8/6/2019 Column Inter at Ion Diagram

1/7

Reference Calculation Output

Column Interaction Diagram

Es = 200 kN/mm2

fcu = 25 N/mm2

fy = 460 N/mm2

m-s = 1.15

m-c = 1.5

Bending about Major Axis Bending about Minor Axis

16 mm 16 mmNo of Bars 2 No of Bars 2

16 mm 16 mmNo of Bars 2 No of Bars 2

30 mm 30 mm30 mm 30 mm

b = 225 mm b = 225 mm

h = 225 mm h = 225 mm

Mx = 20 kN.m My = 20 kN.m

N = 250 kN N = 250 kN= 3000 mm = 3000 mm

= 1.2 = 1.2

Place 1 to relevent one for following

Braced = 0 -- Braced = 0 --

Unbraced = 1 Unbraced Unbraced = 1 Unbraced

Biaxial Bending

N / Nuz 0.282

1.14

0.721

0.721

1.379 > 1 not ok

1.724 > 1 not ok

Increase Column Size,Reinforcement

Goal seek

Set following value of axial force to 250 kN by changing the cell here

Do not Change Axial Force Change following Cell

Major Axis Minor Axis Major Axis Minor Axis

250.00 250.00 110.56 110.56

Diameter of Links = 6 mm

As1 = 402.1 mm2 As1 = 402.1 mm2As2 = 402.1 mm2 As2 = 402.1 mm2

a1 = 44 mm a1 = 44 mm

a2 = 44 mm a2 = 44 mm

Doc. No.DESIGN UNIT Designed DateEPC DIVISION Checked DateC Job Code Page

As1 As1

As2 As2

Cover- C1 Cover- C1

Cover- C2 Cover- C2

lox loy x y

n(Mx/Mux)

(My/Muy)

(Mx/Mux)n+(My/Muy)

n

(Mx/Mux)+(My/Muy)+(N/N uz)

DEC

h

b

a2

a1

x

y

y

x

As1

As2

My

Mx

8/6/2019 Column Inter at Ion Diagram

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Reference Calculation Output

Column Interaction Diagram

As1 = 402.12 mm2 Es = 200 kN/mm2As2 = 402.12 mm2 fcu = 25 N/mm2a1 = 44 mm fy = 460 N/mm2a2 = 44 mm m-s = 1.15

b = 225 mm m-c = 1.5

h = 225 mm

fs = 400 N/mm2

Bending about Major Axis

Braced = 0 Place 1 to

Unbraced = 1 Relevent

Mx = 20 kN.m

N = 250 kN

= 3000 m

= 1.2

= 3600.00 mm

= 0.128

= 28.80

= 11.25

= 887.01 kN= 260.45 kN

K = 1.00

N M Fs1 Fs2 Fc x y

887 0.00 0.70 0.00 225

762 10.39 0.60 0.04 92.79 160.85 508.78 0.0012 0.0029 270 225

745 11.56 0.59 0.04 75.6 160.8 508.8 0.001 0.003 247.5 225

725 12.97 0.57 0.05 55.0 160.8 508.8 0.001 0.003 225 225

649 18.76 0.51 0.07 29.9 160.8 457.9 0.000 0.003 202.5 202.5

566 23.95 0.45 0.08 -1.6 160.8 407.0 0.000 0.003 180 180

475 28.72 0.38 0.10 -42.0 160.8 356.1 -0.001 0.003 157.5 157.5

370 33.39 0.29 0.12 -95.9 160.8 305.3 -0.001 0.002 135 135

254 37.78 0.20 0.13 -160.8 160.8 254.4 -0.002 0.002 112.5 112.5

187 35.53 0.15 0.12 -160.8 143.9 203.5 -0.004 0.002 90 90

90 30.27 0.07 0.11 -160.8 98.0 152.6 -0.006 0.001 67.5 67.5

-53 20.83 -0.04 0.07 -160.8 6.3 101.8 -0.011 0.000 45 45

-379 -2.20 -0.30 -0.01 -160.8 -269.0 50.9 -0.025 -0.003 22.5 22.5

0 24.46 0.00 0.09

Calculation of M at no axial force

1. When the fs2 has not resched to maximum tensile strength while the fs1

reaches to its maximum

fs.As = fs2.As2 +0.405.fcu.b.x

s2 = (x-a2) .(0.0035)

x

a = 0.045.fcu.b = 2261.25

b = E.(0.0035).As2-fs1.As1 = 120637.16

c = (-)E.a2.As2.-0.0035 = -12385414.88

X can be calculated by solving above second order polinomial

Doc. No.DESIGN UNIT Designed DateEPC DIVISION Checked DateC Job Code Page

lox x

ley aa ue minN

uz

Nbal

N/bhf cu

M/bh 2f cu

s1 s2

DEC

8/6/2019 Column Inter at Ion Diagram

3/7

Reference Calculation Output

When, s2 >=0.002 When , s2 < 0.002

x = 0.00 mm x = 51.99 mm

s2 = 0.002 s2 = 0.00054

Fc = 0.00 kN Fc = 117.57 kN

Fs1 = 160.85 kN Fs1 = 160.85 kN

Fs2 = 160.85 kN Fs2 = 43.28 kN

M1 = 22.04 kN.m M1 = 24.46 kN.m

Calculation of Maximum N

Maximum Value of N = 891.23 k N

Balanced failure condition

x = 115.18 mm = 400

y = 103.66 mm = 400

= 0.002 = 160.85 kN

= 0.002 = 160.85 kN= 260.45 kN

= 260.45 kN= 37.84 kNm

Bending about Mojor Axis of Column Bending about Minor Axis of Column

Calculation of additional Moment Calculation of additional moment

= 16.00

Type of Column = Slender Column

= 28.80

= 7.20 kN.m

= 11.25 mm= 2.81 kN.m

M = 27.20 kN.m

Construction of Column Capacity Line

N = 250 kNM = 27.20 kN.m

= 0.198

= 0.096

Length of demand l ine = 0.219Length of capacity line = 0.284Demand Capacity Ratio = 0.773

Doc. No.DESIGN UNIT Designed DateEPC DIVISION Checked DateC Job Code Page

f s1 N/mm 2

f s2 N/mm 2

s1 F s1s2 F s2

F cN balanceM balance

ley /h

a uMadd

e minM (emin .N)

N/bhf cuM/bh 2f cu

DEC

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Reference Calculation Output

m' c m f

0.70 0.000 -2.70 0.7 2.07 0.147 0.304 0.11 8.77 0 0.00 0.00

0.60 0.036 -3.28 0.72 2.07 0.135 0.279 0.10 10.41 0 0.00 0.00

0.59 0.041 -3.28 0.72 2.07 0.135 0.279 0.10 10.41 0 0.00 0.00

0.57 0.046 -2.96 0.71 2.07 0.141 0.291 0.10 9.56 0 0.00 0.00

0.51 0.066 -3.57 0.75 2.07 0.133 0.274 0.09 10.78 0 0.00 0.00

0.45 0.084 -4.30 0.81 2.07 0.127 0.263 0.09 11.75 0 0.00 0.00

0.38 0.101 -5.05 0.88 2.07 0.124 0.257 0.08 12.27 0 0.00 0.00

0.29 0.117 -5.93 0.99 2.07 0.123 0.255 0.08 12.43 12.43 0.26 0.12

0.20 0.133 6.78 -0.7 2.07 0.148 0.307 0.12 8.63 0 0.00 0.00

0.15 0.125 4.14 -0.37 2.07 0.178 0.368 0.17 5.97 0 0.00 0.000.07 0.106 3.40 -0.29 2.07 0.218 0.451 0.25 3.99 0 0.00 0.00

-0.04 0.073 3.19 -0.27 2.07 0.246 0.508 0.32 3.14 0 0.00 0.00

-0.30 -0.008 3.20 -0.27 2.07 0.243 0.503 0.312 3.21 0 0.00 0.00

0.00 0.086

= 12.426

Coordinates of Intersection point of Column intreraction diagram and

Capacity Line

= 0.26

= 0.12

For the Section subjected to byaxial bending

x = 110.56 mm N = 250.00 kN

= -0.0022 = 37.72 kN.m

= 0.0021 = 0.721

= 400.00 N/mm2 = 0.28

= 400.00 N/mm2

y = 99.50 mm

= 160.85 kN

= 160.85 kN= 250.00 kN

N = 250.00 kN

M = 37.72 kN.m

= 0.198

= 0.132

Doc. No.DESIGN UNIT Designed DateEPC DIVISION Checked DateC Job Code Page

Reference Calculation Output

N/bhf cu

M/bh 2f cu x1 y1 x1

2

+y 12 1/

x12+y1 2N/

bhf cuM/

bh 2f cu

Max (x 12+y1

2)

N/bhf cuM/bh 2f

cu

s1 Muxs2 Mx/ Muxf s2 N/N uzf s1

F s2F s1F c

N/bhf cuM/bh 2f cu

DEC

8/6/2019 Column Inter at Ion Diagram

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Reference Calculation Output

Column Interaction Diagram

As1 = 402.12 mm2 Es = 200 kN/mm2As2 = 402.12 mm2 fcu = 25 N/mm2a1 = 44 mm fy = 460 N/mm2a2 = 44 mm m-s = 1.15

b = 225 mm m-c = 1.5

h = 225 mm

fs = 400 N/mm2

Bending about Major Axis

Braced = 0 Place 1 to

Unbraced = 1 Relevent

My = 20 kN.m

N = 250 kN

= 3000 m

= 1.2

= 3600.00 m

= 0.128

= 28.80

= 11.25

= 887.01 kN= 260.45 kN

K = 1.00

N M Fs1 Fs2 Fc x y

887 0.00 0.70 0.00 225

762 10.39 0.60 0.04 92.79 160.85 508.8 0.0012 0.003 270.00 225

745 11.56 0.59 0.04 75.6 160.8 508.8 0.001 0.003 247.5 225

725 12.97 0.57 0.05 55.0 160.8 508.8 0.001 0.003 225 225

649 18.76 0.51 0.07 29.9 160.8 457.9 0.000 0.003 202.5 202.5

566 23.95 0.45 0.08 -1.6 160.8 407.0 0.000 0.003 180 180

475 28.72 0.38 0.10 -42.0 160.8 356.1 -0.001 0.003 157.5 157.5

370 33.39 0.29 0.12 -95.9 160.8 305.3 -0.001 0.002 135 135

254 37.78 0.20 0.13 -160.8 160.8 254.4 -0.002 0.002 112.5 112.5

187 35.53 0.15 0.12 -160.8 143.9 203.5 -0.004 0.002 90 90

90 30.27 0.07 0.11 -160.8 98.0 152.6 -0.006 0.001 67.5 67.5

-53 20.83 -0.04 0.07 -160.8 6.3 101.8 -0.011 0.000 45 45

-379 -2.20 -0.30 -0.01 -160.8 -269.0 50.9 -0.025 -0.003 22.5 22.5

0 24.46 0.00 0.09

Calculation of M at no axial force

1. When the fs2 has not resched to maximum tensile strength while the fs1

reaches to its maximum

fs.As = fs2.As2 +0.405.fcu.b.x

s2 = (x-a2) .(0.0035)

x

a = 0.045.fcu.b = 2261.25

b = E.(0.0035).As2-fs1.As1 = 120637.16

c = (-)E.a2.As2.-0.0035 = -12385414.88

X can be calculated by solving above second order polinomial

Doc. No.DESIGN UNIT Designed DateEPC DIVISION Checked DateC Job Code Page

loy y

ley aa ue minN

uzNbal

N/bhf cu

M/bh 2f cu

s1 s2

DEC

8/6/2019 Column Inter at Ion Diagram

6/7

Reference Calculation Output

When, s2 >=0.002 When , s2 < 0.002

x = 0.00 mm x = 51.99 mm

s2 = 0.002 s2 = 0.00054

Fc = 0.00 kN Fc = 117.57 kN

Fs1 = 160.85 kN Fs1 = 160.85 kN

Fs2 = 160.85 kN Fs2 = 43.28 kN

M1 = 22.04 kN.m M1 = 24.46 kN.m

Calculation of Maximum N

Maximum Value of N = 891.23 kN

Balanced failure condition

x = 115.18 mm = 400

y = 103.66 mm = 400

= 0.002 = 160.85 kN

= 0.002 = 160.85 kN= 260.45 kN

= 260.45 kN= 37.84 kNm

Bending about Mojor Axis of Column Bending about Minor Axis of Column

Calculation of additional Moment Calculation of additional moment

= 16.00

Type of Column = Slender Column

= 28.80

= 7.20 kN.m

= 11.25 mm= 2.81 kN.m

M = 27.20 kN.m

Construction of Column Capacity Line

N = 250 kNM = 27.20 kN.m

= 0.198

= 0.096

Length of demand l ine = 0.219Length of capacity line = 0.284Demand Capacity Ratio = 0.773

Doc. No.DESIGN UNIT Designed DateEPC DIVISION Checked DateC Job Code Page

f s1 N/mm 2

f s2 N/mm 2

s1 F s1s2 F s2

FcN balanceM balance

lex /h

a uMadd

e minM (emin .N)

N/bhf cuM/bh 2f cu

DEC

8/6/2019 Column Inter at Ion Diagram

7/7

Reference Calculation Output

m' c m f

0.701 0.000 -2.70 0.70 2.07 0.147 0.304 0 .114 8.768 0 .00 0.00 0.00

0.602 0.036 -3.28 0.72 2.07 0.135 0.279 0.096 10.410 0.00 0.00 0.00

0.589 0.041 -3.28 0.72 2.07 0.135 0.279 0.096 10.410 0.00 0.00 0.00

0.573 0.046 -2.96 0.71 2.07 0.141 0.291 0 .105 9.564 0 .00 0.00 0.00

0.513 0.066 -3.57 0.75 2.07 0.133 0.274 0.093 10.777 0.00 0.00 0.00

0.447 0.084 -4.30 0.81 2.07 0.127 0.263 0.085 11.746 0.00 0.00 0.00

0.375 0.101 -5.05 0.88 2.07 0.124 0.257 0.081 12.272 0.00 0.00 0.00

0.293 0.117 -5.93 0.99 2.07 0.123 0.255 0.080 12.426 12.43 0.26 0.12

0.201 0.133 6.78 -0.70 2.07 0.148 0.307 0 .116 8.627 0 .00 0.00 0.00

0.147 0.125 4.14 -0.37 2.07 0.178 0.368 0 .168 5.969 0 .00 0.00 0.00

0.071 0.106 3.40 -0.29 2.07 0.218 0.451 0 .250 3.993 0 .00 0.00 0.00

-0.042 0.073 3.19 -0.27 2.07 0.246 0.508 0 .319 3.135 0 .00 0.00 0.00

-0.299 -0.008 3.20 -0.27 2.07 0.243 1 0 .312 3.209 0 .00 0.00 0.00

0.000 0.086

= 12.426

Coordinates of Intersection point of Column intreraction diagram and

Capacity Line

= 0.26

= 0.12

For the Section subjected to byaxial bending

x 110.56 mm N = 250.00 kN

-0.00223 = 37.72 kN.m

0.00211 = 0.721

400.00 N/mm2 = 0.28

400.00 N/mm2

y 99.50 mm

160.85 kN

160.85 kN

250.00 kN

N 250.00 kN

M 37.72 kN.m

0.198

0.132

Doc. No.DESIGN UNIT Designed DateEPC DIVISION Checked DateC Job Code Page

Reference Calculation Output

N/bhf cuM/bh 2f cu

x1 y1 x12+y12 1/x12+y12N/bhf cu

M/bh 2f cu

Max (x 12+y1

2)

N/bhf cuM/bh 2f cu

s1 Muys2 My/ Muyf s2 N/N uzf s1

F s2F s1F c

N/bhf cuM/bh 2f cu

DEC