•
COLUMNS: BUCKLING (DIFFERENT
ENDS)
Slide No. 1
Buckling of Long Straight
Columns
„
Example 4
A simple pin-connected truss is loaded and
supported as shown in Fig. 12. All
members of the truss are WT102 × 43
sections made of structural steel with a
modulus of elasticity of 200 GPa and a
yield strength of 250 MPa. Determine (a)
the factor of safety with respect to failure
by slip, and (b) the factor of safety with
respect to failure by buckling.
1
Slide No. 2
Buckling of Long Straight
Columns
„
Example 4 (cont’d)
Figure 12
D
E
3m
A
C
B
4m
15 kN
4m
30 kN
Slide No. 3
Buckling of Long Straight
Columns
„
Example 4 (cont’d)
Free-body diagram:
D
E
3m
A
B
θ
4m
15 kN
θ
C
RE
RCy
RCx
4m
30 kN
2
Slide No. 4
Buckling of Long Straight
Columns
„
Example 4 (cont’d)
Find the reactions RE and RC:
∑M
+ ∑M
+↑ ∑F
+
C
= 15(8) + 30(4) − RE (3) = 0; ⇒ RE =80 kN
E
= 15(8) + 30(4 ) + RCx (3) = 0; ⇒ RCx = − 80 kN
y
= 0; RCy − 30 − 15 = 0; ⇒ RCy = 45 kN
Note that,
FDE = RE = 80 kN (T)
θ = tan −1
3
= 36.87 0
4
Slide No. 5
Buckling of Long Straight
Columns
„
Example 4 (cont’d)
FDC
At pin C:
FBC
45
36.87
+ ↑ ∑ Fy = 0; FDC sin 36.87 + 45 = 0
C
80
FDC = −75 kN = 75 kN (C)
+ → ∑ Fx = 0; − FBC − 80 − FDC cos 36.87 = 0
⇒ − FBC − 80 − (− 75)(0.8) = 0
⇒ FBC = −20 kN = 20 kN (C)
3
Slide No. 6
Buckling of Long Straight
Columns
„
FDB
Example 4 (cont’d)
At joint B:
B
FAB
FDB = 30 kN
FAB = −20 kN = 20 kN (C)
20
30
At Joint A:
FAD
+ ↑ ∑ Fy = 0;
A
− 15 + FAD sin 36.87 = 0
∴ FAD = 25 kN (T)
36.87
20
15
Slide No. 7
Buckling of Long Straight
Columns
„
Example 4 (cont’d)
– Thus, the forces in the truss are as follows:
D
3m
A
kN
25
(
T)
20 kN (C)
80 kN (T)
75
30 kN (T)
B
4m
15 kN
kN
E
(C
)
C
20 kN (C)
4m
30 kN
4
Slide No. 8
Buckling of Long Straight
Columns
„
Example 4 (cont’d)
(a) Factor of safety with respect to slip:
• For WT102 × 43, A = 5515 mm2 (Appendix B)
• For member DE, the critical member that gives
the largest force:
(
)
Pmax = σ yield A = 250 ×10 6 5515 × 10 −6 = 1379 × 103 N = 1379 kN
∴ FS =
Pmax 1379
=
= 17.24
FDE
80
Slide No. 9
Buckling of Long Straight
Columns
„
Example 4 (cont’d)
(b) Factor of safety with respect to failure by buckling:
• For WT102 × 43, rmin = 26.2 mm (Appendix B)
• For member DC, the critical member that gives
the largest compressive force:
L
5,000
=
= 190.84 (slender)
rmin
26.2
Pcr =
π 2 EA
(L / rmin )2
∴ FS =
=
π 2 (200 ×109 )(5515 ×10 −6 )
(190.84)2
= 298.9 kN
298.9
Pcr
=
= 3.99 ≅ 4
75
FDC
5
Slide No. 10
Effects of Different Idealized
End Conditions
„
General Notes On Column Buckling
1. Boundary conditions other than simplysupported will result in different critical loads and
mode shapes.
2. The buckling mode shape is valid only for small
deflections, where the material is still within its
elastic limit.
3. The critical load will cause buckling for slender,
long columns. In contrast, failure will occur in
short columns when the strength of material is
exceeded. Between the long and short column
Slide No. 11
Effects of Different Idealized
End Conditions
„
General Notes On Column Buckling
limits, there is a region where buckling occurs
after the stress exceeds the proportional limit but
is still below the ultimate strength. These
columns are classified as intermediate and their
failure is called inelastic buckling.
4. Whether a column is short, intermediate, or long
depends on its geometry as well as the stiffness
and strength of its material. This concept is
addressed in the columns introduction page.
6
Slide No. 12
Effects of Different Idealized
End Conditions
„
The Concept of Effective Length
– The Euler buckling formula, namely Eqs. 9
or 12 was derived for a column with
pivoted ends.
– The Euler equation changes for columns
with different end conditions, such as the
four common ones found in Figs.13 and 14.
– While it is possible to set up the differential
equation with appropriate boundary
Slide No. 13
Effects of Different Idealized
End Conditions
Figure 13
7
Slide No. 14
Effects of Different Idealized
End Conditions
Figure 14
(Beer and Johnston 1992)
Slide No. 15
Effects of Different Idealized
End Conditions
„
ENES 220 ©Assakkaf
The Effective Length Concept
– Conditions to determine the Euler buckling
formula for each case, a more common
approach makes use of the concept of an
“effective length”.
– The pivoted ended column, by definition,
has zero bending moments at each end.
– The length L in the Euler equation,
therefore, is the distance between
8
Slide No. 16
Effects of Different Idealized
End Conditions
„
The Effective Length Concept
– Successive points of zero bending
moment.
– All that is needed to modify the Euler
column formula for use with other end
conditions is to replace L by L.′
′ defined as the effective length of the
– Lis
column.
Slide No. 17
Effects of Different Idealized
End Conditions
„
The Effective Length Concept
Definition:
The effective length L′ (or Le) of a column
is defined as the distance between
successive inflection points or points of
zero moment.
9
Slide No. 18
Effects of Different Idealized
End Conditions
„
The Effective Length Concept
Based on the effective length concept, the
Euler Buckling load formula becomes
π 2 EI
(16a)
Pcr =
L2e
or
Pcr =
Le = L′ = effective length
π 2 EA
(Le / r )
2
(16b)
Slide No. 19
Effects of Different Idealized
End Conditions
„
Example 5
What is the least thickness a rectangular
wood plank 4 in. wide can have, if it is used
for a 20-ft column with one end fixed and
one end pivoted, and must support an axial
load of 1000 lb? Use a factor of safety
(FS) of 5. The modulus of elasticity of
wood is 1.5 × 106 psi.
10
Slide No. 20
Effects of Different Idealized
End Conditions
„
Example 5 (cont’d)
The rectangular cross section is
b = 4 in
t?
I=
bt 3
12
For the end conditions specified in the
problem, Fig. 13d (or Fig. 14c) gives
Le = 0.7 L
Slide No. 21
Effects of Different Idealized
End Conditions
„
Example 5 (cont’d)
Figure 13
11
Slide No. 22
Effects of Different Idealized
End Conditions
„
Example 5 (cont’d)
Since FS = 5 is required,
but
Pcr = (FS )P = 5(1000) = 5,000 lb
 4t 3 

π (1.5 ×10 )
12
π 2 EI

 = 5,000
Pcr =
=
2
Le
[0.7(20 ×12)]
From which
2
6
t = 3.06 in
Slide No. 23
Effects of Different Idealized
End Conditions
„
Example 6
An L102 × 76 × 6.4-mm aluminum alloy (E
= 70 GPa) angle is used for fixed-end,
pivoted-end column having an actual
length of 2.5 m. Determine the maximum
safe load for the column if a factor of safety
1.75 with respect to failure by buckling is
specified.
12
Slide No. 24
Effects of Different Idealized
End Conditions
„
Example 6 (cont’d)
For fixed-end, pivoted-end column, Fig. 13d
(or Fig. 14c) gives the equivalent length as
Le = 0.7 L = 0.7(2.5) = 1.75 m = 1,750 mm
For L102 × 76 × 6.4 section (see Fig. 15 or
Appendix B of textbook):
A = 1090 mm 2
rmin = 16.5 mm
Slide No. 25
Effects of Different Idealized
End Conditions
Figure 15
„
Example 6 (cont’d)
13
Slide No. 26
Effects of Different Idealized
End Conditions
„
Example 6 (cont’d)
Therefore,
L′ 1750
=
= 106.06 (slender)
rmin 16.5
π 2 EA
Pmax =
=
2
Pcr (L′ / rmin )
π 2 EA
=
=
2
FS
FS
FS (L′ / rmin )
π 2 (70 × 109 )(1090 ×10 −6 )
1.75(106.06)
2
= 38,254.54 N = 38.3 kN
Slide No. 27
Effects of Different Idealized
End Conditions
„
Example 7
Determine the maximum load that a 50mm × 75-mm × 2.5-m long aluminum alloy
bar (E = 73 GPa) can support with a factor
of safety of 3 with respect to failure by
buckling if it is used as a fixed-end, pivoted
column.
For fixed-end, pivoted column, Fig. 13d (or Fig. 14c) gives
effective length, L′ = 0.7 L = 0.7(2.5) = 1.75 m
14
Slide No. 28
Effects of Different Idealized
End Conditions
„
Example 7 (cont’d)
Computation of the section properties:
75 mm
A = 75(50 ) = 3,750 mm 2
ba 3 75(50 )
=
= 0.7813 ×106 mm 4
12
12
3
I min =
50 mm
rmin
0.7813 × 106
I min
=
= 14.43 mm
3750
A
Slide No. 29
Effects of Different Idealized
End Conditions
„
Example 7 (cont’d)
Therefore, the slenderness ratio of the
column can be obtained:
L′ 1.75 ×103
=
= 121.28 (slender)
rmin
14.43
and
Pmax =
(
)(
)
π 2 EA
π 2 73 ×109 3750 ×10 −6
Pcr
=
=
= 61.2 kN
2
FS FS (L′ / rmin )
3(121.28)
15
Slide No. 30
Effects of Different Idealized
End Conditions
„
Example 8
A structural steel (E =29,000 ksi) column
20 ft long must support an axial
compressive load of 200 kip. The column
can be considered pivoted at one end and
fixed at the other end for bending about
one axis and fixed at both ends for bending
about the other axis. Select the lightest
wide-flange or American standard section
that can be used for the column.
Slide No. 31
Effects of Different Idealized
End Conditions
„
Example 8 (cont’d)
First case: fixed-end, pivoted column
Lex = Lx′ = 0.7(20 ) = 14 ft
Pcr =
π 2 EI x
Lx′2
Pcr Lx′2 200(14 ×12 )
= 19.72 in 2
Ix = 2 = 2
π E
π (29,000)
2
16
Slide No. 32
Effects of Different Idealized
End Conditions
„
Example 8 (cont’d)
Second case: fixed ends column
Ley = L′y = 0.5(20 ) = 10 ft
200(10 × 12 )
= 10.06 in 2
Iy = 2 = 2
π E
π (29,000 )
Pcr L′y2
2
Use a W254 × 33 section
Slide No. 33
Effects of Different Idealized
End Conditions
„
Example 9
A 25 mm-diameter tie rod AB and a pipe
strut AC with an inside diameter of 100 mm
and a wall thickness of 25 mm are used to
support a 100-kN load as shown in Fig. 16.
Both the tie rod and the pipe strut are
made of structural steel with modulus of
elasticity of 200 GPa and a yield strength
of 250 MPa. Determine
17
Slide No. 34
Effects of Different Idealized
End Conditions
„
Example 9 (cont’d)
(a) the factor of safety with respect to
failure by slip.
(b) the factor of safety with respect to
failure by buckling
Slide No. 35
Effects of Different Idealized
End Conditions
„
Example 9 (cont’d)
B
Figure 16
2.5 m
A
4.5 m
100 kN
C
6m
18
Slide No. 36
Effects of Different Idealized
End Conditions
„
Example 9 (cont’d)
FAB
y
At joint A:
A
2.5
4.5
θ = tan −1
= 22.62 0 , α = tan −1
= 36.87 0
6
6
θ
x
α
+ → ∑ Fx = 0; FAB cosθ + FAC cos α = 0
FAC
FAB cos 22.62 + FAC cos 36.87 = 0
FAB = −0.8667 FAC
(17)
+ ↑ ∑ Fx = 0; FAB sin θ − FAC sin α − 100 = 0
FAB sin 22.62 − FAC sin 36.87 − 100 = 0
FAB = 1.56 FAC + 260
(18)
Slide No. 37
Effects of Different Idealized
End Conditions
„
Example 9 (cont’d)
Substituting for FAB in Eq. 17 into Eq. 18,
gives
− 0.8667 FAC = 1.56 FAC + 260
Thus,
FAC = −107.14 kN = 107.14 kN (C)
From Eq. 18, with FAC known, we have
FAB = 1.56 FAC + 260
= 1.56(−107.14) + 260 = 92.86 kN (T)
19
Slide No. 38
Effects of Different Idealized
End Conditions
Example 9 (cont’d)
„
Calculate the cross-sectional areas for AB
and AC:
AAB =
π
4
(25)2 = 490.9 mm 2 ,
AAC =
π
4
[(150) − (100) ] = 9817 mm
2
2
2
(a) FS due to Failure by Slip:
σ AB =
FAB 92.86 × 103
= 189.16 × 10 6 N/m 2 = 189.2 MPa
=
−6
AAB 490.9 ×10
FSAB =
250
= 1.32
189.16
Slide No. 39
Effects of Different Idealized
End Conditions
„
Example 9 (cont’d)
σ AC =
FAC − 107.14 ×103
=
= −10.914 × 106 N/m 2 = 10.914 MPa (C)
9817 ×10 −6
AAC
FSAB =
250
= 22.9
10.914
(b) FS due to Failure by Buckling:
• Member AC is the compression member with a
compressive force of 107.14 kN:
π
(150)2 − (100)2 = 19.942 ×106 mm 4
I AC =
64
[
]
20
Slide No. 40
Effects of Different Idealized
End Conditions
„
Example 9 (cont’d)
(b) FS for Buckling (cont’d):
r=

L 
=
r
19.942 × 106
I
=
= 45.07
9817
A
(6)2 + (4.5)2 ×1000 
 = 166.41 (slender)
45.07
π 2 EA π 2 200 ×109 9817 × 10 −6
=
= 699.8 × 103 N ≅ 700 kN
Pcr =
(L / r )2
(166.41)2
700
P
FSAC = cr =
= 6.53
FAC 107.14
(
)(
)
21