Gross Section versus Net Section: Which One Governs a Bolted Tension Member

Take two tension members cut from the same plate, the same steel, carrying the same pull. Bolt the first through a clean, full cross-section and it will stretch, yield and finally tear only after the whole bar has gone plastic. Drill a line of bolt holes through the second and it can snap at a single cross-section, at a load the first one shrugs off, with almost no warning. Same steel, same force, two completely different failure stories.

That is the entire subject of this article. A bolted tension member is checked against two limit states at once: yielding on the gross section, spread over the full length, and rupture on the net section, concentrated where the bolt holes remove material. The design strength is the smaller of the two, and which one wins is not obvious. It depends on the steel grade, on how many holes you punch, on whether they are staggered, and on how the member is attached. Get the wrong one and you either waste steel or, far worse, miss the mode that actually breaks.

Below we define both limit states, compute them by hand for a plate and an angle to the third decimal, derive the exact crossover rule that tells you which governs before you even open a calculator, and compare how AISC 360, NBR 8800 and Eurocode 3 draw the same line with different safety factors.

AISC 360 Chapter D states the two checks explicitly. For a member in axial tension the design tensile strength is the lesser of:

• Yielding of the gross section: φ_tP_n = 0.90 · F_y · A_g
• Rupture of the net section: φ_tP_n = 0.75 · F_u · A_e

Three areas appear, and keeping them straight is half the battle:

• A_g, gross area: the full cross-section, holes ignored. This is the area you read off a profile table or the CalcSteel section engine.
• A_n, net area: the gross area minus the material removed by the bolt holes on the critical failure path, A_n = A_g − ∑ d_h · t.
• A_e, effective net area: the net area reduced further by the shear-lag factor U, A_e = U · A_n, to account for the fact that not every part of the section is fully engaged at the connection.

Yielding uses F_y (the yield stress) and the resistance factor 0.90; rupture uses F_u (the ultimate stress) and the lower factor 0.75. Both the stresses and the factors differ, which is exactly why the winner is never a foregone conclusion. The phrase gross section versus net section is shorthand for this competition: F_yA_g against F_uA_e, weighted by 0.90 against 0.75.

The gap between 0.90 and 0.75 is not arbitrary bookkeeping, it encodes how the two failures behave. Gross-section yielding is ductile. When the full length of the bar reaches F_y, it stretches visibly, sheds load into neighbouring members, and gives ample warning long before it tears. A member that yields is not a member that has collapsed, so the code treats yielding as a serviceability-flavoured limit with a generous factor of 0.90.

Net-section rupture is brittle. Fracture initiates at the edge of a bolt hole, where stress concentrates, and races across the reduced section with little plastic warning. There is no reserve past it: the member is in two pieces. Because the consequence is sudden and total, and because F_u itself is less certain than F_y, the code caps rupture with the lower factor of 0.75.

So the design is deliberately biased to force the ductile mode to govern where it can. When rupture wins anyway, the code is telling you the holes have taken away too much section, and the fix is in the connection geometry, not the steel grade.

The net area is the gross area minus the material the holes remove along the critical straight path across the member:

A_n = A_g − n · d_h · t

where n is the number of holes crossed, t is the plate thickness, and d_h is the hole width used for net-area calculations. The single most common error here is using the bolt diameter for d_h. The hole is always larger than the bolt, and the code adds even more on top for punching damage:

• AISC 360: the standard hole is the bolt diameter plus about 2 mm (1/16 in), and for net-area calculations you add a further 2 mm (1/16 in) for damage around the hole. An M20 bolt in a 22 mm standard hole is therefore computed with d_h = 24 mm.
• NBR 8800: the same idea, standard hole equal to bolt plus 1.5 mm, plus 2 mm added for the net section, giving effectively bolt plus 3.5 mm.

The concept goes back a century. Once engineers moved from solid eyebars to riveted and bolted plates, they realised the reduced section at the fastener line, not the full bar, sets the fracture load. Salmon and Johnson trace the modern net-area and shear-lag treatment straight into today's AISC provisions. Throughout this article we use d_h = 24 mm for M20 bolts, the AISC convention.

When bolts are staggered, the failure path can zig-zag from one hole to the next diagonally. A diagonal leg is longer than a straight cut, so it removes proportionally less area, and the code credits that back with the classic s²/4g term, one addition for every diagonal segment in the path:

A_n = A_g − ∑ d_h · t + ∑ (s² / 4g) · t

Here s is the longitudinal spacing (pitch) between the two holes measured along the axis, and g is the transverse gage between the two lines. You evaluate every plausible path and the one giving the smallest net area governs. A zig-zag path crosses more holes than a straight one, but the s²/4g credits may or may not compensate.

Take the 200 × 12 mm plate from later in this article, with a gage g = 65 mm and a stagger pitch s = 50 mm. The straight path across two holes gives A_n = 2400 − 2 · 24 · 12 = 1824 mm². The zig-zag path crosses three holes but picks up two diagonal credits of s²/4g = 50²/(4 · 65) = 9.615 mm each: A_n = 2400 − 3 · 24 · 12 + 2 · 9.615 · 12 = 1766.8 mm². The zig-zag path is smaller, so it governs. Staggering the bolts made the critical section worse here, not better, which is precisely why you must check every path.

Net area assumes the whole reduced section resists uniformly. It does not, unless every element of the cross-section is connected. Bolt a single angle through one leg only and the force has to funnel from the outstanding, unconnected leg into the connected one through shear. Near the first bolts, the outstanding leg lags behind and carries less than its share. This is shear lag, and the code captures it with the reduction factor U:

A_e = U · A_n, with U = 1 − x̄/L

x̄ is the distance from the connected face to the centroid of the whole section, and L is the length of the connection (first bolt to last bolt). The further the centroid sits from the connection plane, and the shorter the connection, the more the section lags and the smaller U becomes. AISC Table D3.1 also lets you use simple lower-bound values, for example U = 0.80 for angles with four or more bolts per line, but the computed 1 − x̄/L is usually more generous and worth the arithmetic.

Two design levers fall straight out of this equation. Lengthening the connection (more bolts in the line, larger L) raises U toward 1.0. Connecting more of the section (both legs, or a gusset each side) raises U as well. A member with a generous gross area can still be governed by rupture purely because a short, one-sided connection throttled its effective area, as the worked angle below shows.

A flat plate 200 × 12 mm in ASTM A36 steel (F_y = 250 MPa, F_u = 400 MPa) is spliced with two M20 bolts across its width in a single transverse line. The plate is connected across its full width, so there is no shear lag: U = 1.0. Work the two limit states.

Areas. A_g = 200 · 12 = 2400 mm². Two holes at d_h = 24 mm: A_n = 2400 − 2 · 24 · 12 = 1824 mm². With U = 1.0, A_e = 1824 mm². The holes have removed 24 percent of the area, so A_e/A_g = 0.760.

Yielding of the gross section: φ_tP_n = 0.90 · 250 · 2400 = 540 000 N = 540.0 kN.

Rupture of the net section: φ_tP_n = 0.75 · 400 · 1824 = 547 200 N = 547.2 kN.

Yielding governs, at 540.0 kN, but only by 1.3 percent. Losing a quarter of the area to holes was almost, but not quite, enough to hand the member to rupture. This razor-thin margin is the whole lesson: you cannot eyeball which limit state wins, you have to compute both. In the next section, pull the gross area of your own profile and repeat the two lines.

The gross area A_g is the starting point for both checks, and for a real rolled or cold-formed profile it is not a number you want to derive by hand. The calculator below is the live CalcSteel section engine, the same one behind the 3D editor. Pick a section, read its cross-sectional area A directly, and that is your A_g. Subtract n · d_h · t for your bolt line to get A_n, apply the shear-lag U, and you are one multiplication away from both limit states.

For the flat-plate example above the area is trivial (200 × 12 = 2400 mm²), but the moment you reach for an angle, a channel or a built-up section, the gross area is worth reading straight off the engine rather than risking an arithmetic slip in the very first term.

Now a single angle L102 × 102 × 9.5 (A_g = 1845 mm², centroid x̄ = 28.7 mm) in ASTM A36, bolted through one leg with three M20 bolts at 75 mm pitch. Same steel as the plate, similar area, but a one-sided connection.

Shear lag. Connection length L = 2 · 75 = 150 mm. U = 1 − x̄/L = 1 − 28.7/150 = 0.809.

Areas. One hole through the connected leg: A_n = 1845 − 24 · 9.5 = 1617 mm². Effective net area A_e = 0.809 · 1617 = 1307.6 mm². Now A_e/A_g = 0.709, below the plate's 0.760.

Yielding: φ_tP_n = 0.90 · 250 · 1845 = 415.1 kN. Rupture: φ_tP_n = 0.75 · 400 · 1307.6 = 392.3 kN. This time rupture governs, at 392.3 kN.

The angle has almost the same gross area as many plates, yet shear lag pushed A_e low enough to flip the governing mode. And the fix is telling: add a fourth bolt, L grows to 225 mm, U rises to 0.872, rupture climbs to 423.2 kN, and yielding takes back control at 415.1 kN. One extra bolt, a different limit state, 6 percent more capacity, without changing the steel at all.

Every major code runs the same two-limit-state check, they only dress the safety factors differently. The gross-yielding term always uses the yield stress and the net-rupture term always uses the ultimate stress; what changes is where the factor sits and how big it is.

The three look different but land close together. AISC's 0.90 on yielding is nearly EN's γ_M0 = 1.00 (a factor of 1.0 on strength) and NBR's 1/1.10 = 0.909. On rupture, AISC's 0.75 sits between NBR's 1/1.35 = 0.741 and EN's combined 0.9/1.25 = 0.72. Eurocode folds an extra 0.9 onto the net term and treats shear lag through separate connection rules rather than a U factor inside A_e, but the philosophy is identical: a ductile check on the full section and a brittle check on the reduced one, with the reduced one penalised harder.

For the CalcSteel audience this matters because the same member can be verified against NBR 8800, AISC 360 or EN 1993 in the editor, and the governing limit state can shift slightly between them. The engineering is the same; only the partial factors move.

You can settle which limit state governs with one inequality, before any number-crunching. Rupture governs when 0.75 F_u A_e < 0.90 F_y A_g, which rearranges to a pure ratio of areas:

Rupture governs when A_e/A_g < 1.2 · F_y/F_u

The right-hand side is a property of the steel alone. Evaluate it once per grade:

• ASTM A36 (F_y/F_u = 0.625): rupture governs when A_e/A_g < 0.750. You can lose up to 25 percent of the section before rupture takes over.
• ASTM A572 Gr.50 (F_y/F_u = 0.767): rupture governs when A_e/A_g < 0.920. Now even an 8 percent loss tips it to rupture.

This exposes a genuinely counter-intuitive result: higher-strength steel makes net rupture govern more often, not less. As F_y climbs toward F_u, the ductile yielding limit rises faster than the brittle rupture limit, so the brittle mode catches up and controls at ever-smaller hole losses. A single bolt line already removes more than 8 percent of most sections, so on Grade 50 and higher, rupture at the connection is the mode to watch by default. Our A36 plate sat at A_e/A_g = 0.760, a hair above 0.750, which is exactly why yielding scraped the win by 1.3 percent.

• Using the bolt diameter as the hole width. The hole is the bolt plus clearance plus a damage allowance. An M20 bolt is computed with d_h = 24 mm, not 20. This alone can shift the net area by several percent.
• Forgetting shear lag. Using A_n in the rupture equation instead of A_e = U A_n overstates the capacity, badly for angles, tees and channels bolted through part of the section.
• Checking only the straight hole path. With staggered bolts a zig-zag path can govern even though it crosses more holes. Evaluate every path and take the smallest net area.
• Mixing the resistance factors. 0.90 belongs to yielding, 0.75 to rupture. Swapping them, or carrying an AISC factor into an NBR or EN calculation, quietly moves the governing limit state.
• Assuming the bigger gross area is safe. A generous A_g means nothing if a short, one-sided connection throttles A_e. The angle example had ample gross area and still failed by rupture.

A bolted tension member is only as strong as the smaller of two numbers: 0.90 F_y A_g for ductile yielding over its length, and 0.75 F_u A_e for brittle rupture at the holes. Compute the net area with the hole width, not the bolt; add the s²/4g credit on staggered paths; reduce to the effective net area with the shear-lag U. Then let the crossover rule, A_e/A_g against 1.2 F_y/F_u, tell you which mode to expect, and remember that higher-strength steel biases the answer toward rupture. Check both, every time, because as the plate showed, the winner can come down to a single percent.