Presentation on theme: "2.2 STRUCTURAL ELEMENT Reinforced Concrete Column"— Presentation transcript:
12.2 STRUCTURAL ELEMENT Reinforced Concrete Column 2.0 ANALYSIS AND DESIGN2.2 STRUCTURAL ELEMENTReinforced Concrete ColumnRearrangement by :-NOR AZAH BINTI AIZIZKOLEJ MATRIKULASI TEKNIKAL KEDAH
2INTRODUCTIONColumn 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.
3INTRODUCTIONAxially 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.
4Types of column i) Braced column(tiang dirembat) – the lateral loads are resisted by wallor some other form of bracingii) Unbraced column(tiang tidak dirembat)– the lateral loads are resisted by thebending of the columniii) Short Column (tiang pendek)lex/h and ley/b < 15 for a braced columnlex/h and ley/b < 10 for an unbraced columniv) Slender(Long) Column ( tiang langsing)ley/b >15 for a braced columnley/b >10 for an unbraced column
5Failure ModeShort column (p> pcr)usually fail by crushing when the material achieved its ultimate strength.Slender column is liable to fail by bucklingThe end moment on slender column causeit to deflect sideways exceed a criticalvalues until the column buckles.
6Failure Mode 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 / L2Where; E = Modulus of ElasticityI = second moment of areaL = length between pins
8Reinforcement detail 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 lessthan 10mmLinks- Minimum size = ¼ x size of the largest compression barbut not less than 6mm- Maximum spacing = 12 x size of the smallestcompression bar
9Types of ReinforcedConcrete ColumnCircularcolumnRectangular column
12Short Axially Loaded Column Design EquationN = 0.4 fcu Aconc+0.8 fy AscWhereN – Ultimate Axial loadfcu – Characteristic of concrete strengthAconc – area of concrete cross sectionfy – Characteristic of reinforcement strengthAsc – area of reinforcement cross sectionN
13Example:A column with 200 mm x 200 mm resists a factored axial load 500kN.Calculate the area of steel required iffy = 460 N/mm2 and fcu = 30 N/mm2Table 1Diameters and areas of reinforcing barsBar dia.(mm)C/s area (mm2)
14Formula for design column N = 0.4 fcu Aconc fy Asc500x103 = 0.4 (30) ( Asc) (460) Asc= – 12 Asc Asc= AscAsc = mm2Asc (for each bar) = 56.18mm2/4 = 14mm2Asc = Πj2= ΠD2/4=14mm2D = 14 x 4 / Π= 4.22mmStandard required 4 number of steel reinforcement bars size 10mm minimum) . So size column = 4 T 104T10