Upcoming SlideShare
×

TALAT Lecture 2301: Design of Members Example 4.4: Bending moment resistance of welded hollow section with outstands. Class 4 cross section

846 views

Published on

This example provides calculations on bending moment of members based on Eurocode 9.

Published in: Education
0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
846
On SlideShare
0
From Embeds
0
Number of Embeds
2
Actions
Shares
0
51
0
Likes
0
Embeds 0
No embeds

No notes for slide

TALAT Lecture 2301: Design of Members Example 4.4: Bending moment resistance of welded hollow section with outstands. Class 4 cross section

1. 1. TALAT Lecture 2301 Design of Members Bending Moment Example 4.4 : Bending moment resistance of welded hollow section with outstands. Class 4 cross section 10 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm Date of Issue: 1999  EAA - European Aluminium Association TALAT 2301 – Example 4.4 1
2. 2. Example 4.4. Bending moment resistance of welded hollow section with outstands. Class 4 cross section Half top flange a 200 . mm Half hollow width c 96 . mm fo 200 . MPa Half bottom flange b 96 . mm Width of HAZ haz 30 . mm γ M1 1.1 99 . mm MPa 10 . Pa 6 Section dept h HAZ reduction ρ haz 0.8 Flange thickness d 6 . mm haz .( c bz c b) Web thickness w 8 . mm 2 2 (c b) h Bottom flange t. t 8 . mm Type of element haz .h hz - Internal element: T = 0 2 2 - Outstand: T > 0 (c b) h o 0 . mm Nodes, co-ordinates (y, z), thickness (t) and types (T) 0 a o o 0 Comments: 1 c haz o d 1 Node 10 and 13 indicate shift in neutral axis. 2 c o d 1 Element 1-2, 2-3, 4-5, 5-6, 8-9 3 c haz o d 0 and 14-15 define HAZs. 4 c haz o d 0 5 c o d 0 This solution of the example is 6 c haz o d 1 comprehensive because: 1. Calculations are shown in every 7 a o d 1 detail i y z t T 8 c o o 0 2. It covers all classes of cross section 9 bz hz w 0 3. The possibility to increase the 10 0.5 .( c b ) 0.5 . h w 0 effective thickness for elements not fully stressed is shown. The iteration 11 b h w 0 procedure is then mandatory. 12 b h t 0 However, in this example, the result is almost exactly the same as if a 13 0.5 . ( c b) 0.5 . h w 0 simplified method was used, that's because there is no reduction of the 14 bz hz w 0 web thickness due to local buckling. 15 c o w 0 The beam is composed by two extrusions. The weld is close to the neutral axis why HAZ softening does not 0 influence the moment resistance. Nodes i 1 .. rows ( y ) 1 rows ( y ) = 16 Area of cross ti . 2 2 100 section dAi yi yi 1 zi zi 1 200 100 0 100 200 elements TALAT 2301 – Example 4.4 2
3. 3. Cross rows ( y ) 1 section A dAi area A = 5.52 . 10 mm 3 2 i =1 First rows ( y ) 1 dAi Sy moment Sy zi zi 1 . z gc of area 2 A Gravity centre i =1 z gc = 41.752 mm z gc.gr z gc Second rows ( y ) 1 dAi zi . zi 1 . A . z gc moment 2 2 2 Iy zi zi 1 Iy Iy of area 3 i =1 I y = 1.061 . 10 mm 7 4 I y.gr Iy Elastic section Iy Iy modulus W el if max z z gc > min z z gc , , max z z gc min z z gc W el = 1.853 . 10 mm 5 3 Torsion rows ( y ) 1 2 ti dAi . I tu = 9.536 . 10 mm constant for 4 4 I tu open parts 3 i =1 Hollow part, (nodes 2, 5, 11, 12 and 15) y2 z2 t1 96 0 6 y5 z5 t4 96 0 6 yh y11 y h = 96 mm zh z11 zh = 99 mm th t 11 th= 8 mm y12 96 z12 99 t 12 8 96 0 8 y15 z15 t 14 rows y h 1 Area within 0.5 . y h . z A tu = 1.901 . 10 mm 4 2 mid-line A tu yh hi zh i i 1 i 1 i =1 Sum of l/t rows y h 1 2 2 yh yh zh zh i i 1 i i 1 22 Dn if t i > 0 , , 10 Dn = 80.75 th i =1 i 4 .A 2 Torsion constant tu I t = 1.79 . 10 mm 7 4 for closed part It Dn Torsion constant I t = 1.799 . 10 mm 7 4 whole section It It I tu Torsion 2 . A tu . min t h W v = 2.281 . 10 mm 5 3 resistance Wv min t h = 6 mm z gc = 41.752 mm TALAT 2301 – Example 4.4 3
4. 4. Effective cross section. First iteration z8 z gc zd zi z gc 5.4.3 (1) ψi 1 j 9 .. 11 ψj ψ j 4 ψ j i z11 z gc max z max z d 0.8 (5.7 or 5.8) gi if ψ i > 1 , 0.7 0.30 . ψ i , Outstand gi if Ti > 0 , 1 , gi 1 ψ i Figure 5.2 y2 y0 0.5 . t 13 5.4.3 (1) c) k 1 .. 2 βk βk 5 β k y2 = 96 mm t2 y0 = 200 mm y2 y0 y5 y2 t 13 = 17.333 l 3 .. 5 βl t2 t4 2 2 y11 y8 z11 z8 β j if t 11> 0 , gj. ,1 βj 4 β j t 11 250 . MPa max z ε 5.4.4 (5) ε o εi ε o. εi if zi z gc . zi 1 z gc , ε i , 99 ε o 1.118 = fo z gc 5.4.5 (3) 2 2 a) or c) β ε ε β ε ε > 4 , 8. 16 . > 15 , 25 . 150 . i i i i i i non heat ρ c if Ti > 0 , if , 1.0 , if , 1.0 i ε β β ε β β treated, welded i i i i i i Effective thickness t eff ρ .c t Heat Affected Zone Flange t haz.f .t ρ haz 2 t haz.f = 4.8 mm Web .t t haz.w ρ haz 9 t haz.w = 6.4 mm n 2 .. 3 t eff if t eff > t haz.f , t haz.f , t eff t eff t eff n n n 8 n n t eff if t eff > t haz.w , t haz.w , t eff t eff t eff 9 9 9 15 9 βi t eff = i ε ρ c= = ψi = gi = βi = εi = i i mm 1 1 16.667 1.309 12.731 0.53 3.178 1 1 16.667 1.309 12.731 0.53 3.178 Comment: 1 1 30.667 1.309 23.425 0.794 4.763 εi was given a large value (99) 1 1 30.667 1.309 23.425 0.794 4.763 for elements entirely on 1 1 30.667 1.309 23.425 0.794 4.763 the tension side 1 1 16.667 1.309 12.731 0.53 3.178 1 1 16.667 1.309 12.731 0.53 3.178 1 1 0 1.309 0 1 0 -0.729 0.481 5.955 1.309 4.549 1 6.4 -0.729 0.481 5.955 99 0.06 1 8 -0.729 0.481 5.955 99 0.06 1 8 Area of effective 1 1 0 99 0 1 8 thickness -0.729 0.481 5.955 99 0.06 1 8 t eff . elements 2 2 dAi yi yi 1 zi zi 1 -0.729 0.481 5.955 99 0.06 1 8 i -0.729 0.481 5.955 1.309 4.549 1 6.4 TALAT 2301 – Example 4.4) rows ( y 1 4
5. 5. Cross rows ( y ) 1 section A eff dAi A = 5.52 . 10 mm 3 2 area i =1 A eff = 4.6 . 10 mm 3 2 First moment rows ( y ) 1 dAi of area Sy zi zi 1 . Sy 2 z gc z gc = 49.794 mm i =1 A eff Gravity centre Effective cross section. Second iteration Node 10 and 13 is moved to the neutral z10 z gc 0.01 . mm z13 z gc 0.01 . mm axis from the first iteration z10 y10 y5 . y y11 y13 y10 5 z11 z8 z gc zd zi z gc 5.4.3 (1) ψi 1 j 9 .. 11 ψj ψ j 4 ψ j i z11 z gc max z max z d 0.8 (5.7 or 5.8) gi if ψ i > 1 , 0.7 0.30 . ψ i , Outstand gi if Ti > 0 , 1 , gi 1 ψ i Figure 5.2 y2 y0 0.5 . t 13 y5 y2 t 13 5.4.3 (1) c) k 1 .. 2 βk βk 5 β k l 3 .. 5 βl t2 t4 2 2 y11 y8 z11 z8 β j if t 11> 0 , gj. ,1 βj 4 β j t 11 250 . MPa max z 5.4.4 (5) ε o εi ε o. εi if zi z gc . zi 1 z gc , ε i , 99 ε o 1.118 = fo z gc 5.4.5 (3) 2 2 a) or c) β ε ε β ε ε > 4 , 8. 16 . > 15 , 25 . 150 . i i i i i i non heat ρ c if Ti > 0 , if , 1.0 , if , 1.0 i ε β β ε β β treated, welded i i i i i i Effective thickness t eff ρ .c t TALAT 2301 – Example 4.4 5
6. 6. Heat Affected Zone t eff if t eff > t haz.f , t haz.f , t eff t eff t eff n n n 8 n n t eff if t eff > t haz.w , t haz.w , t eff t eff t eff 9 9 9 15 9 βi t eff = i ε ρ c= = ψ i = Ti = i = gi = βi = εi = i i mm 1 1 1 1 16.667 1.118 14.907 0.465 2.788 2 1 1 1 16.667 1.118 14.907 0.465 2.788 3 0 1 1 30.667 1.118 27.429 0.712 4.272 4 0 1 1 30.667 1.118 27.429 0.712 4.272 5 0 1 1 30.667 1.118 27.429 0.712 4.272 6 1 1 1 16.667 1.118 14.907 0.465 2.788 7 1 1 1 16.667 1.118 14.907 0.465 2.788 8 0 1 1 0 1.118 0 1 0 9 0 -1.012 0.398 4.921 1.118 4.401 1 6.4 10 0 -1.012 0.398 4.921 1.118 4.401 1 8 11 0 -1.012 0.398 4.921 99 0.05 1 8 β i Max slender- β e if Ti > 0 , 0 , 12 0 1 1 0 99 0 1 8 ness i ε i 13 0 -1.012 0.398 4.921 99 0.05 1 8 - Internal 14 0 -1.012 0.398 4.921 1.118 4.401 1 8 elements β Imax max β e 15 0 -1.012 0.398 4.921 1.118 4.401 1 6.4 β Imax = 27.429 β i Max slender- β e if Ti > 0 , ,0 ness i ε i - Outstands β Omax max β e β Omax = 14.907 Cross section class I if β Imax 7 , 1 , if β Imax 11 , 2 , if β Imax 15 , 3 , 4 class I = 4 class class O if β Omax 2 , 1 , if β Omax 3 , 2 , if β Omax 4 , 3 , 4 class O = 4 (5.15) class if class I > class O , class I , class O class = 4 Area of effective t eff . thickness 2 2 dAi yi yi 1 zi zi 1 elements i rows ( y ) 1 Area of A eff dAi A = 5.52 . 10 mm effective 3 2 cross section i =1 A eff = 4.424 . 10 mm 3 2 First rows ( y ) 1 moment dAi Sy of area. Sy zi zi 1 . z gc 2 A eff Gravity centre i =1 z gc = 51.768 mm TALAT 2301 – Example 4.4 6
7. 7. Second rows ( y ) 1 moment of dAi zi . zi 1 . A eff . z gc 2 2 2 area of effective Iy zi zi 1 I eff Iy 3 cross section i =1 I eff = 8.344 . 10 mm 6 4 Section I eff W eff = 1.612 . 10 mm 5 3 modulus zd zi z gc max z.eff max z d W eff i max z.eff Compare I y.gr W gr = 1.853 . 10 mm 5 3 gross zd zi z gc.gr max z.gr max z d W gr i max z.gr section Heat Affected Zone Reduced thickness in HAZ h 2 .. 3 th .t ρ haz h th 3 th t9 .t ρ haz h t 15 .t ρ haz h ti . 2 2 Area dAi yi yi 1 zi zi 1 rows ( y ) 1 Area of A ele dAi A = 5.52 . 10 mm effective 3 2 cross section i =1 A ele = 5.126 . 10 mm 3 2 First rows ( y ) 1 moment dAi Sy of area. Sy zi zi 1 . z gc.e 2 A ele Gravity centre i =1 z gc.e = 44.228 mm Second rows ( y ) 1 moment of dAi zi . zi 1 . A ele. z gc.e 2 2 2 area of effective Iy zi zi 1 I ele Iy 3 cross section i =1 I ele = 1.013 . 10 mm 7 4 I ele W ele = 1.849 . 10 mm 5 3 Section zd zi z gc.e max z.ele max z d W ele modulus i max z.ele W el = 1.853 . 10 mm 5 3 TALAT 2301 – Example 4.4 7
8. 8. Plastic section modulus A = 5.52 . 10 mm 3 2 rows ( y ) 1 Area of ti . 2 2 A pl yi yi 1 zi zi 1 A pl = 5.126 . 10 mm effective 3 2 cross section i =1 Cross section 7 ti . 2 . t9. area of upper 2 2 2 2 A up yi yi 1 zi zi 1 y9 y8 z9 z8 part i =1 A up = 2.486 . 10 mm 3 2 A z9 = 30 mm Move node 8 A up 2 to plastic z10 z9 z13 z10 z10 = 47.1 mm neutral axis t 11 t 13 z10 y10 y5 . y y11 y13 y10 y10 = 96 mm 5 z11 rows ( y ) 1 Plastic section zi zi 1 W ple . t. yi yi 1 2 zi zi 1 2 W ple = 2.267 . 10 mm 5 3 modulus 2 i i =1 W ple = 1.224 W el Shape factor TALAT 2301 – Example 4.4 8
9. 9. W el Shape factor W ele 15 β Imax W ple W ele 4 β Omax W ple (5.15) α 3I . 1 . 1 α 3O . 1 . 1 W el 15 11 W ele W el 4 3 W ele α 3I 0.296 = α 3O 1.465 = α 3 if α 3I < α 3O , α 3I , α 3O α 3 1.465 = W ple W eff α if class< 3 , , if class< 4 , α 3 , class = 4 α = 0.87 W el W el 0 50 100 200 150 100 50 0 50 100 150 200 Elastic neutral axis: - first iteration dashed-dotted line - second iteration dashed line Plastic neutral axis dotted line TALAT 2301 – Example 4.4 9
10. 10. Bending moment resistance fo W el = 1.853 . 10 mm 5 3 (5.14) M y.Rd α . W el. γ M1 W ple = 2.267 . 10 mm 5 3 M y.Rd = 29.3 kN . m Cross section class = 4 α = 0.87 class A eff = 4.424 . 10 mm 3 2 I eff = 8.344 . 10 mm 6 4 0 50 100 200 150 100 50 0 50 100 150 200 TALAT 2301 – Example 4.4 10