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# TALAT Lecture 2301: Design of Members Example 7.1: Concentrated force

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This example provides calculations on concentrated load of members based on Eurocode 9.

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### TALAT Lecture 2301: Design of Members Example 7.1: Concentrated force

1. 1. TALAT Lecture 2301 Design of Members Concentrated Load Example 7.1 : Concentrated force 4 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm Date of Issue: 1999  EAA - European Aluminium Association TALAT 2301 – Example 7.1 1
2. 2. Example 7.1. Concentrated force Dimensions and material properties Beam: Purlin (p): Section height: h 570 . mm Section height: hp 180 . mm Flange width: b 160 . mm Flange width: bp 12 . mm Flange thickness: tf 16 . mm Flange thickness: t fp 12 . mm Web thickness: tw 5 . mm Web thickness: t wp 5 . mm Fillet radius: r 4 . mm Fillet radius: rp 5 . mm Overall length: L 10 . m Distance between purlins: c p 1.2 . m Table 3.2b Alloy: EN AW-6082 T6 EP/O t > 5 mm f 0.2 260 . MPa fu 310 . MPa (5.4), (5.5) fo f 0.2 fa fu E 70000 . MPa Partial safety factors: γ M11.10 S.I. units kN 1000 . newton kNm kN . m MPa 1000000 . Pa TALAT 2301 – Example 7.1 2
3. 3. Concentrated force resistance of beam 5.12.8 bf b f of fo f ow fo hw h 2 .t f h w = 538 mm a L Figure 5.24 Length of stiff bearing g r p. 2 rp . 2 g = 2.9 mm and above ss t wp 2 .g 2 . t fp s s = 34.9 mm f of . b f (5.109) Parameter m 1 m1 m 1 = 32 f ow . t w 2 hw Figure 5.24 Buckling coefficient kF 6 2. k F = 6.01 a 4 . t f . h w . f ow 2 ss hw (5.110) Parameter m 2 m2 if > 0.2 , 0.02 . ,0 m 2 = 22.6 k F .E .t w 2 tf (5.111) Effective loaded lengthly ly ss 2 .t f . 1 m1 m2 l y = 303.3 mm k .l .f .E (5.108) Design resistance F Rd . t 2 . F y ow . 1 0.57 w F Rd = 101.7 kN hw γ M1 f ow f ow F Rd if F Rd> t w . l y . , t w .l y . , F Rd F Rd = 101.7 kN γ M1 γ M1 5.12.9 Flange induced buckling Elastic moment resistance utilised k 0.55 f of fo . hw k .E . h w t w (5.115) LS = = 107.6 RS = = 151.8 LS < RS OK! tw f of b .t f TALAT 2301 – Example 7.1 3
4. 4. Concentrated force resistance of purlin (upward) 5.12.8 bf bp f of fo f ow fo h wp hp 2 . t fp h wp = 156 mm ap 10 m Figure 5.24 Length of stiff bearing g r. 2 r . 2 g = 2.3 mm and above ss tw 2 .g 2 .t f s s = 41.7 mm f of . b f (5.109) Parameter m 1 m1 m 1 = 2.4 f ow . t wp 2 h wp Figure 5.24 Buckling coefficient kF 6 2. kF=6 ap 4 . t fp . h wp . f ow 2 ss h wp (5.110) Parameter m 2 m2 if > 0.2 , 0.02 . ,0 m 2 = 3.4 k F . E . t wp 2 t fp (5.111) Effective loaded lengthly ly ss 2 . t fp . 1 m1 m2 l y = 123.4 mm k F . l y . f ow . E 1 0.57 . t 2. . (5.108) Design resistance F Rd wp F Rd = 120.4 kN h wp γ M1 f ow f ow F Rd if F Rd> t wp . l y . , t wp . l y . , F Rd F Rd = 120.4 kN γ M1 γ M1 TALAT 2301 – Example 7.1 4