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Bài giảng Strength of materials - Chương 11: Columns - Nguyễn Sỹ Lâm

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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  
P
    
all  
A
- deformation falls within specifications  
PL  
  
  
spec  
AE  
• After these design calculations, may discover  
that the column is unstable under loading and  
that it suddenly becomes sharply curved or  
buckles.  
11.1 STABILITY OF STRUCTURES  
STABILITY OF STRUCTURES  
• Consider model with two rods and torsional  
spring (stiffness is K). After a small perturbation,  
K
2  
restoring moment  
L
L
P sin P destabilizing moment  
2
2
• Column is stable (tends to return to aligned  
orientation) if  
L
P K  
2  
2
4K  
P P   
cr  
L
L
L
P sin P   
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.  
L
P sinK  
2  
2
PL P  
sin  
4K P  
cr  
• Noting that sin< , the assumed  
configuration is only possible if P > Pcr.  
L
P sin  
2
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  
2
d y M  
P
  y  
2
EI  
EI  
dx  
2
d y P  
y 0  
2
EI  
dx  
• Solution with assumed configuration  
can only be obtained if  
2
EI  
P P   
cr  
2
L
2
2
2
P
E  
Ar  
E  
    
cr  
2
2
A
L A  
L r  
11.1 STABILITY OF STRUCTURES  
EULER’S FORMULA FOR PIN-ENDED BEAMS  
• The value of stress corresponding to  
the critical load,  
2
EI  
P P   
cr  
2
L
P
P
cr  
A
    
cr  
A
2
2
E  
Ar  
  
cr  
2
L A  
2
E  
critical stress  
2
L r  
L
slenderness ratio  
r
• 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,  
2
EI  
P   
cr  
2
L
e
2
E  
  
cr  
2
L r  
e
L 2L equivalent length  
e
11.1 STABILITY OF STRUCTURES  
EXTENSION OF EULER’S FORMULA  
0.7  
0.5  
1  
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  
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.  
• Buckling in xy Plane:  
3
1
12  
2
ba  
I
a
a
2
z
r    
r   
z
z
A
ab  
12  
12  
L
0.7L  
e,z  
• Most efficient design:  
r
a 12  
z
L
L
e,y  
e,z  
r
r
• Buckling in xz Plane:  
z
y
3
1
12  
2
ab  
I
0.7L  
2L  
b
b
y
2
r    
r   
y
y
a 12 b/ 12  
A
ab  
12  
12  
L
a 0.7  
2L  
b/ 12  
a
b
e,y  
0.35  
b 2  
r
y
11.1 STABILITY OF STRUCTURES  
EXAMPLE 11.01  
• Design:  
L
e
2L  
2
20 in  
138.6  
r b 12 b 12  
b
y
P   
FS  
P   
2.55 kips  
12.5 kips  
cr  
P
12500 lbs  
cr  
  
cr  
A
0.35b b  
2
2
6
E  
10.110 psi  
  
cr  
2
2
L r  
138.6 b  
e
2
6
12500 lbs   
10.110 psi  
L = 20 in.  
2
0.35b b  
138.6 b  
E = 10.1 x 106 psi  
P = 5 kips  
b 1.620 in.  
a 0.35b 0.567 in.  
FS = 2.5  
a/b = 0.35  
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.  
• The deflection become infinite when P = Pcr  
2
d y Py Pe  
2
EI  
dx  
y
2
P  
2 P  
EI  
e sec  
1  
P   
cr  
max  
2
L
cr  
e
• Maximum stress  
P
y
e c  
max  
1  
Remind:  
sec(x) = 1/cos(x)  
cse(x) = 1/sin(x)  
max  
2
A
r
P
ec  
1 P L  
e
2 EA r  
1sec  
2
A
r
11.2 ECCENTRIC LOADING; THE SECANT FORMULA  
ECCENTRIC LOADING; THE SECANT FORMULA  
P
ec  
1 P L  
e
2 EA r  
1sec  
max  
Y
2
A
r
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:  
- Effective length,  
L 2  
8ft 16 ft 192 in.  
e
- Critical load,  
2
2
6
4
EI   
29 10 psi8.0 in  
P   
cr  
2
2
L
192 in  
e
62.1kips  
- Allowable load,  
P
62.1kips  
2
cr  
P 31.1kips  
P   
all  
all  
FS  
P
31.1kips  
all  
8.79 ksi  
  
2
A
3.54 in  
11.2 ECCENTRIC LOADING; THE SECANT FORMULA  
EXAMPLE 11.02  
• Eccentric load:  
- End deflection,  
P  
y e sec  
1  
m
2 P  
cr  
2 2  
0.075 in  
sec  
1  
y 0.939 in.  
m
- Maximum normal stress,  
P
ec  
P  
2 P  
1sec  
m
2
A
r
cr  
31.1 kips  
0.75 in 2 in  
2 2  
1  
sec  
2
2
3.54 in  
1.50 in  
22.0 ksi  
m
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 small Le/r, cr is  
determined by the yield  
strength Y and not E.  
- for intermediate Le/r, cr  
depends on both Y and E.  
11.3 DESIGN OF COLUMNS UNDER CENTRIC LOAD  
DESIGN OF COLUMNS UNDER CENTRIC LOAD  
• For Le/r > Cc  
Structural Steel  
2
E  
cr  
FS  
  
  
cr  
all  
American Inst. of Steel Construction  
2
L / r  
e
FS 1.92  
• For Le/r < Cc  
2
L / r  
cr  
FS  
e
1  
  
all  
cr  
Y
2
2C  
c   
3
5 3 L / r 1 L / r  
e
e
FS    
3 8 C  
8 C  
c
c
• At Le/r = Cc  
2
2E  
2
1
  
C   
c
cr  
Y
2
Y
11.3 DESIGN OF COLUMNS UNDER CENTRIC LOAD  
DESIGN OF COLUMNS UNDER CENTRIC LOAD  
• Alloy 6061-T6  
Aluminum  
Le/r < 66:  
Aluminum Association, Inc.  
 
20.2 0.126  
L / r ksi  
all  
e
139 0.868  
L / r  
 
MPa  
e
Le/r > 66:  
3
51000 ksi 35110 MPa  
  
all  
2
2
L /r  
L /r  
e
e
• Alloy 2014-T6  
Le/r < 55:  
  
30.7 0.23  
L / r ksi  
all  
e
212 1.585  
L / r  
  
MPa  
e
Le/r > 66:  
3
54000 ksi 37210 MPa  
  
all  
2
2
L /r  
L /r  
e
e
11.3 DESIGN OF COLUMNS UNDER CENTRIC LOAD  
EXAMPLE 11.03  
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.  
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  
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