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Diagram Interaksi Kolom Muhammad Fauzi Novrizaldy
DIAGRAM INTERAKSI KOLOM
Diketahui penampang kolom beton bertulang yang mempunyai tulangan di keempat sisinya, kolom ditulangi pada keempat sisinya sebanyak 12D19 dan sengkang D10
Luas Tulangan
f'c = 30 Mpa b = 350 mm d' = p + ½ x 19 + 10 = 59,5 mm
fy = 400 Mpa h = 350 mm d = h-d' = 500-59,5= 290,5 mm
As = As'= (¼ x 3,14 x 192 ) x 6 = 1701,17 mm2 Ast = As'+ As = 1701,17x 2 =3402,34 mm2
Ag = 350 x 350 = 122500 mm2
Beban Terpusat
Pnmax = 0,8 x (0,85 x f'c x Ag + Ag x fy) = 0,8 x (0,85 x 30 x 122500 + 3402,34 x 400) = 3587748 N = 3587,74 kN
[0 ; 3587,74]
Keadaan Seimbang
Cb = 600600+fy x d= 600600+400 x 290,5=174,3 mm
Ɛs' = 0,003 xCb-d'Cb = 0,003 x174,3-59,5174,3 =0,0019 mm/mm
Ɛy = fyEs = 400200000 =0,002 mm/mm
fs' = Ɛs'x Es = 0,0019 x 2 x 105 =395 MPa
ab = β1 x Cb
Menghitung reduksi β1 akibat f'c>28 MPa
β1=0,85 x 27x0,05=0,84 ;
ab = β1 x Cb = 0,84 x 174,3 = 146,412 mm
Pnb = 0,85 f'c x b x ab + As' x fs' -As x fy
= (0,85 x 30 x 350 x 146,12 ) + (1701,17 x 380) – (1701,17 x 400)
= 1295300 N = 1298,53 kN
Mnb =[ (0,85 f'c x b x ab x (y-a2 )) + (As' x fs' x (y-d')) – (As x fy x (d-y)) ]
= [ 0,85 x 30 x 350 x 146,412 x (3502-146,122) + (1701,17 x 395 x ( 3502-59,5)) +(1701,17 x 400 x (290,5-3502)) x 10-6
= 289,26 kNm
eb = MnbPnb= 286,191270,09=0,225 m
[286,19 ; 1270,09]
Lentur Murni
Mno = As x fy (d - a2) ; dimana a = As x fy0,85 x f'c x b= 1701,17 x 4000,85 x 30 x 350=76,24 mm
= 1701,17 x 400 (290,5 - 76,242) x 10-6 = 171,73 kNm
[171,13 ; 0,00]
Compression Control
c > cb ; cb = 174,3 mm
Coba nilai c = 180 mm
Ɛs' = 0,003 xc-d'c = 0,003 x180-59,5180 =0,00201 mm/mm Ɛs'> Ɛy (tulangan leleh)
fs' = fy = 400 MPa
Ɛs = 0,003 x d-cc= 0,003 x 290,5-180180=0,0018 mm/mm
fs = Ɛs x Es = 0,0018 x 200000 = 368,3 MPa
a = β1 x c = 0,84 x 180 = 151,2 mm
Cc = ( 0,85 x f'c x b x a) x 10-3 = ( 0,85 x 30 x 350 x 151,2) x 10-3 =1349,46 kN
Ts = As x fs = 1701,17 x 368,3 x 10-3 = 626,54 kN
Cs = As' x fs' = 1701,17 x 400 = 680468 x 10-3 = 680,46 kN
Pn = Cc + Cs – Ts = 1349,460 + 680,46 - 626,54 = 1403,38 kN
Mn = [Cc (y-a2) + Cs (y-d') + Ts (d-y) ] x 10-6
= [ 1349460 x (175-151,22) + 680460 x ( 175 -59,5) + 626540 x ( 290,5-175) ] x 10-6
= 285 kNm
[285 ; 1403,38]
Tension Control
c < cb ; cb = 174,3 mm
Coba nilai c = 150 mm
Ɛs' = 0,003 xc-d'c = 0,003 x150-59,5150 =0,0018 mm/mm Ɛs'< Ɛy (tulangan leleh)
fs' = Ɛs' x Es = 0,0018 x 200000 = 362 MPa
Ɛs = 0,003 x d-cc= 0,003 x 290,5-150150=0,0028 mm/mm
fs = fy = 400 MPa
a = β1 x c = 0,84 x 150 = 126 mm
Cc = ( 0,85 x f'c x b x a) x 10-3 = ( 0,85 x 30 x 350 x 126) x 10-3 =1124,5 kN
Ts = As x fs = 1701,17 x 400 x 10-3 = 680,46 kN
Cs = As' x fs' = 1701,17 x 362 = 615823,5 x 10-3 = 615,82 kN
Pn = Cc + Cs – Ts = 1124,5 + 615,82 - 680,46 = 1059,86 kN
Mn = [Cc (y-a2) + Cs (y-d') + Ts (d-y) ] x 10-6
= [ 1124500 x (175-1262) + 615820 x ( 175 -59,5) + 680460 x ( 290,5-175) ] x 10-6
= 275,66 kNm
[275,66 ; 1059,86]
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Koordinat X dan Y untuk Diagram Interaksi
Diagram Interaksi Kolom
