Lecture-3-Column-Design.pdf
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About This Presentation
Column
Slide Content
COLUMN DESIGN
Dr.IzniSyahrizalbin Ibrahim
Faculty of Civil Engineering
Universiti Teknologi Malaysia
Email: [email protected]
Dr.IzniSyahrizalbin Ibrahim
Faculty of Civil Engineering
Universiti Teknologi Malaysia
Email: [email protected]
Introduction
•Column: Subjected to axial compressive forces
•Carries load from beams and slabs down to the foundation.
•May also resist bending momentdue to continuity of structure
and loading eccentricity
•EC2 Clause 5.3.1(7):
a)Compression member where the greater cross sectional
dimension does not exceed 4 times the smaller dimension
(h4b)
b)Height is at least 3 times the section depth
•Column: Subjected to axial compressive forces
•Carries load from beams and slabs down to the foundation.
•May also resist bending momentdue to continuity of structure
and loading eccentricity
•EC2 Clause 5.3.1(7):
a)Compression member where the greater cross sectional
dimension does not exceed 4 times the smaller dimension
(h4b)
b)Height is at least 3 times the section depth
Classification of Columns
Column Classification
Braced
Slender
Non-slender
Unbraced
Slender
Non-slender
Lateral stability to the structure
as a whole is provided by walls or
bracing –resist all lateral forces
Lateral loads are resisted
by the bending actionof
the column
Column Classification
Braced
Slender
Non-slender
Unbraced
Slender
Non-slender
Lateral stability to the structure
as a whole is provided by walls or
bracing –resist all lateral forces
Lateral loads are resisted
by the bending actionof
the column
Classification of Columns
•Slender or Non-slender column depending on the
sensitivity to second order effect (P-effect)
•Use slenderness ratio, to measure column
vulnerability by elastic instability or buckling
•Non-Slender:
a)Design action are not significantly affected by
deformation (P-effect is small)
b)P-effect can be ignored if does not exceed a
particular value
c)P-effect can be ignored if 10% of the
corresponding first order moments
•Slender or Non-slender column depending on the
sensitivity to second order effect (P-effect)
•Use slenderness ratio, to measure column
vulnerability by elastic instability or buckling
•Non-Slender:
a)Design action are not significantly affected by
deformation (P-effect is small)
b)P-effect can be ignored if does not exceed a
particular value
c)P-effect can be ignored if 10% of the
corresponding first order moments
Classification of Columns
•Short column –, crushing at ultimate strength
•Slender column–, buckling under low compressive
load
Compression
failure
Buckling
failure
•Short column –, crushing at ultimate strength
•Slender column–, buckling under low compressive
load
Compression
failure
Buckling
failure
Classification of Columns
Major axis
(x-x)
Minor axis
(y-y)
Plane of
bending
Clear height, l
Actual height
Major axis
(x-x)
Minor axis
(y-y)
Plane of
bending
Clear height, l
Actual height
Slenderness Ratio
??????=
�
�
�
=
�
�
??????
�
l
o= Effective length of the column
i= Radius of gyration and the axis consider
I= Section moment of area of the section about the axis
A= Cross sectional area of the column
??????=
�
�
�
=
�
�
??????
�
l
o= Effective length of the column
i= Radius of gyration and the axis consider
I= Section moment of area of the section about the axis
A= Cross sectional area of the column
Effective Length of Column
a) l
o= lb) l
o= 2lc) l
o= 0.7ld) l
o= 0.5le) l
o= l g) l
o2lf) 0.5 l
ol
For constant cross section
a) l
o= lb) l
o= 2lc) l
o= 0.7ld) l
o= 0.5le) l
o= l g) l
o2lf) 0.5 l
ol
For constant cross section
Effective Length of Column
Braced Column:
�
�=�.���+
�
�
�.��+�
�
�+
�
�
�.��+�
�
Unbraced Column:
�
�=�.����+��
�
��
�
�
�+�
�
;�+
�
�
�+�
�
�+
�
�
�+�
�
k
1, k
2Relative flexibilities of rotational restraints at ends 1 & 2, respectively =
�
�
????????????
�
Rotation of restraining members for bending moment, M
EI Bending stiffness of compression member
l Clear height of compression member between end restraints at each end
k
1, k
2=
��������??????������,??????
���
������??????�����,??????
����
=
????????????
����
2
????????????
�����
EC2: Clause 5.8.3.2(3)
Braced Column:
�
�=�.���+
�
�
�.��+�
�
�+
�
�
�.��+�
�
Unbraced Column:
�
�=�.����+��
�
��
�
�
�+�
�
;�+
�
�
�+�
�
�+
�
�
�+�
�
k
1, k
2Relative flexibilities of rotational restraints at ends 1 & 2, respectively =
�
�
????????????
�
Rotation of restraining members for bending moment, M
EI Bending stiffness of compression member
l Clear height of compression member between end restraints at each end
k
1, k
2=
��������??????������,??????
���
������??????�����,??????
����
=
????????????
����
2
????????????
�����
EC2: Clause 5.8.3.2(3)
Effective Length of Column
Table 3.19 & 3.20, BS 8110: Part 1: 1997
End Condition at Top
EndCondition at Bottom
1 2 3
Braced Column
1 0.75 0.80 0.90
2 0.80 0.85 0.95
3 0.90 0.95 1.00
Unbraced Column
1 1.2 1.3 1.6
2 1.3 1.5 1.8
3 1.6 1.8 -
4 2.2 - -
Simplified Method
Table 3.19 & 3.20, BS 8110: Part 1: 1997
End Condition at Top
EndCondition at Bottom
1 2 3
Braced Column
1 0.75 0.80 0.90
2 0.80 0.85 0.95
3 0.90 0.95 1.00
Unbraced Column
1 1.2 1.3 1.6
2 1.3 1.5 1.8
3 1.6 1.8 -
4 2.2 - -
Simplified Method
Effective Length of Column
End Condition at Top
EndCondition at Bottom
1 2 3
Braced Column
1 0.75 0.80 0.90
2 0.80 0.85 0.95
3 0.90 0.95 1.00
Unbraced Column
1 1.2 1.3 1.6
2 1.3 1.5 1.8
3 1.6 1.8 -
4 2.2 - -
Condition 1Column connected monolithically to beams on either side which are
at least as deep as the overall dimension of the column in the plane
considered. Where column connected to a foundation this should
be designed to carry moment
Condition 2Column connected monolithically to beams or slabs on either side
which are shallower than the overall dimension of the column in
the plane considered
Condition 3Column connected to members that do not provide more than
nominal restraint to rotation
Condition 4End of column is unrestrained against both lateral movement and
rotation
Table 3.19 & 3.20, BS 8110: Part 1: 1997
End Condition at Top
EndCondition at Bottom
1 2 3
Braced Column
1 0.75 0.80 0.90
2 0.80 0.85 0.95
3 0.90 0.95 1.00
Unbraced Column
1 1.2 1.3 1.6
2 1.3 1.5 1.8
3 1.6 1.8 -
4 2.2 - -
Condition 1Column connected monolithically to beams on either side which are
at least as deep as the overall dimension of the column in the plane
considered. Where column connected to a foundation this should
be designed to carry moment
Condition 2Column connected monolithically to beams or slabs on either side
which are shallower than the overall dimension of the column in
the plane considered
Condition 3Column connected to members that do not provide more than
nominal restraint to rotation
Condition 4End of column is unrestrained against both lateral movement and
rotation
Table 3.19 & 3.20, BS 8110: Part 1: 1997
Limiting Slenderness Ratio
??????
���=
��∙�∙�∙�
�
A =
1
1+0.2??????
���
: ??????
���= Effective creep ratio
B = 1+2?????? : ??????=
�
��
��
�
��
��
C = 1.7−??????
� : ??????
�=
�
�1
�
�2
n =
�
??????�
�
��
��
N
Ed= Design ultimate axial column load
M
o1, M
o2= First order moments at the end of the column with �
�2≥�
�1
f
yd= Design yield strength of reinforcement
f
cd= Design compressive strength of concrete
If ??????
���, & r
mis not known, A= 0.7, B= 1.1 & C= 0.7 may be used
EC2: Clause 5.8.3.1
??????
���=
��∙�∙�∙�
�
A =
1
1+0.2??????
���
: ??????
���= Effective creep ratio
B = 1+2?????? : ??????=
�
��
��
�
��
��
C = 1.7−??????
� : ??????
�=
�
�1
�
�2
n =
�
??????�
�
��
��
N
Ed= Design ultimate axial column load
M
o1, M
o2= First order moments at the end of the column with �
�2≥�
�1
f
yd= Design yield strength of reinforcement
f
cd= Design compressive strength of concrete
If ??????
���, & r
mis not known, A= 0.7, B= 1.1 & C= 0.7 may be used
EC2: Clause 5.8.3.1
Limiting Slenderness Ratio
Condition apply for C:
(1)If the end moments, M
o1& M
o2give rise to tension on the same
side of the column, then r
mshould be taken +ve(follows C1.7)
(2)If the column is in a state of double curvature, then r
mshould be
taken –ve(follows C1.7)
(3)For braced members in which the first order moment arise only
from or predominantly due to imperfections or transverse
loading, then r
mshould be taken as 1.0 (C= 0.7)
(4)For unbraced member, in general r
mshould be taken as 1.0 (C=
0.7)
If
lim: Short (Non-slender) column
If
lim: Slender column.Second order effects must be considered
in design
EC2: Clause 5.8.3.1
Condition apply for C:
(1)If the end moments, M
o1& M
o2give rise to tension on the same
side of the column, then r
mshould be taken +ve(follows C1.7)
(2)If the column is in a state of double curvature, then r
mshould be
taken –ve(follows C1.7)
(3)For braced members in which the first order moment arise only
from or predominantly due to imperfections or transverse
loading, then r
mshould be taken as 1.0 (C= 0.7)
(4)For unbraced member, in general r
mshould be taken as 1.0 (C=
0.7)
If
lim: Short (Non-slender) column
If
lim: Slender column.Second order effects must be considered
in design
EC2: Clause 5.8.3.1
Example 1
SLENDERNESS
SLENDERNESS
Example 1: Slenderness
4 m
4 m
4 m
250 500
250 500
250 500
250 500
250
400
250
400
250
400
275 350
A B C
1
2
3
4
•Bracedcolumn
•Axial load = 1050 kN
•Bending moment (major axis)= 40 kNm(top) & 12 kNm(bottom)
•Bending moment (minor axis)= 15 kNm(top) & 10 kNm(bottom)
•Concrete grade= C25
•Steel= 500 N/mm
2
l
1= 6 m l
2= 8 m
4 m
4 m
4 m
250 500
250 500
250 500
250 500
250
400
250
400
250
400
275 350
A B C
1
2
3
4
•Bracedcolumn
•Axial load = 1050 kN
•Bending moment (major axis)= 40 kNm(top) & 12 kNm(bottom)
•Bending moment (minor axis)= 15 kNm(top) & 10 kNm(bottom)
•Concrete grade= C25
•Steel= 500 N/mm
2
l
1= 6 m l
2= 8 m
Example 1: Slenderness
l
1= 6 m l
2= 8 m
4 m
5m
1.5 m
250 500
250 500
250 500
275 350
A B C
Roof
1
st
Floor
Ground
l
1= 6 m l
2= 8 m
4 m
5m
1.5 m
250 500
250 500
250 500
275 350
A B C
Roof
1
st
Floor
Ground
Example 1: Slenderness
EC2: Clause 5.8.3.2(3) Method
z
z
y y
Secondary beam
250 400
L= 4 m 5m
Main beam
250 500
L= 6 m
Main beam
250 500
L= 8 m
Secondary beam
250 400
L= 4 m
-12 kNm
40 kNm
N
Ed= 1050 kN
M
z
15 kNm
-10 kNm
M
y
275 mm
M
z
350 mm
M
y
Moment & Axial Force
EC2: Clause 5.8.3.2(3) Method
z
z
y y
Secondary beam
250 400
L= 4 m 5m
Main beam
250 500
L= 6 m
Main beam
250 500
L= 8 m
Secondary beam
250 400
L= 4 m
-12 kNm
40 kNm
N
Ed= 1050 kN
M
z
15 kNm
-10 kNm
M
y
275 mm
M
z
350 mm
M
y
Moment & Axial Force
Example 1: Slenderness
Dimension & Size
Column:
bh= 275 300 mm
Actual length:l
z= 5000 –500 = 4500 mm
l
y= 5000 –400 = 4600 mm
Beam:
Main beam, bh= 250 500 mm
Actual length:l
1= 6000 mm
l
2= 8000 mm
Secondary beam, bh= 250 400 mm
Actual length:l
1= l
2= 4000 mm
Dimension & Size
Column:
bh= 275 300 mm
Actual length:l
z= 5000 –500 = 4500 mm
l
y= 5000 –400 = 4600 mm
Beam:
Main beam, bh= 250 500 mm
Actual length:l
1= 6000 mm
l
2= 8000 mm
Secondary beam, bh= 250 400 mm
Actual length:l
1= l
2= 4000 mm
Example 1: Slenderness
Moment of Inertia, I= bh
3
/12
Column:
??????
��=
275×300
3
12
=0.98×10
9
mm
4
??????
��=
300×275
3
12
=0.61×10
9
mm
4
Beam:
Main beam, ??????
��=
250×500
3
12
=2.60×10
9
mm
4
Secondary beam, ??????
��=
250×400
3
12
=1.33×10
9
mm
4
Moment of Inertia, I= bh
3
/12
Column:
??????
��=
275×300
3
12
=0.98×10
9
mm
4
??????
��=
300×275
3
12
=0.61×10
9
mm
4
Beam:
Main beam, ??????
��=
250×500
3
12
=2.60×10
9
mm
4
Secondary beam, ??????
��=
250×400
3
12
=1.33×10
9
mm
4
Example 1: Slenderness
Stiffness, K= EI/l
Column:
�
��=
0.98×10
9
4500
=2.18×10
5
mm
3
�
��=
0.61×10
9
4600
=1.32×10
5
mm
3
Beam:
Main beam �
��1=
2.60×10
9
6000
=4.34×10
5
mm
3
�
��2=
2.60×10
9
8000
=3.26×10
5
mm
3
Secondary beam�
��1=�
��2=
1.33×10
9
4000
=3.33×10
5
mm
3
Stiffness, K= EI/l
Column:
�
��=
0.98×10
9
4500
=2.18×10
5
mm
3
�
��=
0.61×10
9
4600
=1.32×10
5
mm
3
Beam:
Main beam �
��1=
2.60×10
9
6000
=4.34×10
5
mm
3
�
��2=
2.60×10
9
8000
=3.26×10
5
mm
3
Secondary beam�
��1=�
��2=
1.33×10
9
4000
=3.33×10
5
mm
3
Example 1: Slenderness
Relative Column Stiffness, k= K
col/2(K
beam)
z-axis:
Top end: �
2=
2.18×10
5
24.34×10
5
+3.26×10
5
=0.14>0.1 k
2= 0.14
Bottom end:�
1=
2.18×10
5
24.34×10
5
+3.26×10
5
=0.14>0.1 k
1= 0.14
y-axis:
Top end: �
2=
1.32×10
5
23.33×10
5
+3.33×10
5
=0.10<0.1 k
2= 0.10
Bottom end:�
1=
1.32×10
5
23.33×10
5
+3.33×10
5
=0.10<0.1 k
1= 0.10
Relative Column Stiffness, k= K
col/2(K
beam)
z-axis:
Top end: �
2=
2.18×10
5
24.34×10
5
+3.26×10
5
=0.14>0.1 k
2= 0.14
Bottom end:�
1=
2.18×10
5
24.34×10
5
+3.26×10
5
=0.14>0.1 k
1= 0.14
y-axis:
Top end: �
2=
1.32×10
5
23.33×10
5
+3.33×10
5
=0.10<0.1 k
2= 0.10
Bottom end:�
1=
1.32×10
5
23.33×10
5
+3.33×10
5
=0.10<0.1 k
1= 0.10
Example 1: Slenderness
Effective Length of Column
�
�=0.5�1+
�
1
0.45+�
1
1+
�
2
0.45+�
2
�
�,�=0.5�
�1+
0.14
0.45+0.14
1+
0.14
0.45+0.14
=2795mm
�
�,�=0.5�
�1+
0.10
0.45+0.10
1+
0.10
0.45+0.10
=2718mm
Effective Length of Column
�
�=0.5�1+
�
1
0.45+�
1
1+
�
2
0.45+�
2
�
�,�=0.5�
�1+
0.14
0.45+0.14
1+
0.14
0.45+0.14
=2795mm
�
�,�=0.5�
�1+
0.10
0.45+0.10
1+
0.10
0.45+0.10
=2718mm
Example 1: Slenderness
Radius of Gyration, �=
??????
�
??????
��=
??????
��
??????
=
0.98×10
9
275×350
=101
??????
��=
??????
��
??????
=
0.61×10
9
275×350
=79.4
Slenderness Ratio, = l
o/i
??????
��=
2795
101
=��.�
??????
��=
2718
79.4
=��.�
Radius of Gyration, �=
??????
�
??????
��=
??????
��
??????
=
0.98×10
9
275×350
=101
??????
��=
??????
��
??????
=
0.61×10
9
275×350
=79.4
Slenderness Ratio, = l
o/i
??????
��=
2795
101
=��.�
??????
��=
2718
79.4
=��.�
Example 1: Slenderness
Slenderness Limit, ??????
���=
��∙�∙�∙�
�
A= 0.7 (
effNOT known)
B= 1.1 (NOT known)
C= 1.7 –r
m(where r
m= (M
o1/M
o2)
z-axis:??????
�,�=
−12
40
=−0.30
C
z= 1.7 –(0.30) = 2.00
y-axis:??????
�,�=
−10
15
=−0.67
C
y= 1.7 –(0.67) = 2.37
�=
�
??????�
??????
��
��
�
��=
0.85�
��
??????
�
=
0.85×25
1.5
=14.17N/mm
2
�=
1050×10
3
275×350×14.17
=0.77
??????
�??????�,��=
20×0.7×1.1×2.00
0.77
=35.1>??????
��=27.7 Non-slender about z-axis
??????
�??????�,��=
20×0.7×1.1×2.37
0.77
=41.5>??????
��=34.2 Non-slender about y-axis
Slenderness Limit, ??????
���=
��∙�∙�∙�
�
A= 0.7 (
effNOT known)
B= 1.1 (NOT known)
C= 1.7 –r
m(where r
m= (M
o1/M
o2)
z-axis:??????
�,�=
−12
40
=−0.30
C
z= 1.7 –(0.30) = 2.00
y-axis:??????
�,�=
−10
15
=−0.67
C
y= 1.7 –(0.67) = 2.37
�=
�
??????�
??????
��
��
�
��=
0.85�
��
??????
�
=
0.85×25
1.5
=14.17N/mm
2
�=
1050×10
3
275×350×14.17
=0.77
??????
�??????�,��=
20×0.7×1.1×2.00
0.77
=35.1>??????
��=27.7 Non-slender about z-axis
??????
�??????�,��=
20×0.7×1.1×2.37
0.77
=41.5>??????
��=34.2 Non-slender about y-axis
Example 1: Slenderness
Effective Length, l
o= Factor Clear Height
z-axis:End condition: Top & Bottom = Condition 1
Factor = 0.75
�
�,�=0.75×4500=3375mm
y-axis:End condition: Top & Bottom = Condition 1
Factor = 0.75
�
�,�=0.75×4600=3450mm
Slenderness Ratio, = l
o/i
??????
��=
3375
101
=33.4<??????
�??????�,��=35.1 Non-slender about z-axis
??????
��=
3450
79.4
=43.5>??????
�??????�,��=41.5 Slender about y-axis
Simplified Method
Effective Length, l
o= Factor Clear Height
z-axis:End condition: Top & Bottom = Condition 1
Factor = 0.75
�
�,�=0.75×4500=3375mm
y-axis:End condition: Top & Bottom = Condition 1
Factor = 0.75
�
�,�=0.75×4600=3450mm
Slenderness Ratio, = l
o/i
??????
��=
3375
101
=33.4<??????
�??????�,��=35.1 Non-slender about z-axis
??????
��=
3450
79.4
=43.5>??????
�??????�,��=41.5 Slender about y-axis
Simplified Method
Axial Load & Moment in Column
For analysis without full frame analysis:
a)Axial loads may generally be obtained by increasing the loads
obtained by 10%by assumption that beams & slabs are simply
supported. Higher percentagemay be required when adjacent
spans and/or loadings on them are grossly dissimilar.
b)Bending moments may be calculated using the simplified one
free-joint sub-frame. The arrangement of the design ultimate
variable action should be such as that to cause maximum
moment in the column.
For analysis without full frame analysis:
a)Axial loads may generally be obtained by increasing the loads
obtained by 10%by assumption that beams & slabs are simply
supported. Higher percentagemay be required when adjacent
spans and/or loadings on them are grossly dissimilar.
b)Bending moments may be calculated using the simplified one
free-joint sub-frame. The arrangement of the design ultimate
variable action should be such as that to cause maximum
moment in the column.
Example 2
AXIAL LOAD & MOMENT IN
COLUMN
AXIAL LOAD & MOMENT IN
COLUMN
Example 2: Axial Load & Moment in
Column
Level Beam
Characteristics Action (kN/m)
Permanent Variable
Roof 2/A-C 20.0 6.0
B/1-4 8.0 3.0
FirstFloor 2/A-C 27.0 12.0
B/1-4 15.0 7.0
GroundFloor 2/A-C 7.8 -
B/1-4 7.8 -
Determine axial force and bending moment in column B/2
Column
Level Beam
Characteristics Action (kN/m)
Permanent Variable
Roof 2/A-C 20.0 6.0
B/1-4 8.0 3.0
FirstFloor 2/A-C 27.0 12.0
B/1-4 15.0 7.0
GroundFloor 2/A-C 7.8 -
B/1-4 7.8 -
Determine axial force and bending moment in column B/2
Example 2: Axial Load & Moment in
Column
Column: 275 350 mm
Main Beam: 250 500 mm
�
1�=�
2�=4�
Secondary Beam:250 400 mm
�
1�=8�
�
2�=6�
z
z
y y
x
x
4 m
2/A-C
B/1-4
5m
1.5 m
w
1Z w
2Z
w
2y
w
1y
Roof
1
st
Level
Ground
Column
Column: 275 350 mm
Main Beam: 250 500 mm
�
1�=�
2�=4�
Secondary Beam:250 400 mm
�
1�=8�
�
2�=6�
z
z
y y
x
x
4 m
2/A-C
B/1-4
5m
1.5 m
w
1Z w
2Z
w
2y
w
1y
Roof
1
st
Level
Ground
Example 2: Axial Load & Moment in
Column
Roof w
1z w
2z w
1y w
2y
G
k 20 20 8 8
Q
k 6 6 3 3
Max 36 36 15.3 15.3
Min 27 27 10.8 10.8
1
st
Floor w
1z w
2z w
1y w
2y
G
k 27 27 15 15
Q
k 12 12 7 7
Max 54.5 54.5 30.8 30.8
Min 36.45 36.45 20.25 20.25
Ground w
1z w
2z w
1y W
2y
G
k 7.8 7.8 7.8 7.8
Q
k 0 0 0 0
Max 10.5 10.5 10.5 10.5
Min 10.5 10.5 10.5 10.5
Column
Roof w
1z w
2z w
1y w
2y
G
k 20 20 8 8
Q
k 6 6 3 3
Max 36 36 15.3 15.3
Min 27 27 10.8 10.8
1
st
Floor w
1z w
2z w
1y w
2y
G
k 27 27 15 15
Q
k 12 12 7 7
Max 54.5 54.5 30.8 30.8
Min 36.45 36.45 20.25 20.25
Ground w
1z w
2z w
1y W
2y
G
k 7.8 7.8 7.8 7.8
Q
k 0 0 0 0
Max 10.5 10.5 10.5 10.5
Min 10.5 10.5 10.5 10.5
Example 2: Axial Load & Moment in
Column
Roof to 1
st
Level:
Main Beam: (36 8 0.5) + (36 6 0.5) 252 kN
Sec. Beam: (15.3 4 0.5) + (15.3 4 0.5) 61 kN
Selfweight: 1.35 (0.275 0.35 4 25) 13 kN
N
1st-Roof: 326 kN
1
st
Level to Ground:
Load from above: 326 kN
Main Beam: (54.5 8 0.5) + (54.5 6 0.5) 381 kN
Sec. Beam: (30.8 4 0.5) + (30.8 4 0.5) 123 kN
Selfweight: 1.35 (0.275 0.35 5 25) 16 kN
N
1st-Gnd 847 kN
Ground to Footing:
Load from above: 847 kN
Main Beam: (10.5 8 0.5) + (10.5 6 0.5) 74 kN
Sec. Beam: (10.5 4 0.5) + (10.5 4 0.5) 42 kN
Selfweight: 1.35 (0.275 0.35 1.5 25) 5 kN
N
Gnd-Foot: 967 kN
Axial Loads
Column
Roof to 1
st
Level:
Main Beam: (36 8 0.5) + (36 6 0.5) 252 kN
Sec. Beam: (15.3 4 0.5) + (15.3 4 0.5) 61 kN
Selfweight: 1.35 (0.275 0.35 4 25) 13 kN
N
1st-Roof: 326 kN
1
st
Level to Ground:
Load from above: 326 kN
Main Beam: (54.5 8 0.5) + (54.5 6 0.5) 381 kN
Sec. Beam: (30.8 4 0.5) + (30.8 4 0.5) 123 kN
Selfweight: 1.35 (0.275 0.35 5 25) 16 kN
N
1st-Gnd 847 kN
Ground to Footing:
Load from above: 847 kN
Main Beam: (10.5 8 0.5) + (10.5 6 0.5) 74 kN
Sec. Beam: (10.5 4 0.5) + (10.5 4 0.5) 42 kN
Selfweight: 1.35 (0.275 0.35 1.5 25) 5 kN
N
Gnd-Foot: 967 kN
Axial Loads
Example 2: Axial Load & Moment in
Column
Bending Moments at z-zAxis
Stiffness, K= I/L
Beam:6�:�
�1=
250×500
3
12×6000
=4.34×10
5
��
3
8�:�
�2=
250×500
3
12×8000
=3.26×10
5
��
3
Column:4�:
275×350
3
12×4000
=2.46×10
5
��
3
5�:
275×350
3
12×5000
=1.97×10
5
��
3
1.5�:
275×350
3
12×1500
=6.55×10
5
��
3
Column
Bending Moments at z-zAxis
Stiffness, K= I/L
Beam:6�:�
�1=
250×500
3
12×6000
=4.34×10
5
��
3
8�:�
�2=
250×500
3
12×8000
=3.26×10
5
��
3
Column:4�:
275×350
3
12×4000
=2.46×10
5
��
3
5�:
275×350
3
12×5000
=1.97×10
5
��
3
1.5�:
275×350
3
12×1500
=6.55×10
5
��
3
Example 2: Axial Load & Moment in
Column
Fixed End Moment, FEM
�
1=
��
2
12
=
27×6
2
12
=81.0���
�
2=
��
2
12
=
36×8
2
12
=192.0���
∆���=�
2−�
1=������6 m 8 m
4m
w
min= 27.0 kN/m
w
max= 36.0 kN/m
K
c
0.5K
b1 0.5K
b2
Roof to 1
st
Level
Moment in Column
�=∆���×
??????�
??????�+0.5??????
�1+0.5??????
�2
=111×
2.46
2.46+0.5×4.34+0.5×3.26
=��.�kNm
M
1M
2
Column
Fixed End Moment, FEM
�
1=
��
2
12
=
27×6
2
12
=81.0���
�
2=
��
2
12
=
36×8
2
12
=192.0���
∆���=�
2−�
1=������6 m 8 m
4m
w
min= 27.0 kN/m
w
max= 36.0 kN/m
K
c
0.5K
b1 0.5K
b2
Roof to 1
st
Level
Moment in Column
�=∆���×
??????�
??????�+0.5??????
�1+0.5??????
�2
=111×
2.46
2.46+0.5×4.34+0.5×3.26
=��.�kNm
M
1M
2
Example 2: Axial Load & Moment in
Column
Fixed End Moment, FEM
�
1=
��
2
12
=
36.5×6
2
12
=109.4���
�
2=
��
2
12
=
54.5×8
2
12
=290.4���
∆���=�
2−�
1=������
1
st
Level to Ground
Moment in upper column, M
upper
�
�����=∆���×
??????�,�
??????�,�+??????
�,�+0.5??????
�1+0.5??????
�2
=181×
2.46
2.46+1.97+0.5×4.34+0.5×3.26
=��.�kNm
Moment in lower column, M
lower
�
�����=∆���×
??????�,�
??????�,�+??????
�,�+0.5??????
�1+0.5??????
�2
=181×
1.97
2.46+1.97+0.5×4.34+0.5×3.26
=��.�kNm
M
1M
2
6 m 8 m
4m
5 m
w
min= 36.5 kN/m
w
max= 54.5 kN/m
K
c,l
K
c,u
0.5K
b1 0.5K
b2
Column
Fixed End Moment, FEM
�
1=
��
2
12
=
36.5×6
2
12
=109.4���
�
2=
��
2
12
=
54.5×8
2
12
=290.4���
∆���=�
2−�
1=������
1
st
Level to Ground
Moment in upper column, M
upper
�
�����=∆���×
??????�,�
??????�,�+??????
�,�+0.5??????
�1+0.5??????
�2
=181×
2.46
2.46+1.97+0.5×4.34+0.5×3.26
=��.�kNm
Moment in lower column, M
lower
�
�����=∆���×
??????�,�
??????�,�+??????
�,�+0.5??????
�1+0.5??????
�2
=181×
1.97
2.46+1.97+0.5×4.34+0.5×3.26
=��.�kNm
M
1M
2
6 m 8 m
4m
5 m
w
min= 36.5 kN/m
w
max= 54.5 kN/m
K
c,l
K
c,u
0.5K
b1 0.5K
b2
Example 2: Axial Load & Moment in
Column
Fixed End Moment, FEM
�
1=
��
2
12
=
10.5×6
2
12
=31.6���
�
2=
��
2
12
=
10.5×8
2
12
=56.2���
∆���=�
2−�
1=��.����
Ground to Footing
Moment in upper column, M
upper
�
�����=∆���×
??????�,�
??????�,�+??????
�,�+0.5??????
�1+0.5??????
�2
=24.6×
1.97
1.97+6.55+0.5×4.34+0.5×3.26
=�.�kNm
Moment in lower column, M
lower
�
�����=∆���×
??????�,�
??????�,�+??????
�,�+0.5??????
�1+0.5??????
�2
=24.6×
6.55
1.97+6.55+0.5×4.34+0.5×3.26
=��.�kNm
M
1M
2
6 m 8 m
5 m
1.5 m
w
max= 10.5 kN/m w
max= 10.5 kN/m
K
c,l
K
c,u
0.5K
b1 0.5K
b2
Column
Fixed End Moment, FEM
�
1=
��
2
12
=
10.5×6
2
12
=31.6���
�
2=
��
2
12
=
10.5×8
2
12
=56.2���
∆���=�
2−�
1=��.����
Ground to Footing
Moment in upper column, M
upper
�
�����=∆���×
??????�,�
??????�,�+??????
�,�+0.5??????
�1+0.5??????
�2
=24.6×
1.97
1.97+6.55+0.5×4.34+0.5×3.26
=�.�kNm
Moment in lower column, M
lower
�
�����=∆���×
??????�,�
??????�,�+??????
�,�+0.5??????
�1+0.5??????
�2
=24.6×
6.55
1.97+6.55+0.5×4.34+0.5×3.26
=��.�kNm
M
1M
2
6 m 8 m
5 m
1.5 m
w
max= 10.5 kN/m w
max= 10.5 kN/m
K
c,l
K
c,u
0.5K
b1 0.5K
b2
Example 2: Axial Load & Moment in
Column
Bending Moments at z-zAxis, M
z-z(kNm)
43.6
54.1 43.3
3.9 13.1
6.5
Column
Bending Moments at z-zAxis, M
z-z(kNm)
43.6
54.1 43.3
3.9 13.1
6.5
Example 2: Axial Load & Moment in
Column
Bending Moments at y-yAxis
Stiffness, K= I/L
Beam:4�:�
�1=
250×400
3
12×4000
=3.33×10
5
��
3
4�:�
�2=
250×400
3
12×4000
=3.33×10
5
��
3
Column:4�:
350×275
3
12×4000
=1.52×10
5
��
3
5�:
350×275
3
12×5000
=1.21×10
5
��
3
1.5�:
350×275
3
12×1500
=4.04×10
5
��
3
Column
Bending Moments at y-yAxis
Stiffness, K= I/L
Beam:4�:�
�1=
250×400
3
12×4000
=3.33×10
5
��
3
4�:�
�2=
250×400
3
12×4000
=3.33×10
5
��
3
Column:4�:
350×275
3
12×4000
=1.52×10
5
��
3
5�:
350×275
3
12×5000
=1.21×10
5
��
3
1.5�:
350×275
3
12×1500
=4.04×10
5
��
3
Example 2: Axial Load & Moment in
Column
Fixed End Moment, FEM
�
1=
��
2
12
=
10.8×4
2
12
=14.4���
�
2=
��
2
12
=
15.3×4
2
12
=20.4���
∆���=�
2−�
1=����4 m 4m
4m
w
max= 10.8 kN/m
w
max= 15.3 kN/m
K
c
0.5K
b1 0.5K
b2
Roof to 1
st
Level
Moment in Column
�=∆���×
??????�
??????�+0.5??????
�1+0.5??????
�2
=111×
1.52
1.52+0.5×3.33+0.5×3.33
=�.�kNm
M
1M
2
Column
Fixed End Moment, FEM
�
1=
��
2
12
=
10.8×4
2
12
=14.4���
�
2=
��
2
12
=
15.3×4
2
12
=20.4���
∆���=�
2−�
1=����4 m 4m
4m
w
max= 10.8 kN/m
w
max= 15.3 kN/m
K
c
0.5K
b1 0.5K
b2
Roof to 1
st
Level
Moment in Column
�=∆���×
??????�
??????�+0.5??????
�1+0.5??????
�2
=111×
1.52
1.52+0.5×3.33+0.5×3.33
=�.�kNm
M
1M
2
Example 2: Axial Load & Moment in
Column
Fixed End Moment, FEM
�
1=
��
2
12
=
20.3×4
2
12
=27���
�
2=
��
2
12
=
30.8×4
2
12
=41���
∆���=�
2−�
1=�����
1
st
Level to Ground
Moment in upper column, M
upper
�
�����=∆���×
??????�,�
??????�,�+??????
�,�+0.5??????
�1+0.5??????
�2
=14×
1.52
1.52+1.21+0.5×3.33+0.5×3.33
=�.�kNm
Moment in lower column, M
lower
�
�����=∆���×
??????�,�
??????�,�+??????
�,�+0.5??????
�1+0.5??????
�2
=14×
1.21
1.52+1.21+0.5×3.33+0.5×3.33
=�.�kNm
M
1M
2
4 m 4 m
4m
5 m
w
min= 20.3 kN/m
w
max= 30.8 kN/m
K
c,l
K
c,u
0.5K
b1 0.5K
b2
Column
Fixed End Moment, FEM
�
1=
��
2
12
=
20.3×4
2
12
=27���
�
2=
��
2
12
=
30.8×4
2
12
=41���
∆���=�
2−�
1=�����
1
st
Level to Ground
Moment in upper column, M
upper
�
�����=∆���×
??????�,�
??????�,�+??????
�,�+0.5??????
�1+0.5??????
�2
=14×
1.52
1.52+1.21+0.5×3.33+0.5×3.33
=�.�kNm
Moment in lower column, M
lower
�
�����=∆���×
??????�,�
??????�,�+??????
�,�+0.5??????
�1+0.5??????
�2
=14×
1.21
1.52+1.21+0.5×3.33+0.5×3.33
=�.�kNm
M
1M
2
4 m 4 m
4m
5 m
w
min= 20.3 kN/m
w
max= 30.8 kN/m
K
c,l
K
c,u
0.5K
b1 0.5K
b2
Example 2: Axial Load & Moment in
Column
Fixed End Moment, FEM
�
1=
��
2
12
=
10.5×4
2
12
=14���
�
2=
��
2
12
=
10.5×4
2
12
=14���
∆���=�
2−�
1=����
Ground to Footing
Moment in upper column, M
upper
�
�����=∆���×
??????�,�
??????�,�+??????
�,�+0.5??????
�1+0.5??????
�2
=�.�kNm
Moment in lower column, M
lower
�
�����=∆���×
??????�,�
??????�,�+??????
�,�+0.5??????
�1+0.5??????
�2
=�.�kNm
M
1M
2
4 m 4 m
5 m
1.5 m
w
max= 10.5 kN/m w
max= 10.5 kN/m
K
c,l
K
c,u
0.5K
b1 0.5K
b2
Column
Fixed End Moment, FEM
�
1=
��
2
12
=
10.5×4
2
12
=14���
�
2=
��
2
12
=
10.5×4
2
12
=14���
∆���=�
2−�
1=����
Ground to Footing
Moment in upper column, M
upper
�
�����=∆���×
??????�,�
??????�,�+??????
�,�+0.5??????
�1+0.5??????
�2
=�.�kNm
Moment in lower column, M
lower
�
�����=∆���×
??????�,�
??????�,�+??????
�,�+0.5??????
�1+0.5??????
�2
=�.�kNm
M
1M
2
4 m 4 m
5 m
1.5 m
w
max= 10.5 kN/m w
max= 10.5 kN/m
K
c,l
K
c,u
0.5K
b1 0.5K
b2
Example 2: Axial Load & Moment in
Column
Bending Moments at y-yAxis, M
y-y(kNm)
1.9
3.5 2.8
0.0
Column
Bending Moments at y-yAxis, M
y-y(kNm)
1.9
3.5 2.8
0.0
Design Moments
•Elastic moment from frame action –should NOT
consider any redistribution
•Slender column:
a)Non-linear analysis to determine second order
moment
b)Nominal stiffness method (EC2: Clause 5.8.7), or
c)Nominal curvature method (EC2: Clause 5.8.8)
•Elastic moment from frame action –should NOT
consider any redistribution
•Slender column:
a)Non-linear analysis to determine second order
moment
b)Nominal stiffness method (EC2: Clause 5.8.7), or
c)Nominal curvature method (EC2: Clause 5.8.8)
Design Moments
Slender Column
Nominal Curvature Method (EC2: Clause 5.8.8)
�
??????�=�
�??????�+�
�
M
0Ed= First order moment including effect imperfection
M
Ed= Nominal second order moment
Slender Column
Nominal Curvature Method (EC2: Clause 5.8.8)
�
??????�=�
�??????�+�
�
M
0Ed= First order moment including effect imperfection
M
Ed= Nominal second order moment
Design Moments
Braced Slender Column
�
??????�=����
��,�
�??????+�
�,�
��+�.��
�,�
??????�∙�
�
M
02
M N
Ede
i
M
01
M
0E
First Order
Moment
0.5M
2
0.5M
2
M
2= N
ed e
2
Additional Second
Order Moment
+ =
M
02
M
0E+M
2
M
01+ 0.5M
2
Total Moment
Diagram
Braced Slender Column
�
??????�=����
��,�
�??????+�
�,�
��+�.��
�,�
??????�∙�
�
M
02
M N
Ede
i
M
01
M
0E
First Order
Moment
0.5M
2
0.5M
2
M
2= N
ed e
2
Additional Second
Order Moment
+ =
M
02
M
0E+M
2
M
01+ 0.5M
2
Total Moment
Diagram
Design Moments
Unbraced Slender Column
�
??????�=����
��+�
�,�
��+�
�,�
??????�∙�
�
M
02
M N
Ede
i
M
01
First Order
Moment
M
2
M
2 = N
Ede
2
Additional Second
Order Moment
+ =
M
02+ M
2
M
01+ M
2
Total Moment
Diagram
Unbraced Slender Column
�
??????�=����
��+�
�,�
��+�
�,�
??????�∙�
�
M
02
M N
Ede
i
M
01
First Order
Moment
M
2
M
2 = N
Ede
2
Additional Second
Order Moment
+ =
M
02+ M
2
M
01+ M
2
Total Moment
Diagram
Design Moments
�
01=�??????��
���,�
���+�
??????��
??????
�
02=����
���,�
���+�
??????��
??????
N
Ed= Ultimate axial load
�
�=���
ℎ
30
,20��
�
??????=
�
�
400
M
top, M
bot= Moment at top and bottom of column
�
0??????=0.6�
02+0.4�
01≥0.4�
02(M
01& M
02should have the same sign if they
produce tension on the same side, otherwise opposite sign)
M
2= Nominal second order moment = �
??????��
2
e
2= Deflection =
�
��
�
2
�
l
o= Effective length
c= Factor depending on curvature distribution, normally ??????
2
≈10
�
01=�??????��
���,�
���+�
??????��
??????
�
02=����
���,�
���+�
??????��
??????
N
Ed= Ultimate axial load
�
�=���
ℎ
30
,20��
�
??????=
�
�
400
M
top, M
bot= Moment at top and bottom of column
�
0??????=0.6�
02+0.4�
01≥0.4�
02(M
01& M
02should have the same sign if they
produce tension on the same side, otherwise opposite sign)
M
2= Nominal second order moment = �
??????��
2
e
2= Deflection =
�
��
�
2
�
l
o= Effective length
c= Factor depending on curvature distribution, normally ??????
2
≈10
Design Moments
�
�
= Curvature = �
��
??????
�
�
�
K
r= Axial load correction factor =
�
�−�
�
�−�
���
<1
�=
�
??????�
���
��
�
�=1+?????? �
���=0.4 ??????=
���
��
���
��
K
= Creep correction factor = 1+????????????
��≥1
ef= Effective creep factor =
??????��??????��
�
�??????�
=0if ??????<2,
�
�
>ℎ,�<75)
??????=0.35+
�
��
200
−
??????
150
�
??????
�
=
??????
��
0.45�
=
�
��
�
�
0.45�
EC2: Clause 5.8.8.3: Curvature
�
�
= Curvature = �
��
??????
�
�
�
K
r= Axial load correction factor =
�
�−�
�
�−�
���
<1
�=
�
??????�
���
��
�
�=1+?????? �
���=0.4 ??????=
���
��
���
��
K
= Creep correction factor = 1+????????????
��≥1
ef= Effective creep factor =
??????��??????��
�
�??????�
=0if ??????<2,
�
�
>ℎ,�<75)
??????=0.35+
�
��
200
−
??????
150
�
??????
�
=
??????
��
0.45�
=
�
��
�
�
0.45�
EC2: Clause 5.8.8.3: Curvature
Design Moments
EN 1992-1-2: Figure 3.1: Creep Correction Factor
EN 1992-1-2: Figure 3.1: Creep Correction Factor
Design Moments
EN 1992-1-2: Figure 3.1: Creep Correction Factor
EN 1992-1-2: Figure 3.1: Creep Correction Factor
Design Moments
Non-Slender Column
•Ignored second order moment effects
•Ultimate design moment, M
Ed= M
02
�
??????�=����
��,�
��,�
??????�∙�
�
M
02
M N
Ede
1
M
01
M
0E
First Order
Moment
Non-Slender Column
•Ignored second order moment effects
•Ultimate design moment, M
Ed= M
02
�
??????�=����
��,�
��,�
??????�∙�
�
M
02
M N
Ede
1
M
01
M
0E
First Order
Moment
Design of Longitudinal Reinforcement
Design Chart
Solution of Basic
Equation
Approximate
Method
•For columns having
rectangular or
circularcross
section
•Symmetrical
arrangement of
reinforcement
•Unsymmetricalarrangement of
reinforcement
•Cross section is non-rectangular
Design Chart
Solution of Basic
Equation
Approximate
Method
•For columns having
rectangular or
circularcross
section
•Symmetrical
arrangement of
reinforcement
•Unsymmetricalarrangement of
reinforcement
•Cross section is non-rectangular
Design of Longitudinal Reinforcement
For practical purpose as with BS8110, the rectangular
stress block used for design of beam may also be used
for the design of columns
F
ccF
sc
F
st
st
sc
cc
A
s
A
s’
f
cc=
ccf
ck/
c
x
z
d
2
hd
x
Section Strain Stress
Neutral Axis
For practical purpose as with BS8110, the rectangular
stress block used for design of beam may also be used
for the design of columns
F
ccF
sc
F
st
st
sc
cc
A
s
A
s’
f
cc=
ccf
ck/
c
x
z
d
2
hd
x
Section Strain Stress
Neutral Axis
Design of Longitudinal Reinforcement
•However, unlike with BS8110 the maximum compressive
strain when designing to EC2 will be less than 0.0035if
the whole section is in compression.
•This compressive strain will further fall to 0.00175 (f
ck<
50 N/mm
2
) if the section is subject to pure compression.
•This will affect the steel strains and hence forces which
the reinforcement can carry
•However, unlike with BS8110 the maximum compressive
strain when designing to EC2 will be less than 0.0035if
the whole section is in compression.
•This compressive strain will further fall to 0.00175 (f
ck<
50 N/mm
2
) if the section is subject to pure compression.
•This will affect the steel strains and hence forces which
the reinforcement can carry
Design of Longitudinal Reinforcement
EC2 strain relationship at ULS for f
ck50 N/mm
2
Stress
0.0035 max
hd
x
General
Relationship
When xh
Pure
Compression
Hinge
point
h/2
0.00175 min
x
�.������
�−
�
�0.00175
0.00175
F
ccF
sc
F
st
d
2
x
EC2 strain relationship at ULS for f
ck50 N/mm
2
Stress
0.0035 max
hd
x
General
Relationship
When xh
Pure
Compression
Hinge
point
h/2
0.00175 min
x
�.������
�−
�
�0.00175
0.00175
F
ccF
sc
F
st
d
2
x
Design of Longitudinal Reinforcement
Equilibrium Equations
Equilibrium of Load:
�=�
��+�
��−�
��
Equilibrium of Moment:
�=�
��0.5ℎ−0.5??????�+�
��0.5ℎ−�
2−�
���−0.5ℎ
where
�
��=
��
��
??????
�
??????��; �
��=
�
���
��
??????
�
;�
��=
�
���
��
??????
�
Equilibrium Equations
Equilibrium of Load:
�=�
��+�
��−�
��
Equilibrium of Moment:
�=�
��0.5ℎ−0.5??????�+�
��0.5ℎ−�
2−�
���−0.5ℎ
where
�
��=
��
��
??????
�
??????��; �
��=
�
���
��
??????
�
;�
��=
�
���
��
??????
�
Design of Longitudinal Reinforcement
Rearranged to:
�
��
=�??????;??????;�
��;??????
�;
�
�
;
�
��
��
;
�
��
��
;�
��;??????
�
�
��
�
=�??????;??????;�
��;??????
�;
�
�
;
�
��
��
;
�
��
��
;�
��;??????
�;
�
�
;
�
�
�
These equations form the basis for the N-Minteraction charts
used for the design of columns
Rearranged to:
�
��
=�??????;??????;�
��;??????
�;
�
�
;
�
��
��
;
�
��
��
;�
��;??????
�
�
��
�
=�??????;??????;�
��;??????
�;
�
�
;
�
��
��
;
�
��
��
;�
��;??????
�;
�
�
;
�
�
�
These equations form the basis for the N-Minteraction charts
used for the design of columns
Design of Longitudinal Reinforcement
Area and Number of Reinforcement
??????
�,�??????�=
0.10�
??????�
�
��
�??????0.002??????
�
??????
�,���=0.04??????
�(Outside lap location)
??????
�,���=0.08??????
�(At lap location)
EC2: Clause 9.5.2
??????
�,�??????�=
0.10�
??????�
�
��
�??????0.002??????
�
??????
�,���=0.04??????
�(Outside lap location)
??????
�,���=0.08??????
�(At lap location)
EC2: Clause 9.5.2
Requirements for Links
EC2: Clause 9.5.3
•The diameter of linksshould not be less than 6 mm or one-
quarter of the diameter of the longitudinal bar.
•The maximum spacing, S
maxshould not be less than:
a)20 times the minimum diameter of longitudinal bars
b)the lesser dimension of the column
c)400 mm
•At a distance greater than the larger dimensionof the column
above or below a beam these spacing can increased by a
factor of 1.67
EC2: Clause 9.5.3
•The diameter of linksshould not be less than 6 mm or one-
quarter of the diameter of the longitudinal bar.
•The maximum spacing, S
maxshould not be less than:
a)20 times the minimum diameter of longitudinal bars
b)the lesser dimension of the column
c)400 mm
•At a distance greater than the larger dimensionof the column
above or below a beam these spacing can increased by a
factor of 1.67
Design Procedure
Step Task Standard
1Determine design life, exposure class, fire resistance
EN 1990 Table 2.1
EN 1992-1-1: Table 4.1
EN 1992-1-2: Sec. 5.6
2Determine material strength
BS 8500-1: Table A.3
EN 206-1: Table F1
3Select size of column EN 1992-1-2: Table 5.5
4Calculate min. cover for durability , fire and bond requirementsEN 1992-1-1: Sec. 4.4.1
5Analyze structure to obtain critical moments and axial forcesEN 1992-1-1: Sec. 5
6Check slenderness EN 1992-1-1: Sec. 5.8.3
7Determine final design moment EN 1992-1-1: Sec. 5.8.8
8Determine area of reinforcement required EN 1992-1-1: Sec. 6.1
9Detailing EN 1992-1-1: Sec. 8 & 9
Step Task Standard
1Determine design life, exposure class, fire resistance
EN 1990 Table 2.1
EN 1992-1-1: Table 4.1
EN 1992-1-2: Sec. 5.6
2Determine material strength
BS 8500-1: Table A.3
EN 206-1: Table F1
3Select size of column EN 1992-1-2: Table 5.5
4Calculate min. cover for durability , fire and bond requirementsEN 1992-1-1: Sec. 4.4.1
5Analyze structure to obtain critical moments and axial forcesEN 1992-1-1: Sec. 5
6Check slenderness EN 1992-1-1: Sec. 5.8.3
7Determine final design moment EN 1992-1-1: Sec. 5.8.8
8Determine area of reinforcement required EN 1992-1-1: Sec. 6.1
9Detailing EN 1992-1-1: Sec. 8 & 9
Example 3
DESIGN OF NON-SLENDER
COLUMN BENT ABOUT
MAJOR AXIS
DESIGN OF NON-SLENDER
COLUMN BENT ABOUT
MAJOR AXIS
Classification: Braced non-slender column
Concrete, f
ck: 25 N/mm
2
Reinforcement, f
yk: 500 N/mm
2
Exposure Class: XC1
Fire Resistance: 1 hour
Design Life: 50 years
Effective Length, l
o: 4.2 m
Bar Size:
bar= 20 mm,
links= 6 mm
Example 3: Design of Non-Slender Column
Bent About Major Axis
N
Ed= 1200 kN
M
02= 35 kNm
25 kNm
M
z
M
z
z
y y250
300
Cross Section
Concrete, f
ck: 25 N/mm
2
Reinforcement, f
yk: 500 N/mm
2
Exposure Class: XC1
Fire Resistance: 1 hour
Design Life: 50 years
Effective Length, l
o: 4.2 m
Bar Size:
bar= 20 mm,
links= 6 mm
Example 3: Design of Non-Slender Column
Bent About Major Axis
N
Ed= 1200 kN
M
02= 35 kNm
25 kNm
M
z
M
z
z
y y250
300
Cross Section
Example 3: Design of Non-Slender Column
Bent About Major Axis
Cover: Durability, Bond & Fire Resistance
Minimum concrete cover regard to bond, C
min, bond= 20 mm (EN 1992-1-1: Table 4.2)
Minimum concrete cover regard to durability, C
min, dur= 15 mm (EN 1992-1-1: Table 4.4N)
Minimum required axis distance for R60 fire resistance,
a
sd= 36 mm assumed
fi= 0.50(EN 1992-1-2: Table 5.2a)
Nominal concrete cover regard to fire, C
nom, fire= a
sd-
links-
bar/2
= 36 –6 –20/2 = 20 mm
Allowance in design for deviation, C
dev= 10 mm
Nominal cover, C
nom= max{C
min, bond, C
min, dur} + C
dev= 20 + 10 = 30 mm
Use C
nom= 30 mm
Bent About Major Axis
Cover: Durability, Bond & Fire Resistance
Minimum concrete cover regard to bond, C
min, bond= 20 mm (EN 1992-1-1: Table 4.2)
Minimum concrete cover regard to durability, C
min, dur= 15 mm (EN 1992-1-1: Table 4.4N)
Minimum required axis distance for R60 fire resistance,
a
sd= 36 mm assumed
fi= 0.50(EN 1992-1-2: Table 5.2a)
Nominal concrete cover regard to fire, C
nom, fire= a
sd-
links-
bar/2
= 36 –6 –20/2 = 20 mm
Allowance in design for deviation, C
dev= 10 mm
Nominal cover, C
nom= max{C
min, bond, C
min, dur} + C
dev= 20 + 10 = 30 mm
Use C
nom= 30 mm
Example 3: Design of Non-Slender Column
Bent About Major Axis
Design Moment
For non-slender column:
�
??????�=����
��,�
��,�
??????�∙�
�
�
02=����
���,�
���+�
??????��
??????=35+1200×0.0105=��.����
����
���,�
���=35���
�
??????=
�
�
400
=
4200
400
=10.5��=0.0105�
�
??????�∙�
�=1200×0.002=�����
�
�=���
ℎ
30
,20��=���
300
30
,20��=20��=0.020�
�
??????�=����
��,�
��,�
??????�∙�
�=��.����
Bent About Major Axis
Design Moment
For non-slender column:
�
??????�=����
��,�
��,�
??????�∙�
�
�
02=����
���,�
���+�
??????��
??????=35+1200×0.0105=��.����
����
���,�
���=35���
�
??????=
�
�
400
=
4200
400
=10.5��=0.0105�
�
??????�∙�
�=1200×0.002=�����
�
�=���
ℎ
30
,20��=���
300
30
,20��=20��=0.020�
�
??????�=����
��,�
��,�
??????�∙�
�=��.����
Example 3: Design of Non-Slender Column
Bent About Major Axis
Reinforcement
�
2=�+∅
�??????��+
∅
���
2
=30+6+
20
2
=46mm
�
2
ℎ
=
46
300
=0.15
�
�ℎ�
��
=
1200×10
3
250×300×25
=0.64
�
�ℎ
2
�
��
=
47.6×10
6
250×300
2
×25
=0.08
From design chart:
���
��
�ℎ�
��
=0.35 ??????
�=
0.35�ℎ�
��
�
��
=
0.35×250×300×25
500
=����mm
2
??????
�,�??????�=
0.10�
??????�
�
��
=
0.10×1200×10
3
0.87×500
=���mm
2
or 0.002??????
�=150mm
2
??????
�,���=0.04??????
�=0.04×250×300=����mm
2
Provide 4H20 + 2H12 (A
s= 1483 mm
2
)
0.35
Bent About Major Axis
Reinforcement
�
2=�+∅
�??????��+
∅
���
2
=30+6+
20
2
=46mm
�
2
ℎ
=
46
300
=0.15
�
�ℎ�
��
=
1200×10
3
250×300×25
=0.64
�
�ℎ
2
�
��
=
47.6×10
6
250×300
2
×25
=0.08
From design chart:
���
��
�ℎ�
��
=0.35 ??????
�=
0.35�ℎ�
��
�
��
=
0.35×250×300×25
500
=����mm
2
??????
�,�??????�=
0.10�
??????�
�
��
=
0.10×1200×10
3
0.87×500
=���mm
2
or 0.002??????
�=150mm
2
??????
�,���=0.04??????
�=0.04×250×300=����mm
2
Provide 4H20 + 2H12 (A
s= 1483 mm
2
)
0.35
Example 3: Design of Non-Slender Column
Bent About Major Axis
Links
links, min= the larger of:
i.0.25 20 mm = 5 mmor
ii.6 mm
Use
links= 6 mm
S
max= the lesser of:
i.20 (12 mm) = 240 mm or
ii.the lesser dimension of the column = 250 mm or
iii.400 mm
Use S
max= 240 mm
Provide H6-240
At section 300 mm below & above beam & at lap joints, S
max= 0.6 240 mm = 144 mm
Provide H6-140
250
300
H6
-
240
H6
-
140
Bent About Major Axis
Links
links, min= the larger of:
i.0.25 20 mm = 5 mmor
ii.6 mm
Use
links= 6 mm
S
max= the lesser of:
i.20 (12 mm) = 240 mm or
ii.the lesser dimension of the column = 250 mm or
iii.400 mm
Use S
max= 240 mm
Provide H6-240
At section 300 mm below & above beam & at lap joints, S
max= 0.6 240 mm = 144 mm
Provide H6-140
250
300
H6
-
240
H6
-
140
Example 4
DESIGN OF SLENDER
COLUMN BENT ABOUT
MINOR AXIS
DESIGN OF SLENDER
COLUMN BENT ABOUT
MINOR AXIS
Classification: Braced slender column
Concrete, f
ck: 25 N/mm
2
Reinforcement, f
yk: 500 N/mm
2
Exposure Class: XC1
Fire Resistance: 1 hour
Design Life: 50 years
Effective Length, l
o: 4.13 m
Slenderness ratio, : 52
Bar Size:
bar= 20 mm,
links= 6 mm
Example 4: Design of Slender Column Bent
About Minor Axis
N
Ed= 1200 kN
35 kNm
25 kNm
M
z
M
z
z
y y250
300
Cross Section
Concrete, f
ck: 25 N/mm
2
Reinforcement, f
yk: 500 N/mm
2
Exposure Class: XC1
Fire Resistance: 1 hour
Design Life: 50 years
Effective Length, l
o: 4.13 m
Slenderness ratio, : 52
Bar Size:
bar= 20 mm,
links= 6 mm
Example 4: Design of Slender Column Bent
About Minor Axis
N
Ed= 1200 kN
35 kNm
25 kNm
M
z
M
z
z
y y250
300
Cross Section
Example 4: Design of Slender Column Bent
About Minor Axis
Cover: Durability, Bond & Fire Resistance
Minimum concrete cover regard to bond, C
min, bond= 20 mm (EN 1992-1-1: Table 4.2)
Minimum concrete cover regard to durability, C
min, dur= 15 mm (EN 1992-1-1: Table 4.4N)
Minimum required axis distance for R60 fire resistance,
a
sd= 36 mm assumed
fi= 0.50(EN 1992-1-2: Table 5.2a)
Nominal concrete cover regard to fire, C
nom, fire= a
sd-
links-
bar/2
= 36 –6 –20/2 = 20 mm
Allowance in design for deviation, C
dev= 10 mm
Nominal cover, C
nom= max{C
min, bond, C
min, dur} + C
dev= 20 + 10 = 30 mm
Use C
nom= 30 mm
About Minor Axis
Cover: Durability, Bond & Fire Resistance
Minimum concrete cover regard to bond, C
min, bond= 20 mm (EN 1992-1-1: Table 4.2)
Minimum concrete cover regard to durability, C
min, dur= 15 mm (EN 1992-1-1: Table 4.4N)
Minimum required axis distance for R60 fire resistance,
a
sd= 36 mm assumed
fi= 0.50(EN 1992-1-2: Table 5.2a)
Nominal concrete cover regard to fire, C
nom, fire= a
sd-
links-
bar/2
= 36 –6 –20/2 = 20 mm
Allowance in design for deviation, C
dev= 10 mm
Nominal cover, C
nom= max{C
min, bond, C
min, dur} + C
dev= 20 + 10 = 30 mm
Use C
nom= 30 mm
Example 4: Design of Slender Column Bent
About Minor Axis
Design Moment
For slender column:
�
??????�=����
��,�
�??????+�
�,�
��+�.��
�,�
??????�∙�
�
�
01=�??????��
���,�
���+�
??????��
??????=25+1200×0.0124=��.����
�
02=����
���,�
���+�
??????��
??????=35+1200×0.014=��.����
����
���,�
���=35���
�
??????=
�
�
400
=
4130
400
=12.4��=0.0124�
�
0??????=0.6�
02+0.4�
01≥0.4�
02
�
0??????=0.647.4+0.437.4≥0.447.4=13.5���≥19.0���
M
OE= 19.0 kNm
About Minor Axis
Design Moment
For slender column:
�
??????�=����
��,�
�??????+�
�,�
��+�.��
�,�
??????�∙�
�
�
01=�??????��
���,�
���+�
??????��
??????=25+1200×0.0124=��.����
�
02=����
���,�
���+�
??????��
??????=35+1200×0.014=��.����
����
���,�
���=35���
�
??????=
�
�
400
=
4130
400
=12.4��=0.0124�
�
0??????=0.6�
02+0.4�
01≥0.4�
02
�
0??????=0.647.4+0.437.4≥0.447.4=13.5���≥19.0���
M
OE= 19.0 kNm
Example 4: Design of Slender Column Bent
About Minor Axis
Second Order Moment, M
2
�
2=�
??????��
2=1200×0.0484=��.����
�
2=
�
���
2
�
=
2.84×10
−5
×4130
2
10
=48.4mm = 0.0484 m
l
o= 4130 mm c= 10
�
�
=�
��
??????
�
�
�
=1×1.20×2.37×10
−5
=2.84×10
−5
Effective depth to the minor axis, d= 250 –30 –6 –(20/2) = 204 mm
??????
�=
�
�−�
��−����
=�(�������)<�
�
??????=1+????????????
��≥1=1+(0.128×1.54)=1.20
??????
��=
??????��??????��
�
�??????�
=0.67×2.3=1.54
��??????��
�
�??????�
=0.67(assumed)
(, t
o)= 2.3 (Fig. 3.1: EN 1992-1-1)
??????=0.35+
�
��
200
−
??????
150
=0.35+
25
200
−
52
150
=0.128
�
��
=
�
��
??????�
0.45�
=
0.87×500
200000
0.45×204
=2.37×10
−5
About Minor Axis
Second Order Moment, M
2
�
2=�
??????��
2=1200×0.0484=��.����
�
2=
�
���
2
�
=
2.84×10
−5
×4130
2
10
=48.4mm = 0.0484 m
l
o= 4130 mm c= 10
�
�
=�
��
??????
�
�
�
=1×1.20×2.37×10
−5
=2.84×10
−5
Effective depth to the minor axis, d= 250 –30 –6 –(20/2) = 204 mm
??????
�=
�
�−�
��−����
=�(�������)<�
�
??????=1+????????????
��≥1=1+(0.128×1.54)=1.20
??????
��=
??????��??????��
�
�??????�
=0.67×2.3=1.54
��??????��
�
�??????�
=0.67(assumed)
(, t
o)= 2.3 (Fig. 3.1: EN 1992-1-1)
??????=0.35+
�
��
200
−
??????
150
=0.35+
25
200
−
52
150
=0.128
�
��
=
�
��
??????�
0.45�
=
0.87×500
200000
0.45×204
=2.37×10
−5
Class R
RH = 80%
Age = 3 days2
3
4
2.3
5
1
Example 4: Design of Slender Column Bent
About Minor Axis
RH = 80%
Age = 3 days2
3
4
2.3
5
1
Example 4: Design of Slender Column Bent
About Minor Axis
Example 4: Design of Slender Column Bent
About Minor Axis
�
??????�∙�
�=1200×0.020=�����
�
�=���
ℎ
30
,20��=���
300
30
,20��=20��=0.020�
�
??????�=����
��,�
�??????+�
�,�
��+�.��
�,�
??????�∙�
�
47.4 kNm
77.0 kNm 66.4 kNm 24 kNm
�
??????�=��.����
About Minor Axis
�
??????�∙�
�=1200×0.020=�����
�
�=���
ℎ
30
,20��=���
300
30
,20��=20��=0.020�
�
??????�=����
��,�
�??????+�
�,�
��+�.��
�,�
??????�∙�
�
47.4 kNm
77.0 kNm 66.4 kNm 24 kNm
�
??????�=��.����
Example 4: Design of Slender Column Bent
About Minor Axis
Reinforcement
�
2=�+∅
�??????��+
∅���
2
=30+6+
20
2
=46mm
�
2
�
=
46
���
=0.18≅0.20
�
�ℎ�
��
=
1200×10
3
300×250×25
=0.64
�
�ℎ
2
�
��
=
77.0×10
6
300×250
2
×25
=0.16
From design chart:
�
��
��
�ℎ�
��
=0.72 K
r= 0.58
�
0??????+�
2=19+0.58×58.1=52.6kNm
�
01+0.5�
2=37.4+0.50.58×58.1=54.2kNm
�
�ℎ
2
�
��
=
54.2×10
6
300×250
2
×25
=0.12
First
assumption,
K
r= 1.0
54.2 kNmmax
0.72
K
r= 0.58
About Minor Axis
Reinforcement
�
2=�+∅
�??????��+
∅���
2
=30+6+
20
2
=46mm
�
2
�
=
46
���
=0.18≅0.20
�
�ℎ�
��
=
1200×10
3
300×250×25
=0.64
�
�ℎ
2
�
��
=
77.0×10
6
300×250
2
×25
=0.16
From design chart:
�
��
��
�ℎ�
��
=0.72 K
r= 0.58
�
0??????+�
2=19+0.58×58.1=52.6kNm
�
01+0.5�
2=37.4+0.50.58×58.1=54.2kNm
�
�ℎ
2
�
��
=
54.2×10
6
300×250
2
×25
=0.12
First
assumption,
K
r= 1.0
54.2 kNmmax
0.72
K
r= 0.58
Example 4: Design of Slender Column Bent
About Minor Axis
Reinforcement
�
2=�+∅
�??????��+
∅���
2
=30+6+
20
2
=46mm
�
2
�
=
46
���
=0.18≅0.20
�
�ℎ�
��
=
1200×10
3
300×250×25
=0.64
�
�ℎ
2
�
��
=
77.0×10
6
300×250
2
×25
=0.16
From design chart:
�
��
��
�ℎ�
��
=0.72 K
r= 0.58
�
0??????+�
2=19+�.��×58.1=52.6kNm
�
01+0.5�
2=37.4+0.5�.��×58.1=54.2kNm
�
�ℎ
2
�
��
=
��.�×10
6
300×250
2
×25
=0.12
First
assumption,
K
r= 1.0
54.2 kNmmax
0.55
K
r= 0.58
About Minor Axis
Reinforcement
�
2=�+∅
�??????��+
∅���
2
=30+6+
20
2
=46mm
�
2
�
=
46
���
=0.18≅0.20
�
�ℎ�
��
=
1200×10
3
300×250×25
=0.64
�
�ℎ
2
�
��
=
77.0×10
6
300×250
2
×25
=0.16
From design chart:
�
��
��
�ℎ�
��
=0.72 K
r= 0.58
�
0??????+�
2=19+�.��×58.1=52.6kNm
�
01+0.5�
2=37.4+0.5�.��×58.1=54.2kNm
�
�ℎ
2
�
��
=
��.�×10
6
300×250
2
×25
=0.12
First
assumption,
K
r= 1.0
54.2 kNmmax
0.55
K
r= 0.58
Example 4: Design of Slender Column Bent
About Minor Axis
From design chart:
�����
�ℎ�
��
=0.45 K
r= 0.48
�
0??????+�
2=19+0.48×58.1=46.8kNm
�
01+0.5�
2=37.4+0.50.48×58.1=51.3kNm
�
�ℎ
2
�
��
=
51.3×10
6
300×250
2
×25
=0.11
From design chart:
�
��
��
�ℎ�
��
=0.40 K
r= 0.45
??????
�=
0.40�ℎ�
��
�
��
=
0.40×300×250×25
500
=����mm
2
??????
�,�??????�=
0.10�
??????�
�
��
=
0.10×1200×10
3
0.87×500
=���mm
2
or 0.002??????
�=150mm
2
??????
�,���=0.04??????
�=0.04×250×300=����mm
2
Provide 4H20 + 2H16 (A
s= 1659 mm
2
)
51.3 kNmmax
About Minor Axis
From design chart:
�����
�ℎ�
��
=0.45 K
r= 0.48
�
0??????+�
2=19+0.48×58.1=46.8kNm
�
01+0.5�
2=37.4+0.50.48×58.1=51.3kNm
�
�ℎ
2
�
��
=
51.3×10
6
300×250
2
×25
=0.11
From design chart:
�
��
��
�ℎ�
��
=0.40 K
r= 0.45
??????
�=
0.40�ℎ�
��
�
��
=
0.40×300×250×25
500
=����mm
2
??????
�,�??????�=
0.10�
??????�
�
��
=
0.10×1200×10
3
0.87×500
=���mm
2
or 0.002??????
�=150mm
2
??????
�,���=0.04??????
�=0.04×250×300=����mm
2
Provide 4H20 + 2H16 (A
s= 1659 mm
2
)
51.3 kNmmax
Example 4: Design of Slender Column Bent
About Minor Axis
From design chart:
�����
�ℎ�
��
=0.45 K
r= 0.48
�
0??????+�
2=19+�.��×58.1=46.8kNm
�
01+0.5�
2=37.4+0.5�.��×58.1=51.3kNm
�
�ℎ
2
�
��
=
51.3×10
6
300×250
2
×25
=0.11
From design chart:
�
��
��
�ℎ�
��
=0.40 K
r= 0.45
??????
�=
0.40�ℎ�
��
�
��
=
0.40×300×250×25
500
=����mm
2
??????
�,�??????�=
0.10�
??????�
�
��
=
0.10×1200×10
3
0.87×500
=���mm
2
or 0.002??????
�=150mm
2
??????
�,���=0.04??????
�=0.04×250×300=����mm
2
Provide 4H20 + 2H16 (A
s= 1659 mm
2
)
51.3 kNmmax
About Minor Axis
From design chart:
�����
�ℎ�
��
=0.45 K
r= 0.48
�
0??????+�
2=19+�.��×58.1=46.8kNm
�
01+0.5�
2=37.4+0.5�.��×58.1=51.3kNm
�
�ℎ
2
�
��
=
51.3×10
6
300×250
2
×25
=0.11
From design chart:
�
��
��
�ℎ�
��
=0.40 K
r= 0.45
??????
�=
0.40�ℎ�
��
�
��
=
0.40×300×250×25
500
=����mm
2
??????
�,�??????�=
0.10�
??????�
�
��
=
0.10×1200×10
3
0.87×500
=���mm
2
or 0.002??????
�=150mm
2
??????
�,���=0.04??????
�=0.04×250×300=����mm
2
Provide 4H20 + 2H16 (A
s= 1659 mm
2
)
51.3 kNmmax
Example 4: Design of Slender Column Bent
About Minor Axis
Links
links, min= the larger of:
i.0.25 16 mm = 4 mmor
ii.6 mm
Use
links= 6 mm
S
max= the lesser of:
i.20 (16 mm) = 320 mm or
ii.the lesser dimension of the column = 250 mm or
iii.400 mm
Use S
max= 250 mm
Provide H6-250
At section 300 mm below & above beam & at lap joints, S
max= 0.6 250 mm = 150 mm
Provide H6-150
250
300
H6
-
250
H6
-
150
y y
About Minor Axis
Links
links, min= the larger of:
i.0.25 16 mm = 4 mmor
ii.6 mm
Use
links= 6 mm
S
max= the lesser of:
i.20 (16 mm) = 320 mm or
ii.the lesser dimension of the column = 250 mm or
iii.400 mm
Use S
max= 250 mm
Provide H6-250
At section 300 mm below & above beam & at lap joints, S
max= 0.6 250 mm = 150 mm
Provide H6-150
250
300
H6
-
250
H6
-
150
y y
Biaxial Bending
•EC2: Clause 5.8.9 states that separate design in each principal
direction (disregarding biaxial bending) may be the first step. NO
further checkingis necessary if:
(a)
??????
�
??????
�
≤2 and
??????
�
??????
�
≤2
(b)
��
ℎ��
��
���
≤0.2or
��
���
��
ℎ��
≤0.2
b, h= Width and depth of section
�
��=??????
�12& ℎ
��=??????
�12for an equivalent rectangular section
y,
z= Slenderness ratio with respect to y-and z-axis, respectively
�
�=
�
??????�,�
�
??????�
: Eccentricity along y-axis
�
�=
�
??????�,�
�??????�
: Eccentricity along z-axis
M
ed,y= Design moment about y-axis including second order moment
M
ed,z= Design moment about z-axis including second order moment
N
Ed= Design axial load in the respective load combination
•EC2: Clause 5.8.9 states that separate design in each principal
direction (disregarding biaxial bending) may be the first step. NO
further checkingis necessary if:
(a)
??????
�
??????
�
≤2 and
??????
�
??????
�
≤2
(b)
��
ℎ��
��
���
≤0.2or
��
���
��
ℎ��
≤0.2
b, h= Width and depth of section
�
��=??????
�12& ℎ
��=??????
�12for an equivalent rectangular section
y,
z= Slenderness ratio with respect to y-and z-axis, respectively
�
�=
�
??????�,�
�
??????�
: Eccentricity along y-axis
�
�=
�
??????�,�
�??????�
: Eccentricity along z-axis
M
ed,y= Design moment about y-axis including second order moment
M
ed,z= Design moment about z-axis including second order moment
N
Ed= Design axial load in the respective load combination
Biaxial Bending
z
y
h
b
N
Ed
e
y
e
z
i
y
i
y
i
z i
z
z
y
h
b
N
Ed
e
y
e
z
i
y
i
y
i
z i
z
Biaxial Bending
•If the above conditions are NOTfulfilled, biaxiallybent columns
may be designed to satisfy the following simplified criterion:
�
??????�,�
�
??????�,�
�
+
�
??????�,�
�
??????�,�
�
≤�.�
M
Rd,y= Moment resistance in y-axis
M
Rd,z= Moment resistance in z-axis
a= Exponent;
For circular& ellipticalcross section, a= 2
For rectangularcross section:
N
Rd= A
cf
cd+ A
sf
yd
A
c= Gross area of concrete section
A
s= Area of longitudinal reinforcement
N
Ed/N
Rd 0.1 0.7 1.0
a 1.0 1.5 2.0
•If the above conditions are NOTfulfilled, biaxiallybent columns
may be designed to satisfy the following simplified criterion:
�
??????�,�
�
??????�,�
�
+
�
??????�,�
�
??????�,�
�
≤�.�
M
Rd,y= Moment resistance in y-axis
M
Rd,z= Moment resistance in z-axis
a= Exponent;
For circular& ellipticalcross section, a= 2
For rectangularcross section:
N
Rd= A
cf
cd+ A
sf
yd
A
c= Gross area of concrete section
A
s= Area of longitudinal reinforcement
N
Ed/N
Rd 0.1 0.7 1.0
a 1.0 1.5 2.0
Biaxial Bending
Adaptation from BS 8110
Column may be designed for a single axis bending BUTwith an
increased moment:
(a) If
�
??????�,�
ℎ′
≥
�
??????�,�
�′
then�
??????��
′
=�
??????�,�+??????
ℎ′
�′
�
??????�,�
(b) If
�
??????�,�
ℎ′
<
�
??????�,�
�′
then�
??????��
′
=�
??????�,�+??????
�′
ℎ′
�
??????�,�
??????=1−
�
??????�
�ℎ�
��
(0.3 1.0)
hh’
b
b’
z z
y
y
M
Edz
M
Edy
Adaptation from BS 8110
Column may be designed for a single axis bending BUTwith an
increased moment:
(a) If
�
??????�,�
ℎ′
≥
�
??????�,�
�′
then�
??????��
′
=�
??????�,�+??????
ℎ′
�′
�
??????�,�
(b) If
�
??????�,�
ℎ′
<
�
??????�,�
�′
then�
??????��
′
=�
??????�,�+??????
�′
ℎ′
�
??????�,�
??????=1−
�
??????�
�ℎ�
��
(0.3 1.0)
hh’
b
b’
z z
y
y
M
Edz
M
Edy
Example 5
DESIGN OF NON-SLENDER
COLUMN BENT ABOUT BOTH
AXIS
DESIGN OF NON-SLENDER
COLUMN BENT ABOUT BOTH
AXIS
Classification: Braced non-slender column
Concrete, f
ck: 25 N/mm
2
Reinforcement, f
yk: 500 N/mm
2
Effective Lengthl
oz: 3.17 m
l
oy: 3.00 m
Slenderness ratio
z: 27.7
y: 34.2
Bar Size:
bar= 25 mm,
links= 6 mm
Nominal cover: 30 mm
Example 5: Design of Non-Slender Column
Bent About Both Axis
M
z
z
z
y y
300
350
Cross Section
M
y N
Ed= 1800 kN
M
z= 55 kNm
M
y= 32 kNm
Concrete, f
ck: 25 N/mm
2
Reinforcement, f
yk: 500 N/mm
2
Effective Lengthl
oz: 3.17 m
l
oy: 3.00 m
Slenderness ratio
z: 27.7
y: 34.2
Bar Size:
bar= 25 mm,
links= 6 mm
Nominal cover: 30 mm
Example 5: Design of Non-Slender Column
Bent About Both Axis
M
z
z
z
y y
300
350
Cross Section
M
y N
Ed= 1800 kN
M
z= 55 kNm
M
y= 32 kNm
Example 5: Design of Non-Slender Column
Bent About Both Axis
DESIGN MOMENT
Imperfection moment, �
??????��=�
??????�∙�
??????=�
??????�∙
��
400
�
??????��,�=1800×
3.70
400
=16.7kNm
�
??????��,�=1800×
3.00
400
=13.5kNm
Design moment including the effect of imperfection:
M
Ed,z= 55 + 16.7 = 71.7 kNm
M
Ed,y= 32 + 13.5 = 45.5 kNm
Bent About Both Axis
DESIGN MOMENT
Imperfection moment, �
??????��=�
??????�∙�
??????=�
??????�∙
��
400
�
??????��,�=1800×
3.70
400
=16.7kNm
�
??????��,�=1800×
3.00
400
=13.5kNm
Design moment including the effect of imperfection:
M
Ed,z= 55 + 16.7 = 71.7 kNm
M
Ed,y= 32 + 13.5 = 45.5 kNm
Example 5: Design of Non-Slender Column
Bent About Both Axis
CHECK BIAXIAL MOMENT
�
�=
�
??????�,�
�
??????�
=
71.7×10
6
1800×10
3
=40mm
�
�=
�
??????�,�
�
??????�
=
45.5×10
6
1800×10
3
=25mm
�
�
ℎ
�
�
�
=
25
350
40
300
=0.54>0.2
�
�
�
�
�
ℎ
=
40
300
25
350
=1.84>0.2
Check Biaxial Bending
??????
�
??????
�
=
34.2
27.7
=1.2<2
??????
�
??????
�
=
27.7
34.2
=0.8<2
Ignore Biaxial Bending
Check Biaxial Bending
Bent About Both Axis
CHECK BIAXIAL MOMENT
�
�=
�
??????�,�
�
??????�
=
71.7×10
6
1800×10
3
=40mm
�
�=
�
??????�,�
�
??????�
=
45.5×10
6
1800×10
3
=25mm
�
�
ℎ
�
�
�
=
25
350
40
300
=0.54>0.2
�
�
�
�
�
ℎ
=
40
300
25
350
=1.84>0.2
Check Biaxial Bending
??????
�
??????
�
=
34.2
27.7
=1.2<2
??????
�
??????
�
=
27.7
34.2
=0.8<2
Ignore Biaxial Bending
Check Biaxial Bending
Example 5: Design of Non-Slender Column
Bent About Both Axis
REINFORCEMENT DESIGN
Effective depth:h’= 350 –30 –6 –(0.5 25) = 301.5 mm
b’= 300 –30 –6 –(0.5 25) = 251.5 mm
�
??????�,�
ℎ
′
=
71.7×10
6
301.5
=238kN
�
??????�,�
�
′
=
45.5×10
6
251.5
=181kN
Since
�
??????�,�
ℎ
′
>
�
??????�,�
�
′
�
??????�,�
′
=�
??????�,�+??????
ℎ′
�′
�
??????�,�
�
??????�
�ℎ�
��
=
1800×10
3
300×350×25
=0.69
??????=1−
�
??????�
�ℎ���
=1−0.69=0.31 0.30
�
??????�,�
′
=71.7+0.31
301.5
251.5
45.5=��.�kNm
Bent About Both Axis
REINFORCEMENT DESIGN
Effective depth:h’= 350 –30 –6 –(0.5 25) = 301.5 mm
b’= 300 –30 –6 –(0.5 25) = 251.5 mm
�
??????�,�
ℎ
′
=
71.7×10
6
301.5
=238kN
�
??????�,�
�
′
=
45.5×10
6
251.5
=181kN
Since
�
??????�,�
ℎ
′
>
�
??????�,�
�
′
�
??????�,�
′
=�
??????�,�+??????
ℎ′
�′
�
??????�,�
�
??????�
�ℎ�
��
=
1800×10
3
300×350×25
=0.69
??????=1−
�
??????�
�ℎ���
=1−0.69=0.31 0.30
�
??????�,�
′
=71.7+0.31
301.5
251.5
45.5=��.�kNm
Example 5: Design of Non-Slender Column
Bent About Both Axis
REINFORCEMENT DESIGN
�
2=�+∅
�??????��+
∅
���
2
=30+6+
25
2
=48.5mm
�
2
ℎ
=
48.5
350
=0.14≈0.15
�
�ℎ�
��
=
1800×10
3
300×350×25
=0.69
�
�ℎ
2
�
��
=
88.8×10
6
300×350
2
×25
=0.10
From design chart:
���
��
�ℎ�
��
=0.48 ??????
�=
0.48�ℎ�
��
�
��
=
0.48×300×350×25
500
=����mm
2
??????
�,�??????�=
0.10�
??????�
�
��
=
0.10×1800×10
3
0.87×500
=���mm
2
or 0.002??????
�=210mm
2
??????
�,���=0.04??????
�=0.04×300×350=����mm
2
Provide 4H25 + 2H20 (A
s= 2592 mm
2
)
0.48
Bent About Both Axis
REINFORCEMENT DESIGN
�
2=�+∅
�??????��+
∅
���
2
=30+6+
25
2
=48.5mm
�
2
ℎ
=
48.5
350
=0.14≈0.15
�
�ℎ�
��
=
1800×10
3
300×350×25
=0.69
�
�ℎ
2
�
��
=
88.8×10
6
300×350
2
×25
=0.10
From design chart:
���
��
�ℎ�
��
=0.48 ??????
�=
0.48�ℎ�
��
�
��
=
0.48×300×350×25
500
=����mm
2
??????
�,�??????�=
0.10�
??????�
�
��
=
0.10×1800×10
3
0.87×500
=���mm
2
or 0.002??????
�=210mm
2
??????
�,���=0.04??????
�=0.04×300×350=����mm
2
Provide 4H25 + 2H20 (A
s= 2592 mm
2
)
0.48
Example 5: Design of Non-Slender Column
Bent About Both Axis
LINKS
links, min= the larger of:
i.0.25 25 mm = 6.3 mmor
ii.6 mm
Use
links= 8 mm
S
max= the lesser of:
i.20 (20 mm) = 400 mm or
ii.the lesser dimension of the column = 300 mm or
iii.400 mm
Use S
max= 300 mm
Provide H8-300
At section 300 mm below & above beam & at lap joints, S
max= 0.6 300 mm = 180 mm
Provide H8-175
300
350
H8
-
300
H8
-
175
z
z
Bent About Both Axis
LINKS
links, min= the larger of:
i.0.25 25 mm = 6.3 mmor
ii.6 mm
Use
links= 8 mm
S
max= the lesser of:
i.20 (20 mm) = 400 mm or
ii.the lesser dimension of the column = 300 mm or
iii.400 mm
Use S
max= 300 mm
Provide H8-300
At section 300 mm below & above beam & at lap joints, S
max= 0.6 300 mm = 180 mm
Provide H8-175
300
350
H8
-
300
H8
-
175
z
z
Example 5: Design of Non-Slender Column
Bent About Both Axis
CHECK BIAXIAL BENDING
Steel Area:
All: 4H25 + 2H20 A
s= 2592 mm
2
z-z: 4H25 + 2H20 A
s,z= 2592 mm
2
y-y: 4H25 + 0H20 A
s,y= 1964 mm
2
�2,�
ℎ
=
48.5
350
=0.14 &
�2,�
�
=
48.5
300
=0.16
�
�ℎ�
��
=
1800×10
3
300×350×25
=0.69
??????
�,��
��
�ℎ�
��
=
2592×500
300×350×25
=0.49
�
��,�
�ℎ
2
�
��
=0.10
M
Rd,z= 0.10 300 350
2
25 10
-6
= 91.9 kNm
0.49
0.10
Bent About Both Axis
CHECK BIAXIAL BENDING
Steel Area:
All: 4H25 + 2H20 A
s= 2592 mm
2
z-z: 4H25 + 2H20 A
s,z= 2592 mm
2
y-y: 4H25 + 0H20 A
s,y= 1964 mm
2
�2,�
ℎ
=
48.5
350
=0.14 &
�2,�
�
=
48.5
300
=0.16
�
�ℎ�
��
=
1800×10
3
300×350×25
=0.69
??????
�,��
��
�ℎ�
��
=
2592×500
300×350×25
=0.49
�
��,�
�ℎ
2
�
��
=0.10
M
Rd,z= 0.10 300 350
2
25 10
-6
= 91.9 kNm
0.49
0.10
Example 5: Design of Non-Slender Column
Bent About Both Axis
CHECK BIAXIAL BENDING (continued)
??????
�,��
��
�ℎ�
��
=
1964×500
350×300×25
=0.37
�
��,�
�ℎ
2
�
��
=0.07
M
Rd,z= 0.07 350 300
2
25 10
-6
= 55.1 kNm
N
Rd= A
cf
cd+ A
sf
yd= 0.567f
ckA
c+ 0.87f
ykA
s
= (0.567 25 300 350) + (0.87 500 2592)
= 2616 kN
�
??????�
�
??????�
=
1800
2616
=0.69 a= 1.49 (from Table)
Imperfection only be taken in one direction where they have the most unfavourableeffects:
�
??????�,�
�
??????�,�
�
+
�
??????�,�
�
??????�,�
�
=
71.7
91.9
1.49
+
��
55.1
1.49
=�.��>�.�FAIL
0.37
0.07
Bent About Both Axis
CHECK BIAXIAL BENDING (continued)
??????
�,��
��
�ℎ�
��
=
1964×500
350×300×25
=0.37
�
��,�
�ℎ
2
�
��
=0.07
M
Rd,z= 0.07 350 300
2
25 10
-6
= 55.1 kNm
N
Rd= A
cf
cd+ A
sf
yd= 0.567f
ckA
c+ 0.87f
ykA
s
= (0.567 25 300 350) + (0.87 500 2592)
= 2616 kN
�
??????�
�
??????�
=
1800
2616
=0.69 a= 1.49 (from Table)
Imperfection only be taken in one direction where they have the most unfavourableeffects:
�
??????�,�
�
??????�,�
�
+
�
??????�,�
�
??????�,�
�
=
71.7
91.9
1.49
+
��
55.1
1.49
=�.��>�.�FAIL
0.37
0.07
Example 5: Design of Non-Slender Column
Bent About Both Axis
NEW ARRANGEMENT OF REINFORCEMENT
Steel Area:
All: 4H25 + 4H20 A
s= 3221 mm
2
z-z: 4H25 + 2H20 A
s,z= 2592 mm
2
y-y: 4H25 + 2H20 A
s,y= 2592 mm
2
�2,�
ℎ
=
48.5
350
=0.14 &
�2,�
�
=
48.5
300
=0.16
�
�ℎ�
��
=
1800×10
3
300×350×25
=0.69
??????
�,��
��
�ℎ�
��
=
2592×500
300×350×25
=0.49
�
��,�
�ℎ
2
�
��
=0.10
M
Rd,z= 0.10 300 350
2
25 10
-6
= 91.9 kNm
0.49
0.10
Bent About Both Axis
NEW ARRANGEMENT OF REINFORCEMENT
Steel Area:
All: 4H25 + 4H20 A
s= 3221 mm
2
z-z: 4H25 + 2H20 A
s,z= 2592 mm
2
y-y: 4H25 + 2H20 A
s,y= 2592 mm
2
�2,�
ℎ
=
48.5
350
=0.14 &
�2,�
�
=
48.5
300
=0.16
�
�ℎ�
��
=
1800×10
3
300×350×25
=0.69
??????
�,��
��
�ℎ�
��
=
2592×500
300×350×25
=0.49
�
��,�
�ℎ
2
�
��
=0.10
M
Rd,z= 0.10 300 350
2
25 10
-6
= 91.9 kNm
0.49
0.10
Example 5: Design of Non-Slender Column
Bent About Both Axis
NEW ARRANGEMENT OF REINFORCEMENT
??????
�,��
��
�ℎ�
��
=
2592×500
350×300×25
=0.49
�
��,�
�ℎ
2
�
��
=0.10
M
Rd,z= 0.10 350 300
2
25 10
-6
= 78.8 kNm
N
Rd= A
cf
cd+ A
sf
yd= 0.567f
ckA
c+ 0.87f
ykA
s
= (0.567 25 300 350) + (0.87 500 3221)
= 2889 kN
�
??????�
�
??????�
=
1800
2889
=0.62 a= 1.44 (from Table)
Imperfection only be taken in one direction where they have the most unfavourableeffects:
�
??????�,�
�
??????�,�
�
+
�
??????�,�
�
??????�,�
�
=
71.7
91.9
1.44
+
��
78.8
1.44
=�.��≤�.�PASS
0.49
0.10
Bent About Both Axis
NEW ARRANGEMENT OF REINFORCEMENT
??????
�,��
��
�ℎ�
��
=
2592×500
350×300×25
=0.49
�
��,�
�ℎ
2
�
��
=0.10
M
Rd,z= 0.10 350 300
2
25 10
-6
= 78.8 kNm
N
Rd= A
cf
cd+ A
sf
yd= 0.567f
ckA
c+ 0.87f
ykA
s
= (0.567 25 300 350) + (0.87 500 3221)
= 2889 kN
�
??????�
�
??????�
=
1800
2889
=0.62 a= 1.44 (from Table)
Imperfection only be taken in one direction where they have the most unfavourableeffects:
�
??????�,�
�
??????�,�
�
+
�
??????�,�
�
??????�,�
�
=
71.7
91.9
1.44
+
��
78.8
1.44
=�.��≤�.�PASS
0.49
0.10
Example 5: Design of Non-Slender Column
Bent About Both Axis
H8
-
300
H8
-
175
300
350
z
z
y y
Bent About Both Axis
H8
-
300
H8
-
175
300
350
z
z
y y
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