STRUKTUR BAJA II data-data konstruksi kuda - kuda type : A 45º kemiringan atap : bentang kuda-kuda (B) : 20 m panjang gudang (L) : 40 m jarak antara kolom : 4.5 m tinggi kolom (h) : 5 m bahan penutup atap : seng gelombang sifat dinding : tertutup perhitungan panjang batang batang atas (A) → A1, A2, A3, A4, A5, A6, A7, A8 A₌ 2.5 2.5 = = 3.536 cos 45º 0.707 batang bawah (B) → B1, B2, B3, B4, B5, B6, B7, B8 B₌ 2.5 2.5 = = 3.536 cos 45º 0.707
( 11 kg / m² )
=
3.5
m
=
3.5
m
batang vertical ( C ) → C1, C2, C3, C4, C5, C6, C7, C8, C9 dimana : C1 ₌ C9 C2 ₌ C8 C3 ₌ C7 C4 ₌ C6 C5 C1 ₌ C9
C2 ₌ C8
C3 ₌ C7
C4 ₌ C6
C5
3.5 3.5
X X
Tg 45º 1
=
3.5361
=
3.5
m
3.5 3.5
X X
Tg 45º 1
=
3.5361
=
3.5
m
3.5 3.5
X X
Tg 45º 1
=
3.5361
=
3.5
m
3.5 3.5
X X
Tg 45º 1
=
3.5361
=
3.5
m
3.5 3.5
X X
Tg 45º 1
=
3.5361
=
3.5
m
batang diagonal (D) → D1, D2, D3, D4, D5, D6, D7, D8 D1 ₌ D8
√ 3,5² + C1²
= =
√ 5.0008
3.5 =
X 5.0
3.5
+
3.5361
X
3.5361 m
D2 ₌ D7
√ 3,5² + C1²
= =
√ 5.00076
3.5 =
X 5.0
3.5
+
3.5361
X
3.5361 m
D3 ₌ D6
√ 3,5² + C1²
= =
√ 5.00076
3.5 =
X 5.0
3.5
+
3.5361
X
3.5361 m
D4 ₌ D5
√ 3,5² + C1²
= =
√ 5.00076
3.5 =
X 5.0
3.5
+
3.5361
X
3.5361 m
daftar panjang batang No A 1 3.5 2 3.5 3 3.5 4 3.5 5 3.5 6 3.5 7 3.5 8 3.5 9 -
B 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 -
C 3.5361 3.5361 3.5361 3.5361 3.5361 3.5361 3.5361 3.5361 3.5361
D 5.0 5.0 5.0 5.0 5.0 5.0 5.0 5.0 -
perencanaan gording dipakai profil canal 150 X 75 X 6 Ω L berat (G) WX WY IX IY δ Bj 37 E jarak gording berat atap + pelengkapan berat langit - langit + plafond berat seng gelombang tekanan angin
: : : : : : : : : : : : : :
45º 4.5 13.2 76.4 17.2 573 91.9 1600 2.10 X 106 2,9 : 2 10 18 11 25
penyelesaian Beban sarah sumbu Y a. Akibat beban mati (q) Berat atap + perlengkapan 10 X 1.4 Berat langit - langit plafond Berat seng gelombang Berat profil canal
Momen akibat beban mati mx1 = 1/8 x = 1/8 x = 0.125 x = 22.35 2235 = c. Momen akibat beban hidup (p = 100 kg) mx2 = 1/4 x = 1/4 x = 0.25 x = 79.538 7953.8 = Beban searah sumbu X a. Momen akibat beban mati my1 = 1/8 x = 1/8 x = 0.125 x = 22.35 2235 = b. Momen akibat beban hidup my2 = 1/4 x = 1/4 x = 0.25 x = 79.538 7953.8 =
m kg / m² cm³ cm³ cm4 cm4 kg / m² = kg / m² kg / m² kg / m² kg / m²
1.45
m
= = = = =
14 18 11 13.2 56.2
kg / m² kg / m² kg / m² kg / m² kg / m²
q 56.2 56.2
x x x
cos Ω cos 45º 0.707
x x x
L 4.5 4.5
p 100 100
x x x
cos Ω cos 45º 0.707
x x x
L 4.5 4.5
q 56.2 56.2
x x x
sin Ω sin 45º 0.707
x x x
L 4.5 4.5
p 100 100
x x x
sin Ω sin 45º 0.707
x x x
L 4.5 4.5
b.
sin30
c.
Momen akibat beban angin ( beban gording ) Angin muka = 1.45 x = 1.45 x = 11.60 kg / m²
-
Angin belakang
Kesimpulan : w1 x = w1 x = w2 y = w2 y = mw1 x = mw1 y = = mw2 x = mw2 y = =
0 11.60 0 -23.2 0 1/8 36.25 0 1/8 72.5
Daftar momen momen berat ( kg cm ) sendiri
beban hidup
kiri
= = = =
1.45 x 1.45 x -2320 kg / m² -23.2 kg / m²
30 30
-
0.4 0.4
( (
-40 -40
) )
x x
w 40
kg / m²
1600
kg / m²
11.6 3625
X
(
5
X
5
)
X =
23.2 7250
X
(
5
X
5
)
angin kanan
kombinasi ( primer )
7953.75
-
-
10188.754
my
2235
7953.75
3625
7250
10188.754
check tegangan a. kombinasi I ( beban mati + beban hidup ) mx δ = wx
+
my wy
≤
δ
≤
1600
=
10189 76.4
+
10189 17.2
=
725.73
≤
1600
b. kombinasi II ( beban mati + beban hidup + beban angin ) mx + my+ma δ = wx wy
= = = =
x x
X =
2235
Qy
0.02 0.02
5.800
mx
check lendutan Qx = = = =
( (
kg / m²
≤
=
10189 76.4
+
10189
+ 17.2
=
10533
≤
2080
kg / m²
q 56.2 56.2 39.7334
x x x kg / m1
cos Ω cos 45 0.707 =
0.3973
kg / m1
q 56.2 56.2 39.7334
x x x kg / m1
sin Ω sin 45 0.707 =
0.3973
kg / m1
725.7301
kg / m²
RCTI OKE.............
1.3
δ
3625
≤
RCTI OKE..................
1.3
x
) )
x x
w 40
kg / m²
Px
= = = =
p 100 100 70.7
x x x kg
cos Ω cos 45 0.707
Py
= = = =
p 100 100 70.7
x x x kg
sin Ω sin 45 0.707
x x
Qx E
x x
(ζ/2)4 Iy
+
5 384 0.2240
x x cm
0.397334 (2.1x106)
x x
(500/2)4 91.9
+
5 384
x x
Qy E
x x
( ζ )4 Ix
+
x x cm
0.397334 (2.1x106)
x x
(500)4 573
+
=
5 384 1.0338
=
√
=
√
perhitungan check lendutan fx 5 = 384 = = fy
=
=
f
0.224 ² 1.1188
+
1.034 ²
<
2
=
Px 48
x x
(ζ/2)³ E
x
Iy
70.7 48
x x
(500/2)³ (2.1x106)
x
91.9
Py 48
x x
( ζ )³ E
x
Ix
5 48
x x
70.7 (2.1x106)
x x
(500)³ 573
500 250 1.0577
<
2
cm
RCTI OKE.............
TABEL GAYA BATANG BEBAN ANGIN KIRI PANJANG BESAR GAYA BATANG BATANG TEKAN ( - ) TARIK ( + ) GAYA BATANG ( cm ) ( kg ) ( kg )
A1 A2 A3 A4 A5=A6=A7=A8 B1 B2 B3 B4 B5=B6=B7=B8 C1 C2 C3 C4 C5=C6=C7=C8=C9 D1 D2 D3 D4 D5=D6=D7=D8
0 4.412 7.613 7.223 31.353 6.442 8.433 8.433 15.265 29.752 0.447 1.994 0 0.386 42.755 6.603 3.66 0.297 0.922 30.532
0 44.12 76.13 72.23 4.47 19.94 0 3.86 66.03 -
0 313.53 64.42 84.33 84.33 152.65 297.52 0 427.55 36.6 2.97 9.22 305.32
TABEL GAYA BATANG BEBAN ANGIN KANAN PANJANG BESAR GAYA BATANG BATANG TEKAN ( - ) TARIK ( + ) GAYA BATANG ( cm ) ( kg ) ( kg )
A1=A2=A3=A4 A5 A6 A7 A8 B1=B2=B3=B4 B5 B6 B7 B8 C1=C2=C3=C4=C5 C6 C7 C8 C9 D1=D2=D3=D4 D5 D6 D7
7.223 31.353 27.291 12.845 0 15.265 29.752 24.95 24.95 16.946 10.002 4.754 0 7.932 42.755 14.03 1.219 3.657 14.714
72.23 313.53 272.91 128.45 0 152.65 100.02 0 7.932 140.3 12.19 36.57 -
297.52 249.5 249.5 169.46 47.54 427.55 147.14
PERHITUNGAN PROFILE POLA KUDA - KUDA
Batang Tekan (tekan) A (A1 S/D A8) Gaya batang maksimum Panjang batang ( L ) Di coba profile
A.
186.000 Cm4 26.400 Cm2 10 mm mm
Ix = f = Tebal plat rencana 10
e Total Iy
iy
= = = =
2.810 + 2 ( Ix + f (e)2) 186 2 950.482
=
950.482 2.f
7554.53 3.54 90 90
= = = Ix e
0.5
16
2.660 Cm4 2.810 Cm2
= =
3.310
= 26.4
+
Kg mm
3.310
x
2
950.482 . 26.4
= 2 =
4.243
cm
=
2.660
cm
Maka : ix
l
=
s kr
=
s tk
=
LK
2
.
l2 s KC C
0.354 2.660
=
i min p
adalah, i minimum
E
=
p
2
=
=
0.1330
2.1 . 0.0177
1,171,698,187.704 2.810
10
6
=
1,171,698,187.704
Kg/Cm2
=
416974444
Tegangan yang timbul
s
P
=
=
2F
7554.53 26.400 .
2
=
143.078
143.078
B.
Batang Tekan (tekan) A (A1 S/D A8) Gaya batang maksimum Panjang batang ( L ) Di coba profile Ix f
= = = =
2.810 + 2 ( Ix + f (e)2) 186 2 950.482
=
950.482 2.f
= 2 =
Ix e
= =
2.660 Cm4 2.810 Cm2
10 mm mm
10
iy
416974444
= = =
186.000 Cm4 26.400 Cm2
= =
Tebal plat rencana
e Total Iy
< stk
950.482 . 26.4 4.243
cm
0.5 +
3.310
= 26.4
x
3.310
2
90
2
Kg/Cm
7383.04 Kg 3.54 mm 90 16
RCTI OK
Maka : ix
2.660
=
l
=
s kr
=
s tk
=
cm
LK
2
0 2.660
=
i min p
adalah, i minimum
.
E
l2 s KC C
=
p
2
=
0.1330
2.1 .
10
6
=
0.0177
=
1,171,698,187.704 2.810
=
416974444
1,171,698,187.704
Kg/Cm2
Tegangan yang timbul
s
P
=
=
2F
7383.04 26.400 .
2
=
139.830
139.830
C.
Batang Diagonal
D
Gaya batang maksimum Panjang batang ( L ) Di coba profile Ix = f = Tebal plat rencana 10
=
416974444
7.91
= = =
427.55 Kg 3.54 mm 90 16
90
186.000 Cm4 26.400 Cm2 10 mm mm
< stk
Ix e
= =
2.660 2.810
Cm4 Cm2
x
3.310
2
|e| e Total Iy
iy
= = = = =
2.810 + 2 ( Ix + f (e)2) 2 186 950 950 2.f 950.482
= 2 =
4.243
.
26.4 cm
0.5
=
+
26.4
3.310
2
Kg/Cm
RCTI OKE…….!!!!!
Maka : ix
l
=
2.660
=
LK
cm =
adalah, i minimum 354 2.660
i min
s kr
=
p
2
.
E
=
p
2
2.1
l2
s tk
=
132.9350
=
.
10
6
=
1,172.844
Kg/Cm2
17,671.714
s KC
1,172.844
=
2.810
C =
417
Tegangan yang timbul
s
=
P
427.55
=
2F
2 =
26.400
.
8.098
< stk
8.098 D.
Batang Vertikal (tarik)
C
Gaya batang maksimum Panjang batang ( L ) Di coba profile 186.000 Ix = 26.400 f = 52.800 2f = F Efektif
= = =
=
l
s
=
P
RCTI OKE…….!!!!!
= C1 s/d C9 6030.33
3.3 90
M 90
9
14
MM
1000 186.000
= =
6030.33
=
186.000 32.4
=
6,030
F.Ef
417.382 Kg/Cm2
5.376
Cm2
32.42
s
=
186
=
186
Kg/Cm2
=
1600
JAR URANG BANJAR TU PROFIL KAWA DIPAKAI ……………. !!!!!
Kg/Cm
RCTI OK!
RCTI OK
RCTI OKE…….!!!!!
RCTI OK!
perhitungan baut tebal plat =
10 mm
=
1 cm
d baut = 14 mm = 1.4 diameter lubang = 1.5 cm Ng = 0.25 x 3.14 x 1.5 = 1695.6 Kg Ntp = 1 x 1.5 x 1.5 = 3600 Kg Ng < Ntp 1695.6 < 3600 ………………..
cm 2
x
x
0.6 1600
RCTI OKE…….!!!!!
jumlah baut pada titik simpul 1
batang C1 gaya batang C1 n
=
100.02 100.02 3600
=
=
0.027783
= batang B1 gaya batang B1 n
= 3232.42 3232.42 = 3600 =
=
batang D1 gaya batang D1 n
1
0.897894 1
= 6030.33 6030.33 = 3600 =
=
bh
bh
1.675092 2
bh
jumlah baut titik simpul 2
batang A1 gaya batang A1 n
= =
72.23 95.80 3600
= =
0.026611 1
bh
x
1600
batang C1 gaya batang C1 n
= =
100.02 100.02 3600
=
0.027783
=
1
bh
jumlah baut titik simpul 3
batang A2 gaya batang A2 n =
=
116.35 116.35 3600
= =
0.032320 1
bh
batang C2 gaya batang C2 n =
=
122.20 122.20 3600
= =
0.033944 1
bh
jumlah baut titik simpul 4
batang C2 gaya batang C2 n =
=
122.20 122.20 3600
= =
0.033944 1
bh
batang B2 gaya batang B2 n =
= 5515.33 5515.33 = 3600 =
1.532036 2
bh
batang D2 gaya batang D2 n
= =
4196.60
4196.60 3600
=
1.165722
=
2
jumlah baut titik simpul 5
batang C3 gaya batang C3 n =
=
100.02 100.02 3600
= =
0.027783 1
bh
batang B4 gaya batang B4 n =
= 6940.65 6940.65 = 3600 =
1.927958 2
bh
jumlah baut titik simpul 6
batang A3 gaya batang A3 n =
=
155.15 155.15 3600
= =
batang C3 gaya batang C3 n =
=
1
bh
100.02 100.02 3600
= =
batang D4 gaya batang D4 n =
0.043097
=
0.027783 1
bh
354.22 354.22 3600
= =
0.098394 1
bh
bh
jumlah baut titik simpul 7
batang A4 gaya batang A4 n
= =
7385.46
7385.46 3600
= =
batang C4 gaya batang C4 n =
batang D4 gaya batang D4 n =
3
= 1447.88 1447.88 = 3600 = =
2.051517 bh
0.402189 1
bh
354.22 354.22 3600
= =
0.098394 1
bh
jumlah baut titik simpul 8
batang C4 gaya batang C4 n =
batang B4 gaya batang B4 n =
= 1447.88 1447.88 = 3600 = = 6940.65 6940.65 = 3600 =
0.402189 1
bh
1.927958 2
bh
jumlah baut titik simpul 9
batang C5 gaya batang C5 n =
=
427.55 427.55 3600
= =
batang B4 gaya batang B4 n =
batang D4 gaya batang D4 n =
1
= 6940.65 6940.65 = 3600 = =
0.118764 bh
1.927958 2
bh
354.22 354.22 3600
= =
0.098394 1
bh
jumlah baut titik simpul 9
batang C5 gaya batang C5 n =
=
427.55 427.55 3600
= =
batang A4 gaya batang A4 n =
= 7385.46 7385.46 = 3600 =
0.118764 1
bh
2.051517 3
bh
Menetukan pelat kopel
-
jumlah pelat kopel tebal pelat Gaya max Lebar pelat
= = = =
Gaya lubang ( D ) yang dipikul pelat kopel D = 0.02 x = 0.02 x = 72.7 Kg
8 bh 10 mm 3634.98 18
Kg
p 3634.98
Tegangan Geser digunakan profil 90.90.16 σ
=
D b
A I e
= = =
26.4 cm2 186 cm4 2.81 cm
S
= = =
( ( 370.9
0.5 0.5 cm3
72.7 18
x x
σ
=
=
x x
S I
x x
t 10
x
x x
e x
) 2.81
h2 10.5
3
370.9 186
8.054
syarat petal kopel Ip a
>
I1 L1
dicoba h Ip
=
10.5 cm
= = =
2 2 11576
x x cm4
0.5 0.5
x x
t 10
a
= = =
2 2 2.81
x x cm4
0.5 0.5
x x
e 2.81
I1
= = =
0.083 0.083 1736
x x cm4
b 18
x x
h3 10.5
=
3
x
d
L1
3
x
A )
x
26.4
= = 11576 2.81
3 63
x cm 1736 63
>
4120
>
Perhitungan sambungan a sambungan batang tekan Gaya tekan max tebal pelat profil ┘└ 90.90.16
21
27.56 RCTI OKE..................
= =
72.23 10
Kg mm
untuk Ø baut diambil sekitar 2 kali tebal rata-rata pelat yang disambungkan jadi…….. Ø baut
Ø lubang
90
+ 2
16
e2
=
90
Syarat e2
a
26
mm
18
=
27
mm
1
x
1600
=
26
mm
2
[
16
+ 2
10
]
=
26
+
1
=
27
mm
=
53
mm
-
53
=
37
mm
>
1.5 lubang
=
1.5
x
Ng
= =
0.25 8139
x Kg
3.14
x
1.8
2
x
Ntp
= =
1.8 5760
x Kg
1
x
2
x
1600
>
Ntp
=
0.013
Ternyata Ng n
=
=
72.23 5760
=
, sehingga Ntp yang menentukan 1
=
bh
sambungan batang tekan Gaya tekan max tebal pelat profil ┘└
= =
5978.02 10 mm
Kg
90.90.16
untuk Ø baut diambil sekitar 2 kali tebal rata-rata pelat yang disambungkan jadi……..
Ø baut
Ø lubang
=
=
2
26
[
16
+
1
+
10
]
27
mm
2 =
RCTI OKE.................. x
2
90 e2
+ 2 =
16
=
53
mm
90
-
53
=
37
mm
>
1.5 lubang
=
1.5
x
18
=
27
1
x
Syarat e2
n
Ng
= =
0.25 8139
x Kg
3.14
x
1.8
2
x
Ntp
= =
1.8 5760
x Kg
1
x
2
x
1600
Ternyata Ng
>
Ntp
5978.02 5760
=
=
1.038
, sehingga Ntp yang menentukan =
2
bh
RCTI OKE..................
1600
x
2
Perhitungan kolom Pembebanan 1 Beban mati Beban gording beban penutup atap dan plafond Berat kuda-kuda 2 Beban Hidup total ( P )
= = = = =
288 175.03 429.62 1591.2 2483.9
Kg Kg Kg Kg Kg
+
P.12.5
+
15
+
Perhitungan reaksi kolom
Ra Σma
Rb =
0 P.25
-RB 25
+
+
P.17.5 +
-RB 25
+
P
x
(
25
-RB 25
+
P
x
95
RB
P
=
=
+
x 95 25 3.8
P
Karena RA dan RB adalah simetris maka : RA= RB
= = =
3.8 P 3.8 x 2483.9 9438.6528 Kg
Pendimensian kolom tinggi kolom Bj 37 N dioba
= = =
WF A ix iy Wx Wy
5 m 1600 Kg/cm2 9438.6528 Kg
400x300x10x16 = 136 = 16.9 = 7.28 = 1980 = 481
arah tegak lurus sumbu x K
=
λx
1 =
Lkx ix
=
500 16.9
=
λg
29.586
=
Π
=
3.14
= λs
=
111.015765 =
λx λg
=
29.5857988 111.015765 0.26650088
e x
0.7
2.1 0.7
x x
σl
10 6 2400
P.15 17.5
+
12.5
P.10 +
10
+
P.7.5
+
+
7.5
+
+
P.5 5
+
P.2.5 2.5
)
=
0
untuk
Wx
0.183
1.41 1.593 -
=
111.0157646 1.593 0.2665 83.690794 e x
3.14
2
=
nx -
x λx
A 2
x 2.1 x 10 6 x x 29.5857988 2 9438.6528
136
340.832413
=
1 =
βx
1
λs
Π2 x = N =
nx
0.26650088 <
=
= nx
<
=
340.832 340.8324127 -
1
1.002942627 =
2
0.6
=
0.6
=
0.6
+
0.4
+
Mx =
0.125 x
=
0.125 x
=
589916
0.4
x
x
L
N
Mx1 Mx2
x
9438.6528
0
x
500
Kgcm
arah tegak lurus sumbu x K
=
λx
1 =
Lkx ix
=
500 7.28
=
λg
68.7
=
Π
=
3.14
= λs
2.1 0.7
x x
σl
10 6 2400
111.015765 =
λx λg
=
68.6813187 111.015765
=
0.61866275
untuk
Wy
e x
0.7
0.183
=
= =
<
1.41 1.593 -
1.593
0.61866275 <
1
λs
1.41 - 0.61866
1.44713753
cek persamaan interaksi Wx
+
N
+
βx
x
ny
x
Mx2
<
σ
Wx
+
+
A
83.690794
βx
x
nx
9438.6528 136
+
331.855016 <
σ
1.44713753
x
9438.6528 136
Mx Wx
+
9438.6528 136
589916 1980
1
<
1600
<
RCTI OKE…….!!!!!
1600
σ
<
589916 1980
+
x
σ
RCTI OKE…….!!!!!
100.4340349 <
N A
<
Wx
0.6
Kg/cm2
<
x
x
1
+
1600
N A
Wx
-
1600
<
367.3391316 <
RCTI OKE…….!!!!!
1600
JAR URANG BANJAR TU PROFIL 400x300x10x16 KAWA DIPAKAI ……………. !!!!! pelat dasar kolom pada pondasi
1 kip =
4.535924
x
2
6.2452
cm2
1 in
=
beban aksial ( P ) f'c
10
2
= 9438.6528 Kg 210.908 Kg/cm2
= =
2080.86661 Kip = 3.37712163 Ksi
Menetukan luas landasan pelat perlu : Fp = = =
0.35 x 0.35 x 1.182 Ksi
A
perlu
f'c 3.37712
P Fp
=
2080.86661 1.18199257
=
=
1760.47351 in2 =
Profil yang digunakan adalah WF data profil
:
A d b
= = =
10994.5092 cm2
400x300x10x16
136 cm2 39 cm 30 cm
menetukan lebar ( dan panjang ( N ) pelat dimana harus memenuhi syarat : B
x
N
>
10994.50917 cm2
untuk B dan N sedemikian hingga perbedaan ( selisih ) diantara keduanya mendekati 0.95 0.95
d
x
0.8 39
-
b 1
x
30
=
13.05
cm
dicoba berbagai gaya B dan N sebagai berikut : B 115 113 111 110
N 116 118 120 122
Luas 13340 13334 13320 13420
B
gunakan
=
110
Selisih' 1 5 9 12 dan
N
= ###
tekanan aktual adalah Fp
=
B
P x
9438.6528 x 110 122
=
N =
0.70332733 Kg/cm2
Hitung M dan N M
=
N
-
0.95 2
d
= =
122
42.475
-
0.95 2 cm
x
39
1600
N
=
B
-
0.95 2
b
110
=
-
40.75
=
0.95 2
x
30
cm
Hitung tebal pelat ( tp ) tp
=
= =
fp 0.25
n
fy
0.703327332 2400 0.25 x
40.75
1.39518159 cm
jadi pelat yang digunakan adalah ### cm
1.39518 cm
x
x
### cm
Menetukan jumlah baut
Pada Kolom digunakan profil WF diametert baut
=
=
e2
=
26
= e1 s Ng
= = = = =
Ntp
Ng
145 2 3
x x
> Ntp n
=
10
]
>
36 mm
= =
2 3
x 2.7 4320
s x Kg
x
27 mm
1
+
d d
d
10
]
mm
mm
3.14 3.14 Kg
+ 2
+
2
0.25 x 0.25 x 9156.24 = = =
26
300
[
300
[
2
= diameter lubang
400x300x10x16 16
RCTI OKE…….!!!!!
x x
27 27
= =
54 mm 81 mm
x x
2
x
σ
d
2
2.7
x
1600
σ 1.0
x
1600
, maka Ntp yang menentukan P Ntp
=
9438.65 4320
= =
2.18487333 bh
=
3 bh
Cek kekuatan baut untuk mengetahui kuat atau tidaknya baut menahan suatau beban harus memenuhi syarat berikut : σ
<
σ
=
=
0.8
σ P
2
2
x
0.25
3.14
x
d2
x
9438.65 0.25 3.14 x
x
3
x
=
824.675002
<
0.8
σ
=
824.675002
<
0.8
x
=
824.675002
<
1280
2
1600 Kg/cm2
RCTI OKE…….!!!!!
Tabel: Jumlah Baut Tiap Batang no 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33
batang A1 A2 A3 A4 A5 A6 A7 A8 B1 B2 B3 B4 B5 B6 B7 B8 C1 C2 C3 C4 C5 C6 C7 C8 C9 D1 D2 D3 D4 D5 D6 D7 D8
panjang batang 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.54 3.54 3.54 3.54 3.54 3.54 3.54 3.54 3.54 5.0 5.0 5.0 5.0 5.0 5.0 5.0 5.0
jumlah baut 1buah 1buah 1buah 3buah 5buah 1buah 1buah 1buah 1buah 4buah 1buah 2buah 5buah 4buah 4buah 1buah 1buah 1buah 1buah 1buah 1buah 1buah 1buah 1buah 1buah 2buah 2buah 1buah 1buah 1buah 1buah 2buah 2buah
