2.0 analysis and design

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2.0 ANALYSIS AND DESIGN 2.2 STRUCTURAL ELEMENT Reinforced Concrete Column Rearrangement by :- NOR AZAH BINTI AIZIZ KOLEJ MATRIKULASI TEKNIKAL KEDAH

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2.0 ANALYSIS AND DESIGN. 2.2 STRUCTURAL ELEMENT Reinforced Concrete Column. Rearrangement by :- NOR AZAH BINTI AIZIZ KOLEJ MATRIKULASI TEKNIKAL KEDAH. Column is defined as a structural members subjected to compressive force. Strut – a small compressive member in a framed structure. - PowerPoint PPT Presentation

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Page 1: 2.0 ANALYSIS AND DESIGN

2.0 ANALYSIS AND DESIGN

2.2 STRUCTURAL ELEMENTReinforced Concrete

Column

Rearrangement by :-NOR AZAH BINTI AIZIZ

KOLEJ MATRIKULASI TEKNIKAL KEDAH

Page 2: 2.0 ANALYSIS AND DESIGN

Column is defined as a structural members subjected to compressive force.

Strut – a small compressive member in a framed structure.

Column – a larger member, such as the main support for a beam in a building.

INTRODUCTION

Page 3: 2.0 ANALYSIS AND DESIGN

Axially loaded compressive member can fail in two principal ways:- short fat member fail by crushing or splitting of the material.- long thin members fail by sideways buckling.

INTRODUCTION

Page 4: 2.0 ANALYSIS AND DESIGN

i) Braced column(tiang dirembat)

– the lateral loads are resisted by wall

or some other form of bracing

ii) Unbraced column(tiang tidak dirembat)

– the lateral loads are resisted by the

bending of the column

iii) Short Column (tiang pendek)● lex/h and ley/b < 15 for a braced column● lex/h and ley/b < 10 for an unbraced column

iv) Slender(Long) Column ( tiang langsing)● lex/h @ ley/b >15 for a braced column● lex/h @ ley/b >10 for an unbraced column

Types of column

Page 5: 2.0 ANALYSIS AND DESIGN

Short column (p> pcr)usually fail by crushing when the material achieved its ultimate strength.

Slender column is liable to fail by buckling The end moment on slender column cause

it to deflect sideways exceed a criticalvalues until the column buckles.

Failure Mode

Page 6: 2.0 ANALYSIS AND DESIGN

Slender Column- fails by sideways buckling- The load at which a slender column buckles, is known as its critical buckling load, P crit.

P crit. = ΩEI / L2

Where; E = Modulus of Elasticity I = second moment of area L = length between pins

Failure Mode

Page 7: 2.0 ANALYSIS AND DESIGN

Column detail

Page 8: 2.0 ANALYSIS AND DESIGN

Longitudinal steel bar

- A minimum of 4 bars is required in a rectangular column and 6 bars in a circular column.- The size of the bars should be not less than 10mm

Links- Minimum size = ¼ x size of the largest compression bar

but not less than 6mm- Maximum spacing = 12 x size of the smallest

compression bar

Reinforcement detail

Page 9: 2.0 ANALYSIS AND DESIGN

Rectangular column

Circular

column

Types of Reinforced Concrete Column

Page 10: 2.0 ANALYSIS AND DESIGN

Load transfer on column for design

Page 11: 2.0 ANALYSIS AND DESIGN

Factored Axial Load (N) for Column Design

●Interior Column – 1.25●Edge/face column – 1.5●Corner column – 2

Corner column

Interior column

Edge /face column

Page 12: 2.0 ANALYSIS AND DESIGN

Short Axially Loaded Column Design

Equation

N = 0.4 fcu Aconc+0.8 fy Asc

WhereN – Ultimate Axial loadfcu – Characteristic of concrete strength

Aconc – area of concrete cross section

fy – Characteristic of reinforcement strength

Asc – area of reinforcement cross section

N

Page 13: 2.0 ANALYSIS AND DESIGN

Example:

A column with 200 mm x 200 mm resists a factored axial load 500kN. Calculate the area of steel required if fy = 460 N/mm2 and fcu = 30 N/mm2

Table 1 Diameters and areas of reinforcing bars

Bar dia.(mm) 6 8 10 12 16 20 25 32 40C/s area (mm2) 28 50 79 113 201 314 491 804 1256

Table 1 Diameters and areas of reinforcing bars

Bar dia.(mm) 6 8 10 12 16 20 25 32 40C/s area (mm2) 28 50 79 113 201 314 491 804 1256

Page 14: 2.0 ANALYSIS AND DESIGN

Formula for design column

N = 0.4 fcu Aconc + 0.8 fy Asc

500x103 = 0.4 (30) (40000 - Asc) + 0.8 (460) Asc

= 480000 – 12 Asc + 368 Asc

= 480000 + 356 Asc

Asc = 56.18mm2

Asc (for each bar) = 56.18mm2/4 = 14mm2

Asc = Πj2= ΠD2/4=14mm2

D = 14 x 4 / Π = 4.22mm

Standard required 4 number of steel reinforcement bars size 10mm minimum) . So size column = 4 T 10

4T10