fc 32 MPa fy 500 MPa D 520 mm B 400 mm Ast Ase 3N28 Cover c 60 mm The load-momen

Question

fc 32 MPa fy 500 MPa D 520 mm B 400 mm Ast Ase 3N28 Cover c 60 mm The load-moment interaction diagram for the column (moment about the y-axis) is given: a) Find the maximum permissible axial load that can be applied to this columrn b) Find the maximum axial load that can be applied together with a 220 KN-m moment so that tension controls the design What is the maximum moment that can be applied to this column? c) d) Is this column adequate for a combined action of an axial load N'-2800 KN and a e) Is this column adequate for a combined action of an axial load N 3400 KN and a f) moment M 260 KN-m? (Plot the point in the diagram) moment M 300 KN-m? (Plot the point in the diagram) What is the minimum eccentricity for this column about the y-axis? What is the radius of gyration of this column about the y-axis? g)

Explanation / Answer

(you should have posted your interaction diagram for accurate answer,whatever i am solving the given problem with interaction diagram which i have,answer differ but concept will be same).

f'c = 32 MPa ,fy = 500 MPa

b = 400 mm , D = 520mm ,Cover, c= 60mm

so, d = (D-c) = 520-60 = 460 mm

Gross area of the column, Ag = 520x400 = 208000mm2

Ast = Asc = 3x(Pi/4)x282 = 1847.26 mm2

so, total area of steel in the colummn, As = Ast + Asc = 3694.5 mm2

and area of concrete = Ag - As = 204305.5 mm2

percenrage of steel in the column, p = Asx100/bD = (3694.5x100)/(520x400) = 1.776%

p/f'c = 1.776/32 = 0.0555

cosider factor of safety,F = 1.5

(a) now from interaction diagram for fy = 500

for maximum axial load look for the value of Pu/f'cbD at p/f'c = 0.0555 on Y-axis between the curve for p/f'c = 0.04 and p/f'c = 0.06

we will get approximately, Pu/f'cbD = 0.64

Pu = 0.64x32x400x520 = 4259840 N = 4259.840 KN

so maximum axial load, P = Pu/(factor of safety) = 4259.840/1.5 =2840 KN

(b) M= 220 KN-m

Mu = 220x1.5 = 330 KN-m = 330x106 N-mm

so, Mu/f'cbD2 = 0.095

so in the interaction diagram look for the value of Pu/f'cbD with the help of Mu/f'cbD2 = 0.095 and p/f'c = 0.0555

you will get approximate value of Pu/f'cbD = 0.45

hence, Pu = 0.45x32x400x520 = 2995200 N = 2995.2 KN

so maximum axial load together with 220 KN-m moment, P = 2995.2/1.5 = 1997 KN

(c) for the maximum moment look for the value of Mu/f'cbD2 at p/f'c = 0.0555 on X-axis between the curve for p/f'c = 0.04 and p/f'c = 0.06

you will get approximate value of  Mu/f'cbD2 = 0.115

so, Mu = 0.115x32x400x5202 = 398028800 N-mm = 398 KN-m

so maximum moment that can be applied to the column, M = 398/1.5 = 265.33 KN-m

(d)

P = 2800 KN , M = 260 KN-m

Pu = 1.5x2800 = 4200 KN = 4200x103 N and Mu = 1.5x260 = 390 KN-m = 390x106 N-mm

Pu/f'cbD = (4200x103)/( 32x400x520) = 0.63 and Mu/f'cbD2 = (390x106)/(32x400x5202) = 0.11

so lookout for the value of p/f'c in the interaction diagram with the help of Pu/f'cbD = 0.63 and Mu/f'cbD2 = 0.11

you will get approximate value of p/f'c = 0.10 which is graeter than our value 0.055. so this column is not adequate for the given load and moment combinatios.

(d)

P = 3400 KN , M = 3000 KN-m

Pu = 1.5x3400 = 5100 KN = 5100x103 N and Mu = 1.5x300 = 450 KN-m = 450x106 N-mm

Pu/f'cbD = (5100x103)/( 32x400x520) = 0.77 and Mu/f'cbD2 = (450x106)/(32x400x5202) = 0.13

so lookout for the value of p/f'c in the interaction diagram with the help of Pu/f'cbD = 0.77 and Mu/f'cbD2 = 0.13

you will get approximate value of p/f'c = 0.15 which is graeter than our value 0.055. so this column is not adequate for the given load and moment combinatios.

(f) here length of the column is not given, so minimum eccentricity will be 20 mm.

(if length of the column and end condition is given then find out effective length,Leff of the column and then minimum eccentricity will be maximum of 20mm or [(Leff/500)+(Least lateral dimension/30)] ).

(g) here value of modular ratio (Es/Ec) is not given so we will consider column is full of concrete.

area moment of inertia about Y-axis, I = bD3/12 = (400x5203)/12 = 4687x106 mm2

so for radius of gyration, r ,

Ar2 = I

400x520xr2 = 4687x106

r = 150.11 mm

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