TALAT Lecture 2301: Design of Members Example 4.1: Bending moment resistance of open cross section with closed part
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This 3-sentence summary provides the high-level information about the document: The 7-page document presents Example 4.1 on calculating the bending moment resistance of an open cross-section aluminum member with a closed part, including determining the effective thickness of elements, cross-sectional properties, and section modulus. Nodes, coordinates, thicknesses, and other geometric properties of the beam are defined, and formulas are provided to calculate bending properties based on the effective cross section.
TALAT Lecture 2301: Design of Members Example 4.1: Bending moment resistance of open cross section with closed part
- 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. 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. 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. 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. 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. 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. 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