1. Engr. Ayaz Waseem ( Lecturer/Lab Engr., CED) Columns
LECTURE # 3
1. TRIAL SIZE OF COLUMN:
Approximate formula;
i. For tied columns,
yc
yxu
trialg
ff
MMP
A
008.0'43.0
22
)(
+
++
≥
ii. For spiral columns,
yc
yxu
trialg
ff
MMP
A
01.0'50.0
22
)(
+
++
≥
2. CONCENTRICALLY LOADED SHORT COLUMN:
Column is said to be a short column if it satisfies the following condition;
2
1
1234
M
M
r
lK u
−≤ …………………………………………...
(1)
For rectangular section, r = 0.3 hmin
where, hmin is the least column dimension.
And for concentrically loaded column, k = 0.75 and 0
2
1
=
M
M
Putting these values in equation (1),
34
3.0
75.0
min
≤
h
lu
6.13
min
≤
h
lu
or 6.13≤
DimensionColumnLeast
Length
……………. (1)
Now, for circular section, r = 0.25 D
where, D is the column diameter.
Putting these values in equation (1),
34
25.0
75.0
≤
D
lu
33.11≤
D
lu
or 33.11≤
DiameterColumn
Length
…………………… (2)
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2. Engr. Ayaz Waseem ( Lecturer/Lab Engr., CED) Columns
If above conditions in equation (1) and (2) are satisfied then the column is said to be a
short column.
3. DESIGN PROBLEM:
An axially loaded short column has a length of 4m and carries a factored load (Pu)
of 2000 kN. Other data is as follows;
fc’ = 28 MPa
fy = 420 MPa
Maximum aggregate size = 20 mm
Using US Customary Bars design the column as;
i. Square tied column.
ii. Circular spiral column.
Solution:
i. Design of Square Tied Column:
STEP 1: Area Calculations:
yc
yxu
trialg
ff
MMP
A
008.0'43.0
22
)(
+
++
=
( ) ( )
2
)( 13.129870
420008.02843.0
0)10002000(
mmA trialg =
+
+×
=
Trial size of column = ( )361361 × mm
Say, trial size of column is,
( )375375 × mm
STEP 2: Check for the type of column:
6.137.10
375
4000
<==
DimensionColumnLeast
Length
Therefore, it is a short column.
STEP 3: Reinforcement calculations:
For a tied column,
Pu = φPn = ( 0.65 x 0.8 ) [ 0.85 fc’Ag + ( fy – 0.85 fc’ )Ast ]
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3. Engr. Ayaz Waseem ( Lecturer/Lab Engr., CED) Columns
( 2000 x 1000 ) = (0.65 x 0.8) ( ) ( )[ ]{ }stA2885.04203752885.0 2
×−+××
Ast = 1260.2 mm2
STEP 4: Reinforcement Ratio check:
ρ = 2
375
2.1260
=
hb
Ast
= 0.009 < 0.01 ( Not Satisfied )
Therefore,
ρ = ρmin = 1% = 0.01
So,
Ast = 0.01 x (375)2
≈ 1407 mm2
Selecting,
8 # 16 Bars
STEP 5: Design of ties:
• As maximum diameter of the longitudinal bars is < 32 mm, so,
Diameter of the ties = 10 mm
• Spacing of the ties = Minimum of;
a. 16 x 16 = 352 mm.
b. 48 x 10 = 480 mm.
c. Least column dimension = 375 mm.
d. 300 mm. ( Governs )
Therefore,
Spacing of the ties, S = 300 mm.
Note: In this step shape of the ties is not given. In step 6, shape of the ties is decided
during detailing.
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4. Engr. Ayaz Waseem ( Lecturer/Lab Engr., CED) Columns
STEP 6: Detailing:
ii. Design of Circular Spiral Column:
STEP 1: Area Calculations:
yc
yxu
trialg
ff
MMP
A
01.0'50.0
22
)(
+
++
=
( ) ( )
2
)( 11.109890
42001.02850.0
0)10002000(
mmA trialg =
+
+×
=
Trial diameter of column, D = 374 mm
Say, trial diameter of column is,
D = 375 mm
STEP 2: Check for the type of column:
33.117.10
375
4000
<==
ColumnofDiameter
Length
Therefore, it is a short column.
STEP 3: Reinforcement calculations:
For a spiral column,
Pu = φPn = ( 0.7 x 0.85 ) [ 0.85 fc’Ag + ( fy – 0.85 fc’ )Ast ]
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5. Engr. Ayaz Waseem ( Lecturer/Lab Engr., CED) Columns
( 2000 x 1000 ) = (0.7 x 0.85) ( )( ) ( )[ ]{ }stA2885.0420375
4
2885.0 2
×−+×× π
Ast = 1849.4 mm2
STEP 4: Reinforcement Ratio check:
ρ = 22
375
4
4.1849
4






=





 ππ
D
Ast
= 0.0167 > 0.01 ( Satisfied )
Therefore,
Ast ≈ 1850 mm2
Selecting,
4 # 16 , 4 # 19 Bars
STEP 5: Design of Spiral:
• Diameter of the spiral = dsp = 10 mm
• Minimum pitch of the spiral = Larger of;
a. 1.5 x 20 = 30 mm. ( Governs )
b. 25 mm.
Therefore,
Smin = 30 mm.
• Smax =






−1'45.0
2
c
g
cc
ysp
A
A
fD
fdπ
.
Now, Dc = 375 – ( 40 x 2 ) = 295 mm
Ac = ( )22
295
44
ππ
=cD = 68349.2 mm2
Ag = ( )22
375
44
ππ
=D = 110446.6 mm2
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6. Engr. Ayaz Waseem ( Lecturer/Lab Engr., CED) Columns
Therefore,
Smax =






−1'45.0
2
c
g
cc
ysp
A
A
fD
fdπ
= 57.63 mm < 75 mm ( O.K. )
• Selecting pitch between Smin and Smax ,
S = 45 mm
STEP 6: Detailing:
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7. Engr. Ayaz Waseem ( Lecturer/Lab Engr., CED) Columns
4. INTERACTION CURVE:
It is a graph between load and moment. In order to draw an interaction curve we
require the coordinates of different points.
Fig. Interaction Curve
5. PLASTIC CENTROID:
It is a point through which the resultant of all the internal forces should pass, in
pure axially loaded column case, when no moment is acting on the column at ultimate
stage of failure.
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8. ySySyScc fAfAfAAfR 321'85.0 +++=
Engr. Ayaz Waseem ( Lecturer/Lab Engr., CED) Columns
6. SYMMETRICAL SECTIONS:
Those sections in which plastic centroid coincide with the geometrical centroid
are called symmetrical sections.
7. UNSYMMETRICAL SECTIONS:
Those sections in which plastic centroid do not coincide with the geometrical
centroid are called unsymmetrical sections. Plastic centroid of unsymmetrical sections
can be found out in the following manner;
Fig. Section unsymmetrical about x-axis
R = Cc + F1 + F2 + F3
Lets suppose plastic centroid is at a distance ‘ y ’ from the bottom face. Taking moment
about bottom face,
R y = 





2
h
Cc + F1 L1+ F2 L2+ F3 L3
y =
321
332211
2
FFFC
LFLFLF
h
C
c
c
+++
+++





………………………….. (1)
Equation (1) is used to find out location of plastic centroid.
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