P=400 kNP = 400 , text{kN}
Leff=2.5 m=2500 mmL_{text{eff}} = 2.5 , text{m} = 2500 , text{mm}
fck=20 N/mm2f_{ck} = 20 , text{N/mm}^2
fy=415 N/mm2f_y = 415 , text{N/mm}^2
Step 1: To Find Factored Load
Pu=1.5×PP_u = 1.5 times P
Pu=1.5×400P_u = 1.5 times 400
Pu=600 kNP_u = 600 , text{kN}
Step 2: Assume 1% of Steel in Column
Area of steel AscA_{sc}:
Asc=0.01×AgA_{sc} = 0.01 times A_g
Area of concrete AcA_c:
Ac=Ag−AscA_c = A_g – A_{sc}
Ac=0.99×Ag
A_c = 0.99 times A_g
Step 3: To Find Gross Area
A_g
Factored load formula:
Pu=(0.4×fck×Ac)+(0.67×fy×Asc)P_u = (0.4 times f_{ck} times A_c) + (0.67 times f_y times A_{sc})
Substitute values:
600×103=(0.4×20×0.99×Ag)+(0.67×415×0.01×Ag)600 times 10^3 = (0.4 times 20 times 0.99 times A_g) + (0.67 times 415 times 0.01 times A_g)
Solve for AgA_g:
600×103=7.92×Ag+2.78×Ag600 times 10^3 = 7.92 times A_g + 2.78 times A_g
600×103=10.70×Ag600 times 10^3 = 10.70 times A_g
Ag=600×10310.70A_g = frac{600 times 10^3}{10.70}
Ag=56074.77 mm2A_g = 56074.77 , text{mm}^2
Ag=56.07×103 mm2A_g = 56.07 times 10^3 , text{mm}^2
Assuming a square column:
Side=56.07×103text{Side} = sqrt{56.07 times 10^3}
Side≈236.79 mmtext{Side} approx 236.79 , text{mm}
Step 4: Check for Minimum Eccentricity
emin=L500+D30 OR 20mm whichever is greatere_{text{min}} = frac{L}{500} + frac{D}{30} text{ OR } 20 text{mm whichever is greater}
Given:
-
L=2500 mmL = 2500 , text{mm}
-
D=240 mmD = 240 , text{mm}
emin=2500500+24030e_{text{min}} = frac{2500}{500} + frac{240}{30}
emin=5+8e_{text{min}} = 5 + 8
emin=13mm OR 20mm whichever is greatere_{text{min}} = 13 text{mm OR } 20 text{mm whichever is greater}
emin=20mme_{text{min}} = 20 text{mm}
But, emine_{text{min}} is more than 0.05D.
0.05D=0.05×240=12 mm0.05D = 0.05 times 240 = 12 , text{mm}
So, check for minimum eccentricity is not satisfied.
Increase the Dimension
New dimensions: 320 mm×320 mm320 , text{mm} times 320 , text{mm}
emin=2500500+32030e_{text{min}} = frac{2500}{500} + frac{320}{30}
emin=5+10.67e_{text{min}} = 5 + 10.67
emin=15.67mme_{text{min}} = 15.67 text{mm}
0.05D=0.05×320=16mm0.05D = 0.05 times 320 = 16 text{mm}
emin<0.05De_{text{min}} < 0.05D
15.67mm<16mm15.67 text{mm} < 16 text{mm}
Check for minimum eccentricity is satisfied.
Revised Size of Column
Revised size of column: 320 mm×320 mm320 , text{mm} times 320 , text{mm}
Step 5: Calculate Area of Steel
Asc=0.01AgA_{text{sc}} = 0.01 A_{text{g}}
Ag=320×320A_{text{g}} = 320 times 320
Asc=0.01×320×320A_{text{sc}} = 0.01 times 320 times 320
Asc=1024 mm2A_{text{sc}} = 1024 , text{mm}^2
Provide 4 bars of 20 mm Ø bar.
Step 6: Lateral Ties
Diameter of ties:
Diameter of ties=14×Diameter of longitudinal steel bartext{Diameter of ties} = frac{1}{4} times text{Diameter of longitudinal steel bar}
=14×20= frac{1}{4} times 20
=5 mm= 5 , text{mm}
Since the calculated diameter is less than 6 mm, provide 6 mm diameter lateral ties.
Spacing of Lateral Ties
Spacing should not be greater than:
-
Least lateral dimension of column =320 mm= 320 , text{mm}
-
16 times the diameter of longitudinal steel =16×20=320 mm= 16 times 20 = 320 , text{mm}
-
300 mm
Provide lateral ties of 6 mm Ø at 300 mm center-to-center.
