Design of short columns using helical reinforcement

B.Tech. 3rd Year Civil
RCC 1

1. Design of short columns using helical reinforcement By- Shivam Gautam B.Tech. 3 rd Year Civil 1700668

2. Helical Reinforcement The main longitudinal reinforcement bars are enclosed within closely spaced and continuously wound spiral reinforcement. Circular and octagonal columns are mostly of this type Helical reinforcement is also termed as spiral reinforcement which is used only in the circular column. In case of circular column, number of minimum reinforcement should not be less than 6 bars. The diameter of the reinforcement should not be less than 12 mm and maximum distance between the longitudinal bars should not be greater than 300 mm. Spiral reinforcement is shown in the below figure

3. Advantages of Helical Reinforcement Several important structures collapsed due to stirrups opening when subjected to important seismic actions. This risk is minimized in the case of using spiral stirrups, since it consists of only one wire as transversal reinforcement, throughout the entire length of the element. The circular section concrete columns (or drilled piles) with spiral transversal reinforcement are easier to produce, require a shorter time to assemble, and when subjected to lateral loads the failure by stirrup opening is not an option. These advantages could be obtained for usual rectangular section by using the rectangular spiral reinforcement.

4. The use of  links  for column design is very popular. However, engineers tend to use  helical reinforcement  instead of normal links because helical reinforcement has the potential advantage of protecting columns/piles against seismic loads. Moreover, when the columns reach the failure state, the concrete outside hoops cracks and falls off firstly, followed by the eventual failure of the whole columns. The peeling off of concrete outside helical reinforcement provides a warning signal before the sudden failure of columns. In addition, it can take up a higher working load than normal link reinforcement. For instance, helical reinforcement is adopted in the design of marine piles in Government piers. Why is Helical Reinforcement sometimes designed instead of normal links ? 

5. Design short columns using Helical Reinforcement

6. Assumptions: ( i ) Plane sections normal to the axis remain plane after bending. (ii) The maximum strain in concrete at the outer most compression fibre is taken as 0.0035 in bending. (iii) The acceptable stress-strain curve of concrete is assumed to be parabolic. (iv) The tensile strength of concrete is ignored. (v) The design stresses of the reinforcement are derived from the representative stress-strain curves from Figs. 23A and B of IS 456:2000, for the type of steel used using the partial safety factor γm as 1.15. (vi) The maximum compressive strain in concrete in axial compression is taken as 0.002. (vii) The maximum compressive strain at the highly compressed extreme fibre in concrete subjected to axial compression and bending and when there is no tension on the section shall be 0.0035 minus 0.75 times the strain at the least compressed extreme fibre .

7. Slenderness Limits: The code (Clause 25.3.1) specifies that the ratio of the unsupported length (l) to the least lateral dimension (d) of a column should not exceed a value of 60: Furthermore, in case one end of a column is free in any given plane, the code specifies (Clause 25.3.1) specifies that Where, D- depth of X-section measured in the plane of the cantilever. b- width (in perpendicular direction)  

8. Minimum Eccentricity: In practical construction, columns are rarely truly concentric. Even a theoretical column loaded axially will have accidental eccentricity due to inaccuracy in construction or variation of materials etc. Accordingly, all axially loaded columns should be designed considering the minimum eccentricity as stipulated in cl. 25.4 of IS 456 as given below – e x min ≥ greater of: {l/500 + D/30) or 20 mm } e y min ≥ greater of: {(l/500 + b/30) or 20 mm } where l-the unsupported length, D- larger lateral dimension b-least lateral dimension

9. (a) Pitch: Helical reinforcement shall be of regular formation with the turns of the helix spaced evenly and its ends shall be anchored properly by providing one and a half extra turns of the spiral bar. The pitch of helical reinforcement shall be determined as- The maximum pitch of transverse reinforcement shall be the least of the following: ( i ) the least lateral dimension of the compression members; (ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied; and (iii) 300 mm. The above criteria is valid for all cases except where an increased load on the column is allowed for on the strength of the helical reinforcement.

10. In such cases only, the maximum pitch shall be the lesser of - ( i ) 75 mm and (ii) one-sixth of the core diameter of the column, and the minimum pitch shall be the lesser of ( i ) 25 mm and (ii) three times the diameter of the steel bar forming the helix. (b) Diameter: The diameter of the polygonal links or lateral ties shall be not less than- one-fourth of the diameter of the largest longitudinal bar, and in no case less than 6 mm.

11. Governing Equation of Short Axially Loaded Columns with Helical Ties

12. Columns with helical reinforcement take more load than that of tied columns due to additional strength of spirals in contributing to the strength of columns. Accordingly, cl. 39.4 recommends a multiplying factor of 1.05 regarding the strength of such columns. The code further recommends that the ratio of volume of helical reinforcement to the volume of core shall not be less than 0.36 ( A g /A c – 1) ( f ck / f y ), in order to apply the additional strength factor of 1.05 (cl. 39.4.1).

13. Accordingly, the governing equation of the spiral columns may be written as- P u = 1.05(0.4 f ck A c + 0.67 f y A sc ) where, P u = factored axial load on the member, f ck = characteristic compressive strength of the concrete, A c = area of concrete, f y = characteristic strength of the compression reinforcement, and A sc = area of longitudinal reinforcement for columns The above equation, given in cl. 39.3 of IS 456, has two unknowns Ac and Asc to be determined from one equation. The equation is recast in terms of Ag , the gross area of concrete and p , the percentage of compression reinforcement employing- A sc = pA g /100 ...(10.5) A c = A g (1 – p/ 100) ...(10.6)

14. Earlier observations of several investigators reveal that the effect of containing holds good in the elastic stage only and it gets lost when spirals reach the yield point. Again, spirals become fully effective after spalling off the concrete cover over the spirals due to excessive deformation. Accordingly, the two points should be considered in the design of such columns. ( i ) the enhanced load carrying capacity taken into account by the multiplying factor of 1.05. (ii) maintaining specified ratio of volume of helical reinforcement to the volume of core, as specified in cl.39.4.1 and mentioned earlier.

15. The second point, in fact, determines the pitch p of the helical reinforcement- Volume of helical reinforcement in one loop = …(10.9) Volume of core = … (10.10) where D c = diameter of the core (Fig.10.21.2b) φ sp = diameter of the spiral reinforcement (Fig.10.21.2b) a sp = area of cross-section of spiral reinforcement p = pitch of spiral reinforcement (Fig.10.21.2b) To satisfy the condition of cl.39.4.1 of IS 456, we have – )  

16. which finally gives- … (10.11) Thus, Eqs.10.8 and 11 are the governing equations to determine the diameter of column, pitch of spiral and area of longitudinal reinforcement. It is worth mentioning that the pitch p of the spiral reinforcement, if determined from Eq.10.11, automatically satisfies the stipulation of cl.39.4.1 of IS 456. However, the pitch and diameter of the spiral reinforcement should also satisfy cl. 26.5.3.2 of IS 456:2000.  

17. Numerical 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.

18. Soln. 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 l e = l = 3000 mm. Therefore, l e /D = 7.5 < 12 confirms that it is a short column.

19. Step 2: Minimum eccentricity e min = Greater of (l/500 + D/30) or 20 mm = 20 mm 0.05 D = 0.05(400) = 20 mm 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. So, we can use the equation given in that clause of IS 456 i.e., Eq.10.8 for the design.

20. Step 3: Area of steel From Eq.10.8, we have P u = 1.05(0.4 f ck Ac + 0.67 f y A sc ) … (10.8) A c = A g – A sc = 125714.29 – A sc Substituting the values of Pu, f ck , Ag and f y in Eq.10.8, 1.5(1500)(103) = 1.05{0.4(25)(125714.29 – A sc ) + 0.67(415) A sc } we get the value of A sc = 3304.29 mm 2 . Provide 11 nos. of 20 mm diameter bars (= 3455 mm2) as longitudinal reinforcement giving p = 2.75%. This p is between 0.8 (min.) and 4 (max.) %. Hence o.k.

21. Step 4: Lateral ties It has been mentioned in sec.10.22.4 that the pitch p of the helix determined from Eq.10.11 automatically takes care of the cl.39.4.1 of IS 456. Therefore, the pitch is calculated from Eq.10.11 selecting the diameter of helical reinforcement from cl.26.5.3.2 d-2 of IS 456. However, automatic satisfaction of cl.39.4.1 of IS 456 is also checked here for confirmation. 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. Therefore, with 20 mm diameter bars as longitudinal reinforcement, the diameter of helical reinforcement = 6 mm.

22. From Eq.10.11, we have Pitch of helix p ≤ 11.1(D c - φ sp ) a sp f y /(D 2 – ) f ck Where, Dc = 400 – 40 – 40 = 320 mm, φ = 6mm, a sp = 28 mm 2 , f y = 415 N/mm 2 , D = 400 mm, f ck = 25 N/mm 2 . So, p ≤ 11.1(320 – 6) (28) (415)/(400 2 – 320 2 ) (25) ≤ 28.125 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 18 mm and 53.34 mm. So, provide 6 mm bars @ 25 mm pitch forming the helix.  

23. Checking of cl. 39.4.1 of IS 456 The values of helical reinforcement and core in one loop are obtained from Eqs.10.8 and 9, respectively. Substituting the values of D c , φ sp , a sp and pitch p in the above two equations, we have Volume of helical reinforcement in one loop = 27632 mm 3 and Volume of core in one loop = 2011428.571 mm 3 . Their ratio = 27632/2011428.571 = 0.0137375 0.36(A g /A c – 1) ( f ck / f y ) = 0.012198795 It is, thus, seen that the above ratio (0.0137375) is not less than 0.36(A g /A c – 1) ( f ck / f y ).

24. Hence, the circular column of diameter 400 mm has eleven longitudinal bars of 20 mm diameter and 6 mm diameter helix with pitch p = 25 mm. The reinforcing bars are shown in above Fig.

25. Thank You

26. References https://www.chegg.com/homework-help/definitions/design-of-spiral-reinforcement-8 https://thecivilengineeringdaily.wordpress.com/2015/02/17/for-column-reinforcements-why-is-helical-reinforcement-sometimes-designed-instead-of-normal-links/ https://nptel.ac.in/content/storage2/courses/105105104/pdf/m10l21.pdf https://nptel.ac.in/content/storage2/courses/105105104/pdf/m10l22.pdf https://theconstructor.org/structural-engg/design-of-axially-loaded-column/4699/ https://www.quora.com/For-column-reinforcements-why-is-helical-reinforcement-sometimes-designed-instead-of-normal-links https://encyclopedia2.thefreedictionary.com/Helical+reinforcement https://thecivilengineeringdaily.wordpress.com/2015/02/17/for-column-reinforcements-why-is-helical-reinforcement-sometimes-designed-instead-of-normal-links/ https://www.engineeringcivil.com/for-column-reinforcements-why-is-helical-reinforcement-sometimes-designed-instead-of-normal-links.html