Design of short circular axially loaded column

Design of circular column as per IS 456:2000

1. Design of short circular axially loaded column Nadya Baracho Assistant Professor Don Bosco College of Engineering

2. Governing Equations For short axially loaded columns with LATERAL ties CLAUSE 39.3, IS 456:2000, Pg.71 ..( i ) Individual loops are provided as ties.  

3. Governing Equations For short axially loaded columns with HELICAL TIES CLAUSE 39.4, IS 456:2000, Pg.71 Continuously wound spiral reinforcement is provided as tie. Continuous winding increases capacity by 5%.  

4. Governing Equations  

5. Volume of helical reinforcement in one loop = Volume of core = = diameter of the core = diameter of the spiral reinforcement = area of cross-section of spiral reinforcement = pitch of spiral reinforcement   Source: IITK

6. Pitch of helix - ) /( - ) ……(iii)   Source: IITK

7. DIA & PITCH OF HELICAL REINFORCEMENT CLAUSE 26.5.3.2(c)2, IS 456:2000, Pg.49 Diameter of the tie should not be less than ¼ (dia. Of largest longitudinal bar) 6 mm CLAUSE 26.5.3.2(d)1, IS 456:2000, Pg.49 The pitch of the helical reinforcement shall not be more than 75 mm 1/6 th The pitch of the helical reinforcement shall not be less than 25 mm 3  

8. SOLVED EXAMPLE Design a circular column of 400 mm diameter with helical reinforcement subjected to an axial load of 1500 kN under service load and live load. The column has an unsupported length of 3 m effectively held in position at both ends but not restrained against rotation. Use M 25 concrete and Fe 415 steel . Step 1: To check the slenderness ratio Given data are: unsupported length l = 3000 mm, D = 400 mm. Table 28 of Annex E of IS 456 gives effective length le = l = 3000 mm. Therefore , le/D = 7.5 < 12 confirms that it is a short column. Step 2: To check the slenderness ratio Given data are: Minimum eccentricity = Greater of (l/500 + D/30) or 20 mm = 20 mm 0.05 D = 0.05(400) = 20 mm   Source: IITK

9. As per cl.39.3 of IS 456, should not exceed 0.05D to employ the equation given in that clause for the design. Here, both the eccentricities are the same. Step 3: Area of steel = – = 125663.71 - Substituting the values of , , and Provide 12 nos. Of 20 mm diameter bars ) as longitudinal reinforcement giving p=3% which is between 0.8%(minimum) and 4% (maximum). Hence o.k.  

10. Step 4: Ties Diameter of helical reinforcement (cl.26.5.3.2 d-2) shall be not less than greater of ( i ) one-fourth of the diameter of largest longitudinal bar, and (ii) 6 mm. Use 8 mm tie. Assuming a clear cover of 40 , = 400-40-40=320 ; =  

11. = 25 N/ = 415 N/ =8 mm; =50.24 Substituting in equation (iii) As per cl.26.5.3.2 d-1, the maximum pitch is the lesser of 75 mm and 320/6 = 53.34 mm and the minimum pitch is lesser of 25 mm and 3(6) = 18 mm. We adopt pitch = 25 mm which is within the range of 24 mm and 53.34 mm. So, provide 8 mm bars @ 25 mm pitch forming the helix.  

12. Step 5: Checking of cl. 39.4.1 of IS 456 = 2.787 = 0.0245 = 0.0122 ∴ check is satisfied.  

13. THANK YOU