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Page 1
COMPRESSION MEMBERS
A perfectly straight member of linear elastic material is shown if figure.
The above member has a friction less hinge at each end, its lower end being fixed
in position while its upper end is free to move vertically but prevented from
deflecting horizontally. It is assumed that the deflections of the member remain
small.
The elastic critical load
E
P at which a straight compression member buckles
laterally can be determined by finding a deflected position which is one of
equilibrium.
Basic Strut Theory
y P
dx
y d
EI - =
2
2
(1)
Eulers critical load is obtained as
2
2
l
EI
P
y
E
p
= (2)
Page 2
COMPRESSION MEMBERS
A perfectly straight member of linear elastic material is shown if figure.
The above member has a friction less hinge at each end, its lower end being fixed
in position while its upper end is free to move vertically but prevented from
deflecting horizontally. It is assumed that the deflections of the member remain
small.
The elastic critical load
E
P at which a straight compression member buckles
laterally can be determined by finding a deflected position which is one of
equilibrium.
Basic Strut Theory
y P
dx
y d
EI - =
2
2
(1)
Eulers critical load is obtained as
2
2
l
EI
P
y
E
p
= (2)
In terms of the stress equation is
( )
2
2
/r KL
E
p
E
p
= (3)
Strut with initial curvature
In practice, columns are generally not straight and the effect out of straightness on
strength is studied. Consider a strut with an initial curvature bent in a half sine
curve as shown in Figure.
If the initial deflection, at x from A is y
o
and the strut deflects ‘y” further under
load, P, the equilibrium equation is
( )
o
y y P
dx
y d
EI + =
2
2
(4)
Where deflection
?
?
?
?
?
?
=
l
x
y
p
sin (5)
If
o
d is the deflection at the centre and d the additional deflection caused by P,
then
( ) 1 /
0
-
=
P P
E
d
d (6)
The maximum stress at the centre of the strut is given by
Page 3
COMPRESSION MEMBERS
A perfectly straight member of linear elastic material is shown if figure.
The above member has a friction less hinge at each end, its lower end being fixed
in position while its upper end is free to move vertically but prevented from
deflecting horizontally. It is assumed that the deflections of the member remain
small.
The elastic critical load
E
P at which a straight compression member buckles
laterally can be determined by finding a deflected position which is one of
equilibrium.
Basic Strut Theory
y P
dx
y d
EI - =
2
2
(1)
Eulers critical load is obtained as
2
2
l
EI
P
y
E
p
= (2)
In terms of the stress equation is
( )
2
2
/r KL
E
p
E
p
= (3)
Strut with initial curvature
In practice, columns are generally not straight and the effect out of straightness on
strength is studied. Consider a strut with an initial curvature bent in a half sine
curve as shown in Figure.
If the initial deflection, at x from A is y
o
and the strut deflects ‘y” further under
load, P, the equilibrium equation is
( )
o
y y P
dx
y d
EI + =
2
2
(4)
Where deflection
?
?
?
?
?
?
=
l
x
y
p
sin (5)
If
o
d is the deflection at the centre and d the additional deflection caused by P,
then
( ) 1 /
0
-
=
P P
E
d
d (6)
The maximum stress at the centre of the strut is given by
( )
y
I
h P
A
P
P
d d +
+ =
0
max
(7)
Where h is shown in figure
i.e.
( )
2
0
Ary
h P
p p
c y
d d +
+ = (8)
i.e.
( )
2
0
ry
h
p p p
c c y
d d +
+ = (9)
( )
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
+ + =
1
1
1
0 2
c E
c
c y
p p ry
h p
p p d (10)
Denoting the Perry factor
2
0
ry
h d
? = (11)
( )
( )
?
?
?
?
?
?
+ = -
c E
c
c c y
p p
p
p p p 1 ? (12)
On simplification it gave
( )( )
c E c y c E
p p p p p p ? = - - (13)
The value of p
c
, the limiting strength at which the maximum stress equal the
design strength, can be found by solving this equation and ? is the Perry factor.
The minimum value of p
c
after solving the quadratic equation is obtained as
( )
5 . 0
2
y E c
p p p - ± = f f (14)
which is of the form
Page 4
COMPRESSION MEMBERS
A perfectly straight member of linear elastic material is shown if figure.
The above member has a friction less hinge at each end, its lower end being fixed
in position while its upper end is free to move vertically but prevented from
deflecting horizontally. It is assumed that the deflections of the member remain
small.
The elastic critical load
E
P at which a straight compression member buckles
laterally can be determined by finding a deflected position which is one of
equilibrium.
Basic Strut Theory
y P
dx
y d
EI - =
2
2
(1)
Eulers critical load is obtained as
2
2
l
EI
P
y
E
p
= (2)
In terms of the stress equation is
( )
2
2
/r KL
E
p
E
p
= (3)
Strut with initial curvature
In practice, columns are generally not straight and the effect out of straightness on
strength is studied. Consider a strut with an initial curvature bent in a half sine
curve as shown in Figure.
If the initial deflection, at x from A is y
o
and the strut deflects ‘y” further under
load, P, the equilibrium equation is
( )
o
y y P
dx
y d
EI + =
2
2
(4)
Where deflection
?
?
?
?
?
?
=
l
x
y
p
sin (5)
If
o
d is the deflection at the centre and d the additional deflection caused by P,
then
( ) 1 /
0
-
=
P P
E
d
d (6)
The maximum stress at the centre of the strut is given by
( )
y
I
h P
A
P
P
d d +
+ =
0
max
(7)
Where h is shown in figure
i.e.
( )
2
0
Ary
h P
p p
c y
d d +
+ = (8)
i.e.
( )
2
0
ry
h
p p p
c c y
d d +
+ = (9)
( )
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
+ + =
1
1
1
0 2
c E
c
c y
p p ry
h p
p p d (10)
Denoting the Perry factor
2
0
ry
h d
? = (11)
( )
( )
?
?
?
?
?
?
+ = -
c E
c
c c y
p p
p
p p p 1 ? (12)
On simplification it gave
( )( )
c E c y c E
p p p p p p ? = - - (13)
The value of p
c
, the limiting strength at which the maximum stress equal the
design strength, can be found by solving this equation and ? is the Perry factor.
The minimum value of p
c
after solving the quadratic equation is obtained as
( )
5 . 0
2
y E c
p p p - ± = f f (14)
which is of the form
( )
5 . 0
2
y E c
p p p - - = f f (15)
and
2
0
y
r
h d
? = (16)
the initial deflection
0
d is taken as (1/1000)
th
of length of the column and hence
? is given by
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= ?
?
?
?
?
?
=
y y y
r
h
r
l
r
h l
1000 1000
2
? (17)
and hence
?
?
?
?
?
?
?
?
=
y
r
l
a ? (18)
and
?
?
?
?
?
?
?
?
=
y
r
l
? (19)
a lower value of a was suggested by Robertson as a = 0.003 for column designs.
This approach was suggested in British code. ? is the slenderness ratio. The total
effect of the imperfections (initial curvature, end eccentricity and residual stresses
on strength). They are combined in to the Perry constant ? and is modified as
( )
0
001 . 0 ? ? ? - = a (20)
and
?
?
?
?
?
?
?
?
=
y
f
E
2
0
02 . 0
p
? (21)
the value of
0
? gives the limit to the plateau over which the design strength p
y
controls the strut load. The Robertson’s constant ‘a ’ is assigned different values
to give the different design curves.
As per IS 800-2007;
mo
y
mo
y
cd
f f
f
? ?
? = = (22)
And ? = stress reduction factor for different buckling class, slenderness ratio and
yield stress.
( ) [ ]
5 . 0
2 2
1
? f f
?
- +
= (23)
Page 5
COMPRESSION MEMBERS
A perfectly straight member of linear elastic material is shown if figure.
The above member has a friction less hinge at each end, its lower end being fixed
in position while its upper end is free to move vertically but prevented from
deflecting horizontally. It is assumed that the deflections of the member remain
small.
The elastic critical load
E
P at which a straight compression member buckles
laterally can be determined by finding a deflected position which is one of
equilibrium.
Basic Strut Theory
y P
dx
y d
EI - =
2
2
(1)
Eulers critical load is obtained as
2
2
l
EI
P
y
E
p
= (2)
In terms of the stress equation is
( )
2
2
/r KL
E
p
E
p
= (3)
Strut with initial curvature
In practice, columns are generally not straight and the effect out of straightness on
strength is studied. Consider a strut with an initial curvature bent in a half sine
curve as shown in Figure.
If the initial deflection, at x from A is y
o
and the strut deflects ‘y” further under
load, P, the equilibrium equation is
( )
o
y y P
dx
y d
EI + =
2
2
(4)
Where deflection
?
?
?
?
?
?
=
l
x
y
p
sin (5)
If
o
d is the deflection at the centre and d the additional deflection caused by P,
then
( ) 1 /
0
-
=
P P
E
d
d (6)
The maximum stress at the centre of the strut is given by
( )
y
I
h P
A
P
P
d d +
+ =
0
max
(7)
Where h is shown in figure
i.e.
( )
2
0
Ary
h P
p p
c y
d d +
+ = (8)
i.e.
( )
2
0
ry
h
p p p
c c y
d d +
+ = (9)
( )
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
+ + =
1
1
1
0 2
c E
c
c y
p p ry
h p
p p d (10)
Denoting the Perry factor
2
0
ry
h d
? = (11)
( )
( )
?
?
?
?
?
?
+ = -
c E
c
c c y
p p
p
p p p 1 ? (12)
On simplification it gave
( )( )
c E c y c E
p p p p p p ? = - - (13)
The value of p
c
, the limiting strength at which the maximum stress equal the
design strength, can be found by solving this equation and ? is the Perry factor.
The minimum value of p
c
after solving the quadratic equation is obtained as
( )
5 . 0
2
y E c
p p p - ± = f f (14)
which is of the form
( )
5 . 0
2
y E c
p p p - - = f f (15)
and
2
0
y
r
h d
? = (16)
the initial deflection
0
d is taken as (1/1000)
th
of length of the column and hence
? is given by
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= ?
?
?
?
?
?
=
y y y
r
h
r
l
r
h l
1000 1000
2
? (17)
and hence
?
?
?
?
?
?
?
?
=
y
r
l
a ? (18)
and
?
?
?
?
?
?
?
?
=
y
r
l
? (19)
a lower value of a was suggested by Robertson as a = 0.003 for column designs.
This approach was suggested in British code. ? is the slenderness ratio. The total
effect of the imperfections (initial curvature, end eccentricity and residual stresses
on strength). They are combined in to the Perry constant ? and is modified as
( )
0
001 . 0 ? ? ? - = a (20)
and
?
?
?
?
?
?
?
?
=
y
f
E
2
0
02 . 0
p
? (21)
the value of
0
? gives the limit to the plateau over which the design strength p
y
controls the strut load. The Robertson’s constant ‘a ’ is assigned different values
to give the different design curves.
As per IS 800-2007;
mo
y
mo
y
cd
f f
f
? ?
? = = (22)
And ? = stress reduction factor for different buckling class, slenderness ratio and
yield stress.
( ) [ ]
5 . 0
2 2
1
? f f
?
- +
= (23)
and ( ) [ ]
2
2 . 0 1 5 . 0 ? ? a f + - + = (24)
a = imperfection factor given in Table 7, in P35, IS800:207.
? =non dimensional effective slenderness ratio.
cc
y
f
f
= ? (25)
and
cc
f Euler’s buckling stress =
( )
2
2
/r KL
E p
(26)
and
?
?
?
?
?
?
r
KL
effective slenderness ratio (or) the effective length KL to appropriate
radius of gyration, r,
mo
? = partial safety factor for material strength. It is noted
that the stress reduction factor ? depends on buckling class, slenderness ratio and
yield stress (Table 8, P36- 39, IS800-2007).
Read More
FAQs on Short Notes: Compression Members - Short Notes for Civil Engineering - Civil Engineering (CE)
| 1. What are compression members in structural engineering? |  |
Ans. Compression members in structural engineering are structural elements that are primarily subjected to axial compressive forces. They are designed to resist the compressive loads and maintain their stability under such loads.
| 2. What are the common types of compression members used in construction? | |
Ans. The common types of compression members used in construction include columns, beams, trusses, and struts. These members are designed to withstand the compressive forces and provide support to the overall structure.
| 3. How are compression members designed and analyzed in structural engineering? | |
Ans. Compression members are designed and analyzed in structural engineering using various methods, such as Euler's formula, Rankine's formula, and Perry-Robertson formula. These methods consider factors like material properties, cross-sectional dimensions, and boundary conditions to ensure the stability and strength of the compression members.
| 4. What factors affect the design of compression members? | |
Ans. Several factors affect the design of compression members, including the material properties, cross-sectional dimensions, slenderness ratio, end conditions, and applied loads. These factors determine the maximum compressive load a member can withstand without failure.
| 5. How can compression members be strengthened to increase their load-carrying capacity? | |
Ans. Compression members can be strengthened to increase their load-carrying capacity by adding reinforcement, using thicker sections, increasing the cross-sectional area, or providing additional support through bracing or stiffeners. These strengthening techniques help to improve the stability and strength of the compression members.
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