Structural Steel Design / Column bases and anchorages

Tension, shear, and how fixed a base really is

The tension side is a T-stub. Shear may be transferred by verified friction, a shear key or a designed anchor arrangement. And the base is neither pinned nor fixed, which matters more than either.

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

Combined uplift plus moment retains contact compression where equilibrium permits it; unsupported contact pattern withheld. Friction and shear key are alternative paths unless combined explicitly. Base spring diagram uses delta/delta_fixed=1+3/r; pinned limit is a mechanism.

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.
EN1992-4 steel/concrete anchor failure modes, embedment, edge/group effects, tension-shear interaction, deformation-compatible shear sharing, contact geometry, real moment-rotation curve and construction stability.

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

Moment can produce anchor tension and a compression contact zone. Shear may pass through friction, a key or a designed anchor arrangement, depending on the detail and load combination. Rotational stiffness affects the frame response and needs a stable model. These mechanisms must be checked together.

By the end of this lesson you should be able to

Treat the tension side of a base as a T-stub
Say why the governing combination has the least axial load
Explain how shear leaves a base, and why not through the bolts
Say which fixity assumption is safe and which is not

What you should already know

The T-stub and its three modes (Module 13)
The effective bearing area (previous lesson)
Favourable permanent actions and the least-load combination (Module 2)
Frame stability and α_cr (Module 9)

Plate bending and anchor tension

An equivalent T-stub can help model base-plate bending, but a concrete-supported base is not identical to an end plate bolted to a rigid steel support. Anchor elongation, prying conditions, effective lengths and concrete failure modes need separate treatment.

The supplied plate/bolt mechanism values illustrate how thickness can change the governing component. They do not establish rotation capacity or anchor resistance in concrete. Check anchor steel, embedment, concrete breakout, pull-out, splitting, edge distances, group effects and combined tension/shear under the applicable anchorage method. Where rotation is required, verify that capacity explicitly.

Take care

The governing combination has the LEAST axial load

Axial compression holds a base down. The more of it there is, the less the anchor bolts have to do — and below a certain eccentricity they do nothing at all.

For this base under 150 kNm of moment:

N_Ed	e = M/N	Tension side?	Bolt tension
1500 kN (with 100 kNm)	67 mm	no	0
800 kN	188 mm	yes	148 kN
400 kN	375 mm	yes	316 kN
150 kN	1000 mm	yes	421 kN
−100 kN (uplift)	—	yes	526 kN

The150 kNm cases show increasing bolt tension as axial compression falls. The first row uses100 kNm and is a separate comparison.

A designer who checks holding-down bolts against the maximum-load combination will find them comfortable and will have checked the wrong thing.

This is exactly the favourable-permanent-action case Module 2 introduced: the combination that governs is minimum permanent action with wind, and it is the one that is easiest to leave out of a schedule of load cases.

Try it

Axial load, moment, and where the bolts start working

The base from the worked example. The red ticks are the resistances. Hold the moment steady and take the axial load away — the bolt tension is what to watch.

Axial load800 kN (negative is uplift)

Moment150 kNm

Shear60 kN

Shear key resistance0 kN (0 = none)

Eccentricity e: 188 mm
To the compression centre: 130 mm
Tension side?: YES
Bolt tension F_t: 148 kN
Bolt resistance F_t,Rd: 407 kN
Compression F_c: 948 kN
Moment resistance M_j,Rd: 230 kNm
Base adequate in bending?: yes
Friction available: 160 kN
Shear resistance V_Rd: 160 kN
Shear adequate?: yes
Axial needed for friction alone: 300 kN

e = 188 mm exceeds the 130 mm to the compression centre, so the base HAS a tension side: the anchor bolts carry 148 kN. Note that reducing the axial load makes this worse, not better — the governing case for holding-down bolts is usually the combination with the least permanent action. Credited friction = 160.0 kN; supplied key resistance = 0.0 kN. Only the stronger individual path is credited; the two resistances are not automatically added. This comparison does not verify the key, anchors, grout or concrete.

Things worth trying

At 800 kN compression and 150 kNm moment, the assumed contact model gives approximately148 kN anchor tension.
Reducing compression to150 kN raises tension to421 kN; compare every relevant load combination.
The key and friction capacities are not added unless their compatibility is established.
Net uplift with moment may retain compression contact. Where the assumed contact force becomes negative, this one-contact-zone model is unsupported and its result is withheld.
Anchors can transfer shear when designed for it. Check hole clearance, slip, grout, concrete and combined actions.

Shear transfer and contact

Shear can be transferred by a verified friction interface, a shear key or a designed anchor arrangement. Oversized holes can require slip before anchors engage, so establish both strength and serviceability compatibility.

Using an explicitly assumed credited compression of150kN and friction coefficient0.2 gives30kN. This is insufficient for60kN shear. One coherent alternative is to design a key for the full60kN without adding friction credit; an independently verified150kN key resistance then exceeds that demand. Its resistance is not the fraction of load it necessarily attracts.

Net uplift does not prove zero contact when moment acts. In the example’s fixed-resultant model, N=−100kN and M=150kNm give F_t=525.8kN and F_c=425.8kN. Whether friction may be credited requires the actual contact, interface and load-combination assessment.

Rotational stiffness and a stable model

A cantilever with a base rotational spring S_j under lateral tip force H has

δ=HL³/(3EI)+HL²/S_j.

Relative to a fixed-base cantilever, δ/δ_fixed=1+3/(S_j/(EI/L)). For EI/L=4935kNm/rad, S_j=15 000 gives1.987 times the fixed-base sway; S_j=200 000 gives1.074 times.

A truly pinned base with no other lateral support is a mechanism, so it has no finite pinned-cantilever sway reference. A complete frame may be stable with nominally pinned bases through its other members and bracing. Establish that load path and required base rotation capacity throughout construction. Neither bolt count nor grout cracking demonstrates a safe pin or a fixed support.

Try it

Cantilever sway and base rotation

Tip-loaded elastic cantilever: column bending and base rotation add. A pinned base is a mechanism without another lateral support.

Base rotational stiffness15,000 kNm/rad

Column height6 m

Sj / (EI/L): 3.040
Total / fixed-base sway: 1.987

For a tip-loaded elastic cantilever, δ = HL³/(3EI) + HL²/Sj. This base gives δ/δfixed = 1.987. The fixed-base limit is 1; the pinned limit is unbounded. The stiffness labels are illustrative, not code classifications.

Worked example

The same base under moment, shear and uplift

Given

The 500 × 500 × 30 mm base plate from the last lesson, N_Rd = 4022 kN in compression
Two M24 grade 8.8 holding-down bolts each side, F_t,Rd = 203 kN each
Tension-side T-stub: l_eff = 250 mm, m = 60 mm, n = 60 mm — resistance 407 kN
Lever arms: 130 mm to the compression centre, 180 mm to the bolt line
Two combinations: maximum load, N = 800 kN with M = 150 kNm and V = 60 kN; and minimum load with wind, N = 150 kN with the same M and V

Find

Whether the base works in both, and which combination governs each check.

Assumptions

The T-stub effective length is a calibrated yield-line result and is given
The friction coefficient is calibrated and unverified here
Illustrative component limits only. Anchor design must also check concrete cone, pull-out, splitting, edge/group effects, anchorage and tension–shear interaction; no complete base adequacy verdict is made.

Practice

A base carries N = 800 kN and M = 150 kNm. What is the eccentricity e, in mm?

Practice

With that N and M, a compression centre 130 mm from the axis and a bolt line 180 mm the other side, what is the bolt tension in kN?

Practice

The same base and moment, but N falls to 150 kN. What is the bolt tension now, in kN?

Practice

A base carries 150 kN of axial compression with a friction coefficient of 0.2. What shear can friction alone transfer, in kN?

Check yourself

Why should shear at a column base not be assumed to pass through the anchor bolts?

Summary

The tension side of a base is a T-stub — Module 13 applies unchanged
Bearing wanted 30 mm and T-stub ductility wanted 20 — a real conflict
Less axial load means MORE bolt tension: 148 kN at 800 kN, 421 kN at 150 kN
So the governing combination is minimum permanent action with wind
Shear leaves through FRICTION, then a shear key — not the bolts
Friction is proportional to axial load, so the light case governs there too
At 150 kN axial the friction was 30 kN against 60 kN of shear
A pinned-base cantilever is a mechanism; assess the actual frame’s load path, stiffness and rotation capacity