Structural Steel Design / Connection and joint design

The component method: weakest, softest, and why they differ

Components in series. The weakest sets the resistance, every one sets the stiffness, and improving the wrong one is the most common wasted effort in joint design.

Lesson review: expert verification remains open

This lesson has received a full-context editorial review and a check of its technical explanations. Selected arithmetic has been checked separately. This is not independent engineering sign-off.

Corrections checked in this pass

764 kN unrestricted row sum and 272 kNm are not joint resistance. Demonstrated compatibility allocation respects 610/520 kN shared limits and stated row-group limits, yielding 196.07 kNm for the explicitly assumed distribution.

Checks needed for project use

A steel-design reviewer must check the first-generation calculation profiles, second-order analysis assumptions, connection rules and execution references against the applicable standards and amendments.
Actual component stiffness, equivalent lever arms, permissible force redistribution and group modes. The bounded proportional allocation is not a complete EN1993-1-8 solver.

These are educational models and exercises. Check their assumptions before using the results. The full technical content and every diagram have not yet been independently verified.

Concrete and steel calculations retain their stated first-generation Eurocode teaching basis. Use the standards and amendments specified for your project; do not mix generations without an appropriate design basis.

Read the lesson register and outstanding review notes

Why this matters

Module 3 showed that a joint's stiffness changes where the moment goes, and that a 'simple' connection carrying half its fixed-end moment is a real and common problem. This lesson is how that stiffness is actually obtained — and it turns out to rest on one piece of algebra already met in Module 10. Once a joint is treated as components in series, both of the questions it raises answer themselves, and they answer differently.

By the end of this lesson you should be able to

List the components of a bolted end-plate joint
Say why the weakest sets resistance and every component sets stiffness
Identify the softest component and say what improving it does
Calculate M_j,Rd and S_j,ini, and classify the joint
Say when the initial stiffness is the wrong number to have used

What you should already know

The T-stub and its modes (previous lesson)
Joint classification by stiffness (Module 3)
Engesser's series flexibility for built-up members (Module 10)

Series paths and shared components

For a single tension–compression couple, component deformations along a load path add:1/k_eq=Σ1/k_i. Resistance is limited by the weakest component on that path.

A multi-row end-plate joint is not one unbranched chain. Individual rows have local components, while row groups, the compression zone and column panel can impose shared limits. Satisfy those limits and equilibrium before summing row force times lever arm. The single-path stiffness example below is separate from the three-row force allocation.

Take care

The weakest and the softest are usually different components

For the joint in the worked example:

Component	Resistance	k_i	Share of flexibility
Column web panel in shear	520 kN	4.2	39%
Column web in compression	610 kN	6.8	24%
Column flange in bending	380 kN	9.1	18%
End plate in bending	450 kN	12.4	13%
Bolts in tension	564 kN	31.0	5%

The weakest is the column flange at 380 kN. The softest is the column web panel, supplying 39% of the flexibility while being 37% stronger than the component that governs.

So:

Changing only the flange resistance input in this teaching model raises the limiting resistance to 450 kN. A physical alteration can also change stiffness and must be checked.
Treating the web-panel flexibility as zero in this model raises k_eq by about 64%; any physical stiffener also needs resistance and detailing checks.

Resistance and stiffness are governed by different component combinations. Check both after any physical alteration; one change may affect several components.

Try it

Weakest, softest, and what each change buys

Five components in series. The bars are each component's share of the joint's total flexibility; the highlighted one is whichever sets the resistance. Watch the two move independently.

column web panel in shear — resistance520 kN

column web in compression — resistance610 kN

column flange in bending — resistance380 kN

end plate in bending — resistance450 kN

bolts in tension — resistance564 kN

column web panel in shear — stiffness4.2

column web in compression — stiffness6.8

column flange in bending — stiffness9.1

end plate in bending — stiffness12.4

bolts in tension — stiffness31.0

Lever arm z400 mm

Resistance F_Rd: 380 kN
Weakest component: column flange in bending
Softest component: column web panel in shear
Same component?: NO
Equivalent stiffness k_eq: 1.645 mm
Smallest individual k_i: 4.2 mm
Lever arm z: 400 mm
Initial stiffness S_j,ini: 55268 kNm/rad
As a multiple of EI/L: 6.3
Joint classification: SEMI-RIGID
If the softest were rigid: 90847 kNm/rad
If the joint were 20% deeper: 79586 kNm/rad

The WEAKEST component is column flange in bending at 380 kN, and the SOFTEST is column web panel in shear, contributing 39% of the joint's flexibility. They are different components, which is the usual case and the practical point: strengthening column flange in bending would not stiffen the joint, and stiffening column web panel in shear would not strengthen it.

Things worth trying

Start at the defaults. The weakest component is the column flange at 380 kN; the softest is the web panel, supplying 39% of the flexibility. They are different components, and the verdict says so.
Raise the column flange's RESISTANCE to 700 kN. The resistance rises to 450 kN and k_eq does not move at all — not by a rounding amount, but exactly not at all.
Now raise the web panel's STIFFNESS from 4.2 to 20. k_eq rises by about 45% and the resistance does not move. Two changes, two effects, no overlap.
Compare k_eq with the smallest individual k_i. It is always below it, because flexibilities add — a free sanity check on any component-method calculation.
Watch the classification as you stiffen the web panel. The joint crosses from semi-rigid to rigid on one component's k, which is a large modelling consequence for a single number.
Now leave every component alone and increase the lever arm instead. z goes from 400 to 480 — 20% — and the stiffness rises 44%, because z is the only term that is squared.
Make the bolts far stiffer, from 31 to 40. Almost nothing happens: they supply 5% of the flexibility, so they can contribute at most 5% of any improvement. Effort belongs where the flexibility is.
Try making every component equally strong. The 'same component?' readout can then turn to yes — and the verdict tells you that is the convenient case and the less common one.

Worked example

A bolted end-plate joint, resistance and stiffness

Given

Two separate illustrations: a single equivalent component chain, and a three-row force allocation with explicitly assumed limits
Use the preceding component table only for the single-chain resistance/stiffness calculation
Bolt row lever arms about the compression centre: 460, 370 and 190 mm
Three-row caps282,282,200kN; shared compression610kN; panel-equivalent tension520kN; row-group caps430kN for rows1+2 and290kN for rows2+3
Beam 457 × 191 × 74 UB spanning 8.0 m in a braced frame
For the separate single-chain stiffness illustration, assume an equivalent lever arm z=400mm; it is not derived from the three-row geometry

Find

The joint's moment resistance and stiffness, and how to classify it.

Assumptions

Component stiffness coefficients k_i are calibrated and are given, not derived
The rigid and pinned boundaries are calibrated and unverified here
Three-row forces are proportional to lever arm as an illustrative compatible allocation. Given row/group/shared limits are assumed to include their local checks; they are not derived EC3 resistances.

A separate moment–rotation illustration

To examine the stiffness model alone, take an independent illustrative joint with assumed M_Rd=272kNm and S_ini=55 268kNm/rad. These are inputs;272kNm is not the resistance established by the corrected three-row allocation.

Using the stated teaching softening law, stiffness at M/M_Rd=0.5 is55 268kNm/rad and rotation2.46mrad. At0.9, stiffness is about24 579kNm/rad and rotation9.96mrad. The initial stiffness cannot represent every load level. Actual component stiffness and applicable classification rules must be checked for the real joint.

Try it

When the initial stiffness is the wrong number

The joint from the worked example: M_j,Rd = 272 kNm, S_j,ini = 55 268 kNm/rad. The straight line is what an analysis using the initial stiffness assumes; the curve is what the joint does.

Applied moment160 kNm

Applied moment M_Ed: 160 kNm
Utilisation: 59 %
Region: elastic
Initial stiffness S_j,ini: 55268 kNm/rad
Stiffness at this moment: 55268 kNm/rad
Softening factor μ: 1.00
Rotation: 2.89 mrad
Rotation if it stayed elastic: 2.89 mrad

At 59% of its resistance the joint is still on its initial stiffness. S_j,ini is the right number to have used in the analysis.

Things worth trying

Below about 180 kNm the joint sits on the straight line: S_j,ini is the right number and the two curves are the same.
Cross the dashed two-thirds line and watch them separate. Nothing physical happens at that point — it is where a calibrated softening expression takes over — but the divergence beyond it is real.
At 245 kNm, 90% of the resistance, the joint is 2.25 times less stiff than an analysis using S_j,ini assumed.
Compare the rotation with the 'if it stayed elastic' figure as you go up. The moment roughly doubles between 50% and 100% of the resistance; the rotation rises about six-fold.
This matters wherever the answer depends on how load is shared — a semi-continuous frame where the joint decides the split between span and support moments, or an unbraced frame whose α_cr depends on joint stiffness.
It matters much less in a braced frame with nominally pinned joints, where nothing is being asked of the joint's stiffness in the first place. Knowing which situation you are in is the point.

Practice

Five components have k values of 4.2, 6.8, 9.1, 12.4 and 31.0 mm. What is k_eq?

Practice

With k_eq = 1.645 mm and a lever arm z = 400 mm, what is S_j,ini in kNm/rad? Take E = 210 000 N/mm².

Practice

Three rows have an assumed admissible force allocation of234.510,188.627 and96.863kN at460,370 and190mm. What moment does this allocation transfer, in kNm?

Practice

A joint has S_j,ini = 55 268 kNm/rad and the beam it connects has EI/L = 8796 kNm/rad. What is S_j as a multiple of EI/L?

Check yourself

A joint is 20% too flexible. Its weakest component is the column flange in bending; its softest is the column web panel in shear. What should be changed?

Summary

A joint is components in series, so the force passes through every one
The WEAKEST sets the resistance — 380 kN here, the column flange in bending
1/k_eq = Σ1/k_i, so the SOFTEST sets the stiffness — the web panel, at 39%
They are different components, so a weak AND flexible joint needs two changes
The model varies resistance and stiffness inputs separately; real detail changes require both to be recalculated
Apply row, group and shared limits before summing forces and levers; the stated compatible allocation gives196.1kNm
S_j,ini goes with z², so 20% more depth was worth 44% more stiffness
Above two thirds of M_j,Rd a joint softens: at 90% it was 2.25 times less stiff