Giáo trình Strength of Materials - Chapter 11: Columns

CHAPTER 11: COLUMNS 
11.1 Stability of Structures 
11.2 Eccentric Loadings; The Secant Formula 
11.3 Design of Columns Under Centric Load 
11.4 Design of Columns Under An Eccentric Load 
11.1 STABILITY OF STRUCTURES 
STABILITY OF STRUCTURES 
• In the design of columns, cross-sectional area is 
selected such that 
- allowable stress is not exceeded 
all
A
P
 
• After these design calculations, may discover 
that the column is unstable under loading and 
that it suddenly becomes sharply curved or 
buckles. 
- deformation falls within specifications 
spec
AE
PL
 
11.1 STABILITY OF STRUCTURES 
STABILITY OF STRUCTURES 
• Consider model with two rods and torsional 
spring (stiffness is K). After a small perturbation, 
 
moment ingdestabiliz 
2
sin
2
moment restoring 2




L
P
L
P
K
 
2
sin
2
L
P
L
P
• Column is stable (tends to return to aligned 
orientation) if 
 
L
K
PP
K
L
P
cr
4
2
2

 
11.1 STABILITY OF STRUCTURES 
STABILITY OF STRUCTURES 
• Assume that a load P is applied. After a 
perturbation, the system settles to a new 
equilibrium configuration at a finite 
deflection angle. 
 



sin4
2sin
2


crP
P
K
PL
K
L
P
sin
2
L
P
• Noting that sin <  , the assumed 
configuration is only possible if P > Pcr. 
11.1 STABILITY OF STRUCTURES 
EULER’S FORMULA FOR PIN-ENDED BEAMS 
• Consider an axially loaded beam. 
After a small perturbation, the system 
reaches an equilibrium configuration 
such that 
0
2
2
2
2


y
EI
P
dx
yd
y
EI
P
EI
M
dx
yd
• Solution with assumed configuration 
can only be obtained if 
 
 2
2
2
22
2
2
rL
E
AL
ArE
A
P
L
EI
PP
cr
cr





11.1 STABILITY OF STRUCTURES 
EULER’S FORMULA FOR PIN-ENDED BEAMS 
 
 
s ratioslendernes
r
L
tresscritical s
rL
E
AL
ArE
A
P
A
P
L
EI
PP
cr
cr
cr
cr
2
2
2
22
2
2










• The value of stress corresponding to 
the critical load, 
• Preceding analysis is limited to 
centric loadings. 
11.1 STABILITY OF STRUCTURES 
EXTENSION OF EULER’S FORMULA 
• A column with one fixed and one free 
end, will behave as the upper-half of a 
pin-connected column. 
• The critical loading is calculated from 
Euler’s formula, 
 
length equivalent 2
2
2
2
2



LL
rL
E
L
EI
P
e
e
cr
e
cr



11.1 STABILITY OF STRUCTURES 
EXTENSION OF EULER’S FORMULA 
1 7.0 5.02
11.1 STABILITY OF STRUCTURES 
EXAMPLE 11.01 
An aluminum column of length L and 
rectangular cross-section has a fixed end at B 
and supports a centric load at A. Two smooth 
and rounded fixed plates restrain end A from 
moving in one of the vertical planes of 
symmetry but allow it to move in the other 
plane. 
a) Determine the ratio a/b of the two sides of 
the cross-section corresponding to the most 
efficient design against buckling. 
b) Design the most efficient cross-section for 
the column. 
L = 20 in. 
E = 10.1 x 106 psi 
P = 5 kips 
FS = 2.5 
11.1 STABILITY OF STRUCTURES 
EXAMPLE 11.01 
• Buckling in xy Plane: 
12
7.0
1212
,
23
12
1
2
a
L
r
L
a
r
a
ab
ba
A
I
r
z
ze
z
z
z


• Buckling in xz Plane: 
12/
2
1212
,
23
12
1
2
b
L
r
L
b
r
b
ab
ab
A
I
r
y
ye
y
y
y


• Most efficient design: 
2
7.0
12/
2
12
7.0
,,



b
a
b
L
a
L
r
L
r
L
y
ye
z
ze
35.0
b
a
SOLUTION: 
The most efficient design occurs when the 
resistance to buckling is equal in both planes of 
symmetry. This occurs when the slenderness 
ratios are equal. 
11.1 STABILITY OF STRUCTURES 
EXAMPLE 11.01 
L = 20 in. 
E = 10.1 x 106 psi 
P = 5 kips 
FS = 2.5 
a/b = 0.35 
• Design: 
 
    
 
 
 
 
 
 
 2
62
2
62
2
2
cr
cr
6.138
psi101.10
0.35
lbs 12500
6.138
psi101.10
0.35
lbs 12500
kips 5.12kips 55.2
6.138
12
in 202
12
2
bbb
brL
E
bbA
P
PFSP
bbb
L
r
L
e
cr
cr
y
e











in. 567.035.0
in. 620.1


ba
b
11.2 ECCENTRIC LOADING; THE SECANT FORMULA 
ECCENTRIC LOADING; THE SECANT FORMULA 
• Eccentric loading is equivalent to a centric 
load and a couple. 
• Bending occurs for any nonzero eccentricity. 
Question of buckling becomes whether the 
resulting deflection is excessive. 
2
2
max
2
2
1
2
sec
e
cr
cr L
EI
P
P
P
ey
EI
PePy
dx
yd


















• The deflection become infinite when P = Pcr 
• Maximum stress 
 


















 

r
L
EA
P
r
ec
A
P
r
cey
A
P
e
2
1
sec1
1
2
2
max
maxRemind: 
sec(x) = 1/cos(x) 
cse(x) = 1/sin(x) 
11.2 ECCENTRIC LOADING; THE SECANT FORMULA 
ECCENTRIC LOADING; THE SECANT FORMULA 













r
L
EA
P
r
ec
A
P e
Y
2
1
sec1
2max

11.2 ECCENTRIC LOADING; THE SECANT FORMULA 
EXAMPLE 11.02 
The uniform column consists of an 8-ft section 
of structural tubing having the cross-section 
shown. 
a) Using Euler’s formula and a factor of safety 
of two, determine the allowable centric load 
for the column and the corresponding 
normal stress. 
b) Assuming that the allowable load, found in 
part a, is applied at a point 0.75 in. from the 
geometric axis of the column, determine the 
horizontal deflection of the top of the 
column and the maximum normal stress in 
the column. 
11.2 ECCENTRIC LOADING; THE SECANT FORMULA 
EXAMPLE 11.02 
SOLUTION: 
• Maximum allowable centric load: 
  in. 192 ft 16ft 82 eL
- Effective length, 
  
 
kips 1.62
in 192
in 0.8psi 1029
2
462
2
2




e
cr
L
EI
P
- Critical load, 
2in 3.54
kips 1.31
2
kips 1.62


A
P
FS
P
P
all
cr
all

kips 1.31allP
ksi 79.8
- Allowable load, 
11.2 ECCENTRIC LOADING; THE SECANT FORMULA 
EXAMPLE 11.02 
• Eccentric load: 
in. 939.0my
  



























1
22
secin 075.0
1
2
sec


cr
m
P
P
ey
- End deflection, 
  
 




























22
sec
in 1.50
in 2in 75.0
1
in 3.54
kips 31.1
2
sec1
22
2



cr
m
P
P
r
ec
A
P
ksi 0.22m
- Maximum normal stress, 
11.3 DESIGN OF COLUMNS UNDER CENTRIC LOAD 
DESIGN OF COLUMNS UNDER CENTRIC LOAD 
• Previous analyses assumed 
stresses below the proportional 
limit and initially straight, 
homogeneous columns 
• Experimental data demonstrate 
- for large Le/r, cr follows 
Euler’s formula and depends 
upon E but not Y. 
- for intermediate Le/r, cr 
depends on both Y and E. 
- for small Le/r, cr is 
determined by the yield 
strength Y and not E. 
11.3 DESIGN OF COLUMNS UNDER CENTRIC LOAD 
DESIGN OF COLUMNS UNDER CENTRIC LOAD 
Structural Steel 
American Inst. of Steel Construction 
• For Le/r > Cc 
 
92.1
/
2
2


FS
FSrL
E cr
all
e
cr




• For Le/r < Cc 
 
3
2
2
/
8
1/
8
3
3
5
2
/
1

















c
e
c
e
cr
all
c
e
Ycr
C
rL
C
rL
FS
FSC
rL 

• At Le/r = Cc 
Y
cYcr
E
C



2
2
2
1 2
11.3 DESIGN OF COLUMNS UNDER CENTRIC LOAD 
DESIGN OF COLUMNS UNDER CENTRIC LOAD 
• Alloy 6061-T6 
Le/r < 66: 
  
  MPa /868.0139
ksi /126.02.20
rL
rL
e
eall


 Le/r > 66: 
   2
3
2
/
MPa 10513
/
ksi 51000
rLrL ee
all


• Alloy 2014-T6 
Le/r < 55: 
  
  MPa /585.1212
ksi /23.07.30
rL
rL
e
eall


 Le/r > 66: 
   2
3
2
/
MPa 10273
/
ksi 54000
rLrL ee
all


Aluminum 
Aluminum Association, Inc. 
11.3 DESIGN OF COLUMNS UNDER CENTRIC LOAD 
EXAMPLE 11.03 
Using the aluminum alloy2014-T6, 
determine the smallest diameter rod 
which can be used to support the centric 
load P = 60 kN if a) L = 750 mm, 
b) L = 300 mm 
SOLUTION: 
• With the diameter unknown, the 
slenderness ration can not be evaluated. 
Must make an assumption on which 
slenderness ratio regime to utilize. 
• Calculate required diameter for 
assumed slenderness ratio regime. 
• Evaluate slenderness ratio and verify 
initial assumption. Repeat if necessary. 
11.3 DESIGN OF COLUMNS UNDER CENTRIC LOAD 
EXAMPLE 11.03 
2
4
gyration of radius 
radiuscylinder 
2
4 c
c
c
A
I
r
c





• For L = 750 mm, assume L/r > 55 
• Check slenderness ratio assumption: 
 
553.81
mm 18.44
mm750
2/

c
L
r
L
 assumption was correct 
mm 9.362  cd
• Determine cylinder radius: 
 
mm44.18
c/2
m 0.750
MPa 103721060
rL
MPa 10372
2
3
2
3
2
3












c
c
N
A
P
all


11.3 DESIGN OF COLUMNS UNDER CENTRIC LOAD 
EXAMPLE 11.03 
• For L = 300 mm, assume L/r < 55 
• Determine cylinder radius: 
mm00.12
Pa10
2/
m 3.0
585.1212
1060
MPa 585.1212
6
2
3




























c
cc
N
r
L
A
P
all


• Check slenderness ratio assumption: 
 
5550
mm 12.00
mm 003
2/

c
L
r
L
 assumption was correct 
mm 0.242  cd
11.3 DESIGN OF COLUMNS UNDER CENTRIC LOAD 
PRACTICAL METHOD FOR DETERMINING REQUIRED SECTION 
USING TABULATED SLENDERNESS RATIOS 
A
I
L
r
L 
 
From Lambda, based on material, tabulated slenderness ratios have been 
established. They are called 
Iteration 1: assume 5.00  calculate calculate  0
0
zNA 
from table 0
'
0
A
I
L
r
L 
 
'
0interpolate 
If error between and are smaller than 5% then A0 will be chosen. If not, go 
to iteration 2. 
Iteration 2: let calculate calculate  1
1
zNA 
from table 
1
'
1
A
I
L
r
L 
 
'
1interpolate 
2
'
00
1




0
'
0
If error between and are smaller than 5% then A1 will be chosen. If not, go 
to the next iteration 
1
'
1
11.4 DESIGN OF COLUMNS UNDER AN ECCENTRIC LOAD 
DESIGN OF COLUMNS UNDER AN ECCENTRIC LOAD 
• Allowable stress method: 
all
I
Mc
A
P

• Interaction method: 
   
1
bendingallcentricall
IMcAP

• An eccentric load P can be replaced by a 
centric load P and a couple M = Pe. 
• Normal stresses can be found from 
superposing the stresses due to the centric 
load and couple, 
I
Mc
A
P
bendingcentric


max
