DESIGN OF

CONCRETE
COLUMNS
Content:

• Types of Columns
• Analysis and Design of Short Columns
• Effective Length Factor
• Analysis and Design of Columns under Combined Axial and
Bending Load
• Moment Magnification Factor
• Shear for Columns
Introduction

• Columns act as vertical supports to beams and slabs, and to

transmit the loads to the foundations.
• Columns are primarily compressions members, although
they may also have to resist bending moment transmitted by
beams.
Introduction

• Concrete columns cross sections can be square, circular or

rectangular.
• The greatest dimension should not exceed four times its
smaller dimension (h ≤ 4b).
• For h > 4b, the member should be regarded as a wall for
design purposes.
Failure Modes of Columns
Columns may fail in one of four mechanisms:
1. Compression failure of concrete or steel
reinforcement;
2. Buckling
3. Combination of buckling and compression
failure
4. Shear
• Compression failure is likely to occur with
columns which are short and stocky.
• Buckling is probable with columns which are
long and slender.
Columns Classification
Based on Column Ties

• Tied columns
Composite Columns
• Spiral columns
• Composite columns

Tied Columns Spiral Columns

Columns Classification
Based on Length
• Short Columns – where the columns failure is due to the
crushing of concrete or the yielding of the steel bars under full
load capacity of columns
• Long Columns – where buckling effects and slenderness ratio
must be taken into consideration in the design, thus reducing
the load capacity of the column relative to that of the short
columns, its cross sectional dimensions are shall be compared
with its length. The use of moment magnification factors to
account for its slenderness.
Columns Classification
Based on Loading
Axially loaded columns
 where loads are assumed acting at
the center of the column
 Although a column subjected to pure
axial load does not exist in concrete
buildings, it can be assumed that
axially loaded columns are those
h
relatively small with small eccentricity
“e” of about 15 + 0.03h or less.
Columns Classification
Based on Loading
Eccentrically loaded columns
 The eccentric load will cause
moment in the column

h
Columns Classification
Based on Loading
Bi-axially loaded columns
 Where load is applied at any point
on the column section, causing
moment about the x and y-axes
simultaneously.
Columns Classification
Based on Frame Bracing
Braced against sidesway
 The effect of the slenderness maybe
neglected when
𝐾𝐿 12𝑀
≤ 34 − ≤ 40
𝑟 𝑀
r = 0.30 times the overall dimension in the
direction of stability for rectangular compression M1 = smaller factored end moment on a
members (tied column) compression member
r = 0.25 times the overall dimension in the M2 = larger factored end moment on a
direction of stability for circular compression compression member
members (spiral column) Le – unsupported length of
compression member
Columns Classification
Based on Frame Bracing
Not braced against sidesway
(sway frames)
 The effect of the slenderness maybe
neglected when
𝐾𝐿
≤ 22
𝑟
Where M1/M2 us positive if the column is
bent in a single curvature and negative if
the member is bent in double curvature.
Analysis for Short Columns:
Tied and Spiral Columns

Capacity
ϕ𝑷𝒏 ≥ 𝑷𝒖
Analysis for Short Columns:
Tied and Spiral Columns

Capacity
ϕ𝑷𝒏 ≥ 𝑷𝒖
Analysis for Short Columns:
Tied and Spiral Columns

Capacity
ϕ𝑷𝒏 ≥ 𝑷𝒖

𝑃 = ∅𝑃 = ∅ 0.80 [ 0.85 𝑓 𝐴 − 𝐴 + 𝑓 𝐴 ]-Tied

𝑃 = ∅𝑃 = ∅ 0.85 [ 0.85 𝑓 𝐴 − 𝐴 + 𝑓 𝐴 ]-Spiral
Analysis for Short Columns:
Tied and Spiral Columns
Limits for longitudinal reinforcement:

Minimum number of longitudinal bars in

compression members three for triangular
ties and shall be four within rectangular or
circular ties for tied columns and six for spiral
columns.
Analysis for Short Columns:
Tied and Spiral Columns
Limits for ties:
Spiral
Tied • For cast in place concrete, ties shall not be less than 10
• Use 10 mm dia. ties for longitudinal mm dia.
• Clear spacing between spirals shall not exceed 75 mm
bars 32 mm in dia. or smaller and 12 or be less than 25 mm
mm dia. ties for 36 mm & above and • Anchorage of spiral reinforcement shall be provided by
bundled longitudinal reinforcement one and one-half extra turn of spiral bar at each end of a
spiral unit
• Vertical spacing of ties (except for The steel ratio of spiral reinforcement shall not be less than
section 418: seismic resistant criteria) the value given by:
𝐴 𝑓′
16𝑑 , 48𝑑 𝑜𝑟 𝜌 = 0.45 −1
𝐴 𝑓
𝑙𝑒𝑎𝑠𝑡 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑐𝑜𝑙𝑢𝑚𝑛
Analysis for Short Columns:
Example 01
A reinforced concrete tied column 1. Smallest dimension of the tied column if the steel ratio
carries an axial dead load of 950 kN is 2.5%.
and an axial live load of 1220 kN. Use 𝑃 = ∅ 0.80 [ 0.85 𝑓 𝐴 − 𝐴 + 𝑓 𝐴 ]
fc’ = 28 MPa and fy = 400 MPa.
𝐴
1. Compute the smallest dimension 𝜌=
of the tied column if the steel ratio 𝐴
is 2.5%. 𝐴 = 𝜌𝐴
2. Compute the number of 25 mm 𝑃 = ∅ 0.80 [ 0.85 𝑓 𝐴 − 𝜌𝐴 + 𝑓 𝜌𝐴 ]
diameter bars. 𝑃
𝐴 =
3. Compute the clear spacing ∅ 0.80 [ 0.85 𝑓 1 − 𝜌 + 𝑓 𝜌]
between longitudinal bars. 𝑃 = 𝑃 = 1.2𝐷𝐿 + 1.6𝐿𝐿
4. Compute the spacing of 10 mm 𝑃 = 1.2 950 + 1.6 1220
ties. 𝑃 = 3092 𝑘𝑁
Analysis for Short Columns:
Example 01
A reinforced concrete tied 1. Smallest dimension of the tied column if the steel ratio is
column carries an axial dead
load of 950 kN and an axial 2.5%.
load of 1220 kN. Use fc’ = 28 3092 𝑥 10
MPa and fy = 400 MPa. 𝐴 =
1. Compute the smallest (0.65) 0.80 [ 0.85 (28) 1 − 0.025 + (400)(0.025)]
dimension of the tied 𝐴 = 179074.0505 𝑚𝑚
column if the steel ratio
is 2.5%. 𝑠= 𝐴 = 179074.0505
𝑠 = 423 𝑚𝑚 𝑠𝑎𝑦 425 𝑚𝑚 𝑠𝑞𝑢𝑎𝑟𝑒
Analysis for Short Columns:
Example 01
A reinforced concrete tied 2. Compute the number of 25 mm diameter bars.
column carries an axial dead
load of 950 kN and an axial
load of 1220 kN. Use fc’ = 28 𝐴 = 425 = 180625 𝑚𝑚
MPa and fy = 400 MPa.
1. 425 mm x 425 mm 𝑃 = ∅ 0.80 [ 0.85 𝑓 𝐴 − 𝐴 +𝑓 𝐴 ]
2. Compute the number 3092𝑥10 = (0.65) 0.80 [ 0.85 28 180625 − 𝐴 + 400 𝐴 ]
of 25 mm diameter 𝐴 = 4378.732 𝑚𝑚
bars.
Using 25 mm diameter bars:
4378.732
𝑵 =
1
𝜋 25
4
𝑵 = 8.92 say 10 pcs – as 9 pcs would be impractical but possible
(for design practicality options are to increase the number of bars to 10
or 12 pcs or maybe to use 20 mm or a mix or 20 and 25 mm)
Analysis for Short Columns:
Example 01
A reinforced concrete tied
column carries an axial dead 3. Compute the clear spacing between longitudinal bars.
load of 950 kN and an axial load
of 1220 kN. Use fc’ = 28 MPa and (The number of bars of 10 pcs could be arranged in many ways.
fy = 400 MPa.
1. 425 mm x 425 mm
The clear spacing of the main reinforcement would be dependent
2. 10-25 mm diameter bars. on this arrangement.)
3. Compute the clear
spacing between
longitudinal bars.
Analysis for Short Columns:
Example 01
A reinforced concrete tied
column carries an axial dead 4. Compute the spacing of ties.
load of 950 kN and an axial load 𝑆 = 48 𝑡𝑖𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 = 48(10) = 480 𝑚𝑚
of 1220 kN. Use fc’ = 28 MPa and
fy = 400 MPa. 𝑆 = 16 𝑚𝑎𝑖𝑛 𝑏𝑎𝑟 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 = 16 25 = 400 𝑚𝑚
1. 425 mm x 425 mm
2. 9-25 mm diameter bars.
𝑆 = 𝑙𝑒𝑎𝑠𝑡 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑜𝑓 𝑐𝑜𝑙𝑢𝑚𝑛 = 425 𝑚𝑚
3. (discussed)
4. Compute the spacing of
10 mm ties. Use 400 mm.

Note that this spacing is still need to be checked based on shear

requirement and earthquake resistant structure provision of NSCP
2015 sec. 418.
Analysis for Short Columns:
Example 02
A spiral column carries a dead load 1. Compute the diameter of the spiral
of 1070 kN and a live load of 980 kN. column.
Use fc’ = 27.5 Mpa and fy = 413.5 𝑃
𝐴 =
Mpa. The steel ratio is 0.03. ∅ 0.85 [ 0.85 𝑓 1 − 𝜌 + 𝑓 𝜌]
𝑃 = 𝑃 = 1.2𝐷𝐿 + 1.6𝐿𝐿
1. Compute the diameter of the
𝑃 = 1.2 1070 + 1.6 980 = 2852 𝑘𝑁
spiral column. 2852 𝑥10
𝐴 =
2. Compute the number of 25 mm (0.75) 0.85 [ 0.85 (27.5) 1 − 0.03 + (413.5)(0.03)]
dia. main reinforcement. 𝐴 = 127533.7773 𝑚𝑚
3. Compute the possible spacing 𝜋𝐷 = 127533.7773
of the 10 mm spirals. 𝐷 = 402.96 𝑚𝑚 𝑠𝑎𝑦 405 𝑚𝑚
Analysis for Short Columns:
Example 02
A spiral column carries a dead 2. Compute the number of 25 mm dia. main
load of 1070 kN and a live load of reinforcement.
980 kN. Use fc’ = 27.5 Mpa and fy 1
𝐴 = 𝜋 405 = 128824.9338 𝑚𝑚
= 413.5 Mpa. The steel ratio is 4
𝑃 = ∅ 0.85 [ 0.85 𝑓 𝐴 − 𝐴 + 𝑓 𝐴 ]
0.03. 2852𝑥10^3
1. 405 mm diameter = (0.75) 0.85 [ 0.85 (27.5) 128824.9338 − 𝐴 + (413.5)𝐴 ]
𝐴 = 3748.651 𝑚𝑚
2. Compute the number of 25 1
mm dia. main reinforcement. 𝜋 25 𝑁 = 3748.651
4
𝑁 = 7.637 say 8 pcs
Analysis for Short Columns:
Example 02
3. Compute the possible spacing of the 10 mm spirals.
A spiral column carries a dead D = 405 − 80
load of 1070 kN and a live load of D = 325 𝑚𝑚
𝐴 𝐷
980 kN. Use fc’ = 27.5 Mpa and fy 0.45
𝐴
−1 𝑓 0.45
𝐷
−1 𝑓
ρ = =
= 413.5 Mpa. The steel ratio is 𝑓 𝑓

0.03. 0.45
405
325
− 1 27.5
ρ = = 0.01655
1. 405 mm diameter 413.5
𝜋
4𝐴 𝐷 − 𝑑 4 10 325 − 10
4
2. 8 pcs 𝑠=
𝜌𝐷
=
0.01655 325
𝑠 = 56.61 say 55 mm
𝑠 > 25 𝑚𝑚 𝑜𝑘 𝑎𝑛𝑑 𝑠 < 75 𝑚𝑚 𝑜𝑘, 𝑡ℎ𝑒𝑛
Use s = 55 mm o.c.
Effective Length Factors for
Concrete Columns
Braced against sidesway Use to determine wither
The effect of the slenderness maybe neglected when
𝐾𝐿 12𝑀 the column is short or
𝑟
≤ 34 −
𝑀
≤ 40 slender. If the column is
slender, moment
Not braced against sidesway (sway frames) magnification factor is
 The effect of the slenderness maybe neglected deem necessary.
when
Where, M1 = smaller factored end moment
𝐾𝐿 M2 = larger factored end moment
≤ 22
𝑟 M1/M2 = positive for member bent in single curvature and negative for
double curvature
Effective Length Factors for
Concrete Columns
𝐿 = 𝑢𝑛𝑠𝑢𝑝𝑝𝑜𝑟𝑡𝑒𝑑 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑚𝑒𝑚𝑏𝑒𝑟𝑠
𝐿 = 𝑐𝑙𝑒𝑎𝑟 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑓𝑙𝑜𝑜𝑟 𝑠𝑙𝑎𝑏𝑠, 𝑏𝑒𝑎𝑚𝑠 𝑜𝑟 𝑜𝑡ℎ𝑒𝑟 𝑚𝑒𝑚𝑏𝑒𝑟𝑠
𝑐𝑎𝑝𝑎𝑏𝑙𝑒 𝑜𝑓 𝑝𝑟𝑜𝑣𝑖𝑑𝑖𝑛𝑔 𝑙𝑎𝑡𝑒𝑟𝑎𝑙𝑠 𝑢𝑝𝑝𝑜𝑟𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑏𝑒𝑖𝑛𝑔 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑒𝑑.
𝑇ℎ𝑒 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑔𝑦𝑟𝑎𝑡𝑖𝑜𝑛:
𝑟 = 0.30 𝑡𝑖𝑚𝑒𝑠 𝑡ℎ𝑒 𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑡𝑎𝑏𝑖𝑙𝑖𝑡𝑦. (rectangular)
𝑟 = 0.25 𝑡𝑖𝑚𝑒𝑠 𝑡ℎ𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟𝑠 𝑓𝑜𝑟 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑚𝑒𝑚𝑏𝑒𝑟𝑠.
𝑇ℎ𝑒 𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝐼𝑛𝑒𝑟𝑡𝑖𝑎:
𝐵𝑒𝑎𝑚𝑠 = 0.35𝐼
𝐶𝑜𝑙𝑢𝑚𝑛𝑠 = 0.7𝐼
𝑊𝑎𝑙𝑙:
𝐶𝑟𝑎𝑐𝑘𝑒𝑑 = 0.35𝐼
𝑈𝑛𝑐𝑟𝑎𝑐𝑘𝑒𝑑 = 0.70𝐼
 Jackson and Moreland Alignment Chart

Cracked moment of inertia:

Beam = 0.35𝐼
Columns = 0.7𝐼
Effective Length Factors for
Concrete Columns
Example 03
A
Figure shows an braced frame. Properties
of each beams and columns are tabulated.
Considering only bending in the plane of
the frame, compute the following: B

a. Effective length factor of column AB.

b. Slenderness ratio of column AB.
c. Calculate whether column AB is a
short or slender column, assuming
single curvature with M1 = 100 kN.m ad 6.0 7.2
M2 = 180 kN.m.
Effective Length Factors for
Concrete Columns A

Example 03
B
a. Effective length factor of column AB.

Moment of Inertia:
300 500
𝐼 = = 3125 𝑥 10 𝑚𝑚 − 𝑐𝑜𝑙𝑢𝑚𝑛 𝐴𝐵
12 6.0 7.2
300 600
𝐼 = = 5400𝑥 10 𝑚𝑚 − 𝐿𝑜𝑤𝑒𝑟 𝐺𝑖𝑟𝑑𝑒𝑟
12
300 450
𝐼 = = 2278𝑥 10 𝑚𝑚 − 𝑢𝑝𝑝𝑒𝑟 𝐺𝑖𝑟𝑑𝑒𝑟
12
Effective Length Factors for
Concrete Columns
Example 03
a. Effective length factor of column AB.
𝑈𝑠𝑖𝑛𝑔 𝑡ℎ𝑒 𝑎𝑙𝑖𝑔𝑛𝑚𝑒𝑛𝑡 𝑐ℎ𝑎𝑟𝑡:
ψ=∑ /∑
At point A:
0.7𝐼
𝐿 At point B:
ψ = 0.7𝐼
0.35𝐼 0.35𝐼
𝐿 + 𝐿 𝐿
ψ =
0.35𝐼 0.35𝐼
0.7 3125𝑥10 𝐿
+
𝐿
ψ = 3000
0.7 3125𝑥10 0.7 3125𝑥10
0.35 2278𝑥10 0.35 2278𝑥10 3000 + 3600
6000 + 7200 ψ =
0.35 5400𝑥10 0.35 5400𝑥10
ψ = 2.99 +
6000 7200
ψ = 2.314
Effective Length Factors for
Concrete Columns
Example 03
a. Effective length factor
of column AB.
𝑈𝑠𝑖𝑛𝑔 𝑡ℎ𝑒 𝑎𝑙𝑖𝑔𝑛𝑚𝑒𝑛𝑡 𝑐ℎ𝑎𝑟𝑡:
ψ = 2.99
ψ = 2.314 0.75

Using the alignment chart the

effective length factor, K = 1.75
Effective Length Factors for
Concrete Columns A

Example 03 Lu

B
b. Slenderness ratio of column AB.
𝐿 = 3000 − − = 2475 mm
kL 0.75 2475
= = 12.375
r 0.30 500 6.0 7.2
Effective Length Factors for
Concrete Columns A

Example 03 Lu

B
c. Calculate whether column AB is a short or
slender column, assuming single curvature
with M1 = 100 kN.m ad M2 = 180 kN.m
kL 𝑀
= 34 − 12
r 𝑀
6.0 7.2
12.375 = 34 – 12 (100/180)
12.375 = 27.333

Therefore the column is not a slender column.

Here, the moment carried by the column must not
be multiplied by the moment magnification factor.
ECCENTRICALLY LOADED
COLUMNS
Modes of Failure in Eccentrically Loaded Columns
1. Tension failure by initial yielding of steel at the tension side
2. Compression failure by initial crushing of the concrete at the
compression side.
Two case of Tension Failure
1. Tension steel yields and compression steel yields
2. Tension steel yields and compression steel does not yields
Case of Compression Failure
1. Tension steel will not yields and compression steel yields

In all these cases the strain-compatibility must be maintained.

MOMENT MAGNIFICATION FACTOR
• Without Sidesway (non sway)
𝛿=
.

𝐶 = 0.6 + 0.4 ≥ 0.40

.
𝑃 = ; E = 4700 𝑓 ′ ; 𝐸𝐼 =
𝑀 = 𝛿𝑀
• With Sidesway (sway)
𝛿= ∑
. ∑
𝑀 =𝑀 + 𝛿𝑀
MOMENT MAGNIFICATION FACTOR
Example: 04
The tied column shown is 300 x 375 mm. It is to
be used in a frame braced against sidesway. The
column is bent in single curvature about its y –
axis and has an unsupported length of 4.8 m.
Assumed that K = 0.83. The column carrying an
unfactored dead axial load of 134 kN. Use fc’ =
27.6 Mpa and fy = 414.6 Mpa. The column is
normal weight.
• Compute the critical load of the column.
• Determine the moment magnifying factor.
• Determine the actual eccentricity of the
load.
MOMENT MAGNIFICATION FACTOR
Example: 04
• Compute the critical load of the column.
𝜋 𝐸𝐼
𝑃 =
𝑘𝐿
𝐸 = 4700 27.6
𝐸 = 24692 𝑀𝑃𝑎
300 375
𝐼 = = 1318.4 𝑥 10 𝑚𝑚
12
1.2𝑃 134
𝛽 = = 1.2 = 0.328
𝑃 490
Relative stiffness of the column:
0.40𝐸 𝐼 0.40 (24692)(1318.4 x 10 )
𝐸𝐼 = =
1+𝛽 1 + 0.328
= 9805.4 𝑥 10 𝑁. 𝑚𝑚
MOMENT MAGNIFICATION FACTOR
Example: 04
• Compute the critical load of the
column.
𝜋 𝐸𝐼
𝑃 =
𝑘𝐿
( . )
𝑃 = = 6097 kN
.
MOMENT MAGNIFICATION FACTOR
Example: 04
• Determine the moment magnifying
factor.
𝐾𝐿 𝑀
= 34 − 12
𝑟 𝑀
0.83(4800) 112
= 34 − 12
0.3(375) 116
35.41 > 22.41 so it’s a slender column
MOMENT MAGNIFICATION FACTOR
Example: 04
• Determine the moment • Determine the actual
magnifying factor. eccentricity of the load.
𝑀 𝑀 = 𝛿𝑀 = 1.1 116
𝐶 = 0.60 + 0.40
𝑀 𝑀 = 127.6 𝑘𝑁. 𝑚
𝐶 = 0.60 + 0.40 = 0.986 𝑀 = 𝑒𝑃
127.6 = 𝑒 490
.
𝛿= = = 1.10 𝑒 = 260 𝑚𝑚
. . ( )
MOMENT MAGNIFICATION FACTOR
Example: 05
The tied column for a given frame not
braced against sidesway has a cross • Determine the moment
section of 450 x 450 mm. It has an magnification factor.
unsupported height of 5.4 m. Assume k
= 1.3 and use fc’ = 27.6 Mpa and fy = • Determine the magnified
414.6 Mpa.
Pu = 2326 kN moment.
Mns = 132 kN.m (no appreciable • Determine the minimum
sidesway)
Mns = 122 kN.m (has appreciable factored end moment as
sidesway) per NSCP.
∑ 𝑃 = 72000 kN
∑ 𝑃 = 21400 kN
MOMENT MAGNIFICATION FACTOR
Example: 05
• Determine the moment magnification • Determine the magnified
factor.
𝑘𝐿 1.3 (5400) moment.
= 𝑀 = 𝑀 + 𝛿𝑀
𝑟 0.3(450)
= 52 > 22 (𝑠𝑙𝑒𝑛𝑑𝑒𝑟 𝑐𝑜𝑙𝑢𝑚𝑛 𝑀 = 132 + 1.81 122
1
𝛿=
∑𝑃 𝑀 = 𝟑𝟓𝟐. 𝟖𝟐 𝐤𝐍. 𝐦
1 −
0.75 ∑ 𝑃 • Determine the minimum
𝛿=
1
= 𝟏. 𝟖𝟏 < 2.5 factored end moment as per
1−
72000 NSCP.
0.75 (214000)
The mmf must be lesser than 2.5, so that 𝑀 = 2326 (15 + 0.03𝑥450)
no danger for the frame becoming unstable 𝑀 = 66.29 kN.m
under gravity loads.
DESIGN OF ECCENTRIC COLUMN USING
INTERACTION DIAGRAM (UNIAXIAL)
EXAMPLE 06
A spiral column having a diameter of
500 mm is reinforced with 8 – 28
mm vertical bars. Using fc’ = 28 Mpa
and fy = 420 Mpa.
• Determine the steel ratio
• Determine the eccentricity
• Determine the eccentric Pn
having an eccentricity of 400 mm
using the interaction diagram.
DESIGN OF ECCENTRIC COLUMN USING
INTERACTION DIAGRAM (UNIAXIAL)
EXAMPLE 06
• Determine the steel ratio

𝜌= = =0.0251
• Determine the eccentricity
𝑒 400
= = 0.80
𝐷 500
• Determine the eccentric Pn having an
eccentricity of 400 mm using the interaction
diagram.
K = Pn/(fc’ Ag) = 0.20
1
𝑃 = 0.20 28 𝜋 500 = 𝟏𝟎𝟗𝟗. 𝟓𝟓𝟕 𝒌𝑵
4
DESIGN OF ECCENTRIC COLUMN USING
INTERACTION DIAGRAM (UNIAXIAL)
EXAMPLE 07
DESIGN OF ECCENTRIC COLUMN USING
INTERACTION DIAGRAM (UNIAXIAL)
EXAMPLE 07
DESIGN OF ECCENTRIC COLUMN USING
INTERACTION DIAGRAM (BIAXIAL)
 DESIGN OF ECCENTRIC COLUMN USING
INTERACTION DIAGRAM (BIIAXIAL)
EXAMPLE 07
A 350 mm x 500 mm is reinforced with
8 – 20 mm bars arranged around the 100

column perimeter. A factored load Pu =

87.5

980 kN is to be applied with

eccentricities ex = 87.5 mm and ey =
100 mm. Using fc’ = 21 Mpa and fy =
420 Mpa. Check the adequacy of the
column using reciprocal load method
and load contour method.
 DESIGN OF ECCENTRIC COLUMN USING
INTERACTION DIAGRAM (BIIAXIAL)
EXAMPLE 07
Along y-axis:
𝑃 = 0.80[0.85𝑓 𝐴 − 𝐴 + 𝐴 𝑓 ]
1
𝐴 = 𝜋 20 8 = 2513.27 𝑚𝑚
4
𝐴 = 350 500 = 175000 𝑚𝑚
𝑃 = 0.80[0.85 21 175000 − 2513.27 + 2513.27 420 ]
𝑃 = 3307.57 kN
210
𝛾= = 0.6
350
𝑒 87.5
= = 0.25
ℎ 350
2513.27
𝜌= = 0.0144
175000
𝑘 = = 0.60
𝑃 = 0.60 21 175000 = 2205 𝑘𝑁
DESIGN OF ECCENTRIC COLUMN USING
INTERACTION DIAGRAM (BIIAXIAL)
EXAMPLE 07
Along x-axis:
300
𝛾= = 0.6
500
𝑒 100
= = 0.2
ℎ 500
𝑘 = = 0.64
𝑃 = 0.64 21 175000 = 2352 𝑘𝑁
1 1 1 1
= + −
∅𝑃 ∅𝑃 ∅𝑃 ∅𝑃
1 1 1 1
= + −
(0.75)𝑃 (0.75)2205 (0.75)2352 0.75 (3307.57)
𝑃 = 1735.063 𝑘𝑁
 DESIGN OF ECCENTRIC COLUMN USING
INTERACTION DIAGRAM (BIIAXIAL)
EXAMPLE 07
Along y-axis:
𝑀
𝑅 = = 0.14
𝑓𝐴 ℎ
𝑀 = 0.14(21)(175000)(350)=180.075 kN.m
 DESIGN OF ECCENTRIC COLUMN USING
INTERACTION DIAGRAM (BIIAXIAL)
EXAMPLE 07
Along x-axis:

𝑀
𝑅 = = 0.13
𝑓𝐴 ℎ
𝑀 = 0.13(21)(175000)(500)=238.875 kN.m
 DESIGN OF ECCENTRIC COLUMN USING
INTERACTION DIAGRAM (BIIAXIAL)
EXAMPLE 07

𝑀 = 980(87.5) = 85.75 𝑘𝑁. 𝑚

𝑀 = 980(100) = 98 𝑘𝑁. 𝑚

𝑀 85.75
= = 0.359
𝑀 238.875
𝑀 98
= = 0.544
𝑀 180.075
 = 1.0
 DESIGN OF ECCENTRIC COLUMN USING
INTERACTION DIAGRAM (BIIAXIAL)
EXAMPLE 07

 
∅𝑀 ∅𝑀
+ ≤ 1.0
∅𝑀 ∅𝑀
0.359 + 0.544 = 0.903 < 1.0 (𝑠𝑎𝑓𝑒)
Sample problems.

• Use this link:

https://drive.google.com/drive/folders/17G7SUaPea3O
P82ZWkstNJn8CTktorbbd?fbclid=IwAR2v3BLot4b4nCk0
8-
gidZICevBsfYWCfVGtxmXQKbLRwYLqH7gPrFzlxHw_aem
_AYES9Jc6xKV9GVCzgM5ni2DbXxLAlZ0acwUdaplsjYEH2
pVaw9ImtFLtl8OistyalHzEJ52dBsBMpnf6mfSesHO-
END