buckling and tension field beam for aerospace structures

1

Aero Structures-Column Buckling

By

Dr. Mahdi Damghani

2016-2017

2

Suggested Readings

Reference 1 Reference 2Chapter 9 of Ref [1] Chapters 8 and 9 of Ref [2]

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Topics

• Familiarisation with buckling of columns with various boundary conditions

• Spar function

• Spar loading

• Buckling of web of spars

4

Introduction

• Buckling is a stability issue (not strength issue)

• Compressive stress is high enough to trigger sudden sideways deflection of slender structures

• Buckling stress is less than yield stress of material (so material does not fail but becomes unstable)

• A large proportion of an aircraft’s structure comprises thin webs stiffened by slender stringers (stiffeners, longerons wing skin, spar webs, etc)

5

Introduction

6

Introduction

7

Introduction

8

Introduction

9

Introduction

10

Buckling of columns

• The first significant contribution to the theory of the buckling of columns was made as early as 1744 by Euler

• His classical approach is still valid, and likely to remain so, for slender columns possessing a variety of end restraints.

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Buckling of columns

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Reminder

z2z

1z

The bending moment M causes the length of beam to bend about a centre of curvature C

Element is small in length and a pure moment is applied. The curved shape can be assumed to be circular with a radius of curvature R measured to the neutral plane

Stresses and strains are zero on neutral axis

Stresses and strains are zero on neutral axis so its length does not change

Planes at an angle of

z

13

Reminder

z2z

1z

z

Fibre ST has shortened in length whilst NQ increased in length so they have gone through strains

Remember that the length of neutral axis does not change and remains as z

Length OriginalLengthin Change

z

z

RyRz

Ry

RRyR

z

Positive y gives negative strain, i.e. compression

z zyR

z

zz E RyEz

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Reminder

• Look at the beam cross section now;

dA

Neutral axis

y

RyEz

AAA A

z dAyREydA

RyEydAFyM 2

REIM

dA y

2

21dz

ydEIR

EIM

15

Reminder

• Solving second order homogenous differential equations;

16

Reminder

02

EIPm CR

EIPi

EIPm

EIPi

EIPm

CRCR

CRCR

2

1

EIP

EIP CRCR 2 zBzAev z sincos0

zBzAv sincos

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Buckling of columns (hinged-hinged)

?, BA

0,0@ vlzz,...)3,2,1(;0sin0sin

0

nnlllB

A

2

2

lEIPCR

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Buckling of columns (hinged-hinged)

• These higher values of buckling load cause more complex modes of buckling

• If no restraints are provided, then these forms of buckling are unstable and have little practical meaning

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Buckling of columns (hinged-hinged)

• We like to work with stresses so;

2

2

lEIPCR

APCR

CR

AIl

ECR

2

2

2

2

2

2

2

rl

E

rl

ECR

• r is radius of gyration

• l/r is slenderness ratio

AIr • Taking

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Column buckling (clamped-clamped)

• In practice, columns usually have their ends restrained against rotation so that they are, in effect, fixed

• An axial compressive load that has reached the critical value, PCR, so that the column is in a state of neutral equilibrium

• The ends of the column are subjected to clamping moments, MF, in addition to axial load

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Column buckling (clamped-clamped)

0@00@

vlxvx

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Column buckling (clamped-clamped)

v is indeterminate as we do not know the value of

MF

But we know slope (dv/dx) at x=l is zero

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Boundary condition effects

• So far we have seen that for a hinged-hinged column the buckling load is;

• For a clamped-clamped column is;

• What do you make of this?

2

2

lEIPCR

2

24l

EIPCR

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Column buckling (general)

• In a more general form buckling stress can be written as;

• le is called effective length which is multiple of the length of structure depending on the boundary conditions

klle

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Boundary conditions

1k

5.0k

7.0k

2k

1k

2k

klle

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Example 1

• A 7m long steel tube having the cross section shown is to be used as a pin-ended column. Determine the maximum allowable axial load the column can support so that it does not buckle or yield. Take the yield stress of 250MPa.

27

Solution

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Example 2

• A 2m long pin ended column with a square cross section is to be made of wood. Assuming E=13GPa and allowable stress of 12MPa (σall=12MPa) and using a factor of safety of 2.5 to calculate Euler’s critical load for buckling. Determine the size of the cross section if the column is to safely support a 100 kN load.

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Solution

For a square of side a;

Now let’s check stress in the column;

P kNFSPCR 2501005.2100

ElPI

lEIP CR

CR 2

2

2

2

46

92

23

10794.71013

210250 mPa

mNI

464

3 10794.71212

1 maaaI mmmma 1003.98

MPa

mkN

AP 10

100.0100

2 1210 all

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Example 3

• A uniform column of length L and flexural stiffness EI is simply supported at its ends and has an additional elastic support at mid-span. This support is such that if a lateral displacement vc occurs at this point, a restoring force kvc is generated at the point. Derive an equation giving the buckling load of the column.

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Solution

M

00@

vz

cvvLz

2/@

0/2/@

dzdv

Lz

0A

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Example 4

• A uniform column of length l and bending stiffness EI is built-in at one end and free at the other and has been designed so that its lowest flexural buckling load is P. Subsequently, it has to carry an increased load, and for this, it is provided with a lateral spring at the free end. Determine the necessary spring stiffness k so that the buckling load becomes 4P.

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Solution

• Buckled state of the column and acting force of spring

M

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Solution

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Tutorial 1

• The structural member shown is to be used as a pin-connected column. Determine the largest axial load it can support before it either begins to buckle or the steel yields. Section properties are as;

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Tutorial 2

• A 3m column with the following cross section is constructed of material with E=13GPa and is simply supported at its two ends.

• Determine Euler buckling load

• Determine stress associated with the buckling load

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Tutorial 3

• A uniform, pin-ended column of length l and bending stiffness EI has an initial curvature such that the lateral displacement at any point between the column and the straight line joining its ends is given by

• Show that the maximum bending moment due to a compressive end load P is given by

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Spars

Span-wise members that carry shear loads. Fuel Tank Boundary. Provide mounting for WLG Fittings and Leading and Trailing edge fittings.

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Spars loading

F

F

dM

The distance between the centroid of spar caps

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Spar loading

• Spar caps (boom) is assumed to carry axial load only

• Spar web is assumed to carry constant shear stresses (average shear) only

d

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Spar buckling

• Spar web is so thin that it is susceptible to buckling under shear loading

• The web of the beam buckles under the action of internal diagonal compressive stresses produced by shear

• This leads to a wrinkled web capable of supporting diagonal tension only in a direction perpendicular to that of the buckle

• The beam is then said to be a complete tension field beam

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Reminder

• How does compression come about?

)0,( t

),0(

)0,( c

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Complete diagonal tension

• At any section where shear force is S, average shear stress for spar web is;

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Complete diagonal tension

• Let’s isolate element ABCD at a chosen section

• Tensile stresses σt are present on faces AB and CD

• The angle of diagonal tension is α

• On vertical plane FD we have both shear and direct stresses

• Let’s write equation of equilibrium for element FDC; 0yF cosFDCD

WS

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Complete diagonal tension

• Let’s write equation of equilibrium for element FDC;

0xF cosFDCD

WS

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Complete diagonal tension

• Both tensile stress and shear stress are constant through the depth of the beam

• σz is also constant

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Complete diagonal tension

• Direct forces in the flanges (booms) can be calculated by drawing a free body diagram at length z of the beam;

0xF

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Complete diagonal tension

• The diagonal tensile stress (σt) induces a direct stress (σy) on horizontal planes at any point in the web

• On a horizontal plane HC in the element ABCD, there is a direct stress (σy) and a complementary shear stress

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Reminder

• Why do we have horizontal stress on plane HC?

)0,( t2

),( y

)0,0(

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Complete diagonal tension

0yF

• σy causes compression in the vertical stiffeners so we can find axial compressive force in the stiffener P as of the next slide;

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Complete diagonal tension

bbb

22

y

tantan

dWbPtbP td

W

yy

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Buckling of stiffeners

• If P is high then stiffeners can buckle similar to columns

• Effective length of stiffener is then taken as;

• Load P also brings about another affect. Can any body indicate as to what could happen next?

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Bending of flanges

y

ty

max5.0 MM mid

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Angle α

Adjusts itself for potential energy to become minimum

If stiffeners and flanges are assumed to be rigid then α=45o

In real life nothing is rigid and therefore

38o<α< 45o

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Angle α

• The angle can be calculated as follow;

Uniform direct compressive stress

in the stiffener

Uniform direct compressive stress

in the flange

We know

We get

AF and AS are cross sectional area of flange and

stiffener, respectively

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Example 5

• The beam is assumed to have a complete tension field web. If the cross-sectional areas of the flanges and stiffeners are, respectively, 350mm2 and 300mm2 and the elastic section modulus of each flange is 750mm3

• Determine the maximum stress in a flange

• Determine whether or not the stiffeners will buckle

• The thickness of the web is 2mm, and the second moment of area of a stiffener about an axis in the plane of the web is 2,000mm4; E = 70,000N/mm2.

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Solution

58

Solution

21 /7.50

35017700 mmN

24

2 /9.114750

106.8 mmN

221 /6.165 mmNTotal

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Solution

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Solution

bbb

22

y

tand

WbP

buckling No CRPP

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Tutorial 1

• The beam shown in the next slide is clamped at one end and has a point load of 7.0kN at the other end. The beam is assumed to have a complete tension field web. The cross-sectional areas of the flanges and stiffeners are, respectively, 385mm2 and 300mm2 and the elastic section modulus of each flange is 750mm3. The thickness of the web is 2mm, and the second moment of area of a stiffener about an axis in the plane of the web is 1,500mm4. Assume modulus of elasticity of material E = 71,700N/mm2.

• Determine the maximum stress in a flange

• Determine whether or not the stiffeners will buckle

62

Tutorial 1

63

Tutorial 2

• A simply supported beam has a span of 2.4m and carries a central concentrated load of 10 kN. The flanges of the beam each have a cross-sectional area of 300mm2, while that of the vertical web stiffeners is 280mm2. If the depth of the beam, measured between the centroids of area of the flanges, is 350mm and the stiffeners are symmetrically arranged about the web and spaced at 300mm intervals, determine the maximum axial load in a flange and the compressive load in a stiffener. It may be assumed that the beam web, of thickness 1.5mm, is capable of resisting diagonal tension only.