Steel Truss System Axial tension member design 1 Nominal Strength Of Tension Member • Yield tension behavior • Fracture tension Behavior • Shear Block tension Behavior Minimum value 2 Tension Member On SNI T – 03 – 2005 Batasan Kelangsingan ๏ฌ๏ฝ 3 L r๏ฝ r , I A L = Panjang Batang r = Jari – jari girasi I = Momen Inersia A = Luas penampang batang Steel Truss system 4 Steel Truss system Tension Member Behavior Fracture tension behavior Yield Tension behavior 5 Steel Truss system Source: Dewobroto, 2015 Tension Member On SNI T – 03 – 2005 Øslot hole = Øbolt+ 2mm 6 Steel Truss system Tension Member On SNI T – 03 – 2005 Gambaran untuk mengetahui nilai x dan L ada pada slide berikutnya 7 Steel Truss system 8 Steel truss system 9 Steel truss system Shear Leg Factor (U) l Tu X × U= 1- ๐ ≥ 0,6 (if use 3 bolt) × ๐ U= 1- ≥ 0,8 (if use 4 bolt) xo Tu 10 Yo Steel truss system Effective Netto Section A d=Øbolt+2mm H Tu A Anetto = t x H – 2x(t x d) U = 1,0 Ae = Anetto x U t A-A cross section l t Anetto = Ag – (t x d) × U= 1- ≥ 0,6 (if use 3 bolt) X B U= d=Øbolt+2mm xo Yo Tu 11 B Steel truss system ≥ 0,8 (if use 4 bolt) Ae = Anetto x U × U= 1- ≥ 0,6 (if use 3 bolt) U= B-B cross section l × 1l l × 1l ≥ 0,8 (if use 4 bolt) Effective Netto Section U= 1- ๐ฅ0 ๐ฟ > 0,6 3 ๐๐๐๐ก U= 1- ๐ฅ0 ๐ฟ > 0,8 4 ๐๐๐๐ก L xo Example: double angle profile 2L 50x50x5-5 The center gravity of each angle (Xo) = 14,1 mm Distance end-end bolt are 100 mm (L) xo Yo Tu 12 Then U factor is 14,1 U= 1- 100 =0,859 Steel truss system Staggered Bolt Configuration Source: Dewobroto, 2015 13 Steel truss system Staggered Bolt Configuration s s g g g g g a d s h b s e c a-b-c path f i :Anet = Ag – 2x(dxt) a-d-b-e-f path g-d-b-e-f path : 14 s ๐ 2 : Anet= Ag – 4x(dxt) +σ 4๐ 2 ๐ Anet= Ag – 4x(dxt) +σ x 4๐ xt t Steel truss system g Staggered Bolt Configuration 15 Steel truss system Effective Netto Section Ag 1 angle = 480,2 mm2 t plate = 5 mm d hole = 19 + 2 = 21 mm Xo = 14,1 mm 75 75 75 xo xo 50 A-C Path Anet = 2 x Ag – 2x d x t Anet =2 × 480,2 − 2 × 21 × 5 Anet = 750,4 mm2 A-B-D path Anet = 2 × 480,2 − 4 × 21 × 5 502 + x5= 4×75 Yo U= 1- Tu 75 75 50 A B C 16 14,1 225 582,067 mm2 = 0,937> 0,8 then U= 0,937 75 Ae = Anet x U = 582,067 x 0,937 Ae = 545,59 mm2 fpn = 0,75 AexFu= 151,4 kN D Steel truss system Staggered Bolt Configuration on Difference Plane Source: Dewobroto, 2015 17 Steel truss system Staggered Bolt Configuration on Difference Plane 18 Steel truss system Staggered Bolt Configuration on Difference Plane H 300 x 300 IWF 300X300 Pelat Buhul 18 mm IWF 300X300 18 lubang Ø26 fPn(t)=....? Fy = 240 Mpa Fu = 370 Mpa 19 19 Steel truss system Staggered Bolt Configuration on Difference Plane IWF 300x300 Ag = 11980mm2 dhole = 26 mm tf = 15 mm tw= 10 mm A-B Path Anet = ๐ด๐ − 4 × ๐ × ๐ก = 11980 − 4 × 26 × 15 = 10420 ๐๐2 C-A-B-D Path Anet = ๐ด๐ − 8 × ๐ × ๐ก + σ ๐ 2 4๐ 582 = 11980 − 8 × 26 × 15 + 2 × × 15 4 × 30 = 9701 mm2 C-E-F-D Path Anet = ๐ด๐ − 8 × ๐ × ๐ก + σ 20 20 Steel ๐ 2 4๐ 582 = 1198000 − 8 × 26 × 15 + 2 × × 15 4 × 30 truss system = 9701 mm2 Staggered Bolt Configuration on Difference Plane IWF 300x300 Ag = 1198000 mm2 dhole = 26 mm tf = 15 mm tw= 10 mm 1 2 Anet = 1196640 mm2(A-B path) U calculation: Center gravity of half Shape Ae = Anet x U Ae = 9701 x 0,961 Ae = 9322,66 mm2 Pn = Fy x Ag 240×11980 = =2875,2 kN 1000 Pn = Fu x Ae 370×9322,66 = = 3449,38 kN 1000 fPn= 0,9 x 2875,2 = 2587,68 kN 21 el b h A Xi AxXi 1 15 300 4500 7,5 33750 2 135 1350 67,5 91125 2110 5850 σ ๐ด๐×๐๐ 124875 = =21,346 mm σ ๐ด๐ 5850 21,345 1− = 0,961> 0,9 maka 554 124875 ๐ฅาง = U= Steel truss system U=0,961 BLOCK SHEAR Session 7 Axial tension member design 22 Steel truss system BLOCK SHEAR 23 Steel truss system BLOCK SHEAR STRESS 24 Steel truss system BLOCK SHEAR STRESS 25 Steel truss system BLOCK SHEAR STRESS 26 Steel truss system BLOCK SHEAR EXAMPLE Source: dewobroto, 2015 27 Steel truss system BLOCK SHEAR EXAMPLE (Anv) (Ant) 28 Steel truss system BLOCK SHEAR EXAMPLE 29 Steel truss system BLOCK SHEAR TASK Agv = (4x75+3x75)x 5x2 = 5250 mm2 Ant = 752 + 502 x 5x2 = 901,39 mm2 AnV= (75x4-1,5x21+75x3-1,5x21)x5x2= 462 mm2 Ubs = 1,0 Fy = 240 Mpa; Fu = 370 Mpa xo = 436,08 kN Yo = 1089,51 kN 75 75 75 75 fRn = 0,75 x 436,08 = 327,06 50 Ant 30 Anv Steel truss system LATIHAN 31 LATIHAN dan tentukan kekuatan tarik desainnya (ØPn). 32