Lecture 4

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OPen INteractive Structural Lab

Lecture 4 Buckling and Ultimate Strength of Columns

Reference : Ship Structural Design Ch.09

NAOE

Jang, Beom Seon

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Goal

 In Lecture 02 :

 Euler buckling load

 Load eccentricity → Secant formula

 Elastic and Inelastic column behavior

 Tangent-Modulus Theory

 Reduced-Modulus Theory

 Shanley Theory

 In this Lecture:

 Residual stress effect

 Load eccentricity + column geometric eccentricity (initial deflection)

→ Perry-Robertson formula

 Effect of lateral load

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OPen INteractive Structural Lab OPen INteractive Structural Lab

E

(

L_e : Effective Length)

 The Euler buckling load is the load for which an ideal column will first have an equilibrium deflected shape.

 Mathematically, it is the eigenvalue in the solution to Euler’s differential equation

 The “ultimate load” P_ult is the maximum load that a column can carry.

 It depends on

 initial eccentricity of the column, eccentricity of the load

 transverse load, end condition,

 in-elastic action, residual stress.

 P_ult of a practical column is less than P_E

 Buckling - the sudden transition to a deflected shape - only occurs in the case of "ideal" columns.

2 E 2

e

P EI

L

 

2

0 d w Pw

dx  EI 

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Ideal Columns

 If the compressive stress exceeds proportional limit even in ideal column

→ Actual buckling load < Euler load

due to the diminished slope of the stress-strain curve.

 For an ideal column (no eccentricity or residual stress) buckling will occur at the tangent modulus load, given by

( E

_t

: tangent modulus,

the slope of the stress-stain curve corresponding to P

_t

/A)

 Since E_t depends on P_t modulusthe calculation of P_tis generally an iterative process.

 It is more convenient to deal in terms of stress rather than load

→ The effects of yielding and residual stress to be included.

2 2

t t

e

P E I

L

 

2

= 2

( / )

E E

e

P E

A L

 

  ρ : the radius of gyration , I= ρ²A L_e/ ρ : slenderness ratio

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 Load-deflection behavior for practical columns containing residual stress and eccentricity

Slender Column (P_E<P_Y) Stocky Column (P_E>P_Y)

Q: Explain P_t?

2 2

t t

e

P E I

L

 

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Ideal Columns

 The ultimate strength of a column is defined as the average applied stress at collapse

Stress-strain curve Tangent modulus curve

Typical experiment values for straight rolled sections and welded sections

ult ult

= P

 A

2 t

ult ideal 2

( ) =

e

E L

 



 

 

 

due to entirely the compressive

residual stress σ_r

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 a metal part produced by rolling and having any one of a number of cross- sectional shapes.

Rolled section Welded

section

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Residual Stress-Rolled Sections

 The uneven cooling between Residual stresses will result after the cooling because of the non-uniform temperature distribution through the cross section during the cooling process at ambient temperature.

flange root : tesile residual stresses

flange tip : compressive residual stresses, about 80 MPa for mild steel

Typical residual stresses in rolled section

(9)

may commence yielding when applied stress = σ_Y- σ_r

→ particularly detrimental because it occurs in the flange tips.

 The material is no longer homogeneous, and the simple tangent modulus approach is no longer valid.

 A comparatively simple solution can be achieved for the buckling strength in the primary direction by assuming an elastic-perfectly plastic stress-strain relationship.

 The progressive loss of bending stiffens is linearly proportional to the extent of the yielded zone → the use of an average value of tangent modulus.

Typical stub column stress-strain curve

Structural tangent modulus diagram

Q: What is σ_spl : Structural proportional limit?

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Residual Stress-Rolled Sections

 Structural tangent modulus E_tsby Ostenfeld-Bleich Parabola

σ_spl: Structural proportional limit

 Substituting E_tsin place of E_t

 For rolled wide flange section, typical value of σ_spl= 1/2 σ_{Y .}

Johnson parabola for rolled section (essentially straight and pinned end) )

( ) for (

) (

Y av

spl spl

Y spl

av Y

av ts

E

E   



 _ _



 

E L_e _Y

Y spl Y

spl Y

ult 

 

 



 _



 



 



1 1 ²,

2 for

4 , 1

2 



  



Y

ult , 2

4 2

, 1 1 4

2

1   ²  ² ₂  when σ_ult= 1/2 σ_spl

2 t

ult ideal 2

( ) =

e

E L

 



 

 

 

(11)

Basic column curve (Johnson parabola and Perry-Robertson curve

σ_ult= 1/2 σ_spl

Y e

L E L

E

E L



















 

2 2 2

2 2

) / (

1  



2

= 2

( / )

E E

e

P E

A L

 

 

 2

 ²

1 ) (





  _



Y ult E

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Residual Stress-Welded Sections

 Residual thermal stress is induced by uneven internal temperature distribution

Residual stress close to yield stress

in welding direction

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tensile stress in and near the weld which is equal to approximately equal to the yield stress.

 The extent of the tension yield zone (3~6 times thickness) depends mainly on the total heat input, the cross-sectional area of the weld deposit, the type of welding and the welding sequence

.

Typical residual stresses in welded section

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Residual Stress-Welded Sections

 From the equilibrium

ηt : width of the tension yield zone, b : total flange width

= 2

2

r

Y

b

t

 

  

 For narrow thick sections residual stress will be high, and this will seriously diminish the strength of the section.

 It should be noted that

 the effect will be much worse for open sections than for box sections due to the tensile stress at corners of box sections.

 flame-cutting creates conditions similar to welding → high tensile residual stress is beneficial.

 the effect of residual stress is somewhat diminished for higher yield steels due to narrower tension zone.

Y

r

b  t  t 

 (  2 )  2

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 Pinned column with an initial deflection δ(x) and addition deflection (x) due to axial load. Bending moment is P(δ+w).

 Deflection continues to grow and magnified as long as P increases.

→ no static equilibrium configuration and no sudden buckling.

2

2 + ( +w)=0 d w P

dx EI



1

=

_n

sin

n

n x L



^

 





 Let δ be represented by a Fourier series

 Assume that the additional deflection due to the bending is also a Fourier series

L x w n

w

_n



 sin

Eccentric columns

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Eccentricity: Magnification Factor



2 2 2

2 2

1

=

_n

sin

n

d w n n x

dx L w L

 





 

 The dominant term is n=1. When P

_E

/P→1, w

_n

/δ

_n

→ ∞

and the total deflection is

where ϕ is called the magnification factor.

,...

3 , 2 , 1 ,

1

2 



 where n

P P w n

E n n

 0

sin ) (

1

2 2

2  



 



  



^

 L

x w n

EI w P

L n

n

n n n

 

 ₂ ( ) 0

2

2   

 _n w_n _n

EI w P

L

n  

0 )

( )

( ₂ ²

2

2     

 w_n _n P n P_E w_n _n

L EI

P n   

,

1

 P w P

E







 

 

 





 P P

P P

P w P

w

E E E

T

P_E

 

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case the magnification factor is given by

 The two types of eccentricity (load application and column geometry) can be combined linearly Δ=δ+e.

 Eccentricity can be related to the slenderness ratio

= sec

2 _E

P P

 ^__ ^__

  ^^







 

Pcr

Pe P

M_max sec 2 Secant Formula in Lecture 2

c A r Z

r A

P Z

A A

P Z

P A P

c

c 2

max

radius core

, 1

1





 







 



 



 



 



 



 





ratio 2

ty eccentrici

 c r_c

 

 

ratio ty eccentrici

2  r ec

Secant Formula in Lecture 2

results for test

bound lower

for 0.003 L , 

  

  rc

Column with eccentric load

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Perry-Robertson Formula

 It adopts Robertson`s value for α and assumes that the column will collapse when the maximum compressive stress reaches the yield stress.

 It is a quadratic, and may be rewritten in a nondimensional, factored form where R is the strength ratio of the column

η is the eccentricity ratio

λ is the column slenderness parameter

 For a perfectly straight column, having η=0, reduces to the Euler curve, , providing that λ>1.

ult

1 1

Y

E

 L

 

 

    

ult

( )

1 ^E

Y

E

P L P

A P P

 

  ^   ^

(1  R )(1  

2

R )   R

ult Y

R 

 

L

 ^ 

2

2 2 2

1 1 1 1 1

1 1

2 4

R  

  

 

   

       

R  1

R

ult R

Y ult

Y

2

2 1 1 1

1 ,

1 











 



 



 





 



E L _Y

E

Y 





   

(19)

types of sections

Column design curve for α=0.002 19α=0.0035 α=0.0055

fail by yield

σ_ult

(MPa) σ_ult

(MPa)

σ_ult (MPa)

L_e/ρ L_e/ρ L_e/ρ

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 Plot Secant curve shown below and Perry-Robertson Formula in a graph and discuss the differences of two approach associate with the plots.

Euler curve

Graph of the secant formula for

σ = 250 MPa (σ

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of slenderness ratio less than some threshold value (L/ρ)₀ . To achieve this the value of the eccentricity ratio is defined

for for

where (L/ρ)₀ is given by

 For welded I-and box sections, the yield stress should be reduced by 5% to allow for welding residual stress.

Curve selection(α) table for rolled and welded sections

0

c

L l

  _ r

  

  

 

0 0

c

L l L L

r 

  _  

 

    

      

     

0

0.2

Y

L  E

 

  

  

(22)

Use of Magnification Factor

 For a pinned column subjected to lateral pressure q and P, the w

_max

is

 The maximum bending moment

→ different for deflection, end condition, load

 The simpler magnification factor which was derived for sinusoidal initial eccentricity is used.

 Max. bending moment =due to lateral load(M

₀

)+ due to eccentricity(δ

₀

)

4 2

max 4

5 24

sec 1

384 5 2

w qL

EI

 



  

     

 

  2

L P

 _{    }^{  }EI ^

   

2

max 2

2(1 sec ) 8

M qL 



  

  

 

E E

P P P







max 0 ( 0 )

M  M P

 

  ^max ^P ^M^max

A Z

  

Homework #2 Derive w_max and M_max.

Central deflection of a

laterally loaded member The effect of axial load

(23)

OPen INteractive Structural Lab E

 We assume that the column will collapse when the maximum compressive stress reaches the yield stress. Therefore

 It can be expressed in terms of nondimensional parameters

where

23

max 2.030 0

M  M

2 4 2

max 0 2 0 0

5 5

0.5 2 1 2.028

384 48

EI qL

M M M M

L EI

 

   

        

   

ult 0 ult 0

ult

( )

1

Y

E

P M P

A Z P

P Z

     ^{ }

 

  

 

(1   R  )(1  

2

R )   R

ult Y

R







^Y ^Y

E

L

E

 

 ^  ^  ⁰

( ) A

Z

   ^{ }

0

Y

M

 Z

 

2

max 2

2(1 sec ) 8

M qL 



  

  

  2

L P

 _{    }^{  } EI ^

   

Homework #3 Derive left formula

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Use of Magnification Factor

 The solution becomes

A pinned beam column subjected to a specified lateral load.

( 0 ) A Z

   ^{ }

0

Y

M

 Z

 

Initial deflection and load eccentricity

B.M due to lateral load

2

2 2 2

1 1 1 1 1

1 1

2 4

R     

  

  

   

         

   

(25)

25

( 0 ) A Z

   ^{ }

0

Y

M

 Z

 

Initial deflection and load eccentricity

B.M due to lateral load