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Design of Column

This document summarizes the design of three interior columns (Column 1, 2, and 3) for a building. Column 1 is designed to carry a factored axial load of 285.78 kN between the roof and third floor. It will be a 300x300 mm square reinforced concrete column with 8 #16 reinforcing bars spaced at 300 mm. Column 2 is designed to carry a factored axial load of 271.01 kN between the third and second floors. It has the same dimensions and reinforcement as Column 1. Column 3 is designed to carry the largest factored axial load of 297.30 kN between the second floor and ground floor. Details are provided for checking load calculations and

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Ayreesh Mey Spnt
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190 views8 pages

Design of Column

This document summarizes the design of three interior columns (Column 1, 2, and 3) for a building. Column 1 is designed to carry a factored axial load of 285.78 kN between the roof and third floor. It will be a 300x300 mm square reinforced concrete column with 8 #16 reinforcing bars spaced at 300 mm. Column 2 is designed to carry a factored axial load of 271.01 kN between the third and second floors. It has the same dimensions and reinforcement as Column 1. Column 3 is designed to carry the largest factored axial load of 297.30 kN between the second floor and ground floor. Details are provided for checking load calculations and

Uploaded by

Ayreesh Mey Spnt
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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Design of Column 1

Roof to Third Floor (Interior)

Dimension of Column 1: 300 mm x 300mm


Check for the Slenderness Ration:
For compression members not braced against sideways when Slenderness be permitted to
be neglected if :
KLu
≤ 22 ------------ design as short column (NSCP 410.11.1 EQ. 410-7)
r
Lu = 3400 mm ---- unsupported length of compression member (mm)
K = 0.50 ------------ fixed end column (effective length factor)
r = 900 ------------- 0.30 times overall dimension of the member in the direction
that stability is being considered (ACI 10.11.2) or 410.12.2

0.50 (3400 mm)


≤ 22 ; 18.89 ≤ 22 (design as short column)
(0.30)(300 mm)
Total Loads Carried by the Column:
Due to dead load:
PDL = Reaction from the beam (longitudinal + transverse ) + weight of the column itself
PDL = 51.172 kN + 70.936 kN + 23.54 kN/m3 (0.30m x 0.30 m x 3.4 m) = 128.46 kN
Due to live load:
PLL = Reaction from the beam (longitudinal + transverse )
PLL = 40.622 kN + 41.639 kN = 82.26 kN
Due to Earthquake load:
PEL = Reaction from the beam (longitudinal + transverse )
PEL = 11.721 kN + 10.727 kN = 22.45 kN
Solve for the Factored Axial Load:
Case 1: Pu = 1.4 PDL (Section 203.3.3 NSCP 2010)
Pu = 1.4 (128.46 kN) = 179.85 kN
Case 2: Pu = 1.2PDL + 1.6PLL (Section 203.3.1 NSCP 2010)
Pu = 1.2(128.46 kN) + 1.6(82.26 kN) = 285.78 kN
Case 3: Pu = 1.2PDL + 1.0PLL + 1.0PEL
Pu = 1.2(128.46 kN) + 1.0(82.26 kN) + 1.0(22.45 kN) = 258.86 kN
Note: Choose whichever the results the greater factored axial load in all cases
PU = 285.78 kN
Solve for Gross Area of the Section: Ag = (300mm) (300mm) = 90000 mm2
Compute for Ast: ( Assume 1.7% Longitudinal Steel is Required)
ρ = 0.017 < 0.08 but shall not less than 0.01 (NSCP 2010 401.10.1)
2 2
Ast = 0.017Ag = 0.017 (90000mm ) = 1530 mm
πd
2
π (16 mm )2
Try 16mm φ bar : Abar = 4 = 4 = 201.06 mm2
As 1530 mm2
N= = = 7.61 pcs say 8 pcs ; Use 8 - 16mm∅ bars (A =
Abar 201.06 mm2
1608.5mm2 )
Checking for Axial Load Capacity of Column:
φPn = 0.80φ [0.85f’c(Ag - Ast) + fy Ast] (NSCP 2010 EQ. 410-2)
where: φ = 0.65 (f0r tied column – NSCP 101.10 Sec. 409.4.2.2)
φPn = 0.80(0.65) [0.85(21 Mpa)( 90000mm2 – 1608.5mm2 ) + (276 Mpa) (1608.5mm2)]
φPn = 1039.58 kN > PU = 285.78 kN (SAFE)
For Tied Columns Vertical Spacing of Ties shall be the smallest of the following:
(Section 5.7.10.5.2 ----using 10 mmφ ties)
a. 16 bar φ = 16(16mm) = 256mm
b. 48 tie diameter = 48(10mm) = 480mm
c. Least dimension of the column = 300mm --- use
Use 10 mmφ ties spaced @ 300mm----max.
Check for ACI Code Requirements, Limits of Reinforcement for Tied Column(Sec.
5.10.9)
1. Ast shall not be less than 0.001Ag and shall not be more 0.08Ag
1608.5mm2 > 0.01(90000mm2) = 900 mm2 (OK)
2 2 2
1608.5mm < 0.08 (90000mm ) = 7200 mm (OK)
2. Number of Longitudinal Bars : 8 pcs. > 4pcs. minimum (OK)
3. Clear distance between longitudinal bars
(300 mm – 120 mm- 20 mm – 16 mm)/2 = 72 mm > 40 mm< 150 mm (OK)

Selected Spacing :

a. 6@50mm b. 6@100mm c. rest@300mm

Column Details :
Design of Column 2
Third Floor to Second Floor (Interior)

Dimension of Column 2 : 300 mm x 300mm


Check for the Slenderness Ratio :
For compression members not braced against sideways when Slenderness be permitted to
be neglected if :
KLu
≤ 22 ------------ design as short column (NSCP 410.11.1 EQ. 410-7)
r
Lu = 3400 mm ---- unsupported length of compression member (mm)
K = 0.50 ------------ fixed end column (effective length factor)
r = 900 ------------- 0.30 times overall dimension of the member in the direction
that stability is being considered (ACI 10.11.2) or 410.12.2

0.50 (3400 mm)


≤ 22 ; 18.89 ≤ 22 (design as short column)
(0.30)(300 mm)
Total Loads Carried by the Column :
Due to dead load:
PDL = Reaction from the beam (longitudinal + transverse ) + weight of the column itself
PDL = 71.304 kN + 87.475 kN + 23.54 kN/m3 (0.30m x 0.30 m x 3.0 m) = 165.13 kN
Due to live load:
PLL = Reaction from the beam (longitudinal + transverse )
PLL = 15.768 kN + 16.048 kN = 31.82 kN
Due to Earthquake load:
PEL = Reaction from the beam (longitudinal + transverse )
PEL = 24.899 kN + 32.038 kN = 56.937 kN
Solve for the Factored Axial Load :
Case 1: Pu = 1.4 PDL (Section 203.3.3 NSCP 2010)
Pu = 1.4 (165.13 kN) = 231.19 kN
Case 2: Pu = 1.2PDL + 1.6PLL (Section 203.3.1 NSCP 2010)
Pu = 1.2(165.13 kN) + 1.6(31.82 kN) = 249.072 kN
Case 3: Pu = 1.2PDL + 1.0PLL + 1.0PEL
Pu = 1.2(165.13 kN) + 1.0(31.82 kN) +1.0(56.937kN) = 271.01 kN
Note: Choose whichever the results the greater factored axial load in all cases
PU = 271.01 kN
Solve for Gross Area of the Section : Ag = (300mm)(300mm) = 90000 mm2
Compute for Ast: ( Assume 1.7% Longitudinal Steel is Required)
ρ = 0.017 < 0.08 but shall not less than 0.01 (NSCP 2010 401.10.1)
2 2
Ast = 0.017Ag = 0.017(90000mm ) = 1530 mm
πd
2
π (16 mm )2
Try 16mm φ bar : Abar = 4 = 4 = 201.06 mm2
As 1530 mm2
N= = = 7.61 pcs say 8 pcs ; Use 8 - 16mm∅ bars (A =
Abar 201.06 mm2

1608.5mm2 )

Checking for Axial Load Capacity of Column :


φPn = 0.80φ [0.85f’c(Ag - Ast) + fy Ast] -------- (NSCP 2010 EQ. 410-2)
where: φ = 0.65 (f0r tied column – NSCP 101.10 Sec. 409.4.2.2)
φPn = 0.80(0.65) [0.85(21 Mpa)( 90000mm2 – 1608.5mm2 ) + (276 Mpa) (1608.5mm2)]
φPn = 1039.58 kN > PU = 271.01 kN (SAFE)
For Tied Columns Vertical Spacing of Ties shall be the smallest of the following:
(Section 5.7.10.5.2 ----using 10 mmφ ties)
a. 16 bar φ = 16(16mm) = 256mm
b. 48 tie diameter = 48(10mm) = 480mm
c. Least dimension of the column = 300mm --- use
Use 10 mmφ ties spaced @ 300mm----max.
Check for ACI Code Requirements, Limits of Reinforcement for Tied Column(Sec.
5.10.9)
1. Ast shall not be less than 0.001Ag and shall not be more 0.08Ag
1608.5mm2 > 0.01(90000mm2) = 900 mm2 (OK)
2 2 2
1608.5mm < 0.08 (90000mm ) = 7200 mm (OK)
2. Number of Longitudinal Bars : 8 pcs. > 4pcs. minimum (OK)
3. Clear distance between longitudinal bars
(300 mm – 120 mm- 20 mm – 16 mm)/2 = 72 mm > 40 mm< 150 mm (OK)

Selected Spacing :

a. 6@50mm b. 6@100mm c. rest@300mm

Column Details:
DESIGN OF COLUMN 3
SECOND FLOOR TO GROUND FLOOR (INTERIOR)

DIMENSION OF COLUMN 1:
300 mm x 300mm
CHECK FOR THE SLENDERNESS RATIO:
 For compression members not braced against sideways when Slenderness be
permitted to be neglected if :
KLu
≤ 22 ------------ design as short column (NSCP 410.11.1 EQ. 410-7)
r
Lu = 3400 mm ---- unsupported length of compression member (mm)
K = 0.50 ------------ fixed end column (effective length factor)
r = 900 ------------- 0.30 times overall dimension of the member in the direction
that stability is being considered (ACI 10.11.2) or
410.12.2
0.50 (3400 mm)
≤ 22 ; 18.89 ≤ 22 (design as short column)
(0.30)(300 mm)

TOTAL LOADS CARRIED BY THE COLUMN:


Due to dead load:
PDL = Reaction from the beam (longitudinal + transverse ) + weight of the column
itself
PDL = 71. kN + 87.475 kN + 23.54 kN/m3 (0.30m x 0.30 m x 3.0 m)
PDL = 165.13 kN
Due to live load:
PLL = Reaction from the beam (longitudinal + transverse )
PLL = 15.768 kN + 16.048 kN
PLL = 31.82 kN
Due to Earthquake load:
PEL = Reaction from the beam (longitudinal + transverse )
PEL = 35.317 kN + 47.915 kN
PEL = 83.23 kN
SOLVE FOR THE FACTORED AXIAL LOAD:
Case 1: Pu = 1.4 PDL (Section 203.3.3 NSCP 2010)
Pu = 1.4 (165.13 kN)
Pu = 231.19 kN
Case 2: Pu = 1.2PDL + 1.6PLL (Section 203.3.1 NSCP 2010)
Pu = 1.2(165.13 kN) + 1.6(31.82 kN)
Pu = 249.072 kN
Case 3: Pu = 1.2PDL + 1.0PLL + 1.0PEL
Pu = 1.2(165.13 kN) + 1.0(31.82 kN) + 1.0(83.32 kN)
Pu = 297.30 kN
Note: Choose whichever the results the greater factored axial load in all cases
PU = 297.30 kN
SOLVE FOR GROSS GROSS AREA OF THE SECTION:
Ag = (300mm)(300mm)
Ag = 90000 mm2
COMPUTE FOR Ast: ( Assume 1.7% Longitudinal Steel is Required)
ρ = 0.017 < 0.08 but shall not less than 0.01 (NSCP 2010 401.10.1)
Ast = 0.017Ag
Ast = 0.017(90000mm2)
Ast = 1530 mm2

Try 16mm φ bar

πd
2
π (16 mm )2
Abar = 4 = 4 = 201.06 mm2

As 1530 mm2
N= = = 7.61 pcs say 8 pcs
Abar 201.06 mm2
Use 8 - 16mm∅ bars (A = 1608.5mm2 )

CHECKING FOR AXIAL LOAD CAPACITY OF COLUMN:


φPn = 0.80φ [0.85f’c(Ag - Ast) + fy Ast] -------- (NSCP 2010 EQ. 410-2)
where: φ = 0.65 (f0r tied column – NSCP 101.10 Sec. 409.4.2.2)
φPn = 0.80(0.65) [0.85(21 Mpa)( 90000mm2 – 1608.5mm2 ) + (276 Mpa)
(1608.5mm2)]
φPn = 1039.58 kN > PU = 297.30 kN (SAFE)
FOR TIED COLUMNS VERTICAL SPACING OF TIES SHALL BE THE SMALLEST
OF THE FOLLOWING:
(Section 5.7.10.5.2 ----using 10 mmφ ties)
d. 16 bar φ = 16(16mm) = 256mm
e. 48 tie diameter = 48(10mm) = 480mm
f. Least dimension of the column = 300mm --- use
 Use 10 mmφ ties spaced @ 300mm----max.
CHECK FOR ACI CODE REQUIREMENTS, LIMITS OF REINFORCEMENT FOR
TIED COLUMN
(Sec. 5.10.9)
2.) Ast shall not be less than 0.001Ag and shall not be more 0.08Ag
1608.5mm2 > 0.01(90000mm2) = 900 mm2 (OK)
2 2 2
1608.5mm < 0.08 (90000mm ) = 7200 mm (OK)
2.) Number of Longitudinal Bars
8 pcs. > 4pcs. mimimum (OK)
3.) Clear distance between longitudinal bars
(300 mm – 120 mm- 20 mm – 16 mm)/2 = 72 mm > 40 mm< 150 mm
(OK)

Selected Spacing Selected Spacing

a.6@50mm

b.6@100mm

c. rest@300mm

COLUMN DETAILS:

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