More Related Content
What's hot
What's hot (20)
Similar to Design of short circular axially loaded column
Similar to Design of short circular axially loaded column (20)
Recently uploaded
Recently uploaded (20)
Design of short circular axially loaded column
- 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 𝑃𝑢 = 0.4𝑓𝑐𝑘 𝐴 𝑐 + 0.67𝑓𝑦 𝐴 𝑠𝑐 ..(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 𝑃𝑢 = 1.05 0.4𝑓𝑐𝑘 𝐴 𝑐 + 0.67𝑓𝑦 𝐴 𝑠𝑐 . . (𝑖𝑖) Continuously wound spiral reinforcement is provided as tie. Continuous winding increases capacity by 5%.
- 4. Governing Equations 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 ℎ𝑒𝑙𝑖𝑐𝑎𝑙 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑜𝑟𝑒 ≮ 0.36( 𝐴 𝑔 𝐴 𝑐 − 1)𝑓𝑐𝑘 𝑓𝑦 𝐴 𝑔 = 𝑔𝑟𝑜𝑠𝑠 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑐𝑡𝑖𝑜𝑛, 𝐴 𝑐 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 ℎ𝑒𝑙𝑖𝑐𝑎𝑙𝑙𝑦 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑 𝑐𝑜𝑙𝑢𝑚𝑛 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 ℎ𝑒𝑙𝑖𝑥 𝑓𝑐𝑘 = 𝑐ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑣𝑒 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑓𝑦 = 𝑐ℎ𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 ℎ𝑒𝑙𝑖𝑐𝑎𝑙 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡 𝑏𝑢𝑡 𝑛𝑜𝑡 𝑒𝑥𝑐𝑒𝑒𝑑𝑖𝑛𝑔 415𝑁/𝑚𝑚2
- 5. Volume of helical reinforcement in one loop = π 𝐷𝑐 − φ 𝑠𝑝 𝑎 𝑠𝑝 Volume of core = π 4 𝐷𝑐 2 𝑝 𝐷𝑐 = diameter of the core φ 𝑠𝑝= diameter of the spiral reinforcement 𝑎 𝑠𝑝 = area of cross-section of spiral reinforcement 𝑝 = pitch of spiral reinforcement Source: IITK
- 6. π 𝐷𝑐 − φ 𝑠𝑝 𝑎 𝑠𝑝 π 4 𝐷𝑐 2 𝑝 ≮ 0.36( 𝐴 𝑔 𝐴 𝑐 − 1)𝑓𝑐𝑘 𝑓𝑦 Pitch of helix 𝑝≤11.1(𝐷𝑐 - φ 𝑠𝑝) 𝑎 𝑠𝑝 𝑓𝑦/(𝐷2 -𝐷𝑐 2) 𝑓𝑐𝑘……(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 1/6th 𝐷𝑐 The pitch of the helical reinforcement shall not be less than 25 mm 3φ 𝑠𝑝
- 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 𝑒 𝑚𝑖𝑛 = 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, 𝑒 𝑚𝑖𝑛 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 𝑃𝑢 = 1.05(0.4𝑓𝑐𝑘 𝐴 𝑐 + 0.67𝑓𝑦 𝐴 𝑠𝑐) 𝐴 𝑔 = π 4 × 4002 = 125663.71 𝑚𝑚2 𝐴 𝑐 = 𝐴 𝑔 – 𝐴 𝑠𝑐 = 125663.71 - 𝐴 𝑠𝑐 Substituting the values of 𝑃𝑢, 𝑓𝑐𝑘, 𝐴 𝑔 and 𝑓𝑦 1.5 × 1500 × 103 = 1.05 0.4 × 25 × 125663.71 − 𝐴 𝑠𝑐 + 0.67 × 415𝐴 𝑠𝑐 𝐴 𝑠𝑐 = 3306.17𝑚𝑚2 Provide 12 nos. Of 20 mm diameter bars (= 3768𝑚𝑚2) 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 𝑚𝑚, 𝐷𝑐= 400-40-40=320 𝑚𝑚; 𝐴 𝑐= π 4 × 3202 = 80424.77 𝑚𝑚2
- 11. 𝑓𝑐𝑘= 25 N/𝑚𝑚2 𝑓𝑦= 415 N/𝑚𝑚2 φ 𝑠𝑝=8 mm; 𝑎 𝑠𝑝 =50.24 𝑚𝑚2 Substituting in equation (iii) 𝑝≤50 mm 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 π 𝐷 𝑐−φ 𝑠𝑝 𝑎 𝑠𝑝 π 4 𝐷 𝑐 2 𝑝 = 2.787 = 0.0245 0.36( 𝐴 𝑔 𝐴 𝑐 −1)𝑓 𝑐𝑘 𝑓𝑦 = 0.0122 ∴ π 𝐷 𝑐−φ 𝑠𝑝 𝑎 𝑠𝑝 π 4 𝐷 𝑐 2 𝑝 ≮ 0.36( 𝐴 𝑔 𝐴 𝑐 −1)𝑓 𝑐𝑘 𝑓𝑦 check is satisfied.
- 13. THANK YOU