1 allows the anchor reinforcement to take the load if it is developed on both sides of the breakout plane. That is a different calculation and needs a detail.
Do post-installed anchors use the same numbers?
No. Adhesive and mechanical anchors have their own kc values and qualification requirements under ACI 355.2, and the manufacturer’s data governs.
What is 1.5 hef?
The critical edge distance. Inside it the breakout cone is truncated, and both the area ratio and the edge factor reduce the capacity.
What about shear?
Shear has its own set of modes — steel, concrete edge breakout, pryout — and combined tension and shear uses an interaction diagram.
Methodology last reviewed 2026-10-22. Results are informational and not financial or engineering advice — see our disclaimer.