Module 12 · Lesson 12.2

Preloaded bolts, welds, and what detailing protects

A preloaded bolt does not carry shear at all. A fillet weld is smaller than it looks. And every spacing rule is protecting a number in a calculation you have already done.

Why this matters

Three subjects that look like separate topics and are really one: each is a case where the obvious reading of a detail is wrong. A preloaded bolt is usually weaker than the same bolt in bearing, and is specified anyway. A 6 mm fillet weld is a 4.2 mm weld. And the spacing rules that look like drawing-office convention are inputs to the bearing resistance calculated in the last lesson.

By the end of this lesson you should be able to

Explain what a preloaded bolt does and why it is specified
Say why applied tension reduces slip resistance
Size a fillet weld on its throat and know when transverse loading helps
Say what each detailing rule protects

What you should already know

Bolt shear and bearing (previous lesson)
The von Mises criterion (Module 1)
Ultimate strength and why connections work to f_u (Module 1)

A preloaded bolt does not carry the shear

Tighten a bolt hard enough and it clamps the plies together. The load then passes across the interface by friction, not by the bolt bearing on the side of its hole. The bolt's own shear strength does not enter the calculation at all.

That leads somewhere counter-intuitive. For an M20 grade 10.9:


Preload F_p,C = 0.7 f_ub A_s	172 kN
Slip resistance, one surface	68.6 kN
The same bolt in bearing-type shear	117.6 kN

The preloaded connection is weaker. It is not specified for strength.

It is specified because it does not slip. A bearing-type connection has clearance holes, so it moves a millimetre or two before the bolts bear. That movement is harmless in most structures and unacceptable in some: where fatigue matters, where a joint must not lose its alignment, where a connection is part of a stiffness-critical bracing system, or where reversal would work a bolt back and forth in its hole.

And one interaction catches people out. Applied tension relieves the clamping force, so it reduces the slip resistance:

Applied tension	Slip resistance	Lost
0	68.6 kN	—
40 kN	55.8 kN	19%
80 kN	43.0 kN	37%

A bolt group carrying both shear and tension therefore loses slip capacity exactly when it is busiest.

Try it

What preloading actually buys

A preloaded bolt clamps the plies and friction carries the load. Compare that with the same bolt in a bearing-type connection, and watch what applied tension does to it.

Bolt diameter20 mm

Bolt grade

Friction surfaces1

Slip factor0.50

Applied tension per bolt0 kN

Preload F_p,C: 172 kN
Slip factor μ: 0.50
Friction surfaces: 1
Slip resistance F_s,Rd: 68.6 kN
Same bolt in bearing: 117.6 kN
Slip over bearing: 0.58 ×
Applied tension: 0 kN
Lost to that tension: 0 %

Preload 172 kN, slip factor 0.5, so the slip resistance is 69 kN across 1 surface. The same bolt in bearing-type shear would give 118 kN — MORE. A preloaded connection is not chosen for strength; it is chosen because it does not slip.

Things worth trying

Start at M20 grade 10.9, one surface, μ = 0.5, no tension. The slip resistance is 68.6 kN against 117.6 kN for the same bolt in bearing — a preloaded connection is WEAKER, and it is specified anyway.
That is the point of it. Bearing-type connections have clearance holes and move a millimetre or two before the bolts bear. Preloading buys the absence of that movement, not strength.
Add a second friction surface. The slip resistance doubles, because friction acts on each interface independently.
Take the slip factor from 0.5 down to 0.2 — a painted rather than a blasted surface. The resistance falls in proportion. Surface preparation is a structural specification here, not a finish.
Now add applied tension. At 80 kN the slip resistance has fallen 37%, because the tension is relieving the clamping force the friction depends on.
Keep adding tension until the resistance reaches zero. A bolt group carrying both shear and tension loses slip capacity exactly when it is busiest.
Switch to grade 8.8 and watch both bars fall together — the preload and the shear resistance both go with f_ub, so the comparison between them barely moves.

A fillet weld is smaller than it looks

A fillet weld's strength is taken on its throat — the shortest distance through the weld — not on its leg. For an equal-leg fillet the throat is s/√2:

Leg	Throat
4 mm	2.83 mm
6 mm	4.24 mm
8 mm	5.66 mm
10 mm	7.07 mm

So a 6 mm fillet is structurally a 4.2 mm weld. That is not a deduction or an allowance — it is where the failure plane is.

Two methods, and the difference is real.

The simplified method takes the resultant force on the throat, whatever direction it acts in, against a single design shear strength. Easy, and always safe.

The directional method resolves the force into components on the throat and applies the von Mises criterion. It gives more, and how much more depends on the angle:

Angle to the weld axis	Gain
0° (along the weld)	1.000
30°	1.044
45°	1.095
60°	1.155
90° (across the weld)	1.225

A transversely loaded fillet really is about 22.5% stronger than one loaded along its length, and the simplified method discards that. The factor is derivable, not fitted: resolving a force at θ onto the 45° throat and applying von Mises gives a resistance proportional to 1/√(3 − sin²θ), which is exactly √3 : √2 between the two extremes.

Note

The full-strength fillet, and a rule worth carrying

A weld that develops the full strength of the thinner plate it joins cannot be the weak link, which removes a whole class of question from a connection.

For S355 plate with matching electrodes, working to the longitudinal (worst) case, that happens at a double-sided fillet whose leg is about equal to the plate thickness:

Plate	Full-strength leg, double-sided
8 mm	8.0 mm
12 mm	12.0 mm
20 mm	20.0 mm

Beyond that, extra weld metal buys nothing — the plate is the weak link and always was going to be.

The neatness of that result is a coincidence of the S355 numbers, not a law. It depends on the plate grade, on the correlation factor β_w for the parent metal, and on the partial factors — all of which are calibrated and unverified in this course. Treat it as a sense-check, not as a substitute for the calculation.

Try it

Throat, direction, and the full-strength leg

A double-sided fillet joining a plate to a support. Compare the two design methods, then find the leg at which the weld stops being the weak link.

Weld leg6 mm

Angle to the weld axis90 degrees

Run length each side250 mm

Plate thickness12 mm

Weld leg s: 6 mm
Throat a: 4.24 mm
Design shear strength f_vw,d: 251 N/mm²
Simplified method: 1067 N/mm
Directional method: 1307 N/mm
Directional gain: 22.5 %
Capacity, simplified: 533 kN
Capacity, directional: 653 kN
The plate's own capacity: 1065 kN
Full-strength leg needed: 12.0 mm
As a fraction of the plate: 1.00 × t
Weld metal, relative to 6 mm: 1.00 ×

The weld is the weak link: 653 kN against the plate's 1065 kN. A 12.0 mm leg would make it full strength, at 4.0 times the weld metal.

Things worth trying

Start at a 6 mm leg. The throat is 4.24 mm — a fillet weld is always smaller structurally than it looks, because strength is taken on the shortest path through it.
Take the angle from 90° down to 0°. The directional method's advantage disappears entirely: 22.5% across the weld, nothing along it.
At 90° with a 250 mm run each side, the two methods give 654 and 534 kN. If the applied load were 600 kN they would disagree about whether it passes — that is not a rounding difference.
Set the plate to 12 mm and raise the leg until the weld bar passes the plate bar. It happens at about 12 mm — for S355, a double fillet with leg equal to the plate thickness is roughly full strength.
Change the plate to 20 mm and repeat. The full-strength leg moves with it, still at about 1.0 × t. The neatness is a coincidence of the S355 numbers, not a law.
Watch the weld-metal figure as you raise the leg. Volume goes with the square, so a 12 mm fillet costs four times a 6 mm one — strength doubles, cost quadruples.
Try a very short run with a large leg against a long run with a small one, at the same weld metal. Length is usually the cheaper way to buy capacity.

Worked example

A bracket welded to a column flange

Given

A 12 mm S355 plate welded to a column flange with a double-sided fillet
Parent metal f_u = 490 N/mm², β_w = 0.9, γ_M2 = 1.25
The connection carries 600 kN acting across the weld run
Weld run length 250 mm each side

Find

The leg size needed, and whether a full-strength weld would be sensible instead.

Assumptions

β_w and γ_M2 are calibrated and unverified here
The weld is treated as loaded uniformly along its length; a real bracket also has a moment, which Module 13 covers

Detailing rules protect calculations you have already done

The minimum spacings are not drawing-office convention. Each one is protecting a number.

Rule	Typical minimum	What it protects
End distance e₁	1.2 d₀	Below this the bolt tears out through the end of the plate, and the bearing calculation does not apply at all
Edge distance e₂	1.2 d₀	The k₁ coefficient in the bearing resistance, and splitting of the plate sideways
Pitch p₁	2.2 d₀	Bolts too close share one failure surface and the group tears out as a block
Gauge p₂	2.4 d₀	The same across the load — plus it has to be physically possible to get a spanner in
Maximum pitch	the lesser of 14t and 200 mm	The plies separating between fasteners: water gets in, and the plate can buckle locally

Notice that the first four are all measured against the hole, not the bolt. That is why going up a bolt size without opening the pattern out tightens every one of them at once — the effect the predict block in the last lesson turns on.

The minima are what make the bearing calculation valid. They are an input to a resistance, not a separate tidiness requirement.

Practice

What is the throat thickness of an 8 mm equal-leg fillet weld, in mm?

Practice

A weld metal has f_u = 490 N/mm², β_w = 0.9 and γ_M2 = 1.25. What is f_vw,d in N/mm²?

Practice

By what factor is a transversely loaded fillet weld stronger than a longitudinally loaded one, by the directional method?

Practice

A preloaded M20 grade 10.9 bolt has A_s = 245 mm². What is its preload F_p,C, in kN? Take F_p,C = 0.7 f_ub A_s.

Check yourself

Why is a preloaded bolted connection specified, given that its slip resistance is usually LOWER than the same bolts in bearing?

Summary

A preloaded bolt clamps; friction carries the load and the bolt's shear strength is irrelevant
M20 10.9: 172 kN preload, 68.6 kN slip resistance — against 118 kN in bearing
It is specified for the absence of movement, not for strength
Applied tension relieves the clamp: 80 kN of tension cost 37% of the slip resistance
A fillet weld's strength is on its THROAT — a 6 mm leg is a 4.24 mm weld
The directional method beats the simplified one by up to 22.5%, and it is derivable
For S355 a double fillet with leg ≈ plate thickness is full strength — a coincidence, not a law
Every minimum spacing is measured against the HOLE, and protects a resistance you have calculated

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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