Module 6 · Lesson 6.1
Tension, compression, and the holes you drilled
A tension member is the simplest thing in structural steel, right up to the point where you connect it.
Why this matters
A tension member has no stability problem at all — you cannot buckle something you are pulling. It should be the easiest member in the frame, and it is, until you have to attach it. Every bolt hole removes material exactly where the force is highest, and it changes the failure from a ductile stretch to a sudden fracture. The margin between those two outcomes is far narrower than most people expect.
By the end of this lesson you should be able to
- Calculate gross yielding and net fracture, and say which governs
- Explain why they carry different partial factors
- Compute how much hole loss a grade will tolerate
- Calculate block tearing and read what its two terms mean
What you should already know
- Net area, including staggered holes (Module 1 library; Structural Analysis Fundamentals, sections)
- Cross-section classification (Module 5)
- The von Mises criterion, at least as τy = fy/√3 (Module 1)
Two limit states with different characters
A tension member can fail in two ways, and they are not variants of each other.
Gross yielding. The whole member reaches fy and stretches.
Npl,Rd = A fy / γM0
This is ductile. The member elongates visibly — remember from Module 1 that steel goes to about 15% strain — so it sags, the cladding cracks, and somebody notices. It also redistributes: a yielding member sheds load to stiffer paths.
Net-section fracture. The reduced section at the holes reaches fu and tears.
Nu,Rd = 0.9 Anet fu / γM2
This is brittle. It happens at one cross-section, at a strain the eye cannot see, and there is no warning and no redistribution.
The 0.9 is calibrated — it allows for the stress concentration around the holes and the eccentricity a bolt group introduces. It is not derivable.
Why the partial factors differ
γM0 = 1.0 for yielding and γM2 = 1.25 for fracture, and that gap is deliberate. A limit state you can see coming needs less margin than one you cannot. The code is arranged so that yielding governs wherever possible, and the extra 25% on the fracture side is part of how it arranges it.
Worked example
A bolted tie, and what the holes cost
Given
- Flat tie, 200 × 25 mm, gross area 5000 mm²
- S355: fy = 355 N/mm², fu = 510 N/mm²
- Two 22 mm holes in one line across the member
- γM0 = 1.0, γM2 = 1.25
Find
Which limit state governs, and by how much.
Assumptions
- Holes in a single line, so no stagger credit
- Partial factors in general UK use, not verified against the National Annex here
Try it
Yield or fracture?
Two resistances, and the smaller one decides both the capacity and the character of the failure. Add a single hole and watch how quickly the brittle case takes over.
Steel grade
- Gross area A
- 5000 mm²
- Net area Anet
- 3900 mm²
- Area lost to holes
- 22.0 %
- Crossover: max loss allowed
- 3.3 %
- Yield ratio fu/fy
- 1.44
- Gross yielding Npl,Rd
- 1775 kN
- Net fracture Nu,Rd
- 1627 kN
- Design resistance Nt,Rd
- 1627 kN
- Governed by
- net fracture
Net fracture governs at 1627 kN, below the 1775 kN gross yielding load. The member will fail at the holes, suddenly and without warning. Consider adding material at the connection so that yielding governs instead.
Things worth trying
- Set the holes to zero. Gross yielding governs, as it must — there is no net section to fracture. Now add ONE hole and watch how far the fracture bar drops.
- Read the two percentages under the chart. The crossover for S355 is at 96.7% of the gross area, so you may remove only 3.3% before the brittle case takes over. A single line of two holes usually removes far more.
- Change the grade and watch the crossover move. It is governed by the yield ratio fu/fy, not by the grade number — S275 at 1.56 tolerates the most, S355 at 1.44 the least. Specifying a stronger steel makes this WORSE.
- Widen the plate while keeping the holes the same. The proportion lost falls, and eventually yielding governs again. This is exactly why tension members are widened locally at their connections.
- Increase the hole diameter from 22 to 26 mm. A small change in the bolt size moves a substantial amount of area, because the loss scales with the diameter directly.
- Note that the DESIGN RESISTANCE is always the lower bar — but the character of the failure changes with which bar that is. Two members of the same capacity can behave completely differently.
Block tearing: a lump comes out
A third failure is possible at a bolted connection, and it is neither of the above. Instead of the member failing across its width, a block of plate containing the bolts tears out of it.
The block is bounded by two kinds of face, and they fail differently:
- the tension face, across the end, which fractures at fu;
- the shear faces, running along the bolt lines, which yield at fy/√3.
Veff,Rd = fu Ant/γM2 + (fy/√3) Anv/γM0
That expression is worth pausing on because it is unusual: it adds an ultimate term to a yield term, with different partial factors. Most interaction expressions do nothing of the sort. It is not a fudge — it is a statement that the two faces really do reach different limit states at the same instant, and the √3 comes straight from von Mises.
Block tearing tends to govern when the bolts are close to the end of the member, because the shear faces are then short. It is also easy to miss, because neither the member check nor the bolt check will find it — it is a failure of the plate around the bolts, and it needs its own calculation.
Practice
A 5000 mm² tie in S355 has 3900 mm² net area. What is the net-fracture resistance, in kN? Take fu = 510 N/mm², γM2 = 1.25.
Practice
For S275 (fy = 275, fu = 430), what fraction of the gross area may be lost before net fracture governs? Take γM0 = 1.0 and γM2,net = 1.10 — EN 1993-1-1 §6.1 as set by the UK National Annex, not the 1.25 that applies to bolts.
Practice
A block of plate has 900 mm² net tension area and 2400 mm² net shear area, in S355. What is the block tearing resistance, in kN? Take fu = 510, fy = 355, γM0 = 1.0, γM2 = 1.25.
Check yourself
Why does net-section fracture carry a larger partial factor than gross yielding?
Summary
- Gross yielding: A fy/γM0 — ductile, visible, redistributes
- Net fracture: 0.9 Anet fu/γM2 — brittle, sudden, local
- The 0.9 is calibrated; the different partial factors reflect the consequence
- In S355 about 14.9% of the area may be lost before fracture governs
- It is the yield ratio fu/fy that decides, so a higher grade tolerates LESS
- Almost any bolted tie is governed by net fracture unless material is added
- Block tearing adds a fracturing tension face to yielding shear faces
- Block tearing is found by neither the member check nor the bolt check
This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint