Section Properties

Section Properties Example 001

Section Properties of a Rectangular Column

GEOMETRY AND PROPERTIES

The section properties for a given rectangular section is tested in this example by comparing the results with SAP 2000 v26.

Note: Refer Section Properties Ex001.cdbx

The column section details and loading details are as tabulated below.

Parameters	Column Designer	SAP 2000 v26
Height (in)	36
Width (in)	24
fc’ (psi)	4,000
fy (psi)	40,000
Number of bars	10
Corner Bars	#9
Bars along direction 2 and 3	#8
Rebar Area (in²)	8.71
Rebar Ratio	1.01%
Clear Cover (in)	1.5

SECTION PROPERTIES COMPARISON

Properties	Units	Column Designer	SAP 2000 v26	By hand
Basic Properties
Area	in²	864.00	864.00	864.00
Shear Area SA₂	in²	720.00	720.00	720.00
Shear Area SA₃	in²	720.00	720.00	720.00
Inertia, I₂₂	in⁴	41,472.00	41,472.00	41,472.00
Inertia, I₃₃	in⁴	93,312.00	93,312.02	93,312.00
Inertia, I₃₂	in⁴	0.00	0.00	0.00
Section Bounds
Total Width, W_total	in	24.0	24.0	24.00
Total Height, H_total	in	36.0	36.0	36.00
Centroid, X_o	in	0.00	0.00	0.00
Centroid, Y_o	in	0.00	0.00	0.00
Centroid, x̄	in	12.00	12.00	12.00
Centroid, ȳ	in	18.00	18.00	18.00
Additional Properties
Radius of gyration, R₂	in	6.93	6.93	6.93
Radius of Gyration, R₃	in	10.39	10.39	10.39
Section Modulus, S_2-Left	in³	3,456.01	3,456.00	3,456.00
Section Modulus, S_2-Right	in³	3,456.01	3,456.00	3,456.00
Section Modulus, S_3-Top	in³	5,184.01	5,184.00	5,184.00
Section Modulus, S_3-Bottom	in³	5,184.01	5,184.00	5,184.00

CALCULATIONS BY HAND

Basic Properties
1. Area, \(A = b \times d\) = \(24\ \times 36 = 864\ {in}^{2}\)
2. Shear Area, SA2 \(= \frac{5}{6}\ \times \ A = \ \frac{5}{6}\ \times \ 864 = 720\ {in}^{2}\)
3. Shear Area, SA3 \(= \frac{5}{6}\ \times \ A = \ \frac{5}{6}\ \times \ 864 = 720\ {in}^{2}\)
4. Inertia, I22 \(= \frac{h}{12}\ \times \ w^{3} = \ \frac{36}{12}\ \times \ 24^{3} = 41,472\ {in}^{4}\)
5. Inertia, I33 \(= \frac{w}{12}\ \times \ h^{3} = \ \frac{24}{12}\ \times \ 36^{3} = 93,312\ {in}^{4}\)
6. Inertia, I23 = I32 \(= \\)0
Section Bounds
1. Total Width, W_total \(= w = 24\ in\)
2. Total Height, H_total \(= h = 36\ in\)
3. Centroid, X_o \(= 0\ in\)
4. Centroid, Y_o \(= 0\ in\\)
5. Centroid, x̄ \(= \frac{w}{2} = \ \frac{24}{2} = 12\ in\)
6. Centroid, ȳ \(= \frac{h}{2} = \ \frac{36}{2} = 18\ in\)
Additional Properties
1. Radius of gyration, R₂ \(= \sqrt{\frac{I_{22}}{A}} = \sqrt{\frac{41472}{864}} = 6.93\ in\)
2. Radius of Gyration, R₃ \(= \sqrt{\frac{I_{33}}{A}} = \sqrt{\frac{93312}{864}} = 10.39\ in\)
3. Section Modulus, S_2-Left \(= \frac{I_{22}}{x_{0}} = \ \frac{41472}{12} = 3,456\ {in}^{3}\)
4. Section Modulus, S_2-Right \(= \frac{I_{22}}{{W_{total} - x}_{0}} = \ \frac{41472}{24 - 12} = 3,456\ {in}^{3}\)
5. Section Modulus, S_3-Top \(= \frac{I_{33}}{{H_{total} - y}_{0}} = \ \frac{93312}{36 - 18} = 5,184\ {in}^{3}\)
6. Section Modulus, S_3-Bottom \(= \frac{I_{33}}{y_{0}} = \ \frac{93312}{18} = 5,184\ {in}^{3}\)

Section Properties Example 002

Section Properties of a Circular Column

GEOMETRY AND PROPERTIES

The section properties for a given circular section is tested in this example by comparing the results with SAP 2000 v26.

Note: Refer Section Properties Ex002.cdbx

The column section details and loading details are as tabulated below.

Parameters	Column Designer	SAP 2000 v26
Radius (in)	18
fc’ (psi)	4,000
fy (psi)	40,000
Rebar	#9
Number of bars	12
Rebar Area (in²)	12.03
Rebar Ratio	1.18%
Clear Cover (in)	1.5

SECTION PROPERTIES COMPARISON

Properties	Units	Column Designer	SAP 2000 v26	By Hand
Basic Properties
Area	in²	1017.88	1011.35	1,017.88
Shear Area SA₂	in²	916.09	913.01	916.09
Shear Area SA₃	in²	916.09	913.01	916.09
Inertia, I₂₂	in⁴	82,447.96	81,395.00	82,447.96
Inertia, I₃₃	in⁴	82,447.96	81,395.00	82,447.96
Inertia, I₃₂	in⁴	0.00	0.00	0.00
Section Bounds
Total Width, W_total	in	36.00	36.00	36.00
Total Height, H_total	in	36.00	36.00	36.00
Centroid, X_o	in	0.00	0.00	0.00
Centroid, Y_o	in	0.00	0.00	0.00
Centroid, x	in	18.00	18.00	18.00
Centroid, y	in	18.00	18.00	18.00
Additional Properties
Radius of gyration, R₂	in	9.00	8.97	9.00
Radius of Gyration, R₃	in	9.00	8.97	9.00
Section Modulus, S_2-Left	in³	4,580.45	4,521.92	4,580.44
Section Modulus, S_2-Right	in³	4,580.45	4,521.92	4,580.44
Section Modulus, S_3-Top	in³	4,580.45	4,521.92	4,580.44
Section Modulus, S_3-Bottom	in³	4,580.45	4,521.92	4,580.44

CALCULATIONS BY HAND

Basic Properties

Area, \(A = \pi \times r^{2}\) = \(\pi\ \times \ 18^{2} = 1,017.88\ {in}^{2}\)

Shear Area, SA2 \(= \frac{5}{6}\ \times \ A = \ \frac{5}{6}\ \times \ 656692.18 = 916.09\ {in}^{2}\)

Shear Area, SA3 \(= \frac{5}{6}\ \times \ A = \ \frac{5}{6}\ \times \ 656692.18 = 916.09\ {in}^{2}\)

Inertia, I22 \(= \frac{\pi \times r^{4}}{4}\ = \ \frac{\pi \times 18^{4}}{4} = 82,447.96\ {in}^{4}\)

Inertia, I33 \(= \frac{\pi \times r^{4}}{4}\ = \ \frac{\pi \times 18^{4}}{4} = 82,447.96\ {in}^{4}\)

Inertia, I23 = I32 \(= \\)0 \({in}^{4}\)

Section Bounds

Total Width, W_total \(= r \times 2 = 36\ in\)

Total Height, H_total \(= r \times 2 = 36\ in\)

Centroid, X_o \(= r = 18\ in\)

Centroid, Y_o \(= r = 18\ in\\)

Centroid, x̄ \(= \frac{w}{2} = \ \frac{24}{2} = 12\ in\)

Centroid, ȳ \(= \frac{h}{2} = \ \frac{36}{2} = 18\ in\)

Additional Properties

Radius of gyration, R₂ \(= \sqrt{\frac{I_{22}}{A}} = \sqrt{\frac{82447.96}{1017.88}} = 9\ in\)

Radius of Gyration, R₃ \(= \sqrt{\frac{I_{33}}{A}} = \sqrt{\frac{82447.96}{1017.88}} = 9\ in\)

Section Modulus, S_2-Left \(= \frac{I_{22}}{x_{0}} = \ \frac{82447.96}{18} = 4,580.44\ {in}^{3}\)

Section Modulus, S_2-Right \(= \frac{I_{22}}{{W_{total} - x}_{0}} = \ \frac{82447.96}{36 - 18} = 4,580.44\ {in}^{3}\)

Section Modulus, S_3-Top \(= \frac{I_{33}}{{H_{total} - y}_{0}} = \ \frac{82447.96}{36 - 18} = 4,580.44\ {in}^{3}\)

Section Modulus, S_3-Bottom \(= \frac{I_{33}}{y_{0}} = \ \frac{82447.96}{18} = 4,580.44\ {in}^{3}\)

Section Properties Example 003

Section Properties of a T-Section

GEOMETRY AND PROPERTIES

The section properties for a given T-section is tested in this example by comparing the results with SAP 2000 v26.

Note: Refer Section Properties Ex003.cdbx

The column section details and loading details are as tabulated below.

Parameters	Column Designer	SAP 2000 v26
Height (in)	150
Width (in)	150
Web Width (in)	15
Flange Height (in)	15
fc’ (psi)	6,000
fy (psi)	60,000
Number of bars	58
Corner Bars	#8
Bars along direction 2 and 3	#8
Rebar Area (in²)	45.55
Rebar Ratio	1.07%
Clear Cover (in)	1.5

SECTION PROPERTIES COMPARISON

Properties	Units	Column Designer	SAP 2000 v26	By hand
Basic Properties
Area	in²	4,275.00	4,275.00	4,275.00
Shear Area SA₂	in²	2,012.75	2,006.05	-
Shear Area SA₃	in²	2,259.24	2,298.55	-
Inertia, I₂₂	in⁴	4,256,681.00	4,256,719.00	4,256,718.75
Inertia, I₃₃	in⁴	9,122,700.93	9,112,722.00	9,112,722.04
Inertia, I₃₂	in⁴	0.00	0.00	0.00
Section Bounds
Total Width, W_total	in	150.00	150.00	150.00
Total Height, H_total	in	150.00	150.00	150.00
Centroid, X_o	in	0.00	0.00	0.00
Centroid, Y_o	in	31.97	31.97	31.97
Centroid, x̄	in	75.00	75.00	75.00
Centroid, ȳ	in	106.97	106.97	106.97
Additional Properties
Radius of Gyration, R₂	in	31.55	31.55	31.56
Radius of Gyration, R₃	in	46.17	46.17	46.17
Section Modulus, S_2-Left	in³	56,755.80	56,756.00	56,756.25
Section Modulus, S_2-Right	in³	56,755.80	56,756.00	56,756.25
Section Modulus, S_3-Top	in³	211,794.14	211,794.00	211,794.15
Section Modulus, S_3-Bottom	in³	85,186.50	85,187.00	85,186.58

CALCULATIONS BY HAND

Basic Properties
1. Area, \(A = A_{1} + A_{2} = 2,025 + 2,250 = 4,275\ {in}^{2}\)
  1. \(A_{1} = \left( h - t_{f} \right) \times t_{w}\)

\(= (150 - 15) \times 150\)

\(= 2,025\ {in}^{2}\)

\(A_{2} = w \times t_{f}\)

\(= 150 \times 15\)

\(= 2,250\ {in}^{2}\)

Inertia, \(I22 = I22_{a} + I22_{b} = 37,968.75 + 4,218,750 = 4,256,718.75\ {in}^{4}\)
1. \(I22_{a} = \frac{\left( h - t_{f} \right) \times {t_{w}}^{3}\ }{12} + A_{1} \times \left( x_{0} - x_{1} \right)^{2}\)

\(= \frac{(150 - 15) \times 150^{3}\ }{12} + 2,025 \times (75 - 75)^{2}\)

\(= 37,968.75\ {in}^{4}\)

\(I22_{b} = \frac{t_{f} \times w^{3}\ }{12} + A_{2} \times \left( x_{0} - x_{2} \right)^{2}\)

\(= \frac{15 \times 150^{3}\ }{12} + 2,250 \times (75 - 75)^{2}\)

\(= 4,218,750\ {in}^{4}\)

Inertia, \(I33 = I33_{a} + I33_{b} = 6,230,766.53 + 2,881,955.51 = 9,112,722.04\ {in}^{4}\)
1. \(I33_{a} = \frac{t_{w} \times \left( h - t_{f} \right)^{3}\ }{12} + A_{1} \times \left( y_{0} - y_{1} \right)^{2}\)

\(= \frac{15 \times (150 - 15)^{3}\ }{12} + 2,025 \times (106.97 - 67.5)^{2}\)

\(= 6,230,766.53\ {in}^{4}\)

\(I33_{b} = \frac{w \times {t_{f}}^{3}\ }{12} + A_{2} \times \left( x_{0} - x_{2} \right)^{2}\)

\(= \frac{150 \times 15^{3}\ }{12} + 2,250 \times (106.97 - 142.5)^{2}\)

\(= 2,881,955.51\ {in}^{4}\)

Inertia, I23 = I32 \(= \\)0

Section Bounds
1. Total Width, W_total \(= w = 150\ in\)
2. Total Height, H_total \(= h = 150\ in\)
3. Centroid, X_o \(= 0\ in\)
4. Centroid, Y_o \(= 31.97\ in\\)
5. Centroid, x̄ \(= 75.00\ in\)
6. Centroid, ȳ \(= 106.97\ in\\)
Additional Properties
1. Radius of gyration, R₂ \(= \sqrt{\frac{I_{22}}{A}} = \sqrt{\frac{4,256,718.75}{4,275.00}} = 31.56\ in\)
2. Radius of Gyration, R₃ \(= \sqrt{\frac{I_{33}}{A}} = \sqrt{\frac{9,112,722.04}{4,275.00}} = 46.17\ in\)
3. Section Modulus, S_2-Left \(= \frac{I_{22}}{x_{0}} = \ \frac{4,256,718.75}{75.00} = 56,756.25\ {in}^{3}\)
4. Section Modulus, S_2-Right \(= \frac{I_{22}}{{W_{total} - x}_{0}} = \ \frac{4,256,718.75}{150 - 75} = 56,756.25\ {in}^{3}\)
5. Section Modulus, S_3-Top \(= \frac{I_{33}}{{H_{total} - y}_{0}} = \ \frac{9,112,722.04}{150 - 106.97} = 211,794.15\ {in}^{3}\)
6. Section Modulus, S_3-Bottom \(= \frac{I_{33}}{y_{0}} = \ \frac{9,112,722.04}{106.97} = 85,186.58\ {in}^{3}\)

P-M Interaction

ACI 318-19 PM Example 001

P-M Interaction Check for Rectangular Column

GEOMETRY AND PROPERTIES

The P-M Interaction Check for a given rectangular section is tested in this example by comparing the P-M results with SAP 2000 v26.

Note: Refer PM Ex001.cdbx

The column section details are as tabulated below.

Parameters	Column Designer	SAP 2000 v26
Code	ACI 318-19
Height (in)	36
Width (in)	24
fc’ (psi)	4,000
fy (psi)	40,000
Number of bars	10
Corner Bars	#9
Bars along direction 2 and 3	#8
Rebar Area (in²)	8.71
Rebar Ratio	1.01%
Clear Cover (in)	1.5

P-M INTERACTION COMPARISON WITH Ф FACTOR

P-M INTERACTION COMPARISON WITHOUT Ф FACTOR

P-M INTERACTION COMPARISON WITHOUT Ф FACTOR AND INCREASED FY

ACI 318-19 PM Example 002

P-M Interaction Check for Circular Column

GEOMETRY AND PROPERTIES

The P-M Interaction Check for a given circular section is tested in this example by comparing the results with SAP 2000 v26.

Note: Refer PM Ex002.cdbx

The column section details are as tabulated below.

Parameters	Column Designer v12	SAP 2000 v26
ACI Code	ACI 318-19
Radius (in)	18
fc’ (psi)	4,000
fy (psi)	40,000
Rebar	#9
Number of bars	12
Rebar Area (in²)	12.03
Rebar Ratio	1.18%
Clear Cover (in)	1.5

P-M INTERACTION COMPARISON WITH Ф FACTOR

P-M INTERACTION COMPARISON WITHOUT Ф FACTOR

P-M INTERACTION COMPARISON WITHOUT Ф INCREASED FY

ACI 318-19 PM Example 003

P-M Interaction Check for T-Section

GEOMETRY AND PROPERTIES

The P-M Interaction Check for a given T-section is tested in this example by comparing the P-M results with SAP 2000 v26.

Note: Refer PM Ex003.cdbx

The column section details are as tabulated below.

Parameters	Column Designer	SAP 2000 v26
Code	ACI 318-19
Height (in)	150
Width (in)	150
Web Width (in)	15
Flange Height (in)	15
fc’ (psi)	6,000
fy (psi)	60,000
Number of bars	58
Corner Bars	#8
Bars along direction 2 and 3	#8
Rebar Area (in²)	45.55
Rebar Ratio	1.07%
Clear Cover (in)	1.5

P-M INTERACTION COMPARISON WITH Ф FACTOR

P-M INTERACTION COMPARISON WITHOUT Ф FACTOR

P-M INTERACTION COMPARISON WITHOUT Ф FACTOR AND INCREASED FY

ACI 318-19 PM Example 004

P-M Interaction Check for Inverted T-Section

GEOMETRY AND PROPERTIES

The P-M Interaction Check for a given Inverted T-section is tested in this example by comparing the P-M results with SAP 2000 v26.

Note: Refer PM Ex004.cdbx

The column section details are as tabulated below.

Parameters	Column Designer	SAP 2000 v26
Code	ACI 318-19
Height (in)	150
Width (in)	150
Web Width (in)	15
Flange Height (in)	15
fc’ (psi)	6,000
fy (psi)	60,000
Number of bars	58
Corner Bars	#8
Bars along direction 2 and 3	#8
Rebar Area (in²)	45.55
Rebar Ratio	1.07%
Clear Cover (in)	1.5

P-M INTERACTION COMPARISON WITH Ф FACTOR

P-M INTERACTION COMPARISON WITHOUT Ф FACTOR AND INCREASED FY

Moment Magnification – Non-Sway

ACI 318-19 Moment Magnification Non-Sway- Example 001

Moment Magnification Calculation for Slender Column (Non-Sway)

GEOMETRY, PROPERTIES AND LOADING

The moment magnification calculation for a given rectangular section using ACI 318-19 is tested in this example by comparing the results with manual calculation.

The column section details are as tabulated below.

Note: Refer ACI 318-19 Moment Magnification NS Ex001.cdbx

	Parameters	Value
	Width (in)	18
	Height (in)	20
	Compressive Strength, fc’ (psi)	4,000
	Modulus of Elasticity of Concrete, E_c (ksi)	3,600
	Minimum Yield Stress, f_y (psi)	40,000
	Modulus of Elasticity of Steel, E_s (ksi)	29,000
	Rebar Layout	10-#8
	Rebar Area (in²)	7.85
	Rebar Ratio	2.18%
	Clear Cover (in)	1.5
	Unsupported Length, l_u (ft)	16
	k-factor, braced (XZ Plane)	0.77
	k-factor, braced (YZ Plane)	0.83

Name	Axial Load, P_u (kip)	Moment Top, M_ux (kip-ft)	Moment Top, M_uy (kip-ft)	Moment Bottom, M_ux (kip-ft)	Moment Bottom, M_uy (kip-ft)	Sustained Load
Combination 1	200	82	185	86	157	100

MOMENT MAGNIFICATION COMPARISON

Column Designer reports both the top and bottom magnified moments in X and Y direction. The higher of the top and bottom magnified moments is reported as M₂ while the lower magnified moment is reported as M₁.

Magnified Moments (kip-ft)	Column Designer	By hand	% Difference
M_1x	85.35	85.34	0.01%
M_2x	89.51	89.50	0.01%
M_1y	157.00	157.00	0.00%
M_2y	185.00	185.00	0.00%

MANUAL CALCULATION

Calculation of Magnified Moment about X-axis
Loading Data
M₁ (Lower Moment)	=	984	kip-in
M₂ (Higher Moment)	=	1,032	kip-in
Axial Load, Pu	=	200	kip

Calculation of Critical Buckling Load
k-factor	=	0.83
Unsupported Length, l_u	=	192	in
E_c	=	3,600	ksi
(I_g)column	=	\[\frac{18*20^{3}}{12}\]	in⁴
	=	12,000	in⁴
0.2E_cI_g	=	8,640,000	kip-in²
E_s	=	29,000	ksi
I_se	=	324.42	in⁴
β_dns	=	0.5

(EI)_eff	=	\[\frac{0.2E_{c}I_{g} + E_{s}I_{se}}{1 + \beta_{dns}}\]	kip-in²	[ACI 318-19 6.6.4.4.4]
	=	12,032,197.01	kip-in²

Critical Buckling Load, P_c	=	\[\frac{\pi^{2}{(EI)}_{eff}}{{(kl_{u})}^{2}}\]	kip	[ACI 318-19 6.6.4.4.2]
	=	4,676.12	kip

Calculation of C_m
C_m	=	\[0.6 - 0.4\frac{M_{1}}{M_{2}}\]	≥0.4	\[\lbrack\frac{M_{1}}{M_{2}} = - ve\ for\ single\ curvature\rbrack\] [ACI 318-19 6.6.4.5.3]

	=	0.98

Calculation of Magnification Factor
δ	=	\[\frac{C_{m}}{1 - \frac{P_{u}}{0.75P_{c}}}\]		[ACI 318-19 6.6.4.5.2]

	=	1.04

Calculation of Magnified Moment
M_1x	=	1,024.09	kip-in
	=	85.34	kip-ft
M_2x	=	1,074.05	kip-in
	=	89.50	kip-ft

Calculation of Magnified Moment about Y-axis
Loading Data
M₁ (Lower Moment)	=	1,884	kip-in
M₂ (Higher Moment)	=	2,220	kip-in
Axial Load, Pu	=	200	kip

Calculation of Critical Buckling Load
k-factor	=	0.77
Unsupported Length, l_u	=	192	in
E_c	=	3,600	ksi
(I_g)_column	=	\[\frac{20*18^{3}}{12}\]	in⁴
	=	9,720	in⁴
0.2E_cI_g	=	6,998,400	kip-in²
E_s	=	29,000	ksi
I_se	=	308.37	in⁴
β_dns	=	0.5

(EI)_eff	=	\[\frac{0.2E_{c}I_{g} + E_{s}I_{se}}{1 + \beta_{dns}}\]	kip-in²	[ACI 318-19 6.6.4.4.4]
	=	10,627,361.11	kip-in²

Critical Buckling Load, P_c	=	\[\frac{\pi^{2}{(EI)}_{eff}}{{(kl_{u})}^{2}}\]	kip	[ACI 318-19 6.6.4.4.2]
	=	4,798.90	kip

Calculation of C_m
C_m	=	\[0.6 - 0.4\frac{M_{1}}{M_{2}}\]	≥0.4	\[\lbrack\frac{M_{1}}{M_{2}} = - ve\ for\ single\ curvature\rbrack\] [ACI 318-19 6.6.4.5.3]
	=	0.94

Calculation of Magnification Factor
δ	=	\[\frac{C_{m}}{1 - \frac{P_{u}}{0.75P_{c}}}\]		[ACI 318-19 6.6.4.5.2]

	=	1

Calculation of Magnified Moment
M_1y	=	1,884.00	kip-in
	=	157.00	kip-ft
M_2y	=	2,220.00	kip-in
	=	185.00	kip-ft

ACI 318-19 Moment Magnification Non-Sway- Example 002

Moment Magnification Calculation for Slender Column (Non-Sway)

GEOMETRY, PROPERTIES AND LOADING

The moment magnification calculation for a given circular section using ACI 318-19 is tested in this example by comparing the results with manual calculation.

The column section details are as tabulated below.

Note: Refer ACI 318-19 Moment Magnification NS Ex002.cdbx

	Parameters	Value
	Diameter (in)	24
	Compressive Strength, fc’ (psi)	4,000
	Modulus of Elasticity of Concrete, E_c (ksi)	3,600
	Minimum Yield Stress, f_y (psi)	40,000
	Modulus of Elasticity of Steel, E_s (ksi)	29,000
	Rebar Layout	8-#9
	Rebar Area (in²)	7.99
	Rebar Ratio	1.77%
	Clear Cover (in)	1.5
	Unsupported Length, l_u (ft)	18
	k-factor, braced (XZ Plane)	0.71
	k-factor, braced (YZ Plane)	0.84

Name	Axial Load, P_u (kip)	Moment Top, M_ux (kip-ft)	Moment Top, M_uy (kip-ft)	Moment Bottom, M_ux (kip-ft)	Moment Bottom, M_uy (kip-ft)	Sustained Load
Combination 1	600	210	95	85	70	400

MOMENT MAGNIFICATION COMPARISON

Column Designer reports both the top and bottom magnified moments in X and Y direction. The higher of the top and bottom magnified moments is reported as M₂ while the lower magnified moment is reported as M₁.

Magnified Moments (kip-ft)	Column Designer	By hand	% Difference
M_1x	85.00	85.00	0.00%
M_2x	210.00	210.00	0.00%
M_1y	72.59	72.58	0.01%
M_2y	98.51	98.50	0.01%

MANUAL CALCULATION

Calculation of Magnified Moment about X-axis
Loading Data
M₁ (Lower Moment)	=	1020	kip-in
M₂ (Higher Moment)	=	2520	kip-in
Axial Load, Pu	=	600	kip

Calculation of Critical Buckling Load
k-factor	=	0.84
Unsupported Length, l_u	=	216	in
E_c	=	3600	ksi
(I_g)column	=	\[\frac{\pi*12^{4}}{4}\]	in⁴
	=	16,286.02	in⁴
0.2E_cI_g	=	11,725,931.75	kip-in²
E_s	=	29,000	ksi
I_se	=	395.27	in⁴
β_dns	=	0.67

(EI)_eff	=	\[\frac{0.2E_{c}I_{g} + E_{s}I_{se}}{1 + \beta_{dns}}\]	kip-in²	[ACI 318-19 6.6.4.4.4]

	=	13,913,201.38	kip-in²
Critical Buckling Load, P_c	=	\[\frac{\pi^{2}{(EI)}_{eff}}{{(kl_{u})}^{2}}\]	kip	[ACI 318-19 6.6.4.4.2]

	=	4,171.20	kip

Calculation of C_m
C_m	=	\[0.6 - 0.4\frac{M_{1}}{M_{2}}\]	≥0.4	\[\lbrack\frac{M_{1}}{M_{2}} = - ve\ for\ single\ curvature\rbrack\] [ACI 318-19 6.6.4.5.3]

	=	0.76

Calculation of Magnification Factor
δ	=	\[\frac{C_{m}}{1 - \frac{P_{u}}{0.75P_{c}}}\]		[ACI 318-19 6.6.4.5.2]

	=	1

Calculation of Magnified Moment
M_1x	=	1,020.00	kip-in
	=	85.00	kip-ft
M_2x	=	2,520.00	kip-in
	=	210.00	kip-ft

Calculation of Magnified Moment about Y-axis
Loading Data
M₁ (Lower Moment)	=	840	kip-in
M₂ (Higher Moment)	=	1,140	kip-in
Axial Load, Pu	=	600	kip

Calculation of Critical Buckling Load
k-factor	=	0.71
Unsupported Length, l_u	=	216	in
E_c	=	3,600	ksi
(I_g)_column	=	\[\frac{\pi*12^{4}}{4}\]	in⁴
	=	16,286.02	in⁴
0.2E_cI_g	=	11,725,931.75	kip-in²
E_s	=	29,000	ksi
I_se	=	395.27	in⁴
β_dns	=	0.67

(EI)_eff	=	\[\frac{0.2E_{c}I_{g} + E_{s}I_{se}}{1 + \beta_{dns}}\]	kip-in²	[ACI 318-19 6.6.4.4.4]
	=	13,913,201.38	kip-in²

Critical Buckling Load, P_c	=	\[\frac{\pi^{2}{(EI)}_{eff}}{{(kl_{u})}^{2}}\]	kip	[ACI 318-19 6.6.4.4.2]
	=	5,838.52	kip

Calculation of C_m
C_m	=	\[0.6 - 0.4\frac{M_{1}}{M_{2}}\]	≥0.4	\[\lbrack\frac{M_{1}}{M_{2}} = - ve\ for\ single\ curvature\rbrack\] [ACI 318-19 6.6.4.5.3]
	=	0.89


Calculation of Magnification Factor
δ	=	\[\frac{C_{m}}{1 - \frac{P_{u}}{0.75P_{c}}}\]		[ACI 318-19 6.6.4.5.2]

	=	1.04

Calculation of Magnified Moment
M_1y	=	870.91	kip-in
	=	72.58	kip-ft
M_2y	=	1,181.95	kip-in
	=	98.50	kip-ft

Eurocode 2-2004 Moment Magnification Non-Sway- Example 001

Moment Magnification Calculation for Slender Column (Non-Sway)

GEOMETRY, PROPERTIES AND LOADING

The moment magnification calculation for a given rectangular section using Eurocode 2:2004 is tested in this example by comparing the results with manual calculation.

The column section details are as tabulated below.

Note: Refer Eurocode 2-2004 Moment Magnification NS Ex001.cdbx

	Parameters	Value
	Width, w (mm)	400
	Height, h (mm)	500
	Rebar Layout	12-d25
	Clear Cover, c_c (mm)	40
	Compressive Strength, f_ck (MPa)	40
	Minimum Yield Stress, f_yk (MPa)	500
	Modulus of Elasticity of Concrete, E_c (MPa)	35,000
	Modulus of Elasticity of Steel, E_s (MPa)	200,000
	Concrete Area, A_c (mm2)	200,000
	Rebar Area, A_s (mm2)	5,890.49
	Concrete Inertia, I_g, 22 (mm4)	2.667x109
	Concrete Inertia, I_g, 33 (mm4)	4.167x109
	Rebar Inertia, I_s, 22 (mm4)	9.018 x107
	Rebar Inertia, I_s, 33 (mm4)	1.617x108
	Perimeter, u (mm)	1,800
	Relative Humidity, RH (%)	50
	Ratio SLS to ULS moments, rm	0.8
	Age of concrete at loading, t₀ (days)	28
	Concrete partial safety factor, γ_c	1.5
	Reinforcing partial safety factor, γ_s	1.15
	Long term compressive strength factor, α_cc	1
	Unsupported Length, l_u (m)	3
	Effective Length Factor (Braced), k_XZ	0.89
	Effective Length Factor (Braced), k_YZ	0.94
	Biaxial Loading	Yes

Name	Axial Load, N_Ed (kN)	Moment Top, M_x,top (kN-m)	Moment Top, M_y,top (kN-m)	Moment Bottom, M_x,bot (kN-m)	Moment Bottom, M_y,bot (kN-m)
Combination 1	1000	110	250	120	210

Name

Axial Load, N_Ed (kN)

Moment Top, M_x,top

(kN-m)

Moment Top, M_y,top

(kN-m)

Moment Bottom, M_x,bot

(kN-m)

Moment Bottom, M_y,bot

(kN-m)

Combination 1

1000

110

250

120

210

MOMENT MAGNIFICATION COMPARISON

Column Designer reports both the design moments in X and Y direction. The design moments are reported as the top moments for capacity calculations, while the bottom moments remain unchanged.

Design Moments (kN-m)	Column Designer	By hand	% Difference
\(M_{Ed,x}\)	150.98	150.97	0.01%
\(M_{Ed,y}\)	270.86	270.87	0.01%
\(M_{c}\)	371.03	371.03	0.00%

MANUAL CALCULATION

Calculation of Design Moment about X-axis
Loading Data
Moment Top, M_x, top	=	110,000,000	N-mm
Moment Bottom, M_x, bot	=	120,000,000	N-mm
Axial Load, N_Ed	=	1,000,000	N

Calculation of Additional Parameters
Effective Length, l₀	=	k_YZ × l_u		[EC2 5.8.3.2 (3)]
	=	0.94×3,000
	=	2820	mm

Design Compressive Strength, f_cd	=	\[\alpha_{cc} \times \frac{f_{ck}}{\gamma_{c}}\]		[EC2 3.1.6 (1)]
	=	1×40⁄1.5
	=	26.67	MPa

Mean Value Cylinder Compressive Strength, f_cm	=	f_ck + 8		[EC2 Table 3.1]
	=	40+8
	=	48	MPa

Design Yield Strength, f_yd	=	\[\frac{f_{yk}}{\gamma_{s}}\]		[EC2 3.2.7 (2)]
	=	500⁄1.15
	=	434.78	MPa

Design Strain, ε_yd	=	\[\frac{f_{yd}}{E_{s}}\]		[EC2 5.8.8.3 (1)]
	=	434.78⁄200000
	=	0.002174	mm/mm

Curvature Distribution Factor, c	=	10		[EC2 5.8.8.2 (4)]
Curvature Distribution Factor, c

Radius of Gyration of Concrete Section, r_c	=	\[\sqrt{\frac{I_{g,33}}{A_{c}}}\]
	=	\[\sqrt{\frac{4.167 \times 10^{9}}{200,000}}\]
	=	144.34	mm

Radius of Gyration of Rebar, r_s	=	\[\sqrt{\frac{I_{s,33}}{A_{s}}}\]
	=	\[\sqrt{\frac{1.617 \times 10^{8}}{5,890.49}}\]
	=	165.68	mm

Effective Depth, d	=	\[\frac{h}{2} + r_{s}\]		[EC2 5.8.8.3 (2)]
	=	500⁄2+165.68
	=	415.68	mm

Determine First Order End Moments
Determine M₀₁ and M₀₂ to satisfy \|M₀₂\| ≥ \|M₀₁\|				[EC2 5.8.8.2 (2)]
NOTE: M₀₁ is the numerically smaller end moment M₀₂ is the numerically larger end moment Algebraically, M₀₁ and M₀₂ signs should be maintained. However, if M₀₂ is < 0, they should be opposite signs.				[IStructE Manual EC2 5.5.4.2]
Lower End Moment, M₀₁	=	min {M_x, top, M_x, bot}
	=	110,000,000	N-mm
Higher End Moment, M₀₂	=	max {M_x, top, M_x, bot}
	=	120,000,000	N-mm

Calculate Curvature
Mechanical Reinforcement Ratio, ω	=	\[\frac{A_{s} \times f_{yd}}{A_{c} \times f_{cd}}\\]		[EC2 5.8.8.3 (3)]
	=	(5,890.49×434.78)/(200,000×26.67)
	=	0.4801

Relative Ultimate Axial Load, nu	=	1+ω		[EC2 5.8.8.3 (3)]
	=	1+0.4801
	=	1.4801
Relative Axial Force, n	=	\[\frac{N_{Ed}}{\left( A_{c} \times f_{cd} \right)}\]		[EC2 5.8.8.3 (3)]
	=	1,000,000⁄((200,000×26.67))
	=	0.1875

Relative Balanced Load, n_bal	=	0.4		[EC2 5.8.8.3 (3)]
Relative Balanced Load, n_bal

Axial Load Correction Factor, K_r	=	\[\frac{nu - n}{nu - n_{bal}} \leq 1\]		[EC2 5.8.8.3 (3)]
	=	\[\frac{1.4801 - 0.4}{1.4801 - 0.1875} \leq 1\]
	=	1

Member Notional Size, h₀	=	\[\frac{2 \times A_{c}}{u}\]		[EC2 Annex B.1 (1) Eqn(B.6)]
	=	(2×200,000)/1800
	=	222.22	mm

Influence of Concrete Strength Coefficient, α₁	=	\[\left( \frac{35}{f_{cm}} \right)^{0.7}\]		[EC2 Annex B.1 (1) Eqn(B.8c)]
	=	(35⁄48)^0.7
	=	0.8016

Influence of Concrete Strength Coefficient, α₂	=	\[\left( \frac{35}{f_{cm}} \right)^{0.2}\]		[EC2 Annex B.1 (1) Eqn(B.8c)]
	=	(35⁄48)^0.2
	=	0.9388

Factor for Effect of Relative Humidity on Notional Creep Coefficient, φ_RH				[EC2 Annex B.1 (1) Eqn(B.3)]
NOTE: \(\varphi_{RH} = 1 + \left( \frac{1 - \frac{RH}{100}}{0.1 \times \sqrt[3]{h_{0}}} \right),\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ for\ f_{cm} \leq 35MPa\) \(\varphi_{RH} = \left\lbrack 1 + \left( \frac{1 - \frac{RH}{100}}{0.1 \times \sqrt[3]{h_{0}}} \times \alpha_{1} \right) \right\rbrack \times \alpha_{2},\ \ \ \ \ for\ f_{cm} > 35MPa\)
φ_RH	=	\[\left\lbrack 1 + \left( \frac{1 - \frac{RH}{100}}{0.1 \times \sqrt[3]{h_{0}}} \times \alpha_{1} \right) \right\rbrack \times \alpha_{2}\]
	=	\[\left\lbrack 1 + \left( \frac{1 - \frac{50}{100}}{0.1 \times \sqrt[3]{222.22}} \times 0.8016 \right) \right\rbrack \times 0.9388\]
	=	1.56

Factor for Effect of Concrete Strength on Notional Creep Coefficient, β(f_cm)	=	\[\frac{16.8}{\sqrt{f_{cm}}}\]		[EC2 Annex B.1 (1) Eqn(B.4)]
	=	\[\frac{16.8}{\sqrt{48}}\]
	=	2.425
Factor for Effect of Concrete Age on Notional Creep Coefficient, β(t₀)	=	1/ (0.1+t₀^0.2)		[EC2 Annex B.1 (1) Eqn(B.5)]
	=	1/ (0.1+28^0.2)
	=	0.4884

Notional Creep Coefficient, φ₀	=	φ_RH × β(f_cm) × β(t₀)		[EC2 Annex B.1 (1) Eqn(B.2)]
	=	1.56×2.425×0.4884
	=	1.848

Effective Creep Ratio, φ_ef	=	φ₀ × rm		[EC2 5.8.4 (2)]
	=	1.848×0.8
	=	1.4784

Slenderness Ratio, λ	=	\[\frac{l_{0}}{r}_{c}\]		[EC2 5.8.3.2 (1)]
	=	2820⁄144.34
	=	19.54

Factor, β	=	\[0.35 + \frac{f_{ck}}{200} - \frac{\lambda}{150}\]		[EC2 5.8.8.3 (4)]
	=	0.35+40⁄200-19.54⁄150
	=	0.4197

Creep Factor, K_φ	=	1 + β × φ_ef ≥ 1		[EC2 5.8.8.3 (4)]
	=	1 + 0.4197 × 1.4784 ≥ 1
	=	1.6205

Curvature, \(\frac{1}{r}\)	=	\[K_{r} \times K_{\varphi} \times \frac{\varepsilon_{yd}}{0.45d}\]		[EC2 5.8.8.3 (1)]
	=	1×1.6205×0.002174/0.45(415.68)
	=	0.00001883	1/mm


Calculate Moment due to Geometric Imperfections
Eccentricity due to Geometric Imperfections, e_i	=	\[\max\left\{ \frac{l_{0}}{400},\frac{h}{30},20 \right\}\]		[EC2 5.2 (7)a]
	=	\[\max\left\{ \frac{2820}{400},\frac{500}{30},20 \right\}\]		[EC2 6.1 (4)]
	=	20	mm


Geometric Imperfections Moment, M_i	=	N_Ed × e_i		[EC2 5.2 (7) Fig 5.1a]
	=	1,000,000×20
	=	20,000,000	N-mm

Calculate First Order Moment
First Order Moment, M_0e	=	0.6M₀₂ + 0.4M₀₁ ≥ 0.4M₀₂		[EC2 5.8.8.2 (2)]
	=	0.6 × 120, 000, 000 + 0.4 × 110000, 000 ≥ 0.4 × 120000000
	=	116,000,000	N-mm

Included Effect of Imperfections, M_0Ed	=	M_0e+M_i		[EC2 5.8.8.2 (1)]
	=	116,000,000+20,000,000
	=	136000000	N-mm

Calculate Nominal Second Order Moment
Deflection, e₂	=	\[\left( \frac{1}{r} \right) \times {l_{0}}^{2}/c\]		[EC2 5.8.8.2 (3)]
	=	0.00001883×2820²/10
	=	14.974	mm

Nominal Second Order Moment, M₂	=	N_Ed × e₂		[EC2 5.8.8.2 (3)]
	=	1,000,000×14.974
	=	14,974,000	N-mm

Calculate Design Moment
The design moment is the maximum of:
M₀₂	=	120	kN-m	[EC2 5.8.8.2 (1)]
M_0Ed + M₂	=	136+14.974		[Concise EC2 5.6.2.2]
	=	150.97	kN-m
M₀₁ + 0.5M₂	=	110+0.5×14.974
	=	117.487	kN-m
\(N_{Ed} \times \max\left( \frac{w}{30},20 \right)\)	=	1,000×max(400/30,2)
\(N_{Ed} \times \max\left( \frac{w}{30},20 \right)\)	=	20	kN-m

Design Moment, M_Ed, x	=	\[\max\left\{ M_{02},M_{0Ed} + M_{2},M_{01} + 0.5M_{2},N_{Ed} \times \max\left( \frac{b}{30},20 \right) \right\}\]
Design Moment, M_Ed, x	=	150.97	kN-m
Calculation of Design Moment about Y-axis
Loading Data
Moment Top, M_y, top	=	250,000,000	N-mm
Moment Bottom, M_y, bot	=	210,000,000	N-mm
Axial Load, N_Ed	=	1,000,000	N

Calculation of Additional Parameters
Effective Length, l₀	=	k_XZ × l_u		[EC2 5.8.3.2 (3)]
	=	0.89×3,000
	=	2670	mm

Design Compressive Strength, f_cd	=	\[\alpha_{cc} \times \frac{f_{ck}}{\gamma_{c}}\]		[EC2 3.1.6 (1)]
	=	1×40/1.5
	=	26.67	MPa

Mean Value Cylinder Compressive Strength, f_cm	=	f_ck+8		[EC2 Table 3.1]
	=	40+8
	=	48	MPa

Design Yield Strength, f_yd	=	\[\frac{f_{yk}}{\gamma_{s}}\]		[EC2 3.2.7 (2)]
	=	500/1.15
	=	434.78	MPa

Design Strain, ε_yd	=	\[\frac{f_{yd}}{E_{s}}\]		[EC2 5.8.8.3 (1)]
	=	434.78/200,000
	=	0.002174	mm/mm

Curvature Distribution Factor, c	=	10		[EC2 5.8.8.2 (4)]
Curvature Distribution Factor, c

Radius of Gyration of Concrete Section, r_c	=	\[\sqrt{\frac{I_{g,33}}{A_{c}}}\]
	=	\[\sqrt{\frac{2.667 \times 10^{9}}{200,000}}\]
	=	115.48	mm

Radius of Gyration of Rebar, r_s	=	\[\sqrt{\frac{I_{s,33}}{A_{s}}}\]
	=	\[\sqrt{\frac{9.018 \times 10^{7}}{5,890.49}}\]
	=	123.73	mm

Effective Depth, d	=	w/2+r_s		[EC2 5.8.8.3 (2)]
	=	400/2+123.73
	=	323.73	mm

Determine First Order End Moments
Determine M₀₁ and M₀₂ to satisfy \|M₀₂\| ≥ \|M₀₁\|				[EC2 5.8.8.2 (2)]
NOTE: M₀₁ is the numerically smaller end moment M₀₂ is the numerically larger end moment Algebraically, M₀₁ and M₀₂ signs should be maintained. However, if M₀₂ is < 0, they should be opposite signs.				[IStructE Manual EC2 5.5.4.2]
Lower End Moment, M₀₁	=	min {M_y, top, M_y, bot}
	=	210,000,000	N-mm
Higher End Moment, M₀₂	=	max {M_y, top, M_y, bot}
	=	250,000,000	N-mm

Calculate Curvature
Mechanical Reinforcement Ratio, ω	=	\[\frac{A_{s} \times f_{yd}}{A_{c} \times f_{cd}}\\]		[EC2 5.8.8.3 (3)]
	=	(5,890.49×434.78)/(200,000×26.67)
	=	0.4801

Relative Ultimate Axial Load, nu	=	1+ω		[EC2 5.8.8.3 (3)]
	=	1+0.4801
	=	1.4801

Relative Axial Force, n	=	\[\frac{N_{Ed}}{\left( A_{c} \times f_{cd} \right)}\]		[EC2 5.8.8.3 (3)]
	=	1,000,000/(200,000×26.67)
	=	0.1875

Relative Balanced Load, n_bal	=	0.4		[EC2 5.8.8.3 (3)]
Relative Balanced Load, n_bal

Axial Load Correction Factor, K_r	=	\[\frac{nu - n}{nu - n_{bal}} \leq 1\]		[EC2 5.8.8.3 (3)]
	=	\[\frac{1.4801 - 0.4}{1.4801 - 0.1875} \leq 1\]
	=	1

Member Notional Size, h₀	=	\[\frac{2 \times A_{c}}{u}\]		[EC2 Annex B.1 (1) Eqn(B.6)]
	=	(2×200,000)/1800
	=	222.22	mm

Influence of Concrete Strength Coefficient, α₁	=	\[\left( \frac{35}{f_{cm}} \right)^{0.7}\]		[EC2 Annex B.1 (1) Eqn(B.8c)]
	=	(35⁄48)^0.7
	=	0.8016



Influence of Concrete Strength Coefficient, α₂	=	\[\left( \frac{35}{f_{cm}} \right)^{0.2}\]		[EC2 Annex B.1 (1) Eqn(B.8c)]
	=	(35⁄48)^0.2
	=	0.9388

Factor for Effect of Relative Humidity on Notional Creep Coefficient, φ_RH				[EC2 Annex B.1 (1) Eqn(B.3)]
NOTE: \(\varphi_{RH} = 1 + \left( \frac{1 - \frac{RH}{100}}{0.1 \times \sqrt[3]{h_{0}}} \right),\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ for\ f_{cm} \leq 35MPa\) \(\varphi_{RH} = \left\lbrack 1 + \left( \frac{1 - \frac{RH}{100}}{0.1 \times \sqrt[3]{h_{0}}} \times \alpha_{1} \right) \right\rbrack \times \alpha_{2},\ \ \ \ \ for\ f_{cm} > 35MPa\)
φ_RH	=	\[\left\lbrack 1 + \left( \frac{1 - \frac{RH}{100}}{0.1 \times \sqrt[3]{h_{0}}} \times \alpha_{1} \right) \right\rbrack \times \alpha_{2}\]
	=	\[\left\lbrack 1 + \left( \frac{1 - \frac{50}{100}}{0.1 \times \sqrt[3]{222.22}} \times 0.8016 \right) \right\rbrack \times 0.9388\]
	=	1.56

Factor for Effect of Concrete Strength on Notional Creep Coefficient, β(f_cm)	=	\[\frac{16.8}{\sqrt{f_{cm}}}\]		[EC2 Annex B.1 (1) Eqn(B.4)]
	=	\[\frac{16.8}{\sqrt{48}}\]
	=	2.425

Factor for Effect of Concrete Age on Notional Creep Coefficient, β(t₀)	=	1/(0.1+t₀^0.2)		[EC2 Annex B.1 (1) Eqn(B.5)]
	=	1/ (0.1+28^0.2)
	=	0.4884

Notional Creep Coefficient, φ₀	=	φ_RH × β(f_cm) × β(t₀)		[EC2 Annex B.1 (1) Eqn(B.2)]
	=	1.56×2.425×0.4884
	=	1.848

Effective Creep Ratio, φ_ef	=	φ₀ × rm		[EC2 5.8.4 (2)]
	=	1.848×0.8
	=	1.4784

Slenderness Ratio, λ	=	l₀/r_c		[EC2 5.8.3.2 (1)]
	=	2670/115.48
	=	23.12

Factor, β	=	\[0.35 + \frac{f_{ck}}{200} - \frac{\lambda}{150}\]		[EC2 5.8.8.3 (4)]
	=	0.35+40/200-23.12/150
	=	0.3959



Creep Factor, K_φ	=	1 + β × φ_ef ≥ 1		[EC2 5.8.8.3 (4)]
	=	1 + 0.3959 × 1.4784 ≥ 1
	=	1.5853

Curvature, \(\frac{1}{r}\)	=	\[K_{r} \times K_{\varphi} \times \frac{\varepsilon_{yd}}{0.45d}\]		[EC2 5.8.8.3 (1)]
	=	1×1.5853×0.002174/0.45(323.73)
	=	0.00002366	1/mm

Calculate Moment due to Geometric Imperfections
Eccentricity due to Geometric Imperfections, e_i	=	\[\max\left\{ \frac{l_{0}}{400},\frac{w}{30},20 \right\}\]		[EC2 5.2 (7)a]
	=	\[\max\left\{ \frac{2670}{400},\frac{400}{30},20 \right\}\]		[EC2 6.1 (4)]
	=	20	mm

Geometric Imperfections Moment, M_i	=	N_Ed × e_i		[EC2 5.2 (7) Fig 5.1a]
	=	1,000,000×20
	=	20,000,000	N-mm

Calculate First Order Moment
First Order Moment, M_0e	=	0.6M₀₂ + 0.4M₀₁ ≥ 0.4M₀₂		[EC2 5.8.8.2 (2)]
	=	0.6 × 250, 000, 000 + 0.4 × 210000000 ≥ 0.4 × 250000000
	=	234,000,000	N-mm

Included Effect of Imperfections, M_0Ed	=	M_0e + M_i		[EC2 5.8.8.2 (1)]
	=	234,000,000+20,000,000
	=	254,000,000	N-mm

Calculate Nominal Second Order Moment
Deflection, e₂	=	\[\left( \frac{1}{r} \right) \times {l_{0}}^{2}/c\]		[EC2 5.8.8.2 (3)]
	=	0.00002366×2670²/10
	=	16.867	mm

Nominal Second Order Moment, M₂	=	N_Ed × e₂		[EC2 5.8.8.2 (3)]
	=	1,000,000×16.867
	=	16,867,000	N-mm

Calculate Design Moment
The design moment is the maximum of:
M₀₂	=	250	kN-m	[EC2 5.8.8.2 (1)]
M_0Ed + M₂	=	254+16.867		[Concise EC2 5.6.2.2]
	=	270.87	kN-m
M₀₁ + 0.5M₂	=	210+0.5×16.867
	=	218.43	kN-m


\(N_{Ed} \times \max\left( \frac{h}{30},20 \right)\)	=	1,000×max (500⁄30,2)
\(N_{Ed} \times \max\left( \frac{h}{30},20 \right)\)	=	20	kN-m

Design Moment, M_Ed, y	=	\[\max\left\{ M_{02},M_{0Ed} + M_{2},M_{01} + 0.5M_{2},N_{Ed} \times \max\left( \frac{b}{30},20 \right) \right\}\]
Design Moment, M_Ed, y	=	270.87	kN-m


Calculate Effective Depths				[IStructE Manual EC2 5.5.5]
Effective Height, h^′	=	\[\max\left\{ h - c_{c} - \frac{c_{b}}{2},w - c_{c} - \frac{c_{b}}{2} \right\}\]
	=	\[\max\left\{ 500 - 40 - \frac{25}{2},400 - 40 - \frac{25}{2} \right\}\]
	=	447.5	mm

Effective Width, b^′	=	\[\min\left\{ h - c_{c} - \frac{c_{b}}{2},w - c_{c} - \frac{c_{b}}{2} \right\}\]
	=	\[\min\left\{ 500 - 40 - \frac{25}{2},400 - 40 - \frac{25}{2} \right\}\]
	=	347.5	mm

Calculate Coefficient for Biaxial Bending				[IStructE Manual EC2 5.5.5]
β	=	\[1 - 1.165 \times min\left\{ 0.6,\ \frac{N_{Ed}}{\left( A_{c} \times f_{ck} \right)} \right\}\]
	=	\[1 - 1.165 \times min\left\{ 0.6,\ \frac{1000000}{(200,000 \times 40)} \right\}\]
	=	0.8544

Calculate Increased Design Moment				[IStructE Manual EC2 5.5.5]
NOTE: \(M_{c} = M_{Ed,x} + \beta \times \frac{h'}{b'} \times M_{Ed,y}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ for\frac{M_{Ed,x}}{h'} \geq \frac{M_{Ed,y}}{b'}\) \(M_{c} = M_{Ed,y} + \beta \times \frac{b'}{h'} \times M_{Ed,x}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ for\frac{M_{Ed,x}}{h'} < \frac{M_{Ed,y}}{b'}\)

Final Design Moment, M_c	=	\[M_{Ed,y} + \beta \times \frac{b'}{h'} \times M_{Ed,x}\\]
	=	\[270.87 + 0.8544 \times \frac{347.5}{447.5} \times 150.97\\]
	=	371.03	MPa

BS 8110-97 Moment Magnification Non-Sway- Example 001

Moment Magnification Calculation for Slender Column (Non-Sway)

GEOMETRY, PROPERTIES AND LOADING

The moment magnification calculation for a given rectangular section using BS 8110-97 is tested in this example by comparing the results with manual calculation.

The column section details are as tabulated below.

Note: Refer BS8110-97 Moment Magnification NS Ex001.cdbx

	Parameters	Value
	Width, w (mm)	400
	Height, h (mm)	500
	Rebar Layout	12-d25
	Clear Cover, c_c (mm)	40
	Compressive Strength, f_cu (MPa)	40
	Minimum Yield Stress, f_y (MPa)	500
	Concrete Area, A_c (mm²)	200,000
	Rebar Area, A_sc (mm²)	5,890.49
	Unsupported Length, l_u (m)	3
	Effective Length Factor (Braced), β_XZ	1.00
	Effective Length Factor (Braced), β_YZ	1.00
	Biaxial Loading	Yes

Name	Axial Load, N (kN)	Moment Top, M_x,top (kN-m)	Moment Top, M_y,top (kN-m)	Moment Bottom, M_x,bot (kN-m)	Moment Bottom, M_y,bot (kN-m)
Combination 1	1000	110	250	120	210

Name

Axial Load,

N (kN)

Moment Top, M_x,top

(kN-m)

Moment Top, M_y,top

(kN-m)

Moment Bottom, M_x,bot

(kN-m)

Moment Bottom, M_y,bot

(kN-m)

Combination 1

1000

110

250

120

210

MOMENT MAGNIFICATION COMPARISON

Column Designer reports both the design moments in X and Y direction. The design moments are reported as the top moments for capacity calculations, while the bottom moments remain unchanged.

Design Moments (kN-m)	Column Designer	By hand	% Difference
\(M_{x}\)	124.51	125	0.39%
\(M_{y}\)	250	250	0.00%
\(M_{c}\)	332.61	332.93	0.1%

MANUAL CALCULATION

Calculation of Design Moment about X-axis
Loading Data
Moment Top, M_x, top	=	110,000,000	N-mm
Moment Bottom, M_x, bot	=	120,000,000	N-mm
Axial Load, N	=	1,000,000	N

Calculation of Additional Parameters
Effective Length, l_e	=	β_YZ × l_u		[BS 8110-1 3.8.1.6.1]
	=	1.00×3000
	=	3000	mm

Balanced Section Axial Load, N_bal	=	0.25 × f_cu × A_c		[BS 8110-1 3.8.1.1]
	=	0.25×40×200,000
	=	2,000,000	N

Determine First Order End Moments
Determine M₁ and M₂ to satisfy \|M₂\| ≥ \|M₁\|				[BS 8110-1 3.8.3.2]
NOTE: M₁ is the numerically smaller end moment M₂ is the numerically larger end moment Algebraically, M₁ and M₂ signs should be maintained. However, if M₂ is < 0, they should be opposite signs.
Lower End Moment, M₁	=	min {M_x, top, M_x, bot}
	=	110,000,000	N-mm
Higher End Moment, M₂	=	max {M_x, top, M_x, bot}
	=	120,000,000	N-mm

Calculate Additional Moment
Section Ultimate Capacity to Axial Load, N_uz	=	0.45 × f_cu × A_c + 0.95 × f_y × A_sc		[BS 8110-1 3.8.3.1]
	=	0.45×40×200,000+0.95×500×5,890.49
	=	6,397,982.75	N

Reduction Factor, K	=	\[\frac{N_{uz} - N}{N_{uz} - N_{bal}} \leq 1\]		[BS 8110-1 3.8.3.1]
	=	\[\frac{6397982.75\ - 1,000,000}{6397982.75\ - 2,000,000} \leq 1\]
	=	1

Slenderness Factor, β_a				[BS 8110-1 3.8.3.1]
NOTE: \(\beta_{a} = \frac{1}{2000}\left( \frac{l_{e}}{h} \right)^{2},\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ for\ Biaxial\ loading\) \(\beta_{a} = \frac{1}{2000}\left( \frac{l_{e}}{\min\left\{ h,w \right\}} \right)^{2},\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ for\ Uniaxial\ loading\)				[BS 8110-1 3.8.3.6]
β_a	=	\[\beta_{a} = \frac{1}{2,000}\left( \frac{l_{e}}{h} \right)^{2}\]
	=	\[\beta_{a} = \frac{1}{2,000}\left( \frac{3,000}{500} \right)^{2}\]
	=	0.018

Deflection, a_u	=	β_a × K × h		[BS 8110-1 3.8.3.1]
	=	0.018×1×500
	=	9	mm

Additional Moment, M_add	=	N × a_u		[BS 8110-1 3.8.3.1]
	=	1,000,000×9
	=	9,000,000	N-mm

Calculate Initial Moment
Initial Moment, M_i	=	0.4M₁ + 0.6M₂ ≥ 0.4M₂		[BS 8110-1 3.8.3.2]
	=	0.4 × 110, 000, 000 + 0.6 × 120000000 ≥ 0.4 × 120000000
	=	116,000,000	N-mm

Calculate Design Moment
The design moment is the maximum of:
M₂	=	120	kN-m	[BS 8110-1 3.8.3.2]
M_i + M_add	=	116 + 9
	=	125	kN-m
M₁ + 0.5M_add	=	110 + 0.5 × 9
	=	114.5	kN-m
N × min (0.05 × h, 20)	=	1000 × max (0.05 × 500, 20)
N × min (0.05 × h, 20)	=	25	kN-m

Design Moment, M_x	=	max {M₂, M_i + M_add, M₁ + 0.5M_add, N × min (0.05 × h, 20)}
Design Moment, M_x	=	125	kN-m

Calculation of Design Moment about Y-axis
Loading Data
Moment Top, M_y, top	=	250,000,000	N-mm
Moment Bottom, M_y, bot	=	210,000,000	N-mm
Axial Load, N	=	1,000,000	N

Calculation of Additional Parameters
Effective Length, l_e	=	β_XZ × l_u		[BS 8110-1 3.8.1.6.1]
	=	1.00×3,000
	=	3000	mm

Balanced Section Axial Load, N_bal	=	0.25 × f_cu × A_c		[BS 8110-1 3.8.1.1]
	=	0.25×40×200,000
	=	2,000,000	N

Determine First Order End Moments
Determine M₁ and M₂ to satisfy \|M₂\| ≥ \|M₁\|				[BS 8110-1 3.8.3.2]
NOTE: M₁ is the numerically smaller end moment M₂ is the numerically larger end moment Algebraically, M₁ and M₂ signs should be maintained. However, if M₂ is < 0, they should be opposite signs.
Lower End Moment, M₁	=	min {M_y, top, M_y, bot}
	=	210,000,000	N-mm
Higher End Moment, M₂	=	max {M_y, top, M_y, bot}
	=	250,000,000	N-mm

Calculate Additional Moment
Section Ultimate Capacity to Axial Load, N_uz	=	0.45 × f_cu × A_c + 0.95 × f_y × A_sc		[BS 8110-1 3.8.3.1]
	=	0.45×40×200,000+0.95×500×5890.49
	=	6,397,982.75	N

Reduction Factor, K	=	\[\frac{N_{uz} - N}{N_{uz} - N_{bal}} \leq 1\]		[BS 8110-1 3.8.3.1]
	=	\[\frac{6397982.75\ - 1,000,000}{6397982.75\ - 2,000,000} \leq 1\]
	=	1

Slenderness Factor, β_a				[BS 8110-1 3.8.3.1]
NOTE: \(\beta_{a} = \frac{1}{2000}\left( \frac{l_{e}}{w} \right)^{2},\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ for\ Biaxial\ loading\) \(\beta_{a} = \frac{1}{2000}\left( \frac{l_{e}}{\min\left\{ h,w \right\}} \right)^{2},\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ for\ Uniaxial\ loading\)				[BS 8110-1 3.8.3.6]
β_a	=	\[\beta_{a} = \frac{1}{2000}\left( \frac{l_{e}}{w} \right)^{2}\]
	=	\[\beta_{a} = \frac{1}{2000}\left( \frac{3000}{400} \right)^{2}\]
	=	0.028125

Deflection, a_u	=	β_a × K × w		[BS 8110-1 3.8.3.1]
	=	0.028125×1×400
	=	11.25	mm

Additional Moment, M_add	=	N × a_u		[BS 8110-1 3.8.3.1]
	=	1,000,000×11.25
	=	11250000	N-mm

Calculate Initial Moment
Initial Moment, M_i	=	0.4M₁ + 0.6M₂ ≥ 0.4M₂		[BS 8110-1 3.8.3.2]
	=	0.4 × 210, 000, 000 + 0.6 × 250000000 ≥ 0.4 × 250000000
	=	234,000,000	N-mm

Calculate Design Moment
The design moment is the maximum of:
M₂	=	250	kN-m	[BS 8110-1 3.8.3.2]
M_i + M_add	=	234+11.25
	=	245.25	kN-m
M₁ + 0.5M_add	=	210+0.5×11.25
	=	215.63	kN-m
N × min (0.05 × w, 20)	=	1000×max (0.05×400,20)
N × min (0.05 × w, 20)	=	20	kN-m

Design Moment, M_y	=	max {M₂, M_i + M_add, M₁ + 0.5M_add, N × min (0.05 × w, 20)}
Design Moment, M_y	=	250	kN-m

Increased Design Moment due to Biaxial Bending
Calculate Effective Depths				[BS 8110-1 3.8.4.5]
Effective Height, h^′	=	\[\max\left\{ h - c_{c} - \frac{c_{b}}{2},w - c_{c} - \frac{c_{b}}{2} \right\}\]
	=	\[\max\left\{ 500 - 40 - \frac{25}{2},400 - 40 - \frac{25}{2} \right\}\]
	=	447.5	mm

Effective Width, b^′	=	\[\min\left\{ h - c_{c} - \frac{c_{b}}{2},w - c_{c} - \frac{c_{b}}{2} \right\}\]
	=	\[\min\left\{ 500 - 40 - \frac{25}{2},400 - 40 - \frac{25}{2} \right\}\]
	=	347.5	mm

Calculate Coefficient for Biaxial Bending				[BS 8110-1 3.8.4.5]
β	=	\[1 - 1.165 \times min\left\{ 0.6,\ \frac{N}{\left( A_{c} \times f_{cu} \right)} \right\}\]
	=	\[1 - 1.165 \times min\left\{ 0.6,\ \frac{1000000}{(200000 \times 40)} \right\}\]
	=	0.8544

Calculate Increased Design Moment				[BS 8110-1 3.8.4.5]
NOTE: \(M_{c} = M_{x} + \beta \times \frac{h'}{b'} \times M_{y}\ \ \ \ \ \ \ \ \ for\frac{M_{x}}{h'} \geq \frac{M_{y}}{b'}\) \(M_{c} = M_{y} + \beta \times \frac{b'}{h'} \times M_{x}\ \ \ \ \ \ \ \ \ for\frac{M_{x}}{h'} < \frac{M_{y}}{b'}\)

Final Design Moment, M_c	=	\[M_{y} + \beta \times \frac{b'}{h'} \times M_{x}\\]
	=	\[250 + 0.8544 \times \frac{347.5}{447.5} \times 125\\]
	=	332.93	MPa

IS 456-2000 Moment Magnification Non-Sway- Example 001

Moment Magnification Calculation for Slender Column (Non-Sway)

GEOMETRY, PROPERTIES AND LOADING

The moment magnification calculation for a given rectangular section using IS 456:2000 is tested in this example by comparing the results with manual calculation.

The column section details are as tabulated below.

Note: Refer IS 456-2000 Moment Magnification NS Ex001.cdbx

	Parameters	Value
	Width, w (mm)	400
	Height, h (mm)	500
	Rebar Layout	12-d25
	Clear Cover, c_c (mm)	40
	Compressive Strength, f_ck (MPa)	40
	Minimum Yield Stress, f_y (MPa)	500
	Concrete Area (Gross), A_c (mm²)	200,000
	Rebar Area, A_sc (mm²)	5,890.49
	Unsupported Length, l_u (m)	3
	Effective Length Factor (Braced), β_XZ	0.94
	Effective Length Factor (Braced), β_YZ	0.97
	Is Biaxial?	Yes

Name	Axial Load, N (kN)	Moment Top, M_x,top (kN-m)	Moment Top, M_y,top (kN-m)	Moment Bottom, M_x,bot (kN-m)	Moment Bottom, M_y,bot (kN-m)
Combination 1	1000	110	250	120	210

Name

Axial Load,

N (kN)

Moment Top, M_x,top

(kN-m)

Moment Top, M_y,top

(kN-m)

Moment Bottom, M_x,bot

(kN-m)

Moment Bottom, M_y,bot

(kN-m)

Combination 1

1000

110

250

120

210

MOMENT MAGNIFICATION COMPARISON

Column Designer reports both the design moments in X and Y direction. The design moments are reported as the top moments for capacity calculations, while the bottom moments remain unchanged.

Design Moments (kN-m)	Column Designer	By hand	% Difference
\(M_{ut,x}\)	123.67	124.47	0.64%
\(M_{ut,y}\)	250	250	0.00%
\(M_{c}\)	278.92	279.27	0.13%

MANUAL CALCULATION

Calculation of Design Moment about X-axis
Loading Data
Moment Top, M_x, top	=	110,000,000	N-mm
Moment Bottom, M_x, bot	=	120,000,000	N-mm
Axial Load, N	=	1,000,000	N

Calculation of Additional Parameters
Effective Length, l_ef	=	β_YZ × l_u		[IS 456 25.2]
	=	0.97×3,000		[IS 456 Annex E E-1]
	=	2910	mm

Balanced Section Axial Load, P_b	=	From interaction diagram
Balanced Section Axial Load, P_b	=	1,363,185.47	N

Determine First Order End Moments
Determine M_u1 and M_u2 to satisfy \|M_u2\| ≥ \|M_u1\|				[IS 456 39.7.1]
NOTE: M_u1 is the numerically smaller end moment M_u2 is the numerically larger end moment Algebraically, M_u1 and M_u2 signs should be maintained. However, if M_u2 is < 0, they should be opposite signs.
Lower End Moment, M_u1	=	min {M_x, top, M_x, bot}
	=	110,000,000	N-mm
Higher End Moment, M_u2	=	max {M_x, top, M_x, bot}
	=	120,000,000	N-mm

Calculate Moment due to Minimum Eccentricity
Minimum Moment, M_min	=	\[P_{u} \times \max\left( \frac{l_{u}}{500} + \frac{h}{30},20 \right)\]		[IS 456 25.4]
	=	1,000,000×max (3,000/500+500/30,20)
	=	22,666,666.67	N-mm

Calculate Additional Moment
Ultimate Capacity Axial Load, P_uz	=	0.45 × f_ck × A_c + (0.75 × f_y − 0.45 × f_ck) × A_sc			[IS 456 39.6]
	=	0.45×40×200,000+(0.75×500-0.45×40) ×5890.49
	=	5,702,904.93	N

Modification Factor, k_a	=	\[\frac{P_{uz} - P_{u}}{P_{uz} - P_{b}} \leq 1\]		[IS 456 39.7.1.1]
	=	\[\frac{5,702,904.93\ - 1,000,000}{5,702,904.93\ - 1,363,185.47} \leq 1\]
	=	1

Additional Moment, M_a	=	\[k_{a} \times \frac{P_{u} \times h}{2000}\left\{ \frac{l_{ef}}{h} \right\}^{2}\]		[IS 456 39.7.1]
	=	\[1 \times \frac{1,000,000 \times 500}{2000}\left\{ \frac{2910}{500} \right\}^{2}\]
	=	8,468,100	N-mm

Calculate Primary Moment
Primary Moment, M_ui	=	0.4M_u1 + 0.6M_u2 ≥ 0.4M_u2		[IS 456 39.7.1 Note2]
	=	0.4 × 110, 000, 000 + 0.6 × 12, 000, 0000 ≥ 0.4 × 120, 000, 000
	=	116,000,000	N-mm

Check Minimum Moment, M_ui	=	M_ui ≥ M_min		[IS 456 25.4]
Check Minimum Moment, M_ui	=	116,000,000	N-mm


Calculate Design Moment
The design moment is the maximum of:				[IS 456 39.7.1 Note2]
M_u2	=	120	kN-m
M_ui + M_a	=	116+8.47
	=	124.47	kN-m

Design Moment, M_ut, x	=	max {M_u2, M_ui + M_a}
Design Moment, M_ut, x	=	124.47	kN-m

Calculation of Design Moment about Y-axis
Loading Data
Moment Top, M_y, top	=	250,000,000	N-mm
Moment Bottom, M_y, bot	=	210,000,000	N-mm
Axial Load, N	=	1,000,000	N

Calculation of Additional Parameters
Effective Length, l_ef	=	β_XZ × l_u		[IS 456 25.2]
	=	0.94×3,000		[IS 456 Annex E E-1]
	=	2,820	mm

Balanced Section Axial Load, P_b	=	From interaction diagram
Balanced Section Axial Load, P_b	=	1,363,185.47	N

Determine First Order End Moments
Determine M_u1 and M_u2 to satisfy \|M_u2\| ≥ \|M_u1\|				[IS 456 39.7.1]
NOTE: M_u1 is the numerically smaller end moment M_u2 is the numerically larger end moment Algebraically, M_u1 and M_u2 signs should be maintained. However, if M_u2 is < 0, they should be opposite signs.
Lower End Moment, M_u1	=	min {M_y, top, M_y, bot}
	=	210,000,000	N-mm
Higher End Moment, M_u2	=	max {M_y, top, M_y, bot}
	=	250,000,000	N-mm

Calculate Moment due to Minimum Eccentricity
Minimum Moment, M_min	=	\[P_{u} \times \max\left( \frac{l_{u}}{500} + \frac{w}{30},20 \right)\]		[IS 456 25.4]
	=	1,000,000×max (3,000/500+400/30,20)
	=	20,000,000	N-mm

Calculate Additional Moment
Ultimate Capacity Axial Load, P_uz	=	0.45 × f_ck × A_c + (0.75 × f_y − 0.45 × f_ck) × A_sc			[IS 456 39.6]
	=	0.45×40×200,000+(0.75×500-0.45×40) ×5890.49
	=	5,702,904.93	N

Modification Factor, k_a	=	\[\frac{P_{uz} - P_{u}}{P_{uz} - P_{b}} \leq 1\]		[IS 456 39.7.1.1]
	=	\[\frac{5,702,904.93\ - 1,000,000}{5,702,904.93\ - 1,363,185.47} \leq 1\]
	=	1

Additional Moment, M_a	=	\[k_{a} \times \frac{P_{u} \times w}{2000}\left\{ \frac{l_{ef}}{w} \right\}^{2}\]		[IS 456 39.7.1]
	=	\[1 \times \frac{1,000,000 \times 400}{2000}\left\{ \frac{2820}{400} \right\}^{2}\]
	=	9,940,500	N-mm

Calculate Primary Moment
Primary Moment, M_ui	=	0.4M_u1 + 0.6M_u2 ≥ 0.4M_u2		[IS 456 39.7.1 Note2]
	=	0.4 × 210000000 + 0.6 × 250000000 ≥ 0.4 × 250000000
	=	234,000,000	N-mm

Check Minimum Moment, M_ui	=	M_ui ≥ M_min		[IS 456 25.4]
Check Minimum Moment, M_ui	=	234,000,000	N-mm


Calculate Design Moment
The design moment is the maximum of:				[IS 456 39.7.1 Note2]
M_u2	=	250	kN-m
M_ui + M_a	=	234+9.94
	=	243.94	kN-m

Design Moment, M_ut, y	=	max {M_u2, M_ui + M_a}
Design Moment, M_ut, y	=	250	kN-m

Increased Design Moment due to Biaxial Bending
Final Design Moment, M_c	=	\[\sqrt{{M_{ut,x}}^{2} + {M_{ut,y}}^{2}}\\]
	=	\[\sqrt{{124.47}^{2} + 250^{2}}\\]
	=	279.27	kN-m

AS 3600-2018 Moment Magnification Non-Sway- Example 001

Moment Magnification Calculation for Slender Column (Non-Sway)

GEOMETRY, PROPERTIES AND LOADING

The moment magnification calculation for a given rectangular section using AS 3600-2018 is tested in this example by comparing the results with manual calculation.

The column section details are as tabulated below.

Note: Refer AS 3600-2018 Moment Magnification NS Ex001.cdbx

	Parameters	Value
	Width, w (mm)	400
	Height, h (mm)	500
	Rebar Layout	12-d25
	Clear Cover, c_c (mm)	40
	Compressive Strength, f_c^′ (MPa)	40
	Minimum Yield Stress, f_y (MPa)	500
	Concrete Area (Gross), A_c (mm²)	200,000
	Rebar Area, A_sc (mm²)	5,890.49
	Unsupported Length, l_u (m)	3
	Effective Length Factor (Braced), k_XZ	0.94
	Effective Length Factor (Braced), k_YZ	0.97

Name	Axial Load, *N^ (kN)**	*Moment Top, M_x^_,top (kN-m)**	*Moment Top, M_y^_,top (kN-m)**	*Moment Bottom, M_x^_,bot (kN-m)**	*Moment Bottom, M_y^_,bot (kN-m)**	Loading Factor, β_d
Combination 1	6000	110	250	120	210	0.5

Name

Axial Load,

N^* (kN)

Moment Top, M_x^*_,top

(kN-m)

Moment Top, M_y^*_,top

(kN-m)

Moment Bottom, M_x^*_,bot

(kN-m)

Moment Bottom, M_y^*_,bot

(kN-m)

Loading Factor, β_d

Combination 1

6000

110

250

120

210

0.5

MOMENT MAGNIFICATION COMPARISON

Column Designer reports both the design moments in X and Y direction. The design moments are reported as the top moments for capacity calculations, while the bottom moments remain unchanged.

Design Moments (kN-m)	Column Designer	By hand	% Difference
\(M_{x}^{*}\)	162.8	165.6	1.71%
\(M_{y}^{*}\)	352.5	357.5	1.42%
\(M_{c}\)	352.5	357.5	1.42%

MANUAL CALCULATION

Calculation of Design Moment about X-axis
Loading Data
M₁^* (Lower Moment)	=	110,000,000	N-mm
M₂^* (Higher Moment)	=	120,000,000	N-mm
Axial Load, N*	=	6,000,000	N

Calculation of Additional Parameters
Effective Length, L_e	=	k_YZ × l_u		[AS 3600 10.5.3]
	=	0.97×3,000
	=	2910	mm

Minimum Moment, M_min	=	0.05DN^*				[AS 3600 10.1.2]
Minimum Moment, M_min	=	150,000,000	N-mm

Balanced Section Moment, M_c	=	M_ub (From interaction diagram)		[Guide to Reinforced Concrete Design]
Balanced Section Moment, M_c	=	476,923,076.9	N-mm	[Guide to Reinforced Concrete Design]
Calculation of Critical Buckling Load
Critical Buckling Load, N_c	=	\[\left( \frac{\pi^{2}}{{L_{e}}^{2}} \right)\left\lbrack \frac{182d_{0}\varnothing M_{c}}{\left( 1 + \beta_{d} \right)} \right\rbrack\]		[AS 3600 10.4.4]
	=	20,165,705.56	N

Calculation of Magnification Factor
Minimum Moment, K_m	=	\[0.6 - 0.4\frac{M_{1}^{}}{M_{2}^{}} \geq 0.4\]			[AS 3600 10.3.1] [AS 3600 10.4.2]
Minimum Moment, K_m	=	0.97
δ_b	=	\[\frac{K_{m}}{\left( 1 - \frac{N^{*}}{N_{c}} \right)} \geq 1\]				[AS 3600 10.4.2]
δ_b	=	1.38

Calculate Magnified Moment
M_x^*	=	δ_bM₂^*		[AS 3600 10.4.1]
	=	165.6	kN-m
Calculation of Design Moment about Y-axis
Loading Data
M₁^* (Lower Moment)	=	210,000,000	N-mm
M₂^* (Higher Moment)	=	250,000,000	N-mm
Axial Load, N*	=	6,000,000	N

Calculation of Additional Parameters
Effective Length, L_e	=	k_XZ × l_u		[AS 3600 10.5.3]
	=	0.94×3,000
	=	2820	mm

Minimum Moment, M_min	=	0.05DN^*				[AS 3600 10.1.2]
Minimum Moment, M_min	=	120,000,000	N-mm

Balanced Section Moment, M_c	=	M_ub (From interaction diagram)		[Guide to Reinforced Concrete Design]
Balanced Section Moment, M_c	=	492,307,692.3	N-mm	[Guide to Reinforced Concrete Design]
Calculation of Critical Buckling Load
Critical Buckling Load, N_c	=	\[\left( \frac{\pi^{2}}{{L_{e}}^{2}} \right)\left\lbrack \frac{182d_{0}\varnothing M_{c}}{\left( 1 + \beta_{d} \right)} \right\rbrack\]		[AS 3600 10.4.4]
	=	17,347,389.78	N

Calculation of Magnification Factor
Minimum Moment, K_m	=	\[0.6 - 0.4\frac{M_{1}^{}}{M_{2}^{}} \geq 0.4\]			[AS 3600 10.3.1] [AS 3600 10.4.2]
Minimum Moment, K_m	=	0.936

δ_b	=	\[\frac{K_{m}}{\left( 1 - \frac{N^{*}}{N_{c}} \right)} \geq 1\]				[AS 3600 10.4.2]
δ_b	=	1.43

Calculate Magnified Moment
M_y^*	=	δ_bM₂^*		[AS 3600 10.4.1]
	=	357.5	kN-m

Final Design Moment
Final Design Moment, M_c	=	max (M_x, M_y)
Final Design Moment, M_c	=	357.5	kN-m

Moment Magnification – Sway

ACI 318-19 Moment Magnification Sway- Example 001

Moment Magnification Calculation for Slender Column (Sway)

GEOMETRY, PROPERTIES AND LOADING

The moment magnification calculation for a given rectangular section using ACI 318-19 is tested in this example by comparing the results with manual calculation.

The column section details are as tabulated below.

Note: Refer ACI 318-19 Moment Magnification Sway Ex001.cdbx

	Parameters	Value
	Width (in)	20
	Height (in)	24
	Compressive Strength, fc’ (psi)	4,000
	Modulus of Elasticity of Concrete, E_c (ksi)	3,600
	Minimum Yield Stress, f_y (psi)	40,000
	Modulus of Elasticity of Steel, E_s (ksi)	29,000
	Rebar Layout	10-#9
	Rebar Area (in²)	9.99
	Rebar Ratio	2.08%
	Clear Cover (in)	1.5
	C/C Length, l_c (ft)	17.5
	Unsupported Length, l_u (ft)	17
	k-factor, Unbraced (YZ Plane)	1.5
	k-factor, Braced (YZ Plane)	0.78

Name	Non-Sway Part	Sway Part
Combo1	350	48	-55	250	10	76	-72

Name

Non-Sway Part

Sway Part

Axial Load, P_u (kip)

Moment Top, M_ux (kip-ft)

Moment Bottom, M_ux

(kip-ft)

Sustained Load

(kip)

Axial Load, P_u (kip)

Moment Top, M_ux (kip-ft)

Moment Bottom, M_ux

(kip-ft)

Combo1

350

48

-55

250

10

76

-72

Parameter	Value
Story Axial Load, ∑P_u (kip)	12,000
Story Critical Load, ∑P_c (kip)	40,000
Relative Sway, Δ (in)	0.3
Story Shear Load, V_us (kip)	243

MOMENT MAGNIFICATION COMPARISON

Column Designer reports both the top and bottom magnified moments in X and Y direction. The higher of the top and bottom magnified moments is reported as M₂ while the lower magnified moment is reported as M₁.

Magnified Moments (kip-ft)	Column Designer	By hand	% Difference
M_1x	129.77	129.77	0.00%
M_2x	-132.46	-132.46	0.00%

MANUAL CALCULATION

Calculation of Magnified Moment about X-axis
Loading Data
M_1ns	=	576	kip-in
M_2ns	=	-660	kip-in
Axial Load, Non-sway Part (P_u)	=	350	kip
M_1s	=	912	kip-in
M_2s	=	-864	kip-in
Axial Load, Sway Part (P_u)	=	10	kip

Calculation of δ_s
ΣP_u	=	12,000	kip
ΣP_c	=	40,000	kip
Relative Lateral Deflection, Δ_o	=	0.3	in
Story Shear, V_us	=	243	kip
C/C length of Column, l_c	=	210	in
Q	=	\[\frac{\sum_{}^{}{P_{u}\mathrm{\Delta}_{o}}}{V_{us}l_{c}}\]		[ACI 318-19 6.6.4.4.1]

	=	0.07

δ_s	=	\[\frac{1}{1 - Q}\]	≥ 1	[ACI 318-19 6.6.4.6.2a]

	=	1.08

Calculation of M₁ and M₂
M₁ (Lower Moment)	=	M_1ns + δ_sM_1s		[ACI 318-19 6.6.4.6.1a]

	=	1557.22	kip-in

M₂ (Higher Moment)	=	M_2ns + δ_sM_2s		[ACI 318-19 6.6.4.6.1b]

	=	-1589.58	kip-in

Magnification along Column Length
Calculation of Critical Buckling Load
K factor (Braced)	=	0.78
Unsupported Length, l_u	=	204	in
(I_g)_column	=	\[\frac{20*24^{3}}{12}\]	in⁴

	=	23,040	in⁴
E_c	=	3,600	ksi
0.2E_cI_g	=	16,588,800	kip-in²
E_s	=	29,000	ksi
I_se	=	636.59	in⁴
β_dns	=	0.69

(EI)_eff	=	\[\frac{0.2E_{c}I_{g} + E_{s}I_{se}}{1 + \beta_{dns}}\]	kip-in²	[ACI 318-19 6.6.4.4.4]

	=	20,685,178.28	kip-in²
Critical Buckling Load, P_c	=	\[\frac{\pi^{2}{(EI)}_{eff}}{{(kl_{u})}^{2}}\]	kip	[ACI 318-19 6.6.4.4.2]

	=	8,063.24	kip

Calculation of C_m
C_m	=	\[0.6 - 0.4\frac{M_{1}}{M_{2}}\]	≥0.4	\[\lbrack\frac{M_{1}}{M_{2}} = + ve\ for\ double\ curvature\rbrack\] [ACI 318-19 6.6.4.5.3]
	=	0.4
Calculation of Magnification Factor
δ	=	\[\frac{C_{m}}{1 - \frac{P_{u}}{0.75P_{c}}}\]	≥ 1	[ACI 318-19 6.6.4.5.2]

	=	1

Calculation of Final Magnified Moments
M₁	=	1,557.22	kip-in
	=	129.77	kip ft
M₂	=	-1,589.58	kip-in
	=	-132.46	kip ft

ACI 318-19 Moment Magnification Sway- Example 002

Moment Magnification Calculation for Slender Column (Sway)

GEOMETRY, PROPERTIES AND LOADING

The moment magnification calculation for a given circular section using ACI 318-19 is tested in this example by comparing the results with manual calculation.

The column section details are as tabulated below.

Note: Refer ACI 318-19 Moment Magnification Sway Ex002.cdbx

	Parameters	Value
	Diameter (in)	28
	Compressive Strength, fc’ (psi)	4,000
	Modulus of Elasticity of Concrete, E_c (ksi)	3,600
	Minimum Yield Stress, f_y (psi)	40,000
	Modulus of Elasticity of Steel, E_s (ksi)	29,000
	Rebar Layout	8-#9
	Rebar Area (in²)	7.99
	Rebar Ratio	1.30%
	Clear Cover (in)	1.5
	C/C Length, l_c (ft)	17.5
	Unsupported Length, l_u (ft)	17
	k-factor, Unbraced (YZ Plane)	1.51
	k-factor, Braced (YZ Plane)	0.81

Name	Non-Sway Part	Sway Part
Combo1	500	125	135	342	13	92	108

Name

Non-Sway Part

Sway Part

Axial Load, P_u (kip)

Moment Top, M_ux (kip-ft)

Moment Bottom, M_ux

(kip-ft)

Sustained Load

(kip)

Axial Load, P_u (kip)

Moment Top, M_ux (kip-ft)

Moment Bottom, M_ux

(kip-ft)

Combo1

500

125

135

342

13

92

108

Parameter	Value
Story Axial Load, ∑P_u (kip)	25,000
Story Critical Load, ∑P_c (kip)	83,333
Relative Sway, Δ (in)	0.24
Story Shear Load, V_us (kip)	316

MOMENT MAGNIFICATION COMPARISON

Column Designer reports both the top and bottom magnified moments in X and Y direction. The higher of the top and bottom magnified moments is reported as M₂ while the lower magnified moment is reported as M₁.

Magnified Moments (kip-ft)	Column Designer	By hand	% Difference
M_1x	235.75	235.74	0.01%
M_2x	264.52	264.50	0.01%

MANUAL CALCULATION

Calculation of Magnified Moment about X-axis
Loading Data
M_1ns	=	1,500	kip-in
M_2ns	=	1,620	kip-in
Axial Load, Non-sway Part (P_u)	=	500	kip
M_1s	=	1,104	kip-in
M_2s	=	1296	kip-in
Axial Load, Sway Part (P_u)	=	13	kip

Calculation of δ_s
ΣP_u	=	25,000	kip
ΣP_c	=	83,333	kip
Relative Lateral Deflection, Δ_o	=	0.24	in
Story Shear, V_us	=	316	kip
C/C length of Column, l_c	=	210	in

Q	=	\[\frac{\sum_{}^{}{P_{u}\mathrm{\Delta}_{o}}}{V_{us}l_{c}}\]		[ACI 318-19 6.6.4.4.1]

	=	0.09

δ_s	=	\[\frac{1}{1 - Q}\]	≥ 1	[ACI 318-19 6.6.4.6.2a]
	=	1.10

Calculation of M₁ and M₂
M₁ (Lower Moment)	=	M_1ns + δ_sM_1s		[ACI 318-19 6.6.4.6.1a]

	=	2713.74	kip-in
M₂ (Higher Moment)	=	M_2ns + δ_sM_2s		[ACI 318-19 6.6.4.6.1b]

	=	3,044.83	kip-in

Magnification along Column Length
Calculation of Critical Buckling Load
K factor (Braced)	=	0.81
Unsupported Length, l_u	=	204	in
(I_g)_column	=	\[\frac{\pi*14^{4}}{4}\]	in⁴

	=	30,171.86	in⁴
E_c	=	3600	ksi
0.2E_cI_g	=	21,723,736.21	kip-in²
E_s	=	29,000	ksi
I_se	=	570.13	in⁴
β_dns	=	0.67

(EI)_eff	=	\[\frac{0.2E_{c}I_{g} + E_{s}I_{se}}{1 + \beta_{dns}}\]	kip-in²	[ACI 318-19 6.6.4.4.4]

	=	22,954,420.67	kip-in²
Critical Buckling Load, P_c	=	\[\frac{\pi^{2}{(EI)}_{eff}}{{(kl_{u})}^{2}}\]	kip	[ACI 318-19 6.6.4.4.2]

	=	8,297.28	kip

Calculation of C_m
C_m	=	\[0.6 - 0.4\frac{M_{1}}{M_{2}}\]	≥0.4	\[\lbrack\frac{M_{1}}{M_{2}} = - ve\ for\ single\ curvature\rbrack\] [ACI 318-19 6.6.4.5.3]
	=	0.96

Calculation of Magnification Factor
δ	=	\[\frac{C_{m}}{1 - \frac{P_{u}}{0.75P_{c}}}\]	≥ 1	[ACI 318-19 6.6.4.5.2]

	=	1.04

Calculation of Final Magnified Moments
M₁	=	2,828.91	kip-in
	=	235.74	kip ft
M₂	=	3,174.05	kip-in
	=	264.50	kip ft

Eurocode 2-2004 Moment Magnification Sway- Example 001

Moment Magnification Calculation for Slender Column (Sway)

GEOMETRY, PROPERTIES AND LOADING

The moment magnification calculation for a given rectangular section using Eurocode 2:2004 is tested in this example by comparing the results with manual calculation.

The column section details are as tabulated below.

Note: Refer Eurocode 2-2004 Moment Magnification Sway Ex001.cdbx

	Parameters	Value
	Width, w (mm)	400
	Height, h (mm)	500
	Rebar Layout	12-d25
	Clear Cover, c_c (mm)	40
	Compressive Strength, f_ck (MPa)	40
	Minimum Yield Stress, f_yk (MPa)	500
	Modulus of Elasticity of Concrete, E_c (MPa)	35,000
	Modulus of Elasticity of Steel, E_s (MPa)	200,000
	Concrete Area, A_c (mm²)	200,000
	Rebar Area, A_s (mm²)	5,890.49
	Concrete Inertia, I_g, 22 (mm⁴)	2.667x10⁹
	Concrete Inertia, I_g, 33 (mm⁴)	4.167x10⁹
	Rebar Inertia, I_s, 22 (mm⁴)	9.018 x10⁷
	Rebar Inertia, I_s, 33 (mm⁴)	1.617x10⁸
	Perimeter, u (mm)	1,800
	Relative Humidity, RH (%)	50
	Ratio SLS to ULS moments, rm	0.8
	Age of concrete at loading, t₀ (days)	28
	Concrete partial safety factor, γ_c	1.5
	Reinforcing partial safety factor, γ_s	1.15
	Long term compressive strength factor, α_cc	1
	Unsupported Length, l_u (m)	3
	Effective Length Factor (Unbraced), k_XZ	2.94
	Effective Length Factor (Unbraced), k_YZ	4.26
	Biaxial Loading	Yes

Name

Axial Load, N_Ed (kN)

Moment Top, M_x,top

(kN-m)

Moment Top, M_y,top

(kN-m)

Moment Bottom, M_x,bot

(kN-m)

Moment Bottom, M_y,bot

(kN-m)

Combination 1

1000

110

250

120

210

MOMENT MAGNIFICATION COMPARISON

Column Designer reports both the design moments in X and Y direction. The design moments are reported as the top moments for capacity calculations, while the bottom moments remain unchanged.

Design Moments (kN-m)	Column Designer	By hand	% Difference
\(M_{Ed,x}\)	337.77	337.74	0.01%
\(M_{Ed,y}\)	379.13	379.12	0.01%
\(M_{c}\)	603.23	603.20	0.01%

MANUAL CALCULATION

Calculation of Design Moment about X-axis
Loading Data
Moment Top, M_x, top	=	110,000,000	N-mm
Moment Bottom, M_x, bot	=	120,000,000	N-mm
Axial Load, N_Ed	=	1,000,000	N

Calculation of Additional Parameters
Effective Length, l₀	=	k_YZ × l_u		[EC2 5.8.3.2 (3)]
	=	4.26×3,000
	=	12780	mm

Design Compressive Strength, f_cd	=	\[\alpha_{cc} \times \frac{f_{ck}}{\gamma_{c}}\]		[EC2 3.1.6 (1)]
	=	1×40/1.5
	=	26.67	MPa

Mean Value Cylinder Compressive Strength, f_cm	=	f_ck + 8		[EC2 Table 3.1]
	=	40+8
	=	48	MPa

Design Yield Strength, f_yd	=	\[\frac{f_{yk}}{\gamma_{s}}\]		[EC2 3.2.7 (2)]
	=	500⁄1.15
	=	434.78	MPa

Design Strain, ε_yd	=	\[\frac{f_{yd}}{E_{s}}\]		[EC2 5.8.8.3 (1)]
	=	434.78/200,000
	=	0.002174	mm/mm

Curvature Distribution Factor, c	=	10		[EC2 5.8.8.2 (4)]
Curvature Distribution Factor, c

Radius of Gyration of Concrete Section, r_c	=	\[\sqrt{\frac{I_{g,33}}{A_{c}}}\]
	=	\[\sqrt{\frac{4.167 \times 10^{9}}{200,000}}\]
	=	144.34	mm

Radius of Gyration of Rebar, r_s	=	\[\sqrt{\frac{I_{s,33}}{A_{s}}}\]
	=	\[\sqrt{\frac{1.617 \times 10^{8}}{5,890.49}}\]
	=	165.68	mm

Effective Depth, d	=	\[\frac{h}{2} + r_{s}\]		[EC2 5.8.8.3 (2)]
	=	500/2+165.68
	=	415.68	mm

Determine First Order End Moments
Determine M₀₁ and M₀₂ to satisfy \|M₀₂\| ≥ \|M₀₁\|				[EC2 5.8.8.2 (2)]
NOTE: M₀₁ is the numerically smaller end moment M₀₂ is the numerically larger end moment Algebraically, M₀₁ and M₀₂ signs should be maintained. However, if M₀₂ is < 0, they should be opposite signs.				[IStructE Manual EC2 5.5.4.2]
Lower End Moment, M₀₁	=	min {M_x, top, M_x, bot}
	=	110,000,000	N-mm
Higher End Moment, M₀₂	=	max {M_x, top, M_x, bot}
	=	120,000,000	N-mm

Calculate Curvature
Mechanical Reinforcement Ratio, ω	=	\[\frac{A_{s} \times f_{yd}}{A_{c} \times f_{cd}}\\]		[EC2 5.8.8.3 (3)]
	=	(5,890.49×434.78)/ (200,000×26.67)
	=	0.4801

Relative Ultimate Axial Load, nu	=	1 + ω		[EC2 5.8.8.3 (3)]
	=	1+0.4801
	=	1.4801

Relative Axial Force, n	=	\[\frac{N_{Ed}}{\left( A_{c} \times f_{cd} \right)}\]		[EC2 5.8.8.3 (3)]
	=	1,000,000/ (200,000×26.67)
	=	0.1875

Relative Balanced Load, n_bal	=	0.4		[EC2 5.8.8.3 (3)]
Relative Balanced Load, n_bal

Axial Load Correction Factor, K_r	=	\[\frac{nu - n}{nu - n_{bal}} \leq 1\]		[EC2 5.8.8.3 (3)]
	=	\[\frac{1.4801 - 0.4}{1.4801 - 0.1875} \leq 1\]
	=	1

Member Notional Size, h₀	=	\[\frac{2 \times A_{c}}{u}\]		[EC2 Annex B.1 (1) Eqn(B.6)]
	=	(2×200,000)/1800
	=	222.22	mm

Influence of Concrete Strength Coefficient, α₁	=	\[\left( \frac{35}{f_{cm}} \right)^{0.7}\]		[EC2 Annex B.1 (1) Eqn(B.8c)]
	=	(35⁄48)^0.7
	=	0.8016

Influence of Concrete Strength Coefficient, α₂	=	\[\left( \frac{35}{f_{cm}} \right)^{0.2}\]		[EC2 Annex B.1 (1) Eqn(B.8c)]
	=	(35⁄48)^0.2
	=	0.9388

Factor for Effect of Relative Humidity on Notional Creep Coefficient, φ_RH				[EC2 Annex B.1 (1) Eqn(B.3)]
NOTE: \(\varphi_{RH} = 1 + \left( \frac{1 - \frac{RH}{100}}{0.1 \times \sqrt[3]{h_{0}}} \right),\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ for\ f_{cm} \leq 35MPa\) \(\varphi_{RH} = \left\lbrack 1 + \left( \frac{1 - \frac{RH}{100}}{0.1 \times \sqrt[3]{h_{0}}} \times \alpha_{1} \right) \right\rbrack \times \alpha_{2},\ \ \ \ \ for\ f_{cm} > 35MPa\)
φ_RH	=	\[\left\lbrack 1 + \left( \frac{1 - \frac{RH}{100}}{0.1 \times \sqrt[3]{h_{0}}} \times \alpha_{1} \right) \right\rbrack \times \alpha_{2}\]
	=	\[\left\lbrack 1 + \left( \frac{1 - \frac{50}{100}}{0.1 \times \sqrt[3]{222.22}} \times 0.8016 \right) \right\rbrack \times 0.9388\]
	=	1.56

Factor for Effect of Concrete Strength on Notional Creep Coefficient, β(f_cm)	=	\[\frac{16.8}{\sqrt{f_{cm}}}\]		[EC2 Annex B.1 (1) Eqn(B.4)]
	=	\[\frac{16.8}{\sqrt{48}}\]
	=	2.425

Factor for Effect of Concrete Age on Notional Creep Coefficient, β(t₀)	=	\[\frac{1}{\left( 0.1 + {t_{0}}^{0.2} \right)}\]		[EC2 Annex B.1 (1) Eqn(B.5)]
	=	\[\frac{1}{\left( 0.1 + 28^{0.2} \right)}\]
	=	0.4884

Notional Creep Coefficient, φ₀	=	φ_RH × β(f_cm) × β(t₀)		[EC2 Annex B.1 (1) Eqn(B.2)]
	=	1.56×2.425×0.4884
	=	1.848

Effective Creep Ratio, φ_ef	=	φ₀ × rm		[EC2 5.8.4 (2)]
	=	1.848×0.8
	=	1.4784

Slenderness Ratio, λ	=	\[\frac{l_{0}}{r}_{c}\]		[EC2 5.8.3.2 (1)]
	=	12780/144.34
	=	88.54

Factor, β	=	\[0.35 + \frac{f_{ck}}{200} - \frac{\lambda}{150}\]		[EC2 5.8.8.3 (4)]
	=	0.35+40/200-88.54/150
	=	-0.0403

Creep Factor, K_φ	=	1 + β × φ_ef ≥ 1		[EC2 5.8.8.3 (4)]
	=	1 + (−0.0403) × 1.4784 ≥ 1
	=	1

Curvature, \(\frac{1}{r}\)	=	\[K_{r} \times K_{\varphi} \times \frac{\varepsilon_{yd}}{0.45d}\]		[EC2 5.8.8.3 (1)]
	=	1×1×0.002174⁄0.45(415.68)
	=	0.00001162	1/mm

Calculate Moment due to Geometric Imperfections
Eccentricity due to Geometric Imperfections, e_i	=	\[\max\left\{ \frac{l_{0}}{400},\frac{h}{30},20 \right\}\]		[EC2 5.2 (7)a]
	=	\[\max\left\{ \frac{12780}{400},\frac{500}{30},20 \right\}\]		[EC2 6.1 (4)]
	=	31.95	mm

Geometric Imperfections Moment, M_i	=	N_Ed × e_i		[EC2 5.2 (7) Fig 5.1a]
	=	1,000,000×31.95
	=	31,950,000	N-mm

Calculate First Order Moment
First Order Moment, M_0e	=	0.6M₀₂ + 0.4M₀₁ ≥ 0.4M₀₂		[EC2 5.8.8.2 (2)]
	=	0.6 × 120, 000, 000 + 0.4 × 110, 000, 000 ≥ 0.4 × 120, 000, 000
	=	116,000,000	N-mm

Included Effect of Imperfections, M_0Ed	=	M_0e + M_i		[EC2 5.8.8.2 (1)]
	=	116,000,000+31,950,000
	=	147,950,000	N-mm

Calculate Nominal Second Order Moment
Deflection, e₂	=	\[\left( \frac{1}{r} \right) \times {l_{0}}^{2}/c\]		[EC2 5.8.8.2 (3)]
	=	0.00001162×12780²/10
	=	189.79	mm

Nominal Second Order Moment, M₂	=	N_Ed × e₂		[EC2 5.8.8.2 (3)]
	=	1,000,000×189.79
	=	189,790,000	N-mm

Calculate Design Moment
The design moment is the maximum of:
M₀₂	=	120	kN-m	[EC2 5.8.8.2 (1)]
M_0Ed + M₂	=	147.95+189.79		[Concise EC2 5.6.2.2]
	=	337.74	kN-m
M₀₁ + 0.5M₂	=	110+0.5×189.79
	=	204.90	kN-m
\(N_{Ed} \times \max\left( \frac{w}{30},20 \right)\)	=	1,000×max (400/30,20)
\(N_{Ed} \times \max\left( \frac{w}{30},20 \right)\)	=	20	kN-m

Design Moment, M_Ed, x	=	\[\max\left\{ M_{02},M_{0Ed} + M_{2},M_{01} + 0.5M_{2},N_{Ed} \times \max\left( \frac{b}{30},20 \right) \right\}\]
Design Moment, M_Ed, x	=	337.74	kN-m

Calculation of Design Moment about Y-axis
Loading Data
Moment Top, M_y, top	=	250,000,000	N-mm
Moment Bottom, M_y, bot	=	210,000,000	N-mm
Axial Load, N_Ed	=	1,000,000	N

Calculation of Additional Parameters
Effective Length, l₀	=	k_XZ × l_u		[EC2 5.8.3.2 (3)]
	=	2.94×,3000
	=	8,820	mm

Design Compressive Strength, f_cd	=	\[\alpha_{cc} \times \frac{f_{ck}}{\gamma_{c}}\]		[EC2 3.1.6 (1)]
	=	1×40⁄1.5
	=	26.67	MPa

Mean Value Cylinder Compressive Strength, f_cm	=	f_ck + 8		[EC2 Table 3.1]
	=	40+8
	=	48	MPa

Design Yield Strength, f_yd	=	\[\frac{f_{yk}}{\gamma_{s}}\]		[EC2 3.2.7 (2)]
	=	500⁄1.15
	=	434.78	MPa

Design Strain, ε_yd	=	\[\frac{f_{yd}}{E_{s}}\]		[EC2 5.8.8.3 (1)]
	=	434.78/200,000
	=	0.002174	mm/mm

Curvature Distribution Factor, c	=	10		[EC2 5.8.8.2 (4)]
Curvature Distribution Factor, c

Radius of Gyration of Concrete Section, r_c	=	\[\sqrt{\frac{I_{g,33}}{A_{c}}}\]
	=	\[\sqrt{\frac{2.667 \times 10^{9}}{200,000}}\]
	=	115.48	mm

Radius of Gyration of Rebar, r_s	=	\[\sqrt{\frac{I_{s,33}}{A_{s}}}\]
	=	\[\sqrt{\frac{9.018 \times 10^{7}}{5,890.49}}\]
	=	123.73	mm

Effective Depth, d	=	\[\frac{w}{2} + r_{s}\]		[EC2 5.8.8.3 (2)]
	=	400⁄2+123.73
	=	323.73	mm

Determine First Order End Moments
Determine M₀₁ and M₀₂ to satisfy \|M₀₂\| ≥ \|M₀₁\|				[EC2 5.8.8.2 (2)]
NOTE: M₀₁ is the numerically smaller end moment M₀₂ is the numerically larger end moment Algebraically, M₀₁ and M₀₂ signs should be maintained. However, if M₀₂ is < 0, they should be opposite signs.				[IStructE Manual EC2 5.5.4.2]
Lower End Moment, M₀₁	=	min {M_y, top, M_y, bot}
	=	210,000,000	N-mm
Higher End Moment, M₀₂	=	max {M_y, top, M_y, bot}
	=	250,000,000	N-mm

Calculate Curvature
Mechanical Reinforcement Ratio, ω	=	\[\frac{A_{s} \times f_{yd}}{A_{c} \times f_{cd}}\\]		[EC2 5.8.8.3 (3)]
	=	(5890.49×434.78)/ (200,000×26.67)
	=	0.4801

Relative Ultimate Axial Load, nu	=	1 + ω		[EC2 5.8.8.3 (3)]
	=	1+0.4801
	=	1.4801

Relative Axial Force, n	=	\[\frac{N_{Ed}}{\left( A_{c} \times f_{cd} \right)}\]		[EC2 5.8.8.3 (3)]
	=	1,000,000/ (200,000×26.67)
	=	0.1875
Relative Balanced Load, n_bal	=	0.4		[EC2 5.8.8.3 (3)]
Relative Balanced Load, n_bal

Axial Load Correction Factor, K_r	=	\[\frac{nu - n}{nu - n_{bal}} \leq 1\]		[EC2 5.8.8.3 (3)]
	=	\[\frac{1.4801 - 0.4}{1.4801 - 0.1875} \leq 1\]
	=	1

Member Notional Size, h₀	=	\[\frac{2 \times A_{c}}{u}\]		[EC2 Annex B.1 (1) Eqn(B.6)]
	=	(2×200,000)/1800
	=	222.22	mm

Influence of Concrete Strength Coefficient, α₁	=	\[\left( \frac{35}{f_{cm}} \right)^{0.7}\]		[EC2 Annex B.1 (1) Eqn(B.8c)]
	=	\[\left( \frac{35}{48} \right)^{0.7}\]
	=	0.8016

Influence of Concrete Strength Coefficient, α₂	=	\[\left( \frac{35}{f_{cm}} \right)^{0.2}\]		[EC2 Annex B.1 (1) Eqn(B.8c)]
	=	\[\left( \frac{35}{48} \right)^{0.2}\]
	=	0.9388

Factor for Effect of Relative Humidity on Notional Creep Coefficient, φ_RH				[EC2 Annex B.1 (1) Eqn(B.3)]
NOTE: \(\varphi_{RH} = 1 + \left( \frac{1 - \frac{RH}{100}}{0.1 \times \sqrt[3]{h_{0}}} \right),\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ for\ f_{cm} \leq 35MPa\) \(\varphi_{RH} = \left\lbrack 1 + \left( \frac{1 - \frac{RH}{100}}{0.1 \times \sqrt[3]{h_{0}}} \times \alpha_{1} \right) \right\rbrack \times \alpha_{2},\ \ \ \ \ for\ f_{cm} > 35MPa\)
φ_RH	=	\[\left\lbrack 1 + \left( \frac{1 - \frac{RH}{100}}{0.1 \times \sqrt[3]{h_{0}}} \times \alpha_{1} \right) \right\rbrack \times \alpha_{2}\]
	=	\[\left\lbrack 1 + \left( \frac{1 - \frac{50}{100}}{0.1 \times \sqrt[3]{222.22}} \times 0.8016 \right) \right\rbrack \times 0.9388\]
	=	1.56

Factor for Effect of Concrete Strength on Notional Creep Coefficient, β(f_cm)	=	\[\frac{16.8}{\sqrt{f_{cm}}}\]		[EC2 Annex B.1 (1) Eqn(B.4)]
	=	\[\frac{16.8}{\sqrt{48}}\]
	=	2.425

Factor for Effect of Concrete Age on Notional Creep Coefficient, β(t₀)	=	\[\frac{1}{\left( 0.1 + {t_{0}}^{0.2} \right)}\]		[EC2 Annex B.1 (1) Eqn(B.5)]
	=	\[\frac{1}{\left( 0.1 + 28^{0.2} \right)}\]
	=	0.4884

Notional Creep Coefficient, φ₀	=	φ_RH × β(f_cm) × β(t₀)		[EC2 Annex B.1 (1) Eqn(B.2)]
	=	1.56×2.425×0.4884
	=	1.848

Effective Creep Ratio, φ_ef	=	φ₀ × rm		[EC2 5.8.4 (2)]
	=	1.848×0.8
	=	1.4784

Slenderness Ratio, λ	=	\[\frac{l_{0}}{r}_{c}\]		[EC2 5.8.3.2 (1)]
	=	8,820⁄115.48
	=	76.37

Factor, β	=	\[0.35 + \frac{f_{ck}}{200} - \frac{\lambda}{150}\]		[EC2 5.8.8.3 (4)]
	=	0.35+40/200-76.37/150
	=	0.0409

Creep Factor, K_φ	=	1 + β × φ_ef ≥ 1		[EC2 5.8.8.3 (4)]
	=	1 + 0.0409 × 1.4784 ≥ 1
	=	1.060

Curvature, \(\frac{1}{r}\)	=	\[K_{r} \times K_{\varphi} \times \frac{\varepsilon_{yd}}{0.45d}\]		[EC2 5.8.8.3 (1)]
	=	1×1.060×0.002174/0.45(323.73)
	=	0.00001582	1/mm

Calculate Moment due to Geometric Imperfections
Eccentricity due to Geometric Imperfections, e_i	=	\[\max\left\{ \frac{l_{0}}{400},\frac{w}{30},20 \right\}\]		[EC2 5.2 (7)a]
	=	\[\max\left\{ \frac{8820}{400},\frac{400}{30},20 \right\}\]		[EC2 6.1 (4)]
	=	22.05	mm

Geometric Imperfections Moment, M_i	=	N_Ed × e_i		[EC2 5.2 (7) Fig 5.1a]
	=	1000,000×22.05
	=	22,050,000	N-mm

Calculate First Order Moment
First Order Moment, M_0e	=	0.6M₀₂ + 0.4M₀₁ ≥ 0.4M₀₂		[EC2 5.8.8.2 (2)]
	=	0.6 × 250, 000, 000 + 0.4 × 210, 000, 000 ≥ 0.4 × 250, 000, 000
	=	234,000,000	N-mm

Included Effect of Imperfections, M_0Ed	=	M_0e + M_i		[EC2 5.8.8.2 (1)]
	=	234,000,000+22,050,000
	=	256,050,000	N-mm

Calculate Nominal Second Order Moment
Deflection, e₂	=	\[\left( \frac{1}{r} \right) \times {l_{0}}^{2}/c\]		[EC2 5.8.8.2 (3)]
	=	0.00001582×8,820²/10
	=	123.07	mm

Nominal Second Order Moment, M₂	=	N_Ed × e₂		[EC2 5.8.8.2 (3)]
	=	1,000,000×123.07
	=	123,070,000	N-mm

Calculate Design Moment
The design moment is the maximum of:
M₀₂	=	250	kN-m	[EC2 5.8.8.2 (1)]
M_0Ed + M₂	=	256.05+123.07		[Concise EC2 5.6.2.2]
	=	379.12	kN-m
M₀₁ + 0.5M₂	=	210+0.5×123.07
	=	271.54	kN-m
\(N_{Ed} \times \max\left( \frac{h}{30},20 \right)\)	=	1000×max (500/30,20)
\(N_{Ed} \times \max\left( \frac{h}{30},20 \right)\)	=	20	kN-m

Design Moment, M_Ed, y	=	\[\max\left\{ M_{02},M_{0Ed} + M_{2},M_{01} + 0.5M_{2},N_{Ed} \times \max\left( \frac{b}{30},20 \right) \right\}\]
Design Moment, M_Ed, y	=	379.12	kN-m

Increased Design Moment due to Biaxial Bending
Calculate Effective Depths				[IStructE Manual EC2 5.5.5]
Effective Height, h^′	=	\[\max\left\{ h - c_{c} - \frac{c_{b}}{2},w - c_{c} - \frac{c_{b}}{2} \right\}\]
	=	\[\max\left\{ 500 - 40 - \frac{25}{2},400 - 40 - \frac{25}{2} \right\}\]
	=	447.5	mm

Effective Width, b^′	=	\[\min\left\{ h - c_{c} - \frac{c_{b}}{2},w - c_{c} - \frac{c_{b}}{2} \right\}\]
	=	\[\min\left\{ 500 - 40 - \frac{25}{2},400 - 40 - \frac{25}{2} \right\}\]
	=	347.5	mm

Calculate Coefficient for Biaxial Bending				[IStructE Manual EC2 5.5.5]
β	=	\[1 - 1.165 \times min\left\{ 0.6,\ \frac{N_{Ed}}{\left( A_{c} \times f_{ck} \right)} \right\}\]
	=	\[1 - 1.165 \times min\left\{ 0.6,\ \frac{1000000}{(200000 \times 40)} \right\}\]
	=	0.8544

Calculate Increased Design Moment				[IStructE Manual EC2 5.5.5]
NOTE: \(M_{c} = M_{Ed,x} + \beta \times \frac{h'}{b'} \times M_{Ed,y}\ \ \ \ \ \ \ \ \ for\frac{M_{Ed,x}}{h'} \geq \frac{M_{Ed,y}}{b'}\) \(M_{c} = M_{Ed,y} + \beta \times \frac{b'}{h'} \times M_{Ed,x}\ \ \ \ \ \ \ \ \ for\frac{M_{Ed,x}}{h'} < \frac{M_{Ed,y}}{b'}\)

Final Design Moment, M_c	=	\[M_{Ed,y} + \beta \times \frac{b'}{h'} \times M_{Ed,x}\\]
	=	\[379.12 + 0.8544 \times \frac{347.5}{447.5} \times 337.74\\]
	=	603.20	MPa

BS 8110-97 Moment Magnification Sway- Example 001

Moment Magnification Calculation for Slender Column (Sway)

GEOMETRY, PROPERTIES AND LOADING

The moment magnification calculation for a given rectangular section using BS 8110-97 is tested in this example by comparing the results with manual calculation.

The column section details are as tabulated below.

Note: Refer BS 8110-97 Moment Magnification Sway Ex001.cdbx

	Parameters	Value
	Width, w (mm)	400
	Height, h (mm)	500
	Rebar Layout	12-d25
	Clear Cover, c_c (mm)	40
	Compressive Strength, f_cu (MPa)	40
	Minimum Yield Stress, f_y (MPa)	500
	Concrete Area, A_c (mm²)	200,000
	Rebar Area, A_sc (mm²)	5,890.49
	Unsupported Length, l_u (m)	3
	Effective Length Factor (Unbraced), β_XZ	1.99
	Effective Length Factor (Unbraced), β_YZ	3.23
	Biaxial Loading	Yes

Name

Axial Load,

N (kN)

Moment Top, M_x,top

(kN-m)

Moment Top, M_y,top

(kN-m)

Moment Bottom, M_x,bot

(kN-m)

Moment Bottom, M_y,bot

(kN-m)

Combination 1

1,000

110

250

120

210

MOMENT MAGNIFICATION COMPARISON

Column Designer reports both the design moments in X and Y direction. The design moments are reported as the top moments for capacity calculations, while the bottom moments remain unchanged.

Design Moments (kN-m)	Column Designer	By hand	% Difference
\(M_{x}\)	204.79	209.90	2.43%
\(M_{y}\)	276.13	278.56	0.01%
\(M_{c}\)	412	417.82	0.87%

MANUAL CALCULATION

Calculation of Design Moment about X-axis
Loading Data
Moment Top, M_x, top	=	110,000,000	N-mm
Moment Bottom, M_x, bot	=	120,000,000	N-mm
Axial Load, N	=	1,000,000	N

Calculation of Additional Parameters
Effective Length, l_e	=	β_YZ × l_u		[BS 8110-1 3.8.1.6.1]
	=	3.23×3,000
	=	9690	mm

Balanced Section Axial Load, N_bal	=	0.25 × f_cu × A_c		[BS 8110-1 3.8.1.1]
	=	0.25×40×200,000
	=	2000,000	N

Determine First Order End Moments
Determine M₁ and M₂ to satisfy \|M₂\| ≥ \|M₁\|				[BS 8110-1 3.8.3.2]
NOTE: M₁ is the numerically smaller end moment M₂ is the numerically larger end moment Algebraically, M₁ and M₂ signs should be maintained. However, if M₂ is < 0, they should be opposite signs.
Lower End Moment, M₁	=	min {M_x, top, M_x, bot}
	=	110,000,000	N-mm
Higher End Moment, M₂	=	max {M_x, top, M_x, bot}
	=	120,000,000	N-mm

Calculate Additional Moment
Section Ultimate Capacity to Axial Load, N_uz	=	0.45 × f_cu × A_c + 0.95 × f_y × A_sc		[BS 8110-1 3.8.3.1]
	=	0.45×40×200000+0.95×500×5890.49
	=	6,397,982.75	N

Reduction Factor, K	=	\[\frac{N_{uz} - N}{N_{uz} - N_{bal}} \leq 1\]		[BS 8110-1 3.8.3.1]
	=	\[\frac{6,397,982.75\ - 1,000,000}{6,397,982.75\ - 2,000,000} \leq 1\]
	=	1

Slenderness Factor, β_a				[BS 8110-1 3.8.3.1]
NOTE: \(\beta_{a} = \frac{1}{2000}\left( \frac{l_{e}}{h} \right)^{2},\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ for\ Biaxial\ loading\) \(\beta_{a} = \frac{1}{2000}\left( \frac{l_{e}}{\min\left\{ h,w \right\}} \right)^{2},\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ for\ Uniaxial\ loading\)				[BS 8110-1 3.8.3.6]
β_a	=	\[\beta_{a} = \frac{1}{2,000}\left( \frac{l_{e}}{h} \right)^{2}\]
	=	\[\beta_{a} = \frac{1}{2,000}\left( \frac{9,690}{500} \right)^{2}\]
	=	0.1878

Deflection, a_u	=	β_a × K × h		[BS 8110-1 3.8.3.1]
	=	0.1878×1×500
	=	93.9	mm

Additional Moment, M_add	=	N × a_u		[BS 8110-1 3.8.3.1]
	=	1,000,000×93.9
	=	93,900,000	N-mm

Calculate Initial Moment
Initial Moment, M_i	=	0.4M₁ + 0.6M₂ ≥ 0.4M₂		[BS 8110-1 3.8.3.2]
	=	0.4 × 110, 000, 000 + 0.6 × 120, 000, 000 ≥ 0.4 × 120, 000, 000
	=	116,000,000	N-mm

Calculate Design Moment
The design moment is the maximum of:
M₂	=	120	kN-m	[BS 8110-1 3.8.3.2]
M_i + M_add	=	116+93.9
	=	209.9	kN-m
M₁ + 0.5M_add	=	110+0.5×93.9
	=	156.95	kN-m
N × min (0.05 × h, 20)	=	1,000×max (0.05×500,20)
N × min (0.05 × h, 20)	=	25	kN-m

Design Moment, M_x	=	max {M₂, M_i + M_add, M₁ + 0.5M_add, N × min (0.05 × h, 20)}
Design Moment, M_x	=	209.9	kN-m

Calculation of Design Moment about Y-axis
Loading Data
Moment Top, M_y, top	=	250,000,000	N-mm
Moment Bottom, M_y, bot	=	210,000,000	N-mm
Axial Load, N	=	1,000,000	N

Calculation of Additional Parameters
Effective Length, l_e	=	β_XZ × l_u		[BS 8110-1 3.8.1.6.1]
	=	1.99×3,000
	=	5,970	mm

Balanced Section Axial Load, N_bal	=	0.25 × f_cu × A_c		[BS 8110-1 3.8.1.1]
	=	0.25×40×200,000
	=	2000,000	N

Determine First Order End Moments
Determine M₁ and M₂ to satisfy \|M₂\| ≥ \|M₁\|				[BS 8110-1 3.8.3.2]
NOTE: M₁ is the numerically smaller end moment M₂ is the numerically larger end moment Algebraically, M₁ and M₂ signs should be maintained. However, if M₂ is < 0, they should be opposite signs.
Lower End Moment, M₁	=	min {M_y, top, M_y, bot}
	=	210,000,000	N-mm
Higher End Moment, M₂	=	max {M_y, top, M_y, bot}
	=	250,000,000	N-mm

Calculate Additional Moment
Section Ultimate Capacity to Axial Load, N_uz	=	0.45 × f_cu × A_c + 0.95 × f_y × A_sc		[BS 8110-1 3.8.3.1]
	=	0.45×40×200,000+0.95×500×5890.49
	=	6397982.75	N

Reduction Factor, K	=	\[\frac{N_{uz} - N}{N_{uz} - N_{bal}} \leq 1\]		[BS 8110-1 3.8.3.1]
	=	\[\frac{6,397,982.75\ - 1,000,000}{6,397,982.75\ - 2,000,000} \leq 1\]
	=	1

Slenderness Factor, β_a				[BS 8110-1 3.8.3.1]
NOTE: \(\beta_{a} = \frac{1}{2,000}\left( \frac{l_{e}}{w} \right)^{2},\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ for\ Biaxial\ loading\) \(\beta_{a} = \frac{1}{2,000}\left( \frac{l_{e}}{\min\left\{ h,w \right\}} \right)^{2},\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ for\ Uniaxial\ loading\)				[BS 8110-1 3.8.3.6]
β_a	=	\[\beta_{a} = \frac{1}{2000}\left( \frac{l_{e}}{w} \right)^{2}\]
	=	\[\beta_{a} = \frac{1}{2,000}\left( \frac{5970}{400} \right)^{2}\]
	=	0.1114

Deflection, a_u	=	β_a × K × w		[BS 8110-1 3.8.3.1]
	=	0.1114×1×400
	=	44.56	mm

Additional Moment, M_add	=	N × a_u		[BS 8110-1 3.8.3.1]
	=	1,000,000×44.56
	=	44,560,000	N-mm

Calculate Initial Moment
Initial Moment, M_i	=	0.4M₁ + 0.6M₂ ≥ 0.4M₂		[BS 8110-1 3.8.3.2]
	=	0.4 × 21, 000, 0000 + 0.6 × 250, 000, 000 ≥ 0.4 × 250, 000, 000
	=	234,000,000	N-mm

Calculate Design Moment
The design moment is the maximum of:
M₂	=	250	kN-m	[BS 8110-1 3.8.3.2]
M_i + M_add	=	234+44.56
	=	278.56	kN-m
M₁ + 0.5M_add	=	210+0.5×44.56
	=	232.28	kN-m
N × min (0.05 × w, 20)	=	1,000×max (0.05×400,20)
N × min (0.05 × w, 20)	=	20	kN-m

Design Moment, M_y	=	max {M₂, M_i + M_add, M₁ + 0.5M_add, N × min (0.05 × w, 20)}
Design Moment, M_y	=	278.56	kN-m

Increased Design Moment due to Biaxial Bending
Calculate Effective Depths				[BS 8110-1 3.8.4.5]
Effective Height, h^′	=	\[\max\left\{ h - c_{c} - \frac{c_{b}}{2},w - c_{c} - \frac{c_{b}}{2} \right\}\]
	=	\[\max\left\{ 500 - 40 - \frac{25}{2},400 - 40 - \frac{25}{2} \right\}\]
	=	447.5	mm

Effective Width, b^′	=	\[\min\left\{ h - c_{c} - \frac{c_{b}}{2},w - c_{c} - \frac{c_{b}}{2} \right\}\]
	=	\[\min\left\{ 500 - 40 - \frac{25}{2},400 - 40 - \frac{25}{2} \right\}\]
	=	347.5	mm

Calculate Coefficient for Biaxial Bending				[BS 8110-1 3.8.4.5]
β	=	\[1 - 1.165 \times min\left\{ 0.6,\ \frac{N}{\left( A_{c} \times f_{cu} \right)} \right\}\]
	=	\[1 - 1.165 \times min\left\{ 0.6,\ \frac{1,000,000}{(200,000 \times 40)} \right\}\]
	=	0.8544

Calculate Increased Design Moment				[BS 8110-1 3.8.4.5]
NOTE: \(M_{c} = M_{x} + \beta \times \frac{h'}{b'} \times M_{y}\ \ \ \ \ \ \ \ \ for\frac{M_{x}}{h'} \geq \frac{M_{y}}{b'}\) \(M_{c} = M_{y} + \beta \times \frac{b'}{h'} \times M_{x}\ \ \ \ \ \ \ \ \ for\frac{M_{x}}{h'} < \frac{M_{y}}{b'}\)

Final Design Moment, M_c	=	\[M_{y} + \beta \times \frac{b'}{h'} \times M_{x}\\]
	=	\[278.56 + 0.8544 \times \frac{347.5}{447.5} \times 209.9\\]
	=	417.82	MPa

IS 456-2000 Moment Magnification Sway- Example 001

Moment Magnification Calculation for Slender Column (Sway)

GEOMETRY, PROPERTIES AND LOADING

The moment magnification calculation for a given rectangular section using IS 456:2000 is tested in this example by comparing the results with manual calculation.

The column section details are as tabulated below.

Note: Refer IS 456-2000 Moment Magnification Sway Ex001.cdbx

	Parameters	Value
	Width, w (mm)	400
	Height, h (mm)	500
	Rebar Layout	12-d25
	Clear Cover, c_c (mm)	40
	Compressive Strength, f_ck (MPa)	40
	Minimum Yield Stress, f_y (MPa)	500
	Concrete Area (Gross), A_c (mm²)	200,000
	Rebar Area, A_sc (mm²)	5,890.49
	Unsupported Length, l_u (m)	3
	Effective Length Factor (Unbraced), β_XZ	2.51
	Effective Length Factor (Unbraced), β_YZ	3.59
	Is Biaxial?	Yes

Name

Axial Load,

N (kN)

Moment Top, M_x,top

(kN-m)

Moment Top, M_y,top

(kN-m)

Moment Bottom, M_x,bot

(kN-m)

Moment Bottom, M_y,bot

(kN-m)

Combination 1

1000

110

250

120

210

MOMENT MAGNIFICATION COMPARISON

Column Designer reports both the design moments in X and Y direction. The design moments are reported as the top moments for capacity calculations, while the bottom moments remain unchanged.

Design Moments (kN-m)	Column Designer	By hand	% Difference
\(M_{ut,x}\)	235.99	235.99	0.00%
\(M_{ut,y}\)	320.88	320.88	0.00%
\(M_{c}\)	398.31	398.32	0.01%

MANUAL CALCULATION

Calculation of Design Moment about X-axis
Loading Data
Moment Top, M_x, top	=	110,000,000	N-mm
Moment Bottom, M_x, bot	=	120,000,000	N-mm
Axial Load, N	=	1,000,000	N

Calculation of Additional Parameters
Effective Length, l_ef	=	β_YZ × l_u		[IS 456 25.2]
	=	3.59×3,000		[IS 456 Annex E E-1]
	=	10770	mm

Balanced Section Axial Load, P_b	=	From interaction diagram
Balanced Section Axial Load, P_b	=	1,363,185.47	N

Determine First Order End Moments
Determine M_u1 and M_u2 to satisfy \|M_u2\| ≥ \|M_u1\|				[IS 456 39.7.1]
NOTE: M_u1 is the numerically smaller end moment M_u2 is the numerically larger end moment Algebraically, M_u1 and M_u2 signs should be maintained. However, if M_u2 is < 0, they should be opposite signs.
Lower End Moment, M_u1	=	min {M_x, top, M_x, bot}
	=	110,000,000	N-mm
Higher End Moment, M_u2	=	max {M_x, top, M_x, bot}
	=	120,000,000	N-mm

Calculate Moment due to Minimum Eccentricity
Minimum Moment, M_min	=	\[P_{u} \times \max\left( \frac{l_{u}}{500} + \frac{h}{30},20 \right)\]		[IS 456 25.4]
	=	\[1,000,000 \times \max\left( \frac{3,000}{500} + \frac{500}{30},20 \right)\]
	=	22,666,666.67	N-mm

Calculate Additional Moment
Ultimate Capacity Axial Load, P_uz	=	0.45 × f_ck × A_c + (0.75 × f_y − 0.45 × f_ck) × A_sc			[IS 456 39.6]
	=	0.45 × 40 × 200, 000 + (0.75 × 500 − 0.45 × 40) × 5, 890.49
	=	5,702,904.93	N

Modification Factor, k_a	=	k_a = 1, for sway		[IS SP24 39.7.1]

Additional Moment, M_a	=	\[k_{a} \times \frac{P_{u} \times h}{2,000}\left\{ \frac{l_{ef}}{h} \right\}^{2}\]		[IS 456 39.7.1]
	=	\[1 \times \frac{1,000,000 \times 500}{2,000}\left\{ \frac{10,770}{500} \right\}^{2}\]
	=	115,992,800	N-mm

Calculate Primary Moment
Primary Moment, M_ui	=	max {M_u2, M_min}		[IS 456 25.4]
Primary Moment, M_ui	=	120,000,000	N-mm


Calculate Design Moment
The design moment is the maximum of:				[IS 456 39.7.1 Note2]
M_u2	=	120	kN-m
M_ui + M_a	=	120+115.99
	=	235.99	kN-m

Design Moment, M_ut, x	=	max {M_u2, M_ui + M_a}
Design Moment, M_ut, x	=	235.99	kN-m

Calculation of Design Moment about Y-axis
Loading Data
Moment Top, M_y, top	=	250,000,000	N-mm
Moment Bottom, M_y, bot	=	210,000,000	N-mm
Axial Load, N	=	1,000,000	N

Calculation of Additional Parameters
Effective Length, l_ef	=	β_XZ × l_u		[IS 456 25.2]
	=	2.51×3,000		[IS 456 Annex E E-1]
	=	7,530	mm

Balanced Section Axial Load, P_b	=	From interaction diagram
Balanced Section Axial Load, P_b	=	1,363,185.47	N

Determine First Order End Moments
Determine M_u1 and M_u2 to satisfy \|M_u2\| ≥ \|M_u1\|				[IS 456 39.7.1]
NOTE: M_u1 is the numerically smaller end moment M_u2 is the numerically larger end moment Algebraically, M_u1 and M_u2 signs should be maintained. However, if M_u2 is < 0, they should be opposite signs.
Lower End Moment, M_u1	=	min {M_y, top, M_y, bot}
	=	210,000,000	N-mm

Higher End Moment, M_u2	=	max {M_y, top, M_y, bot}
	=	250,000,000	N-mm

Calculate Moment due to Minimum Eccentricity
Minimum Moment, M_min	=	\[P_{u} \times \max\left( \frac{l_{u}}{500} + \frac{w}{30},20 \right)\]		[IS 456 25.4]
	=	1,000,000×max (3,000/500+400/30,20)
	=	20,000,000	N-mm

Calculate Additional Moment
Ultimate Capacity Axial Load, P_uz	=	0.45 × f_ck × A_c + (0.75 × f_y − 0.45 × f_ck) × A_sc			[IS 456 39.6]
	=	0.45 × 40 × 200, 000 + (0.75 × 500 − 0.45 × 40) × 5, 890.49
	=	5,702,904.93	N

Modification Factor, k_a	=	\[\frac{P_{uz} - P_{u}}{P_{uz} - P_{b}} \leq 1\]		[IS 456 39.7.1.1]
	=	\[\frac{5,702,904.93\ - 1,000,000}{5,702,904.93\ - 1,363,185.47} \leq 1\]
	=	1

Additional Moment, M_a	=	\[k_{a} \times \frac{P_{u} \times w}{2000}\left\{ \frac{l_{ef}}{w} \right\}^{2}\]		[IS 456 39.7.1]
	=	\[1 \times \frac{1,000,000 \times 400}{2000}\left\{ \frac{7530}{400} \right\}^{2}\]
	=	70,876,125	N-mm

Calculate Primary Moment
Primary Moment, M_ui	=	max {M_u2, M_min}		[IS 456 25.4]
Primary Moment, M_ui	=	250,000,000	N-mm

Calculate Design Moment
The design moment is the maximum of:				[IS 456 39.7.1 Note2]
M₂	=	250	kN-m
M_ui + M_a	=	250+70.88
	=	320.88	kN-m

Design Moment, M_ut, y	=	max {M₂, M_ui + M_a}
Design Moment, M_ut, y	=	320.88	kN-m

Increased Design Moment due to Biaxial Bending
Final Design Moment, M_c	=	\[\sqrt{{M_{ut,x}}^{2} + {M_{ut,y}}^{2}}\\]
	=	\[\sqrt{{235.99}^{2} + {320.88}^{2}}\\]
	=	398.32	MPa

AS 3600-2018 Moment Magnification Sway- Example 001

Moment Magnification Calculation for Slender Column (Sway)

GEOMETRY, PROPERTIES AND LOADING

The moment magnification calculation for a given rectangular section using AS 3600-2018 is tested in this example by comparing the results with manual calculation.

The column section details are as tabulated below.

Note: Refer AS 3600-2018 Moment Magnification Sway Ex001.cdbx

	Parameters	Value
	Width, w (mm)	400
	Height, h (mm)	500
	Rebar Layout	12-d25
	Clear Cover, c_c (mm)	40
	Compressive Strength, f_c^′ (MPa)	40
	Minimum Yield Stress, f_y (MPa)	500
	Concrete Area (Gross), A_c (mm²)	200,000
	Rebar Area, A_sc (mm²)	5,890.49
	Unsupported Length, l_u (m)	3
	Effective Length Factor (Braced), k_XZ	0.94
	Effective Length Factor (Braced), k_YZ	0.97
	Effective Length Factor (Braced), k_XZ	1.98
	Effective Length Factor (Braced), k_YZ	2.69

Name

Axial Load,

N^* (kN)

Moment Top, M_x^*_,top

(kN-m)

Moment Top, M_y^*_,top

(kN-m)

Moment Bottom, M_x^*_,bot

(kN-m)

Moment Bottom, M_y^*_,bot

(kN-m)

Loading Factor, β_d

Combination 1

6000

110

250

120

210

0.5

Parameter	Value
Story Axial Load, ∑N^* (kN)	12,000
Story Critical Load, ∑N_c (kN)	40,000

MOMENT MAGNIFICATION COMPARISON

Column Designer reports both the design moments in X and Y direction. The design moments are reported as the top moments for capacity calculations, while the bottom moments remain unchanged.

Design Moments (kN-m)	Column Designer	By hand	% Difference
\(M_{x}^{*}\)	168.6	171.4	1.66%
\(M_{y}^{*}\)	352.5	357.5	1.42%
\(M_{c}\)	352.5	357.5	1.42%

MANUAL CALCULATION

Calculation of Design Moment about X-axis
Loading Data
M₁^* (Lower Moment)	=	110,000,000	N-mm
M₂^* (Higher Moment)	=	120,000,000	N-mm
Axial Load, N*	=	6,000,000	N

Calculation of Additional Parameters
Effective Length, L_e	=	k_YZ × l_u		[AS 3600 10.4.3(a)] [AS 3600 10.5.3]
	=	0.97×3,000
	=	2910	mm

Minimum Moment, M_min	=	0.05DN^*				[AS 3600 10.1.2]
Minimum Moment, M_min	=	150,000,000	N-mm

Balanced Section Moment, M_c	=	M_ub (From interaction diagram)		[Guide to Reinforced Concrete Design]
Balanced Section Moment, M_c	=	476,923,076.9	N-mm	[Guide to Reinforced Concrete Design]
Calculation of Critical Buckling Load
Critical Buckling Load, N_c	=	\[\left( \frac{\pi^{2}}{{L_{e}}^{2}} \right)\left\lbrack \frac{182d_{0}\varnothing M_{c}}{\left( 1 + \beta_{d} \right)} \right\rbrack\]		[AS 3600 10.4.4]
	=	20,165,705.56	N

Calculation of Magnification Factor
Minimum Moment, K_m	=	\[0.6 - 0.4\frac{M_{1}^{}}{M_{2}^{}} \geq 0.4\]			[AS 3600 10.3.1] [AS 3600 10.4.2]
Minimum Moment, K_m	=	0.97
δ_b	=	\[\frac{K_{m}}{\left( 1 - \frac{N^{*}}{N_{c}} \right)} \geq 1\]				[AS 3600 10.4.2]
δ_b	=	1.38
δ_s	=	\[\frac{1}{\left( 1 - \frac{\sum_{}^{}N^{*}}{\sum_{}^{}N_{c}} \right)} \geq 1\]		[AS 3600 10.4.3(1)]
	=	1.43
Calculate Magnified Moment
M_x^*	=	max (δ_b, δ_s) × M₂^*		[AS 3600 10.4.3]
	=	171.4	kN-m

Calculation of Design Moment about Y-axis
Loading Data
M₁^* (Lower Moment)	=	210,000,000	N-mm
M₂^* (Higher Moment)	=	250,000,000	N-mm
Axial Load, N*	=	6,000,000	N

Calculation of Additional Parameters
Effective Length, L_e	=	k_XZ × l_u		[AS 3600 10.4.3(a)] [AS 3600 10.5.3]
	=	0.94×3,000
	=	2820	mm

Minimum Moment, M_min	=	0.05DN^*				[AS 3600 10.1.2]
Minimum Moment, M_min	=	120,000,000	N-mm

Balanced Section Moment, M_c	=	M_ub (From interaction diagram)		[Guide to Reinforced Concrete Design]
Balanced Section Moment, M_c	=	492,307,692.3	N-mm	[Guide to Reinforced Concrete Design]
Calculation of Critical Buckling Load
Critical Buckling Load, N_c	=	\[\left( \frac{\pi^{2}}{{L_{e}}^{2}} \right)\left\lbrack \frac{182d_{0}\varnothing M_{c}}{\left( 1 + \beta_{d} \right)} \right\rbrack\]		[AS 3600 10.4.4]
	=	17,347,389.78	N

Calculation of Magnification Factor
Minimum Moment, K_m	=	\[0.6 - 0.4\frac{M_{1}^{}}{M_{2}^{}} \geq 0.4\]			[AS 3600 10.3.1] [AS 3600 10.4.2]
Minimum Moment, K_m	=	0.936

δ_b	=	\[\frac{K_{m}}{\left( 1 - \frac{N^{*}}{N_{c}} \right)} \geq 1\]				[AS 3600 10.4.2]
δ_b	=	1.43
δ_s	=	\[\frac{1}{\left( 1 - \frac{\sum_{}^{}N^{*}}{\sum_{}^{}N_{c}} \right)} \geq 1\]		[AS 3600 10.4.3(1)]
	=	1.43

Calculate Magnified Moment
M_y^*	=	max (δ_b, δ_s) × M₂^*		[AS 3600 10.4.3]
	=	357.5	kN-m

Final Design Moment
Final Design Moment, M_c	=	max (M_x, M_y)
Final Design Moment, M_c	=	357.5	kN-m

Demand/Capacity Ratio

DC - Example 001

Capacity Ratio Check for Rectangular Column

GEOMETRY, PROPERTIES AND LOADING

The Capacity Ratio Check for a given rectangular section is tested in this example by comparing the results with hand calculations.

The column section details and loading details are as tabulated below.

Note: Refer DC Ex001.cdbx

*Parameters*	*Column Designer*
Height (in)	36
Width (in)	24
fc’ (psi)	4,000
fy (psi)	40,000
Number of bars	10
Corner Bars	#9
Bars along direction 2 and 3	#8
Rebar Area (in²)	8.71
Rebar Ratio	1.01%
Clear Cover (in)	1.5

Name	Axial Load, Pu (kip)	Moment Top, Mux (kip-ft)	Moment Top, Muy (kip-ft)	Moment Bottom, Mux (kip-ft)	Moment Bottom, Muy (kip-ft)
Combination 1	1100	420	310	300	210
Combination 2	-60	110	65	100	45

CAPACITY RATIO COMPARISON

Column Designer reports 4 types of capacity ratios namely: Moment sum at P, Moment Vector at P, Axial Compression Capacity and Axial Tension Capacity. The values obtained are tabulated below followed by the detailed hand calculation.

Combination 1 (Top End)

Capacity Ratios	Column Designer	By hand
Moment sum at P	0.98	0.98
Moment Vector at P	0.73	0.74
Axial Compression Capacity	0.64	0.64
Axial Tension Capacity	0.00	0.00

Capacity Ratios	Column Designer	By hand
Moment sum at P	0.61	0.61
Moment Vector at P	0.36	0.36
Axial Compression Capacity	0.00	0.00
Axial Tension Capacity	0.19	0.19

Combination 2 (Top End)

CALCULATIONS BY HAND

Combination 1 (Top End)

Moment Sum at P

Moment sum D/C = \(\frac{M_{2u}}{{Ф\ M}_{2n,max}}\ + \ \frac{M_{3u}}{{Ф\ M}_{3n,max}}\) = \(\frac{310}{604}\ + \ \frac{420}{904}\) = \(0.98\\)

Moment Vector at P

\(C_{u} = \ \sqrt{M_{2u}^{2} + M_{3u}^{2}}\) = \(\sqrt{420^{2} + \ 310^{2}\ }\)=\(\ 522.01\ kip - ft\)

\(C_{n} = \ \sqrt{M_{2n}^{2} + M_{3n}^{2}}\) = \(\sqrt{565^{2} + \ 425^{2}\ }\)=\(\ 707.00\ kip - ft\)

Moment Vector D/C = \(\frac{C_{u}}{C_{n}}\) = \(\frac{522.01}{707} = \ 0.74\)

Axial Compression Capacity

Axial Compression D/C = \(\frac{P_{u}}{{Ф\ P}_{n,max}}\) \(= \frac{1100}{1709} = \ 0.64\)

Axial Tension Capacity

Axial Tension D/C = \(0.00\)

Combination 2 (Top End)

Moment Sum at P

Moment sum D/C = \(\frac{M_{2u}}{{Ф\ M}_{2n,max}}\ + \ \frac{M_{3u}}{{Ф\ M}_{3n,max}}\) = \(\frac{65}{227}\ + \ \frac{110}{338}\) = \(0.61\)

Moment Vector at P

\(C_{u} = \ \sqrt{M_{2u}^{2} + M_{3u}^{2}}\) = \(\sqrt{65^{2} + \ 110^{2}\ }\)=\(\ 127.77\ kip - ft\)

\(C_{n} = \ \sqrt{M_{2n}^{2} + M_{3n}^{2}}\) = \(\sqrt{177^{2} + \ 303^{2}\ }\)=\(\ 350.91\ kip - ft\)

Moment Vector D/C = \(\frac{C_{u}}{C_{n}}\) = \(\frac{127.77}{350.91} = \ 0.36\)

Axial Compression Capacity

Axial Compression D/C = \(0.00\)

Axial Tension Capacity

Axial Tension D/C = \(\frac{P_{u}}{{Ф\ P}_{n,\max}}\) \(= \frac{60}{313} = \ 0.19\)

DC - Example 002

Capacity Ratio Check for Circular Column

GEOMETRY, PROPERTIES AND LOADING

The Capacity Ratio Check for a given circular section is tested in this example by comparing the results with hand calculations.

The column section details and loading details are as tabulated below.

Note: Refer DC Ex002.cdbx

*Parameters*	*Column Designer*
Radius (in)	16
fc’ (psi)	4,000
fy (psi)	40,000
Rebar	#8
Number of bars	8
Rebar Area (in²)	6.28
Rebar Ratio	0.78%
Clear Cover (in)	1.5

Name	Axial Load, Pu (kip)	Moment Top, Mux (kip-ft)	Moment Top, Muy (kip-ft)	Moment Bottom, Mux (kip-ft)	Moment Bottom, Muy (kip-ft)
Combination 1	1200	300	105	200	105
Combination 2	-70	-100	60	-75	55

CAPACITY RATIO COMPARISON

Column Designer reports 4 types of capacity ratios namely: Moment sum at P, Moment Vector at P, Axial Compression Capacity and Axial Tension Capacity. The values obtained are tabulated below followed by the detailed hand calculation.

Combination 1 (Top End)

Capacity Ratios	Column Designer	By hand
Moment sum at P	0.75	0.75
Moment Vector at P	0.59	0.59
Axial Compression Capacity	0.77	0.77
Axial Tension Capacity	0.00	0.00

Combination 2 (Top End)

Capacity Ratios	Column Designer	By hand
Moment sum at P	0.88	0.88
Moment Vector at P	0.65	0.65
Axial Compression Capacity	0.00	0.00
Axial Tension Capacity	0.31	0.31

CALCULATIONS BY HAND

Combination 1 (Top End)

Moment Sum at P

Moment sum D/C = \(\frac{M_{2u}}{{Ф\ M}_{2n,max}}\ + \ \frac{M_{3u}}{{Ф\ M}_{3n,max}}\) = \(\frac{105}{537}\ + \ \frac{300}{537}\) = \(0.75\\)

Moment Vector at P

\(C_{u} = \ \sqrt{M_{2u}^{2} + M_{3u}^{2}}\) = \(\sqrt{105^{2} + \ 300^{2}\ }\)=\(\ 317.84\ kip - ft\)

\(C_{n} = \ \sqrt{M_{2n}^{2} + M_{3n}^{2}}\) = \(\sqrt{165^{2} + \ 515^{2}\ }\)=\(\ 540.79\ kip - ft\)

Moment Vector D/C = \(\frac{C_{u}}{C_{n}}\) = \(\frac{317.84}{540.79} = \ 0.59\)

Axial Compression Capacity

Axial Compression D/C = \(\frac{P_{u}}{{Ф\ P}_{n,max}}\) \(= \frac{1,200}{1,550} = \ 0.77\)

Axial Tension Capacity

Axial Tension D/C = \(0.00\)

Combination 2 (Top End)

Moment Sum at P

Moment sum D/C = \(\frac{M_{2u}}{{Ф\ M}_{2n,max}}\ + \ \frac{M_{3u}}{{Ф\ M}_{3n,max}}\) = \(\frac{60}{181.4}\ + \ \frac{100}{181.4}\) = \(0.88\)

Moment Vector at P

\(C_{u} = \ \sqrt{M_{2u}^{2} + M_{3u}^{2}}\) = \(\sqrt{60^{2} + \ 100^{2}\ }\)=\(\ 116.62\ kip - ft\)

\(C_{n} = \ \sqrt{M_{2n}^{2} + M_{3n}^{2}}\) = \(\sqrt{92^{2} + \ 153^{2}\ }\)=\(\ 178.53\ kip - ft\)

Moment Vector D/C = \(\frac{C_{u}}{C_{n}}\) = \(\frac{116.62}{178.53} = \ 0.65\)

Axial Compression Capacity

Axial Compression D/C = \(0.00\)

Axial Tension Capacity

Axial Tension D/C = \(\frac{P_{u}}{{Ф\ P}_{n,max}}\) \(= \frac{70}{226.19} = \ 0.31\)

DC - Example 003

Capacity Ratio Check for T-Section

GEOMETRY, PROPERTIES AND LOADING

The Capacity Ratio Check for a given T-section section is tested in this example by comparing the results with hand calculations.

Note: Refer DC Ex003.cdbx

The column section details and loading details are as tabulated below.

*Parameters*	*Column Designer*
Height (in)	150
Width (in)	150
Web Width (in)	15
Flange Height (in)	15
fc’ (psi)	6,000
fy (psi)	60,000
Number of bars	58
Corner Bars	#8
Bars along direction 2 and 3	#8
Rebar Area (in²)	45.55
Rebar Ratio	1.07%
Clear Cover (in)	1.5

Name	Axial Load, Pu (kip)	Moment Top, Mux (kip-ft)	Moment Top, Muy (kip-ft)	Moment Bottom, Mux (kip-ft)	Moment Bottom, Muy (kip-ft)
Combination 1	10,000	0	0	15,000	0
Combination 2	10,000	0	0	-15,000	0

CAPACITY RATIO COMPARISON

Column Designer reports 4 types of capacity ratios namely: Moment sum at P, Moment Vector at P, Axial Compression Capacity and Axial Tension Capacity. The values obtained are tabulated below followed by the detailed hand calculation.

Combination 1 (Bottom End)

Capacity Ratios	Column Designer	By hand
Moment sum at P	0.40	0.40
Moment Vector at P	0.40	0.40
Axial Compression Capacity	0.79	0.79
Axial Tension Capacity	0.00	0.00

Capacity Ratios	Column Designer	By hand
Moment sum at P	0.85	0.85
Moment Vector at P	0.85	0.85
Axial Compression Capacity	0.79	0.79
Axial Tension Capacity	0.00	0.00

Combination 2 (Bottom End)

CALCULATIONS BY HAND

Combination 1 (Bottom End)

Moment Sum at P

Moment sum D/C = \(\frac{M_{2u}}{{Ф\ M}_{2n,max}}\ + \ \frac{M_{3u}}{{Ф\ M}_{3n,max}}\) = \(\frac{0}{14,181}\ + \ \frac{15,000}{37,086}\) = \(0.40\\)

Moment Vector at P

\(C_{u} = \ \sqrt{M_{2u}^{2} + M_{3u}^{2}}\) = \(\sqrt{0^{2} + \ {15,000}^{2}\ }\)=\(\ 15,000\ kip - ft\)

\(C_{n} = \ \sqrt{M_{2n}^{2} + M_{3n}^{2}}\) = \(\sqrt{0^{2} + \ {37,086}^{2}\ }\)=\(\ 37,086\ kip - ft\)

Moment Vector D/C = \(\frac{C_{u}}{C_{n}}\) = \(\frac{15,000}{37,086} = \ 0.40\)

Axial Compression Capacity

Axial Compression D/C = \(\frac{P_{u}}{{Ф\ P}_{n,max}}\) \(= \frac{10,000}{12,645} = \ 0.79\)

Axial Tension Capacity

Axial Tension D/C = \(0.00\)

Combination 2 (Bottom End)

Moment Sum at P

Moment sum D/C = \(\frac{M_{2u}}{{Ф\ M}_{2n,max}}\ + \ \frac{M_{3u}}{{Ф\ M}_{3n,max}}\) = \(\frac{0}{14,181}\ + \ \frac{- 15,000}{- 17,682}\) = \(0.85\)

Moment Vector at P

\(C_{u} = \ \sqrt{M_{2u}^{2} + M_{3u}^{2}}\) = \(\sqrt{0^{2} + \ {- 15,000}^{2}\ }\)=\(\ 15,000\ kip - ft\)

\(C_{n} = \ \sqrt{M_{2n}^{2} + M_{3n}^{2}}\) = \(\sqrt{0^{2} + \ {- 17,682}^{2}\ }\)=\(\ 17,682\ kip - ft\)

Moment Vector D/C = \(\frac{C_{u}}{C_{n}}\) = \(\frac{15,000}{17,682} = \ 0.85\)

Axial Compression Capacity

Axial Compression D/C = \(\frac{P_{u}}{{Ф\ P}_{n,max}}\) \(= \frac{10,000}{12,645} = \ 0.79\)

Axial Tension Capacity

Axial Tension D/C = \(0.00\)

DC - Example 004

Capacity Ratio Check for T-Section

GEOMETRY, PROPERTIES AND LOADING

The Capacity Ratio Check for a given T-section section is tested in this example by comparing the results with hand calculations.

Note: Refer DC Ex004.cdbx

The column section details and loading details are as tabulated below.

*Parameters*	*Column Designer*
Height (in)	150
Width (in)	150
Web Width (in)	15
Flange Height (in)	15
fc’ (psi)	6,000
fy (psi)	60,000
Number of bars	58
Corner Bars	#8
Bars along direction 2 and 3	#8
Rebar Area (in²)	45.55
Rebar Ratio	1.07%
Clear Cover (in)	1.5

Name	Axial Load, Pu (kip)	Moment Top, Mux (kip-ft)	Moment Top, Muy (kip-ft)	Moment Bottom, Mux (kip-ft)	Moment Bottom, Muy (kip-ft)
Combination 1	10,000	0	0	15,000	0
Combination 2	10,000	0	0	-15,000	0

CAPACITY RATIO COMPARISON

Column Designer reports 4 types of capacity ratios namely: Moment sum at P, Moment Vector at P, Axial Compression Capacity and Axial Tension Capacity. The values obtained are tabulated below followed by the detailed hand calculation.

Combination 1 (Bottom End)

Capacity Ratios	Column Designer	By hand
Moment sum at P	0.85	0.85
Moment Vector at P	0.85	0.85
Axial Compression Capacity	0.79	0.79
Axial Tension Capacity	0.00	0.00

Capacity Ratios	Column Designer	By hand
Moment sum at P	0.40	0.40
Moment Vector at P	0.40	0.40
Axial Compression Capacity	0.79	0.79
Axial Tension Capacity	0.00	0.00

Combination 2 (Bottom End)

CALCULATIONS BY HAND

Combination 1 (Bottom End)

Moment Sum at P

Moment sum D/C = \(\frac{M_{2u}}{{Ф\ M}_{2n,max}}\ + \ \frac{M_{3u}}{{Ф\ M}_{3n,max}}\) = \(\frac{0}{14,181}\ + \ \frac{15,000}{17,682}\) = \(0.85\)

Moment Vector at P

\(C_{u} = \ \sqrt{M_{2u}^{2} + M_{3u}^{2}}\) = \(\sqrt{0^{2} + \ {15,000}^{2}\ }\)=\(\ 15,000\ kip - ft\)

\(C_{n} = \ \sqrt{M_{2n}^{2} + M_{3n}^{2}}\) = \(\sqrt{0^{2} + \ {17,682}^{2}\ }\)=\(\ 17,682\ kip - ft\)

Moment Vector D/C = \(\frac{C_{u}}{C_{n}}\) = \(\frac{15,000}{17,682} = \ 0.85\)

Axial Compression Capacity

Axial Compression D/C = \(\frac{P_{u}}{{Ф\ P}_{n,max}}\) \(= \frac{10,000}{12,645} = \ 0.79\)

Axial Tension Capacity

Axial Tension D/C = \(0.00\)

Figure 7 : MM Curve for Combination 1

Combination 1 (Bottom End)

Moment Sum at P

Moment sum D/C = \(\frac{M_{2u}}{{Ф\ M}_{2n,max}}\ + \ \frac{M_{3u}}{{Ф\ M}_{3n,max}}\) = \(\frac{0}{14,181}\ + \ \frac{- 15,000}{- 37,086}\) = \(0.40\\)

Moment Vector at P

\(C_{u} = \ \sqrt{M_{2u}^{2} + M_{3u}^{2}}\) = \(\sqrt{0^{2} + \ {- 15,000}^{2}\ }\)=\(\ 15,000\ kip - ft\)

\(C_{n} = \ \sqrt{M_{2n}^{2} + M_{3n}^{2}}\) = \(\sqrt{0^{2} + \ {- 37,086}^{2}\ }\)=\(\ 37,086\ kip - ft\)

Moment Vector D/C = \(\frac{C_{u}}{C_{n}}\) = \(\frac{15,000}{37,086} = \ 0.40\)

Axial Compression Capacity

Axial Compression D/C = \(\frac{P_{u}}{{Ф\ P}_{n,max}}\) \(= \frac{10,000}{12,645} = \ 0.79\)

Axial Tension Capacity

Axial Tension D/C = \(0.00\)

Moment Curvature

Moment Curvature Example 001

Moment Curvature Check for Rectangular Column

GEOMETRY AND PROPERTIES

The Moment Curvature for a given rectangular section is tested in this example by comparing the results with SAP2000 v26.

Note: Refer Moment Curvature Ex001.cdbx

The column section details are as tabulated below.

Parameters	Column Designer	SAP2000 v26
Code	ACI 318-19
Height (in)	36
Width (in)	24
fc’ (psi)	4,000
fy (psi)	40,000
Number of bars	10
Corner Bars	#9
Bars along direction 2 and 3	#8
Rebar Area (in²)	8.71
Rebar Ratio	1.01%
Clear Cover (in)	1.5

CONCRETE MATERIAL PROPERTIES

Parameters	Value
Concrete Material Properties
Specified Concrete Compressive Strength, fc’ (psi)	4,000
Modulus of Elasticity, E (psi)	3,605,000
Concrete Stress Strain Model Properties
Stress Strain Model	Mander’s Unconfined
Strain at Unconfined Compressive Strength (fc’)	0.002
Ultimate Unconfined Strain Capacity	0.005

REBAR MATERIAL PROPERTIES

Parameters	Value
Rebar Material Properties
Minimum Yield Stress, fy (psi)	40,000
Minimum Tensile Stress, fu (psi)	60,000
Modulus of Elasticity, E (psi)	29,000,000
Rebar Stress Strain Model Properties
Stress Strain Model	Park Strain Hardening
Strain at Onset of Strain Hardening (ε_sh)	0.02
Ultimate Strain Capacity (ε_su)	0.12

MOMENT CURVATURE DIAGRAM PROPERTIES

Parameters	Value
Angle	0
Maximum Compression Strain (10^-3 in/in)	100
Maximum Tension Strain (10^-3 in/in)	500
Maximum Curvature (10^-3 rad/in)	100

MOMENT CURVATURE COMPARISON WITH AXIAL LOAD (P) = 0 Kips

Moment Curvature Example 002

Moment Curvature Check for Rectangular Column

GEOMETRY AND PROPERTIES

The Moment Curvature for a given rectangular section is tested in this example by comparing the results with SAP2000 v26.

Note: Refer Moment Curvature Ex002.cdbx

The column section details are as tabulated below.

Parameters	Column Designer	SAP2000 v26
Code	ACI 318-19
Height (mm)	900
Width (mm)	600
fc’ (MPa)	27.58
fy (MPa)	413.69
Number of bars	10
Corner Bars	d 25
Bars along direction 2 and 3	d 25
Rebar Area (mm²)	4908.74
Rebar Ratio	0.91%
Clear Cover (mm)	40

CONCRETE MATERIAL PROPERTIES

Parameters	Value
Concrete Material Properties
Specified Concrete Compressive Strength, fc’ (MPa)	27.58
Modulus of Elasticity, E (MPa)	24855.58
Ultimate Strain Capacity (Unconfined)	0.005
Concrete Stress Strain Model Properties
Stress Strain Model	Mander’s Confined
Strain at Compressive Strength (fc’)	0.00219
Maximum Strain	0.05
Width of confinement zone (mm)	520
Height of confinement zone (mm)	820
Tie Bar Strength, fy (MPa)	413.69
Tie Spacing Along X (mm)	545
Tie Diameter Along X (mm)	25
Tie Spacing Along Y (mm)	845
Tie Diameter Along Y (mm)	25
Vertical Spacing of Ties (mm)	250
Total Main Steel (mm²)	4908.74

REBAR MATERIAL PROPERTIES

Parameters	Value
Rebar Material Properties
Minimum Yield Stress, fy (MPa)	413.69
Minimum Tensile Stress, fu (MPa)	620.53
Modulus of Elasticity, E (MPa)	199948.04
Rebar Stress Strain Model Properties
Stress Strain Model	Park Strain Hardening
Strain at Onset of Strain Hardening (ε_sh)	0.01
Ultimate Strain Capacity (ε_su)	0.09

MOMENT CURVATURE DIAGRAM PROPERTIES

Parameters	Value
Angle	0
Maximum Compression Strain (10^-3 mm/mm)	100
Maximum Tension Strain (10^-3 mm/mm)	500
Maximum Curvature (10^-3 rad/mm)	4

MOMENT CURVATURE COMPARISON WITH AXIAL LOAD (P) = 0 kN