Tutorial 5- Flexural Members - Lateral Torsional Buckling

TUTORIAL 5

FLEXURAL MEMBERS

Lateral Torsion Buckling Lateral Torsion Buckling

1X March 2012

Summary of design process

The design process for a beam can be summarised as follows

(a) Determination of all forces and moments on critical section

(b) Selection of UB or UC

(c) Classification of section

(d) Check shear strength; if unsatisfactory return to (b)

(e) Check bending capacity; if unsatisfactory return to (b) (e) Check bending capacity; if unsatisfactory return to (b)

(f) Check deflection; if unsatisfactory return to (b)

(g) Check web bearing and buckling at supports or concentrated load; if unsatisfactory provide web stiffener

(h) Check lateral torsional buckling; if unsatisfactory return to (b) or provide lateral restraints

(i) Summarise results

Problem 1:

Prepare a design in Grade S355 Steel for a beam the above beam. Prepare a design in Grade S355 Steel for a beam the above beam.

Note:

All the loads are characteristics loads,

Lateral torsional restraints exists at A, B, C and D.

SOLUTION:

STEP 1: Factored Loads at ULS

At B: 1.35 x 40 + 1.5 x 70 = 159 kN

At C: 1.35 x 20 + 1.5 x 30 = 72 kN

STEP 2: Bending Moment and Shear Force Diagrams:

STEP 3: Section Selection:

The critical section for the design is BC as the moment gradient is the least The critical section for the design is BC as the moment gradient is the least

Try a 406 x 178 UB 74

Section properties: h = 412.8 ; b = 179.5 ; tw = 9.5 ; tf = 16.0 ; r = 10.2 ; d = 360.4 ; Iel,y = 27300 ; Iel,z = 1550 ; ry = 17 ; rz = 4.04 ; Wel,y = 1320 ;Wel,z = 172 ; Wpl,y = 1500 ; Wpl,z = 267

Moment Resistance:

Section is Safe; Mpl,Rd = 533 kNm > MEd =390 kNm

kNmM

yfyplWRdplM 533

6100.1

35515000000

,,

===

STEP 3: Section Classification:

Compression Flange:

c/tf = 74.8/ 16 = 4.68 < 9 = 9*0.814=7.32 Flange is CLASS 1

Web:

c/t = d/tw = 37.9 < 72 = 72*0.814= 58.6 Web CLASS 1

SECTION is CLASS 1 (PLASTIC)

mmw

trbc 8.74]5.92.1025.179[5.0]2[5.0 ===

SECTION is CLASS 1 (PLASTIC)

STEP 4: Shear Check

Moment capacity is not reduced.

kNMf

AV yvRdpl 873100.1

35542603

103

1 6,

===

5.015.0873130

,

==Rdpl

Ed

VV

STEP 5: Calculation of Mcr:

E = 210Gpa; G = 81GPa, Iw = 0.608 dm6 ; It = 62.8 cm

4

System(BC) Length = 3.0m

28

6

03923.010155010608.0

mII

z

w=

=

kNLEI

z 35700.3

101550102102

862

2

2

=

=

pipi

Mcr = 826kNm

L 0.3

2862

862

2

2

0143.010155010210108.6210810.3

mEIGIL

z

t=

=

pipi

[ ] kNmmmkNEIGIL

II

LEIM

z

t

z

wz

cr8260143.003923.03570 2/122

2/1

2

2

2

2

=+=

+=

pi

pi

STEP 6: Determination of C1 value :

Moment ratio:

Mcr to used in the calculation of normalised slenderness ratio is is C1Mcr

STEP 7: Determination of

777.0390303

,

,

===

BEd

CEd

MM

106.1777.052.0777.04.188.152.04.188.1 221 =+=+= C

LT

STEP 8: Determination of Mb,Rd( lateral torsional buckling resistance):

Note: There are 2 methods to determine the strength reduction factors due to

the lateral torsional buckling:

LT

763.010826106.1

35515000006 =

==

cr

yyLT

MfW

Method 1: General Case( EN 1993-1-1 cl 6.3.2.2)

1. Ratio h/b = 412.8/179.5=2.3 >2, Table 6.4, LT = 0.34( curve b, Table 6.3)

2. Determine LT:

3. Determine LT(eq. 6.56):

( ) ( )[ ]22.015.0 LTLTLTLT ++=( ) ( )[ ] 887.0763.02.0763.034.015.0 2 =++=LT

[ ] [ ] 747.011

===

4. LTB Resistance(EN 1993-1-1 cl 6.3.2.2 , eq. 6.55):

Mb,Rd = 398kNm is greater than MEd,B = 390 kNm

( )[ ] [ ] 747.0763.0887.0887.011

2/1222/122=

+=

+=

LTLTLT

LT

kNmMf

WM yyplLTRdb 398100.13551500000747.0

16

,,===

Method 2: Rolled Sections( EN 1993-1-1 cl 6.3.2.3)

1. Ratio h/b = 412.8/179.5=2.3 >2, Table 6.5, LT = 0.49( curve c, Table 6.3)

2. Determine LT:

=0.75(minimum value) and = 0.4(minimum value)

3. Determine (eq. 6.56):

( )[ ]LTLTLTLTLT 20,15.0 ++=( )[ ] 807.0763.075.04.0763.049.015.0 2 =++=LT

0,LT


4. LTB Resistance(ignoring the correction factor, f):

[ ] [ ] 787.0763.075.0807.0807.011

2/1222/122=

+=

+=

LTLTLT

LT

kNmMf

WM yyplLTRdb 419100.13551500000787.0

16

,,===

Taking into account the correction factor, f:

Determine kc( Table 6.6):

777.0390303

,

,

===

BEd

CEd

MM

931.0777.033.033.1

133.033.1

1=

=

= ck

( ) ( )[ ] 0.18.00.2115.01 2 = LTckf

3. LTB Resistance(now with f) becomes:

( ) ( )[ ] 965.08.0763.00.21931.015.01 2 ==f

kNmkNmM Rdb 434965.0419

,==

Problem 2:The 7.5m long 610x229UB125 of S275 steel as shown bellow is simply supported at both ends

where lateral deflections are effectively prevented and twist rotations are partially restrained.

Check the adequacy of the beam for a central concentrated at the top flange load caused by an

unfactored dead load 60kN(which includes an allowance for self weight) and an unfactored

imposed load of 100kN.

Section Properties: h = 612.2 ; bf = 229.0 ; tw = 11.9 ; tf = 19.6 ; r = 12.7 ; Iz = 3932 ; Wpl,y = 3676 ; It = 154 ; Iw = 3.45

SOLUTION:

STEP 1: Design Bending Moment:

MEd = {(1.35 x 60 + 1.5 x 100)x7.5/4 = 433 kNm

Bending Moment Diagram:

STEP 2: Plastic Resistance Moment:

Section is Safe; Mpl,Rd = 1011 kNm > MEd =433 kNm

kNmM

yf

yplWRdplM 1011610

0.12753676000

0,,

===

STEP 3: Calculation of Mcr:

The critical length of system, to allow the partial torsional end restrain is as given:

Thus Lcr = kcr x L = 1.04 x 7500mm = 7806mm

2/1

2

2

2

2

+= tcrwzcr EI

GILII

LEIM

pi

pi

2/6.192.612

0.22919.112.19

750066.192.61212/

61

33

+

+=

+=

f

f

w

ffcr d

bt

t

Ld

k

04.1=cr

k

22

+=zzcr

cr EIILM

pi

2/1

42

42

4

12

2

42

10393221000010154810007806

1039321045.3

7806103932210000

+

=

pi

picr

M

kNmMcr

6.568=


Moment ratio:

C1 =1.365 ( Table 6.12)

Mcr to used in the calculation of normalised slenderness ratio is is C1Mcr


777.0390303

,

,

===

BEd

CEd

MM

LT

STEP 6: Determination of Mb,Rd( lateral torsional buckling resistance)

General Case( EN 1993-1-1 cl 6.3.2.2)

1. Ratio h/b = 612.2/229.0=2.67 >2, Table 6.4, LT = 0.34( curve b, Table 6.3)

14.1106.568365.1

27536760006 =

==

cr

yyLT

MfW

2. Determine LT:


( ) ( )[ ]22.015.0 LTLTLTLT ++=( ) ( )[ ] 31.114.12.014.134.015.0 2 =++=LT

( )[ ] [ ] 511.014.131.131.111

2/1222/122=

+=

+=

LTLTLT

LT


Mb,Rd = 517kNm is greater than MEd,B = 433 kNm

( )[ ] [ ]14.131.131.1 ++ LTLTLT

kNmMf

WM yyplLTRdb 6.516100.12753676000511.0

16

,,===

Problem 3:The 9.0m long 254x146UB37 of S275 steel as shown bellow, has a central concentrated

design load of 70kN(which includes an allowance for self weight) and a design end

moment of 70kNm. Lateral deflections and twist rotations are effectively prevented at

both ends and by a brace at mid-span. Check the adequacy of the braced beam.

SOLUTION:


R3={( 70 x 4.5) 70 }/9 = 27.22kN;

Med,2 = 27.22 x 4.5 = 122.5 kNm


STEP 2: Section Resistance: STEP 2: Section Resistance:

tf = 10.9 mm ; fy = 275 N/mm2 ; = (235/275)1/2 = 0.924

cf /(tf ) = (146.4/2-6.3/2-7.6)/(10.9x0.924)=6.20 < 9 ----Flange CLASS 1

cw /(tw ) = (256-2x10.9-2x7.6)/(6.3x0.924)= 37.6 < 72 ----Web CLASS 1

Section is CLASS 1Plastic

kNmM

yf

yplWRdplM 133610

0.1275483000

0,,

===

STEP 3: Calculation of Mcr(considering the section 2-3):

Length of system L = 4.5 m

2/1

2

2

2

2

+=

z

tcr

z

w

cr

z

cr EIGIL

II

LEIM

pi

pi

2/1

42

42

4

12

2

42

10571210000103.15810004500

10571100857.0

450010571210000

+

=

pi

picr

M


Moment ratio:

C1 =1.88 ( Table 6.11)

kNmMcr

21.111=

0.05.122

02,

3,===

Ed

Ed

MM


STEP 6: Determination of Mb,Rd( lateral torsional buckling resistance):General Case( EN 1993-1-1 cl 6.3.2.2)

1.Ratio h/b = 256/146.4 = 1.75 < 2, Table 6.4, LT = 0.21( curve a, Table 6.3)2.Determine LT:

LT

797.01021.11188.1

2754830006 =

==

cr

yyLT

MfW

2.Determine LT:

3.Determine LT(eq. 6.56):

( ) ( )[ ]22.015.0 LTLTLTLT ++=( ) ( )[ ] 88.0797.02.0797.021.015.0 2 =++=LT

( )[ ] [ ] OKLTLTLTLT _1798.0797.088.088.011

2/1222/122=

+=

+=


Mb,Rd = 106 kNm is less than MEd,B = 122.5 kNm ----

SECTION is INADEQUATE

kNmMf

WM yyplLTRdb 106100.1275483000798.0

16

,,===

Method 2: Rolled Sections( EN 1993-1-1 cl 6.3.2.3)

1. Ratio h/b = 256/146.4 = 1.75 < 2, Table 6.5, LT = 0.34( curve b, Table 6.3)

2. Determine LT:

=0.75(minimum value) and = 0.4(minimum value)

3. Determine (eq. 6.56):

( )[ ]LTLTLTLTLT 20,15.0 ++=( )[ ] 806.0797.075.04.0797.034.015.0 2 =++=LT

0,LT



[ ] [ ] 818.0797.075.0806.0806.011

2/1222/122=

+=

+=

LTLTLT

LT

kNmMf

WM yyplLTRdb 109100.1275483000818.0

16

,,===

Taking into account the correction factor, f:

Determine kc( Table 6.6):


82.0=c

k

( ) ( )[ ] 0.18.00.2115.01 2 = LTckf ( ) ( )[ ] 91.08.0797.00.2182.015.01 2 ==f


SECTION is STILL INADEQUATE

kNmkNmM Rdb 12091.0109

,==

Problem 4:The 8.0m long 475x191UB82 of S275 steel as shown bellow. The cantilever has lateral,

torsional, and warping restraints at the support, is free at the tip, and has a factored

upwards design uniformly distributed load of 12kN.m(which includes an allowance for

self weight) acting on top flange. Check the adequacy of the cantilever.

SOLUTION:


MEd = 12x82/2 = 384kNm


STEP 2: Section Resistance:

tf = 16.0 mm ; fy = 275 N/mm2 ; = (235/275)1/2 = 0.924

c /(t ) = (191.3/2 - 9.9/2 10.2)/(16 x0.924)=5.44 < 9 ---Flange CLASS 1cf /(tf ) = (191.3/2 - 9.9/2 10.2)/(16 x0.924)=5.44 < 9 ---Flange CLASS 1

cw /(tw ) = (460.2 - 2x16.0-2x10.2)/(9.9x0.924)= 44.6 < 72 ---Web CLASS 1

Section is CLASS 1Plastic

Section is Safe; Mpl,Rd = 503.3 kNm > MEd =384 kNm

kNmM

yf

yplWRdplM 3.503610

0.12751830000

0,,

===

STEP 3: Calculation of Mcr(considering the section):

Length of system Lcr = 8.0m

2/1

2

2

2

2

+=

z

tcr

z

w

cr

z

cr EIGIL

II

LEIM

pi

pi

2/1

42

42

4

12

2

42

101870210000102.698100016000

10187010922.0

16000101870210000

+

=

pi

picrM


C1 =1.132 ( Table 6.11)

kNmM cr 1.06.98=



1.Ratio h/b = 460.0/191.3 = 2.40 >2, Table 6.4, LT = 0.34( curve a, Table 6.3)2.Determine LT:

LT

0673.01006.98132.1

27518306 =

==

cr

yyLT

MfW

2.Determine LT:

3.Determine LT(eq. 6.56):

Take

( ) ( )[ ]22.015.0 LTLTLTLT ++=( ) ( )[ ] 48.00673.02.00673.034.015.0 2 =++=LT

( )[ ] [ ] OKNotLTLTLTLT __105.10673.048.048.011

2/1222/122=

+=

+=

00.1=LT


Mb,Rd = 503.25 kNm is less than MEd,B = 348 kNm ----

SECTION is ADEQUATE

kNmMf

WM yyplLTRdb 25.503100.1275183000000.1

16

,,===

Problem 5:Determine a suitable UB of S275 for simply supported beam, if twist rotations are effectively

prevented at the ends and is a brace is added which effectively prevents lateral deflections and

twist rotation at mid Span.

SOLUTION:STEP 1: Design Bending Moment:

MEd = {(1.35 x 60 + 1.5 x 100)x7.5/4 = 433 kNm

STEP 2: Selecting a Trial Section, for fy = 275N/mm2

36

,1750

2759.010433

9.0cmf

MWy

Edypl =

=

Try a 457x191 UB 82 with Wpl,y = 1830 cm3 > 1750 cm3

STEP 3: Section Resistance:

Section is Safe; Mpl,Rd = 503.3 kNm > MEd =433 kNm

STEP 4: Elastic Buckling Moment

kNmM

yf

yplWRdplM 3.503610

0.12751830000

0,,

===

2/122 GILIEIpiLength of system Lcr = 3750mm 22

2

2

+=

z

tcr

z

w

cr

z

cr EIGIL

II

LEIM

pi

pi

2/1

42

42

4

12

2

42

101870210000102.69810003750

10187010922.0

3750101870210000

+

=

pi

picr

M

kNmMcr

3.727=

STEP 5: Determination of C1 value :C1 =1.879 ( Table 6.11)

STEP 6: Determination of LT606.0

103.727879.12751830000

61

=

==

cr

yyLT

MCfW


1.Ratio h/b = 460.0/191.3 = 2.40 >2, Table 6.4, LT = 0.34( curve a, Table 6.3)

2.Determine LT:

( ) ( )[ ]22.015.0 LTLTLTLT ++=( ) ( )[ ] 753.0606.02.0606.034.015.0 2 =++=LT



( )[ ] [ ] OKLTLTLTLT _1833.0606.0753.0753.011

2/1222/122=

+=

+=

kNmMf

WM yyplLTRdb 2.419100.12751830000833.0

16

,,===

Mb,Rd = 419.2 < 433 kNm= MEd LTB Resistance INADEQUATE

M 0.11

Less Conservative Method: Rolled Sections( EN 1993-1-1 cl 6.3.2.3)

1. Ratio h/b = 460.0/191.3 = 2.40 >2, Table 6.5, LT = 0.49( curve c, Table 6.3)

2. Determine LT: =0.75 and = 0.4


( )[ ]LTLTLTLTLT 20,15.0 ++=( )[ ] 688.0606.075.04.0606.049.015.0 2 =++=LT

0,LT


Mb,Rd = 444.4 > 433 kNm= MEd LTB Resistance is ADEQUATE

[ ] [ ] 883.0606.075.0688.0688.011

2/1222/122=

+=

+=

LTLTLT

LT

kNmMf

WM yyplLTRdb 4.444100.12751830000883.0

16

,,===

Documents

Tutorial 5- Flexural Members - Lateral Torsional Buckling