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1Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
BUCKLING(Tekukan)
2Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
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.
2
3Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
Stability of Structures
• 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
22
4Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
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
2sin2
crPP
KPL
KLP
• Noting that sin < , the assumed configuration is only possible if P > Pcr.
3
5Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
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
6Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
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.
4
7Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
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
8Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
Extension of Euler’s Formula
5
9Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
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
10Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
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.
6
11Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
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
12Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
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
7
13Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
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
14Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
Eccentric Loading; The Secant Formula
rL
EAP
rec
AP e
Y 21sec1 2max
8
15Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
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
16Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
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,
9
17Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
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,
18Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
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.
10
19Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
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
20Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
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
11
21Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
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
22Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
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.
12
23Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
Problem
24
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
24Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
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
13
25Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
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
26Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
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.
14
27Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
27
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
28Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
28
15
29Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
Ref. Product Information : Technical Data of DicoatPunch.
30Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
30
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]
16
31Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
31
32Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
32
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
17
33Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
33
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.
34Departemen Teknik MesinFakultas Teknik– Universitas Indonesia
DTM FTUI
34
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.