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Design of short circular axially
loaded column
Nadya Baracho
Assistant Professor
Don Bosco College of Engineering
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.
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%.
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
 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
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
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
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
 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.
 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
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.
 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.
THANK YOU

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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.