tutorial 5- flexural members - lateral torsional buckling
DESCRIPTION
Design of steel lateral torsionalTRANSCRIPT
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TUTORIAL 5
FLEXURAL MEMBERS
Lateral Torsion Buckling Lateral Torsion Buckling
1X March 2012
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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
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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
3. Determine LT(eq. 6.56):
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
,,===
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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
,==
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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=
-
STEP 4: Determination of C1 value :
Moment ratio:
C1 =1.365 ( Table 6.12)
Mcr to used in the calculation of normalised slenderness ratio is is C1Mcr
STEP 5: Determination of
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:
3. Determine LT(eq. 6.56):
( ) ( )[ ]22.015.0 LTLTLTLT ++=( ) ( )[ ] 31.114.12.014.134.015.0 2 =++=LT
( )[ ] [ ] 511.014.131.131.111
2/1222/122=
+=
+=
LTLTLT
LT
4. LTB Resistance(EN 1993-1-1 cl 6.3.2.2 , eq. 6.55):
Mb,Rd = 517kNm is greater than MEd,B = 433 kNm
( )[ ] [ ]14.131.131.1 ++ LTLTLT
kNmMf
WM yyplLTRdb 6.516100.12753676000511.0
16
,,===
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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:
STEP 1: Design Bending Moment:
R3={( 70 x 4.5) 70 }/9 = 27.22kN;
Med,2 = 27.22 x 4.5 = 122.5 kNm
-
Bending Moment Diagram:
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
STEP 4: Determination of C1 value :
Moment ratio:
C1 =1.88 ( Table 6.11)
kNmMcr
21.111=
0.05.122
02,
3,===
Ed
Ed
MM
-
STEP 5: Determination of
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=
+=
+=
-
4. LTB Resistance(EN 1993-1-1 cl 6.3.2.2 , eq. 6.55):
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
3. Determine LT(eq. 6.56):
4. LTB Resistance(ignoring the correction factor, f):
[ ] [ ] 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):
3. LTB Resistance(now with f) becomes:
82.0=c
k
( ) ( )[ ] 0.18.00.2115.01 2 = LTckf ( ) ( )[ ] 91.08.0797.00.2182.015.01 2 ==f
3. LTB Resistance(now with f) becomes:
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:
STEP 1: Design Bending Moment:
MEd = 12x82/2 = 384kNm
-
Bending Moment Diagram:
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
STEP 4: Determination of C1 value :
C1 =1.132 ( Table 6.11)
kNmM cr 1.06.98=
-
STEP 5: Determination of
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 = 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
-
4. LTB Resistance(EN 1993-1-1 cl 6.3.2.2 , eq. 6.55):
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
STEP 7: Determination of Mb,Rd( lateral torsional buckling resistance):General Case( EN 1993-1-1 cl 6.3.2.2)
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
-
3. Determine LT(eq. 6.56):
4. LTB Resistance(EN 1993-1-1 cl 6.3.2.2 , eq. 6.55):
( )[ ] [ ] 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
3. Determine LT(eq. 6.56):
( )[ ]LTLTLTLTLT 20,15.0 ++=( )[ ] 688.0606.075.04.0606.049.015.0 2 =++=LT
0,LT
4. LTB Resistance(ignoring the correction factor, f):
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
,,===