Columns lecture#3

2

Compartir

Próximo SlideShare
Columns lecture#2
Cargando en…3
×
1 de 8
1 de 8

Columns lecture#3

2

Compartir

Más de Irfan Malik

Gratis con una prueba de 14 días de Scribd

Ver todo

Gratis con una prueba de 14 días de Scribd

Ver todo

Columns lecture#3

1. 1. Engr. Ayaz Waseem ( Lecturer/Lab Engr., CED) Columns LECTURE # 3 1. TRIAL SIZE OF COLUMN: Approximate formula; i. For tied columns, yc yxu trialg ff MMP A 008.0'43.0 22 )( + ++ ≥ ii. For spiral columns, yc yxu trialg ff MMP A 01.0'50.0 22 )( + ++ ≥ 2. CONCENTRICALLY LOADED SHORT COLUMN: Column is said to be a short column if it satisfies the following condition; 2 1 1234 M M r lK u −≤ …………………………………………... (1) For rectangular section, r = 0.3 hmin where, hmin is the least column dimension. And for concentrically loaded column, k = 0.75 and 0 2 1 = M M Putting these values in equation (1), 34 3.0 75.0 min ≤ h lu 6.13 min ≤ h lu or 6.13≤ DimensionColumnLeast Length ……………. (1) Now, for circular section, r = 0.25 D where, D is the column diameter. Putting these values in equation (1), 34 25.0 75.0 ≤ D lu 33.11≤ D lu or 33.11≤ DiameterColumn Length …………………… (2) 1
2. 2. Engr. Ayaz Waseem ( Lecturer/Lab Engr., CED) Columns If above conditions in equation (1) and (2) are satisfied then the column is said to be a short column. 3. DESIGN PROBLEM: An axially loaded short column has a length of 4m and carries a factored load (Pu) of 2000 kN. Other data is as follows; fc’ = 28 MPa fy = 420 MPa Maximum aggregate size = 20 mm Using US Customary Bars design the column as; i. Square tied column. ii. Circular spiral column. Solution: i. Design of Square Tied Column: STEP 1: Area Calculations: yc yxu trialg ff MMP A 008.0'43.0 22 )( + ++ = ( ) ( ) 2 )( 13.129870 420008.02843.0 0)10002000( mmA trialg = + +× = Trial size of column = ( )361361 × mm Say, trial size of column is, ( )375375 × mm STEP 2: Check for the type of column: 6.137.10 375 4000 <== DimensionColumnLeast Length Therefore, it is a short column. STEP 3: Reinforcement calculations: For a tied column, Pu = φPn = ( 0.65 x 0.8 ) [ 0.85 fc’Ag + ( fy – 0.85 fc’ )Ast ] 2
3. 3. Engr. Ayaz Waseem ( Lecturer/Lab Engr., CED) Columns ( 2000 x 1000 ) = (0.65 x 0.8) ( ) ( )[ ]{ }stA2885.04203752885.0 2 ×−+×× Ast = 1260.2 mm2 STEP 4: Reinforcement Ratio check: ρ = 2 375 2.1260 = hb Ast = 0.009 < 0.01 ( Not Satisfied ) Therefore, ρ = ρmin = 1% = 0.01 So, Ast = 0.01 x (375)2 ≈ 1407 mm2 Selecting, 8 # 16 Bars STEP 5: Design of ties: • As maximum diameter of the longitudinal bars is < 32 mm, so, Diameter of the ties = 10 mm • Spacing of the ties = Minimum of; a. 16 x 16 = 352 mm. b. 48 x 10 = 480 mm. c. Least column dimension = 375 mm. d. 300 mm. ( Governs ) Therefore, Spacing of the ties, S = 300 mm. Note: In this step shape of the ties is not given. In step 6, shape of the ties is decided during detailing. 3
4. 4. Engr. Ayaz Waseem ( Lecturer/Lab Engr., CED) Columns STEP 6: Detailing: ii. Design of Circular Spiral Column: STEP 1: Area Calculations: yc yxu trialg ff MMP A 01.0'50.0 22 )( + ++ = ( ) ( ) 2 )( 11.109890 42001.02850.0 0)10002000( mmA trialg = + +× = Trial diameter of column, D = 374 mm Say, trial diameter of column is, D = 375 mm STEP 2: Check for the type of column: 33.117.10 375 4000 <== ColumnofDiameter Length Therefore, it is a short column. STEP 3: Reinforcement calculations: For a spiral column, Pu = φPn = ( 0.7 x 0.85 ) [ 0.85 fc’Ag + ( fy – 0.85 fc’ )Ast ] 4
5. 5. Engr. Ayaz Waseem ( Lecturer/Lab Engr., CED) Columns ( 2000 x 1000 ) = (0.7 x 0.85) ( )( ) ( )[ ]{ }stA2885.0420375 4 2885.0 2 ×−+×× π Ast = 1849.4 mm2 STEP 4: Reinforcement Ratio check: ρ = 22 375 4 4.1849 4       =       ππ D Ast = 0.0167 > 0.01 ( Satisfied ) Therefore, Ast ≈ 1850 mm2 Selecting, 4 # 16 , 4 # 19 Bars STEP 5: Design of Spiral: • Diameter of the spiral = dsp = 10 mm • Minimum pitch of the spiral = Larger of; a. 1.5 x 20 = 30 mm. ( Governs ) b. 25 mm. Therefore, Smin = 30 mm. • Smax =       −1'45.0 2 c g cc ysp A A fD fdπ . Now, Dc = 375 – ( 40 x 2 ) = 295 mm Ac = ( )22 295 44 ππ =cD = 68349.2 mm2 Ag = ( )22 375 44 ππ =D = 110446.6 mm2 5
6. 6. Engr. Ayaz Waseem ( Lecturer/Lab Engr., CED) Columns Therefore, Smax =       −1'45.0 2 c g cc ysp A A fD fdπ = 57.63 mm < 75 mm ( O.K. ) • Selecting pitch between Smin and Smax , S = 45 mm STEP 6: Detailing: 6
7. 7. Engr. Ayaz Waseem ( Lecturer/Lab Engr., CED) Columns 4. INTERACTION CURVE: It is a graph between load and moment. In order to draw an interaction curve we require the coordinates of different points. Fig. Interaction Curve 5. PLASTIC CENTROID: It is a point through which the resultant of all the internal forces should pass, in pure axially loaded column case, when no moment is acting on the column at ultimate stage of failure. 7
8. 8. ySySyScc fAfAfAAfR 321'85.0 +++= Engr. Ayaz Waseem ( Lecturer/Lab Engr., CED) Columns 6. SYMMETRICAL SECTIONS: Those sections in which plastic centroid coincide with the geometrical centroid are called symmetrical sections. 7. UNSYMMETRICAL SECTIONS: Those sections in which plastic centroid do not coincide with the geometrical centroid are called unsymmetrical sections. Plastic centroid of unsymmetrical sections can be found out in the following manner; Fig. Section unsymmetrical about x-axis R = Cc + F1 + F2 + F3 Lets suppose plastic centroid is at a distance ‘ y ’ from the bottom face. Taking moment about bottom face, R y =       2 h Cc + F1 L1+ F2 L2+ F3 L3 y = 321 332211 2 FFFC LFLFLF h C c c +++ +++      ………………………….. (1) Equation (1) is used to find out location of plastic centroid. 8