5.3 column design

7/30/2019 5.3 Column Design

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5.3.Reinforced Concrete Column DesignColumns are vertical members loaded in compression and bending at the same time.

Columns from structures must be designed to fail due to bending moments. If plastic hinges

occur at the end of one column, the structure as a whole can still carry vertical loads and

general failure appears. In case of failure due to compression, general failure occurs because

all the members above the column fail.

5.3.1. Concrete Cover to reinforcementEurocode 2 has special requirements for calculating of the minimum concrete cover.

The minimum cover, Cmin should be provided in order to ensure:

o the safe transmission of bond forces;o the protection of the steel against corrosion;o an adequate fire resistance.

The nominal cover should be specified on the drawings, and it is defined as the

minimum cover, plus an allowance in design for deviation .

The greater value of cmin satisfying both environmental and bond conditions should be

used:

, where:o

- minimum cover due to bond requirement;

o - minimum cover due to environmental conditions;o - additive safety element;o - reduction of minimum cover for use of stainless steel;

o - reduction of minimum cover for use of additional protection5.3.1.1. Concrete cover of the transversal reinforcement

We impose the stirrups diametersw=10 mm.


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In case of stirrups the minimum concrete cover is:

5.3.1.2. Concrete cover of the longitudinal reinforcement

5.3.2. Design compressive and tensile strengthsThe concrete class chosen for the girder, according to the exposure class of the

element XC1 is C20/25.

The value of the design compressive strength is defined as:

fcd=cc*fck/c=1*20/1.5=13.33 N/mm2, where:

o c - partial safety factor for concreteo cc - coefficient taking into account of long term effects on the compressive strength

and of unfavorable effects resulting from the way the load is applied. This value is

defined in the National Annex of each country. The recommended value is 1.0.

The value of the design tensile strength is defined as:

fctd=ct*fck,0.05/c=1*1.5/1.5=1 N/mm2, where:

o c - partial safety factor for concreteo ct - coefficient taking into account of long term effects on the tensile strength and of

unfavorable effects resulting from the way the load is applied. This value is defined in

the National Annex of each country. The recommended value is 1.0.

5.3.3. PC52 (S345) Reinforcement propertiesThe value of the design yield strength is defined as:

fyd=fyk/s=345/1.15=300 N/mm2, where:

o fykcharacteristic yield strength;o fyddesign yield strength;o spartial safety factor for steel.


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5.3.4. The useful height of the transversal section

5.3.5. Minimum and Maximum reinforcement areasThe minimum area of longitudinal reinforcement in columns is given by:

, where:o

Cross sectional area of concrete;

o Ultimate axial load;In Eurocode 2 the maximum nominal reinforcement area for columns is 4%.

However, this area can be increased provided that the concrete can be placed and compacted

sufficiently.

From the above conditions:

The diameter of the transverse reinforcement should not be less than 6 mm or 1/4 of

the maximum diameter of the longitudinal bars.

5.3.6. Minimum spacing of reinforcementThe minimum clear distance between bars should be the greater of:

o Bar diameter;o Aggregate size(16 mm) plus 5mm;o 20 mm;


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5.3.7. Critical lengthAccording to P100, in order to prevent plastic hinges that occur in the column, it is

necessary to have a smaller spacing between stirrups near the nodes, on the critical length.

At the first 2 stories, the critical length is increased with 50%: 5.3.8. Maximum spacing of transverse reinforcementThe maximum spacing of the stirrups in columns, according to Eurocode 2, should

not be greater than:

o 12 times the minimum diameter of the longitudinal bars;o 60% of the lesser dimension of the column;o 240mm

No longitudinal bar should be further than 150 mm from transverse reinforcement in

the compression zone.

According to P100/2006, the spacing between the stirrups should not exceed:

o 33% of the lesser dimension of the column;o 125 mmo 7 times the minimum diameter of the longitudinal bars;

At the first two levels of buildings with more than 5 stories and at the first story in

case of smaller buildings, the spacing will be smaller besides the critical zone, with 50% of

its length.


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5.3.9. Design of the Column B1The maximum bending moments in the columns and design bending moments in the

girders reduced at the column`s edge were extracted from ETABS.

Frame structures are more safety if the plastic hinges appear first at the ends of the

girders. To assure this requirement, weaker columns must be avoided and transversal

reinforcement must be assured on the column to prevent the failure in the node .

5.3.9.1. Maximum Bending MomentsThe bending moments MULS and axial forces NED were determined using ETABS.

Fig.5.3.1. Maximum Bending Moments on X-direction


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Fig.5.3.2 Maximum Bending Moments on Y-direction


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Table.5.3.1. Maximum Bending Moments in ULS3 (Earthquake on X-direction)

Table.5.3.2. Maximum Bending Moments in ULS4 (Earthquake on Y-direction)

Story Column Loc P V2 M3

STORY1 B1 0 -1482.98 83.63 260.219

STORY1 B1 3 -1494.99 -79.45 -19.286

STORY2 B1 0 -1304.79 95.18 153.86

STORY2 B1 2.35 -1319.43 -85.35 -71.159

STORY3 B1 0 -1144.42 91.04 113.177

STORY3 B1 2.35 -1152.36 -80.96 -102.782

STORY4 B1 0 -983.97 86.3 91.433

STORY4 B1 2.35 -985.09 -74.71 -114.228

STORY5 B1 0 -823.31 78.28 72.824

STORY5 B1 2.35 -817.88 -65.69 -114.853

STORY6 B1 0 -662.33 68.33 55.809

STORY6 B1 2.35 -650.9 -54.98 -109.186STORY7 B1 0 -501.02 56.91 39.483

STORY7 B1 2.35 -484.2 -42.54 -99.565

STORY8 B1 0 -310.54 42 65.891

STORY8 B1 2.35 -317.83 -28.88 -81.658

TERRACE B2 0 -177.35 32.25 18.217

TERRACE B3 2.35 -151.71 -11.73 -66.271

Story Column Loc P V3 M2

STORY1 B1 0 -1796.07 -72.24 -241.656

STORY1 B1 3 -1181.9 52.31 55.504

STORY2 B1 0 -1581.8 -71.65 -128.73

STORY2 B1 2.35 -1042.43 30.1 43.38

STORY3 B1 0 -1376.65 -66.65 -89.364

STORY3 B1 2.35 -920.13 29.84 70.828

STORY4 B1 0 -1169.92 -63.52 -70.688

STORY4 B1 2.35 -799.14 24.13 83.721

STORY5 B1 0 -964.95 -59.41 -57.864

STORY5 B1 2.35 -676.24 19.5 88.764

STORY6 B1 0 -763.72 -54.45 -48.157

STORY6 B1 2.35 -549.51 14.15 88.994

STORY7 B1 0 -567.57 -49.07 -41.307

STORY7 B1 2.35 -417.65 6.91 86.968

STORY8 B1 0 -377.41 -39.75 -37.912

STORY8 B1 2.35 -279.75 2.59 75.691

TERRACE B1 0 -193.54 -37.41 -45.331

TERRACE B1 2.35 -135.52 -19.46 70.9


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5.3.9.2. Design Bending Moments Computation

Frame structures are more safety if the plastic hinges appear first at the ends of the

girders. To assure this requirement, the design bending moment are computed with the

formula:

, for Marginal Columns;

, for Central Columns, where:o overstrength factor due to steel consolidation effect;o maximum bending moment on the column in ULS;o capable moment of the girder in the node i;o

- designbending moment of the girder in the node i.


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Story

kN*m kN*m kN*m kN*m

STORY1 241.656 - - - 241.66

STORY2 128.73 100.374 126.48 1.3 210.87

STORY3 89.364 117.08 126.48 1.2 115.85STORY4 70.688 120.805 126.48 1.2 88.81

STORY5 57.864 118.143 126.48 1.2 74.34

STORY6 48.157 111.597 126.48 1.2 65.50

STORY7 41.307 102.531 126.48 1.2 61.15

STORY8 37.912 91.674 126.48 1.2 62.77

Table.5.3.3. Design Bending Moments on the column (Y-direction)

Story

kN*m kN*m kN*m kN*m kN*m kN*m

STORY1 260.219 - - - - - 260.219

STORY2 153.86 66.518 73.58 83.45 126.48 1.3 299.717

STORY3 113.177 68.387 83 83.45 126.48 1.2 188.333

STORY4 91.433 62.883 83.35 83.45 126.48 1.2 157.512

STORY5 72.824 53.152 78.1 83.45 126.48 1.3 151.421

STORY6 55.809 40.777 69.84 83.45 126.48 1.2 127.098

STORY7 39.483 26.859 59.71 83.45 126.48 1.2 114.896

STORY8 65.891 12.66 43.83 83.45 126.48 1.2 93.840

Table.5.3.3. Design Bending Moments on the column (X-direction)

5.3.9.3. Longitudinal Reinforcement Computation

additional eccentricity


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If

, the concrete does not crack and the column is reinforced from

the minimum reinforcement percentage.

Table.5.3.4 Longitudinal Reinforcement of the column

Story

x

Chosen

reinforcement

kNm kN mm mm mm mm mm2

mm2

mm2

STORY1 260.219 1531.74 23.33 193.21 498.21 164.12 -4800 1666 320+222 1702.7

STORY2 299.717 1348.22 23.33 245.64 550.64 144.46 -4298 1666 320+222 1702.7

STORY3 188.333 1181.15 23.33 182.78 487.78 126.55 -3822 1666 320+222 1702.7

STORY4 157.512 1013.88 23.33 178.69 483.69 108.63 -3331 1666 320+222 1702.7

STORY5 151.421 846.67 23.33 202.17 507.17 90.72 -2823 1666 320+222 1702.7

STORY6 127.098 679.68 23.33 210.33 515.33 72.82 -2299 1666 320+222 1702.7

STORY7 114.896 512.99 23.33 247.30 552.30 54.96 -1761 1666 320+222 1702.7

STORY8 93.84 346.62 23.33 294.06 599.06 37.14 -1207 1666 320+222 1702.7

Because of the small eccentricity of the axial force, the concrete does not crack and

can support alone the eccentric compression. Therefore, the minimum reinforcement

percentage is placed from constructive reason.

The capable bending moment of the column is computed using the formula:


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5.3.9.4.

Biaxial bending

The effects of biaxial bending may be checked using the expression:

, where:o - Design moment in the respective direction including second order effects in

a column

o

- Moment of resistance in the respective direction

o the capable axial force;o the design axial force;o , a1.098

5.3.9.5. Design shear force computation

Eurocode 2 introduces the strut inclination method for shear capacity checks. In this

method the shear is resisted by concrete struts acting in compression and shear reinforcement

acting in tension.


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Having the capable moments, in the bearing and in the field, we determine the shear

force corresponding to the capable bending moments, from static equilibrium on the girder.

, where:

o o

5.3.9.6. Longitudinal Reinforcement ComputationThe design capable shear force without transversal reinforcement:

* + , where:o

o o

where d is in mm;

o longitudinal reinforcement coefficient;o for normal concrete;

o o * +

( ) , where:o for normal concrete; ( )


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If VEd>Vrd,c The computation of transversal reinforcement must be done so that:

The capable shear force of stirrups reinforced elements is the minimum value given

by:

, where:

o transversal reinforcement area (forn=4 branches of the stirrup and d=10mm);o design strength of the transversal

reinforcement;

o .9*d=0.9*655=589.5 mm;o strength reduction factor of cracked concrete due to shear;

We impose: s=100mm

Minimum shear reinforcement ratio:

For the Ground Floor we impose s=75 mm


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Resulting maximum distance between stirrups:

o For Story 1-8:On the critical length , we provide 10/100 mm

stirrups with 4 branches.

On the rest of the column, we provide 10/125 mm stirrups with 4 branches.

o For Ground Floor:We provide 10/75 mm stirrups with 4 branches.

5.3.9.7. Horizontal design shear force in the nodes

The design shear force in the nodes is computed using the following formula: for Edge nodes; for Central nodes, where:

o - overstrength factor;o

the real areas of reinforcement from the superior and inferior part of

the girders;

o design shear force from the column under the node;


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Table.5.3.5.Horizontal design shear force in the nodes (on X and Y direction)

Story

On X-direction On Y-direction

kN mm2 mm2 kN mm2 kNSTORY8 1.2 552.9 616 1017.9 35.304 616 -331.14

STORY7 1.2 552.9 616 1017.9 35.304 616 -331.14

STORY6 1.2 552.9 616 1017.9 35.304 616 -331.14

STORY5 1.2 552.9 616 1017.9 35.304 616 -331.14

STORY4 1.2 552.9 616 1017.9 35.304 616 -331.14

STORY3 1.2 552.9 616 1017.9 35.304 616 -331.14

STORY2 1.2 552.9 616 1017.9 35.304 616 -331.14

STORY1 1.3 651.6 616 1017.9 -14.379 616 -411.36

5.3.9.8. Horizontal design shear force checkingFor Central nodes:

For Edge nodes:

, where:o ;o - normalized axial force in the column above;o the width of the

node;


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Table.5.3.6. Horizontal capable shear

forces in the nodes (on X and Y direction)

Story kN kN kN kN kN

STORY1 1531.74 0.2345 3735.1 2988.1 -14.379 -411.36STORY2 1348.22 0.2064 3668.5 2934.8 35.304 -331.14

STORY3 1181.15 0.1808 3606.8 2885.4 35.304 -331.14

STORY4 1013.88 0.1552 3543.9 2835.1 35.304 -331.14

STORY5 846.67 0.1296 3479.9 2783.9 35.304 -331.14

STORY6 679.68 0.1040 3414.8 2731.8 35.304 -331.14

STORY7 512.99 0.0785 3348.6 2624.9 35.304 -331.14

STORY8 346.62 0.0531 3281.1 2624.9 35.304 -331.14

5.3.9.9. Transversal reinforcement checking in the nodesFor Central nodes:

For Edge nodes:

, where:

o normalized axial force of the inferior column;o the real areas of reinforcement from the superior and inferior part ofthe girders;

o -total area of horizontalstirrups in the node;

o area of a branch of a stirrup;o

number of horizontal stirrups in the node;

o - number of bars


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Table.5.3.7.Transversal reinforcement checking in the nodes (on X and Y direction)

Story On X direction On Y direction

mm

2

mm

2

kN mm

2

kN kNSTORY8 0.2345 616 1017.9 118.5851 616 120.1104 192.265

STORY7 0.2064 616 1017.9 127.3974 616 123.4328 192.265

STORY6 0.1808 616 1017.9 135.4197 616 126.4573 192.265

STORY5 0.1552 616 1017.9 143.4517 616 129.4854 192.265

STORY4 0.1296 616 1017.9 151.4807 616 132.5125 192.265

STORY3 0.1040 616 1017.9 159.4992 616 135.5355 192.265

STORY2 0.0785 616 1017.9 167.5033 616 138.5532 192.265

STORY1 0.053055 616 1017.9 175.4921 616 141.565 192.265

The condition is checked for the nodes at all levels.

5.3.9.10. Longitudinal reinforcement checking in the node , where:

o vertical longitudinalreinforcement passing through the node, including the longitudinal reinforcement

of the column;

o total area of horizontal stirrups in the nodeo - interaxial distance between the

edge reinforcements of the columns;

o - interaxial distance betweenthe reinforcement from the superior and inferior part of the girders;

5.3.9.11. Anchorage Lengths

, where:o - coefficients considering the bars shape, the concrete cover and

confinement

o


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o o o

o o o o o

- minimum anchorage length for compressed bars

o , We impose 5.3.9.12. Overlapping Lengths

The scarfed bar proportion is greater than 50%, therefore: .

5.3 column design

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