buckling 19

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1Departemen Teknik MesinFakultas Teknik– Universitas Indonesia

DTM FTUI

BUCKLING(Tekukan)


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Stability of Structures

• In the design of columns, cross-sectional area is selected such that

- allowable stress is not exceeded

allAP

- deformation falls within specifications

specAEPL

• After these design calculations, may discover that the column is unstable under loading and that it suddenly becomes sharply curved or buckles.

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• Consider model with two rods and torsional spring. After a small perturbation,

moment ingdestabiliz 2

sin2

moment restoring 2

LPLP

K

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

LKPP

KLP

cr4

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• Assume that a load P is applied. After a perturbation, the system settles to a new equilibrium configuration at a finite deflection angle.

sin4

2sin2

crPP

KPL

KLP

• Noting that sin < , the assumed configuration is only possible if P > Pcr.

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Euler’s Formula for Pin-Ended Beams• Consider an axially loaded beam.

After a small perturbation, the system reaches an equilibrium configuration such that

02

2

2

2

yEIP

dxyd

yEIP

EIM

dxyd

• Solution with assumed configuration can only be obtained if

2

2

2

22

2

2

rLE

ALArE

AP

LEIPP

cr

cr


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Euler’s Formula for Pin-Ended Beams

s ratioslendernesrL

tresscritical srLE

ALArE

AP

AP

LEIPP

cr

crcr

cr

2

2

2

22

2

2

• The value of stress corresponding to the critical load,

• Preceding analysis is limited to centric loadings.

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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

rLE

LEIP

e

ecr

ecr


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Extension of Euler’s Formula

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ExampleAn 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


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Example

• Buckling in xy Plane:

127.0

1212

,

23121

2

aL

rL

araabba

AIr

z

ze

zz

z

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.

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Example

• Buckling in xz Plane:

12/2

1212

,

23121

2

bL

rL

brbabab

AI

r

y

ye

yy

y

• Most efficient design:

27.0

12/2

127.0

,,

ba

bL

aL

rL

rL

y

ye

z

ze

35.0ba


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Example

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

2cr

cr

6.138psi101.10

0.35lbs 12500

6.138psi101.10

0.35lbs 12500

kips 5.12kips 55.2

6.13812in 202

122

bbb

brLE

bbAP

PFSP

bbbL

rL

e

cr

cr

y

e

in. 567.035.0

in. 620.1

ba

b

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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

2max

2

2

12

sece

crcr L

EIPPPey

EIPePy

dxyd

• The deflection become infinite when P = Pcr

• Maximum stress

rL

EAP

rec

AP

rcey

AP

e21sec1

1

2

2max

max


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Eccentric Loading; The Secant Formula

rL

EAP

rec

AP e

Y 21sec1 2max

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ProblemThe 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.

.psi1029 6E

SF=2


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ProblemSOLUTION:

• Maximum allowable centric load:

in. 192 ft 16ft 82 eL

- Effective length,

kips 1.62

in 192in 0.8psi 1029

2

462

2

2

ecr

LEIP

- Critical load,

2in 3.54kips 1.312kips 1.62

AP

FSPP

all

crall

kips 1.31allP

ksi 79.8

- Allowable load,

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Problem• Eccentric load:

in. 939.0my

122

secin 075.0

12

sec

crm P

Pey

- End deflection,

22sec

in 1.50in 2in 75.01

in 3.54kips 31.1

2sec1

22

2

crm P

Prec

AP

ksi 0.22m

- Maximum normal stress,


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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, crdepends on both Y and E.

- for small Le/r, cr is determined by the yield strength Y and not E.

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Design of Columns Under Centric LoadStructural Steel

American Inst. of Steel Construction

• For Le/r > Cc

92.1

/ 2

2

FS

FSrLE cr

alle

cr

• For Le/r > Cc

3

2

2

/81/

83

35

2/1

c

e

c

e

crall

c

eYcr

CrL

CrLFS

FSCrL

• At Le/r = Cc

YcYcr

EC

22

21 2


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Design of Columns Under Centric LoadAluminum

Aluminum Association, Inc.

• Alloy 6061-T6Le/r < 66:

MPa /868.0139

ksi /126.02.20

rL

rL

e

eall

Le/r > 66:

23

2 /MPa 10513

/ksi 51000

rLrL eeall

• Alloy 2014-T6Le/r < 55:

MPa /585.1212

ksi /23.07.30

rL

rL

e

eall

Le/r > 66:

23

2 /MPa 10273

/ksi 54000

rLrL eeall

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Problem

Using the aluminum alloy 2014-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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Problem

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.

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Problem

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gyration of radius

radiuscylinder

2

4 cc

cAI

r

c

• For L = 750 mm, assume L/r > 55

• Determine cylinder radius:

mm44.18

c/2m 0.750MPa 103721060

rLMPa 10372

2

3

2

3

2

3

cc

N

AP

all

• Check slenderness ratio assumption:

553.81

mm 18.44mm750

2/

cL

rL

assumption was correct

mm 9.362 cd


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Problem• For L = 300 mm, assume L/r < 55

• Determine cylinder radius:

mm00.12

Pa102/m 3.0585.12121060

MPa 585.1212

62

3

c

ccN

rL

AP

all

• Check slenderness ratio assumption:

5550

mm 12.00mm 003

2/

cL

rL

assumption was correct

mm 0.242 cd

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Design of Columns Under an Eccentric Load

• Allowable stress method:

allIMc

AP

• 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,

IMc

AP

bendingcentric

max


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t

Calculation of Punch Strength (example)

1. Calculating of punching force

)σstrength x tensile0.8(τ material of resistance Shearing : τ

[mm] thicknessMaterial : t [mm]length profile Punching : tτ Pmax

[kgf] Pmax. force punching Maximum

B

mm 37.7 12 3.14 d π tτ Pmax ln

: mm 1 of thicnessa with (hard) plate steelsilicon ain punched be tois mm 12 ofdiameter a with hole round

a when force punching maximum thecalculate To 1] [Example

kgf 2110 56 1 37.7 Pmax Thus, .2kgf/mm 56 is τresistance shearing that theis 1 Table fromresult The

Ref. Product Information : Technical Data of Dicoat Punch.

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2. Fracture of punch tip

d / t 4 P/A ) 4/ d : punch tip round afor ( mm punch tip of area sectional Cross A

) t d : punch tip round afor ( kgf force, punching Maximum PP/A

][kgf/mm punch tipon exerted Stress

2

2

2


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Ref. Product Information : Technical Data of DicoatPunch.


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Minimum punching diameter

dmin :diameter punching Minimum

][kgf/mm steel toolofstrength Fatigue : / t 4 dmin

2

2kgf/mm 97 is punching, of shots 100,000 tosubjected

SKH51, of strength fatigue that theis Fig.1 fromresult The mm. 2.1

26/97 2 4 / t 4 dmin

:punch SKH51 a using moreor shots 100,000 mm 2 of thicknessa with plates

SPCCpunch topossibleit makes that diameter punching minimum obtain the To [Example]

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As expected from Euler’s formula, to improve buckling strength P, it is recommended that a stripper guide and amaterial with higher Young’s modulus ( SKD, SKH, and HAP in increasing order of Young’s modulus ) be used,and that the punch tip length be reduced.The buckling load P represents the value of the load when a buckled punch fractures. When selecting a punch,therefore, the safety factor should be taking account.When selecting a punch for small holes, special attention should be paid to buckling load and stress exerted on the punch.[Example] To obtain the buckling load P for SHAL5 – 60 – P2.00 – BC30, the punch is used without a stripper guide:

)4(B / I E n [kgf] P load Buckling 22

56000 : V3023000 : HAP4022000 : SKH5121000 : SKD11

][kgf/mm modulus sYoung' : E 4

30 Blength Punch tip 2.0 ddiameter Punch tip kgf 190

kgf/mm 22000 E 30/785.0220001 SKH51 : materialPunch B / I E n P

222

22

If the tip is too long, a punch will be buckle under a relatively small load.(when B = 13 [standard punch tip length], the buckling load is 1010 kgf.)

[mm]length Punch tip : B64/d Ipunch round afor

][mm inertia ofmoment Secondary : Iguidestripper With ; 2 n

guidestripper Without ; 1 n t Coefficien :n

4

4

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Buckling and traverse tests• Test conditions

As shown in [Fig.3], buckling loads and traverse rupture loads were applied to test pieces at velocity of 1 mm/min. (the

traverse rupture loads were applied to the point 0.5 mm fromthe end of a punch tip, using a knife-edge-shaped presser), and respective maximum load under which the punch rupture was found.

• Test resultsBoth buckling strength and traverse rupture strength increasein the order of SKD11, SKH51, and HAP40. Especially, HAP40 maintains high hardness, and thus it is excellent in

compression resistance. The metal structure of HAP40 is very fine, and it contains high alloy components (W, V, Co, etc), which makes the HAP40 punch excellent in toughness. Therefore, the HAP40 punch is most suitable for punching that may cause fracture of chipping.

Since the TD processed punches are degraded in base material hardness, they are also slightly low in buckling strength and traverse rupture strength.

Ref. Product Information : Technical Data of Dicoat Punch.


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References

1. Ferdinand P. Beer, E. Russell Johnston & John T. DeWolf ; Mechanics of Material, McGraw-Hill; 2002

2. Bernard J. Hamrock. Fundamental of Machine Elements. Mc-Graw Hill.1999

3. R.S Khurmi & J.K Gupta. A Text Book of Machine Element. Eurasia Publishing House. 1987

4. Popov. Mechanics of Material. Prentice-Hall, 1976.

5. Timoshenko. Strength of Material. 1965

6. Product Information : Technical Data of Dicoat Punch.

buckling 19

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