Cấu tạo thép tòa nhà - P5

Every building consists of a load bearing frame, usually created by reinforced con-crete, or by steel, or by the combination of both areas with usual earthquake activity. Although the load bearing frame is not visible after the completion of the construc

3 The reinforcement of structural elements

3.1 Columns

The above column has an overall dimension of 30/30 which is the smallest dimensions that can be used for columns in an earthquake resistant building, according to common building regulations This column is rarely used nowadays but is selected because of its simplicity in order to explain the several different reinforcement options and charac-teristics This column is reinforced with four (4) steel bars and one simple stirrup The height of the column’s body is 2,70m and the height of the node area is 0,5m (it is equal to the height of the adjacent beam) The transverse reinforcement is common stirrups with Ø10 diameter The longitudinal reinforcement consists of four (4) steel bars with Ø20 diameter The covering of reinforcement is 2.5cm

In general, columns are the most important earthquake resistant structural elements in a building and the stirrups are the most important characteristic of a column The col-umn’s body is divided into the critical areas (due to the influence of earthquakes), the intermediate area and the node areas (which is the common area of the column and the adjacent beam)

COLUMN 30 x 30cm (1st reinforcement method)

In the first reinforcement method of the column we use stirrups to reinforce the mediate area The distance between the stirrups is 15cm and their diameter is Ø10

inter-We need 10 stirrups to cover the 1.50m height of this area Hence, the onomatology used for the stirrups will be 10Ø10/15

The critical areas are reinforced with stirrups Ø10/10 The total number of stirrups needed is 6 Hence the onomatology in the critical areas will be 6Ø10/10

In the node area we use Ø10 stirrups for every 10cm The total number of stirrups needed is 5 and the onomatology is 5Ø10/10

The tablet of the column’s reinforcement details contains all data required in order for the worker to successfully reinforce the column

COLUMN 30 x 30cm (2nd reinforcement method)

In this method we assume that the entire height of the column is considered critical and is reinforced with Ø10 stirrups for every 10cm For the total height of 2.70m we need 27 stirrups, thus the onomatology of the reinforcement will be 27Ø10/10

The node area is covered with 5 Ø10 stirrups for every 10cm This is displayed in the drawings as 5Ø10/10

3.1.1 Reinforcement bars splice

A multistory building would be ideally forced with continuous single steel bars for the total height of the building However, it is not possible to produce steel bars of such length Thus, the columns are reinforced separately for each floor and are then con-nected together

rein-In order for the steel bars of a column in a specific level to function and ‘cooperate’ sat-isfactory with the steel bars of the same col-umn in the lower or the above floor, it should be welded together However this is a time consuming and unpractical solution and is only implemented in special cases The most common solution that is usually selected is to simply splice the bars to-gether.

The steel bars of the column are fitted into their position with an extra length called ‘dowel length’ This length is equal or higher from the cooperation length of the continu-ing steel bars of two adjacent floors The length of cooperation is calculated as the diameter of the steel bar multiplied by a co-operation coefficient (This coefficient varies between 40 and 70) The cooperation coef-ficient is proportional to the steel’s strength and inversely proportional to the concrete’s strength

For example, when we have B500 steel and C20/25 concrete then a Ø20 steel bar needs a splice length of 67*2.0=134cm If we use a C25/30 concrete, then for a Ø20 steel bar, the coefficient is 58 and the bar will need a splice length of 58*2 = 116cm For a C30/37 concrete, the coefficient of cooperation is 51 and for a Ø20 steel bar, we will need a splice length of 51*2 = 102cm

If we use a Ø14 steel bar the appropriate splice lengths would be 67*1.4=94cm, 58*1.4=81cm και 51*1.4=71cm What is more, for a Ø25 steel bar the adequate splice lengths would 67*2.5=168cm, 58*2.5=145cm και 51*2.5=128cm

The following table analytically illustrates the adequate splice lengths for all concrete categories and for various bar diameters

It is fundamental to examine in detail the procedure of the appropriate splice tion The engineer should be aware of the fact that the stirrups will only function satis-factory when there is a steel bar firmly fitted in each of the stirrup’s corners Since it is difficult to achieve the above requirement in the connection areas of the steel bars, we should use the appropriate techniques to ensure that the stirrups will function as ex-pected If the splice connection is performed on site, then the technique illustrated in method A should be implemented The dowel bars are straight and the bars from the above floor must be bended in the connection area The bending length should be equal to one or two stirrups However, this can only be achieved with the use of the appropriate bending machine For bigger bar diameters (e.g Ø20 or Ø25) the on site bending of bars is extremely difficult or practically impossible and thus the bending ma-chine should always be used

connec-The bending of bars during the splice connection can be performed in either direction as illustrated in the following pictures

For columns that do not have antiseismic requirements it is more functional to use a lot of small diameter bars than a small number of large diameter steel bars in the perime-ter of the column However, for the columns which this book examines that need high earthquake resistance, it is more adequate to use steel bars only in the corners of the

stirrups in order to ensure high buckling strength Thus it is suggested to use a small number of steel bars with large diameter What is more, in earthquake resistant struc-tures that the necessary use of excessive reinforcement may cause congestion in the node areas of the structure, the application of a small number of large diameter bars is almost unavoidable in order to ensure the proper reinforcement of the structural ele-ments

As a result of the above, a 40/40 orthogonal column with stirrups implemented in a diamond layout (that means 4+4 corners) which requires 30.0 cm2 of reinforcement would be reinforced as 4Ø20+12Ø14 when using Ø20 steel bars (according to the ap-propriate reinforcement tables 4*3.14+12*1.54=12.56+18.48=31.04cm2) On the other hand, if we use Ø25 steel bars then the best possible combination would be 4Ø25+4 Ø20 which commensurate to 19.64+12.56=32.2cm2

Reinforcement with 16 bars

4Ø20+12Ø14

Reinforcement with 8 bars

(c) A crane is needed to move the bars into their position since the weight of a single 4.65m long Ø25 steel bar can be up to 18kg

The first requirement depends on the production from the concrete suppliers The ond and third requirements depend on the industry that produces and implements the steel reinforcement

sec-The usage of high strength concrete (e.g C25/30 or C30/37) is a solution easy to plement nowadays since:

im-(a) All major concrete suppliers have the means to produce concrete of high strength, (b) The need for less reinforcement in combination with the small price difference be-tween regular and high strength concrete results in a quite economical solution

(c) Finally, high strength concrete mixtures contain larger quantities of pure concrete, thus expanding the overall life span of the frame The life span of the frame is a signifi-cant parameter, especially for sea side buildings, which are buildings situated within a 1km range from the sea

Concluding, in most of the countries with advanced and sophisticated construction dustries, it is quite common to use concrete of even higher strength even for simplified structures

in-The industrialization of the construction process and the evolution in reinforcement plementation procedures have resulted in techniques such as pre-cast stirrup cages or even entire pre-cast columns These columns can then be placed in to their position with the use of cranes Both the pre-casting technology of reinforcement and the me-chanical implementation are advancing and evolving in parallel

im-The weight of the column examined in this chapter is 60kg (stirrups and steel bars) The weight of a 40/40 column with Ø10/10 stirrups placed in a ‘diamond’ arrangement with 8Ø20 steel bars is 150kg A 50/50 column with stirrups placed in a ‘cross’ ar-rangement and 12Ø20 steel bars, has a mass of 230kg

The reinforcement of a 40/40 column has a mass

of 150kg

The reinforcement of a 50/50 column has a mass

of 230kg

3.1.2 The Anchorage of reinforcement in the highest floor level

The reinforcement bars of the column should be properly anchored in the last floor level in order to function satisfactory

The most appropriate anchorage technique is the creation of a ‘hat’ above the column in order for the steel bars to be anchored without bending

The anchorage of steel bars in the higher floor with the use of an additional ‘hat’ is a simple and efficient solution

In cases that the construction of the ‘hat’ is not possible (e.g the flat roof will be used for specific purposes) then there are other anchorage solutions that can be imple-mented as illustrated below:

- (a) implementation of 180 degrees anchorage

- (b) Implementation of 90 degrees anchorage (method A)

- (c) Implementation of 90 degrees anchorage (method B)

- (d) Implementation of 90 or 180 degrees with additional ‘hairpins’

- (e) Usage of high strength concrete that might allow for the anchorage of bars side the node, for small diameter steel bars