Design of Column
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Column design |
Working Stress Method<
Slenderness ratio (λ)
![image039](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171628/image039.png)
If λ > 12 then the column is long.
Load carrying capacity for short column
![image040](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171632/image040.png)
where, AC = Area of concrete, ![image041](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171635/image041.png)
![image041](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171635/image041.png)
σSC Stress in compression steel
σCC Stress in concrete
Ag Total gross cross-sectional area
ASC Area of compression steel
Load carrying capacity for long column
![image046](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171647/image046.png)
where,
Cr = Reduction factor
![image047](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171651/image047.png)
where, leff = Effective length of column
B = Least lateral dimension
imin = Least radius of gyration and ![image048](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171654/image048.png)
![image048](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171654/image048.png)
where, l = Moment of inertia and A = Cross-sectional area
Effective length of column
Effective length of Compression Members
![image049](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171658/image049.png)
![image050](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171704/image050.png)
![image051](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171711/image051.png)
Column with helical reinforcement
Strength of the column is increased by 5%
![image052](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171715/image052.png)
![image053](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171719/image053.png)
Longitudinal reinforcement
(a) Minimum area of steel = 0.8% of the gross area of column
(b) Maximum area of steel
(i) When bars are not lapped Amax = 6% of the gross area of column
(ii) When bars are lapped Amax = 4% of the gross area of column
Minimum number of bars for reinforcement
For rectangular column 4
For circular column 6
Minimum diameter of bar = 12 mm
Maximum distance between longitudinal bar = 300 mm
Pedestal: It is a short length whose effective length is not more than 3 times of lest lateral dimension.
Transverse reinforcement (Ties)
![image054](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171722/image054.png)
where
dia of main logitudnal bar
![image055](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171724/image055.png)
φ = dia of bar for transverse reinforcement
Pitch (p)
![image057](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171731/image057.png)
where, φmin = minimum dia of main longitudinal bar
Helical reinforcement
(i) Diameters of helical reinforcement is selected such that
![image058](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171735/image058.png)
(ii) Pitch of helical reinforcement: (p)
![image059](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171737/image059.png)
where,
dC = Core diameter = dg – 2 × clear cover to helical reinforcement
AG = Gross area ![image060](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171741/image060.png)
![image060](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171741/image060.png)
dg = Gross diameter
Vh = Volume of helical reinforcement in unit length of column
φh = Diameter of steel bar forming the helix
![image062](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171747/image062.png)
![image063](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171749/image063.jpg)
dh = centre to centre dia of helix
= dg – 2 clear cover - φh
φh = diameter of the steel bar forming the helix
![image064](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171753/image064.jpg)
Some others IS recommendations
(a) Slenderness limit
- Unsupported length between end restrains
60 times least lateral dimension.
- If in any given plane one end of column is unrestrained than its unsupported lengthAll column should be designed for a minimum eccentricity of
![image067](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171801/image067.png)
Limit state method
- Slenderness ratio (λ)
if
λ<12 Short column
- Eccentricity
Ifthen it is a short axially loaded column.
where, Pu = axial load on the column - Short axially loaded column with helical reinforcement
- Some others IS code Recommendations
(a) Slenderness limit
(i) Unsupported length between end restrains
60 times least lateral dimension.
![image065](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171755/image065.png)
(ii) If in any given plane one end of column is unrestrained than its unsupported length
![image074](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171824/image074.png)
![image065](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171755/image065.png)
![image074](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171824/image074.png)
(b) All column should be designed for a minimum eccentricity of
![image075](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171826/image075.png)
Concentrically Loaded Columns
Where e = 0, i.e., the column is truly axially loaded.
![image076](https://gs-blog-images.grdp.co/gate-exam/wp-content/uploads/2016/10/06171828/image076.png)
This formula is also used for member subjected to combined axial load and bi- axial bending and also used when e > 0.05 D.
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