1

Pelarknäckning,

i-planet

knäckning,

böjknäckning.

In-plane column

buckling.

Vridknäckning

.

Torsional

buckling.

Böjvridknäckning.

Bending-torsional

buckling?

Vippning, kantring

Lateral-torsional

buckling. Tilting?

Course Stabilty of Structures

Lecture notes 2015.03.06 about

3D beams, some preliminaries (1:st order theory)

Torsion, 1:st order theory

3D beams 2:nd order theory

Torsional buckling

Coupled buckling modes, examples

Numerical calculation method

/Per J. G.

2

3D beams, some preliminaries

Coordinate system (right hand side orientation):

Stresses in beam (as considered in governing equations):

0τσσ yzzy

0τand0τ0,σ xzxyx (in general)

Cross section reference points and directions:

x

y

z

x

y

z

Centre of gravity

Principal directions

Shear centre

Rotation centre (yo, zo)

3

1:st order governing equations for 3D Bernoulli/Euler/StVenant/Vlasov beam

xvω

yoyz

zozy

ox

m)(GK)(EI

Mmq)w(EI

Mmq)v(EI

N-q)u(EA

The equations are linear and uncoupled.

v=v(xo,yo) and w=(xo,yo). Thus v(x,y)=v-(z-zo)ϕ and w(x,y)=w+(y-yo)ϕ.

x, u

y,

z,

w

mx

x, u

y,

z,

Centre of gravity

Principal directions

Shear centre

Rotation centre (yo, zo)

ϕ

bar-action (1)

bending in x-y plane (2)

bending in x-z plane (3)

torsion (4)

4

Torsion

Kinematics:

1) In the y-z plane, the beam cross-section moves as a rigid body rotating around the rotation center

2) Displacement u is not equal in different parts of a cross section: warping

x, u

y, v

z, w TL

Rotation center (yo, z

o)

ϕ

TL

5

Two torsion deformation modes:

The magnitude of warping may be constant or vary along the beam.

These two cases corresponds to different stress and deformation performance.

a) (“St Venant theory”) The warping is constant along the beam. This is the case

if the torque is constant and there is nothing preventing warping at the ends

of the beam.

In this case σx=εx=0. Twisting and warping is due to γxy≠0 and γxz≠0.

vGK

T

dx

d

L

(0)(L)

For distributed load, equilibrium for a part dx gives dT+mxdx=0, i.e. T´=-mx :

xv m)(GK (5)

To be strict, mx≠0 should not be allowed since it gives varying T and warping, and therefor contradicts the theory.

The case of constant warping is the “St Venants torsion theory”. Cross section geometry parameter Kv and shear stress distribution are in this theory determined by an Poisson’s equation, derived from definition of shear strain, τ=Gγ and σx=0.

x

T T

L

6

b) (“Vlasov theory”) The warping is varying along the beam. This is the case if

the torque is varying or warping is prevented at an end of the beam.

In “pure Vlasov torsion” is γxy=0 and γxz=0, and twisting is due to εx≠0.

Prevention of warping is of great importance the torsional stiffness of beams with

certain thin-walled cross-sections, eg:

For most other cross-sections is the influence (on torsional stiffness) of prevented warping disregarded. The pure Vlasov torsion gives:

xm)(EI (6)

x

T

L

7

An illustration

Superposition of the two stress-systems gives

“the governing equation for mixed torsion” acc. to 1:st order theory:

xv m)(GK-)(EI (7)

Boundary conditions if Iω=0

ϕ=0 (or other known value)

T known, giving ϕ’=T/(GKv) Boundary conditions if Iω≠0

ϕ=0 (or other known value)

ϕ’=0 (prevented warping, clamped)

ϕ’’=0 (free warping (or known normal stress giving some ϕ’’≠0 ) )

T known, giving -(EIωϕ’’)’+GKvϕ’=T

Prevented warping (“Vlasov”). Twist of beam due to normal stress σx

and strain εx giving bending deformation of upper and lower flanges.

Free warping (“St Venant”). Twist of beam due to shear stress and strain

giving “spiral shape” of the web and each of the two flanges.

8

2:nd order governing equations for 3D Bernoulli/Euler/-StVenant/Vlasov beam

Assumptions

Linear elastic material

Constant EA, EI, GKv, EIω and N along beam

No initial stress and no initial curvature (imperfections)

No distributed moment loads

N known and no influence of shear on normal axial force

)terms order nd:(2m''GK''''EI

')'(M)''y'(w'Nq'''w'EI

')'(M)''z'(v'Nq'''v'EI

quEA

xvω

ozozy

oyoyz

x

where (2:nd order terms)=

)oy1(yyq)oz1(zzqwyqvzq

...oy)'oz(M2oz)'o

y(M2

...w)'oy(V)'w'o

z(Mv)'oz(V)'v'o

y(M

...)'v'oz'w'oN(y''/A)oN(I

Notations: see next page.

Reference: ”Byggnadsmekanik. Knäckning” (Runesson, Samuelsson, Wiberg, 1992)

x

y

z

bar-action (8)

bending in x-y plane (9)

bending in x-z plane (10)

torsion (11)

9

Notations

x

y

z

(yo,zo) is the location of the rotation center

(y1,zo) is the location of load qy

(yo,z1) is the location of load qz

Io=Iy+Iz+(yo2+zo

2)A is the polar moment of inertia with respect to

the rotation center

Vyo, Vz

o, Myo, Mz

o are section shear forces and moments acc. to 1:st

order theory

10

Outline of derivation of governing equations for 3D beam, 2:nd order

The three “beam stress” components σx, τxy and τxz, acting on a small volume

dxdydz in the beam are considered:

Equilibrium in y- and z-directions:

0)w'(σUτ

0)v'(σUτ

xzxz

xyxy

(12a,b)

Uy and Uz är load/volume in the y- and z-directions.

2:nd order effects of the shear stresses are not considered, just as in the 2D

beam analysis.

The 2:nd order equations can obtained from the 1:st order equation by adding

the second order component of the normal stress to the loads in the y- and z-

directions (loads/length qy and qz) :

(In case of torsion with prevented warping additional considerations are needed.)

Note: σx is the normal stress in direction (1, v’, w’), that is in the direction along a

line in the beam which in the unloaded beam was oriented parallel with x-axis.

This direction of σx is thus not the x-direction and in general neither

perpendicular to the surface on which σx acts. The latter is of relevance for St

Venant torsion.

x, u

y, v

z, w

τxy

τxz

σx

Position when beam is unloaded

Position when

beam is loaded

replacing load qy with load

replacing load qz with load

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Equations (8-11) in the special case of double symmetric and massive

(thickwalled) beam cross sections:

Double symmertric: yo=zo=0

Massive (thick walled): yo≈zo≈0

)terms order nd:(2m''GK''''EI

')'(M'w'Nq'''w'EI

')'(M'v'Nq'''v'EI

quEA

xvω

ozzy

oyyz

x

(13a,b,c,d)

where (2:nd order terms)=

1yyq1zzqwyqvzq

...w)'oy(V)'w'o

z(Mv)'oz(V)'v'o

y(M

...''/A)oN(I

For the I-beam section is Iω≠0.

For all other is Iω≈0.

12

If moreover no distributed load (qx=qy=qz=mx=0):

w)'oy(V)'w'o

z(Mv)'oz(V)'v'o

y(M''/A)oN(I''GK''''EI

')'(M'w'N'''w'EI

')'(M'v'N'''v'EI

0uEA

vω

ozy

oyz

…..(14a,b,c,d)

And for stability analysis with respect only to compressive force –N:

0)(

0

0

''/AoNIGK''''EI

'w'N'''w'EI'v'N'''v'EI

vω

y

z

(15a,b,c)

Eq (15a,b,c) have the nice feature of being uncoupled.

13

Torsional buckling

Only double symmetric and/or thickwalled cross sections considered.

Governing equation and solution for beams with Iω≠0:

0)( ''/AoNIGK''''EI vω (15c)

ωovov

4321

ωovov

4321

/A)/EINI(GKλ compr.), smal l or (tens ion0/A)NI(GKfor

CxCλx)s inh(Cλx)cosh(C

/A)/EINI(GK-λ on),(compress i0/A)NI(GKfor

CxCλx)s in(Cλx)cos(C

(16a,b)

One solution for homogeneous boundary conditions is C=0.

From the boundary conditions: AC=0.

By investigation of det(A)=0 can possible Ncr for the actual boundary conditions

be determined.

Governing equation and solution for beams with Iω=0:

0)( ''/AoNIGKv (17)

21 CxCv (18)

One solution for homogeneous boundary conditions (ϕ=0 and/or ϕ’=0) is C=0.

Other solutions are possible only if

0)( /AoNIGKv (19)

Thus the critical load for torsional buckling is

oA/IvGKcrN (20)

The curve ϕ(x) may have any continuous shape in the case torsional buckling of a beam with Iω=0! (Provided that the boundary values at x=0 and X=L are fulfilled.)

14

Further studies of torsional buckling:

Thin walled pipe

Potential energy, π

For γ=0 is π´=0 (equilibrium) and π´´=0 (neutral equilibrium) found for

σcr = G (a simple and useful relation!?) (21)

For the pipe this means

GAcrN (22)

Compare with Eq (20) for a thin walled pipe:

GAoA/IoGIoA/IvGKcrN

σ

Radius = R

Wall thickness = t<<R

σ σ a

γ G

Cross section area =A

h h h(1-cosγ)

Gγcosσπ

Gγγs inσπ

GγτandVahVγτ(1/2))γcosh(1aσπ where

15

The result σcr = G should be possible to derive also by means of equilibrium

analysis. How?

σ

Radius = R

Wall thickness = t<<R

σ σ a

γ G

Cross section area =A

h h h(1-cosγ)

16

Further studies of torsional buckling:

Derivation of buckling load at St Venant torsion (beams with Iω=0)

Potential energy, π

/2γ1γcosand/LGKT,/Lrγwhere

2

T)γcos(1LdAσπ

2v

A

L

GKI

L

σπ

L

GKI

L

σπ

2L

GKdAr

2L

σ

2L

GKdA

2

/L)(rσLπ

vo

vo

2v

A

222

v

A

2

For ϕ=0 is π´=0 (equilibrium) and π´´=0 (neutral equilibrium) found for

o

vcr

I

GKσ giving

oA/IvGKcrN which is Eq (20)

The similar energy based derivation of critical load for the case Iω≠0 can be made,

but then variation of ϕ along the beam must be considered (by means of Eq

(15a)) .

x σ φ

r

rφ A

dA

x

L

γ

17

18

19

20

21

continued next page….

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