Module 14 · Lesson 14.2

Tension, shear, and how fixed a base really is

The tension side is a T-stub. The shear goes out through friction, not the bolts. And the base is neither pinned nor fixed, which matters more than either.

Why this matters

Once a base carries moment, three things happen that the bearing calculation says nothing about. One side of the plate lifts, and the holding-down bolts have to hold it — which is a T-stub, so Module 13's work carries straight over. The shear has to get out somehow, and it is not the bolts that carry it. And the base has a rotational stiffness that is neither zero nor infinite, which is the assumption a whole frame analysis has rested on.

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)

The tension side is a T-stub you have already met

A base plate under moment lifts on one side. What holds it down is a length of plate bending about the column flange, with anchor bolts pulling on it — which is precisely the T-stub of Module 13, with the base plate as the flange and the column flange as the web.

So everything from that module applies unchanged, including the mode question. For this base, with two M24 grade 8.8 anchor bolts:

Plate thickness	Mode	Resistance
20 mm	2	351 kN	ductile
25 mm	3	407 kN	brittle
30 mm	3	407 kN	brittle

And here is the tension that makes base design interesting:

Bearing wanted a 30 mm plate. T-stub ductility wanted 20 mm or less.

They are genuinely in conflict, and it is a real design decision rather than a calculation. The usual resolution is to accept the thicker plate and larger anchor bolts, so that the T-stub returns to a ductile mode at the thickness bearing demands — which is a good illustration of why the two lessons belong in one module.

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

The moment never changed. The bolt tension went from zero to 526 kN purely by removing axial load.

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

Friction alone carries it: 160 kN against 60 kN applied. But friction is proportional to the axial load, so check the combination with the LEAST compression before relying on it.

Things worth trying

Start at 800 kN and 150 kNm. The eccentricity is 188 mm, past the 130 mm to the compression centre, so there is already a tension side — 148 kN, comfortably inside the 407 kN available.
Now take the axial load down to 150 kN and change NOTHING else. The bolt tension jumps to 421 kN and passes the red tick. The moment never moved.
That is the whole point of the lesson: the governing combination for a base is the one with the LEAST axial load, which is the opposite of every other check on the column.
Watch the shear bar at the same time. At 800 kN the friction was 160 kN; at 150 kN it is 30 kN against 60 applied, and that check fails too. Two independent failures from one change.
Add a shear key. 150 kN of key takes the total to 180 kN and the shear check passes — but the bolt tension is untouched, because they are separate load paths.
Take the axial load negative, into uplift. The friction goes to exactly zero: there is nothing pressing the plate onto the grout, so nothing to rub. On a portal frame column at uplift, this is the real condition.
Now raise the axial load past about 1150 kN at 150 kNm. The eccentricity falls below 130 mm, the tension side disappears entirely, and the bolts carry nothing at all.

Shear does not go out through the bolts

This is the part most often got wrong, and the reason is that it looks as though it should.

Anchor bolts sit in holes that are several millimetres oversized — they have to, because a bolt cast into concrete cannot be placed to steel tolerances. Around them is grout, which is far weaker than either the concrete or the steel. Before an anchor bolt can bear on anything, the base has to move.

So the load path is:

1.Friction between the plate and the grout, proportional to the axial compression. Nothing else is needed if it is enough.
2.A shear key — a stub of section welded under the plate and cast into a pocket — where friction is not enough.
3.The anchor bolts, only where the design has explicitly accounted for the movement needed to mobilise them.

For this base with 60 kN of shear:

N_Ed	Friction at C_f,d = 0.2	Adequate?
1500 kN	300 kN	yes
400 kN	80 kN	yes
150 kN	30 kN	no
−100 kN (uplift)	0	no

Friction alone would need 300 kN of axial compression for this shear. Adding a 150 kN shear key at the 150 kN axial case gives 180 kN in total, of which the key supplies 83%.

The same pattern as the bolts: friction is proportional to axial load and wind is not, so the governing case is again the lightly loaded one. On a portal frame column at uplift, friction is exactly zero.

How fixed is the base, really?

Every frame analysis has assumed something about this. For a 254 UC over 6 m, EI/L is 4935 kNm/rad, and:

Base stiffness S_j	S_j/(EI/L)	Character	Sway, relative to pinned
1 000	0.20	effectively pinned	94%
5 000	1.01	semi-rigid	75%
15 000	3.04	semi-rigid	50%
50 000	10.1	semi-rigid	23%
200 000	40.5	effectively fixed	7%

A real nominally pinned base — a plate with four bolts inside the column profile — sits near the top of that table. A real nominally fixed base, with the bolts well outside the column and a thick plate, sits in the middle: stiff, but nowhere near rigid.

The asymmetry is what matters:

Assuming pinned when the base has stiffness overestimates sway. That is conservative for frame stability. It also means the base is designed for no moment, and a base that attracts moment it was not designed for will crack its grout and rotate until it genuinely is pinned — which is usually acceptable, and is why the assumption survives.
Assuming fixed when it is not underestimates sway. That is the dangerous direction, and it is the assumption people reach for when they see four holding-down bolts on a drawing.

Four bolts do not make a fixed base. A fixed base needs the bolts well outside the column flanges, a plate thick enough to deliver the lever arm, and a foundation stiff enough not to rotate — and even then it is semi-rigid.

Try it

How much fixity a base really has

A 254 UC on a base of the stiffness you choose. The bars compare its sway with a truly pinned base and a truly fixed one — the two things a frame analysis usually assumes.

Base rotational stiffness15,000 kNm/rad

Column height6.0 m

Base stiffness S_j: 15000 kNm/rad
Column EI/L: 4935 kNm/rad
S_j as a multiple of EI/L: 3.04
Character: semi-rigid
Sway, relative to pinned: 50 %
Safe assumption: pinned

S_j = 3.0 EI/L for this column, which is semi-rigid. A cantilever column on this base sways 50% of what it would on a truly pinned one. Assuming PINNED is the safe simplification for stability, because it overestimates sway. Assuming fixed would underestimate it, which is the dangerous direction — and it is the assumption people make when they see four holding-down bolts.

Things worth trying

Start at 15 000 kNm/rad — a realistic nominally fixed base with bolts outside the flanges. It reads 3.0 EI/L and sways 50% of the pinned case: stiff, and nowhere near rigid.
Drop to 1000, a plate with four bolts inside the column profile. It reads 0.2 EI/L and sways 94% — effectively pinned, which is what such a base is.
Now go to 200 000. Only there does it read as effectively fixed, and even then it sways 7% rather than 0%.
Note the gap between the middle bar and the bottom one at any realistic stiffness. That gap is the error in assuming a fixed base, and it is in the direction that OVERSTATES α_cr — the dangerous one.
The gap to the top bar is the error in assuming pinned, and it goes the other way: more sway than the frame will have, so second-order effects are overstated. Conservative.
Change the column height instead and watch the ratio move without touching the base. A base of fixed stiffness is relatively stiffer under a taller column, because the column it is competing with has become more flexible.
That last point is why base fixity cannot be judged from the base alone. The same detail is nearly fixed under one column and nearly pinned under another.

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
The foundation itself is assumed adequate — this is the steel side of the interface only

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
Assuming pinned overestimates sway and is safe; assuming fixed is the dangerous direction

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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