11 columns 2015 bách khoa

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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

- deformation falls within specifications

spec

AE

PL 

destabiliz 2

sin 2

moment restoring

L P K

L P

L P

• Column is stable (tends to return to aligned orientation) if

L

K P

P

K

L P

cr

4

2 2

STABILITY OF STRUCTURES

• Assume that a load P is applied After a

perturbation, the system settles to a new equilibrium configuration at a finite

2

sin 2

PL

K

L P

 sin 2

L P

• Noting that sin <  , the assumed configuration is only possible if P > P cr

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

P dx

y d

y EI

P EI

M dx

y d

• Solution with assumed configuration can only be obtained if

 

 2

2 2

2 2

2 2

r L

E A

L

Ar E A

P

L

EI P

EULER’S FORMULA FOR PIN-ENDED BEAMS

 

 

s ratio

slendernes r

L

tress

critical s r

L E

A L

Ar E

A

P A

P

L

EI P

P

cr

cr cr

cr

2 2 2

2 2

2 2

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

r L E L

EI P

e

e cr

e cr







EXTENSION OF EULER’S FORMULA

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

EXAMPLE 11.01

• Buckling in xy Plane:

12

7 0

12 12

,

2

3 12

1 2

a

L r

L

a r

a ab

ba A

I r

z

z e

z

z z

12 12

,

2

3 12

1 2

b

L r

L

b r

b ab

ab A

I r

y

y e

y

y y

12 /

2 12

7 0

, ,







b a b

L a

L

r

L r

L

y

y e z

z e

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

6 2

2

2 cr

cr

6 138

psi 10 1 10 0.35

lbs 12500

6 138

psi 10 1 10 0.35

lbs 12500

kips 5 12 kips

5 5 2

6 138 12

in 20 2 12 2

b b

b

b r

L E

b b A

P

P FS P

b b

b

L r

L

e

cr cr

y e

0

in.

620 1







b a

b

• 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

P

P e

y

EI

Pe Py

dx

y d

P r

ec A

P

r

c e y

A P

e

2

1 sec 1

1

2

2

max max



Remind:

sec(x) = 1/cos(x)

cse(x) = 1/sin(x)

P r

ec A

Y

2

1 sec 1

2



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

in 192

in 0 8 psi 10

29

2

4 6

2 2

- Critical load,

2

in 3.54

kips 1 31

2

kips 1 62

P P

all

cr all



kips 1 31



all

P

ksi 79 8





- Allowable load,

• Eccentric load:

in.

939 0

2

sec in

075 0

1 2

sec in

1.50

in 2 in 75 0 1 in 3.54

kips 31.1

2

sec 1

2 2

ec A

P

ksi 0 22

• Previous analyses assumed stresses below the proportional limit and initially straight,

homogeneous columns

• Experimental data demonstrate

- for large L e /r, cr follows Euler’s formula and depends upon E but not Y

- for intermediate L e /r, cr

depends on both Y and E

- for small L e /r, cr is determined by the yield strength Y and not E

all e

2

/ 8

1 /

8

3 3 5

2

/ 1

e

cr all

c

e Y

cr

C

r L C

r L FS

FS C

cr

E C

126 0 2 20

r L

r L

e

e all

/

MPa 10

51 3 /

ksi 51000

r L r

23 0 7 30

r L

r L

e

e all

/

MPa 10

2 7 3 /

ksi 54000

r L r

Using the aluminum alloy2014-T6,

determine the smallest diameter rod

which can be used to support the centric

slenderness ratio regime to utilize

• Calculate required diameter for assumed slenderness ratio regime

• Evaluate slenderness ratio and verify initial assumption Repeat if necessary

2 4

gyration of

radius

radius cylinder

2

4

c c

c A

I r



c

L r

L

assumption was correct

mm 9 36

c/2

m 0.750

MPa 10

372 10

60

r L

MPa 10

372

2

3 2

3

2 3

• For L = 300 mm, assume L/r < 55

• Determine cylinder radius:

mm 00 12

Pa

10 2

/

m 3 0 585 1 212 10

60

MPa 585

1 212

6 2

N

r

L A

/   



c

L r

L

assumption was correct

mm 0 24

2 

d

USING TABULATED SLENDERNESS RATIOS

A I

L r



  

From Lambda, based on material, tabulated slenderness ratios have been

established They are called 

Iteration 1: assume 0  0.5 calculate 0 0  calculate

A I

L r



' 0

A I

L r



' 1



interpolate

2

' 0 0

If error between and are smaller than 5% then A1 will be chosen If not, go

to the next iteration

1

1



• Allowable stress method:

all

I

Mc A

• Interaction method:

all Pcentric A  all Mcbending I 1

• 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

bending centric