CONTOH 3Terdapat PLAT yang dijadikan BASE dan disambung menggunakan ANGKUR
kedalam BETON
F’c
BJ - 41
IWF 250 x 125
KETERANGAN GAMBAR
• B = 350 mm
• N = 350 mm
Menggunakan,
BJ – 41
F’c = 30 MpaF’c
BJ - 41
IWF 250 x 125
tp = 10 mm
PEMBEBANAN
• DEAD LOAD
Berat sendiri kolom,
29.6 kg/m x 5 = 148 kg
Berat Sendiri penyangga,
Double Angel
39.88 kg/m x 2 = 79.76 kg/m
PEMBEBANAN
• WIND ( ISAP )
BI angin = 25 kg/m²
Lebar plat = 125 mm
= 0.125 m
Maka.
qWI = 25 x 0.125
= 3. 13 kg/m
PEMBEBANAN
• WIND ( TEKAN )
BI angin = 25 kg/m²
Lebar plat = 125 mm
= 0.125 m
Maka.
qWT = 25 x 0.125
= 3. 13 kg/m
KOMBINASI PEMBEBANAN
KOMBINASI
• COMB 1 : 1.4 DL
• COMB 2 : 1.2 DL + 0.8 W
• COMB 3 : 1.2 DL + 1.3 W + LL
• COMB 4 : 0.9 DL + 1.3 W
HASIL GAYA TERBESAR
• Pu = 9.3302 KN
= 9330.2 N
• Vu = 0.234 KN
= 234 N
• Mu = 14.6521 KN m
= 14652100 N mm
LUASAN
• A1 = Pendestel ( Beton )
= B x N
= 350 x 350
= 122500 mm²
• A2 = Base Plate
= 122500 mm²
350
350
TEGANGAN YANG TERJADI
• Fpp1 = 0.6 x 0.85 x f ’c x √(A2/A1)
= 0.6 x 0.85 x 30
= 15.3 Mpa
• Fpp2 = 0.6 x 1.7 x f ’c
= 0.6 x 1.7 x 30
= 13.6 Mpa
Diambil yang terkecil,
Fpp = 15.3 Mpa
CEK TEGANGAN
• Fc1 = 𝑃𝑢
𝐵 . 𝑁+
𝑀𝑢 (𝑁
2)
1
12𝐵 . 𝑁³
= 9930
122500+
14652100 (350
2)
1
12350 . 350³
= 2.126 Mpa
• Fc2 = 𝑃𝑢
𝐵 . 𝑁−
𝑀𝑢 (𝑁
2)
1
12𝐵 . 𝑁³
=9930
122500−
14652100 (350
2)
1
12350 . 350³
= - 1.947 Mpa
Fpp = 15.3 Mpa
Syarat, Fc1 & Fc2 < Fpp
Memenuhi Syarat ( OK )
DIMENSI m DAN n
• m = 𝑁 −0.95 𝑑
2
= 350 −0.95 (250)
2
= 56.25 mm
• n = 𝐵 −0.8 𝑏𝑓
2
=350 −0.8 (125)
2
= 125 mm
d = Panjang IWF
bf = Lebar IWF
CEK TEBAL PLAT
• a = 𝐹𝑐1 . 𝑁
𝐹𝑐1+𝐹𝑐2
= 2.126 . 350
2.126+1.947= 181.5 mm
• Fcx = 𝐹𝑐1 . 𝑎 − 𝐹𝑐1 . 𝑚
𝑎
= 2.126 𝑥 181.5 −2.216 𝑥 56.25
181.5
= 1.4675 Mpa
• Mpl = 𝐹𝑐𝑥 . 𝑚²
2+
𝐹𝑐1−𝐹𝑐𝑥 . 𝑚²
3
= 1.4675 . 56.25²
2+
2.126−1.4675 .56.25²
3
= 3016.798 Nmm
• tp min. = √4 . 𝑀𝑝𝑙
∅ 𝐹𝑦= √
4 . 3016.798
0.9 250
= 7.328 mm ( MINIMAL )
tp analisis = 10 mm ( OK )
DESIGN ANGKUR
• D angkur = 16 mm
• Jumlah Baut = 4 buah
• Fy = 742 Mpa
• Fu = 825 Mpa
• x1 = 𝑁
2+
𝑎
3=
350
2+
181.5
3
= 114.5 mm
• x2 = N −𝑚
2−
𝑎
2
= 350 −56.25
2−
181.5
2
= 261.4 mm
CEK KEKUATAN ANGKUR
• Tu = 𝑀𝑢 −( 𝑃𝑢 . 𝑥1 )
𝑥2
= 14652100 −( 9330.2 𝑥 114.5 )
261.4
= 51970.548 N
• Ag = ¼ . 𝜋 . d²
= ¼ . 𝜋 . 16²
= 201.062 mm²
• Ae = 0.75 ¼ . 𝜋 . d² ( ULIR )
= 0.75 ¼ . 𝜋 . 16²
= 150.796 mm²
CEK KEKUATAN ANGKUR
• Tu < ∅b . Ag . Fy . n/2
51970.548 < 0.9 201.062 742 4/2
51970.548 < 268538.31 N
• Tu < ∅v . Ae . Fu . n/2
51970.548 < 0.75 150.796 825 4/2
51970.548 < 186610.60 N
OK OK
MENCARI PANJANG ANGKUR
• Tu < ∅ . 𝜋 . Id . D . Fcl . n/2
Dimana,
• Fcl = √f’c
= √30
= 5.477 Mpa
• Id > 𝑇𝑢
∅ . 𝜋 . 𝐷 . 𝑓𝑐𝑙 . 𝑛/2
> 51970.548
0.75 . 𝜋 . 16 . 5.477 . 4/2
Id > 125.844 mm
CEK KUAT GESER
Syarat,
• Vu < 0.75 Fu Ab n
234 < 0.75 825 201.062 4
234 < 497628 N
Dimana,
Ab = Ag
F’c
BJ - 41Vu = 234 N
OK
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