failure of slender and stocky columns (2nd year)

Failure of slender and stocky columns

Dr Alessandro Palmeri <[email protected]>

Teaching schedule Week Lecture 1 Staff Lecture 2 Staff Tutorial Staff 1 Beam Shear Stresses 1 A P Beam Shear Stresses 2 A P --- --- 2 Shear centres A P Basic Concepts J E-R Shear Centre A P 3 Principle of Virtual

forces J E-R Indeterminate Structures J E-R Virtual Forces J E-R

4 The Compatibility Method

J E-R Examples J E-R Virtual Forces J E-R

5 Examples J E-R Moment Distribution -Basics

J E-R Comp. Method J E-R

6 The Hardy Cross Method

J E-R Fixed End Moments J E-R Comp. Method J E-R

7 Examples J E-R Non Sway Frames J E-R Mom. Dist J E-R 8 Column Stability 1 A P Sway Frames J E-R Mom. Dist J E-R 9 Column Stability 2 A P Unsymmetric Bending 1 A P Colum Stability A P 10 Unsymmetric Bending 2 A P Complex Stress/Strain A P Unsymmetric

Bending A P

11 Complex Stress/Strain A P Complex Stress/Strain A P Complex Stress/Strain

A P

Christmas Holiday

12 Revision 13 14 Exams 15 2

Mo@va@ons (1/5)

•  Load-‐carrying structures may fail in a variety of ways, depending upon: –  Type of structure (truss, frame, …) –  Condi@ons of support (pinned, fixed, …) –  Loads applied (sta@c, dynamic, …) –  Materials used (briQle, duc@le, …)

•  Failures are prevented by designing structures so that maximum stresses (strength criterion) and maximum displacements (s,ffness criterion) remain within admissible limits

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Mo@va@ons (2/5)

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•  For the fans of The Big Bang Theory: –  Sheldon and Howard have got this seriously wrong!

•  You can’t use the Young’s modulus to quan@fy the strength of material, but its s,ffness!

Mo@va@ons (3/5)

•  S@ffness and strength of materials –  In the stress-‐strain curve for a duc@le material (e.g. steel), the Young’s modulus E defines the s@ffness, while the yield stress σy represents the strength

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Mo@va@ons (4/5)

•  S@ffness criterion: “Slender Column”

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•  Strength criterion: “Short Column”

Mo@va@ons (5/5)

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Coventry Cathedral

ç Slender column

Detail of the support è

Learning Outcomes

•  When we have completed this unit (2 lectures + 1 tutorial), you should be able to:

– Derive the Euler’s cri@cal load for slender pinned-‐pinned columns in compression

– Predict the mode of failure for both short and slender columns in compression

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Further reading

•  R C Hibbeler, “Mechanics of Materials”, 8th Ed, Pren@ce Hall – Chapter 13 on “Buckling of Column”

•  T H G Megson, “Structural and Stress Analysis”, 2nd Ed, Elsevier – Chapter 21 on “Structural Instability” (eBook)

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Short and Slender Struts

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•  Increasing the length of a strut reduces its buckling load –  For instance, a matchs,ck is reasonably strong in compression (lek), but a longer s,ck, with the same cross sec@on and the same material, would be weaker and buckles in compression (right)

–  The slenderness of a strut plays an important role in its mode of failure in compression

Buckling, i.e. Lateral Instability (1/2)

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•  That is, if a column is rela@vely slender, it may deflect laterally when subjected to a compressive force P (Fig (a)) and fail by bending (Fig (b)), rather than failing by direct compression of the material

Buckling, i.e. Lateral Instability (2/2)

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•  Pcrit is the so-‐called cri,cal buckling load –  If the axial load P is less than Pcrit, bending is caused by lateral loads only –  If P is greater than Pcrit, the ruler bends even without lateral loads

Euler’s Cri@cal Load for Pinned-‐Pinned Slender Columns

•  One of the Learning Outcomes of this Unit is for you to become able to mathema@cally derive (and remember as well) the expression of the cri@cal load Pcrit for pinned-‐pinned slender column

•  Pcrit=PE is oken called Euler’s buckling load –  Aker the Swiss mathema@cian Leonhard Euler (1707-‐1783)

!!Pcrit =

π 2EImin

L2

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Mathema@cal Deriva@on: Bending Equa@on

•  What’s the equa@on ruling the beam’s downward deflec@on, uz(x), for a given bending moment diagram, My(x)?

•  We used this second-‐order differen@al equa@on in part A to calculate the beam’s deflec@on under transverse loads…

•  where, as usual: –  E= Young’s modulus –  Iyy= Second moment of area about the horizontal neutral axis

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EIyyd2uz (x)dx2

= −My (x)

Mathema@cal Deriva@on: Sign Conven@on

•  Do you remember from last year?

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Mathema@cal Deriva@on: P-‐Delta (1/2)

•  What’s the bending moment My in this circumstance?

•  We don’t have transverse loads this @me –  the column is subjected to the axial load P only

•  How can we have a bending moment?

•  In order to derive the expression of the Euler’s buckling load, we need to assume that –  a disturbance/imperfec@on exists in the column, –  therefore the buckling occurs –  and My can be consistently evaluated by using the equilibrium

equa@ons in the deformed shape 16

Mathema@cal Deriva@on: P-‐Delta (1/2)

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Deformed shape

Equilibrium condi@on P

P My= P uz

EIyy

z

uz

Mathema@cal Deriva@on: Buckling Equa@on (1/2)

•  Knowing the bending moment My in the deformed shape:

•  we can subs@tute it within the deflec@on equa@on:

•  This equa@on can be rewriQen as:

•  Where α is a posi@ve quan@ty, given by:

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My = Puz

EIyyd2uzdx2

= −Puz

d2uzdx2

+α 2uz =0

α = PEIyy

Mathema@cal Deriva@on: Buckling Equa@on (2/2)

•  What do we do in order to solve an ordinary differen@al equa@on?

•  First, we find the general solu,on, which contains as many integra@on constants as the order of the differen@al equa@on (two, in this case)

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uz= C

1cos(αx) + C

2sin(αx)

Mathema@cal Deriva@on: Boundary Condi@ons (1/2)

•  Second, we apply the boundary condi,ons (BCs) to get the values of the integra@ons constants for the par@cular case

–  For two unknown constants, C1 and C2, two BCs are needed!

•  For a pinned-‐pinned column, the BCs read:

•  uz=0 @ x=0 (i.e. the transverse transla@on is prevented at the lek-‐hand side end)

•  uz=0 @ x=L (i.e. the transverse transla@on is prevented at the right-‐hand side end as well)

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EIyy

z

Mathema@cal Deriva@on: Boundary Condi@ons (2/2)

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•  The applica@on of the first BC is quite straighrorward

uz(x) = C

1cos(αx) + C

2sin(αx)

uz= 0@ x = 0 1 20 1 0C C⇒ = × + ×

1 0C⇒ =

General solu5on

Boundary condi5on

Mathema@cal Deriva@on: Non-‐Trivial Solu@on (1/3)

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•  The second BC does require more effort

•  Trivial solu@on: –  It would follow y=0 for any value of the abscissa x –  No transverse displacements would occur (straight column) –  This solu@on is therefore unacceptable

•  Non-‐trivial solu@on:

uz(x) = C

1cos(αx) + C

2sin(αx)

uz= 0@ x = L 20 sin( )C Lα⇒ =

sin( ) 0Lα⇒ = L nα π⇒ =

nnLπα α⇒ = =

2 0C⇒ =


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•  Recalling now the expression of the parameter a, one obtains:

•  The associated modes of instability, for n= 1, 2, 3, …, are sinusoidal func@ons, having a total number n of peaks and valleys

PEIyy

= nπL

⇒ PEIyy

= n2 π 2

L2⇒ P

n= n2

π 2 EIyy

L2

n= 3

n= 2

n= 1


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•  Larger values of the buckling load are associated to more complicated modes of instability

•  Theore@cally, these modes could be achieved if roller supports are applied at the points of contraflexure

•  However, in prac@ce, the lower value P1 is never exceeded

Mathema@cal Deriva@on: Euler’s Buckling Load

•  The actual cri,cal load, i.e. the so-‐called Euler’s buckling load, is the “engineering solu@on”, which is the minimum among the mathema@cal solu@ons P1, P2, P3, …, and is obtained for n=1

•  Moreover, in order to be truly the minimum, you must use the minimum value of the second moment of area, which might not be Iyy

•  The laQer expression is very important in Structural Engineering –  You are requested to remember it –  You must be able to derive this expression as well

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Pcrit

= PE= P

1=π 2 EI

yy

L2

2min

E 2EIPL

π=

FEM-‐Computed Modes of Instability

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•  Euler’s buckling load (PE= P1= 181 kN)

•  Higher buckling load in the orthogonal direc@on

•  (P4= 3,518 kN)

XY X

Z

Y

ZXY X

Z

Y

Z

4min 330 cmI = = 4

max 6572 cmI

Horizontal sway

Ver0cal deflec0on

Effects of the Boundary Condi@ons (1/2)

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The more the column’s ends are restrained, the higher is the buckling load

Similar sinusoidal shapes are

observed for different BCs

Effects of the Boundary Condi@ons (2/2)

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(a)  Pinned-‐pinned (b)  Can@levered (c)  Fixed-‐fixed (d)  Propped

L0 is the distance between two consecu@ve crosses of the horizontal axis

Effec@ve Length •  It is useful to introduce the concept of equivalent length, Le=k L as

the length of a pinned-‐pinned column having the same Euler’s cri@cal load

•  We therefore must know the value of the coefficient k for different BCs

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k=2

k=1 k=0.7 k=0.5

Can5levered Pinned-‐ pinned Propped

Fixed-‐ fixed

2min

E 2e

EIPL

π=

Stocky Columns (1/2) •  If we divide the Euler’s cri@cal load PE by the cross sec@onal area A, we get

the so-‐called Euler’s cri@cal stress σcrit:

•  This is the maximum normal stress which is allowable to prevent buckling instability, and is inversely propor@onal to the square of the equivalent length Le

•  If we introduce the parameter ρmin as the minimum radius of gyra@on of the cross sec@on, and then the slenderness ra@o λ=Le/ρmin, the above equa@on can be rewriQen as:

30 σcrit

=π 2 E I

minA( )

Le2

=π 2 E ρ

min2

Le2

= π 2 E

Le

ρmin( )2

= π 2

λ2Eρ

min=Imin

A

2min

crit 2e

EP EIA AL

πσ = =

Stocky Columns (2/2)

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•  The s@ffer the material, i.e. the larger the Young’s modulus E, the higher is σcrit:

•  The larger the slenderness ra@o λ, the lower is σcrit, i.e. very slender columns will have very low values of σcrit

•  Conversely, stocky columns, with a small slenderness ra@o λ, will not experience the buckling failure, as the yielding of the material is likely to happen first:

2

crit 2 Eπσλ

=

crit y Material’s yield stressfσ > =

Strength and S@ffness Criteria (1/2)

•  “Strength” criterion

•  Stocky columns tend to fail because the elas@c limit of the material is reached

•  The safety checks is:

•  “S,ffness” criterion

•  Slender columns tend to fail because the elas@c configura@on is unstable

•  The safety check is:

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2min

E 2e

EIP PL

π< =y yP P f A< =

Both must be sa0sfied

Strength and S@ffness Criteria (2/2)

•  For briQle materials such as concrete, the yielding stress fy is replaced with the crushing stress fc

•  The safety check then reads:

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2min

E 2e

EIP PL

π< =P < Pc= f

cA

Both must be sa0sfied

fc

Strength and S@ffness Criteria

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•  The Rankine’s failure load PR combines these two different criteria, therefore taking into account both material and geometrical nonlineari@es

•  PR=Py for λ=0

•  PR approaches PE as λ goes to infinity

0 100 200 300 4000.0

0.5

1.0

1.5

2.0

l

PêPy Py

PE PR

PR=PyPE

Py+ P

Eλ

P/P y

Rankine (1820-‐1872) was a Scoush civil engineer, physicist and mathema@cian

Ul@mate Normal Stress

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•  … experimentally derived (dots) for wide-‐flange steel columns

•  … as a func@on of the slenderness λ= k L/ρmin

λ

Key Learning Points 1.  Columns in compression may fail because

–  Insufficient bending s@ffness: è Lateral buckling –  Insufficient axial capacity: è Yielding/Crushing

2.  Euler’s buckling load PE depends on: –  Minimum second moment of area, Imin

–  Length of the column, L –  Boundary condi@ons

3.  Interac@on between lateral buckling and axial capacity can be taken into account through the (approximate) Rankine’s formula

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è Effec@ve length, Le

PR=PyPE

Py+ P

E

PE =π 2EImin

Le2

failure of slender and stocky columns (2nd year)

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