## 9

## Column Splices and Bases

#### 9.1 Introduction

This chapter deals with column splices and column bases. A column splice means the joining of two parts of a column whereas a column base transfers forces and moments at the lower end of a column to a foundation. The common thing between these is the use of bearing plates since to splice columns of different sizes bearing plates are used as is also the case with column bases.

#### 9.2 Column Splices

In practice, the splicing of columns is to be done to get the required height. In multistorey buildings, it is very convenient to splice the column just above the floor and it is economical to provide a splice near the inflection point in the column. The parts of the column to be spliced may be of the same size or of a different size. The splicing may be done using welding or bolting.

For splicing columns using welding, it is economical to directly butt weld the parts of the column. The ends of the column to be spliced are milled to have perfect plane surfaces. To keep the parts of the column in position i.e. for alignment, erection angles are shop welded. The outstanding legs of erection angles have holes to connect them for erection purposes. Before doing butt welding at the site, the outstanding legs of splice angles are connected by bolts as shown in Figure 9.1. For full penetration butt weld, the stresses in the weld are the same as in the column.

Figure 9.1(a) shows a column splice of the same column section whereas Figure 9.1(b) shows a column splice for different column sections. To connect the parts of column of different sizes, a bearing plate is used. The plate is shop butt welded to the lower portion of the column whereas the upper portion is butt welded in the field to the bearing plate.

In the case of column splices with bolting, splice plates are to be used on the flanges as well as on the web as shown in Figure 9.2. Flange splice plates and their connections to the column are designed to resist an axial compression and the bending moment. Splice plates on the web and their connections with the web of the column are designed for the shear force. To splice parts of a column of different size, a bearing plate and filler plates are used as shown in Figure 9.2(b).

When the ends of a column to be spliced are not milled, the connections are to be designed for all forces to which they may be subjected to. If the ends of the column are milled, the axial load directly transfers through the bearing and the splice material and the connections are to be designed to keep the parts in position and to resist any other forces like the shear force and the bending moment. It is a usual practice to design the splice for 50% of the axial load if it is only the force acting on the column when the ends of the column are milled for complete bearing.

**Example 9.1**

Design a splice for a beam-column using high strength bolts of the property class 8.8.

The factored axial force = 750 kN

Factored bending moment = 150 kNm

Factored shear force = 75 kN

The section of the column is SC 250. Assume that the ends of the column are milled and the connections are of bearing type. *f*_{y} = 250 MPa.

As the ends of the column are milled, the splice is designed for 50% of the axial load. Let the thickness of the splice plates on the flanges be 10 mm.

The compressive force in each flange splice plate due to axial load

The compressive force in each splice plate due to bending moment

The total compressive force in each flange splice plate = 187.5 + 577 = 764.5 kN

*f*=

_{cd}*f*

_{y}/γ

_{m0}= 250/1.1 = 227 MPa

The required area of cross-section of each splice plate

The width of the splice plate = Width of flange of SC 250 = 250 mm

The thickness of flange splice plate

The thickness of the flange splice plate may be kept equal to or more than the thickness of the flange of the column section. Hence, 250 mm × 18 mm splice plates may be provided.

Using M16 bolts

*k*_{b} is minimum of

Assuming *e* = 30 mm, *p* = 60 mm, *k*_{b} = 0.55

*V*_{nsb} = 2.5 *k*_{b} *d* *t* *f*_{u} = 2.5 × 0.55 × 16 × 17 × 410 = 1,53,340 N

*V*_{dpb} = 15,53,340/1.25 = 1,22,672 N

The least design strength of the bolt = 74,293 N = 74 kN

The number of bolts needed =

Twelve bolts may be provided to connect each flange as shown in Figure 9.4(a).

To transfer the shear force by the splice, another set of splice plates are provided on the web. The required shearing area of these splice plates is given by

A 120 mm wide and 6mm thick plate may be provided on either side of web. Bolts connecting these plates and the web are, therefore, subjected to double shear.

The number of bolts required

Two bolts may be provided as shown in Figure 9.4(b).

**Example 9.2**

HB 350 @ 67.4 kg/m is to be spliced to HB 450 @ 87.2 kg/m using a butt weld. The factored axial force is 500 kN and the factored bending moment where the column is spliced is 100 kNm.

Since two different cross-sections of the column are to be spliced, a bearing plate of size of 250 mm × 450 mm is used shown in Figure 9.5(a). It is assumed that the axial load is resisted by the entire area of the cross-section of the column HB 350 @ 67.4 kg/m whereas only its flanges resist the bending moment.

The stress in the column HB 350 due to the axial load

(*P* = axial load, *A* = sectional area of HB 350)

The stress in the flanges of HB 350 due to the bending moment

(*M* = bending moment, *b _{f}* = width of the flange,

*t*= thickness of the flange,

_{f}*d*= centre to centre distance between the flanges = 350 – 11.6 = 338.4 mm)

The resultant stress in the left flange = −58 + 102 = 44 MPa

The resultant stress in the right flange = 58 + 102 = 160 MPa

Considering a longitudinal strip of a 1 mm wide plate, the forces acting through the flanges of HB 350 are as follows (Figure 9.5(b))

Force at *C* = 11.6 × 1.0 × 44 = 510 N

Force at *D* = 11.6 × 1.0 × 160 = 1,856 N

The reactions from the flanges of HB 450, viz. *R _{A}* and

*R*are calculated from the equilibrium equations.

_{B}

*F*= 0 ⇒

_{y}*R*+

_{A}*R*+ 510 –1,856 = 0

_{B}*M*= 0 ⇒

_{A}*R*× 436.3 + 510 × 49 – 1,856 × 387.3 = 0

_{B}*R*= – 244N and

_{A}*R*= 1,590N

_{B}

The maximum bending moment is at *D*, i.e., *M _{D}* = 1,590 × 49 = 77,910 Nmm

or

*t*= 37.5 mm

Therefore, a bearing plate of the thickness of 40 mm may be provided and the two parts of the column may be butt welded as shown in Figure 9.5(a).

#### 9.3 Column Bases

A column base consists of a horizontal steel plate which is provided at the bottom most end of a column and placed on a concrete pedestal. The column base ensures that the bearing pressure between the column end and the concrete pedestal is within the permissible limit so that no crushing of concrete occurs. The base plate is usually butt welded to the column and is connected to the concrete pedestal using anchor bolts. The simplest of column bases is a slab base which may be directly butt welded to the column as shown in Figure 9.6. The slab base is connected by 2 anchor bolts in the case of an axially loaded column (Figure 9.6(a)) and 4 anchor bolts in the case of a beam-column (Figure 9.6(b)).

For columns subjected to heavy axial loads with or without a bending moment, the slab base becomes uneconomical since the required thickness of the slab base is very high. In such cases, it is economical to use a gusseted base as shown in Figure 9.7. In gusseted bases, vertical plates known as gussets are provided which are connected to the base plate usually by butt welds. These gusset plates stiffen the base plate thereby reducing the bending of the base plates. These bases may be used for axially loaded columns and beam-columns. In the case of axially loaded columns, anchor bolts (*A*) may be provided in the middle of the base plate whereas in the case of beam-columns, anchor bolts (*B*) are provided as shown in Figure 9.7.

#### 9.3.1 Design of Column Bases as per IS800:2007

Column bases should be designed such that they transfer the axial load, the bending moment and the shear force, usually, to the concrete pedestal. The shear resistance between the column base and the concrete pedestal may be calculated by taking the coefficient of friction as 0.45. If this calculated value of the shear resistance is less than the factored shear force, shear keys have to be provided. The maximum bearing pressure below the base plate should not exceed the bearing strength of concrete, i.e. 0.6 *f*_{ck} where *f*_{ck} is the characteristic compressive strength of concrete at 28 days.

For an axially loaded column, the variation of the bearing pressure below a base plate may be assumed as uniform. The thickness of the rectangular slab base *t*_{s} supporting axially loaded columns is given by

where *w* = actual uniform bearing pressure

*a* = larger projection of the slab base

*b* = smaller projection of the slab base

In no case should the thickness of the slab base be less than the thickness of the flange of the column section.

**Example 9.3**

Design a slab base for a column section SC 200 which carries a factored axial compression of 1000 kN. The grade of the steel is E250 and the grade of the concrete pedestal is M20.

The bearing strength of the concrete = 0.6 *f** _{ck}* = 0.6 × 20 = 12 MPa

The required area of the slab base =

A 300 mm × 300 mm slab base may be provided as shown in Figure 9.8.

*a*=

*b*= 50 mm

A 15 mm thick base plate may be provided.

The slab base is directly connected to the column section using a full penetration butt weld.

**Example 9.4**

Design a slab base for a beam-column SC 250 to transfer a factored axial compression of 750 kN and a factored bending moment of 75 kNm. The grade of the steel is E250 and the grade of the concrete pedestal is *M*_{30}.

Eccentricity, *e* = 75 / 750 = 0.1 m

The length of the base plate is kept equal to or more than 6*e* so that the entire plate is subjected to downward pressure and no tension develops in the anchor bolts.

∴ The length of the base plate, *L* = 6 × 0.1 = 0.6 m

The bearing strength of the concrete = 0.6 *f*_{ck} = 0.6 × 30 = 18 MPa

It is assumed that the bearing pressure varies linearly below the base plate.

The maximum bearing pressure,

or

*B*= 139 mm

The width of the base plate to be provided

= the width of the flange of SC 250 + the projections on either side

= 250 + 2 × 100 = 450 mm

Therefore, a rectangular base plate of 600 mm × 450 mm as shown in Figure 9.9 may be provided.

For this base plate,

The variation bearing pressure is shown in Figure 9.9.

The maximum bending moment in the base plate at Sec. X–X

or *t* = 41 mm

A base plate of 42 mm thick may be provided. The column section may be directly welded to the base plate by a full penetration butt weld. Four anchor bolts may be provided to keep the beam-column in position as shown in Figure 9.9.

**Example 9.5**

Design a gusseted base for the data in Example 9.4. The grade of the steel is E250 and the grade of the concrete pedestal is *M*_{30}.

The same size of base plate is used as in Example 9.4, i.e. 600 mm × 450 mm. Gusset plates are provided as shown in Figure 9.10. The column section and gusset plates are connected to the base plate by full penetration butt welds.

The vertical shear force in the gusset plate at Sec. X–X

= upward force acting on the hatched area as shown in Figure 9.10(a)

The bending moment in the gusset in the vertical plane at Sec. X-X

The shear capacity of a gusset plate =

As the shear force in the gusset plate 185 kN is less than 0.6 × 367 = 220 kN, the moment capacity of the gusset plate is not reduced due to the combined action of the shear force and the bending moment.

The vertical cross section of gusset plate is considered as semi-compact.

The moment capacity of the gusset plate = *β*_{b}*Z _{p}*

*f*

*/*

_{y}*γ*

_{m0}

Consider a 1 mm width of the base plate along A–A as shown in Figure 9.11.

The bending moment at *B* in the base plate

The bending moment at *C* in the base plate

The moment capacity of the base plate per unit width = 1.2*Z _{e}*

*f*

*/*

_{y}*γ*

_{m}_{0}

*t*

^{2}= 24,131

or

*t*= 23 mm

Hence, the thickness of the base plate may be 24 mm.

#### Problems

*For the following problems, consider the grade of the steel as E250*

- A WPB 200 × 200 × 74.01 column is subjected to a factored axial force of 500 kN. The effective length of the column is 6 m. Design a splice for this column using bolts. Assume that the ends of column are milled for complete bearing.
- A column is to be reduced from HB 350 to HB 250. The factored axial load is 600 kN. The effective length of the column is 5 m. Design the splice using a butt weld.
- Design a splice using bolts for a beam-column 5 m high subjected to a factored axial load of 600 kN at an eccentricity of 125 mm along the minor axis. Assume that the ends of the beam-column are milled for complete bearing. The sections of the beam-column is HB 400.
- Design a slab base for a built-up column consisting of 2MC 250 placed back to back separated by a distance of 160 mm. The factored axial load on the column is 1200 kN.
- Design a slab base for a beam-column SC 250 subjected to a factored axial compression of 750 kN and a factored bending moment of 75 kNm about the major axis.
- Design a gusseted base for the data in Problem 9.4.
- Design a gusseted base for the data in Problem 9.5.