CHAPTER 5
STRUTS
1
2
3
TOPICS TO BE COVERED
5.1 Introduction
5.2 Euler’s Theory
5.3 Equivalent strut length
5.4 Comparison of Euler’s theory with
experimental results
5.5 Euler’s validity limit
5.6 Rankine or Rankine-Gordon formula
5.7 Struts with Initial Curvature
5.8 Struts with Eccentric Load
5.9 Laterally Loaded Struts
4
5.1 INTRODUCTION
• Structural members which carry compressive
loads may be divided into two broad categories
depending on their relative lengths and crosssectional dimensions.
• Short members are generally termed
Columns and these usually fail by crushing
when the stress of the material in compression
is exceeded.
• Long, slender columns or Struts, however, fail
by buckling some time before the yield stress in
compression is reached.
5
• The Buckling occurs owing to one or more of the
following reasons:
A) The strut may not be perfectly straight
initially;
B) The load may not be applied exactly along the
axis of the strut;
C) One part of the material may yield in
compression more readily than others owing
to some lack of uniformity in the material
properties throughout the strut.
6
• At values of load below the buckling load a strut
will be in stable equilibrium where the
displacement caused by any lateral disturbance
will be totally recovered when the disturbance is
removed.
• At the buckling load the strut is said to be in a
state of neutral equilibrium, and theoretically
it should then be possible to gently defect the
strut in to a simple sine wave provided that the
amplitude of the wave is kept small.
Example: Thin Metal strips
7
• Theoretically, it is possible for struts to achieve a
condition of unstable equilibrium with loads
exceeding the buckling load, any slight lateral
disturbance then then causing failure by Buckling;
this condition is never achieved in practice under
static load conditions.
• Buckling occurs immediately at the point where
the buckling load is reached owing to the reasons
stated earlier.
8
5.2 EULER’S THEORY
A. Strut with pinned ends
Consider the axially loaded strut shown in Figure 5.1 subjected to the
crippling load Pe producing a deflection y at a distance x from one end.
Assume that the ends are either pin-jointed or rounded so that there is
no moment at either end.
d 2y
B.M at C = EI 2 = _ Pe y
dx
d 2y
EI 2 + Pe y = 0
dx
d 2 y Pe
y=0
2 +
EI
dx
Fig.5.1 Strut with axial load and pinned ends
9
• Let D=d/dx,
( D + n ) y = 0, where n = P / EI
2
2
2
e
• This is a second-order differential equation which has a solution of the
form
y=A cos nx+B sin nx
i.e
P
P
y = A cos (
)x + B sin (
)x
EI
EI
Now at x=0, y=0
And at x=L, y=0
e
e
A=0
BsinL√(Pe/EI)=0
Either B=0 or sinL√(Pe/EI)=0
10
• If B=0 then y=0 and the strut has not yet Buckled. Thus the solution
required is:
Pe
Pe
sin L ( ) = 0 ∴ L ( ) = π
EI
EI
2
π EI
Pe = 2 .......... .......... ...EQ.1
L
• It should be noted that other solutions(infinite number of solutions)
exist for the equation
Pe
sin L ( ) = 0 i.e sin nL = 0
EI
nL = π,2π,3π,5π, etc
11
• The above selected value in equation 1 is the so-called
Fundamental mode value and is the lowest critical
load producing the single-bow buckling condition.
• The solution nL=2π produces in two half waves, 3π in
three half-waves etc as shown in Figure 5.2.
• If the load is applied sufficiently to the strut, it is
possible to pass through the fundamental mode and
to achieve at least one of the other modes which are
theoretically possible.
12
Fig 5.2: Strut Failure Modes
13
B. One end fixed, the other free
Consider now the strut of Figure 5.3 with the origin at the fixed end.
d 2y
B.M at C = EI 2 = + P(a _ y )
dx
d 2 y Py Pa
=
2 +
EI EI
dx
( D2 + n 2 )y = n 2a.......... ......Eq.2
14
Fig. 5.3: Fixed-free strut
N.B: It is always convenient to arrange the diagram and origin such
that the differential equation is achieved in the above form since the
solution will then always be of the form.
y=A cos nx+B sin nx + (Particular solution)
The particular solution is a particular value of y which satisfies
equation 2, and in this case can be shown to be y=a.
y=A cos nx+B sin nx+a
Now when x=0, y=0
A=-a
When x=0, dy/dx=0
B=0
y= -acosnx+a
But when x=L, y=a
a= -acosnL+a
0= cosnL
15
16
• The fundamental mode of buckling in this case therefore is given
when nL=(1/2)π.
P
π
L ( )=
EI
2
π 2EI
Pe =
2 .......... .......... .......Eq. 3
4L
17
C. Fixed Ends
Consider the strut of Figure 5.4 with the origin at the centre.
Fig 5.4: Strut with fixed ends
In this case the B.M at C is given by:
d 2y
EI 2 = M _ Py
dx
d 2y P
M
y=
2 +
EI
EI
dx
( D2 + n 2 )y = M / EI
18
• Here the particular solution is:
M
M
y=
=
n EI
P
y = A cos nx + B sin nx + M / P
2
• Now when x=0, dy/dx=0 Therefore, B=0 and when x=1/2L, y=0
_M
nL
∴ A=
sec
P
2
M
nL
M
y=_
sec
cos nx +
P
2
P
• But when x=1/2L,dy/dx is also zero,
nM
nL
nL
0=
sec
sin
P
2
2
nM
nL
0=
tan
P
2
19
• The fundamental buckling mode is then given when nL/2=π.
L P
( ) =π
2 EI
2
4π EI
Pe =
.......... .......... ..Eq.4
2
L
20
D. One end fixed, the other Pinned
In order to maintain the pin-joint on the horizontal axis of the unloaded
strut, it is necessary in this case to introduce a vertical load F at the Pin
(Figure 5.5). The moment of F about the built-in end then balances the
fixing moment.
Fig 5.5: Strut with one end pinned,
d 2y
EI 2 = _ Py + F( L _ x )
dx
d 2y P
F
y=
(L _ x )
2 +
EI
EI
dx
F
2
2
( D + n )y =
(L _ x )
EI
the other fixed
21
• The particular solution is:
F
F
y = 2 (L _ x ) = (L _ x )
P
n EI
• The full solution is therefore
F
y = A cos nx + B sin nx + ( L _ x )
P
• When x=0, y=0, Therefore, A=_FL/P
• When x=0, dy/dx=0 , B=F/Np
FL
F
F
y=_
cos nx +
sin nx + ( L _ x )
P
nP
P
F
=
[_ nL cos nx + sin nx + n( L _ x )]
nP
22
• But when x=L, y=0
nL cos nL = sin nL
tan nL = nL
• The lowest value of nL (neglecting zero) which satisfies this condition
and which therefore produces the fundamental buckling condition is
nL=4.5 radians.
P
L
= 4 .5
EI
20.25EI
Pe =
.......... .......... .......... ..Eq.5
2
L
or approximately
2π 2EI
Pe =
.......... .......... .......... ...Eq.6
2
L
23
5.3 EQUIVALENT STRUT LENGTH
• Having derived the result for the buckling load of a strut, the Euler
loads for other end conditions may all be written in the same form,
i.e
π 2EI
Pe = 2 .......... .......... ..........Eq.7
l
Where l is the equivalent length of the strut and can be related to the
actual length of the strut depending on the end conditions. The
equivalent length is the length of a single bow (half sine wave) in each
of the strut deflection curves shown in the figure below.
24
Figure 5.6: “Equivalent length” of struts with different end conditions.
In each case l is the length of a single bow.
25
5.4 COMPARISON OF EULER’S THEORY WITH
EXPERIMENTAL RESULTS
• Between L/k=40 and L/k=100 , the Euler results are not close to the
experimental values, suggesting a critical load which is in excess of
what is actually required for failure - a very unsafe situation!
• Other formulae have therefore been derived in an attempt to obtain
closer agreement between the actual failing load and the predicted
value in this range of slenderness ratio.
26
a) Straight-line formula
P = σ y A[1 _ n( L / k )].................. .......... ......Eq.8)
The value of n depending on the material used and the end condition.
b) Johnson Parabolic formula
P = σ y A[1 _ b( L / k )2 ]................... .......... .....Eq.9)
The value of b depending also on the end condition.
Neither of the above formulae proved to be very successful, and they
were replaced by:
C. Rankine-Gordon formula
1
1
1
=
+
.......... .......... .......... ........Eq.10
PR Pe PC
Where Pe is the Euler buckling load
crushing(compressive yield) load= yA.
and
Pc
is
the
27
Fig 5.7: Comparison of experimental results with Euler curve
28
5.5 EULER “VALIDITY LIMIT”
• From the graph of Fig. 5.7 and the comments above, it is evident that
the Euler theory is unsafe for small L/k ratios. It is useful, therefore,
to determine the limiting value of L/k below which the Euler theory
should not be applied; this is termed the validity limit.
• The validity limit is the point where the Euler stress e equals the
yield or crushing stress y, i.e the point where the strut load
• Now the Euler load can be written in the form
•
P = σyA
• Where C is a constant depending on the end condition of
j
the strut.
π 2EI
π 2EAk 2
Pe = C 2 = C
L
L2
29
• Therefore in the limiting condition
π 2EAk 2
σyA = C
L2
L
Cπ 2E
= (
)
k
σy
• The value of this expression will vary with the type of end condition; as
an example, low carbon steel struts with pinned ends give L/k ≈ 80.
30
5.6 Rankine or Rankine-Gordon formula
• The Rankine formula is a combination of the Euler and crushing loads
for a strut
1
1
1
=
+
PR Pe PC
• For very short struts Pe is very large; 1/Pe can therefore be neglected
and PR=PC. For very long struts Pe is very small and 1/Pe is very large
so that 1/Pc can be neglected. Thus PR=Pe.
• The Rankine formula is therefore valid for extreme values of L/k. It is
also found to be fairly accurate for the intermediate values in the range
under consideration.
31
• Thus, re-writing the formula in terms of stresses,
1
1
1
=
+
σA σ e A σ y A
i.e
• For a strut with both ends pinned
σe + σy
1 1
1
=
+
=
σ σe σy
σ eσ y
σ=
σ eσ y
σe + σy
=
σy
[1 + ( σ y / σ e )]
π 2E
σe =
( L / k )2
σy
σ=
σy
1 + 2 ( L / k )2
π E
Rankine stress
σR =
σy
1 + a( L / k )2
.......... ..Eq.11
• Where a= y/π²E, theoretically, but having a value normally found by
experiment for various materials.
32
• Therefore, The Rankine load is :
PR =
σyA
2
1 + a( L / k )
.......... .......... .......... .Eq.12
• Typical values of a for use in the Rankine formula are given in Table
5.1 below. However, since the values of a are not exactly equal to the
theoretical values, the Rankine loads for long struts will not be
identical to those estimated by the Euler theory as suggested earlier.
33
Table 5.1 Value of ‘a’ for different materials
34
5.7 STRUTS WITH INITIAL CURVATURE
Perry-Robertson Formula
• The Perry-Robertson proof is based on the assumption that any
imperfections in the strut, through faulty workmanship or material
or eccentricity of loading, can be allowed for by giving the strut an
initial curvature. For ease of calculation this is assumed to be a cosine
curve, although the actual shape assumed has very little effect on the
result.
• Consider, therefore, the strut AB of figure 5.8, of length L and pinjointed at the ends. The initial curvature yo at any distance x from the
centre is then given by :
πx
y O = C0 cos
L
35
Figure 5.8: Strut with initial curvature
• If a load P is now applied at the ends, this deflection will be increased to
y+y0.
d 2y
πx
BMC = EI 2 = _ P( y + Co cos )
L
dx
d 2y P
πx
( y + Co cos ) = 0
2 +
EI
L
dx
The solution of which is
2
36
P
P
PCO
πx π
P
y = A sin ( )x + B cos ( )x + [(
cos ) /( 2 _ )
EI
EI
EI
L
L EI
• Where A and B are the constants of integration.
Now when x = ±L/2, y = 0
A=B=0
PC0
πx π 2 P
πx π 2EI
y = [(
cos ) /( 2 _ )] = [( PC0 cos ) /( 2 _ P)]
EI
L
L
L EI
L
• Therefore dividing through, top and bottom, by A,
P
πx π 2EI P
y = [( C0 cos ) /( 2 _ )]
A
L
LA A
• But P/A = σ and (π²EI)/(L²A) = σe (the Euler stress for pin-ended
struts)
σ
πx
y=
CO cos
(σ e _ σ )
L
37
• Therefore total deflection at any point is given by :
σ
πx
πx
y + yO = [
]CO cos
+ CO cos
(σ e _ σ )
L
L
σe
πx
=[
]C0 cos .......... .......... .......... ........Eq.13
(σ e _ σ )
L
• Maximum deflection (when x = 0), Bending Moment (BM) and Maximum
stress due to bending are:
σe
Deflection = [
]C .......... .......... ........Eq.14
(σ e _ σ ) O
B.M = P[
σe
]CO .......... .......... .......... .....Eq.15
(σ e _ σ )
My P
σe
Stress =
= [
]C h
I
I (σ e _ σ ) 0
where h is the dis tan ce of the outside
fibre from the N.A of the strut
38
• Therefore the maximum stress owing to combined bending and axial is
given by:
σ max
P
σe
P
= [
]C0h + .......... .......... .....Eq.16
I (σ e _ σ )
A
P
σe
P
=
]C0h +
2[
A
Ak ( σ e _ σ )
ησ e
CO h
= σ[
+ 1] where η = 2
(σ e _ σ )
k
• If σmax=σy, the compressive yield stress for the material of the strut, the
above equation when solved for σ gives
σ=
[ σ y + ( η + 1)σ e ]
2
_ {[
σ y + ( η + 1)σ e
2
( Perry-Robertson formula.(Eq.17) )
]2 _ σ y σ e }
39
• Thus for an initial curvature with a central deflection Co,
σe
Pe
Maximum Deflection = [
]CO = [
]CO ..........Eq.19
(σ e _ σ )
Pe _ P
σe
PPe
Maximum B.M = P[
]CO = [
]CO .......... ........Eq.20
(σ e _ σ )
Pe _ P
σ max
P
P σ e C0 h P
PPe hCO
= ±[
]
= ±[
]
........Eq.21
A (σ e _ σ ) I
A Pe _ P I
where h is the dis tan ce of the outside fibre from the
N.A of the strut
40
5.8 STRUTS WITH ECCENTRIC LOAD
• For eccentric loading at the ends of a strut, Ayrton and Perry suggest
that the Perry-Robertson formula can be modified by replacing Co by
(Co+1.2e) where e is the eccentricity.
eh
η = η + 1.2 2 .......... .......... .......... .........Eq.22
k
'
• And η’ replaces η in the original Perry-Robertson Equation.
41
A.
Pinned Ends -- The Smith-south well Formula
• Consider a strut loaded as shown in Figure 5.9. In this case there is
strictly no ‘buckling’ load as previously described since the strut will
bend immediately when the load is applied, bending taking place
about the other principal axis.
42
Figure 5.9 : Strut with eccentric load(Pinned Ends)
• Applying a similar procedure to that of the previous:
Maximum deflection,
nL
sin
nL
nL
2
δ+e =e
+ e cos
= e sec
.......... .......Eq.23
nL
2
2
cos
2
nL
Maximum B.M = P( δ + e) = Pe sec
.......... ...Eq.24
2
Maximum stress owing to bending
2
My
nL h
=
= Pe sec
×
I
2
I
Where h is the dis tan ce from the N.A to the highest
stressed fibre.
43
• Therefore the total maximum compressive stress to combined bending and
thrust, assuming a ductile material, is given by [Smith-Southwell
formula]
σ max
P
nL h
= + ( Pe sec
)
A
2 I
eh
nL
= σ[1 + 2 sec
]
2
k
eh
L P
= σ[1 + 2 sec
]...............Eq.25
2 EI
k
• For a brittle material (i.e weak in tension it is the maximum tensile
stress which becomes the criterion for failure and the bending and direct
stress components are opposite in sign.
σ max
eh
L
P
= σ[1 + 2 sec
( 2 ) ]................... ....Eq.26
2 Ek
k
44
B. One End Fixed, The Other Free
• Consider the strut shown in Figure 2.10 below:
Figure 2.10: Strut with eccentric load (One fixed and free
• In such case the Smith-Southwell formula can be written in the form
σ max
eh
L
σ
= σ[1 + 2 sec
( 2 ) ].................Eq.27
2 Ek
k
45
5.9 LATERALLY LOADED STRUTS
a. Central concentrated Load
With the origin at the centre of the strut as shown in Figure 5.11.
d 2y
W L
B.M at C = EI 2 = _ Py _ ( _ x )
2 2
dx
d 2y
_W L
2
( _ x)
2 +n y =
2EI 2
dx
Figure 5.11
46
• The maximum deflection occurs where x is zero,
y max
W
nL nL
=
[tan
_
]......... .......... ......Eq.28
2nP
2
2
• The maximum B.M acting on the strut is at the same position and is
given by
WL
Mmax = _ Py max _
2 2
_W
nL
=
tan
.......... .......... .......... ..Eq.29
2n
2
47
b. Uniformly distributed load
Consider now the uniformly loaded strut of Figure 2.12 with the origin
again selected at the centre but y measured from the maximum
deflected position.
48
• The maximum moment and also can be given by :
• Refer Mechanics of Materials, E.J. Hearn. Page 45-48 for the
derivation of the above equations.
49