Home | ENGINEERING | Design Problem, Important Question And Answer: Civil - Limit State Design Of Columns

Civil - Design of Reinforced Concrete Elements - Limit State Design Of Columns

Design Problem, Important Question And Answer: Civil - Limit State Design Of Columns

   Posted On :  17.07.2016 10:38 am

Civil - Design of Reinforced Concrete Elements - Limit State Design Of Columns






Design Problems

 

1.Determine the load carrying capacity of a column of size 300 x 400 mm reinforced with six rods of 20 mm diameter i.e, 6-#20. The grade of concrete and steel are M20 and Fe 415 respectively. Assume that the column is short.

 

fck = 20 MPa, fy= 415 MPa

 

Area of steel ASC = 6 x ? x 202/4 = 6 x 314 = 1884 mm2 Percentage of steel = 100Asc/bD = 100x1884/300x400 = 1.57 % Area of concrete Ac = Ag – Asc = 300 x 400 – 1884 = 118116 mm2 Ultimate load carried by the column

Pu     = 0.4 fck Ac + 0.67 fy Asc

0.4x20x118116 + 0.67x415x1884

944928 + 523846 = 1468774 N = 1468. 8 kN

Therefore the safe load on the column = 1468.8 /1.5 = 979.2 kN

 

2.Determine the steel required to carry a load of 980kN on a rectangular column of size 300 x 400 mm. The grade of concrete and steel are M20 and Fe 415 respectively. Assume that the column is short.

fck = 20 MPa, fy= 415 MPa, P = 980 kN Area of steel ASC = ?

 

Area of concrete Ac = Ag – Asc = (300 x 400 –  ASC)

 

Ultimate load carried by the column

Pu     = 0.4 fck Ac + 0.67 fy Asc

980 x 1.5 x 1000 = 0.4x20x (300 x 400 –  ASC) + 0.67x415 ASC

= 960000 - 8 ASC + 278.06 ASC

ASC  =1888.5 mm2,

 

Percentage of steel = 100Asc/bD = 100x1888.5 /300x400 = 1.57 % which is more than 0.8% and less than 6% and therefore ok.

 

Use 20 mm dia. bas, No. of bars = 1888.5/314 = 6.01 say 6

 

3.Design a square or circular column to carry a working load of 980kN. The grade of concrete and steel are M20 and Fe 415 respectively. Assume that the column is short.

 

Let us assume 1.0% steel (1 to 2%)

 

Say ASC = 1.0% Ag =1/100 Ag = 0.01Ag fck = 20 MPa, fy= 415 MPa, P = 980 kN

 

Area of concrete Ac = Ag – Asc =  Ag -0.01Ag = 0.99 Ag

 

Ultimate load carried by the column Pu = 0.4 fck Ac + 0.67 fy Asc

 

980 x 1.5 x 1000 = 0.4x20x 0.99 Ag + 0.67x415 x 0.01Ag = 7.92 Ag + 2.78 Ag =10.7Ag

Ag  =  137383 mm2

 

Let us design  a square column:

B = D = ? Ag =370.6 mm say 375 x 375 mm

 

This is ok. However this size cannot take the minimum eccentricity of 20 mm as emin/D = 20/375 =0.053 > 0.05. To restrict the eccentricity to 20 mm, the required size is 400x 400 mm.

 

Area of steel required is Ag = 1373.8 mm2. Provide 4 bar of 22 mm diameter. Steel provided is 380 x 4 = 1520 mm2

 

Actual percentage of steel = 100Asc/bD = 100x1520 /400x400 = 0.95 % which is more than 0.8% and less than 6% and therefore ok.

 

Design of Transverse steel:

 

Diameter of tie = ผ diameter of main steel = 22/4 =5.5mm or 6 mm, whichever is greater. Provide 6 mm.

 

Spacing: < 300 mm, < 16 x22 = 352mm, < LLD = 400mm. Say 300mm c/c

 

Design of circular column:

 

Here Ag = 137383 mm2

? x D2/4 = Ag, D= 418.2 mm say 420 mm. This satisfy the minimum eccentricity of 20m Also provide 7 bars of 16 mm, 7 x 201 = 1407 mm2

 

 

Design of Transverse steel:

 

Dia of tie = ผ dia of main steel = 16/4 = 4 mm or 6 mm, whichever is greater. Provide 6 mm.

 

Spacing: < 300 mm, < 16 x16 = 256 mm, < LLD = 420mm. Say 250 mm c/c

 

4.Design a rectangular column to carry an ultimate load of 2500kN. The unsupported length of the column is 3m. The ends of the column are effectively held in position and also restrained against rotation. The grade of concrete and steel are M20 and Fe 415 respectively.

 

Given:

 

fck = 20 MPa, fy= 415 MPa,  Pu = 2500kN

 

Let us assume 1.0% steel (1 to 2%)

Say  ASC = 1.0% Ag =1/100 Ag = 0.01Ag

 

Area of concrete Ac = Ag – Asc =  Ag -0.01Ag = 0.99 Ag

 

Ultimate load carried by the column Pu = 0.4 fck Ac + 0.67 fy Asc

 

2500 x 1000 = 0.4x20x 0.99 Ag + 0.67x415 x 0.01Ag = 7.92 Ag + 2.78 Ag =10.7Ag

Ag  =  233645 mm2

 

If it is a square column:

 

B = D = ? Ag =483 mm. However provide rectangular column of size 425 x 550mm. The area provided=333750 mm2

 

Area of steel = 2336 mm2, Also provide 8 bars of 20 mm, 6 x 314 = 2512 mm2

 

 

Check for shortness: Ends are fixed. lex = ley = 0.65 l = 0.65 x 3000 = 1950 mm

 

lex /D= 1950/550 < 12, and  ley /b = 1950/425 < 12, Column is short

 

Check for minimum eccentricity:

 

In the direction of longer direction

 

emin, x = lux/500 + D/30 = 3000/500 + 550/30 = 24.22mm or 20mm whichever is greater.

 

emin, x =  24.22 mm < 0.05D = 0.05 x 550 =27.5 mm. O.K

 

In the direction of shorter direction

 

emin, y= luy/500 + b/30 = 3000/500 + 425/30 = 20.17 mm or 20mm whichever is greater.

 

emin, x =  20.17 mm < 0.05b = 0.05 x 425 =21.25 mm. O.K

 

Design of Transverse steel:

 

Dia of tie = ผ dia of main steel = 20/4 = 5 mm or 6 mm, whichever is greater. Provide 6 mm or 8 mm.

 

Spacing: < 300 mm, < 16 x20 = 320 mm, < LLD = 425mm. Say 300 mm c/c

 

5.Design a circular column with ties to carry an ultimate load of 2500kN. The unsupported length of the column is 3m. The ends of the column are effectively held in position but not against rotation. The grade of concrete and steel are M20 and Fe 415 respectively.

 

 

Given:

 

fck = 20 MPa, fy= 415 MPa,  Pu = 2500kN

 

Let us assume 1.0% steel (1 to 2%)

Say  ASC = 1.0% Ag =1/100 Ag = 0.01Ag

 

Area of concrete Ac = Ag – Asc =  Ag -0.01Ag = 0.99 Ag

 

Ultimate load carried by the column Pu = 0.4 fck Ac + 0.67 fy Asc

 

2500 x 1000 = 0.4x20x 0.99 Ag + 0.67x415 x 0.01Ag = 7.92 Ag + 2.78 Ag =10.7Ag

Ag  =  233645 mm2

? x D2/4  = Ag, D = 545.4 mm say 550 mm.

 

Area of steel = 2336 mm2, Also provide 8 bars of 20 mm, 6 x 314 = 2512 mm2

 

Check for shortness: Ends are hinged lex = ley =  l = 3000 mm

 

lex /D= 3000/550 < 12, and  ley /b = 3000/425 < 12, Column is short

 

Check for minimum eccentricity:

 

Here, emin, x = emin, y = lux/500 + D/30 = 3000/500 + 550/30 = 24.22mm or 20mm whichever is greater.

 

emin  =  24.22 mm < 0.05D = 0.05 x 550 =27.5 mm. O.K

 

 

Design of Transverse steel:

 

Diameter of tie = ผ dia of main steel = 20/4 = 5 mm or 6 mm, whichever is greater. Provide 6 mm or 8 mm.

 

Spacing: < 300 mm, < 16 x20 = 320 mm, < LLD = 550mm. Say 300 mm c/c

 

Similarly square column can be designed.

 

If the size of the column provided is less than that provided above, then the minimum eccentricity criteria are not satisfied. Then emin is more and the column is to be designed as uni axial bending case or bi axial bending case as the case may be. This situation arises when more steel is provided ( say 2% in this case).

 

 

Try to solve these problems by using SP 16 charts, though not mentioned in the syllabus.

 

6.Design the reinforcement in a column of size 450 mm ื 600 mm, subject to an axial load of 2000 kN under service dead and live loads. The column has an unsupported length of 3.0m and its ends are held in position but not in direction. Use M 20 concrete and Fe 415 steel.

 

 

Solution:

 

Given:  lu= 3000 mm, b = 450 mm, D = 600 mm, P =2000kN, M20, Fe415

 

 

Check for shortness: Ends are fixed. lex = ley = l =  3000 mm

 

lex /D= 3000/600 < 12, and  ley /b = 3000/450< 12, Column is short

 

Check for minimum eccentricity:

 

In the direction of longer direction

 

emin, x = lux/500 + D/30 = 3000/500 + 600/30 = 26 mm or 20mm whichever is greater. emin, x = 26 mm < 0.05D = 0.05 x 600 =30 mm. O.K

 

In the direction of shorter direction

 

emin, y= luy/500 + b/30 = 3000/500 + 450/30 = 21 mm or 20mm whichever is greater. emin, x = 21 mm < 0.05b = 0.05 x 450 =22.5 mm. O.K

 

Minimum eccentricities are within the limits and hence code formula for axially loaded short columns can be used.

 

Factored Load

 

P

= service load ื partial load factor

U

 

 

= 2000 ื 1.5 = 3000 kN

Design of Longitudinal Reinforcement

Pu

= 0.4 fck Ac + 0.67 fy Asc   or

Pu

= 0.4 fck Ac + (0.67 fy  - 0.4fck) Asc

 

3

3000 ื 10  = 0.4 ื 20 ื (450 ื 600) + (0.67 ื 415–0.4 ื 20)Asc

3

= 2160ื10  + 270.05Asc

 

? Asc = (3000–2160) ื 10 /270.05 = 3111 mm

In view of the column dimensions (450 mm, 600 mm), it is necessary to place intermediate

bars, in addition to the 4 corner bars:

2

Provide 4–25? at corners ie, 4 ื 491 = 1964 mm

2

and 4–20? additional ie, 4 ื 314 = 1256 mm

2             2

? Asc = 3220 mm  > 3111 mm

 

? p = (100ื3220) / (450ื600) = 1.192 > 0.8 (minimum steel), OK.

 

Design of transverse steel

 

Diameter of tie = ผ diameter of main steel = 25/4 =6.25 mm or 6 mm, whichever is greater. Provide 6 mm.

 

Spacing: < 300 mm, < 16 x 20 = 320 mm, < LLD = 450mm. Say 300 mm c/c Thus provide ties 8mm @ 300 mm c/c

 

Sketch:

 

 

7.Determine the reinforcement to be provided in a square column subjected to uniaxial bending with the following data:

Size of column 450 x 450 mm

Concrete mix M 25

Characteristic strength of steel 415 N/mm2

Factored load 2500 kN

Factored moment 200 kN.m

 

Arrangement of reinforcement:

(a)On two sides

(b)             On four sides

 

Assume moment due to minimum eccentricity to be less than the actual moment Assuming 25 mm bars with 40 mm cover, d = 40 + 12.5 = 52.5 mm d1/D = 52.5/450- 0.12

 

Charts for d1/D = 0.15 will be used

 

Pu/fckbD = (2500 x 1000)/ (25 x 450 x 450) = 0.494

 

Mu/fckbD2 =200 x 106 /(25 x 450 x 4502) = 0.088

 

a) Reinforcement on two sides,

 

Referring to Chart 33, p/fck = 0.09

 

Percentage of reinforcement, p = 0.09 x 25 = 2.25 %

 

As = p bD/100 = 2.25 x 450 x 450/100 = 4556 mm2

 

b) Reinforcement on four sides from Chart 45,

p/fck = 0.10

p = 0.10 x 25 = 2.5 %

As = 2.5 x 450 x 450/100 = 5063 mm2

 

8.Example:  Circular Column with Uniaxial Bending

 

Determine the reinforcement to be provided in a circular column with the following data:

 

Diameter of column 500 mm Grade of concrete M20 Characteristic strength 250 N/mm2 Factored load 1600 kN

Factored moment 125 kN.m

 

Lateral reinforcement :

(a)Hoop reinforcement

(b)             Helical reinforcement

 

(Assume moment due to minimum eccentricity to be less than the actual moment). Assuming 25 mm bars with 40 mm cover,

 

d1 = 40 + 12.5 = 52.5 mm d1/D – 52.5/50 = 0.105

Charts for d’/D = 0.10 will be used.

 

(a) Column with hoop reinforcement

 

Pu/fck D D = (1600 x 1000)/ (20 x 500 x 500) = 0.32

 

Mu/fck D x D2 =125 x 106 /(20 x 500 x 5002) = 0.05

Referring to Chart 52, for fy = 250 N/mm2 p/fck = 0.87

 

Percentage of reinforcement, p = 0.87 x 20 = 1.74 %

As = 1.74 x (? x 5002/4)/100 = 3416 mm2

 

 

(b) Column with Helical Reinforcement

 

According to 38.4 of the Code, the strength of a compression member with helical reinforcement is 1.05 times the strength of a similar member with lateral ties. Therefore, the, given load and moment should be divided by 1.05 before referring to the chart.

 

Pu/fck D D = (1600/1.05 x 1000)/ (20 x 500 x 500) = 0.31

 

Mu/fck D x D2 =125/1.05 x 106 /(20 x 500 x 5002) = 0.048

 

Hence, From Chart 52, for fy = 250 N/mm2,

 

p/fck = 0.078

 

p = 0.078 x 20 = 1.56 %

As = 1.56 x( ? x 500 x 500/4 )/100 = 3063 cm2

 

According to 38.4.1 of the Code the ratio of the volume of helical reinforcement to the volume of the core shall not be less than

 

0.36 (Ag/Ac - 1) x fck /fy

 

where Ag is the gross area of the section and Ac is the area of the core measured to the outside diameter of the helix. Assuming 8 mm dia bars for the helix,

Core diameter = 500 - 2 (40 - 8) = 436 mm

 

Ag/AC  = 500/436  = 1.315

 

0.36 (Ag/Ac - 1) x fck /fy = 0.36(0.315) 20/250 =0.0091

 

Volume of helical reinforcement / Volume of core

 

= Ash ? x 428 /( ?/4 x 4362) sh

 

0.9     Ash / sh

 

where, Ash is the area of the bar forming the helix and sh is the pitch of the helix. In order to satisfy the coda1 requirement,

 

0.09  Ash / sh  ? 0.0091

For 8 mm dia bar,

sh ? 0.09 x 50 / 0.0091 = 49.7 mm. Thus provide 48 mm pitch

 

Example:  Rectangular column with Biaxial Bending

 

9.Determine the reinforcement to be provided in a short column subjected to biaxial bending, with the following data:

 

size of column = 400 x 600 mm Concrete mix = M15

 

Characteristic strength of reinforcement = 415 N/mm2 Factored load, Pu = 1600 kN

 

Factored moment acting parallel to the larger dimension, Mux =120 kNm Factored moment acting parallel to the shorter dimension, Muy = 90 kNm Moments due to minimum eccentricity are less than the values given above.

 

Reinforcement is distributed equally on four sides.

 

As a first trial assume the reinforcement percentage, p = 1.2% p/fck = 1.2/15 = 0.08

 

Uniaxial moment capacity of the section about xx-axis :

 

d1/D = 52.5 /600 = 0.088

 

Chart for d’/D = 0.1 will be used.

 

Pu/fck b D = (1600 x 1000)/ (15 x 400 x 600) = 0.444

 

Referring to chart 44

 

Mu/fck b x D2 = 0.09

Mux1 = 0.09 x 15 x 400 x 6002) = 194.4 kN.m

 

Uni-axial moment capacity of the section about yy axis :

 

d1/D = 52.5 /400 = 0.131

Chart for d1/D =0.15 will be used. Referring to Chart 45,

Mu/fck b x D2 = 0.083

Mux1 = 0.083 x 15 x 600 x 4002) = 119.52 kN.m

 

Calculation of Puz :

 

Referring to Chart 63 corresponding to p = 1.2, fy = 415 and fck = 15,

Puz/Ag = 10.3

 

Puz = 10.3 x 400 x 600 = 2472 kN

 

Mux/Mux1 = 120/194.4 =0.62

Muy/Muy1=90/119.52 = 0.75

Pu /Puz =1600/2472 = 0.65       Referring to Churn 64, the permissible value of Mux/Mux1 corresponding to Muy/Muy1 and Pu /Puz is equal to 0.58

The actual value of 0.62 is only slightly higher than the value read from the Chart.

 

This can be made up by slight increase in reinforcement.

 

Using Boris load contour equation as per IS:456-2000

 

Pu /Puz = 0.65 thus, ?n = 1 + [(2-1) / (0.8 - 0.2)]  (0.65-0.2) = 1.75

 

[0.62 ]1.75 + [0.75]1.75 = 1.04 slightly greater than 1 and slightly unsafe. This can be made up by slight increase in reinforcement say 1.3%

 

Thus provide As = 1.3x400x600/100 = 3120 mm2

 

 

Provide 1.3 % of steel p/fck = 1.3/15 = 0.086

d1/D = 52.5 /600 = 0.088 = 0.1

 

From chart 44

Mu/fck b x D2 = 0.095

Mux1 = 0.095 x 15 x 400 x 6002) = 205.2  kN.m

 

Referring to Chart 45,

Mu/fck b x D2 = 0.085

Mux1 = 0.085 x 15 x 600 x 4002) = 122.4 kN.m

 

Chart 63 : Puz/Ag = 10.4

Puz = 10.4 x 400 x 600 = 2496 kN

 

Mux/Mux1 = 120/205.2 =0.585

Muy/Muy1=90/122.4  = 0.735

Pu /Puz =1600/2496 = 0.641

 

Referring to Chart 64, the permissible value of Mux/Mux1 corresponding to Muy/Muy1 and Pu /Puz is equal to 0.60

 

Hence the section is O.K.

 

Using Boris load contour equation as per IS:456-2000

 

Pu /Puz = 0.641 thus, ?n = 1 + [(2-1) / (0.8 - 0.2)]  (0.641-0.2) = 1.735

 

[120/205.2]1.735 + [90/122.4]1.735 = 0.981 ? 1 Thus OK

 

As = 3120 mm2. Provide 10 bars of 20 mm dia. Steel provided is 314 x 10 = 3140 mm2

Design of transverse steel: Provide 8 mm dia stirrups at 300 mm c/c as shown satisfying the requirements of IS: 456-2000

10.Verify the adequacy of the short column section 500 mm x 300 mm under the following load conditions:

 

Pu = 1400 kN, Mux = 125 kNm, Muy = 75 kNm. The design interaction curves of SP 16

 

should be used. Assume that the column is a ‘short column’ and the eccentricity due to moments is greater than the minimum eccentricity.

 

 

Solution:

 

2

Given: D  = 500 mm, b = 300 mm, A = 2946 mm  M   = 125 kNm, M   = 75 kNm, f  = 25

x                                s                  ux                 uy              ck

MPa, f  = 415 MPa

y

 

Applied eccentricities

3

ex = Mux/Pu = 125 ื 10 /1400 = 89.3 mm ? ex/Dx = 0.179

3

ey = Muy/Pu = 75 ื 10 /1400 = 53.6 mm ? ey/Dy = 0.179

 

These eccentricities for the short column are clearly not less than the minimum eccentricities specified by the Code.

 

Uniaxial moment capacities: Mux1, Muy1

As determined in the earlier example, corresponding to P  = 1400 kN,

u

M       = 187 kNm

ux1

 

Muy1 = 110 kNm Values of Puz and ?n

Puz = 0.45fck Ag + (0.75fy – 0.45fck)Asc

= (0.45 ื 25 ื 300 ื 500) + (0.75 ื 415 – 0.45 ื 25)ื2946

= (1687500 + 883800)N = 2571 kN

 

? Pu/Puz = 1400/2571 = 0.545 (which lies between 0.2 and 0.8)

? ?n = 1.575

 

Check safety under biaxial bending

 

[125/187]1.575 + [75/110]1

 

= 0.530 + 0.547

 

1.077 > 1.0 Hence, almost ok.

 

 

Tags : Civil - Design of Reinforced Concrete Elements - Limit State Design Of Columns
Last 30 days 50 views
Related words :

What is Design Problem, Important Question And Answer: Civil - Limit State Design Of Columns Define Design Problem, Important Question And Answer: Civil - Limit State Design Of Columns Definition of Design Problem, Important Question And Answer: Civil - Limit State Design Of Columns where how meaning of Design Problem, Important Question And Answer: Civil - Limit State Design Of Columns lecturing notes for Design Problem, Important Question And Answer: Civil - Limit State Design Of Columns lecture notes question and answer for Design Problem, Important Question And Answer: Civil - Limit State Design Of Columns answer Design Problem, Important Question And Answer: Civil - Limit State Design Of Columns study material Design Problem, Important Question And Answer: Civil - Limit State Design Of Columns assignment Design Problem, Important Question And Answer: Civil - Limit State Design Of Columns reference description of Design Problem, Important Question And Answer: Civil - Limit State Design Of Columns explanation of Design Problem, Important Question And Answer: Civil - Limit State Design Of Columns brief detail of Design Problem, Important Question And Answer: Civil - Limit State Design Of Columns easy explanation solution Design Problem, Important Question And Answer: Civil - Limit State Design Of Columns wiki Design Problem, Important Question And Answer: Civil - Limit State Design Of Columns wikipedia how why is who were when is when did where did Design Problem, Important Question And Answer: Civil - Limit State Design Of Columns list of Design Problem, Important Question And Answer: Civil - Limit State Design Of Columns school assignment college assignment Design Problem, Important Question And Answer: Civil - Limit State Design Of Columns college notes school notes kids with diagram or figure or image difference between Design Problem, Important Question And Answer: Civil - Limit State Design Of Columns www.readorrefer.in - Read Or Refer

​TOP MENU

OTHER SUGEST TOPIC

Recent New Topics :