Column math

Lecture note of Dr.H.M.A.Mahzuz on RCC Column
Page- 1
RCC Column
What is Column?
Column may be defined as a vertical structural member that carry loads chiefly in compression. It can carry moment
as well, about one or both axes of the cross section. Bending action may produce tensile forces over a part of the
cross-section. Even in such case columns are referred as compression members because compression force
dominates its behavior. Its ratio of height to least lateral dimension is 
t
L
3.
Types of Column:
A) Based on reinforcement column can be divided into the following was:
1) Column reinforced with longitudinal bar and lateral ties.
2) Column reinforced with longitudinal bar and continuous spirals.
3) Composite compression members reinforced longitudinal with structural steel shapes,pipe or tubing.
B) Based on length:
1. Short column: Short columns those whose ratio of effective length to the minimum thickness is less then 12
(i.e. 12 ). Material strength is the main factor for its failure.
2. Long Column: Short columns those whose ratio of effective length to the minimum thickness is greater
then 12 ( 12). Material strength and buckling are the main factors for its failure.
C) Depending on position at a building or loading pattern:
1) Axially loaded column
2) Uni-axial column
3) Bi-axial column.
Preliminary size of column:
The initial stage in column design involves estimating the required of the column. There is no simple rule to do this
as the axial load capacity of a given cross-section varies depending on the moment acting on the section.The
approximate relation between axial load and gross area for tied column is,
)(45.0 ',
gyc
U
trialg
ff
P
A


The approximate relation between axial load and gross area for tied column is,
)(55.0 ',
gyc
U
trialg
ff
P
A


Depending on span length an initial approximation for gross area of column can also be made. For columns
supporting heavily loaded floors minimum overall dimensions of one-fifteenth the average span of the panel is
considered satisfactory.Roof columns may be somewhat lighter; one-eighteenth the average span is specified by
some codes as a minimum diameter.
Function of Tie/ spiral/ Transverse reinforcement:
1. To hold longitudinal bars in position in forms while the concrete is being placed. For this purpose both of
them are wired togetherto form a cage of reinforcement.
2. To prevent the highly stressed slenderlongitudinal bars from buckling outward bursting the thin concrete
cover.

Page- 2
Fig: Tied column Fig: Spiral column
Serial No. Contents Tied Spiral
a) Least Dimension Minm 10 in, in EQ regions 12 in. Minm 12 in
b) Least area -- ---------
c) Steel ratio, min,g 1% 1%
d) Steel ratio, max,g 8% 8%
e) Minm No. of longitudinal bar 4 6
f) Minm longitudinal bar size #5 bar #5 bar
g) Minm No. of tie bar #3 (for bundle bars or longitudinal
bar size>#10 use #4 bar as tie)
#3
h) Minm Clear cover 1.5 in 1.5 in
Note: Generally max,g is preferable between 1%- 4% to avoid steel congestion.
Guideline for Tie/ spiral reinforcement:
1) All bars of tied columns shall be enclosed by lateral ties, at least No. 3 in size for longitudinal bars up to No. 10,
and at least No. 4 in size for Nos. 11, 14, and 18 and bundled longitudinal bars.
2) The spacing of the ties shall not exceed 16 diameters of longitudinal bars (16D), 48 diameters of tie bars
(48d), nor the least dimension (tmin) of the column.
3) No bar shall be farther than 6 in. clear on either side from such a laterally supported bar.
4) The ties shall be so arranged that every corner and alternate longitudinal bar shall have lateral support pro -
vided by the corner of a tie having an included angle of not more than 135°.
5) Spirals shall be continuous bar of not less than #3. The c lear spacing between turns of the spiral must not
exceed 3 inch (but not less than 1 inch).
Design of axially loaded column:
WSD:

Page- 3
For concentrically loaded spirally reinforced column Code provides the following formula for maximum allowable
load:
)25.0( '
gscg ffAP 
For concentrically loaded tied reinforced column Code provides the following formula for maximum allowable load:
)25.0(85.0 '
gscg ffAP 
Ag = Gross concrete area,
g
s
g
A
A
 .
USD:
Let us consider an axially loaded column where the load is taken by both concrete and steel. Therefore,
sycc AfAfP  '
. But full strength of
'
cf is not used rather 0.85
'
cf . So we can say:
sycc AfAfP  '
85.0 . Rearranging the equation we get,
g
g
s
ysgc A
A
A
fAAfP  )(85.0 '
})1(85.0{ '
gygcg ffAP  
For concentrically loaded tied reinforced column Code provides the following formula for maximum allowable
load:
})1(85.0{ '
gygcg ffAP  
85.0 and 75.0 for spiral column
80.0 and 70.0 for tied column
Recommendation for tie:
Fig: Minimum details for frames in regions of moderate risk

Page- 4
Problem 1:
Design a tied column for load 500 kip both in WSD and USD.
Suggestedsolution:
WSD
Considering, 02.0g
)25.0(85.0 '
gscg ffAP 
)02.0604.0325.0(85.0500  gA
2
24.478 inAg  [ "87.2124.4782
 hh ]
Therefore select a square section 22''22''.
As = gg A = 0.022222 = 9.68 in2. So provide 10 No. # 9 bar as main bar.
Tie reinforcement:
1) 16D= 16 18
8
9
 '' 2) 48d= 18
8
3
48  '' 3) tmin= 22'', So provide #3 @18''c/c as tie reinforcement.
USD
})1(85.0{ '
As 80.0 and 70.0 for tied column. Considering, 01.0g
)01.060)01.01(385.0(70.08.0500  gA
2
76.285 inAg  [ "7.1676.2852
 hh ]
Select a square section 18''18''.
As = gg A = 0.011818 = 3.2420.28 in2
So provide 6 No. # 6 bar as main bar.
Tie reinforcement:
1) 16D= 16 12
8
6
 '' 2) 48d= 18
8
3
Note: This problem is done assuming an arbitrary g value. If anyone does this problem assuming an arbitrary gA
then this way is also ok but remember that the g value must have to be greater than 0.01.
Problem 2:
A plan showing the beams and columns of a 10 storied market building is shown in Fig: 1(a). Here brick walls are
used only for partition. Design the Column A for Ground floor and 4th floor in both WSD and USD having the
following information:
a) Slab thickness = 6.5'',
b) Beam cross-section = 18''12''
c) Live load on the floor = 100 psf
d) Live load in the roof = 60 psf.
e) Brick wall thickness = 5''.

Page- 5
f) Story clear height = 10'
g)
'
cf =3 ksi, and yf = 60 ksi
SuggestedSolution:
WSD
Item No Item Calculation Load
1. Slab self load calculation =




 10)
12
5.6
2
1720
2
2225
(15.0
353.3 kip
2. Beam weight =





 10)
1212
2218
2
17202225
(15.0
173.25
3. Wall weight =


 10)10
12
5
2
17202225
(12.0
210 kip
4. Floor Live load =
0.1 



9)
2
1720
2
2225
(
391.3 kip
5. Roof Live load =
0.06 



1)
2
1720
2
2225
(
26.1 kip
1153.95 kip
Trial gross area of column,
)(45.0 ',
gyc
trialg
ff
P
A

 , considering, 03.0g
2
, 24.534
)03.0603(45.0
95.1153
inA trialg 

 [ "11.2324.5342
 hh ]
Select a square section = 24''24''.
Self load of column = 0.15' 60100
12
24
12
24
 kip
Total load = 1153.95 + 60 = 1214 kip
)25.0(85.0 '
gscg ffAP 

Page- 6
)03.0604.0325.0(85.012.1032  gA
2
03.826 inAg  [ "74.2803.8262
 hh ]
Self load of column = 0.15'  100
12
30
12
30
93.8 kip
Total load = 981.7+93.8=1075.45 kip
)25.0(85.0 '
gscg ffAP 
)03.0604.0325.0(85.045.1075  gA
2
7.860 inAg  [ "34.297.8602
 hh ]
Therefore the selected square section 30''30'' is ok!!!!
As = gg A = 0.033030 = 27 in2, So provide 28 No. # 9 bar as main bar.
Note: Sometimes extra bar has to be used to make the section symmetric.
Tie reinforcement:
1) 16D= 16 18
8
9
 '' 2) 48d= 18
8
3
USD
Item No Item Calculation Load
6. Slab self load calculation =




 4.110)
12
5.6
2
1720
2
2225
(15.0
494.62 kip
7. Beam weight =


 4.110)10
12
5
2
17202225
(12.0
294 kip
8. Floor Live load =
0.1 



7.19)
2
1720
2
2225
(
665.21
9. Roof Live load =
0.06 



7.11)
2
1720
2
2225
(
44.37 kip
1498.2 kip

Page- 7
Trial gross area of column,
)(45.0 ',
gyc
trialg
ff
P
A

 , considering, 03.0g
2
, 96.1046
)03.0603(45.0
2.1498
inA trialg 

 [ "36.3296.10462
 hh ]
Self load of column = 0.15'  4.1100
12
33
12
33
158.82 kip
Total load = 1498.2+158.82=1657.1 kip
})1(85.0{ '
80.0 and 70.0 for tied column
)03.060)03.01(385.0(70.08.01.1657  gA
2
43.692 inAg  [ "31.2643.6932
 hh ]
Self load of column = 0.15'  4.1100
12
27
12
27
106.31 kip
Total load = 1498.2+106.31=1604.6 kip
})1(85.0{ '
as 80.0 and 70.0 for tied column
)03.060)03.01(385.0(70.08.06.1604  gA
2
5.670 inAg  [ "89.255.6702
 hh ]
Select a square section 26''26''.
As = gg A = 0.032626 = 20.28 in2. So provide 26 No. # 8 bar as main bar.
Note: Sometimes extra bar has to be used to make the section symmetric.
Tie reinforcement:
1) 16D= 16 16
8
8
 '' 2) 48d= "18
8
3
48  3) tmin= 26'', So provide #3 @16''c/c as tie reinforcement.
Short column: Its strength is governed by strength of materials (i.e. concrete and steel) nd geometry of
cross-section.
𝜌𝑠 can be selected by the equation, 𝜌𝑠 = 0.45(
𝐴 𝑔
𝐴 𝑐
− 1)
𝑓𝑐
′
𝑓𝑦
Problem :3 Design a column for 474 kip load.
Answer:
Assume the missing data, 𝒇 𝒄
′ = 4 𝑘𝑠𝑖, 𝒇 𝒚 = 60 𝑘𝑠𝑖, 𝐴 𝑔= 15x15 in2
.

Page- 8
})1(85.0{ '
gygcgn ffAPP  
80.0 and 70.0 for tied column, 85.0 and 75.0 for spiral column,
})1(85.0{ '
)60)1(485.0(151570.08.0474 gg  
01.00064.0  g , use 01.0min 
2
25.2151501.0 inbhA gs  
Answer:
′ = 4 𝑘𝑠𝑖, 𝒇 𝒚 = 60 𝑘𝑠𝑖, 𝐴 𝑔= 15x15 in2
.
})1(85.0{ '
})1(85.0{ '
)60)1(485.0(151570.08.0700 gg  
00381.0 g
2
57.815150381.0 inbhA gs  
Answer:
′ = 4 𝑘𝑠𝑖, 𝒇 𝒚 = 60 𝑘𝑠𝑖, 𝐴 𝑔= 15x15 in2
.
})1(85.0{ '
})1(85.0{ '
)60)1(485.0(151570.08.01200 gg  
08.011.0  g
Change the material property/ section/ both and redesign
′ = 4 𝑘𝑠𝑖, 𝒇 𝒚 = 60 𝑘𝑠𝑖, 𝐴 𝑔= 24x18 in2
.
})1(85.0{ '
})1(85.0{ '
)60)1(485.0(182470.08.01200 gg  
01.00276.0  g
2
91.1118240276.0 inbhA gs  
Spacing of transverse reinforcement should be not less
then (NON-EQ zone)
𝑠 = (
16 diameter of Longitudinal steel (16D)
48 diameter of transverse steel (48d)
Least dimension of column (tmin)
)
Spacing of transverse reinforcement should be not less
then (EQ zone)
𝑠 = (
8 diameter of Longitudinal steel (8D)
24 diameter of transverse steel (24d)
1
2
of least dimension of column
)

Page- 9
and 2s
Problem :6 Design a square column at middle for a DL=1 kip/ft and LL= 0.7 kip/ft. Consider that the
building 10 storied. Also Consider, 𝒇 𝒄
′ = 3 𝑘𝑠𝑖, 𝒇 𝒚 = 60 𝑘𝑠𝑖,

Page- 10
Load at each over the column=(1 × 1.4 + 0.7 × 1.7) × (
25
2
+
25
2
+
18
2
+
18
2
) = 2.59 × 43 = 111.37 kip
So, 𝑃10 = 111.37 × 10 = 1113.7 kip
Considering, 𝜌𝑔 = 0.03,
𝑃 = 𝜑𝑃𝑛 = 𝛼𝜑𝐴 𝑔(0.85𝑓𝑐
′(1 − 𝜌𝑔) + 𝑓𝑦 𝜌𝑔)
=> 1113.7 = 0.8 × 0.7𝐴 𝑔(0.85 × 3 × (1 − 0.03) + 60 × 0.03)
=> 𝐴 𝑔 = 465.4 𝑖𝑛2 = 21.57 × 21.57,take − 24" × 24"
𝐴 𝑠 = 0.03 × 24" × 24" = 17.28 𝑖𝑛2 = 30 #7
Provide this column at 1st
, 2nd, 3rd
and 4th
story.
So, 𝑃6 = 111.37 × 6 = 668.2 kip
′(1 − 𝜌𝑔 ) + 𝑓𝑦 𝜌𝑔)
=> 668.2 = 0.8 × 0.7𝐴 𝑔(0.85 × 3 × (1 − 0.02) + 60 × 0.02)
=> 𝐴 𝑔 = 322.6 𝑖𝑛2 = 17.96 × 17.96, take − 20" × 20"
𝐴 𝑠 = 0.02 × 20" × 20" = 8 𝑖𝑛2 = 20 #6
Provide this column at 5th
, 6th
and 7th
story.
So, 𝑃3 = 111.37 × 3 = 334.11 kip
′(1 − 𝜌𝑔 ) + 𝑓𝑦 𝜌𝑔)
=> 334.11 = 0.8 × 0.7𝐴 𝑔(0.85 × 3 × (1 − 0.01) + 60 × 0.01)
=> 𝐴 𝑔 = 190.95 𝑖𝑛2 = 13.82 × 13.82, take − 15" × 15"
𝐴 𝑠 = 0.01 × 15" × 15" = 2.25 𝑖𝑛2 = 8 #5
, 9th
and 10th
story.
Check the column considering column self weight.
So, 𝑃10 = 111.37 × 10 + 1.4 × {(0.15 ×
22
12
×
22
12
× 10) × 4 + (0.15 ×
18
12
×
18
12
× 10) × 3

Page- 11
+(0.15 ×
15
12
×
15
12
× 10) × 3} = 111.37 + 52.25 = 1166 kip
′(1 − 𝜌𝑔 ) + 𝑓𝑦 𝜌𝑔)
=> 1166 = 0.8 × 0.7𝐴 𝑔(0.85 × 3 × (1 − 0.03) + 60 × 0.03)
=> 𝐴 𝑔 = 487.2 𝑖𝑛2 = 22.07 × 22.07 < 24×24(ok)
So, 𝑃6 = 111.37 × 6 = 668.2 kip
′(1 − 𝜌𝑔 ) + 𝑓𝑦 𝜌𝑔)
=> 668.2 = 0.8 × 0.7𝐴 𝑔(0.85 × 3 × (1 − 0.02) + 60 × 0.02)
=> 𝐴 𝑔 = 322.6 𝑖𝑛2 = 17.96 × 17.96, take − 18" × 18"
𝐴 𝑠 = 0.02 × 18" × 18" = 6.48 𝑖𝑛2 = 16 #6
, 6th
and 7th
story.
So, 𝑃3 = 111.37 × 3 = 334.11 kip
′(1 − 𝜌𝑔 ) + 𝑓𝑦 𝜌𝑔)
=> 334.11 = 0.8 × 0.7𝐴 𝑔(0.85 × 3 × (1 − 0.01) + 60 × 0.01)
=> 𝐴 𝑔 = 190.95 𝑖𝑛2 = 13.82 × 13.82, take − 15" × 15"
𝐴 𝑠 = 0.01 × 15" × 15" = 2.25 𝑖𝑛2 = 8 #5
, 9th
and 10th
story.
Interaction Diagram (ID):
For a column subjected to concentrated load or the loading condition where 10.
h
e  the following equations are
the dominating equations for design.
)25.0(85.0 '
gscg ffAP  (WSD) , })1(85.0{ '
gygcg ffAP   (USD)
But if 10.
h
e  then it is an easy practice to take the help of Interaction Diagram.
Interaction diagram is the graphical presentation of the different safe combinations
of load (P) and moment (M). The following equations are necessary this diagram.
sssscn fAfAabfP  '''
85.0
)
2
()'
2
()
22
(85.0 ''' h
dfAd
h
fA
ah
abfM sssscn 
These equations will be able to give us the Nominal strength curve only. To make
that usable an ACI design strength curve has to be made. Depending on column
geometry, tie/ spiral values of  and  are important for that. The shaded portion

Page- 12
is the final ID for a typical column. By the help of ID any type of column even shear walls can be designed.
Design of column subjected to Uni-axial moment:
Problem: 7
In a two storied structure an exterior column is to be designed for a service dead load of 142 kips, maximum live
load of 213 kips, dead load moment of 83 kip-ft, live load moment of 124 kip-ft. The minimum live load compatible
with the full the full live load moment is 106 kips, obtained when no live load is placed on the roof but a full live
load is placed on the 2nd floor. Consider column dimension b = 16 in, and h = 20 in.
'
cf =4 ksi, and yf = 60 ksi
(a) Find the column RF for the full live load condition.
(b) Check to ensure that the column is adequate for the condition of no live load on the roof.
Answer:
(a)
kipPu 5612137.11424.1 
ftkipMu  3271247.1834.1
Now, ksi
A
P
g
u
75.1
320
561
 and
ksi
hA
M
g
u
61.0
20320
12327



 . With a cover 2.5 in
75.0
20
520



Having these main information from the graph
039.0g
Therefore,
2
48.122016039.0 inbhA gs  
Select 10 No.10 bars.
(b)
kipPu 3791067.11424.1 
ftkipMu  327 (as before)
Now, ksi
A
P
g
u
18.1
320
379
 and ksi
hA
M
g
u
61.0
20320
12327



 . With a cover 2.5 in 75.0
20
520



Having these main information from the graph 032.0g

Page- 13
Therefore,
2
24.102016032.0 inbhA gs  
< the result of case (a). So no modification is required.
Problem: 8
A column carries a factored load 518 kip and factored
moment 530 kip-ft. Design the column.
Answer:
As no other information are given so let us assume
03.0g , h = 24 inch and a concrete cover 3 inch
75.0
24
624


 .
Now, eccentricity, inch
P
M
e
u
u
3.12
518
12530


 ,
51.0
24
3.12

h
e
. Using g and e from graph
ksi
A
P
g
u
35.1
2
52.11241603.0 inbhA gs  
Select 8 No.11 bars.
What will happen if interaction diagram is not given?????????
Design of column subjected to Bi-axial moment:
Problem: 9
The column shown below has to carry a factored load 275 kips. The is acting with eccentricities "3ye and
"6xe . The material strength
'
cf = 4 ksi, and yf = 60 ksi. Check the adequacy of the trial section.
Answer:
Let us follow the Reciprocal load method.
Considering bending about the Y-axis Considering bending about the Y-axis
;
75.0
20
520



3.0
20
6

h
e
033.0
2012
8



g
s
A
A
;
58.0
12
512


 (say = 0.6)
25.0
12
6

h
e
033.0
2012
8



g
s
A
A
inchb
P
bhA u
g
1698.15
2435.1
518
35.1





Page- 14
From Graph A.7
kipPksi
A
P
yOu
g
yOu
40875.1 ,
,
 

kipPksi
A
P
o
g
o
87665.3  

From Graph A.7
kipPksi
A
P
xOu
g
xOu
43282.1 ,
,
 

kipPksi
A
P
o
g
o
87665.3  

Known,
OyOnxOnn PPPP 
1111
,,

876
1
408
1
432
11

nP
92.275 un PP kip> the design load (275 kip). Therefore the selection of column section is ok!!
Design for long/slender column:
Problem: 10
A 22X14 inch2 (trial section)column carries an axial load Pd = 125 Md = 105 k-ft, Pl=112 kip, and Ml =96 kip-ft
due to live load. The column is part of a frame that is braced against side-sway ant bent in single curvature about its
major axis. The unsupported column length is lu = 19 ft. The moments at both ends of the column are equal. Design
the column. The material strength
'
cf = 4 ksi, and yf = 60 ksi.
Answer:

Page- 15
kipPu 4.3651127.11254.1 
ftkipMu  2.310967.11054.1
Now, ksi
A
P
g
u
2.1
1422
4.365


 and ksi
hA
M
g
u
55.0
221422
122.310



 . Considering 75.0 from graph
026.0g
Now let us check whether it is slender or not……



 )5.34
223.0
12191
(
r
klu
)2211234(1234
2
1

M
M
Therefore it is a slender column.
478.0
4.365
1254.1
7.14.1
4.1





LD
D
d
2
12119435
478.01
1242236054.0
1
4.0
inkip
IE
EI
d
gc







kip
kl
EI
P
u
c 3313
)( 2
2


)(4.014.06.0
4.0
6.0
2
1
ok
M
M
Cm 
Moment magnifier 17.1
331375.0
4.365
1
1
75.0
1






c
u
m
nx
P
P
C

Design ftkipMM unsc  3632.31017.1
Again,
Now, ksi
A
P
g
u
2.1
1422
4.365


 and ksi
hA
M
g
u
65.0
221422
12363



 .
From the same graph 036.0g
2
1.111422036.0 inbhA gs   Provide 12 No. 9 bars

Page- 16
Construction of interaction diagram:
Basic equations Manipulation:
s
s
y
E
f

yu
u
b dc




ca 85.0 (At balanced condition bcc  )
ysus f
c
cd
Ef 

 
ysus f
c
dc
Ef 


''

sssscn fAfAabfP  '''
85.0
)
2
()'
2
()
22
(85.0 ''' h
dfAd
h
fA
ah
abfM sssscn 
s
s
u
u
b
E
f
dc




ysus f
c
cd
Ef 

 
ysus f
c
dc
Ef 


''

sssscn fAfAcbf.P  '''2
850
)
2
()'
2
()
2
85.0
(85.0 '''2 h
dfAd
h
fA
ch
bfM sssscn 


Problem: 11
a. Construct an interaction diagram of the column section shown at right.
b. Construct an ACI design strength diagram for the same section.
c. Is the column is good choice for a load 540 kips applied at an eccentricity e = 4.44''?
Answer:
At first let us simplify the process:
ssn
ssn
sssscnn
ffccM
ff
c
cM
h
dfAd
h
fA
ch
cbfePM
5.225.22)425.08(8.57
)
2
16
13(25.22)3
2
16
(25.22)
2
85.016
(16585.0
)
2
()'
2
()
2
85.0
(85.0
'
'2
'''2







ssn
ssn
sssscn
ffcP
ffcP
fAfAcbf.P
5.45.480.57
25.2225.2216585.0
850
'
'2
'''2




Page- 17
Table for Interaction diagram:
Condition
c
sf '
s
f ssn ffcP 5.45.480.57 '
 ssn ffccM 5.225.22)425.08(8.57 '

B 7.69'' 53.06 413.52 4647.03
T 6.69 47.99 332.64 4423.8
T 5.69 41.13 243.97 4111.16
T 4.69 31.35 142.16 3683.7
T 3.69 16.27 16.5 3087.85
C 9 38.67 616.19 4391.91
C 11 15.82 834.61 3819.99
C 13 0.00 1021.4 3209.72
C 15 -11.60 1189.2 2497.88
Note: The more the pointscan be plotted the better the shape of ID will be.
Interaction diagram and ACI design strength diagram
If the column is subjected to a load 540 kips applied at an eccentricity e = 4.44'' then the corresponding moment is
inchkip 23764.4540 . Form the figure it is seen that the point lies in the ACI recommended safe zone.
There it can be said that the column is a good choice for a load 540 kips applied at an eccentricity e = 4.44''
Column Interaction Diagram0; 1628
2497.88; 1189.2
3209.72; 1021.4
3819.99; 834.61
4391.91; 616.19
4674.03; 413.52
4423.8; 332.64
4111.16; 243.97
3683.7; 142.16
3087.85; 16.5
0; 1139.6
2246.804; 832.44
2673.993; 584.227
3074.337; 431.333
3271.821; 289.464
3096.66; 232.848
2877.812; 170.779
2578.59; 99.512
2161.495; 11.55
2376; 540
1748.516; 832.44
0; 911.68
0
200
400
600
800
1000
1200
1400
1600
0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000
Moment (Kip-inch)
Load(Kip)

Page- 18
Few questions:
Why  is lower for column?
What is the significance of distributed and un-symmetric reinforcement on column?
Lecture Summery for column:
1) Introduction.
2) Several considerations in design of column.
3) Short column:
i) Concentrically loaded column.
ii) Eccentrically loaded column:
a) Uni-axial
b) Bi-axial
4) Long column.
5) Construction of Strength Interaction Diagram.
The End

Column math

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