Load Contour Method
If the interaction diagram is horizontally sliced at a
particular load level as done in Fig. 14.27 the resulting
slice is actually a graph between M
nx
and M
ny
at a constant
load and is called a load contour.
The values of the moments M
nx
and M
ny
are usually
normalized with M
nxo
when M
ny
0 and M
nyo
when M
nx
0 respectively.
In this method a curve of the following form is passed
through the failure line of a load contour:

slide 2:

P
n
M
ny
M
nx
Load
Contour
M
ny
M
nx
a Isometric View b Plan View
Fig. 14.27. A Typical Load Contour.

slide 3:

0 . 1
2
1
+
α
α
nyo
ny
nxo
nx
M
M
M
M
The constants α
1
and α
2
are the exponents depending on
• column dimensions
• amount and distribution of steel reinforcement
• stress-strain characteristics of steel and concrete
• amount of concrete cover and
• size of lateral ties or spirals.
Generally the values of α
1
and α
2
equal to a constant
value α give satisfactory results reducing the equation
to:
0 . 1
+
α
α
nyo
ny
nxo
nx
M
M
M
M

slide 4:

The range of values of α for square and rectangular
columns is between 1.15 and 1.55 and the lower values of
α are more conservative.
A value of α 1.5 is reasonably accurate for the most
square and rectangular sections having uniformly
distributed reinforcement.
According to some researchers the value of α may be
more accurately obtained in the following form:
α log 0.5 / log β

slide 5:

Values of β are given in the form of charts on pages 7-17
to 7-19 of PCA Notes on ACI-02 against the values of P
u
/
P
o
and the reinforcement index.
Reinforcement index
Separate curves are available for various bar arrangements.
The column must be rectangular with larger to shorter
sides ratio of less than 4.0 f
c
′ must be between 12 and 41
MPa and γ must be between 0.6 and 1.0.
c
y
t
f
f
′
ρ

slide 6:

Example 14.5: Select a square tied column cross-
section to resist P
u
1600 kN M
ux
95 kN-m and M
uy
110 kN-m. f
c
′ 20 MPa f
y
300 MPa and clear
cover 40 mm. Use the following two methods:
a Equivalent uniaxial eccentricity method
b Reciprocal load method.
Solution:
a Equivalent Uniaxial Eccentricity Method
y c
uy ux u
g
f f
M M P
A
008 . 0 43 . 0
2 2
trial
+
′
+ +
300 008 . 0 20 43 . 0
1000 110 2 95 2 1600
× + ×
× + × +
182727 mm
2

slide 7:

Try 450 × 450 mm column.
450
40 2 25 10 2 450
× − − × −
γ
0.72
e
x
68.8 mm
u
uy
P
M
1600
1000 110 ×
e
y
59.4 mm
u
ux
P
M
1600
1000 95 ×
e
x
/ b 68.8 / 450 ≥ e
y
/ h 59.4 / 450
0.395 ≤ 0.4
g c
u
A f
P
′
2
450 20
1000 1600
×
×

slide 8:

α
6 . 0
720
300
5 . 0 ≥
+
′
+
y
g c
u
f
A f
P
0.746
720
300 300
395 . 0 5 . 0
+
+
e
ox
e
x
+ α b
h
e
y
68.8 + 0.746 × 59.4 × 450 / 450 113.1 mm
Equivalent uniaxial moment is:
M
oy
P
u
e
ox
1600 × 113.1 / 1000 180.98 kN-m
Use uniaxial interaction diagrams with bars in all the four
faces to determine the total steel ratio ρ
t
.

slide 9:

1600000 / 450
2
7.9 MPa
g
u
A
P
180.98 ×10
6
/ 450
3
2.0 MPa
h A
M
g
oy
For γ 0.6 ρ
t
0.023
For γ 0.75 ρ
t
0.018
For γ 0.72
ρ
t
0.018 + 0.005 / 0.15 × 0.75 − 0.72 0.019
A
st
ρ
t
× A
g
0.019 × 450
2
3848 mm
2
Use 8 − 25 bars

slide 10:

Reciprocal Load Method
Trial Size: 450 × 450 mm column
Use 8 − 25 bars as the first try.
ρ
t
8 × 510 / 450
2
0.02
110 ×10
6
/ 450
3
1.21 MPa
h A
M
g
uy
h A
M
g
ny
φ
1600000 / 450
2
7.90 MPa
g
n
A
P φ
Join this point with the origin and extend to the 2 steel
curve to get the value of the required capacity.

slide 11:

For γ 0.60 9.8 MPa
g
nx
A
P φ
For γ 0.75 10.2 MPa
g
nx
A
P φ
For γ 0.72 10.12 MPa
g
nx
A
P φ
⇒ φP
nx
2049 kN
95 ×10
6
/ 450
3
1.04 MPa
h A
M
g
ux
h A
M
g
nx
φ
1600000 / 450
2
7.90 MPa
g
n
A
P φ

slide 12:

For γ 0.60 10.6 MPa
g
ny
A
P φ
For γ 0.75 10.75 MPa
g
ny
A
P φ
10.72 MPa for γ 0.72 ⇒ φP
ny
2171 kN
g
ny
A
P φ
The point to calculate φ P
no
is located on the diagram
where the interaction curve for ρ 0.02 intersects the
vertical load line.
14.70 MPa ⇒ φP
no
2977 kN
g
no
A
P φ
As the loads φ P
nx
and φ P
ny
are quite closer to φ P
no
φ-
factor of 0.65 seems reasonable for all the loads.

slide 13:

o ny nx ni
P P P P φ φ φ φ
1 1 1 1
− +
2977
1
2171
1
2049
1
− +
φ P
ni
1632 kN P
u
1600 kN
Design is OK according to the Reciprocal Load Method.
Example 14.6: Check the adequacy of a rectangular
tied column X-section of size 300 × 450 mm to resist P
u
1000 kN acting at e
x
125 mm and e
y
50 mm as
shown in Fig. 14.28. f
c
′ 25 MPa f
y
420 MPa and
cover to centroid of bars 60 mm. The reinforcement is
arranged around the perimeter of the column consisting of
8 − 25 bars.

slide 14:

Use the following two methods:
a Reciprocal load method
b Load contour method.
P
n
x
y
Fig. 14.28.Column For Example 14.28.
Solution:
ρ
t
4080 / 300 × 450 0.03
a Reciprocal Load Method
i Considering bending about y-axis:
γ 330 / 450 0.73 ≅ 0.75
1000 × 1000 / 300 × 450 7.41 MPa
g
u
A
P

slide 15:

1000 × 1000 × 125 / 300 × 450
2
2.06 MPa
h A
M
g
u
h A
e P
g
x u
×
Join the point P
u
/A
g
M
u
/A
g
h with the origin and
extend to ρ 0.03 to get the following:
10.8 MPa ⇒ φP
nx
1458 kN
g
nx
A
P φ
ii Considering bending about x-axis:
γ 180 / 300 0.60
1000 × 1000 / 300 × 450 7.41 MPa
g
u
A
P

slide 16:

1000 × 1000 × 50 / 450 × 300
2
1.23 MPa
h A
M
g
u
h A
e P
g
y u
×
Join the point P
u
/A
g
M
u
/A
g
h with the origin and
extend to ρ 0.03 to get the following:
13.6 MPa ⇒ φP
ny
1836 kN
g
ny
A
P φ
iii No eccentricity case:
φP
no
21.4 × 300 × 450 2889 kN
Assuming same φ-factors for all the loads we have

slide 17:

o ny nx ni
P P P P φ φ φ φ
1 1 1 1
− +
2889
1
1836
1
1458
1
− +
φ P
ni
1131 kN P
u
1000 kN OK
a Load Contour Method
i Considering bending about x-axis:
γ 180 / 300 0.60
1000 × 1000 / 300 × 450 7.41 MPa
g
u
A
P
ρ
t
0.03

slide 18:

2.92 MPa ⇒
h A
M
g
nx
φ
φM
nxo
2.92 × 450 × 300
2
/ 10
6
118.26 kN-m
ii Considering bending about y-axis:
γ 320 / 450 0.71 ≅ 0.75
1000 × 1000 / 300 × 450 7.41 MPa
g
u
A
P
ρ
t
0.03
3.47 MPa ⇒
h A
M
g
ny
φ
φM
nyo
3.47 × 300 × 450
2
/ 10
6
210.80 kN-m

slide 19:

M
ux
P
u
× e
y
1000 × 50 / 1000 50 kN-m
M
uy
P
u
× e
x
1000 × 125 / 1000 125 kN-m
Selecting the conservative value of α 1.15 we have
α
α
+
nyo
ny
nxo
nx
M
M
M
M
α
α
φ φ
+
nyo
uy
nxo
ux
M
M
M
M
15 . 1 15 . 1
80 . 210
00 . 125
26 . 118
00 . 50
+
0.92 1.0 OK
Hence the column is safe according to the load contour
method. This method is relatively more exact for portion
of curve below the balanced condition.

slide 20:

SLENDER COLUMNS
There are three methods of design of slender columns
described below:
1. Perform exact P −∆ analysis to find P
u
and M
umax
and then use the standard interaction diagram for the
short columns. This method is called the second order
analysis of frames.
2. Find P
u
and approximately magnified moment M
u
and then use the standard interaction diagram for short
columns. This method is called Moment Magnification
Method. In this method the moments obtained from the
first order analysis are multiplied with an empirical
moment magnifier.

slide 21:

3. Modify the interaction diagram to account for the
slenderness effects having the coordinates P
u
and
unmagnified moment M
u
just like a regular diagram.
Elastic Buckling Load For Concentrically
Loaded Columns
P
c
2
2
u
t
k
I E
l
π
2
2
/ r k
A E
u
t
l
π
Concrete is an inelastic material and hence the modulus
of elasticity varies all along the stress-strain curve as
shown in Fig. 14.29.
By replacing the E-value with the tangent modulus of
elasticity E
t
Euler’s formula may be used for materials
like concrete.

slide 22:

e
f
E
t
Fig. 14.29.Modulus of Elasticity For Concrete.

slide 23:

The tangent modulus of elasticity is different at all points
of the stress-strain curve and is difficult to estimate
precisely.
Hence approximate methods are used to calculate the
effective E-value along with the reduced moment of
inertia due to cracking and long-term effects.
A typical load − slenderness ratio curve is shown in
Fig.14.30 to represent the buckling behavior.
For concrete columns the chances of elastic buckling are
usually very less and only moment magnification and at
the most inelastic buckling are the important parameters.

slide 24:

Elastic Buckling
Inelastic Buckling
With Crushing
r
k
u
l
P
Short
Column
Intermediate
Column
Long
Column
Fig. 14.30. Buckling Load Versus Slenderness Ratio For Columns.

slide 25:

Effective Length Factor
This factor gives the ratio of length of half sine wave
portion of defected shape after buckling distance between
two points of contra-flexure to full-unsupported length of
column.
This depends upon the end conditions of the column and
the fact that whether side-sway is permitted or not. Greater
the k-value greater is the effective length and slenderness
ratio and hence smaller is the buckling load.
The value of k-factor in case of no side-sway is between
0.5 and 1.0 whereas in case of appreciable side-sway it is
always greater than or equal to 1.0.

slide 26:

Any appreciable lateral or sideward movement of top of
a vertical column relative to its bottom is called side-
sway sway or lateral drift.
If side-sway is possible k-value increases by a greater
degree and column buckles at a lesser load.
Side-sway in a frame takes place due to the following
factors shown in Fig. 24.31:-
1. Lengths of different columns are unequal.
2. Sections of columns have different cross-sectional
properties.
3. Loads are un-symmetrical.
4. Lateral loads are acting.

slide 27:

2I
I
I
Fig. 14.31. Chances of Side-sway.

slide 28:

Side-sway can be prevented in a frame by:-
1. Providing shear or partition walls.
2. Fixing the top of frame with adjoining rigid
structures.
3. Provision of properly designed lift well in a building
which may act like backbone of the structure reducing
the lateral deflections.
4. Provision of lateral bracing which may be of
following two types:
A. Diagonal bracing.
B. Longitudinal bracing.
Effective length factor and the buckled shape of columns
having well-defined end conditions are given in Fig. 14.32.

slide 29:

Theoretical k 0.5
Practical k 0.65
No side-sway
Theoretical k 1.0
Practical k 1.0
No side-sway
Inflection
points
l
e
k l
l
e
k l

slide 30:

Theoretical k 1.0
Practical k 1.2
Side-sway present
Theoretical k 0.7
Practical k 0.8
No side-sway
l
e
k l
l
e
k l

slide 31:

Theoretical k 2.0
Practical k 2.10
Side-sway present
Theoretical k 2.0
Practical k 2.0
Side-sway present
l
e
k l
l
e
k l
Fig. 14.32. Values of k-factor For Various End Conditions.

slide 32:

Consider the example of column AB shown in Fig.
14.33.
The ends are not free to rotate.
However these ends are also not perfectly fixed.
Instead the ends are partially fixed with the fixity
determined by the ratio of relative flexural stiffness of
columns meeting at a joint to the flexural stiffness of
beams meeting at the joint.
This ratio is denoted by ψ or G using the expression
given below:-

slide 33:

A
B
B
A
G
A
or ψ
A
G
B
or ψ
B
Columns
Beams
Fig. 14.33. Partially Fixed Column Ends.

slide 34:

ψ or G at each end
EI of columns
EI of beams
l
l
∑
∑
This value is calculated at both ends of the columns
denoted by points A and B and summation is taken for all
members meeting at a particular end.
The lower columns of Fig. 14.33 have their top ends
partially fixed but the bottom ends may have well-defined
end conditions.
The ψ or G value at these ends are decided as follows:

slide 35:

Hinged Support
G or Ψ 10.0 for braced columns
20.0 for sway columns
Fixed Support
G or Ψ 0.5 for braced columns
1.0 for sway columns
To find out the k-value the alignment charts given in
Figs. 14.34 and 14.35 are used. The alignment chart of
Fig. 14.34 is for columns without any side-sway and the
alignment chart of Fig. 14.35 is for columns having
appreciable side-sway.

Restraint Provided By Footings
For Calculation Of Ψ Factor
In case of footing resting on soil the Ψ value is
calculated by using the foundation stiffness K
f
in place
of the beam stiffness in the usual expression as under:
ψ
∑
∑
footing of
columns of
l
l
EI
EI
f
c
K
K
To calculate the stiffness of the footing a moment M is
applied to the footing of Fig. 14.36 and the
corresponding rotation θ
f
settlement at extreme
compression end due to moment alone leaving the
uniform downward settlement ∆ and the contact stress
under the footing due to the rotation σ are observed.

slide 39:

θ
f
∆
y
Fig. 14.36. Rotation and Stiffness of a Foundation.
The extreme edge contact stress may be calculated as
follows:
σ M / S
f
M y / I
f
where S
f
Section modulus of the contact surface
and I
f
Moment of inertia of the contact surface.

slide 40:

θ
f
y
∆
y
K
s
/ σ
s f
K I
M
where K
s
is the modulus of sub-grade reaction pressure
corresponding to 1mm settlement which may
approximately be found from Table 14.1.
Table 14.1. Soil Sub-grade Reaction.
Allowable Soil Bearing Capacity Sub-grade Modulus
kN/m
2
or kPa N/mm
2
/mm
60 0.0136
100 0.0272
180 0.0450
240 0.0504
360 0.0654
480 0.0736

slide 41:

∴ K
f
f
M
θ
s f
K I
The value of Ψ at a footing to column joint is estimated
as follows:
ψ
f
c
K
K
s f
c c c
K I
I E l / 4
After calculation of Ψ factors at both ends of the
column the value of k may also be calculated from the
following ACI equations:
For braced members k should be the lesser of the
following two values:

slide 42:

k 0.7 + 0.05 Ψ
A
+ Ψ
B
≤ 1.0
k 0.85 + 0.05 Ψ
min
≤ 1.0
where Ψ
min
smaller of Ψ
A
and Ψ
B
For unbraced compression members k may be evaluated
as follows:
For Ψ
m
2k
2
B A
ψ ψ +
m
m
ψ
ψ
+
+
1
20
20
For unbraced compression members hinged at one end
and partially fixed at the other end k-value is obtained
as follows:
k 2.0 + 0.3 Ψ

You do not have the permission to view this presentation. In order to view it, please
contact the author of the presentation.

Send to Blogs and Networks

Processing ....

Premium member

Use HTTPs

HTTPS (Hypertext Transfer Protocol Secure) is a protocol used by Web servers to transfer and display Web content securely. Most web browsers block content or generate a “mixed content” warning when users access web pages via HTTPS that contain embedded content loaded via HTTP. To prevent users from facing this, Use HTTPS option.