Ch 11 - Column Bases and Anchorage Into Concrete Aug 2011

1

11. Column bases and anchorage into concrete 11.1 General remarks on column bases

Regardless of the nature of the foundations of a structure – spread footings,

concrete piles or even steel piles – it is common practice to construct that part of a

structure which is just below and just above the ground in concrete. The most

important reason for doing so is that the region which is exposed to both oxygen

from the air and moisture from the soil can be quite corrosive. Lifting the steelwork

above this level makes sense. In other cases the steelwork may be built on top of a

concrete structure, or supported by it.

The base of a steel column can be placed directly on top of a concrete foundation,

or on top of a plinth or short concrete column built on the foundation. A plinth or

stub column lifts the steel away from the splash zone where it can be exposed to

water during floor washing and other activities. It can also put the steelwork out of

harm’s way where vehicles, forklift trucks and other moving objects can accidentally

damage it.

Whichever way the concrete surface on which the steel base will sit is built, the

concrete contractor is likely to experience difficulty in getting its level accurate to

within the tolerance on this level according to SANS 2001-CS1 (see Table 11.1). This

problem is addressed as shown in Figure 11.1 by placing the bottom of the steel

base some 40 mm above the specified level of the top of the concrete foundation,

supporting the base on steel packs (or on nuts on the holding down (HD) bolts, if the

base plate and HD bolts are designed for this situation) and inserting a flowable wet

grout in the space between the concrete and the steel. We recommend that a

non-shrink grout be used, for which a number of proprietary products are available,

or alternatively a grout consisting of a CEM/II/V-B cement and clean well-graded

sand with a low water demand. If the base plate is larger than about 700x700 mm,

50 mm diameter holes should be made in it to facilitate making sure that the grout

fills the whole space under the plate. The grout should be given time to cure before

it is loaded. The holding down bolts will be discussed under 11.3 below.

Figure 11.1 and Table 11.1 illustrate typical base plate details and dimensions, with

the minimum projections of the plate beyond the faces of the column or the

minimum distance from the HD bolts to the edge of the plate. At least four HD bolts

should be used for a column, with the exception of a 152x152H-section, for which

two may be adequate. Even if the base will not be required to resist bending

moment in the final structure, it may have to keep the column upright when the

wind blows and other forces act on it during construction. This is dealt with under

11.6 below in the introduction to the resistance tables.

Base plates for hot-rolled sections should be welded to the column, with 6 mm welds

up to a flange thickness of 20 mm and 8 mm thereafter. A normal, flat plate and a

sawn column end should easily be within the tolerances of SANS2001-CS1. However,

with bigger columns welded up from plate it is common practice not to weld the

column to the base plate, to machine the column end and, if necessary, the base

plate for full contact, and to use tensile bolts to tie down the column. This is depicted

in Figure 11.1( d ) and (e). A box column can also be handled as in (d).

2

Some vertical elements, with base plates, are called ‘posts’ rather than ‘columns’

because they are intended to resist wind and other loads in bending, rather than

axial compression. Provided that a post will be laterally supported at its top from the

moment it is erected, it can be provided with a base plate of minimal size.

Tables 11.6(a) and (b) are intended not only to provide the resistances of axially-

loaded base plates, but also to give guidance on standard sizes. For each universal

beam sizes the first size listed is intended for posts. The tables also show how tall a

column with the smallest base plate listed may be, without exceeding the

resistances under the action of construction loads.

3

Figure 11.1 – Typical base plate details

TOLERANCES

DIMENSION Tolerance (mm)

Top of concrete 20,10

Top of base plate 5

Position of column 3

DETAILS

Bolt hole diameters

d = diameter of HD bolt

hd diameter of hole

mmddmmd h 6:24

mmddmmd h 10:4024

mmddmmd h 15:40

Washers (may be plates)

If 6 ddh mm: use heavy duty washers

If 6 ddh mm: 2

bhw

ddt

If column base subject to shear, weld washer to base plate.

Note: See 11.3 below for details and tolerances for HD bolts

Table 11.1 - Details and tolerances of base plates and foundations

It is permissible to support a base plate on levelling nuts, provided that this is

approved by the designer of the structure and concrete foundation, who should

base his/her decision on the following considerations:

The base plate must be strong enough to resist the forces it will be subjected

to while it rests on the bolts.

The bolts must be strong enough to resist these forces. They should not yield,

buckle or deflect sideways. The strength of the supporting concrete in

providing anchorage to the bolts must also be adequate.

If the intent is that the base plate should only sit on the nuts during a certain

stage of the erection process, steps must be taken to ensure that grouting

(and curing of the grout) will be complete before further loads are placed on

the column.

4

There are basically two types of bases for steel columns: pinned bases and moment

bases. The distinction between the two is simple: it depends on how the structure

was analysed. If it was assumed in the analysis that there is a hinge at the bottom of

a column, it will be a ‘pinned’ base, essentially regardless of its details or the

behaviour of the structure. If it was assumed that the support at the base of a

column is rigid, it will be a ‘moment’ or ‘rigid’ base which has to be able to resist the

moments and other forces emerging from the structural analysis. It may appear

rather arbitrary to just assume that a base is pinned whenever that suits us, but there

are two arguments that underlie this approach:

Rigid bases tend to be a lot more expensive than pinned bases. Not only is

their design and fabrication time consuming, but more material will be

required and the concrete foundation may also have to be bigger. Rigid

bases should only be used where they are either inevitable or demonstrably

beneficial.

On the basis of the Lower Bound Theorem as discussed under 1.5 above, it is

almost always safe to assume that a base is pinned, even if that is not true.

Rotation in a column can be accommodated by elastic deformation and

yielding of the base plate and bolts, slight localised crushing of the grout and

concrete, or tilting of the concrete foundation. In some situations, especially

where the concrete base is very stiff, the capacity of the base plate to

absorb deformation may have to be checked. It is, in general, erring on the

safe side to assume that the base of a column is pinned when it can actually

resist a bending moment.

The design of pinned column bases is discussed under 11.4 below, and that of rigid

bases under 11.5

11.2 Cast–in or embedded elements

The interface between steelwork and concrete supporting it is, of course, not limited

to the bases of columns. Steelwork is often attached to concrete walls or columns by

casting steel elements into the wet concrete as shown in Figure 11.2 (a) and (b) and

attaching the steelwork to these elements. Another strategy is to leave a pocket in

the concrete, to place, for example, the end of a beam in this pocket, and to cast

concrete around this beam end. This is illustrated in Figure 11.2(c).

5

Figure 11.2 – Steel element embedded in concrete.

Note that in each case shown in Figure 11.2 rods or shear connectors are connected

to the steel elements and cast into the concrete to ensure that the steel will not part

company with the concrete. The design of these anchorages will be discussed

under 11.3 below.

No tolerances relating to the position of embedded steelwork are given in SANS

2001-CS1, except that the level must be within 5 mm of the specified value. British

and European specifications require the embedded plate in (a) to be within 10mm,

horizontally, vertically and out of plane. The angle in (b) is really a trimming on the

concrete structure, and thus the concrete tolerances should generally apply to it.

The connection of steelwork to a cast-in plate as shown in (a) is a situation where site

welding may make eminent sense. It is in any case advisable to allow for

adjustability in the steelwork to be attached to a cast-in element, as it may be very

difficult to fix any positional inaccuracies that are noted once the concrete has

hardened, despite any tolerances that may have been prescribed.

11.3 Anchoring steelwork to concrete

11.3.1 Holding down bolts

The standard item holding a base plate firmly in position on a foundation is the

holding down (HD) bolt, called ‘anchor rod’ in American practice.

Typical details for HD bolts are shown in Figure 11.3. In (b) a pocket is provided to

allow moving the bolt sideways if it proves not to be in exactly the right position.

Suggested tolerances on the position of a HD bolt are given Table 11.2. Note that

these tolerances are more liberal than those in Table 11 of SANS 2001-CC1, which

are rather impracticably tight.

6

Figure 11.3(c) shows an anchor plate intended to help prevent pulling the HD bolt

from the concrete. This can be replaced by a frame as shown in (d) to help keep

the bolts in position relative to each other during casting of concrete. A more

effective position for this frame is near the top of the HD bolts, as shown in (c), but in

this position it will not help much in resisting tensile forces in the bolts and the anchor

plate is still required.

The anchor plate can be deleted and replaced with a single nut if the HD bolt will

not be subjected to a tensile force. However, this will require checking that the HD

bolts will have sufficient strength to resist wind and other forces acting on the column

during construction. The SAISC recommends that all HD bolts be detailed as shown in

Figure 11.3.

Dimensions of pocket (mm)

7

d 20 24 30 > 30

ph 150 200 250 Pocket not

recommended pd 75 75 100

Figure 11.3 – Dimensions and symbols for HD bolts

Tolerances for positions of HD bolts

Top of HD bolt: 20 mm (high), 5 mm (low)

Horizontal position of HD bolt without pocket: 3 mm

Horizontal position of HD bolt with pocket: 8 mm

Table11.2 –Tolerances for HD bolts

Note that in Figure 11.3 the anchor plates and frames are shown as being attached

to the HD bolts by nuts. The reason for doing so is that it is easy to do and that the

quality problems often experienced with welding are obviated. The issue of the

weldability of the steel also disappears.

The question may arise: what is the appropriate torque for a HD bolt? The answer is

that HD bolts should be tightened to the snug tight condition after the grout

supporting the base plate has hardened sufficiently. The possibility of preloading is

discussed later in this section.

The SAISC recommends that Commercial Quality steel (i.e. any steel, regardless of

grade) be specified for all HD bolts that will not be subjected to any significant

forces, i.e. for bases subject to vertical compressive force only. These bolts are

primarily intended to keep the base plate in position until the structure has been

erected, and the main criterion is that they should not easily deform when subjected

to the normal activities on a construction site. If stronger bolts are needed, a steel of

appropriate quality can be chosen or ‘Class 8.8 equivalent’ can be obtained by

using material as commonly used for producing Class 8.8 bolts and giving it the

appropriate heat treatment after manufacture, or by using EN 19 (BS970 Grade 709

M40) steel and doing heat treatment to achieve Hardness T. See also 2.1.6 above.

We suggest the minimum diameters and lengths for HD bolts listed in Table 11.3. Note

that the SAISC recommends a minimum HD bolt diameter of 20 mm, to ensure

robustness during construction.

Base plate

thickness bt

(mm)

Minimum

HD bolt

diameter

(mm)

Minimum HD bolt

overall length

(mm)

50

5020

20

b

b

b

t

t

t

20

24

36

300

400

500

Table 11.3 – Minimum sizes of HD bolts

8

The resistance of a bolt to failure in tension or being pulled out of the concrete will

now be addressed.

Tensile resistance of the bolt: unbr fAT (11.1)

(from Clause 25.2.2.1 in SANS 10162-1)

where 67,0b

4

938,02

pdAn

d = diameter of bolt on thread

p = pitch of thread

uf = 365 MPa for Commercial Quality steel, 800 MPa for Class 8.8 equivalent.

For simplicity, take 4

75,02d

An

. (11.2)

The resistance of the bolt to pull-out will depend on the bond strength between the

bolt shank and the concrete cuf28,0 and bearing on the anchor plate .6,0 cuf

Thus the resistance rcT of a bolt as determined by the concrete is given by

(assuming the anchor plate to be dxd 5,35,3 ):

hdcubcurc AdfdfT 2

5,36,028,0 (11.3)

where 4

2dAhd

Having ensured that the HD bolt will neither yield nor pull out of the concrete, the

next step is to make sure the concrete will not fail. In unreinforced concrete, failure

of the concrete will happen in the form of a cone pulled out of the concrete as

shown in Figure 11.4 for single and grouped HD bolts.

The following simplified equations can be used to solve the problem.

For a single HD bolt as shown in Figure 11.4(a) the effective surface area of

the cone can be approximated by:

75,18 bcA (11.4)

For a single HD bolt near an edge as shown in Figure 11.5(a) with edge

distance bad .6 it is conservative to say:

75,14 bcA (11.5)

For a group of HD bolts as shown in Figure 11.5(b) the following effective area

can be used:

9

75,175,1

2

2

88.8

88bb

b

pbb

c nA

(11.6)

where n is the number of HD bolts in the group

The ultimate stress on the surface may not exceed cuf25,0

Thus:

cucr fAT 25,0. (11.7)

An alternative approach to anchoring a HD bolt is to let the force in it be transferred

by bond to the reinforcing in the concrete base, just as one would for a normal

rebar, in accordance with SANS 10100-1.

10

Figure 11.4 – Conical failure of concrete

A HD bolt can also act in shear, as shown in Figure 11.5. This situation need only be

considered if the shear force acting on the column concurrently with a (minimum)

axial force uC exceeds uC , where is the minimum coefficient of friction on any of

the relevant surfaces. The value of can conservatively be taken as 0,3. Figure 11.6

shows how a shear key can be used to resist shear rather than the HD bolts. Other

options may be to tie the base to another object, or to tilt the top of the concrete so

that the forces in the column act perpendicular to it.

Figure 11.5 – HD bolts in shear

Figure 11.6 – Shear key to resist high shear forces on base plate

As for normal bolts, SANS 10162-1 Clause 25.2.3.3 specifies that the shear resistance

of a HD bolt is given by:

uhdbr fAxV 6,07,0 (11.8)

where 8,0b

4

2dAhd

d diameter of bolt

uf tensile strength of bolt steel

11

Bearing of the base plate against the bolt should never be a problem.

Clause 25.2.3.2 is intended to prevent crushing of the concrete when the HD bolt

pushes horizontally against it. The resistance is given by:

cucrr fABV 12,1 (11.9)

where 6,0c

25dA

cuf concrete cube strength.

Note that this assumes that the grout layer will be solid, at least as strong as the

concrete, and will not slide on top of the concrete. The resistances of HD bolts to

shear forces are given in the last row in Table 11.5. For 25 MPa concrete, the

concrete controls, regardless of the quality of the HD bolt. All HD bolts must be at

least d.7 away from any edges for these equations to apply.

For HD bolts that are loaded in combined shear and tension the following interaction

equation is recommended.

0,13

53

5

r

n

r

u

V

V

T

T (11.10)

Table 11.5 lists resistances for HD bolts.

A situation where a combination of tensile and shear forces act on HD bolts can of

course cause problems if a bolt is close to the edge of the concrete. Such bolts must

be constrained with stirrups and hairpins and the requisite care must be taken to

ensure that they will not fail. Moreover in cases where the concrete is cracked, all

capacities must be reduced by 20%.

When HD bolts are subjected to repetitive tensile loading, such as can be expected

when they secure a mast, chimney or post subject to wind loads, there may be

concern about fatigue in the bolts. The solution would be to apply the requirements

of SANS 10162 Clause 13.12.1.3, and to limit the stress in a Class 8.8 equivalent bolt to

214 MPa under the worst combination of specified loads. It is not advisable to use

Commercial Quality bolts in such a situation.

One may think about following the approach discussed under 3.8 above and

pretension the HD bolts so that they will not experience the fluctuating loads. The

problem is that the concrete will tend to shrink and creep, so that the bolts will lose

their preload. The anchorage of the HD bolts may also not be sufficient to resist the

preload force. These problems can be solved by using a sleeve around the bolt, as

shown in Figure 11.4. The base plate rests on the sleeve, which rests, in turn, on a

thick anchor plate. Preload in the bolt will put the sleeve in compression, and

shrinkage or creep of the concrete will play no role. The cross-sectional area of the

sleeve must, of course, at least equal yp fT / , where pT is the preload in the bolt.

The problem with this solution is that it will be very difficult to have the tops of all the

sleeves under a base plate on exactly the right level so that the plate will rest on all

of them. The attractive aspect, on the other hand, is that horizontal adjustability of

12

the bolt will be greatly enhanced, making the detail an option even if the sleeves

are not taken to the bottom of the base plate, thus ruling out preload.

Figure 11.7 – HD bolt with sleeve

11.32 Studs for cast-in elements

It is difficult to generalise about the design of steelwork embedded in concrete.

Obviously, any steel plate or other element must be able to resist the forces acting

on it while spanning between the points where the anchoring elements support it.

The anchoring elements themselves must be strong enough, attached to the

embedded steelwork by strong enough welds, and be properly anchored by bond,

bends, attachment to rebar, or other elements, so as to be able to resist the forces

acting on them. Where the anchoring elements are studs, the shear resistance of

each stud is given by SANS 10162-1 Clause 17.7.2.1 as rsq , where:

uscscccuscscrsr fAEfAqV 45,0 (11.11)

where uf = 415 MPa for commonly-available studs

8,0sc

4

2dAsc

d = diameter of stud

cE = short term E of concrete to SANS 10100-1

Assuming that there is steel reinforcement in close proximity to the stud, the distance

to the edge of the concrete must be greater than d.7 to preclude concrete failure

due to shear in the stud.

The weld connecting the stud to the embedded steel must be strong enough in

tension or shear, and the tensile resistance of the stud must not be exceeded. The

concrete should also not fail under the action of tensile force, and here it is best to

use the model of conical failure as depicted in Figure 11.7. It is advisable to keep the

stud a minimum of d.6 away from edges in order to preclude side face blow out

due to tension in the stud.

13

Figure 11.8 – Conical failure in concrete

The surface area of the cone is given by

22 44,42 bbcA (11.12)

The stress on this surface under ultimate load may not exceed cuf25,0 , thus the

resistance is given by:

cubr fT 25,0*44,4 2

cubr fT 211,1 (11.13)

Thus the required embedded length of the stud is given by:

cu

u

bf

T

11,1 (11.14)

This method is appropriate for a stud length mmb 110 . For longer or studs refer to

the method for HD bolts under 11.3.1 above.

11.4 Design of pinned (simple) bases

If the engineer assumed during the analysis of the structure that a base is pinned

and there is no shear, only axial compression needs to be transmitted between the

column end and the base plate. If the end of the column is sawn and the base

plate is flat, the tolerance required in SANS 2001: CS1 Item 3 of Table 3 and Item 5 in

Table 4 should be easily attainable and no forces need to be transmitted through

the welds connecting the column to the base plate.

However, these welds have to be strong enough to keep the base plate attached

to the column during transport and erection, and to resist any shear forces and

moments that may occur at the bottom of the column during the lifetime of the

structure, because the base is not actually an ideal pin. At a minimum, welding must

be applied over the width of the outside of each flange and over half the length of

the web on either side of it. The welds should have a leg size of 6 mm for flanges up

to 20 mm thick and 8 mm for thicker flanges. An alternative is to weld cleats to the

base plate and to bolt the column to these. If the value fb tt 2 exceeds 60 mm the

need for preheating the welded area must be investigated, where bt thickness of

base plate, ft thickness of column flange. (See Chapter 4 above on preheating.)

14

The following steps can be followed for the design of base plates:

a) Calculate the area of base plate required, by dividing the factored axial load

uC by cuf6,0 , according to SANS 10100, Clause 6.2.4.4.4 (b)

b) Choose an effective area equal to the required area around the column. It

may be possible to find an adequate area that is similar to the shape of the

column and projects everywhere a distance c from the face of the column,

as shown in Figure 11.9(a). The value of the required c can be obtained by

solving the following quadratic equation:

24c (column perimeter) c column area = cuu fC 6,0/ (11.15)

If the calculated c exceeds half the distance between the flanges, there will be an

overlap area as shown in Figure 11.8 (b), and a new value of c has to be obtained

by doing a second calculation, which involves solving the following equation for c :

cuu fCcbch 6,0/22 (11.16)

where h = depth of column

b = width of column

Figure 11.9 – Concentrically-loaded base plates

If the real edge of the base plate is closer to the column at any point than the value

of c from Equation 11.16, as shown in Figure 11.8(c), a new effective area and value

of c have to be found. This can be obtained by solving the following equation for c :

wff tccthcbcta 22222 = cuu fC 6,0/ (11.17)

15

c) Determine the required plate thickness from

42

12

2 yb

ru

ftMcM (11.18)

cf

ty

b

2 (11.19)

where = pressure under plate uC / (effective area) cuf6,0

yf = yield stress of plate

(Note that the plastic section modulus is used)

d) The thickness of the plate should not be less than the flange thickness of the

column.

Values of the minimum thickness mmtb of the base plate depending on the cube

strength of the concrete, the yield stress of the steel and the dimension c are listed in

Table 11.4, assuming that the pressure on the concrete equals cuf6,0 .

Concrete cube

strength cuf

( MPa)

20

25

30

35 40

Steel yield

stress

yf

(MPa)

275

21,3c

87,2c 62,2

c 43,2

c 27,2

c

355

65,3c

26,3c

98,2c

76,2c

58,2c

Table 11.4 – Required base plate thickness bt as a proportion of minimum value of c .

Table 11.6 provides the resistances of base plates for various column sizes, for

different values of the concrete cube strength and the yield stress of the base plate

steel.

The minimum diameter and length of HD bolts for a base plate are also given (in

accordance with Table 11.3), and the requirements stated in Figure 11.1 are

observed.

11.4 Design of column bases under combined compression and moment

Some column bases are required to resist simultaneously both the axial forces and

end moments in columns. Figure 11.10 shows a number of common bases of this

kind, subject to increasing axial force and moment. It is common to make such

16

bases symmetric, even when the moment will only act in one direction, but that may

not necessary be advisable.

Figure 11.10 – Column bases to resist bending moment

As long as the moment is small, compressive stress can be present over the whole

area of the base plate, varying from a minimum at one edge of the plate to a

maximum at the other. However, when the moment is large, stresses cannot be

present over the whole base because tensile stresses cannot be developed

between the base plate and the concrete foundation. As the eccentricity increases,

a tensile force becomes necessary to maintain equilibrium, and this is provided by

the holding down bolts. The situation is analogous to a reinforced concrete beam;

the bolts in this case corresponding to the steel reinforcement.

17

There is no quick way to find the required size of a base plate with bending moment,

such as we have for concentrically-loaded bases in the form of Equations 11.15,

11.16 and 11.17. The size must be found by trial and error.

What part of the base plate will be effective in carrying compressive stress will

depend on the thickness of the base plate, the shape of the column, and the

loading. The thickness pt will determine the distance maxc from the face of the web or

flange of the column to the edge of the effective bearing area, shown hatched in

Figure 11.11. The relationship between maxc and pt can be read from Table 11.4 with

the relevant cuf and yf . Any part of the base plate more than maxc away from the

nearest column edge (except at the corners) must be ignored. Figure 11.11 also

gives an idea of the shape of the area in compression under the base, depending

on the load condition. The required thickness of the base plate may also be

controlled by tension in the HD bolts.

Let us consider first the base plate in Figure 11.11(a), where the whole width of the

plate will be effective. The forces and stresses acting on the plate, and the

necessary dimensions, are shown in (b). We can say that there will be no tension in

the HD bolts if:

cub

ubuu

fb

CdCM

6,022 (11.20)

In this case the resistance rC of the base plate is given by:

u

u

bbcurC

MdbfC

26,0 (11.21)

If ur CC failure will not occur.

18

Figure 11.11 – Pressures and forces on moment bases

If uM is bigger than the value calculated in Equation 11.20 there will be tension in the

HD bolts and we can say:

02

6,0 12

xdfxbdCM cubuu (11.22)

We can solve this equation for x , and then get bT , the force in the HD bolts, from

the following equation:

ucubb CfxbT 6,0 (11.23)

19

If the collective resistance of the HD bolts that will be in tension exceeds bT the base

plate will be able to resist uC and uM acting together. The signal that the base plate

is too small to resist uP and uM will be that a sensible solution for x cannot be found.

If this happens, or if the tension in the HD bolts is too large, the size of the base plate

must be increased, which will require making it thicker. We need to make sure that

the base plate can resist the force bT the group of HD bolts acting in tension will

exert on it. Let:

mTM uu . (11.24)

Where uT = load in the HD bolt group

x = distance from the centroid of the bolt group to the centre of the

nearest flange.

4

.2

yp

b

u

b

uft

b

mT

b

M (11.25)

yb

up

fb

mTt

.

.2 (11.26)

The problem in Figure 11.11(c) and (d) is exactly the same as that in (a) and (b)

if 1cx where max1 ccc end . But if x is larger the problem can be handled as

follows:

21 xccbA b (11.27)

where wtcc max2 2

cuc fAC 6,0 (11.28)

012 xdCdCM cuu (11.29)

xcbcxb bb 12

b

b

cb

cbxcx 12 (11.30)

Entering all the factors we get the equation that has to be solved for x :

02

6,0 121212

b

bcubuu

b

cbxcdfxccbdCM (11.31)

With x known, A and cC can be obtained and then:

20

ucb CCT (11.32)

As before, if a sensible solution for x cannot be found, the base is too small.

We present next an entirely different approach that can be followed in the case

where the full width of the base plate is effective, as in Figure 11.11(a) and (b).

Let brT be the combined resistance of the group of HD bolts that will resist any tensile

forces, as defined by Equations 11.1, 11.3 or 11.7.

Then the maximum force rC that can be resisted at a given excentricity u

u

C

Me

can be demonstrated to be predicted conservatively by:

2

42 rC (11.33)

where bcu bf

6,02

1

ebf

Td

bcu

brb 6,02

bcu

br

brbf

TdT

6,021

However, if the actual load in the column bbcuu dbfC 6,01,0 we must also check

that

bcu

br

brroubf

TdTMM

6,021 (11.34)

Where roM = resistance of base plate in absence of axial load.

The difference between the two approaches is clear: in the approach that leads up

to Equation 11.32 we work with the actual values uC and uM and calculate the

required tensile resistance bT of the group of HD bolts in tension. In Equation 11.33

we assume a resistance brT for the applied group of HD bolts in tension and work

with a relationship uu CMe / of the forces, which gives us the maximum force rC

that can resisted at an eccentricity e .

The reason for introducing the last approach is that it lends itself to the handling of

baseplates with biaxial bending and axial force. Let the force in the column be uC

and the moments about the x and y axes respectively uxM and uyM . The

eccentricities are:

21

u

uxy

C

Me (11.35)

u

uy

xC

Me (11.36)

Looking at the group of bolts along the edge where they can best resist uxM , their

collective resistance to tensile force is defined as brxT . Equally, the resistance of the

bolts resisting ryM is bryT . (A corner bolt may be a member of both groups.)

Using Equation 11.33 with the appropriate value of e and brT , and switching bd , bb

and 1d to the appropriate values, we can calculate rxC and ryC .

Furthermore, define:

bbcu dbfC 6,00 (11.37)

Then the base plate can resist a force rC at eccentricity yx ee , , such that

0

1111

CCCC ryrxr

(11.38)

If 01,0 CCu the equation may be unconservative, and we also need to check:

0,1ry

uy

rx

ux

M

M

M

M (11.39)

where

bcu

brbrr

bf

TdTM

6,021 (11.40)

with the appropriate values of brT , 1d , and bb for each axis.

11.5 Examples

11.5.1 HD bolt examples

Example 11.1

What embedment length b is required for a 24 mm Class 8.8 equivalent HD bolt to

carry an ultimate tensile load uT of 150 kN. Assume 25cuf MPa and an anchor

plate of size 3,5d (square). Assume uf = 800 MPa for the bolt.

(i) Check that the HD bolt can carry the imposed load.

According to Equation 11.1:

22

8004

75,067,02

xd

fAT unbr

= 182 kN > 150 kN OK

(ii) Check that the HD bolt cannot be pulled out.


nbrc AddT 2

5,3154,1

1042206,105 b N

Letting NTT urc 000150 we get:

b 434 mm

(iii) Check that the HD bolt tension does not cause concrete cone breakout.

(a) Assuming that there is no edge close to the HD bolt 434a :

Equations 11.4 and 11.7 give us:

75,175,1 1025.08 bcubr fT

Letting ur TT we get

b 244 mm

(b) If the anchor is a distance a = 190 mm away from an edge:

Check if this is an edge condition by comparing

maxba

190 mm < 434 mm therefore this is an edge condition.

Check to preclude sideface blow out if da 6 : ( 24.6190 =144 mm OK)

According to Equations 11.5 and 11.7:

75,175,1 525,0.4 bcubr fT

Letting ur TT we get

b 362 mm

Therefore b =434 mm

23

From Table 11.5 it is clear that the HD bolt strength can be fully developed in

tension if HD bolt pull out is precluded by increasing b to 800 mm. Then rT = 182

kN.

Example 11.2

Evaluate whether the HD bolt in Example 11.1 can carry a tensile load uT = 100 kN

and a shear load uV = 35 kN simultaneously. Assume the edge distance a = 190 mm

(i) Check to make sure that the edge distance dc 7 to preclude side concrete

breakout

190 mm 247 = 168 OK

(ii) Check concrete crushing due to shear. This limit is always more critical than

the shear resistance of the bolts for Class 8.8 equivalent HD bolts.


2556,012,112,1 2 dfAV cucr = 48,4 kN

(iii) Check interaction of tension and shear.

From Example 11.1 rT =150 kN for b = 434 mm


1103

53

5

r

u

r

u

V

V

T

T

0,109,14,48

35

150

100 35

35

no good !

If b is increased to 800 mm, then rT = 182 kN (as per Table 11.5)

Then:

0,195,04,48

35

182

100 35

35

OK

Example 11.3

Evaluate the maximum tension that can be imposed on an cast in plate that is

anchored to concrete ( cuf = 25 MPa) using a single standard headed stud ( d = 20

mm and = 110 mm). Assume the plate already carries a 25 kN shear load and is not

24

close to an edge. Also assume reinforcing around the stud and uncracked

concrete.

(i) Check the shear capacity of the stud.

If the stud were purely in shear we would limit our check to Equation 11.11,

but because the stud is carrying shear and tension we will be conservative and

also check Equation 11.9 for concrete crushing assuming dA

251106,012,1 rV = 37 kN OK


uscscccuscscr fAEfAV .45,0

cE = 26 000 MPa from SANS 10100 for normal density concrete

uf = 415 MPa for commonly available studs

26000254

8,045,02

d

Vr

=91 kN 415

48,0

2

d

=104 kN

We work with the lower value:

rV = 37 kN

(ii) Check the tensile capacity of the stud

According to Equation 11.13: ( b 110 mm)

2511011,111,1 22 cubr fT = 67 kN

(iii) Check the interaction Equation 11.10 and solve for uT

0.137

25

67

35

35

uT

uT 43 kN

11.5.2 Base plate examples

Example 11.4

Design a base plate for an axially loaded 60203203 xx H column with an axial load

uC of 824 kN (equal to the resistance for an effective length of 6m).

Assume a concrete strength of 25 MPa. The dimensions of the column are: h = 209,6

b = 205,2, ft = 14,2, wt = 9,3, A = 7600 mm2

25

Assume a base plate size of 350350x

Then according to Equation 11.17:

24c + column perimeter c + column area cu

u

f

C

6,0

Column perimeter = 1221,4 mm

256,0

82400076004,12214 2

cc = 55 000 mm2

Solving for c

c = 34,8 mm

The distance between the flanges 2,1426,209 =180,8mm

Check: 8,1802 c . Therefore the value for c is OK.

Also check 6,2093502 c and 2,2053502 c . Both are OK.

Using Table 11.4 and base plate yf = 355 MPa, the required plate thickness is

26,3

8,3426,3

ct p = 10,7 mm

However the thickness should not be less than ft , so use pt = 16mm

Conclusion:

Use 16350350 xx plate

with 4-16 mm HD bolts embedded 250 mm

and arranged as per Figure 11.1(b).

Note that according to Table 11.6(a) this column and base plate combination can

resist a load of 966 kN.

Example 11.5

Design a base plate for a 57171356 xx column that carries an axial load of 200 kN

and a moment of 75 kN.m. Assume cuf =25 MPa and plate yf = 355 MPa

(i) Make a preliminary base plate size selection of 350550x mm, and check if

there will be tension in the HD Bolts. According to Equation 11.20 if:

26

cub

ub

uufb

CdCM

6,022 there will be no tension in the bolts

But

6,501000

1

256,03502

000,200

2

55020075

uM

(i) Therefore there is tension in the HD bolts. To calculate the tension in the bolts

we can use Equations 11.22 and 11.23 if x is less than 1c or Equations 11.31

and 11.32 if x is greater than 1c . Solving equation 11.22 we get (assuming

that the HD bolts are 60 mm from the edges)

2490256,0305215000,2001075 6 x

xx =0

This yields x = 48,2 mm 1c therefore Equation 11.23 can be used to solve for bT

000,200256,02,483506.0 ucubb CfxbT =53,05 kN

(ii) Determine the size and anchorage length of two HD bolts to carry bT

assuming

Class 8.8 equivalent bolts, and no edge condition.

Checking for concrete cone breakout;

The centres of the two HD bolts are 602350 =230 mm .

Assuming 24 mm bolts and using Equation 11.6 and 11.7, for the group of bolts:

75,1

2

2

88

230882525,0 b

b

bb

rT

and solving for b

b 2,4 mm

Checking against Equation 11.6 for a single bolt:

75,1822525,0 brT and solving for b

90,4 mm

Checking for HD bolt pull out according to Equation 11.3

bbrb AddTT 2

5,3154,12/05,53 for d5.3 square anchor plates.

990546,10552,26 b

b 0 mm

27

It is recommended to use the anchor lengths and details given in Table 11.3 and

Figure 11.5.

therefore use b = 400 mm > 90,4 mm

(iii) Determine the required thickness of the base plate, with the information

already calculated illustrated below.

5,352/)359550( a mm

4,712

2,48

2

359550

x mm

The bearing force 2,48350256,0 cC = 253 kN

Clearly, the compressive force and lever arm are much bigger than the values for

the HD bolts.

Thus

6101,184,71253 xxM u N.mm

2

22

000284

3553509,0

4p

p

y

pp

r txt

ftb

M N.mm

Setting ru MM yields

4,25Bt mm

According to Equation 11.26 with the necessary changes:

3559,0350

4,7100025322 2

x

fb

xFt

yb

u

=25,4 mm

Conclusion

Use a 30350550 xx base plate with 4-24 mm Class 8.8 equivalent HD bolts with

400 mm anchor length.

Example 11.6

If the anchor length of the M24 HD bolts is limited to 200 mm, would the base plate

in Example 11.5 be able to carry a minor axis bending uyM of 20 kN.m in addition to

the 75 kN.m major axis bending and 200 kN axial load?

28

From Equations 11.4 and 11.7:

75,120082525.02 brT = 212 kN

From Equation 11.3:

990542006,1052 brT = 240 kN

Therefore brT = 212 kN, uxM =75 kN.m, uyM = 20 kN.m and uC =200 kN.

20000020

u

uyx C

Me = 100 mm (Equation 11.35)

20000075

u

uxy C

Me = 375 mm (Equation 11.36)

In order to check the interaction equations given in Equations 11.38 and 11.39 we

have to evaluate rxC , ryC , oC , rxM and ryM


2

42 rxC

10500

1

350256,02

1

x

140375350256,0

212000

2

550

x

71096,9

350256,02

000212490000212 xx

50010

12

10500/1096,94140140 72

rxC = 524 kN

xrxM =99,6 kN.m

Calculate ryC using Equation 11.33

50016

1

550256,02

1

y

100550256,0

212000

2

356

Y = 49,3

29

6108,58

550256,02

212000212000 xy

5001612

1088,53,493,49 72

ryC = 1472 kN

yryM = 58,8 kN.m

It is easy to to calculate that, for a concentrically-loaded base plate of the given

dimensions:

oC = 2611 kN

Now we can check the interaction equations.

According to equation 11.40:

453

1

2611

1

1472

1

524

11111

oryrxr CCCC

rC = 453 kN > uC = 200 kN

But ou CC 1,0 = 261,1 kN so we have also to check Equation 11.39

0.109,18,58

20

6,99

75

ry

uy

rx

ux

M

M

M

M no good.

Therefore at an anchor length of 200 mm the maximum minor axis bending that

can be imposed on the base plate is 14 kN.m which will give an interaction of 1,0.

Even then one has to check the thickness of the plate using Equation 11.26:

3559.0350

4221200022

yb

uy

ptfb

Tt

= 17,8 mm < 30 mm ok

When considering the thickness required due to bearing it is conservative to that

work with Table 11.4, noting that the maximum projection of the base plate

beyond the column face is 95,5 mm:

26,3

5,95pt = 29,3 mm < 30 mm OK.

Conclusion

Therefore the maximum bending moments that this base plate can carry together

with uC = 200 kN are uxM = 75 kN.m and uyM = 14 kN.m. Larger bending moments

can be resisted with higher values of uC .

30

11.6 Resistance tables

Table 11.5 provides the resistances of individual HD bolts for different values of the

anchor length b and diameter d with 25 MPa concrete. The resistances listed take

the anchorage in the concrete (Equation 11.3) and the resistance of the concrete

to conical failure (Equations 11.4, 11.5 and 11.7) into account, i.e the values for

single bolts. Grouping of bolts as covered by Equation 11.6 was not considered. The

tensile force on a HD bolt may, however, not exceed the resistance of the bolt in

tension as determined by Equation 11.1, thus the maximum values of Commercial

Quality (CQ) bolts and Class 8.8 equivalent bolts were also listed. Lastly, the shear

resistance as controlled by Equations 11.8 and 11.9 was listed.

Note the limitations in the notes at the bottom of Table 11.5. For concrete with a

cube strength higher than 25 MPa higher resistances can be found on the basis of

the equations listed above, roughly proportional to the increase in the cube

strength.

Table 11.6 provides the resistances of base plates for columns consisting of universal

beams or columns, for different values of the cube strength of the concrete. Several

base plate sizes and thicknesses are listed for each column, the first of which for

each column size is intended to be for ‘bolts inside’ as shown in Figure 11.1 and for

use with relatively small loads. The values are based on the approach discussed in

11.4 above. Interpolation can be done.

The minimum HD bolt diameters and lengths in Table 11.6 are based on Table 11.3.

The ‘maximum column height’ values indicate the maximum length of column of the

particular size and with the minimum size base plate and associated bolts (of

Commercial Quality steel) that can stand on the base plate and resist a wind

pressure of 1 kPa coming from any direction during the construction period. Longer

columns may need lateral support.

31

Table 11.5 – Resistances (kN) of individual HD bolts to tensile and shear forces

Anchorage

length b

(mm)

Bolt diameter, d (mm)

Edge Distance ≥ b d7 Edge Distance < b

20 24 30 36 20 24 30 36

R

esi

sta

nc

e t

o t

en

sio

n

200 86 106 106 106 53 53 53 -

300 95 131 194 216 95 108 108 108

400 104 141 208 286 104 141 179 179

500 113 152 221 302 113 152 221 264

600 122 162 234 318 122 162 234 318

700 126 173 247 334 126 173 247 334

800 126 182 260 350 126 182 260 350

900 126 182 274 365 126 182 274 365

1000 126 182 284 381 126 182 284 381

Max value

CQ bolts 57,7 8,30 130 187 57,7 83,0 130 187

Max value for

8.8 bolts 126 182 284 410 126 182 284 410

Shear

resitance 34 48 76 109 34 48 76 109

Notes:

1. b for any HD bolt must exceed d5

2. HD must be at least d7 from edge of concrete

3. Bolt shanks must be degreased before embedment in concrete.

4. Assumed concrete strength = 25 MPa.

32

Notes:

1. Base plate made of S355 steel

2. See 11.6 above for explanation

Table 11.6(a) – Capacities of base plates for axially-loaded columns, fy = 355 Mpa

33

Notes:

1. Base plate made of S275 steel

2. See 11.6 above for explanation of table

Table 11.6(b) – Capacities of base plates for axially-loaded columns fy = 275 Mpa