chapter 5 torsion buckling
TRANSCRIPT
-
8/10/2019 Chapter 5 Torsion Buckling
1/68
Flexual buckling deforms by bending.
Twisting buckling---tensional stiffness of the member
is very small. Thin-walled open sections
Combination of bending and twisting bending and
twisting are coupled so that one necessarily producesthe other. (1)Axially loaded members whose shear
center axis and centroidal axis do not coincide shcu as
angles and channels. (2)Also transversedly loaded
beams.Firstly review some fundamental relationships of
torsional behavior in general.
Chapter 5: Torsion Buckling
-
8/10/2019 Chapter 5 Torsion Buckling
2/68
5.1 Elastic torsional and torsional-flexural buckling
of axially loaded columns.
Elastic torsional and torsional-flexural buckling of axially
loaded columns mostly take place in thin-walled columns.Thin-walled columns are divided into two parts:Open
section,Close section
5.1.1 Fundamental relationships of torsional behavior
of thin-walled open section columns.
-
8/10/2019 Chapter 5 Torsion Buckling
3/68
Torsional characteristics of thin-walled open section
columns:section warping
As we all know,noncircular sections of the torsional column are
no longer plane during twisting,but warping knaggy sections).Inanother word,the section has a displacement along the axial
direction.
1 The sorts of torsion
Uniform torsion
Nonuniform torsion
Uniform torsion--The member is allowed to warp freely,then
the applied torque is resisted solely by St.Venant shearingstresses.This type of behavior is referred to as uniform
torsion,pure torsion,mean torsion or St.Venant torsion,as
shown in Fig.1-42;
-
8/10/2019 Chapter 5 Torsion Buckling
4/68
-
8/10/2019 Chapter 5 Torsion Buckling
5/68
If a couple of torques with opposite direction are applied to
the both ends of the thin-walled open section (I section)
column,uniform torsion will occur,as shown in Fig.1-42.Uniform torsion has two characteristics:
a.the same magnitude of twisting in every section.Thus,the
longitudinal fibers do not have axial strain,and there is no
normal stress but only shear stress caused by torsion in thesection.The distribution of the shear stresses relates to the shape
of the section and it is the same in each section.
b. the longitudinal fibers do not bend,viz.the longitudinal
fibers of the flanges and the web are still in line,there is just anangle (torsional angle )caused by torsion between the upper
and lower flanges.
-
8/10/2019 Chapter 5 Torsion Buckling
6/68
Known as the mechanics of material,to a noncircular sectionbar,the relationships between angle of twist and torque
is similar to a circular section bar, and the torsional rate is
constant,
1.89
t
sv
GI
M
dz
d
==
svM
2 The relationships between andtM
In which torsional rigidity of the section;
G shearing modulus of elasticity;
torsional constant
tGI
tI
-
8/10/2019 Chapter 5 Torsion Buckling
7/68
For a thin-walled open section made of rectangular
elements,such as I [ T and Z section the torsional constant
can be approximated by
1.90( )= 33
1iit tbI
In which i the number of the rectangular elements in the
section;
the length and thickness of the i element.ib it
From Eq. 1.89 ,the relationship between the torsionalangle on the left end and the torque in an arbitrary
section z position)is
svM
zGI
M
t
sv= 1.91
-
8/10/2019 Chapter 5 Torsion Buckling
8/68
-
8/10/2019 Chapter 5 Torsion Buckling
9/68
For certain support or loading conditions will prevent
torsional sections from twisting freely,nonuniform torsion occurs.
see Fig.1-44
Fig.1-44
1 Characteristics
As shown in Fig.1-44,the characteristics of nonuniform torsion
are
a. Each section of the column is knaggily ( ) viz. thelongitudinal fibers change in length.b. Torsional rate changes along the z axes.
As shown in Fig.1-44a,the fixed end of the cantilever beam will
restrain other sections from warping freely,and the closer to the
fixed end the stronger the restraint is.For the different warpingconditions in each section, the longitudinal fibers of the column are
no longer in line but bending.
-
8/10/2019 Chapter 5 Torsion Buckling
10/68
c. Besides shear stress caused by St.Venant torsion, there
are normal stress and shear stress caused by warping
torsion in the section which are called warping normal stressand warping shear stress, respectively. Sector normal stress
and sector shear stress, respectively because of the sector
coordination
s
2 The relationships between andtM Restrained torsion St.Venant Torsion warping torsion
Restrained torsional tresses
Free torsion shear stress Fig.1-43
Warping normal stress and warping shear stress
s
-
8/10/2019 Chapter 5 Torsion Buckling
11/68
St.Venant Torsionacts to resist St.Venant Torsions svM
Msv
s
Nonuniform torsion
Warping normal stress flange bending moment Mf
Mf flange shear force VfV
f torsional moment M=Vfh
-
8/10/2019 Chapter 5 Torsion Buckling
12/68
Mf Vf
Mf
Warping normal stressflange bending moment M
f
Mf flange shear force V
f
Vfh
M
Vf torsional moment M
M=Vfh 1.93
Mf constitute
double moment B
B=Mfh
-
8/10/2019 Chapter 5 Torsion Buckling
13/68
Relationships between Mfand the displacement u of thef lange along the axis x
VfMf
x
y
z
dz
u
2
hu=
2
2
2
2
2 dz
dh
dz
ud = a moment of inertia of theflange about axis yf
I
2
2
2
2
2 dz
dEI
h
dz
udEIM fff
== 1.94
xh/2
-
8/10/2019 Chapter 5 Torsion Buckling
14/68
VfMf
x
y
z
dz
From the relationships between the bending moment in the upper
flange Mfand the shear force Vfthat
dz
dMV
f
f =
Substitution of the Eq. 1.94)gives
3
3
2 dz
dEI
hV ff
= 1.95
-
8/10/2019 Chapter 5 Torsion Buckling
15/68
Eq 1.93 becomes
3
32
2 dz
dhEIM
f
= 1.96
or
1.97
in which
3
3
dz
dEIM
=
2442
2322 htbhIhII
yf === (1.98)
is a significant property of the cross section in the
calculation of the restrained torsion called the warpingconstant or sector moment of inertia. Eq. 1.98 can beused for a doubly symmetric I section.
I
-
8/10/2019 Chapter 5 Torsion Buckling
16/68
Introducing the notation as the applied torque of the restrained
torsion,and for the equilibrium conditiontM
MMM svt += 1.99a
Substitution of Eq. 1.89 and Eq. 1.97 into theequation above gives
1.99b = EIGIM tt
Eq. 1.99b) is the general expression for any restrained
torsioned open thin-walled section.In which and are called
the free torsional rigidity and the warping rigidity of the section.
tGI EI
-
8/10/2019 Chapter 5 Torsion Buckling
17/68
In Eq. 1.99b ,the first term represent the resistance of thesection to twist and the second term represent the resistance to
warping.But it is important to point out that the second term iscaused not by the warping of the member but by its resistance to
warping.
4 Strain energy of torsion
The strain energy stored in a twisted member can be
broken down into two parts
St.Venant strain energy of torsion
warping strain energy of torsion
The first part is caused by St.Venant torsion,and the secondpart is caused by warping torsion.
1)St.Venant strain energy of torsion Usv
-
8/10/2019 Chapter 5 Torsion Buckling
18/68
The increment of strain energy stored in an element dz of a
twisted member due to St.Venant torsion is equal to one half
the product of the torque and the change in angle of
twist.Hence
dMdU svsv2
1=
Substitution of Eq. 1.89 gives
dzGI
MdU
t
svsv
2
2
1=
dzdGIM tsv =
or
dzdz
dGIdU tsv
2
2
1
=
-
8/10/2019 Chapter 5 Torsion Buckling
19/68
Integrating over the length gives the strain energy in the entire
member due to St.Venant torsion.Thus
( )
=
t
tsv dzGIU0
2
2
1 1.100
2)Warping TorsionFor an I beam,the strain energy stored in the member due
to its resistance to warping is assumed to be equal to the bending
energy of the flanges,and the shear energy is assumed to be
negligible.The bending energy stored in an element dz of one ofthe flanges is equal to the product of one half the moment
EIfu) and the rotation(u).Thus
( ) dzuEIdU f2
2
1 =
Notice that and so the expression becomes2
hu = ,
2
2hII
f=
-
8/10/2019 Chapter 5 Torsion Buckling
20/68
( ) dzEIdU 24
1 =
Integrating over the length and multiplying by 2 to account
for both flanges.That is
( ) = l
dzEIU0
2
2
1
1.101
The total strain energy is
UUU sv+=
( ) ( ) += l
o
l
ot dzEIdzGI
22
2
1
2
1
1.102
The shear center It is a special point in the cross section,whichis also the center of rotation when a pure torque is applied.when
there are symmetrical axes the shear center must be on the
symmetrical axes.
-
8/10/2019 Chapter 5 Torsion Buckling
21/68
5.1.2 Elastic torsional and torsional-flexural buckling of
axially loaded columns.
1 Torsional bucklingThe mode of buckling is torsional deformation.
It usually takes place in the axially loaded columns with
lower torsional stiffness and doubly symmetric section (Eg.+
section).This kind of compressive columns may occur
torsional buckling before the load reaches to Euler load.
A axially compressive column with doubly symmetric I
section,shown as Fig.1-48,is fixed by simply supports at the
ends of the member.The coordinates x y z are shown asgiven.
-
8/10/2019 Chapter 5 Torsion Buckling
22/68
x
y
z
N
o s
xs
N
s
N
Fig.1-48
-
8/10/2019 Chapter 5 Torsion Buckling
23/68
2 The property of the problem
restrained torsion symmetric section).
3)The buckling differential equation
Introducing the applied torque Mt1, the
equation stand by the condition of
equilibrium is
01 = MMM svt a
Equilibrium equation
1 simply supported
zero warping, zero rotation
x
y
z
o s
N
N
-
8/10/2019 Chapter 5 Torsion Buckling
24/68
The applied torsion Mt1The applied torsion Mt1is caused by axial force N.So lets
analyze how does N cause Mt1
Pick out an element dz from Fig.1-48 c) analyze the fiber
DE on dz.
01 = MMM svt a
1tM
svM
M
the applied torsion
St.Venant torsion
warping torsion
Substitution of Eq.(1.89)and Eq. 1.97)gives
01 =+ EIGIM tt b
-
8/10/2019 Chapter 5 Torsion Buckling
25/68
D
x
y
z
N
o s x
dAD
s
N
z
dzD
E
dA
s
D
N
E
D
Fi g. 1-48
a) b) c)
-
8/10/2019 Chapter 5 Torsion Buckling
26/68
dz
s
s
D
E
D
E
d
E
dAF1
F2
Torsion of fiber DE DE
D D torsional angle
E E torsional angle +d
The angle between DE and the
plumb line DE) .
dz
d=
Force on line DE decomposed by F1 and F2 along
DE and EE F2 along s creates torsion torque dMt1
dA
dAdzddAFdMt 221 ===
Integrating,we have
-
8/10/2019 Chapter 5 Torsion Buckling
27/68
In which i0 is the polar radius of gyration.
AIIAdAi yxA
/)(/220 +== Buckling differential equation
Substitution of Eq. 1.103 into a ,gives the buckling
differential equation
==== 2021 NidzdI
AIdz
ddA
dz
dM
At (1.103)
0)( 20 = NiGIEI t 1.105The boundary conditions
) z=0, =0, no torsion
) z=0, =0, free warping ) z=l, =0, no torsion
) z=l, =0 free warping
-
8/10/2019 Chapter 5 Torsion Buckling
28/68
The explanation about free warping
normal stress without torsion
the flange bending moment
0=
0=fM
02
== ff EIh
M
thus 0=
The critical load NSolving Eq. 1.105 with the boundary condition gives
the critical load of torsional buckling:
2
02
2
/)( iGIl
EIN t+=
242
232 htbhII
f ==
-
8/10/2019 Chapter 5 Torsion Buckling
29/68
Exampl e 14
I section compressive column with simple supports at the
ends, l=8m the section is shown in Fig.1-49
x
y
b=250
h=218 tw=6
t=10
h0=200
t=10
E=206 103N/mm2,
G=79 103N/mm2
Calculate the bending critical load and
torsional critical load
Solving2
2
l
EINcr
=
2
02
2
/)( iGIl
EIN t+=
A=2 25 1+20 0.6=62cm2
Ix=0.6 203/12+2 25 1 10.52=5913 cm4
Iy=2 1 253/12=2604cm4
-
8/10/2019 Chapter 5 Torsion Buckling
30/68
22
0 37.13762/8517/)( cmAIIi yx ==+=cmAIi yy 48.662/2604/ ===
433
11.183/)6.0201252( cmIt =+= 4232 4.287109)212/(212512/ cmhII f ===
Bending critical load
222
432
2
2
, /4.13310628000
10260410206mmNl
EIyycr = ==
Torsional critical load
22
0
2
2
, /6.2758517/)1430690912083( mmNAi
GI
l
EIt
cr =+=+
=
,, crycr < Bending buckling first
-
8/10/2019 Chapter 5 Torsion Buckling
31/68
2 Torsional-flexural bucklingIt mostly takes place in singly symmetric thin-walled
compressive columns.There is bending deformation besidetorsional deformation when the element is buckling. seeFig.1.51 .
1)Torsional buckling differential equation
Torque by St.Venant shear stressThe warping torsion
The applied torque
Torsional differential equation
= tsv GIM = EIM
= 201 NiMt
0)( 20 = NiGIEI t 1.105 2 Because of the bending axial load N
produces torque Mt2
-
8/10/2019 Chapter 5 Torsion Buckling
32/68
y
x os a
u
au0
y
xo
s
z
N N
z
=du/dz
z dz
x
N N
u
Bending moment My by
axial force in plane xz
NuMy
dz
dMQ
y=Shear force Q acting on the centroid andcreating torsion torque Mt2
Q
-
8/10/2019 Chapter 5 Torsion Buckling
33/68
adz
duNM t =2 In the same direction with
torsional angle
Thus the equilibrium equation becomes
021 =+ MMMM svtt 3 Differential equation of torsional-flexural buckling
On the basis of Eq. 1.105),we introduce the notation Mt2,thus
0)( 20 =+ uNaNiGIEI t 1.108
1.108 two unknown u so we need one more equationwhich is the differential equation of bending deformation:
00=+ NuuEIy Deformation of the shapecenter
-
8/10/2019 Chapter 5 Torsion Buckling
34/68
auu +=00=++ aNNuuEIy
Conclusion:
0)( 20 =+ uNaNiGIEI t 0=++ NauNuEIy 1.111
1.110
4 Critical loadAssume:
l
zA
sin=
l
zBu
sin=
Thus
1)())((22
0
= crcrcry Ni
aNNNN
-
8/10/2019 Chapter 5 Torsion Buckling
35/68
2
2
l
EIN
y
y
= Critical bending load
along axis y
)(12
2
2
0
tGIlEI
iN += Critical load of the
torsional buckling
IIai
yx++= 220
So we can obtain the critical load Ncr
5 Discussion Double symmetric section
a=0 use the smaller one between Ncr=Ny and Ncr=N
Simple symmetric sectionNcr is smaller than both Ny and N ,the larger a/i0 the
smaller Ncr is.
5 1 3 Design of steel columns in Chinese code
-
8/10/2019 Chapter 5 Torsion Buckling
36/68
5.1.3 Design of steel columns in Chinese code
Base on three column strength-slenderness ratio curve, modify it
(curve C) to design torsion buckling steel column.
y yf
-
8/10/2019 Chapter 5 Torsion Buckling
37/68
5.2 Lateral buckling of beams
A lateral buckling of a beam is a combination of twisting
and lateral bending brought about by the instability of thecompression flange.
5.2.1 Lateral buckling of rectangular beams in pure
bending
Assume:
the material obeys Hookes law
the deformations remain small
the geometry of the cross section does not changeduring buckling
Consider the rectangular beam in pure bending.
-
8/10/2019 Chapter 5 Torsion Buckling
38/68
The ends of the member are assumed to be simply
supported as far as bending about the x and y axes is
concerned.
Hence
at2 2
2 20
d u d vu v
dz dz = = = =
In addition, the ends of the member are prevented fromrotating about the z axis but are free to warp.
Thus
at2
20
d
dz
= = 0,z l=
0,z l=
-
8/10/2019 Chapter 5 Torsion Buckling
39/68
-
8/10/2019 Chapter 5 Torsion Buckling
40/68
The differential equations of bending and twisting are2
2
2
2
x x
y
z
uEI M
GJ M
d vEI
dz
d
dz
d
dz
=
=
=
z
The third equation is an analogous expression for twisting
about the axis.
The moment about the and axes, denoted by vectors
in the figure, are given byx xy
cos
cos( 90) sinx x x
y x x x
M
M M M M
= == + = =
(5.59)
(5.60)
(5.61)
(5.62)
(5.63)
-
8/10/2019 Chapter 5 Torsion Buckling
41/68
cos(90 ) sinz x x xM M Mdu du du
Mdz dz dz
= = = (5.64)
Substitution of the expression in (5.62), (5.64) into Eqs.(5.59),
(5.60) leads to the following differential equations:
(5.65)
(5.66)
(5.67)
2
2
2
2
0
0
0
x x
y x
x
M
EI M
GJ M
d v
EI dz
d u
dz
d du
dz dz
+ =
+ =
=
H th fi t ti d ib b di i th ti l
-
8/10/2019 Chapter 5 Torsion Buckling
42/68
Hence the first equation describes bending in the vertical
plane. The second and third equations describe lateral bending
and twisting.
The variable can be eliminated between Eqs.(5.66) and (5.67)and one obtains
u
22
20x
y
M
EI
dGJ
dz
+ =
or
22
20k
d
dz
+ =
Where The solution of Eq.(5.69) is2 2 / .x yGJEIk =
sin cosA kz B kz= +
(5.68)
(5.69)
(5.70)
S b tit ti f th b d diti t i t
-
8/10/2019 Chapter 5 Torsion Buckling
43/68
Substitution of the boundary condition at into
Eq.(5.70) gives0= 0z=
0=
And from the condition at one obtains0= 0z=sin 0kl= (5.71)
Equilibrium in a deformed configuration is possible only when
sin 0kl=This gives kl =
For whichcr yGJEI
lM
= (5.72)
Obviously, the critical moment is proportional to the
torsional stiffness GJ and the bending stiffnessyI
The extreme fiber stress:
-
8/10/2019 Chapter 5 Torsion Buckling
44/68
The extreme fiber stress:
crcr
xS =
cr y
x
GJEIlS =or (5.73)
For the rectangular cross section being considered,
3
3
hb
J=
2 xx hS =
3
12x
bh
I =In view of these expressions, Eq.(5.73) becomes
23
3
12
3 4 /
y y
cr
x x
I h Ihb GE GE
l I bh l b I
= =
(5.74)
-
8/10/2019 Chapter 5 Torsion Buckling
45/68
/ xII
ratio of span to width
critical stress
ratio of the principal
rigidities
5 22 B kli f I b b h d
-
8/10/2019 Chapter 5 Torsion Buckling
46/68
5.22 Buckling of I beams by energy method
1 Uniform Bending-Simple SupportsConsider the simply supported I beam, subject to a uniform
bending moment M.
boundary conditions:
2 2
2 2
2
2
0
0
d u d vu v dz dz
d
dz
= =
= =
= = at
at
0,z l=
0,z l=
The condition indicates that the section is free towarp tat the supports.
2 2/ 0dzd =
-
8/10/2019 Chapter 5 Torsion Buckling
47/68
The strain energy: due to bending about the y axis and the
-
8/10/2019 Chapter 5 Torsion Buckling
48/68
The strain energy: due to bending about the y axis and the
energy due to twisting about the z axis.
22 2
2 2
2 20 0 01 1 1( ) ( ) ( )2 2 2
l l ly
d u d d I dz GJ dz E dzdz dz dz
U + + = (5.76)
And the potential energy is
2V M= Where is the angle of rotation about the axis at each end
of the beam.
x
(5.77)
0
1
2
ldu ddz
dz dz
= (5.85)
0
ldu dV M dz
dz dz
= (5.86)
Thus the potential energy of the external loads given byEq.(5.77) becomes
-
8/10/2019 Chapter 5 Torsion Buckling
49/68
Thus the total potential energy of the system
222
20 0
1 1( ) ( )
2 2
l l
y
d u dU V EI dz GJ dz
dz dz
+ = +
22
20 0
1( )
2
l ld du d dz M dz
dz dz dz
+
(5.87)
The boundary conditions will be satisfied if and areu
-
8/10/2019 Chapter 5 Torsion Buckling
50/68
The boundary conditions will be satisfied if and are
approximated by
u
sin
sin
zu A
l
zB
l
=
=
(5.88)
(5.89)
Using (5.89) and (5.92), the total potential energy becomes
2 2 2
2 2 220 0
1 1sin cos2 2
l l
y
M z zU V dz GJB dz I l l l
+ = + 4 2 2
2 2 2
4 0 0
1sin cos
2
l l
y
z M B zB dz dz
l l EI l
+
(5.93)
-
8/10/2019 Chapter 5 Torsion Buckling
51/68
and since2 2
0 0
1sin cos
2
l lz zdz dz
l l
= =
Eq.(5.93) reduces to
2 2 2 4 2 2
3
1
4 y
GJB E B M B l U V
l l EI
+ = +
The critical moment is reached when neutral equilibrium is
possible. Hence
( ) 2 4 23
02 y
d U V B GJ E M l
dB l l EI
+ = + =
(5.95)
(5.94)
Thus
-
8/10/2019 Chapter 5 Torsion Buckling
52/68
2 4 2
30
y
GJ E M l
l l EI
+ =
from which2
2cr yM EI GJ E
l l
= +
(5.96)
2 Uniform Bending-Fixed Ends
-
8/10/2019 Chapter 5 Torsion Buckling
53/68
2 Uniform Bending-Fixed Ends
Figue.Boundary condition:
0
0
0
v v
u u
= == == =
at
at
at
0,
0,
0,
z l
z l
z l
===
(5.97)
The conditions in (5.97)will be satisfied if and are
approximated by
u
21 cos
21 cos
zu A
l
zB
l
=
=
(5.98)
S b tit ti f th h i t th i i
-
8/10/2019 Chapter 5 Torsion Buckling
54/68
Substitution of these shapes into the energy expression given
by Eq.(5.87) leads to
2 42
4 0
1 16 2cos
2
l
y
A zU V EI dz
l l
+ =
2 22
2 0
1 4 2sin
2
l zGJ dz
l l
+
2 4 2
4 0
1 16 2cos2
lB z dzl l
+ 2
2
2 0
4 2sin
lAB zdz
l l
(5.99)
To determine the critical moment, the equilibrium is
-
8/10/2019 Chapter 5 Torsion Buckling
55/68
, q
( )
( )
2 2
2
2 2
2
8 2 0
2 8 2 0
y
U V
EI A MBA l l
U VGJBV E B MA
B l l
+ = =
+ = + =
(5.101)
or
( )
( )
2
2
2
2
4 0
4 0
yEI A M Bl
M A GJ E Bl
=
+ + =
(5.102)
Hence
-
8/10/2019 Chapter 5 Torsion Buckling
56/68
2 22
2 24 4 0yEI GJ E M
l l
+ =
from which
2
2
24cr yM EI GJ El l
= + (5.103)
Compare a fixed beam with a hinged beam, it is evident that the
critical moment of the fixed beam can be anywhere from two to
four times as large as the critical moment of the hinged beam.
3 Concentrated Load-Simple Supports
-
8/10/2019 Chapter 5 Torsion Buckling
57/68
-
8/10/2019 Chapter 5 Torsion Buckling
58/68
The strain energy stored in the member during buckling
has the same form it had in the preceding computations.Consider an element of the beam located a distance
from the right support, as shown in Fig.5-14c.dz
2 2/ 2
00 2
l
y
z dzw I
= (5.106)
2 2 2/ 2
0 0 2
l
y
P zV Pw dz
EI
= =
(5.107)
the potential energy of P is
the total potential energy of the system
-
8/10/2019 Chapter 5 Torsion Buckling
59/68
22
20 0
2 2
l lyEI d u GJ d U V dz dz
dz dz
+ = +
22 2 2 2/ 2
20 02 2
l l
y
E d P zdz dz
dz EI
+
(5.108)
Substitution of as given by (5.105) into Eq.(5.60) leads toy
-
8/10/2019 Chapter 5 Torsion Buckling
60/68
g y ( ) q ( )y
2
2 2y
d u Pz
EI dz
=or 2
2 2 y
d u Pz
dz EI
= (5.109)
Making use of this relationship to rewrite the first term in
Eq.(5.108) in terms of , one obtains
222 2 2 2/ 2
20 0 04 2 2
l l l
y
P z GJ d E dU V dz dz dz
EI dz dz
+ = + +
(5.110)
assume that :
-
8/10/2019 Chapter 5 Torsion Buckling
61/68
sin z
Bl
= (5.111)
Thus2 2 2 2
/ 22 2 2
20sin cos
4 2
l
y
B z GJB zU V z dz dz
I l l l
+ = +
2 42
4 0sin
2
l
y
B zdz
l l
+
(5.112)
Using the definite integrals
3 2/ 2
2 2
0
2 2
0 0
sin 18 6
sin cos2
l
l l
z lz dz
l
z z ldz dz l l
= +
= =
(5.113)
Hence
-
8/10/2019 Chapter 5 Torsion Buckling
62/68
e ce
( ) 2 3 2 2 42 3
1 0
2 8 6y
d U V B P l GJ E
dB EI l l
+ = + + + =
from which
2 2
2 2 2
4 3
6cr yP EI GJ El l
= + +
(5.115)
5.2.4 Design simplifications for lateral buckling
-
8/10/2019 Chapter 5 Torsion Buckling
63/68
for a simply supported I beam subject to uniform bending:
2
2cr y
EM EI GJ
l l
= +
for the same beam, bent by a concentrated load at midspan:
2 2
2 2 24 3
6cr y
EP EI GJl l
= + +
(5.139)
(5.140)
A critical internal bending moment for that of a critical
applied loading
2
21.36
4cr y
Pl EM EI GJ
l l
= = +
(5.141)
a suitable design relationship is
2
-
8/10/2019 Chapter 5 Torsion Buckling
64/68
2
1 2cr y
EM C EI GJ
l l
= +
Where loading coefficient
1C
(5.142)
1.0 for uniform bending
1.04 for concentrated loads at the third points
1.13 for a uniformly distributed load
1.36 for a concentrated load at midspan
applicable to different boundary conditions.
-
8/10/2019 Chapter 5 Torsion Buckling
65/68
( )
2
1 2cr yEM C EI GJ
kl kl = +
where is an effective-length factor.k
(5.144)
k
1.0 for simply supported ends
0.5 for fully fixed ends.
Stress was assumed to be proportional to strain. Inelastic
critical moment becomes a smaller.
Chinese code
-
8/10/2019 Chapter 5 Torsion Buckling
66/68
(1)If there are slabs connected with the beam, no
stability need to be calculated.(2)Otherwise for welded simple supported I beam:
1 2
2
,
4320 235[ 1 ( ) ]
4.4
x
b x
y
b b b
y x y
fW
tAh
W h f
= + +
Where stability coefficient, coefficients relatedloading and loading point, coefficient related to
symmetric of section
bb
b
(3)For rolled I beam can be get directly from the table.b
-
8/10/2019 Chapter 5 Torsion Buckling
67/68
(4)For rolled channel beam,a simple equation is deducted:
(5)When is greater than 0.6, inelastic parametershould be used:b
'
3/ 2
0.4646 0.12691.1 ( )b
b b= +
For simple supported box beams no stability needed to
calc late if the follo ing condition be satisfied:
-
8/10/2019 Chapter 5 Torsion Buckling
68/68
calculate if the following condition be satisfied:
0
1 0
(1) / 6(2) / : 95,65,57
h bl b