-Page 1- November 2006

Topic: Reinforced Concrete (II)- Design in RC Beam and slab-2; Design in Column. Shear force in concrete beam In the previous notes, the bending force in the beam has been discussed. However, when the beam in loaded, it is also subjected to the shear force. Since shear is usually associated with tensile stress and since also concrete is not strong in tension, additional reinforcement would also be required to resist the effect of the shear force. (1) Shear stress in the concrete Shear stress in a concrete beam is determined as:

dbVv⋅

=

where V is the shear force at the section under consideration, v is the shear stress b is the width of the beam d is the effective depth of the beam Concrete has some shear resisting capacity, cv , and this capacity is not a constant value; rather it is a function of amount of bending reinforcement used and the effective depth of the beam as shown in the table below: (Table 1)

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Table 1:- Values of cv , design concrete shear stress (for 2/25 mmNfcu = )

(2) Shear Reinforcement Shear reinforcement in beam is usually provided in the form of link (or stirrup in US terms). Once the size (diameter) of the link is determined, its spacing vs can be determined as below:-

-Page 3- November 2006

( )c

yvsvv vvb

fAs

−

⋅=

87.0

Shear resistance of the beam is:

dbvfsA

bdv cyvv

sv ⋅⎟⎟⎠

⎞⎜⎜⎝

⎛+=⋅ 87.0

where V is total shear resistance stress of the beam Asv is the area of shear reinforcement sv is the spacing of shear reinforcement fyv is the strength of shear reinforcement vc Is the shear resistance strength of concrete b is the width of beam d is the effective depth of beam In most case, links are made of mild steel (fyv=250N/mm2) mainly because the amount of bending required to form a link. Links are also used at locations where the shear stress due to the applied load is less than shear resistance of the concrete; it is used to tie the reinforcing steel together. On the other hand, it should be noted that maximum shear stress in a beam can not

exceed cuf8.0 or 5N/mm2 regardless of shear reinforcement is provided. If shear

stress is greater than this value, the size of beam has to be increased. The shear reinforcement in BS8110 is as shown in Table 2.

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Table 2:- Form and area of shear reinforcement in beams

Example:- Design for Shear Reinforcement (refer to Lecture 3 Example 2) Maximum Shear Stress at the support

23

/02.1310210106.66 mmN

xx

dbVv ==⋅

=

sA (steel area) near the support is 1890/2 = 945mm2

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Remarks

Design Concrete Shear Stress )( cv - (using the formula from Table 1)

( ) ( )( ) ( )[ ] [ ] 25.1/310/400310210/94510079.0

//400/10079.0

41

31

41

31

xx

ddbAv mvsc

=

= γ

2/74.0 mmNvc =→

Refer to Table 2,

( ) ( ) ( ) 221

21

21

2

2

/4258.08.08.0

/14.14.074.04.0

/37.0)74.0(5.05.0

mmNff

mmNv

mmNv

cucu

c

c

===

=+=+

==

4.05.0 +<< cc vvv

0.37 < 1.023 < 1.14

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From Table 2, minimum links should be provided for the whole length of the Beam And providing 8mm diameter TWO-Leg vertical Links or stirrups (Grade 250 steel)

(Area of links, ( ) 22 10084/2 mmAsy =⋅= π )

y

vvys f

sbA

87.04.0

>

Then spacing of the vertical links (or stirrups), vysyv bfAs 4.0/87.0<

< ( )( ) ( )2504.0/25087.0100 = 259mm

Or by table (extracted from BS8110)

However, spacing of Links should be < 0.75d = 0.75 x 330 = 247mm Then, for the shear reinforcement, providing 8mm diameter links at a spacing of 240mm for the whole length of the beam. Other issues concerned in reinforced concrete design (1) Bond It is very important that steel and concrete work together to form the material called reinforced concrete. This is achieved by the property call the bond between steel and concrete. One of the assumptions in reinforced concrete design is that perfect bond must exist between steel and concrete.

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(2) Deflection This provision in the BS code is to ensure the serviceability of the structure to be maintained. It is often found that a much shallower member may be used from strength point of view. However, to avoid excessive deflection, much deeper member could be required. The tabulated value has including also the creep and shrinkage effects of concrete under sustained load condition. Table 3:- Permissible value of span/depth ratio beams and slabs

(3) Minimum reinforcement areas Thermal and shrinkage cracking may be controlled within acceptable limits by use of minimum reinforcement quantities specified by BS8110 for general structures except water-retaining structures. The principle requirements for rectangular section are summarized as Table 4.

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Table 4:- Minimum Reinforcement Areas

(4) Cover Subject to fire resistance requirements or Code of Practise: Fire Resistance Construction 1995 (issued by Buildings Department, HKSAR), reinforcement should have concrete cover and the thickness of such cover (exclusive of plaster or other decorative finish) should be:

(a) for each end of a reinforcing bar, not les than 25mm nor less than 2 times the diameter of such bar;

(b) for a longitudinal reinforcing bar in a beam, not less than 25mm nor less than the diameter of such bar;

(c) for any other reinforcement not less than 15mm nor less than the diameter of such reinforcement.

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Table 5:-Fire resistance Requirement for Columns and Beams (Extract from COP fire resistance constriction 1995, issued by Buildings Department, HKSAR)

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(5) Distance between reinforcement bars The horizontal distance between two parallel steel reinforcements in reinforced concrete should usually, except at splices, be not less than the greatest of three following distances:- (a) the diameter of either bar if their diameter be equals; (b) the diameter of the larger bar if their diameters be unequal; (c) 5mm more than the nominal maximum size of the coarse aggregate used in the

concrete. A greater distance should be provided where convenient. Where immersion vibrators are intended to be used, however, the horizontal distance between bars of a group may be reduced to 2/3 of the nominal maximum size of the coarse aggregate provided that a sufficient space is left between groups of bars to enable the vibrator to be inserted; this would normally be a space of 75mm. The vertical distance between two horizontal main steel reinforcements, or the corresponding distance at right angles to two inclined main steel reinforcements, should not less than 15 mm or the nominal size of aggregate, whichever is the greater, except at splices or where one of such reinforcement is transverse to the other. (6) Maximum reinforcement areas An excessive amount of reinforcement usually indicates that a section is undersized and it may cause difficulty in steel fixing and concreting. Consequently, it results poor compaction of the concrete around reinforcement. The maximum reinforcement area specified in BS 8110 is summarized below:- Table 6:- Maximum Reinforcement Areas

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Reinforced Concrete Columns In most reinforced concrete buildings, the load on the building is eventually transmitted by columns to the building foundation. Concrete columns can have any shape. However, in most buildings, concrete column are either square or round or rectangular. Relative to design of beams, column design is more complex. Unlike beam, columns can be subjected to both compression and bending simultaneously which will make the analysis more difficult. In addition since columns are not supported along it height, columns could deflect sideways which in combining with the compressive load will cause additional bending. Since the deflection is a function of column height and dimension and the supporting condition at both column ends, load carrying capacity of two columns with the same reinforcement can be very different. Columns can be braced and unbraced A braced column in given direction is one where deflection at the ends of that column is complete eliminated by structural elements such as wall or bracing. An braced column is given direction is one where the ends of column can moved laterally with respect to other. therefore, in a building where the lateral force, such as wind load is resisted by walls or lift shaft, the column in that building are regarded as braced column. Column in Figure 1 are Unbraced column whereas those in Figure 2 and Figure 3 are Braced column.

Figure 1 Figure 2 Figure 3

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Long Column and Short column Columns of the same size could behave quite differently under the same loading if the conditions at the ends of column are different. Since columns are structural elements which support direct compression, their mode of failure could be different to other types of structural elements, such as beams. Column, especially the slender one (slender means long and thin) frequently failed by buckling rather than by material failure like beams. Long column will fail by buckling before material reaches it ultimate capacity. To define whether a column is a long column, the effective length of a column must be considered. Effective length The effective length, CL , of column can be defined as the height of a column that may be considered equal to the height of similar column with pin pinned ends. The effective length of columns with different end conditions is as shown in the Figure 4. below.

Figure 4

A column is classified as a short column if the ratio of effective length to the dimension of the column, in both directions, is less than 15 in a braced column and less than 10 in an unbraced column. Since the column in multi-storey buildings do not have exactly pinned or fixed ends, the end conditions are determined based on the number and size of beams connected to that column; the bigger the beams connecting to the column at the top and bottom,

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the more the ends like “fixed” and similarly for the number of beams. Since detailed discussion of column is beyond the scope of this module, only design braced short column will be discussed in this lecture. Design of short square columns – Short Braced Axially Loaded Column This type of column is very common in current precast concrete construction. The maximum a column can carry is:

scyccu AfxAxfN 75.04.0 +=

For rectangular column, the area of concrete can be allowed to be displaced by the longitudinal reinforcement, the above equation may be modified to:

( )cuysccu ffAbhfN 40.075.04.0 −+=

where N is ultimate axial load fcu is the characteristic strength of concrete Ac is the area of concrete Asc is the area of longitudinal steel fy is the characteristic strength of the steel B is the size of the column Reinforcement details The rule governing the minimum and maximum amount of reinforcement are as follow:- Longitudinal Steel: (a) a minimum of four bars is required for a rectangular column (and six bars for a

circular column)

(b) 4.0100

≥col

s

AA

where As is the total area of longitudinal steel and Acol is the

cross-section area of the column. Links: When the longitudinal bars in a column are subjected to compressive load, they tend to buckle. To prevent the buckling of main steel which could lead to premature failure, links are provided to tie these longitudinal bars according to the following rules:

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(a) minimum size should be one quarter of the largest bar but not less than 6mm; (b) maximum spacing equal to 12 times the smallest size of main bars (c) the links should be arranged so that every corner bar and alternate bar or group in

an outer layer of longitudinal steel is supported by a link passing round the bar and having an inclined angle of not greater than 135 degree;

(d) all other bars or groups not restrained by a link should be within 150mm of a restrained bar.

The arrangement of links in columns is shown in the following figures.

Figure 5

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Figure 6 Other considerations In the design of column, it should be bear in mind the sequence of constructing columns in the field. The space of bars should not be too close to allow room for concrete be placed and vibrated. The size of column and the cover thickness is also governed by the Code of Practise Fire Resistance Period. Example – Axially Loaded Column Design the longitudinal reinforcement for a 300mm square column which supports an axial load of 1700kN at the ultimate limit state. The characteristic material strength

2/460 mmNf y = for the reinforcement and 2/30 mmNfcu = for the concrete.

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Solution:

( )cuysccu ffAbhfN 40.075.04.0 −+=

( )304.046075.0300304.0101700 23 xxAxxx sc −+=

Therefore, ( ) 23

1892333

1010801700 mmAsc =⋅−

=

Provide four T25 bars, area = 1960mm2

-End-