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• Rizwan Mirza

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Project

Job Ref.

Pearl Project

Q09077

Section

Sheet no./rev.

DARGROUP

Medina Centrale Calc. by

ENG.IC

Date

Chk'd by

09-Nov-07

11 Date

App'd by

Date

DR.AE

600 mm

542 mm

RC RECTANGULAR COLUMN DESIGN (ACI 318-05)

350 mm

Geometry of column Depth of column (larger dimension of column);

h = 600 mm

W idth of column (smaller dimension of column);

b = 350 mm

Clear cover to reinforcement (both sides);

c c = 40 mm

Unsupported height of column;

lu = 4500 mm

Effective height factor;

k = 1.00

Check for overall column dimensions

h < 4b, column dimensions are OK Reinforcement of column Numbers of bars of longitudinal steel;

N=8

Longitudinal steel bar diameter number;

D bar_num = 6

Diameter of longitudinal bar;

D long = 19 mm

Stirrup bar diameter number;

D stir_num = 3

Diameter of stirrup bar;

D stir = 9 mm

Specified yield strength of reinforcement;

fy = 415 MPa

Specified compressive strength of concrete;

f’c = 40 MPa

Modulus of elasticity of bar reinforcement;

E s = 200000 MPa

Modulus of elasticity of concrete (cl. 8.5);

E c = 4700  (f’c  1 MPa) = 29725 MPa

Ultimate design strain;

 c = 0.003 mm/mm

Check for minimum area of steel (ACI 318-05, cl. 10.9) Gross area of column;

A g = h  b = 210000 mm 2

Area of steel provided;

A st = N  (/ 4)  D long 2 = 2268 mm 2

Minimum required area of steel;

A st_min = 0.01  A g = 2100 mm 2

Project

Job Ref.

Pearl Project

Q09077

Section

Sheet no./rev.

DARGROUP

Medina Centrale Calc. by

ENG.IC

Date

Chk'd by

09-Nov-07

21 Date

App'd by

Date

DR.AE

PASS - A st > A st_min , provided area of steel is greater than minimum required area of steel Check for maximum area of steel (ACI 318-05, cl. 10.9) Permissible maximum area of steel;

A st_max = 0.08  A g = 16800 mm 2

PASS - A st < A st_max , provided area of steel is less than permissible maximum area of steel Braced column slenderness check (ACI 318-05, cl. 10.12) Maximum slenderness ratio limit;

s r_max = 100

Permissible slenderness ratio;

s r_perm = 40

Slenderness check for braced column Radius of gyration;

rx = 0.3  h = 180 mm ry = 0.3  b = 105 mm rmin = min(rx, ry) = 105 mm

Actual slenderness ratio;

s r_act = k  lu / r min = 42.86

Column slenderness limit OK, column is braced slender column Design load and moments for biaxially loaded slender column Ultimate axial force acting on column;

P u_act = 2750.00 kN

Ultimate moment about major (X) axis;

M ux_act = 120.00 kNm

Ultimate moment about minor (Y) axis;

M uy_act = 25.00 kNm

Contour beta factor;

 = 0.50

Ratio of DL moment to total moment;

 d = 0.65

Magnified moments for biaxial slender column (ACI 318-05, cl. 10.12) Assuming strength reduction factor;

 = 0.65

Moment of inertia of section @ X axis;

Igx = (b  h 3 ) / 12 = 6300000000 mm 4

Moment of inertia of section @ Y axis;

Igy = (h  b 3 ) / 12 = 2143750000 mm 4

Euler’s buckling load @ X axis;

P cx = ( 2  0.4  E c  Igx) / ((1 +  d )  (k  lu )2 ) = 22126.83 kN

Euler’s buckling load @ Yaxis;

P cy = ( 2  0.4  E c  Igy) / ((1 +  d )  (k  lu )2 ) = 7529.27 kN

Correction factor for actual to equiv. mmt.diagram; C m = 1 Moment magnifier for M @ X axis;

 nsx1 = C m / (1 - (P u_act/ (0.75  P cx ))) = 1.199

Moment magnifier for M @ X axis;

 nsx =  nsx1 = 1.199

Moment magnifier for M @ Y axis;

 nsy1 = C m / (1 - (P u_act / (0.75  P cy))) = 1.949

Moment magnifier for M @ Y axis;

 nsy =  nsy1 = 1.949

Ultimate magnified uniaxial M @ X axis;

M cx =  nsx  M ux_act = 143.84 kNm

Ultimate magnified uniaxial M @ Y axis;

M cy =  nsy  M uy_act = 48.73 kNm

Net magnified uniaxial M @ X axis;

M nx = M cx = 221.28 kNm

Net magnified uniaxial M @ Y axis;

M ny = M cy = 74.97 kNm

Required eccentricities;

e x = M cx / P u_act = 52 mm e y = M cy / P u_act = 18 mm

Axial load capacity of biaxially loaded column assuming no M uy_act (ACI 318-05, cl 10.3.6) c/d t ratio;

rxb = 1.233

Effective cover to reinforcement;

d’ = c c + D stir + (D long / 2) = 59 mm

Depth of tension steel;

d t = h - d’ = 542 mm

Depth of NA from extreme compression face;

c x = rxb  d t = 668 mm

Factor of depth of comp. stress block (cl.10.2.7.3);  1 = 0.764 Depth of equivalent rectangular stress block;

a x = min(( 1  c x), h) = 510 mm

Stress in compression reinforcement;

f’sx = E s   c  (1 - (d’ / c x)) = 547 MPa

Project

Job Ref.

Pearl Project

Q09077

Section

Sheet no./rev.

DARGROUP

Medina Centrale Calc. by

ENG.IC

Date

Chk'd by

09-Nov-07

31 Date

App'd by

Date

DR.AE

Since abs(f' sx ) > f y , hence f' csx = f y f’csx = 415 MPa Stress in tension reinforcement;

fsx = E s   c  ((d t / c x) - 1) = -114 MPa

Since abs(f sx ) < f y , f sx = f tsx Capacity of concrete in compression;

C cx = 0.85  f’c  b  a x = 6074.92 kN

Strength of steel in compression;

C sx = A’s  f’csx = 470.72 kN

Strength of steel in tension;

T sx = A s  ftsx = 128.83 kN

Nominal axial load strength;

P nx = C cx + C sx + T sx = 6674.46 kN

Strength reduction factor;

x = 0.65 = 0.650

Ultimate axial load carrying capacity of column;

P u1 = x  P nx = 4338.40 kN

PASS - column is safe in axial loading Uniaxial moment capacity of column Centroid of column along larger dimension;

y x = h  0.5 = 300 mm

Nominal moment strength;

M ox = = C cx  (y x - (0.5  a x)) + C sx  (y - c cx) - T sx  (d t - y x ) = 363.136

kNm x = (M nx / M ox) = 0.609 Ultimate moment strength;

M u1 = M ox  x = 236.04 kNm

PASS - column is safe for bending Eccentricity ratio Actual eccentricity;

e x = 52 mm

Allowable eccentricity;

e all_x = M u1 / P u1 = 54 mm

Eccentricity ratio;

e rx = e x / e all_x = 0.961

Biaxially loaded column about minor axis Details of column cross-section c/d t ratio;

ryb = 1.142

Effective cover to reinforcement;

d’ = c c + D stir + (D long / 2) = 59 mm

Area of each layer of steel;

A st_l = 2  (D long 2 ) / 4 = 567 mm 2

Spacing between bars;

s = ((b - (2  d’))) / ((N / 2) -1) = 78 mm

Depth of tension steel;

b t = b - d’ = 292 mm

Depth of NA from extreme compression face;

c y = ryb  b t = 333 mm

Depth of equivalent rectangular stress block;

a y = min(( 1  c y), b) = 255 mm

Yield strain in steel;

 sy = fy / E s = 0.002

Strength reduction factor;

y = 0.650

Details of concrete block Force carried by concrete Forces carried by concrete;

C cy = 0.85  f’c  h  a y = 5192.55 kN

Moment carried by concrete Moment carried by concrete;

M ccy = C cy  ((b / 2) - (a y / 2)) = 247.85 kNm

Details of steel layers Details of first steel layer Depth of first layer;

x1 = d’ = 59 mm

Strain of first layer;

 1 = 0.003  (1 - (x1 / c y)) = 0.00247

Stress in first layer;

 1 = 415 MPa

Project

Job Ref.

Pearl Project

Q09077

Section

Sheet no./rev.

DARGROUP

Medina Centrale Calc. by

Date

ENG.IC

Chk'd by

09-Nov-07

41 Date

App'd by

Date

DR.AE

Force carried by first layer;

F1 =  1  A st_l = 235.33 kN

Moment carried by first steel layer;

M1 = F1  ((b / 2) - x1 ) = 27.42 kNm

Details of second steel layer Depth of second layer;

x2 = x1 +s = 136 mm

Strain of second layer;

 2 = 0.003  (1 - (x2 / c y)) = 0.00177

Stress in second layer;

 2 = 355 MPa

Force carried by second layer;

F2 =  2  A st_l = 201.13 kN

Moment carried by second steel layer;

M2 = F2  ((b / 2) - x2 ) = 7.81 kNm

Details of third steel layer Depth of third layer;

x3 = 214 mm

Strain of third layer;

 3 = 0.00107

Stress in third layer;

 3 = 215 MPa

Force carried by third layer;

F3 =  3  A st_l = 121.78 kN

Moment carried by third steel layer;

M3 = F3  ((b / 2) - x3 ) = -4.73 kNm

Details of fourth steel layer Depth of fourth layer;

x4 = 292 mm

Strain of fourth layer;

 4 = 0.00037

Stress in fourth layer;

 4 = 75 MPa

Force carried by fourth layer;

F4 =  4  A st_l = 42.44 kN

Moment carried by fourth steel layer;

M4 = F4  ((b / 2) - x4 ) = -4.94 kNm

Tensile force carried by steel Sum of tensile forces by steel;

T sy = 0.00 kN

Compressive force carried by steel Sum of compressive forces by steel;

C sy = 600.67 kN

Total force carried by column Nominal axial load strength;

P ny = 5793.22 kN

Strength reduction factor;

y = 0.65 = 0.650

Ultimate axial Load carrying capacity of column;

P u2 = y  P ny = 3765.59 kN

PASS - column is safe in axial loading Moment carried by biaxial column minor axis Nominal moment strength;

M oy = 273.40 kNm

Contour beta factor Contour beta factor;

 = 0.500 M nx_upon_M ox = x = 0.609

From Contour beta factor chart for rectangular columns in biaxial bending M ny_upon_M oy = 0.391 Net moment along minor axis resisted by column; M ny1 = M oy  (M ny_upon_M oy) = 106.90 kNm Ultimate moment along minor axis;

M u2 = M ny1  y = 69.49 kNm

Check for moment capacity about minor axis

PASS - column is safe for bending Eccentricity ratio Actual eccentricity;

e y = 18 mm

Allowable eccentricity;

e all_y = M u2 / P u2 = 18 mm

Project

Job Ref.

Pearl Project

Q09077

Section

Sheet no./rev.

DARGROUP

Medina Centrale Calc. by

ENG.IC Eccentricity ratio;

Date

09-Nov-07

Chk'd by

51 Date

App'd by

Date

DR.AE

e ry = e y / e all_y= 0.960

Design of column ties (ACI 318-05, cl. 7.10) 16 times longitudinal bar diameter;

s v1 = 16  D long = 304 mm

48 times stirrup bar diameter;

s v2 = 48  D stir = 432 mm

Least column dimension;

s v3 = min(h, b) = 350 mm

Maximum allowable stirrup spacing;

s = min(s v1 , s v2 , s v3 ) = 304 mm

Design summary Column is 350 mm wide and 600 mm deep with 40 MPa concrete and 415 MPa steel. Longitudinal reinforcement is 8 No.6 and lateral reinforcement for shear is 2 legs No.3 stirrup @ 304 mm center to center Design status PASS - column is safe