# Design and Detailing of Column Base Plates

Column base plates are provided beneath steel columns in order to transmit the applied design forces safely to the foundations.  Steel columns are heavily loaded and their cross-sections are typically small. Applying the loads directly on the foundation could possibly result in a punching failure i.e. the column punching through the foundation. Thus, a steel plate must be provided beneath the column in order to spread the column load over a large base area.

This post concerns typical base plates encountered in braced multi-storey steel-framed building structures. It focuses on column base-plates that do not transmit bending moments (pinned bases). Two types of actions are normally transmitted to pinned column base plates, this includes axial forces and shear forces in certain instances.

The principles and procedures presented in this post are designed for readers working to the current BS EN 1993-1-18. However, readers working to other codes of practice will no doubt gain a good understanding of the underlying principles.

#### Detailing a Base Plate

The components and detailing of a base plate must be well understood before it can be designed. Typically a base plate consist of a single plate, fillet welded to a steel column and then attached to the foundation with four hold-down bolts. The bolts are normally cast into the foundations with anchor plates to avoid pull out. High strength grout is also poured in between the interface of the base plate and foundation.

Base plates are usually cut from S275 or S355 steel. Fillet welds of 8mm or 10mm are typically provided along the side of the flanges and a short distance either side of the web. Four hold-down bolts are typically provided, these are usually property class 8.8 with a diameter size of 20mm up to 40mm. The embedded length of the bolt is usually within the range of 16-18 times the bolt diameter with a 100mm allowance to allow tolerance.

##### Recommended Layout

As mentioned earlier in the post, pinned base-plates do not transmit bending moments. To achieve this, the base-plate must be detailed correctly. However, there is no base rule, but there are three main considerations that can be used to achieve the objective:

• The plate size must be of sufficient size to spread the loads from the column to the foundation and accomodate the anchor bolts
• The setting out dimentions for the anchor bolts should be on a regular and simple geometry
• The base plate and holding down system should be sufficiently robust to withstand loads experienced during construction. E.g. from wind, non verticallity etc.

In view of this considerations, column bases are made at least 100mm wide, round-about the steel column with a thickness greater than or equal to the flange of the column and having 4 holding down bolts.

#### Designing a Base Plate

Pinned bases typically only resist axial forces, however, there are instances where they’re also required to resist shear forces. Such actions are primarily resisted by friction between the base plates and the foundation.

The procedure for designing a column base plate is given in Clause 6.2.5  and 6.2.8.2(1) of BS-EN-1993-1-8. The underlying principle behind the design of base plates is that the applied stress does not exceed the bearing strength of the foundation which is typically concrete. These clauses describe how the applied stresses can be modelled using an equivalent area known as ” Equivalent T-stub in compression. Thus, this models the whole section of the steel column which exerts stress over a defined area based on the geometry of the column section (figure 2).

The procedure in designing a pinned base can be summarized as follows:

• Find the required are Areq
• Determine the effective area Aeff in terms of the projection width c
• By equating Areq and Aeff determine c
• Calculate the plate thickness assuming the projection width c is a uniformly loaded cantilever.
##### Required Area

A starting point in the design of a base plate is to determine the area required to spread the load safely on the foundation Areq and then provide a base area Ap greater or equal to the Areq. This can be expressed mathematically as:

{ A }_{ p }\ge { A }_{ req }
{ A }_{ p }={ h }_{ p }{ d }_{ p }(square\quad plate)\quad ;\quad =\frac { \pi { d }^{ 2 } }{ 4 } (circular\quad columns
{ A }_{ req }=\frac { { N }_{ Ed } }{ { f }_{ jd } }

Where : NEd is the design axial actions ; fjd is the concrete bearing stress

{ f }_{ jd }={ \beta }_{ i }\alpha { \alpha }_{ cc }\frac { { f }_{ ck } }{ { \gamma }_{ c } }

Where:

• βj is a factor usually taken equal to 0.67
• α is a coefficient of diffusion of the vertical load being applied to the foundation. Conservatively this can be taken as 1.5
• αcc is the coefficient that allows for long term effects on the compressive strength of concrete vs applied actions. Taken as 0.85 in the U.K National Annex
• γc is the partial factor of safety of concrete taken as 1.5 in the U.K National Annex.
• fck is the characteristics compressive strength of concrete
##### Effective Area

The magnitude of the effective area Aeff is dependent on the manner in which the steel column transmits actions to the steel base-plate. If the load from the column is concentric with the base plate, the whole column section applies the load and the effective area can be determined thus.

{ A }_{ eff }={ 4c }^{ 2 }+{ cP }_{ col }+{ A }_{ col }\quad (for \quad UKC\quad column)

Where: Pcol= perimeter of the column cross-section; Acol = is the cross-sectional area of the column section; t=thickness of hollow section; c= projection width

The equations given above assumes that there’s no overlap within the T stub area. However, this is not always the case. In order to confirm, the width of the flange T-stub has to be less than half the depth between flanges. Mathematically this can be expressed as:

c\le \frac { h-2{ t }_{ f } }{ 2 }

When there is an overlap within the T-sub areas, the effective area needs to be recalculated. This can be obtained from the following equations.

Where: h= depth of column section in the major axis; tf = is the thickness of the flange.

##### Plate Thickness

The thickness of the base plate is based on the assumption that there’s a bending stress under the column caused by bearing pressure. The magnitude of this bending stress assumes that the bearing pressure between the base plate and concrete foundation is the same. As a result, the thickness of the base plate can be determined on the same basis and expressed as:

{ t }_{ p }\ge c\times \sqrt { \frac { 3{ f }_{ jd }{ \gamma }_{ M0 } }{ { f }_{ yp } } }

Where: tf = is the design thickness of the base plate ; fyp = is the design strength of the steel base plate; γM0 = is the partial factor for cross-sectional resistance of steel elements taken as 1.0 in the UK National Annex.

##### Welds

As mentioned earlier any applied shear is resisted by the welds and friction between the base plate and the concrete. Thus, the basic requirement for the welds is as follows.

{ V }_{ Ed }\le { F }_{ V,Rd }\cdot { l }_{ w,eff }

Where: VEd= is the design shear force; FV,Rd = is the resistance of the fillet weld per unit length; lw,eff = is the effective length of weld in the direction of shear

{ F }_{ v,Rd }={ f }_{ vw,d }a
{ f }_{ vw,d }=\frac { { f }_{ u }/\sqrt { 3 } }{ { \beta }_{ w }{ \gamma }_{ M2 } }

Where: a= throat thickness = 0.7s; s= length of weld; βw = 0.85 for S275 and 0.9 for S355; γM2 is the material partial factor of safety given as 1.25 in the U.K National Annex; fu = is the ultimate strength of the steel plate: 410N/mm2 for S275 and 470 for S355 steel

#### Worked Example

A 305 × 305 × 198 UKC column in a multi-storey steel building carries a design axial action of 5200kN only. Design a base plate for this column in grade S275 steel. Assuming the base plate sits on a pile cap made from class C40/50 concrete.

##### Required Area

Try a base plate of (600× 600× 50mm) Ap = (360,000 mm2)

{ A }_{ req }=\frac { { N }_{ Ed } }{ { f }_{ jd } }
{ f }_{ jd }={ \beta }_{ i }\alpha { \alpha }_{ cc }\frac { { f }_{ ck } }{ { \gamma }_{ c } } =0.67\times 1.5\times 0.85\frac { 40 }{ 1.5 } =22.78Mpa
{ A }_{ req }=\frac { 5200\times { 10 }^{ 3 } }{ 22.78 } =228,270.4{ mm }^{ 2 }
\quad { A }_{ p }(360000{ mm }^{ 2 })>\quad { A }_{ req }(228270.4{ mm }^{ 2 })\quad o.k
##### Effective Area
{ A }_{ eff }=4{ c }^{ 2 }+c{ P }_{ col }+{ A }_{ col }
{ P }_{ col }=1938mm\quad ;\quad { A }_{ col }=25200{ mm }^{ 2 }
{ A }_{ eff }=4c^{ 2 }+1938c+25200={ A }_{ req }=228270.4
{ 4c }^{ 2 }+1938c-203070.4=0

Verify that there is no overlap

c<\frac { h-2{ t }_{ f } }{ 2 }
=\frac { 339.9-2(31.4) }{ 2 } =138.55>88.6\quad o.k

Verify that the effective area fits on the column base

h+2c=339.9+2(88.6)=517.7mm<600mm
b+2c=314.5+2(88.6)=491.7mm<600mm
##### Plate Thickness
{ t }_{ p,min }=c\sqrt { \frac { 3{ f }_{ jd }{ \gamma }_{ M0 } }{ { f }_{ yp } } } \quad { f }_{ yp }={ R }_{ eH }=255Mpa
=88.6\times \sqrt { \frac { 3\times 22.78\times 1.0 }{ 255 } } =45.86<50\quad o.k

Hence, provide a 600 × 600 × 50mm base plate with M24 property class 8.8 bolts and 10mm fillet welds. Note that the weld check has been ignored, this is because the column does not transmit shear forces to the base plate.

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## 50 Replies to “Design and Detailing of Column Base Plates”

1. Kunlex says:

Nice work Sir… Is the concrete footing a sort of concrete column being projected from the pad base?

1. Omotoriogun Victor says:

Thankyou for reading. The pad base is connected directly to the foundation, when carrying out the concreting of the pad base, the anchor bolts are carefully placed with an allowance of 6-10mm to allow for tolerance. After casting the foundation, the plate connected to the steel column can then be put in place.

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