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Column math
1. 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 ο³
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. 12ο£ ). 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 ( ο³12). 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
ο²ο«
ο³
The approximate relation between axial load and gross area for tied column is,
)(55.0 ',
gyc
U
trialg
ff
P
A
ο²ο«
ο³
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.
2. Lecture note of Dr.H.M.A.Mahzuz on RCC Column
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,gο² 1% 1%
d) Steel ratio, max,gο² 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,gο² 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 135Β°.
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:
3. Lecture note of Dr.H.M.A.Mahzuz on RCC Column
Page- 3
For concentrically loaded spirally reinforced column Code provides the following formula for maximum allowable
load:
)25.0( '
gscg ffAP ο²ο«ο½
For concentrically loaded tied reinforced column Code provides the following formula for maximum allowable load:
)25.0(85.0 '
gscg ffAP ο²ο«ο½
Ag = Gross concrete area,
g
s
g
A
A
ο½ο² .
USD:
Let us consider an axially loaded column where the load is taken by both concrete and steel. Therefore,
sycc AfAfP ο«ο½ '
. But full strength of
'
cf is not used rather 0.85
'
cf . So we can say:
sycc AfAfP ο«ο½ '
85.0 . Rearranging the equation we get,
g
g
s
ysgc A
A
A
fAAfP ο«οο½ )(85.0 '
})1(85.0{ '
gygcg ffAP ο²ο² ο«οο½
For concentrically loaded tied reinforced column Code provides the following formula for maximum allowable
load:
})1(85.0{ '
gygcg ffAP ο²ο²ο‘οͺ ο«οο½
85.0ο½ο‘ and 75.0ο½οͺ for spiral column
80.0ο½ο‘ and 70.0ο½οͺ for tied column
Recommendation for tie:
Fig: Minimum details for frames in regions of moderate risk
4. Lecture note of Dr.H.M.A.Mahzuz on RCC Column
Page- 4
Problem 1:
Design a tied column for load 500 kip both in WSD and USD.
Suggestedsolution:
WSD
Considering, 02.0ο½gο²
)25.0(85.0 '
gscg ffAP ο²ο«ο½
)02.0604.0325.0(85.0500 ο΄ο΄ο«ο΄ο½ο gA
2
24.478 inAg ο½ο [ "87.2124.4782
ο½οο½ hh ]
Therefore select a square section 22''ο΄22''.
As = gg Aο² = 0.02ο΄22ο΄22 = 9.68 in2. So provide 10 No. # 9 bar as main bar.
Tie reinforcement:
1) 16D= 16 18
8
9
ο½ο΄ '' 2) 48d= 18
8
3
48 ο½ο΄ '' 3) tmin= 22'', So provide #3 @18''c/c as tie reinforcement.
USD
})1(85.0{ '
gygcg ffAP ο²ο²ο‘οͺ ο«οο½
As 80.0ο½ο‘ and 70.0ο½οͺ for tied column. Considering, 01.0ο½gο²
)01.060)01.01(385.0(70.08.0500 ο΄ο«οο΄ο΄ο΄ο½ο gA
2
76.285 inAg ο½ο [ "7.1676.2852
ο½οο½ hh ]
Select a square section 18''ο΄18''.
As = gg Aο² = 0.01ο΄18ο΄18 = 3.2420.28 in2
So provide 6 No. # 6 bar as main bar.
Tie reinforcement:
1) 16D= 16 12
8
6
ο½ο΄ '' 2) 48d= 18
8
3
48 ο½ο΄ '' 3) tmin= 18'', So provide #3 @12''c/c as tie reinforcement.
Note: This problem is done assuming an arbitrary gο² value. If anyone does this problem assuming an arbitrary gA
then this way is also ok but remember that the gο² 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''ο΄12''
c) Live load on the floor = 100 psf
d) Live load in the roof = 60 psf.
e) Brick wall thickness = 5''.
5. Lecture note of Dr.H.M.A.Mahzuz on RCC Column
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 =
ο½ο΄ο΄
ο«
ο΄
ο«
ο΄ 10)
12
5.6
2
1720
2
2225
(15.0
353.3 kip
2. Beam weight =
ο½ο΄
ο΄
ο΄
ο΄
ο«ο«ο«
ο΄ 10)
1212
2218
2
17202225
(15.0
173.25
3. Wall weight =
ο½ο΄ο΄ο΄
ο«ο«ο«
ο΄ 10)10
12
5
2
17202225
(12.0
210 kip
4. Floor Live load =
0.1ο΄ ο½ο΄
ο«
ο΄
ο«
9)
2
1720
2
2225
(
391.3 kip
5. Roof Live load =
0.06ο΄ ο½ο΄
ο«
ο΄
ο«
1)
2
1720
2
2225
(
26.1 kip
1153.95 kip
Trial gross area of column,
)(45.0 ',
gyc
trialg
ff
P
A
ο²ο«
ο³ , considering, 03.0ο½gο²
2
, 24.534
)03.0603(45.0
95.1153
inA trialg ο½
ο΄ο«
ο½ [ "11.2324.5342
ο½οο½ hh ]
Select a square section = 24''ο΄24''.
Self load of column = 0.15' 60100
12
24
12
24
ο½ο΄ο΄ο΄ kip
Total load = 1153.95 + 60 = 1214 kip
For concentrically loaded tied reinforced column Code provides the following formula for maximum allowable load:
)25.0(85.0 '
gscg ffAP ο²ο«ο½
6. Lecture note of Dr.H.M.A.Mahzuz on RCC Column
Page- 6
)03.0604.0325.0(85.012.1032 ο΄ο΄ο«ο΄ο½ο gA
2
03.826 inAg ο½ο [ "74.2803.8262
ο½οο½ hh ]
Select a square section = 30''ο΄30''.
Self load of column = 0.15' ο½ο΄ο΄ο΄ 100
12
30
12
30
93.8 kip
Total load = 981.7+93.8=1075.45 kip
)25.0(85.0 '
gscg ffAP ο²ο«ο½
)03.0604.0325.0(85.045.1075 ο΄ο΄ο«ο΄ο½ο gA
2
7.860 inAg ο½ο [ "34.297.8602
ο½οο½ hh ]
Therefore the selected square section 30''ο΄30'' is ok!!!!
As = gg Aο² = 0.03ο΄30ο΄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
ο½ο΄ '' 2) 48d= 18
8
3
48 ο½ο΄ '' 3) tmin= 30'', So provide #3 @18''c/c as tie reinforcement.
USD
Item No Item Calculation Load
6. Slab self load calculation =
ο½ο΄ο΄ο΄
ο«
ο΄
ο«
ο΄ 4.110)
12
5.6
2
1720
2
2225
(15.0
494.62 kip
7. Beam weight =
ο½ο΄ο΄ο΄ο΄
ο«ο«ο«
ο΄ 4.110)10
12
5
2
17202225
(12.0
294 kip
8. Floor Live load =
0.1ο΄ ο½ο΄ο΄
ο«
ο΄
ο«
7.19)
2
1720
2
2225
(
665.21
9. Roof Live load =
0.06ο΄ ο½ο΄ο΄
ο«
ο΄
ο«
7.11)
2
1720
2
2225
(
44.37 kip
1498.2 kip
7. Lecture note of Dr.H.M.A.Mahzuz on RCC Column
Page- 7
Trial gross area of column,
)(45.0 ',
gyc
trialg
ff
P
A
ο²ο«
ο³ , considering, 03.0ο½gο²
2
, 96.1046
)03.0603(45.0
2.1498
inA trialg ο½
ο΄ο«
ο½ [ "36.3296.10462
ο½οο½ hh ]
Select a square section = 33''ο΄33''.
Self load of column = 0.15' ο½ο΄ο΄ο΄ο΄ 4.1100
12
33
12
33
158.82 kip
Total load = 1498.2+158.82=1657.1 kip
For concentrically loaded tied reinforced column Code provides the following formula for maximum allowable load:
})1(85.0{ '
gygcg ffAP ο²ο²ο‘οͺ ο«οο½
80.0ο½ο‘ and 70.0ο½οͺ for tied column
)03.060)03.01(385.0(70.08.01.1657 ο΄ο«οο΄ο΄ο΄ο½ο gA
2
43.692 inAg ο½ο [ "31.2643.6932
ο½οο½ hh ]
Select a square section = 27''ο΄27''.
Self load of column = 0.15' ο½ο΄ο΄ο΄ο΄ 4.1100
12
27
12
27
106.31 kip
Total load = 1498.2+106.31=1604.6 kip
})1(85.0{ '
gygcg ffAP ο²ο²ο‘οͺ ο«οο½
as 80.0ο½ο‘ and 70.0ο½οͺ for tied column
)03.060)03.01(385.0(70.08.06.1604 ο΄ο«οο΄ο΄ο΄ο½ο gA
2
5.670 inAg ο½ο [ "89.255.6702
ο½οο½ hh ]
Select a square section 26''ο΄26''.
As = gg Aο² = 0.03ο΄26ο΄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
ο½ο΄ '' 2) 48d= "18
8
3
48 ο½ο΄ 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.
ππ can be selected by the equation, ππ = 0.45(
π΄ π
π΄ π
β 1)
ππ
β²
ππ¦
Problem :3 Design a column for 474 kip load.
Answer:
Assume the missing data, π π
β² = 4 ππ π, π π = 60 ππ π, π΄ π= 15x15 in2
.
8. Lecture note of Dr.H.M.A.Mahzuz on RCC Column
Page- 8
})1(85.0{ '
gygcgn ffAPP ο²ο²ο‘οͺοͺ ο«οο½ο½
80.0ο½ο‘ and 70.0ο½οͺ for tied column, 85.0ο½ο‘ and 75.0ο½οͺ for spiral column,
})1(85.0{ '
gygcg ffAP ο²ο²ο‘οͺ ο«οο½
)60)1(485.0(151570.08.0474 gg ο²ο² ο΄ο«οο΄ο΄ο΄ο΄ο΄ο½ο
01.00064.0 οΌο½ο gο² , use 01.0min ο½ο²
2
25.2151501.0 inbhA gs ο½ο΄ο΄ο½ο½ο ο²
Problem :4 Design a column for 700 kip load.
Answer:
Assume the missing data, π π
β² = 4 ππ π, π π = 60 ππ π, π΄ π= 15x15 in2
.
})1(85.0{ '
gygcgn ffAPP ο²ο²ο‘οͺοͺ ο«οο½ο½
80.0ο½ο‘ and 70.0ο½οͺ for tied column, 85.0ο½ο‘ and 75.0ο½οͺ for spiral column,
})1(85.0{ '
gygcg ffAP ο²ο²ο‘οͺ ο«οο½
)60)1(485.0(151570.08.0700 gg ο²ο² ο΄ο«οο΄ο΄ο΄ο΄ο΄ο½ο
00381.0ο½ο gο²
2
57.815150381.0 inbhA gs ο½ο΄ο΄ο½ο½ο ο²
Problem :5 Design a column for 1200 kip load.
Answer:
Assume the missing data, π π
β² = 4 ππ π, π π = 60 ππ π, π΄ π= 15x15 in2
.
})1(85.0{ '
gygcgn ffAPP ο²ο²ο‘οͺοͺ ο«οο½ο½
80.0ο½ο‘ and 70.0ο½οͺ for tied column, 85.0ο½ο‘ and 75.0ο½οͺ for spiral column,
})1(85.0{ '
gygcg ffAP ο²ο²ο‘οͺ ο«οο½
)60)1(485.0(151570.08.01200 gg ο²ο² ο΄ο«οο΄ο΄ο΄ο΄ο΄ο½ο
08.011.0 ο³ο½ο gο²
Change the material property/ section/ both and redesign
Assume the missing data, π π
β² = 4 ππ π, π π = 60 ππ π, π΄ π= 24x18 in2
.
})1(85.0{ '
gygcgn ffAPP ο²ο²ο‘οͺοͺ ο«οο½ο½
80.0ο½ο‘ and 70.0ο½οͺ for tied column, 85.0ο½ο‘ and 75.0ο½οͺ for spiral column,
})1(85.0{ '
gygcg ffAP ο²ο²ο‘οͺ ο«οο½
)60)1(485.0(182470.08.01200 gg ο²ο² ο΄ο«οο΄ο΄ο΄ο΄ο΄ο½ο
01.00276.0 ο³ο½ο gο²
2
91.1118240276.0 inbhA gs ο½ο΄ο΄ο½ο½ο ο²
Spacing of transverse reinforcement should be not less
then (NON-EQ zone)
π = (
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)
π = (
8 diameter of Longitudinal steel (8D)
24 diameter of transverse steel (24d)
1
2
of least dimension of column
)
9. Lecture note of Dr.H.M.A.Mahzuz on RCC 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, π π
β² = 3 ππ π, π π = 60 ππ π,
11. Lecture note of Dr.H.M.A.Mahzuz on RCC Column
Page- 11
+(0.15 Γ
15
12
Γ
15
12
Γ 10) Γ 3} = 111.37 + 52.25 = 1166 kip
Considering, ππ = 0.03,
π = πππ = πΌππ΄ π(0.85ππ
β²(1 β ππ ) + ππ¦ ππ)
=> 1166 = 0.8 Γ 0.7π΄ π(0.85 Γ 3 Γ (1 β 0.03) + 60 Γ 0.03)
=> π΄ π = 487.2 ππ2 = 22.07 Γ 22.07 < 24Γ24(ok)
So, π6 = 111.37 Γ 6 = 668.2 kip
Considering, ππ = 0.02,
π = πππ = πΌππ΄ π(0.85ππ
β²(1 β ππ ) + ππ¦ ππ)
=> 668.2 = 0.8 Γ 0.7π΄ π(0.85 Γ 3 Γ (1 β 0.02) + 60 Γ 0.02)
=> π΄ π = 322.6 ππ2 = 17.96 Γ 17.96, take β 18" Γ 18"
π΄ π = 0.02 Γ 18" Γ 18" = 6.48 ππ2 = 16 #6
Provide this column at 5th
, 6th
and 7th
story.
So, π3 = 111.37 Γ 3 = 334.11 kip
Considering, ππ = 0.01,
π = πππ = πΌππ΄ π(0.85ππ
β²(1 β ππ ) + ππ¦ ππ)
=> 334.11 = 0.8 Γ 0.7π΄ π(0.85 Γ 3 Γ (1 β 0.01) + 60 Γ 0.01)
=> π΄ π = 190.95 ππ2 = 13.82 Γ 13.82, take β 15" Γ 15"
π΄ π = 0.01 Γ 15" Γ 15" = 2.25 ππ2 = 8 #5
Provide this column at 8th
, 9th
and 10th
story.
Interaction Diagram (ID):
For a column subjected to concentrated load or the loading condition where 10.
h
e ο£ the following equations are
the dominating equations for design.
)25.0(85.0 '
gscg ffAP ο²ο«ο½ (WSD) , })1(85.0{ '
gygcg ffAP ο²ο²ο‘οͺ ο«οο½ (USD)
But if 10.
h
e οΎ 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 οο«ο½ '''
85.0
)
2
()'
2
()
22
(85.0 ''' h
dfAd
h
fA
ah
abfM sssscn οοοο«οο½
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 ο‘ and οͺ are important for that. The shaded portion
12. Lecture note of Dr.H.M.A.Mahzuz on RCC Column
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 ο½ο΄ο«ο΄ο½
ftkipMu οο½ο΄ο«ο΄ο½ 3271247.1834.1
Now, ksi
A
P
g
u
75.1
320
561
ο½ο½ and
ksi
hA
M
g
u
61.0
20320
12327
ο½
ο΄
ο΄
ο½ . With a cover 2.5 in
75.0
20
520
ο½
ο
ο½ο¬
Having these main information from the graph
039.0ο½gο²
Therefore,
2
48.122016039.0 inbhA gs ο½ο΄ο΄ο½ο½ ο²
Select 10 No.10 bars.
(b)
kipPu 3791067.11424.1 ο½ο΄ο«ο΄ο½
ftkipMu οο½ 327 (as before)
Now, ksi
A
P
g
u
18.1
320
379
ο½ο½ and ksi
hA
M
g
u
61.0
20320
12327
ο½
ο΄
ο΄
ο½ . With a cover 2.5 in 75.0
20
520
ο½
ο
ο½ο¬
Having these main information from the graph 032.0ο½gο²
13. Lecture note of Dr.H.M.A.Mahzuz on RCC Column
Page- 13
Therefore,
2
24.102016032.0 inbhA gs ο½ο΄ο΄ο½ο½ ο²
< 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
03.0ο½gο² , h = 24 inch and a concrete cover 3 inch
75.0
24
624
ο½
ο
ο½ο¬ .
Now, eccentricity, inch
P
M
e
u
u
3.12
518
12530
ο½
ο΄
ο½ο½ ,
51.0
24
3.12
ο½ο½
h
e
. Using gο² and e from graph
ksi
A
P
g
u
35.1ο½
2
52.11241603.0 inbhA gs ο½ο΄ο΄ο½ο½ ο²
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 "3ο½ye and
"6ο½xe . 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
ο½
ο
ο½ο¬
3.0
20
6
ο½ο½
h
e
033.0
2012
8
ο½
ο΄
ο½
g
s
A
A
;
58.0
12
512
ο½
ο
ο½ο¬ (say = 0.6)
25.0
12
6
ο½ο½
h
e
033.0
2012
8
ο½
ο΄
ο½
g
s
A
A
inchb
P
bhA u
g
1698.15
2435.1
518
35.1
ο»ο½
ο΄
ο½ο
ο½ο½ο
14. Lecture note of Dr.H.M.A.Mahzuz on RCC Column
Page- 14
From Graph A.7
kipPksi
A
P
yOu
g
yOu
40875.1 ,
,
ο½οο½ ο¦
ο¦
kipPksi
A
P
o
g
o
87665.3 ο½οο½ ο¦
ο¦
From Graph A.7
kipPksi
A
P
xOu
g
xOu
43282.1 ,
,
ο½οο½ ο¦
ο¦
kipPksi
A
P
o
g
o
87665.3 ο½οο½ ο¦
ο¦
Known,
OyOnxOnn PPPP ο¦ο¦ο¦ο¦
1111
,,
οο«ο½
876
1
408
1
432
11
οο«ο½ο
nPο¦
92.275ο½ο½ο un PPο¦ 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:
15. Lecture note of Dr.H.M.A.Mahzuz on RCC Column
Page- 15
kipPu 4.3651127.11254.1 ο½ο΄ο«ο΄ο½
ftkipMu οο½ο΄ο«ο΄ο½ 2.310967.11054.1
Now, ksi
A
P
g
u
2.1
1422
4.365
ο»
ο΄
ο½ and ksi
hA
M
g
u
55.0
221422
122.310
ο½
ο΄ο΄
ο΄
ο½ . Considering 75.0ο½ο¬ from graph
026.0ο½gο²
Now let us check whether it is slender or notβ¦β¦
ο³ο½
ο΄
ο΄ο΄
ο½ )5.34
223.0
12191
(
r
klu
)2211234(1234
2
1
ο½ο΄οο½ο
M
M
Therefore it is a slender column.
478.0
4.365
1254.1
7.14.1
4.1
ο½
ο΄
ο½
ο«
ο½
LD
D
dο’
2
12119435
478.01
1242236054.0
1
4.0
inkip
IE
EI
d
gc
οο½
ο«
ο΄ο΄
ο½
ο«
ο½
ο’
kip
kl
EI
P
u
c 3313
)( 2
2
ο½ο½
ο°
)(4.014.06.0
4.0
6.0
2
1
ok
M
M
Cm ο³ο½ο«ο½ο«ο½
Moment magnifier 17.1
331375.0
4.365
1
1
75.0
1
ο½
ο΄
ο
ο½
ο
ο½
c
u
m
nx
P
P
C
ο€
Design ftkipMM unsc οο½ο΄ο½ο½ 3632.31017.1ο€
Again,
Now, ksi
A
P
g
u
2.1
1422
4.365
ο»
ο΄
ο½ and ksi
hA
M
g
u
65.0
221422
12363
ο»
ο΄ο΄
ο΄
ο½ .
From the same graph 036.0ο½gο²
2
1.111422036.0 inbhA gs ο½ο΄ο΄ο½ο½ ο² Provide 12 No. 9 bars
16. Lecture note of Dr.H.M.A.Mahzuz on RCC Column
Page- 16
Construction of interaction diagram:
Basic equations Manipulation:
s
s
y
E
f
ο½ο₯
yu
u
b dc
ο₯ο₯
ο₯
ο«
ο½
ca 85.0ο½ (At balanced condition bcc ο½ )
ysus f
c
cd
Ef ο£
ο
ο½ ο₯
ysus f
c
dc
Ef ο£
ο
ο½
''
ο₯
sssscn fAfAabfP οο«ο½ '''
85.0
)
2
()'
2
()
22
(85.0 ''' h
dfAd
h
fA
ah
abfM sssscn οοοο«οο½
s
s
u
u
b
E
f
dc
ο«
ο½
ο₯
ο₯
ysus f
c
cd
Ef ο£
ο
ο½ ο₯
ysus f
c
dc
Ef ο£
ο
ο½
''
ο₯
sssscn fAfAcbf.P οο«ο½ '''2
850
)
2
()'
2
()
2
85.0
(85.0 '''2 h
dfAd
h
fA
ch
bfM sssscn οοοο«
ο
ο½
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
ο«ο«οο½ο
οο΄ο΄ο«οο΄ο΄ο«
ο
ο΄ο΄ο΄ο½ο
οο«οο«
ο
ο½ο½
ssn
ssn
sssscn
ffcP
ffcP
fAfAcbf.P
5.45.480.57
25.2225.2216585.0
850
'
'2
'''2
οο«ο½ο
ο΄οο΄ο«ο΄ο΄ο½ο
οο«ο½
17. Lecture note of Dr.H.M.A.Mahzuz on RCC Column
Page- 17
Table for Interaction diagram:
Condition
c
sf '
s
f ssn ffcP 5.45.480.57 '
οο«ο½ ssn ffccM 5.225.22)425.08(8.57 '
ο«ο«οο½
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
inchkipοο½ο΄ 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)
18. Lecture note of Dr.H.M.A.Mahzuz on RCC Column
Page- 18
Few questions:
Why οͺ 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