Design of Members. Shear Force. Example : Shear resistance of webs without and with stiffeners

Transcription

1 TALAT Lecture 0 Design o Members Ser Force Exmple : Ser resistnce o ebs itout nd it stieners pges Advnced Level prepred by Torsten Höglund, Royl Institute o Tecnology, Stockolm Dte o Issue: 999 EAA - Europen Aluminium Assocition TALAT 0 Exmples

2 Exmple Ser resistnce o ebs itout nd it stieners Dimensions nd strengt o mteril Dt input in igligted regions Web dept 000. MP P Web tickness t 5. kn 000. neton Totl lengt o eb pnel l , proo strengt o eb plte mteril o 55. MP Ultimte strengt o eb plte mteril u 470. MP Elstic modulus E MP Prtil coeicient. spv.4. kg (Mss) ( 0. m) u Tble 5. η η = o ) No intermedite stiener I "rigid end post" ten endpost = else endpost = 0 endpost Lengt o eb pnel l = 4000 Re. to Eurocode 9 (5.97) i > ,, = = 6.4 (5.96) λ.. t o E λ =.055 Tble 5. ρ v i λ > ,,.66 λ λ Tble 5. ρ v i ρ v > η, η, ρ v Tble 5. ρ v i endpost 0 i ρ , v >,, ρ v, ρ v λ λ ρ v = 0.8 ρ v = 0.8 ρ v = 0.8 (5.95) V.Rd ρ. v t.. o V.Rd =.7. 0 kn TALAT 0 Exmples Weigt o eb Mss. t. l. spv Mss = 88 kg V Rd V.Rd

3 ) Eqully spced lexible trnsverse stieners I "rigid end post" ten endpost = else endpost = 0 endpost Number o stieners n st Single plte stiener t st 5. b st Wole eb pnel. Buckling o stiener 0. b st = 8 t st Figure 5. A st t. st b st 0. t A st = t. st b st C.G. e st Second moment o re. A st t. st b st I st e st =.6 A. st e st I st = Pnel lengt l = 40. (5.99) st 9. 4 I. st t. st =.8. I (5.99) k. st τ stmin k t τ stmin =.5 (5.97) (5.98) i st i st < stmin, stmin, st st =.8 > ,, k τ = 6.4 st = 8.7 (5.96) λ.. o λ t E =.605 λ tot λ TALAT 0 Exmples

4 Sub pnel Lengt l n st = 0. = 0.5 (5.97 nd 5.98) i > ,, k τ = 5.6 (5.96) λ.. o λ sub t E λ λ sub =.57 λ tot =.605 Te lrger o te slenderness prmeter λ tot or te totl pnel nd λ sub or te sub pnels is used. I λ tot > λ sub ten te stieners re lexible else te stieners re rigid. λ i λ tot > λ sub, λ tot, λ sub λ =.605 Tble 5. ρ v i λ > ,, ρ v.66 λ λ i ρ v > η, η, ρ v ρ v = 0. Tble 5. ρ v i endpost 0 i ρ , v >,, ρ v, ρ v ρ v λ λ = 0. (5.95) V Rd ρ. v t.. o V Rd = kn Alterntive = trnsverse stieners V Rd V Rd Weigt Mss. t. l n. st b. st t. st. spv Mss = 4 kg TALAT 0 Exmples

5 ) Trnsverse intermedite, rigid stieners Number o trnsverse stieners n st endpost = Web pnel lengt l n st = 0. = 0.5 (5.97) i > ,, k τ = 5.6 (5.96) λ.. o λ t E =.57 Tble 5. ρ v i λ > ,, ρ v.66 λ λ = 0.44 Tble 5. ρ v i ρ v > η, η, ρ v ρ v = 0.44 (5.95) V Rd ρ. v t.. o V Rd = kn Ceck rigidity o stiener t st 8. b st 0. Figure 5. A st t. st b st 0. t A st = t. st b st C. G. e st Second moment o re (5.04) or (5.05) Axil orce in stiener. A st t. st b st I st I limit i e st = A. st e st I st = t <, ,. t I limit = N st V Rd.4. t. E.. o TALAT 0.06 N st (6.0) σ st σ A st = 4 MP st I st > I limit OK! o = MP σ st < o OK! Alterntive = trnsverse intermedite, rigid stieners V Rd V Rd TALAT 0 Exmples Weigt Mss. t. l n. st b. st t. st. spv Mss = 45 kg

6 4) Rigid trnsverse stieners nd longitudinl stieners Number o longitudinl stieners n sl Longitudinl plte stiener t sl 5. b sl 0. Number o rigid trnsverse stieners n st = I "rigid end post" ten endpost = else endpost = 0 endpost b sl = 8 t sl Buckling o longitudinl stiener Are A sl t. sl b sl 40. t A sl = t. sl b sl GC e sl. A sl e sl = 0 Second moment o re t. sl b sl I sl A. sl e sl I sl = Pnel lengt l n st = 0. (5.99) st 9. n. 4 sl I. sl t. st = n. sl I (5.99) k. sl τ stmin k t τ stmin =.748 (5.99) st i st < stmin, stmin, st st = (5.97) st = (5.96) λ.. o λ t E = 0.8 Wole pnel λ tot λ λ tot = 0.8 TALAT 0 Exmples

7 Sub pnel Dept = 667 (5.98) k τ = 6.7 (5.96) λ.. t o E λ = 0.8 Sub-pnel λ sub λ λ sub =.06 compre λ tot = 0.8 Te lrger o te slenderness prmeter λ tot or te totl pnel nd λ sub or te sub pnels is used. I λ tot > λ sub ten te stieners re lexible else te stieners re rigid. λ i λ tot > λ sub, λ tot, λ sub λ =.06 Tble 5. ρ v i λ > ,, ρ v.66 λ λ = 0.49 Tble 5. ρ v i ρ v > η, η, ρ v Tble 5. ρ v i endpost 0 i ρ , v >,, ρ v, ρ v ρ v λ λ = 0.49 (5.95) V Rd ρ. v t.. o V Rd = kn Alterntive 4 = trnsverse nd longitudinl intermedite stieners V Rd4 V Rd Weigt Mss 4. t. l n. st b. st t. st n. sl t. sl b. sl l. spv Mss 4 = 80 kg TALAT 0 Exmples

8 5) Ser resistnce contribution o te lnges dded to girder 4 Te pnel is t te end o te plte girder. Ten te bending moment is neglected Flnge tickness nd idt t 50. b 750. Yield strengt o lnge plte o 55. MP Design ser orce V Ed kn Design bending moment M Ed V. Ed l M Ed = kn. m (5.0) M Rd b. t. t (5.0) c b. t. o. t.. o (5.0) V Rd i M Ed < M Rd. o M Ed γ = M M M Rd = kn. m Rd = 0. c = 7.5 b. t. o M Ed,., 0 V c. M Rd = kn Rd Sum V Rd5 V Rd V Rd Exmple 5 = inclusive ser resistnce contribution o te lnges V Rd5 = kn Weigt Mss 5. t. l n. st b. st t. st n. sl t. sl b. sl l. spv Mss 5 = 80 kg TALAT 0 Exmples

9 6) Corrugted eb Web dept 000. MP P Web tickness t. kn 000. neton Totl lengt o eb pnel l , proo strengt o eb plte mteril o 55. MP Ultimte strengt o eb plte mteril u 470. MP Elstic modulus E MP Trpezoidl eb: b o 40. b u 40. b d 400. c 00. s b d b o b. u 0.5 c s = 6.69 Prtil coeicient. Tble 5. η 0.4 u 0.. o η = Locl buckling o b o, b u nd b u Mx idt b m i b o < b u, b u, b o b m i b m > s, b m, s b m = 40 b m = 40 b m (5.96) λ t o E λ = 0.9 Tble 5. Non-rigid end post ρ v i λ 0.48 <, η η 0.48, η = ρ v λ = (5.7) V Rd 0.7. ρ. vt.. o V Rd = kn TALAT 0 Exmples

10 Trpezoidl eb, idt b d Are A b. o t b. u t. s. t A = Grvity centre e gc b. o t. c. s. t c A e gc = 50 Second moment I x b. o t. c. s. t. c Ae. o re gc b d I x = b d t (5.) I. z b u b o. s 0.9 I z = E 4 (5.) V. o.cr I. z I x V o.cr = kn. t. o (5.0) λ o λ V o = 0.76 o.cr 0.7. ρ v= (5.9) χ o 0.8 λ o χ o i χ o> 0.7. ρ v, 0.7. ρ v, χ o χ o = 0.45 (5.8) V o.rd χ. o. t. o V o.rd = kn Min V Rd6 i V Rd > V o.rd, V o.rd, V Rd V Rd6 = kn b Weigt Mss 6 t. l. u b o. s.. spv Mss b 6 = 96 kg d TALAT 0 Exmples

11 Sury ) No intermedite stieners t _5 = 5 ) Trnsverse lexible stieners ) Trnsverse rigid stieners 4) Trnsverse rigid + longitudinl stieners 5) Trnsverse rigid + longitudinl stieners + contribution o lnges 6) Trpezoidl eb t = i Increse in V Rdi resistnce = = 000. kn i = Mss i = Mss Weigt / resistnce Mss i. = Mss i "No buckling" V 0.Rd η.. t. o _5 V 0.Rd = kn Coents Te second column gives te increse in ser resistnce compred to girder en dding stieners. By dding bot trnsversl nd orizontl stieners (5) te resistnce cn be lmost doubled. Te lst column, "eigt per resistnce" so tt stieners give ligter girders, but te comprison does no py regrd to te cost or elding o te stieners. Te contribution o te lnges (girder 5) increse te resistnce it =.7 % compred to girder 4 Torsten Höglund TALAT 0 Exmples