Civil Engineering Reference
In-Depth Information
diagram two design equations can be developed: one gives
the area of reinforcement required to resist bending, the other
the reinforcement required give the column suffi cient axial
resistance. These two equations are solved iteratively to deter-
mine the minimum area of reinforcement required for a par-
ticular column. In practice, this is too long-winded for hand
calculations and either computer software is used, or column
charts are used. Column charts are available from a number of
sources, including the website www.eurocode2.info. A design
fl ow chart is provided in
Figure 17.3
.
M
M
01
15
41
.
41
41
.7
41
41
41
+
.
41
41
41
.
02
λ
lim
λ
lim
λ
=
N
A
f
Ed
N
Ed
cc
A
cc
Af
d
When
λ
exceeds
λ
lim
the second order moments,
M
2
should
be calculated and Eurocode 2 gives a number of approaches.
The design moment is then:
M
Ed
= max(
M
02
;
M
0e
+
M
2
;
M
01
+ 0.5
M
2
)
where
M
0e
= the mid-height moment which can be taken as
0.6
M
02
+ 0.4
M
01
> 0.4
M
02
, provided there are no moments
applied between the ends of the column.
For a braced column this simplifi es to
M
Ed
=
M
02
17.5.5 Shear
Unlike fl exural design, there are several theories for determin-
ing the shear strength of concrete. One widely used approach
is to assumed that the shear strength of concrete comprises a
truss (see
Figure 17.4
), with a concrete strut at an angle to the
vertical, shear reinforcement acting in tension and the longitu-
dinal reinforcement acting in tension. This basic theory has to
17.5.4.1 Column resistance
Figure 6.1 in Eurocode 2 describes the strain limits to be used
in determining the resistance of a column section. From this
Design for axial forces
El
c
El
c
2
El
b
El
b
Σ
l
b
kkk
=
kkk
=
l
c
Determine the axial force,
N
Ed
N
E
N
and
moments:
M
top
M
to
M
= moment at top of element
M
bot
Where
M
bo
M
= moment at bottom of element
III
= column second moment of area
III
= beam second moment of area
Determine joint stiffness at top and
bottom of element,
kkk
and
kkk
lll
= column length
lll
= beam length
Determine effective length,
lll
For braced members,
1
1
kkk
0.45 +
kkk
kkk
0.45 +
kkk
Determine the design moments,
M
01
M
0
M
,
1 +
1 +
1
1
1
1
lll
= 0.5
o
= 0.5
l
=
and
M
02
M
02
M
0
M
= max{|
M
top
M
to
M
|;|
M
bot
M
bo
M
|} +
e
i
N
Ed
M
02
N
E
N
≥
e
0
N
Ed
N
Ed
Determine slenderness limit,
λ
li
λ
l
λ
m
M
0
M
= min {|
M
top
M
to
M
|;|
M
bo
M
b
M
t
|}
M
01
e
i
=
lll
/400
e
0
=
h
/30
Determine slenderness,
λ
≥ 20mm
Ye s
N
λ
λ
No
Is
lim
λ
lim
li
m
li
= 15.4
m
= 15.4
C
≤
Is
λ
≤
λ
lim
N
Ed
N
Ed
N
A
c
f
cd
f
cd
MMM
= 0
Calculate,
MMM
where
Determine
M
Ed
M
Ed
C = 1.7 +
M
01
M
0
M
/
M
02
M
02
λ
=
λ
=
λ
lll
/√(
I
///
c
)
To determine
A
s
:
M
E
M
= MAX{
M
0Ed
M
0
M
}
M
Ed
M
0Ed
M
+
MMM
;
M
02
M
0
M
;
M
01
M
0
M
+ 0.5
M
02
• Using column design chart
• Use spreadsheet
• Solve by iteration
M
0Ed
M
0E
M
=
M
mi
M
m
M
d
where there is a moment applied between
the ends
= (0.6
M
02
M
0
M
) ≥ 0.4
M
02
M
0
M
for other cases
M
0
M
+ 0.4
M
01
Carry out reinforcement detailing
Figure 17.3 Design process braced for axially loaded elements
