The document details the structural calculations for a building's base plate, including specifications for various base plate examples, load combinations, and stress checks. It outlines the necessary parameters for the design of anchors and verifies their strength against applied loads. All calculated stresses comply with the relevant safety requirements.

1. 3 GEDUNG 2 SORE
IVAN KRISTIANTO
SATRIAN FAJAR
YASMIN SABILA

2. BAJA 2
BASE PLATE

3. CONTOH 1
Terdapat PLAT yang dijadikan BASE dan disambung menggunakan ANGKUR
kedalam BETON

4. CONTOH 2
Terdapat PLAT yang dijadikan BASE dan disambung menggunakan ANGKUR
kedalam BETON

5. CONTOH 3
Terdapat PLAT yang dijadikan BASE dan disambung menggunakan ANGKUR
kedalam BETON

6. CONTOH 3
Terdapat PLAT yang dijadikan BASE dan disambung menggunakan ANGKUR
kedalam BETON
F’c
BJ - 41
IWF 250 x 125

7. KETERANGAN GAMBAR
• B = 350 mm
• N = 350 mm
Menggunakan,
BJ – 41
F’c = 30 Mpa
F’c
BJ - 41
IWF 250 x 125
tp = 10 mm

8. 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

9. 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

10. 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

11. 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

12. PU – COMB 1
Satuan Kg, m, C

13. VU – COMB 3
Satuan Kg, m, C

14. MU – COMB 1
Satuan Kg, m, C

15. LUASAN
• A1 = Pendestel ( Beton )
= B x N
= 350 x 350
= 122500 mm²
• A2 = Base Plate
= 122500 mm²
350
350

16. 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

17. CEK TEGANGAN
• Fc1 =
𝑃𝑢
𝐵 . 𝑁
+
𝑀𝑢 (
𝑁
2
)
1
12
𝐵 . 𝑁³
=
9930
122500
+
14652100 (
350
2
)
1
12
350 . 350³
= 2.126 Mpa
• Fc2 =
𝑃𝑢
𝐵 . 𝑁
−
𝑀𝑢 (
𝑁
2
)
1
12
𝐵 . 𝑁³
=
9930
122500
−
14652100 (
350
2
)
1
12
350 . 350³
= - 1.947 Mpa
Fpp = 15.3 Mpa
Syarat, Fc1 & Fc2 < Fpp
Memenuhi Syarat ( OK )

18. CEK TEGANGAN
350
350
Fc1
Fc2
350

19. 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

20. 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 )

21. 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

22. 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

23. 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

24. 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

25. 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

26. CEK INTERAKSI GESER
Syarat,
•
𝑇𝑢 +
𝑉𝑢
𝜂
∅ . 𝑅𝑛
< 1
•
51970.548+
234
0.70
0.75 . 663504.37
< 1
• 0.10443 < 1
F’c
BJ - 41
𝜂 = 0.70
• Rn = 0.75 Fu An n
= 0.75 825 201.062 4
= 663504.37 N
OK