Transcript
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COLUMN – TOPANGCOLUMN – TOPANG
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CRITICAL LOADCRITICAL LOAD• Long slender members subjected to
an axial compressive force are called columnscolumns.
• The lateral deflection that occurs is called bucklingbuckling.
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CRITICAL LOAD, PCRITICAL LOAD, Pcrcr
• The maximum axial load that a column can support when it is on the verge of buckling is called the CRITICAL LOAD, Pcr or sometimes called as Euler Load, Pe.
• Any additional load will cause the column to buckle and therefore deflect laterally
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To Understand The Concept:To Understand The Concept:
• When the bar are in the vertical position, the spring having the stiffness k, is unstretched.
• FBD, the bar are displaced by pin at A which produce force F=k∆.
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The spring will produce The spring will produce the force, the force,
Restoring spring force Restoring spring force become:become:
Applied load P will Applied load P will develops two horizontal develops two horizontal
components,components,
Since Since is small, is small,and tanand tan==
2L
kF
P2P2 x
tanPx
P
2
LkF
7
4kLPcr
•Stable equilibriumThe force developed by the spring would be adequate to restore the bars back to their vertical position.
•Unstable equilibriumIf the load P is applied and a slight displacement occurs at A, the mechanism will tend to move out of equilibrium and not be restored to its original positions.
•Neutral equilibriumAny slight disturbance given to the mechanism will not cause it to move further out of equilibrium, nor will it be restored to its original position. Instead, the bars will remain in the deflected position.
4kLP
4kLP
•If the restoring force is greater than disturbing force
PLk 22
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• When the critical load Pcr is reached, the column is on the verge of becoming unstable, so that a small lateral force F will cause the column to remain in the deflected position when F is removed.
• In order to determine the critical load and buckled shape of the column, the followed equation is used:-
• Deflection y and internal moment M are in positive direction
Mdx
ydEI
2
2
1.Column With Pin Support
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• With M=-Py
2
1
2
2
2
2
2
2
2
2
EI
Por
EI
Pwith
)3_________(0ydx
yd
or
)2_______(0yEI
P
dx
yd
)1_________(Pydx
ydEI
• Equation 3 is homogeneous, second order, linear differential equation
which is similar to simple harmonic equation.
Using methods of differential equation or by direct substitution,general solution for equation 3 can be written as follows:
)4____()x(sinB)x(cosAy
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• At x=0, y=0, then from equation (4), A=0 and equation (4) become as:-
y=B sin (x)_____(5)
• At x=L, y=0, then from equation (5):-
0= B sin (L)
• If B=0, It means no deflection occur in the column. Therefore, B0, but:-
sin (L)=0 (L)= , 2, 3, 4, …….n, if , n0
• The smallest value of P is obtained when n=1, • (L)= or
• so critical load for this column is:
L.EI
P 2
1
)6(__________L
EIP
2
2
cr
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Least Moment of InertiaLeast Moment of Inertia
• A column will buckle about the principal axis of the cross section having the least moment of inertia (the weakest axis).
• As in picture, the column will buckle at the a-a axis not the b-b axis.
• Pcr=critical or max axial load on the column just before it begins to buckle. This loads must not cause the stress in the column to exceed the proportional limit.
• E=modulus of elasticity material.
• I=least moment of inertia for the column cross-sectional area.
2
2
L
EIPcr
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5.Column With Various Types of Supports
Based on all Euler’s Formula for various types of supports, the formula can be written as:
2
e
2
2
2
cr )L(
EI
)KL(
EIP
With K = constant depends on the end support types
= 1, 2, 0.5, and 0.7
EI = column minimum stiffness (kNm2)
L = Column actual length (m)
Le = effective length
P = Pcr=Buckling @ Critical Load (kN@MN)
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2
2
rKL
AEPcr
2
2
rKL
AEPcr
2
2
rKL
AEPcr
2
e
2
2
2
2
2
cr
)L(
EI
)KL(
EIr
KL
AEP
2
2
2
2
2
2
25.0
4
2
L
EI
L
EI
L
EIPcr
2
2
2
2
2
2
4
25.0
5.0
L
EI
L
EI
L
EIPcr
2
2
2
2
2
2
2
49.0
7.0
L
EI
L
EI
L
EIPcr
2
2
2
2
2
2
cr
L
EI1L
EI
L
EIP
Effective Length :Le=KLEffective Length :Le=KL
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Column Buckling Stress
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• Curve hyperbolic valid for critical stress below yield point.
• Eg:
• The smallest acceptance slenderness ratio for steel.(L/r=89)
• If (L/r>89) euler’s formula can be used however if (L/r<89) euler formula not valid.
Ycr
steelY MPa
250
SLENDERNESS RATIO,L/r
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Example:Example:
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Example:Example: The aircraft link is made from an A-36 steel rod. Determine The aircraft link is made from an A-36 steel rod. Determine the smallest diameter of the rod, to the nearest mm, that the smallest diameter of the rod, to the nearest mm, that will support the load of P=4kN without buckling. The ends will support the load of P=4kN without buckling. The ends are pin connected. are pin connected. E=210GPa, E=210GPa, σσyy=250MPa=250MPa
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SOLUTION:SOLUTION:
m8m71.7d
300)1(
64
d10210
)10(4
)300(1
64
d10210
)KL(
EIP
,formulas'EulerApplying
.columnendsportedsuprollerfor1Kand64
d
2
d
4I
:LoadBucklingCritical
2
432
3
2
432
2
2
cr
44
I
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Check!:Check!:
validisformulas'Euler,Therefore
MPa250MPa6.798
4
)10(4
A
PY
2
3
crcr
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22
23
24
25
26
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EXERCISE 1:EXERCISE 1:
SolutionSolution
Solution (Contd.)Solution (Contd.)
EXERCISE 2:EXERCISE 2:
SolutionSolution
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• The Euler formula was derived with the assumptions that the load P is always applied through the centroid of the columns’s cross-sectional area and that the column is perfectly straight.
• This is quite unrealistic since manufactured columns are never perfectly straight.
• In reality, columns never suddenly buckle; instead they begin to bend although ever so slightly , immediately upon application of the load.
• Therefore, load P will be applied to the column at a short eccentric distance e from the centroid of the cross section.
The Secant Formula
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The Secant Formula
yePM
Mdx
ydEI
2
2
Internal moment in the column:
Differential equation for the deflection curve:
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GENERAL SOLUTIONGENERAL SOLUTION
1cossin2tan
as; written curve deflection thehence,2tan1
2sin2cos2sin2sin
2sin2cos1
:ince
sin
cos11
0,
.2
0,0
cos2sin1
2
2
xEIPxEIPLEIPey
LEIPeC
LEIPLEIPLEIP
and
LEIPLEIP
s
LEIP
LEIPeC
yLxwhen
eC
yxwhen
ConditionBoundary
exEI
PCx
EI
PCy
0y 0,e
12sec
:yy ,2
Lwhen x
Therefore, midpoint. scolumn' at theoccur
stress maximum and deflection Maximum
:Deflection Maximum
max
max
max
if
LEI
Pey
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The Secant FormulaThe Secant Formula
2sec
max
LEI
PPeM
yePM
EAP
r
L
r
ec
A
P
ArI
LEI
PI
Pec
A
PI
Mc
A
P
2sec1
2sec
;
2max
2
max
max
Max stress in the column is caused by both the axial load and the moment:-
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Exercise 1Exercise 1• The W250x18 structural A-36
steel column is used to support a load of 4 kN. If the column is fixed at the base and free at the top, determine:
i. The deflection at the top of the column due to the loading.
ii. The maximum stress in the column.
E=210GPa, y=250MPa.
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Solution (i):Solution (i):
mmy
KL
EI
P
KKL
EI
Pey
deflectionMaximum
mmdmmrmmImmA
Wforpropertiestion
xx
136.21)1455.0(sec200
1455.02
)5000(0.2
)10)(5.22)(10(210
)10(4
2
0.21)2
(sec
:
2513.99)10(5.222280
;18250sec
max
63
3
max
462
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.OKMPa252.6
)1455.0(sec5455.2175.1
1455.0)2280)(10(210
)10(4
)3.99(2
)5000)(0.2(
EA
P
r2
KL
5455.23.99
2
251200
r
ec
MPa75.12280
104
A
P
)EA
P
r2
KL(sec
r
ec1
A
P
mm251dmm3.99rmm)10(5.22Imm2280A
;18250Wforpropertiestionsec
Ymax
max
3
3
22
3
2max
x
46
x
2
Solution (ii):Solution (ii):
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Exercise 2aExercise 2a
• The W360 X 39 structural A-36 steel member is used as a column that is assumed to be fixed at its top and its bottom. If the 15 kN load is applied at an eccentric distance of 250 mm. Determine the MAXIMUM STRESS in the column. E = 210 GPa, σy = 250 MPa.
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Solution:Solution:
.OKMPa250MPa72.9
)03317.0(sec2189.2102.3
03317.0)4960)(10(210
)10(15
)143(2
)1000)(5(5.0
EA
P
r2
KL
2189.2)143(2
363250
r
ec;MPa02.3
4960
)10(15
A
P
5.0KEA
P
r2
KLsec
r
ec1
A
P
:axisxxaboutYielding
mm143rmm)10(102Imm363dmm4960A
:39360WforpropertiesSection
Ymax
max
3
3
22
3
2max
x
46
x
2
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• Solve the problem if the column is fixed at its top and pinned at its bottom.
Exercise 2bExercise 2b
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Solution:Solution:
OKMPa250MPa73.9
)04644.0(sec2189.2102.3
04644.0)4960)(10(210
)10(15
)143(2
)1000)(5(7.0
EA
P
r2
KL
2189.2)143(
2
363250
r
ec;MPa02.3
4960
)10(15
A
P
7.0KEA
P
r2
KLsec
r
ec1
A
P
:axisxxaboutYielding
mm143rmm)10(102Imm363dmm4960A
:39360WforpropertiesSection
Ymax
max
3
3
22
3
2max
x
46
x
2
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Exercise 3aExercise 3a• A W360 x 45 structural A-36 steel
column is pin connected at its ends and has a length L = 5 m. Determine the maximum eccentric load P that can be applied so the column does not buckle or yield. Compare this value with an axial critical load P applied through the centroid of the column. E = 210 GPa, σy = 250 MPa.
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Solution:Solution:
OKMPa250MPa9.26
)5710)(10)(210(
68544
)146(2
5000sec
146
2
352150
15710
68544
EA
P
r2
KLsec
r
ec1
A
P
,formulaantsectheApplying:axisxxaboutYielding
.OKMPa250MPa0.125710
68544
A
P
ifvalidonlyisformulas'Euler:stressCritical
kN54.68N68544)5000(
)10)(16.8)(10)(210(
)KL(
EIPP'P
,FormulaEulerApplying:axisyyaboutBuckling
mm5000)1000)(5(1)KL()KL(
:1K,endsbothatpinnedcolumnaFor
mm352dmm)10(16.8Imm146rmm5710A
:45360WforpropertiesSection
Ymax
32max
2max
ycr
cr
Ycr
2
632
y2
2
cr
xy
46
yx
2
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Exercise 3bExercise 3b
• Solve the problem if the column is fixed connected at its ends.
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