TALAT Lecture 2301

                        Design of Members

                           Bending Moment


   Example 4.1 ...
Example 4.1. Bending moment resistance of open cross section
with closed part
                     fo     300 . MPa       ...
First                     rows ( y )         1
                                                                  dAi      ...
Effective cross section.                                           First iteration
                                       ...
Effective cross section.                                       Second iteration
Node 8 is moved to the neutral axis       ...
rows ( y )             1
Area of
                         A eff                                         dAi
              ...
Bending moment resistance                    300


                                       fo
(5.14)            M Rd    α ....
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TALAT Lecture 2301: Design of Members Example 4.1: Bending moment resistance of open cross section with closed part

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

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TALAT Lecture 2301: Design of Members Example 4.1: Bending moment resistance of open cross section with closed part

  1. 1. TALAT Lecture 2301 Design of Members Bending Moment Example 4.1 : Bending moment resistance of open cross section with closed part 7 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm Date of Issue: 1999  EAA - European Aluminium Association TALAT 2301 – Example 4.1 1
  2. 2. Example 4.1. Bending moment resistance of open cross section with closed part fo 300 . MPa γ M1 1.1 Half flange width b 150 Half section depth h 240 Half closed part c 75 Flange thickness t 8 Web thickness d 4 Zero o 0 Type of element - Internal element: T = 0 - Outstand: T > 0 Comment: Node 8 is added to indicate shift in neutral axis Nodes, co-ordinates (y, z), thickness (t) and types (T) 0 b h o o 1 c h t 2 2 c h t o MPa 10 . Pa 6 3 b h t 1 4 c h o o kN 1000 . newton 5 o h c d o i y . mm z . mm t . mm T 6 c h o o 7 o h c d o 8 o o 1 d o 300 9 o h d o 10 b h o o 200 11 b h t 1 100 The beam is composed by two extrusions. The weld is close to the neutral axis why HAZ softening does not 0 z influence the moment resistance. 0 mm 100 Nodes i 1 .. rows ( y ) 1 rows ( y ) = 12 Area of cross 200 ti . section 2 2 dAi yi yi 1 zi zi 1 elements 300 200 100 0 100 200 Cross rows ( y ) 1 y A = 7.269 . 10 mm section 3 2 A dAi mm area i =1 TALAT 2301 – Example 4.1 2
  3. 3. First rows ( y ) 1 dAi Sy moment Sy zi zi 1 . z gc of area 2 A Gravity centre i =1 z gc = 15.282 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 = 3.344 . 10 mm 8 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 = 1.31 . 10 mm 6 3 Torsion rows ( y ) 1 2 ti dAi . I tu = 1.156 . 10 mm constant for 5 4 I tu open parts 3 i =1 Hollow part in top flange y1 z1 t1 75 240 8 y2 75 z2 240 t2 8 yh yh= mm zh zh = mm th th= mm y5 0 z5 165 t5 4 y6 75 z6 240 t7 4 rows y h 1 Area within 0.5 . y h . z A tu = 5.625 . 10 mm 3 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 = 71.783 th i =1 i 4 .A 2 Torsion constant tu I t = 1.763 . 10 mm 6 4 for closed part It Dn Torsion constant I t = 1.879 . 10 mm 6 4 whole section It It I tu Torsion 2 . A tu . min t h W v = 4.5 . 10 mm 4 3 resistance Wv min t h = 4 mm TALAT 2301 – Example 4.1 3
  4. 4. Effective cross section. First iteration zi 1 z gc zi z gc zd zi z gc 5.4.3 (1) ψi if zi 1 z gc < zi z gc , , i zi z gc zi 1 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 , 1 ψ i Outstand Figure 5.2 gi if Ti > 0 , 1 , gi 2 2 yi yi 1 zi zi 1 250 . MPa 5.4.3 (1) c) βi if t i > 0 , gi . ,1 ε o ε o 0.913 = ti fo max z max z 5.4.4 (5) εi if zi z gc . zi 1 z gc , if zi 1 z gc < zi z gc , ε o . , ε o. , 99 zi z gc zi 1 z gc 5.4.5 (3) 2 2 a) or c) β ε ε β ε ε > 6 , 10 . 24 . > 22 , 32 . 220 . i i i i i i heat-treated, ρ c if Ti > 0 , if , 1.0 , if , 1.0 i ε β β ε β β unwelded i i i i i i Effective βi t eff t eff ρ .c t = i thickness ε ρ c= = ψi = gi = βi = εi = i i mm 1 1 9.375 0.973 9.635 0.779 6.235 1 1 18.75 0.973 19.271 1 8 Comment: 1 1 9.375 0.973 9.635 0.779 6.235 εi was given a large value (99) 1 1 1 0.973 1.028 1 0 for elements entirely on 0.666 0.9 23.862 0.973 24.524 0.939 3.756 the tension side 0.666 0.9 1 0.973 1.028 1 0 0.666 0.9 23.862 0.973 24.524 0.939 3.756 -0.109 0.667 27.696 99 0.28 1 4 0.064 0.719 42.968 99 0.434 1 4 1 1 1 99 0.01 1 0 1 1 37.5 99 0.379 1 8 Area of effective thickness t eff . elements 2 2 dAi yi yi 1 zi zi 1 i Cross rows ( y ) 1 A = 7.269 . 10 mm 3 2 section A eff dAi area A eff = 6.952 . 10 mm 3 2 i =1 First moment rows ( y ) 1 dAi Sy of area Sy zi zi 1 . z gc 2 A eff Gravity centre i =1 z gc = 5.329 mm TALAT 2301 – Example 4.1 4
  5. 5. Effective cross section. Second iteration Node 8 is moved to the neutral axis z8 z gc 0.01 . mm from the first iteration zi 1 z gc zi z gc zd zi z gc 5.4.3 (1) ψi if zi 1 z gc < zi z gc , , i zi z gc zi 1 z gc max z max z d 0.8 max z = 245.329 mm (5.7 or 5.8) gi if ψ i > 1 , 0.7 0.30 . ψ i , 1 ψ i Outstand gi if Ti > 0 , 1 , gi element yi yi 1 2 zi zi 1 2 z gc = 5.329 mm 5.4.3 (1) c) βi if t i > 0 , gi . ,1 ti z8 = 5.339 mm z9 = 240 mm max z max z 5.4.4 (5) εi if zi z gc zi 1 z gc , if zi 1 z gc < zi z gc , ε o . , ε o. , 99 zi z gc zi 1 z gc 5.4.5 (3) a) or c) 2 2 β ε ε β ε ε > 6 , 10 . 24 . > 22 , 32 . 220 . i i i i i i heat-treated, ρ c if Ti > 0 , if , 1.0 , if , 1.0 unwelded i ε i β i β i ε i β i β i Effective βi t eff t eff ρ .c t = i thickness ε ρ c= = ψ i = Ti = i = gi = βi = εi = i i mm 1 2 1 1 9.375 0.933 10.044 0.758 6.062 β i Max slender- β e if Ti > 0 , 0 , 2 0 1 1 18.75 0.933 20.088 1 8 ness i ε i 3 1 1 1 9.375 0.933 10.044 0.758 6.062 - Internal 4 0 1 1 1 0.933 1.071 1 0 elements β Imax max β e 5 0 0.68 0.904 23.974 0.933 25.686 0.912 3.65 β Imax = 25.686 6 0 0.68 0.904 1 0.933 1.071 1 0 7 0 0.68 0.904 23.974 0.933 25.686 0.912 3.65 β i 8 06.263·10 -5 0.7 27.941 1.132 24.693 0.935 3.74 Max slender- β e if Ti > 0 , ,0 9 0 -1·10 30.01 0.613 0.913 0.672 1 4 ness i ε i 10 0 1 1 1 99 0.01 1 0 - Outstands β Omax max β e 11 1 1 1 37.5 99 0.379 1 8 β Omax = 10.044 Cross section class I if β Imax 11 , 1 , if β Imax 16 , 2 , if β Imax 22 , 3 , 4 class I = 4 class class O if β Omax 3 , 1 , if β Omax 4.5 , 2 , if β Omax 6 , 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 TALAT 2301 – Example 4.1 5
  6. 6. rows ( y ) 1 Area of A eff dAi A = 7.269 . 10 mm effective 3 2 cross section i =1 A eff = 6.862 . 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 = 3.31 mm 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 = 3.158 . 10 mm 8 4 Section I eff W eff = 1.298 . 10 mm 6 3 modulus zd zi z gc max z.eff max z d W eff i max z.eff Compare I y.gr W gr = 1.31 . 10 mm 6 3 gross zd zi z gc.gr max z.gr max z d W gr i max z.gr section Plastic section modulus A = 7.269 . 10 mm 3 2 Cross section 7 ti . A up = 3.249 . 10 mm area of upper 2 2 3 2 A up yi yi 1 zi zi 1 part i =1 A Move node 8 A up 2 to plastic z8 z7 z8 = 68.566 mm neutral axis t8 rows ( y ) 1 Plastic section zi zi 1 W pl . t. yi yi 1 2 zi zi 1 2 W pl = 1.475 . 10 mm 6 3 modulus 2 i i =1 W pl = 1.126 W el Shape factor 22 β Imax W pl 6 β Omax W pl (5.15) α 3I 1 . 1 α 3O 1 . 1 22 16 W el 6 4.5 W el α 3 if α 3I < α 3O , α 3I , α 3O α 3 0.661 = W pl W eff α if class< 3 , , if class< 4 , α 3 , class = 4 α = 0.991 W el W el TALAT 2301 – Example 4.1 6
  7. 7. Bending moment resistance 300 fo (5.14) M Rd α . W el. γ M1 200 M Rd = 354 kN . m 100 Cross section class = 4 class Elastic neutral axis: 0 - first iteration dashed-dotted line - second iteration dashed line Plastic neutral axis dotted line 100 200 300 200 100 0 100 200 TALAT 2301 – Example 4.1 7

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