DATA PERENCANAAN KUDA - KUDA
Dimensi
- Bentang kuda-kuda (L) = 15 m
- Jarak antar kuda-kuda (s) = 3 m
- Jarak antar skur kuda-kuda (x) = 2.5 m
- Jarak overlap = 30 cm
- Jarak efektif gording (Lk) = 1.5 m
- Kemiringan atap = 22 o
Beban Perencanaan
- Berat penutup atap seng BJLS = 10 kg
- Beban pekerja + alat = 100 kg
- Beban air hujan = 10 kg
- Beban angin = 25 kg
Kayu kelas I
- Elastisitas = 125000
- Berat Jenis (BJ) = 810 kg/m3
- Tegangan Ijin σlt = 150 kg/cm2
σtk // = σtr // = 130 kg/cm2
σtk ┴ = 40 kg/cm2
τ// = 20 kg/cm2
Ukuran
L1 = 2.425 m
L2 = 2.425 m
L3 = 2.650 m
L4 = 2.650 m
L5 = 2.420 m
L6 = 2.420 m
H1 = 1.080 m
H2 = 1.080 m
H3 = 1.390 m
H4 = 0.520 m
H5 = 0.560 m
H6 = 0.520 m
H7 = 1.950 m
Sudut-sudut
αo = 22 o γo = o
tan αo = 0.404 tan γo = s13 - 2 tan 12 0.037019098
sin αo = 0.375 sin γo =
cos αo = 0.927 cos γo =
βo = 11 o δo = o
tan βo = 0.194 tan δo = s15 - 2 tan 12 #############
sin βo = 0.191 sin δo =
cos βo = 0.982 cos δo =
Panjang Batang
L1 L1
cos αo cos βo
L2 S13 = L1 tan αo - L1 tan βo
cos αo = 0.508 m
L3 L2
cos αo cos γo
L4 S15 = (L1 + L2) x tan αo - (L1 + L2) x tan βo
cos αo = 1.017 m
L5 L3
cos αo cos δo
L6 S17 = (L1 + L2 + L3 ) x tan αo - (L1 + L2 + L3) x tan βo
cos αo = 1.570 m
L6 L4
cos βo cos δo
L5 S19 = (L5 + L6) x tan αo - (L5 + L6) x tan βo
cos βo = 1.015 m
L4 L5
cos βo cos γo
L3 S21 = L6 tan αo - L6 tan βo
cos βo = 0.507 m
L2
cos βo
m
S16 = = 2.716 m
S12 = = 2.470 m
m
S14 = = 2.425 m
S18 = = 2.716
S11 = = 2.470 m
S20 = = 2.420
S10 = = 2.70 m
S9 = = 2.70 m
S8 = = 2.4653 m
S7 = = 2.4653 m
S6 = = 2.610 m
S5 = = 2.610 m
S4 = = 2.858 m
S3 = = 2.610 m
S2 = = 2.858 m
=S1 = 2.615 m
0.875
0.015
0.015
1.000
12.676
0.225
0.219
0.976
Tabel Panjang Batang
(3)
meter
meter
meter
meter
meter
meter
meter
meter
meter
meter
meter
meter
meter
meter
meter
meter
meter
meter
meter
meter
meter
(1) (2)
S1
S2
S3
S4
2.615
2.858
2.610
2.858
S14
S15
S16
S17
S18
S19
S20
S21
S5
S6
S7
S8
S9
S10
S11
S12
S13
2.425
1.017
2.716
1.570
2.716
1.015
2.420
0.507
2.610
2.610
2.465
2.465
2.700
2.700
2.470
2.470
0.508
Btg. Rangkap
Btg. Tunggal
Btg. Tunggal
Btg. Tunggal
Btg. Rangkap
Btg. Tunggal
Btg. Rangkap
Btg. Tunggal
Btg. Tunggal
Btg. Tunggal
Btg. Tunggal
Btg. Tunggal
Btg. Rangkap
46.328Total Panjang meter
Kode Batang Panjang Batang Satuan Keterangan
(4)
Btg. Tunggal
Btg. Tunggal
Btg. Tunggal
Btg. Tunggal
Btg. Tunggal
Btg. Tunggal
Btg. Tunggal
Btg. Tunggal
PERENCANAAN GORDING
PERHITUNGAN PEMBEBANAN PADA GORDING
Beban Mati (DL)
Berat penutub atap seng BJLS = berat BJLS x jarak efektif gording = 15 kg/m
Berat gording kayu 8/15 = luas penampang balok x BJ kayu = 5.832 kg/m
Jumlah berat = 20.832 kg/m
Berat alat penyambung 10% = Jumlah berat x 10% = 2.0832 kg/m
Total berat = 22.915 kg/m
Beban Mati Total (q) = 22.915 ≈ 23 kg/m
qy = q beban mati x cos α = 21.325 kg/m
qx = q beban mati x sin α = 8.616 kg/m
M.qy = 1/8 x qx x 2.52 = 16.660 kg.m
M.qx = 1/8 x qy x (2.5/2)2 = 4.165 kg.m
Beban Hidup (LL)
Beban Hidup Total (P) = 100 kg/m
Py = P beban hidup x cos α = 92.718 kg/m
Px = P beban hidup x sin α = 37.461 kg/m
M.Py = 1/4 X Px X L = 23.413 kg.m
M.Px = 1/4 X Py X (L/2) = 28.974 kg.m
Beban Air Hujan (GL)
Beban Air Hujan Total (G) = 10 kg/m
Gy = G beban hidup x cos α = 9.272 kg/m
Gx = G beban hidup x sin α = 3.746 kg/m
M.Gy = 1/4 X Gx X L = 2.341 kg.m
M.Gx = 1/4 X Gy X (L/2) = 2.897 kg.m
qy
qx
x y
q
α°
py
px
x
y
p
α°
Gy
Gx
x
y
G
α°
Beban Angin (WL)
Beban Angin Total (WL) = 25 kg/m
1. Angin tekan
WL tekan = (0.02 x α - 0.4) x WL x jrk. gording = 1.5 kg/m
M.WL tekan = 1/8 x WL x L2 = 1.172 kg.m
2. Angin hisap
WL hisap = – 0.4 x WL x jarak gording = -15 kg/m
M.WL hisap = 1/8 x WL x L2 = -11.719 kg.m
Tabel Pembebanan Momen
Beban
Mati A. Tekan A. Hisap
1 4 5 7
36.037
42.415
6
33.140
40.073
Pembebanan
Tetap Sementara
kg.m
Beban AnginMomen
Mx
My
4.165
16.660
28.974
23.413
2.897
2.341
-
1.172
-
-11.719
Beban
Air Hujan
3
Beban
Hidup
2
W tekan
y α°
Whisap
y
α°
Kontrol tegangan lentur
Karena konstruksi terlindung dan tegangan akibat beban tetap dan sementara
maka tegangan lentur dinaikkan 25%
τlt = 150 x 1.25
= 187.5 kg/cm2
Coba-coba pakai balok dimensi 6/12
Dik:
b = 6 cm
h = 12 cm
Momen sementara
Mx = 3603.703 kg.cm
My = 4241.454 kg.cm
Ix = 1/12 x b x h3 Iy = 1/12 x b3 x h
= 864 cm4 = 216 cm4
Wx = Ix Wy = Iy
0.5h 0.5h
= 144 cm3 = 36 cm3
τmax = ( Mx / Wx ) + ( My / Wy )
= 142.844 kg/cm2 < kg/cm2 OK…!
Kontrol lendutan
ḟ =
= 1 cm
fx =
= +
= 0.852 cm
fy =
= +
= 0.213 cm
f = {( fx 2 ) + ( fy
2 )}0,5
= 0.879 cm < cm OK…!
Kesimpulan : Balok dengan dimensi tersebut bisa digunakan.
0.208 0.005
+
187.50
jarak antar kuda-kuda
300
+1 x P cos α x s3
48 x E x Iy
0.019
1 x P cos α x s3
48 x E x Iy384 x E x Iy
5 x q cos α x s4
384 x E x Iy
5 x q cos α x s4
0.833
1
PERENCANAAN KUDA -KUDA
Perencanaan Pembebanan pada kuda - kuda
Beban Mati Akibat Beban Atap
Beban Mati Atap = Beban q x Jarak Kuda-kuda = 69 kg
Berat Gording = Berat Gording x Jarak Kuda-kuda = 17.496 kg
Berat Sendiri Kuda-kuda = L. Penmpng x P. Total kayu x BJ kayu = 270.1838 kg
Berat Alat Penyambung 10% = Berat Sendiri Kuda-kuda x 10% = 27.01838 kg
Total berat = 383.698 kg
Beban hidup akibat Beban pekerja
P = 100 kg
Beban angin
1. Angin tekan = WL tekan x Jarak kuda-kuda = 4.5 kg
2. Angin hisap = WL hisap x Jarak kuda-kuda = -45 kg
Beban air hujan = GL x jarak gording x jarak kuda-kuda = 45 kg
Tabel Perencanaan Pembebanan
Kode Beban Beban
Beban Mati Hidup
Beban Sin α
1 2 3 5 6
P1 31.975 50 -22.5 -8.429
P2 63.950 100 -45 -16.857
P3 63.950 100 -45 -16.857
P4 63.950 100 -22.5 -8.429
P5 63.950 100 - -
P6 63.950 100 - -
P7 31.975 50 - -
Beban Angin
Angin Tekan
-
kg383.698
6
Angin Hisap
-
Sin α
9
0.843
1.686
1.686
0.843
Jadi Beban Untuk Satu Titik Buhul = = 63.950
Beban
Air Hujan
4
22.5
-
-
-
45
2.25
Cos α
7
-20.862
-41.723
-41.723
-20.862
-
45
Cos α
10
2.086
4.172
4.172
2.086
-
-
-
-
-
-
45
45
45
22.5
Beban
8
2.25
4.5
4.5
• Panjang Jarak dan Beban
L1 = 2.425 m H1 = 0.980 m P1 = kg
L2 = 2.425 m H2 = 0.980 m P2 = kg
L3 = 2.650 m H3 = 1.071 m P3 = kg
L4 = 2.650 m H4 = 0.470 m P4 = kg
L5 = 2.420 m H5 = 0.509 m P5 = kg
L6 = 2.420 m H6 = 0.039 m P6 = kg
H7 = 1.570 m P7 = kg
L7 = L1 + L2 + L3 + L4 + L5 + L6 = m
L8 = L2 + L3 + L4 + L5 + L6 = m
L9 = L3 + L4 + L5 + L6 = m
L10 = L4 + L5 + L6 = m
L11 = L5 + L6 = m
L12 = L1 + L2 = m
L13 = L1 + L2 + L3 = m
L14 = L1 + L2 + L3 + L4 = m
L15 = L1 + L2 + L3 + L4 + L5 = m
• Sudut - sudut :
*) αo = 22.0 o *) γo = 0.875 o
tan αo = 0.40 tan γo = 0.015
sin αo = 0.37 sin γo = 0.015
cos αo = 0.93 cos γo = 0.9999
*) βo = 11.0 o *) δo = 12.676 o
tan βo = 0.19 tan δo = 0.225
sin βo = 0.19 sin δo = 0.219
cos βo = 0.98 cos δo = 0.976
31.975
63.950
63.950
63.950
63.950
63.950
31.975
12.570
PERHITUNGAN GAYA-GAYA BATANG AKIBAT BEBAN MATI
14.990
12.565
10.140
7.490
4.840
4.850
7.500
10.150
A
B
C D
E
F
G
H
I
J
K
L
δo
γo
αo βo
P7
P6
P5
P4
P3
P2
P1 S1
S2
S3 S4
S5
S6
S7
S8
S9 S10
S11
S12
S13
S14
S15
S16
S17
S18 S19
S20
S21
L1 L2 L3 L4 L5 L6
H3
H2
H1
H7
H6
H4
H5
RAV RLV
• Reaksi Perletakan
Σ ML = 0
Σ MA = 0
Kontrol Σ V = 0
= 0
= 0 ............ Ok
• Gaya - Gaya Batang
◊ Joint A
Σ V = 0
RAV - P1 + S1 sin α + S12 sin β = 0
0, 41 S1 + 0,21 S12 = ..Pers. 1
Σ H = 0
S1 cos α + S12 cos β = 0
0, 91 S1 + 0,98 S12 = ..Pers. 2
Eliminasi Pers. 1 dan Pers. 2
= x
= x
=
= -
-0.19 S12 =
S12 = kg
Subtitusi ke Pers. 1
=
= kg
0, 41 S1 + 0,21 S12
S1
0.93
0.37
-148.154
776.450
-159.79
0, 41 S1 + 0,21 S12
0, 91 S1 + 0,98 S12
0,37 S1 + 0,19 S12
0,37 S1 + 0,40 S12
-159.789
0.000
-148.154
0.000
-822.043
RLV =P7 ( L7 ) + P6 ( L15 ) + P5 ( L14 ) + P4 ( L13 ) + P3 ( L12 ) + P2 (L1)
=L7
RAV =L7
P1 ( L7 ) + P2 ( L8 ) + P3 ( L9 ) + P4 ( L10 ) + P5 ( L11 ) + P6 ( L6 )
0.000
-159.789
RAV + RLV - P1 - P2 - P3 - P4 - P5 - P6 - P7
0
kg
191.93 kg
191.76=
A
αo βo
P1
S1
S12
RAV
◊ Joint B
Σ H = 0
- S12 cos β + S11 cos β = 0
S11 = kg
Σ V = 0
- S12 sin β + S13 + S11 sin β = 0
S13 = S12 sin β - S11 sin β
S13 = kg
◊ Joint C
Σ V = 0
- S1 sin α - S13 - P2 + S2 sin α - S14 sin γ = 0
0,41 S2 - 0,02 S14 = ..Pers. 1
Σ H = 0
- S1 cos α + S2 cos α + S14 cos γ = 0
0,91 S2 + 0,9998 S14 = ..Pers. 2
Eliminasi Pers. 1 dan Pers. 2
= x
= x
=
= -
-0.39 S14 =
S14 = kg
Subtitusi ke Pers. 1
=
= kg
◊ Joint D
Σ H = 0
- S11 cos β - S14 cos γ + S10 cos β = 0
S10 = kg
-243.993
-762.185
0, 41 S2 - 0,02 S14 -243.993 0.93
776.450
=S11 cos β
S12 cos β
0.000
0,37 S1 + 0,41 S14 -285.519
59.293
-152.535
0, 41 S2 - 0,02 S14 -243.99
0, 91 S2 + 0,9998 S14 -762.185 0.37
0,37 S1 - 0,02 S14 -226.226
621.078
S2
cos β
S11 cos β + S14 cos γ=S10
-657.547
B
βo
S11
S12
S13
C γo
αo
P2
S1
S2
S14
S13
D
S10
S11
S14
S15
βo γo
Σ V = 0
- S11 sin β + S14 sin γ + S10 sin β + S15 = 0
S15 = S11 sin β - S14 sin γ - S10 sin β
S15 = kg
◊ Joint E
Σ V = 0
- S2 sin α - S15 - P3 + S3 sin α - S16 sin δ = 0
0,41 S3 - 0,25 S16 = ..Pers. 1
Σ H = 0
- S2 cos α + S3 cos α + S16 cos δ = 0
0,91 S3 + 0,97 S16 = ..Pers. 2
Eliminasi Pers. 1 dan Pers. 2
= x
= x
=
= -
-0.57 S16 =
S16 = kg
Subtitusi ke Pers. 1
=
= kg
◊ Joint F
Σ H = 0
- S3 cos α + S4 cos α = 0
S4 = kg
Σ V = 0
- S3 sin α - P4 - S17 - S4 sin α = 0
S17 = kg
31.975
-150.40
-609.67
0,37 S3 - 0,22 S16 -139.45
0,37 S3 + 0,39 S16 -228.39
88.94
0, 41 S3 - 0,25 S16 -150.40 0.93
0, 91 S3 + 0,97 S16 -609.67 0.37
S3 cos α=S4
-156.329
0, 41 S3 - 0,22 S16 -150.40
S3 -493.051
-493.051
cos α
305.450
E
δo
αo
P3
S2
S3
S15
S16
F
P4
S3 S4
S17
αo αo
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
S1
S2
S3
-822.043
-657.547
-493.051
-493.051
-657.547
-822.043
2.615
2.858
2.610
2.858
2.610
2.610
S9
S10
Kode Batang Panjang Batang
S4
S5
S6
S7
S8
Gaya Batang
1 2 3
2.465
2.465
2.700
-156.329
31.975
-152.535
0.000
31.975
-156.329
305.450
2.700
2.470
2.470
0.508
2.425
1.017
2.716
1.570
2.716
776.450
776.450
S21
1.015
2.420
0.507
621.078
621.078
776.450
776.450
Tarik
Tarik
-
Tekan
0.000
-152.535
S20
S11
S12
S13
S14
S15
S16
S17
S18
S19
Tabel Rekapitulasi Gaya - Gaya Batang Akibat Beban Mati
Tarik
Tekan
Tarik
Tekan
Tarik
Tekan
-
Keterangan
4
Tekan
Tekan
Tekan
Tekan
Tekan
Tekan
Tarik
Tarik
Tarik
Tarik
• Panjang Jarak dan Beban
L1 = 2.425 m H1 = 0.980 m P1 = kg
L2 = 2.425 m H2 = 0.980 m P2 = kg
L3 = 2.650 m H3 = 1.071 m P3 = kg
L4 = 2.650 m H4 = 0.470 m P4 = kg
L5 = 2.420 m H5 = 0.509 m P5 = kg
L6 = 2.420 m H6 = 0.039 m P6 = kg
H7 = 1.570 m P7 = kg
L7 = L1 + L2 + L3 + L4 + L5 + L6 = m
L8 = L2 + L3 + L4 + L5 + L6 = m
L9 = L3 + L4 + L5 + L6 = m
L10 = L4 + L5 + L6 = m
L11 = L5 + L6 = m
L12 = L1 + L2 = m
L13 = L1 + L2 + L3 = m
L14 = L1 + L2 + L3 + L4 = m
L15 = L1 + L2 + L3 + L4 + L5 = m
• Sudut - sudut :
*) αo = 22.0 o *) γo = 0.875 o
tan αo = 0.40 tan γo = 0.015
sin αo = 0.37 sin γo = 0.015
cos αo = 0.93 cos γo = 0.9999
*) βo = 11.0 o *) δo = 12.676 o
tan βo = 0.19 tan δo = 0.225
sin βo = 0.19 sin δo = 0.219
cos βo = 0.98 cos δo = 0.976
100.000
50.000
14.990
12.565
10.140
7.490
PERHITUNGAN GAYA-GAYA BATANG AKIBAT BEBAN HIDUP
50.000
100.000
100.000
100.000
100.000
4.840
4.850
7.500
10.150
12.570
A
B
C D
E
F
G
H
I
J
K
L
δo
γo
αo βo
P7
P6
P5
P4
P3
P2
P1 S1
S2
S3 S4
S5
S6
S7
S8
S9 S10
S11
S12
S13
S14
S15
S16
S17
S18 S19
S20
S21
L1 L2 L3 L4 L5 L6
H3
H2
H1
H7
H6
H4
H5
RAV RLV
• Reaksi Perletakan
Σ ML = 0
Σ MA = 0
Kontrol Σ V = 0
= 0
= 0 ............ Ok
• Gaya - Gaya Batang
◊ Joint A
Σ V = 0
RAV - P1 + S1 sin α + S12 sin β = 0
0, 41 S1 + 0,21 S12 = ..Pers. 1
Σ H = 0
S1 cos α + S12 cos β = 0
0, 91 S1 + 0,98 S12 = ..Pers. 2
Eliminasi Pers. 1 dan Pers. 2
= x
= x
=
= -
-0.19 S12 =
S12 = kg
Subtitusi ke Pers. 1
=
= kg
kgL7
RLV =P7 ( L7 ) + P6 ( L15 ) + P5 ( L14 ) + P4 ( L13 ) + P3 ( L12 ) + P2 (L1)
= 300.13 kg
RAV =P1 ( L7 ) + P2 ( L8 ) + P3 ( L9 ) + P4 ( L10 ) + P5 ( L11 ) + P6 ( L6 )
L7
RAV + RLV - P1 - P2 - P3 - P4 - P5 - P6 - P7
0
-249.867
0.000
0, 41 S1 + 0,21 S12 -249.867 0.93
= 299.87
-231.672
1214.158
0, 41 S1 + 0,21 S12 -249.87
S1 -1285.452
0, 91 S1 + 0,98 S12 0.000 0.37
0,37 S1 + 0,19 S12 -231.672
0,37 S1 + 0,40 S12 0.000
A
αo βo
P1
S1
S12
RAV
◊ Joint B
Σ H = 0
- S12 cos β + S11 cos β = 0
S11 = kg
Σ V = 0
- S12 sin β + S13 + S11 sin β = 0
S13 = S12 sin β - S11 sin β
S13 = kg
◊ Joint C
Σ V = 0
- S1 sin α - S13 - P2 + S2 sin α - S14 sin γ = 0
0,41 S2 - 0,02 S14 = ..Pers. 1
Σ H = 0
- S1 cos α + S2 cos α + S14 cos γ = 0
0,91 S2 + 0,9998 S14 = ..Pers. 2
Eliminasi Pers. 1 dan Pers. 2
= x
= x
=
= -
-0.39 S14 =
S14 = kg
Subtitusi ke Pers. 1
=
= kg
◊ Joint D
Σ H = 0
- S11 cos β - S14 cos γ + S10 cos β = 0
S10 = kg
-381.539
-1191.850
0, 41 S2 - 0,02 S14 -381.539 0.93
0, 91 S2 + 0,9998 S14 -1191.85 0.37
S11 =S12 cos β
cos β
1214.158
0.000
0, 41 S2 - 0,02 S14 -381.54
S2 -1028.226
S10 =S11 cos β + S14 cos γ
cos β
0,37 S1 - 0,02 S14 -353.757
0,37 S1 + 0,41 S14 -446.475
92.718
-238.524
971.198
B
βo
S11
S12
S13
C γo
αo
P2
S1
S2
S14
S13
D
S10
S11
S14
S15
βo γo
Σ V = 0
- S11 sin β + S14 sin γ + S10 sin β + S15 = 0
S15 = S11 sin β - S14 sin γ - S10 sin β
S15 = kg
◊ Joint E
Σ V = 0
- S2 sin α - S15 - P3 + S3 sin α - S16 sin δ = 0
0,41 S3 - 0,25 S16 = ..Pers. 1
Σ H = 0
- S2 cos α + S3 cos α + S16 cos δ = 0
0,91 S3 + 0,97 S16 = ..Pers. 2
Eliminasi Pers. 1 dan Pers. 2
= x
= x
=
= -
-0.57 S16 =
S16 = kg
Subtitusi ke Pers. 1
=
= kg
◊ Joint F
Σ H = 0
- S3 cos α + S4 cos α = 0
S4 = kg
Σ V = 0
- S3 sin α - P4 - S17 - S4 sin α = 0
S17 = kg
0, 91 S3 + 0,97 S16 -953.35 0.37
0,37 S3 - 0,22 S16 -218.06
0,37 S3 + 0,39 S16 -357.13
50.000
-235.18
-953.35
0, 41 S3 - 0,25 S16 -235.18 0.93
S4 =S3 cos α
cos α
-770.998
477.642
139.08
-244.456
0, 41 S3 - 0,22 S16 -235.18
S3 -770.998
E
δo
αo
P3
S2
S3
S15
S16
F
P4
S3 S4
S17
αo αo
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
Tabel Rekapitulasi Gaya - Gaya Batang Akibat Beban Hidup
S1 2.615 -1285.452 Tekan
S2 2.858 -1028.226 Tekan
Kode Batang Panjang Batang Gaya Batang Keterangan
1 2 3 4
S5 2.610 -1028.226 Tekan
S6 2.610 -1285.452 Tekan
S3 2.610 -770.998 Tekan
S4 2.858 -770.998 Tekan
S9 2.700 971.198 Tarik
S10 2.700 971.198 Tarik
S7 2.465 1214.158 Tarik
S8 2.465 1214.158 Tarik
S13 0.508 0.000 -
S14 2.425 -238.524 Tekan
S11 2.470 1214.158 Tarik
S12 2.470 1214.158 Tarik
S17 1.570 477.642 Tarik
S18 2.716 -244.456 Tekan
S15 1.017 50.000 Tarik
S16 2.716 -244.456 Tekan
S21 0.507 0.000 -
S19 1.015 50.000 Tarik
S20 2.420 -238.524 Tekan
• Panjang Jarak dan Beban
L1 = 2.425 m H1 = 0.980 m P1 = kg
L2 = 2.425 m H2 = 0.980 m P2 = kg
L3 = 2.650 m H3 = 1.071 m P3 = kg
L4 = 2.650 m H4 = 0.470 m P4 = kg
L5 = 2.420 m H5 = 0.509 m P5 = kg
L6 = 2.420 m H6 = 0.039 m P6 = kg
H7 = 1.570 m P7 = kg
L7 = L1 + L2 + L3 + L4 + L5 + L6 = m
L8 = L2 + L3 + L4 + L5 + L6 = m
L9 = L3 + L4 + L5 + L6 = m
L10 = L4 + L5 + L6 = m
L11 = L5 + L6 = m
L12 = L1 + L2 = m
L13 = L1 + L2 + L3 = m
L14 = L1 + L2 + L3 + L4 = m
L15 = L1 + L2 + L3 + L4 + L5 = m
• Sudut - sudut :
*) αo = 22.0 o *) γo = 0.875 o
tan αo = 0.40 tan γo = 0.015
sin αo = 0.37 sin γo = 0.015
cos αo = 0.93 cos γo = 0.9999
*) βo = 11.0 o *) δo = 12.676 o
tan βo = 0.19 tan δo = 0.225
sin βo = 0.19 sin δo = 0.219
cos βo = 0.98 cos δo = 0.976
45.000
22.500
14.990
12.565
10.140
7.490
PERHITUNGAN GAYA-GAYA BATANG AKIBAT BEBAN AIR HUJAN
22.500
45.000
45.000
45.000
45.000
4.840
4.850
7.500
10.150
12.570
A
B
C D
E
F
G
H
I
J
K
L
δo
γo
αo βo
P7
P6
P5
P4
P3
P2
P1 S1
S2
S3 S4
S5
S6
S7
S8
S9 S10
S11
S12
S13
S14
S15
S16
S17
S18 S19
S20
S21
L1 L2 L3 L4 L5 L6
H3
H2
H1
H7
H6
H4
H5
RAV RLV
• Reaksi Perletakan
Σ ML = 0
Σ MA = 0
Kontrol Σ V = 0
= 0
= 0 ............ Ok
• Gaya - Gaya Batang
◊ Joint A
Σ V = 0
RAV - P1 + S1 sin α + S12 sin β = 0
0, 41 S1 + 0,21 S12 = ..Pers. 1
Σ H = 0
S1 cos α + S12 cos β = 0
0, 91 S1 + 0,98 S12 = ..Pers. 2
Eliminasi Pers. 1 dan Pers. 2
= x
= x
=
= -
-0.19 S12 =
S12 = kg
Subtitusi ke Pers. 1
=
= kg
kgL7
RLV =P7 ( L7 ) + P6 ( L15 ) + P5 ( L14 ) + P4 ( L13 ) + P3 ( L12 ) + P2 (L1)
= 135.06 kg
RAV =P1 ( L7 ) + P2 ( L8 ) + P3 ( L9 ) + P4 ( L10 ) + P5 ( L11 ) + P6 ( L6 )
L7
RAV + RLV - P1 - P2 - P3 - P4 - P5 - P6 - P7
0
-112.440
0.000
0, 41 S1 + 0,21 S12 -112.440 0.93
= 134.94
-104.253
546.371
0, 41 S1 + 0,21 S12 -112.44
S1 -578.453
0, 91 S1 + 0,98 S12 0.000 0.37
0,37 S1 + 0,19 S12 -104.253
0,37 S1 + 0,40 S12 0.000
A
αo βo
P1
S1
S12
RAV
◊ Joint B
Σ H = 0
- S12 cos β + S11 cos β = 0
S11 = kg
Σ V = 0
- S12 sin β + S13 + S11 sin β = 0
S13 = S12 sin β - S11 sin β
S13 = kg
◊ Joint C
Σ V = 0
- S1 sin α - S13 - P2 + S2 sin α - S14 sin γ = 0
0,41 S2 - 0,02 S14 = ..Pers. 1
Σ H = 0
- S1 cos α + S2 cos α + S14 cos γ = 0
0,91 S2 + 0,9998 S14 = ..Pers. 2
Eliminasi Pers. 1 dan Pers. 2
= x
= x
=
= -
-0.39 S14 =
S14 = kg
Subtitusi ke Pers. 1
=
= kg
◊ Joint D
Σ H = 0
- S11 cos β - S14 cos γ + S10 cos β = 0
S10 = kg
-171.692
-536.333
0, 41 S2 - 0,02 S14 -171.692 0.93
0, 91 S2 + 0,9998 S14 -536.33 0.37
S11 =S12 cos β
cos β
546.371
0.000
0, 41 S2 - 0,02 S14 -171.69
S2 -462.702
S10 =S11 cos β + S14 cos γ
cos β
0,37 S1 - 0,02 S14 -159.190
0,37 S1 + 0,41 S14 -200.914
41.723
-107.336
437.039
B
βo
S11
S12
S13
C γo
αo
P2
S1
S2
S14
S13
D
S10
S11
S14
S15
βo γo
Σ V = 0
- S11 sin β + S14 sin γ + S10 sin β + S15 = 0
S15 = S11 sin β - S14 sin γ - S10 sin β
S15 = kg
◊ Joint E
Σ V = 0
- S2 sin α - S15 - P3 + S3 sin α - S16 sin δ = 0
0,41 S3 - 0,25 S16 = ..Pers. 1
Σ H = 0
- S2 cos α + S3 cos α + S16 cos δ = 0
0,91 S3 + 0,97 S16 = ..Pers. 2
Eliminasi Pers. 1 dan Pers. 2
= x
= x
=
= -
-0.57 S16 =
S16 = kg
Subtitusi ke Pers. 1
=
= kg
◊ Joint F
Σ H = 0
- S3 cos α + S4 cos α = 0
S4 = kg
Σ V = 0
- S3 sin α - P4 - S17 - S4 sin α = 0
S17 = kg
0, 91 S3 + 0,97 S16 -429.01 0.37
0,37 S3 - 0,22 S16 -98.12
0,37 S3 + 0,39 S16 -160.71
22.500
-105.83
-429.01
0, 41 S3 - 0,25 S16 -105.83 0.93
S4 =S3 cos α
cos α
-346.949
214.939
62.59
-110.005
0, 41 S3 - 0,22 S16 -105.83
S3 -346.949
E
δo
αo
P3
S2
S3
S15
S16
F
P4
S3 S4
S17
αo αo
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
Tabel Rekapitulasi Gaya - Gaya Batang Akibat Beban Air Hujan
S1 2.615 -578.453 Tekan
S2 2.858 -462.702 Tekan
Kode Batang Panjang Batang Gaya Batang Keterangan
1 2 3 4
S5 2.610 -462.702 Tekan
S6 2.610 -578.453 Tekan
S3 2.610 -346.949 Tekan
S4 2.858 -346.949 Tekan
S9 2.700 437.039 Tarik
S10 2.700 437.039 Tarik
S7 2.465 546.371 Tarik
S8 2.465 546.371 Tarik
S13 0.508 0.000 -
S14 2.425 -107.336 Tekan
S11 2.470 546.371 Tarik
S12 2.470 546.371 Tarik
S17 1.570 214.939 Tarik
S18 2.716 -110.005 Tekan
S15 1.017 22.500 Tarik
S16 2.716 -110.005 Tekan
S21 0.507 0.000 -
S19 1.015 22.500 Tarik
S20 2.420 -107.336 Tekan
PERHITUNGAN GAYA-GAYA BATANG AKIBAT BEBAN ANGIN TEKAN
• Panjang Jarak dan Beban
L1 = 2.425 m H1 = 0.980 m P1 = 2.250 kg
L2 = 2.425 m H2 = 0.980 m P2 = 4.500 kg
L3 = 2.650 m H3 = 1.071 m P3 = 4.500 kg
L4 = 2.650 m H4 = 0.470 m P4 = 2.250 kg
L5 = 2.420 m H5 = 0.470 m
L6 = 2.420 m H6 = 0.515 m
H7 = 1.570 m
L7 = L1 + L2 + L3 + L4 + L5 + L6 = m
L8 = L2 + L3 + L4 + L5 + L6 = m
L9 = L3 + L4 + L5 + L6 = m
L10 = L4 + L5 + L6 = m
L11 = L5 + L6 = m
L12 = L1 + L2 = m
L13 = L1 + L2 + L3 = m
L14 = L1 + L2 + L3 + L4 = m
L15 = L1 + L2 + L3 + L4 + L5 = m
• Sudut - sudut :
*) αo = 22.00 o *) γo = 0.875 o
tan αo = 0.404 tan γo = 0.015
sin αo = 0.375 sin γo = 0.015
cos αo = 0.927 cos γo = 0.9999
*) βo = 11.00 o *) δo = 12.676 o
tan βo = 0.194 tan δo = 0.225
sin βo = 0.191 sin δo = 0.219
cos βo = 0.982 cos δo = 0.976
12.570
14.990
12.565
10.140
7.490
4.840
4.850
7.500
10.150
A
B
C D
E
F
G
H
I
J
K
L
δo
γo
αo βo
P4
P3
P2
P1 S1
S2
S3 S4
S5
S6
S7
S8
S9 S10
S11
S12
S13
S14
S15
S16
S17
S18 S19
S20 S21
L1 L2 L3 L4 L5 L6
H3
H2
H1
H7
H6
H4
H5
RAV RLV
RAH
• Reaksi Perletakan
Σ H = 0
= cos αo ( P1 + P2 + P3 + P4 )
= kg
Σ ML = 0
Σ MA = 0
Kontrol Σ V = 0
= 0.000
= 0.000 ............ Ok !
• Gaya - Gaya Batang
◊ Joint A
Σ V = 0
RAV - P1 sin α + S1 sin α + S12 sin β = 0
0, 41 S1 + 0,21 S12 =
Σ H = 0
P1 cos α + S1 cos α + S12 cos β - RAH = 0
0, 91 S1 + 0,98 S12 =
Eliminasi Pers. 1 dan Pers. 2
= x
= x
=
= -
-0.191 S12 =
S12 = kg
Subtitusi ke Pers. 1
=
= kg
◊ Joint B
Σ H = 0
RAH
RAH
RAV =
0, 91 S1 + 0,98 S12 10.431 0.375
0,37 S1 + 0,19 S12 -1.608
RAV + RLV - sin αo (P1 + P2 + P3 + P4)
0.000
0, 41 S1 + 0,21 S12 -1.735
-1.735
10.431
kg
= 2.48
3.907
0.927
kg
..Pers. 1
..Pers. 2
L7
P2 sin αo (L1) + P3 sin αo (L12) + P4 sin αo (L13) + P2 cos αo (H1)
+ P3 cos αo (H1 + H2) + P4 cos αo (H1 + H2 + H3)
- P2 cos αo (H1) - P3 cos αo (H1 + H2) - P4 cos αo (H1 + H2 + H3)
P1 sin αo (L7) + P2 sin αo (L8) + P3 sin αo (L9) + P4 sin αo (L10)
= 2.578
12.517
-5.516
28.907
0,37 S1 + 0,40 S12
0, 41 S1 + 0,21 S12 -1.735
S1 -19.35481
L7
RLV =
A
αo βo
P1
S1
S12
RAV
RAH
- S12 cos β + S11 cos β = 0
S11 = kg
Σ V = 0
- S12 sin β + S13 + S11 sin β = 0
S13 = S12 sin β - S11 sin β
S13 = kg
◊ Joint C
Σ V = 0
- S1 sin α - S13 - P2 sin α + S2 sin α - S14 sin γ = 0
0,41 S2 - 0,02 S14 = ..Pers. 1
Σ H = 0
- S1 cos α + S2 cos α + S14 cos γ + P2 cos α = 0
0,91 S2 + 0,9998 S14 = ..Pers. 2
Eliminasi Pers. 1 dan Pers. 2
= x
= x
=
= -
-0.389 S14 =
S14 = kg
Subtitusi ke Pers. 1
=
= kg
◊ Joint D
Σ H = 0
- S11 cos β - S14 cos γ + S10 cos β = 0
S10 = kg
S11 =S12 cos β
cos β
28.907
-5.565
0.000
0,37 S1 - 0,02 S14 -5.160
0,37 S1 + 0,41 S14 -8.285
3.126
-8.042
-22.118
0, 41 S2 - 0,02 S14 -5.565 0.93
0, 91 S2 + 0,9998 S14 -22.118 0.37
0, 41 S2 - 0,02 S14 -5.565
S2 -15.18253
S10 =S11 cos β + S14 cos γ
cos β
20.716
B
βo
S11
S12
S13
D
S10
S11
S14
S15
βo γo
C
γo
αo
P2
S1
S2
S14
S13
Σ V = 0
- S11 sin β + S14 sin γ + S10 sin β + S15 = 0
S15 = S11 sin β - S14 sin γ - S10 sin β
S15 = kg
◊ Joint E
Σ V = 0
- S2 sin α - S15 - P3 sin α + S3 sin α - S16 sin δ = 0
0,41 S3 - 0,25 S16 = ..Pers. 1
Σ H = 0
- S2 cos α + S3 cos α + S16 cos δ + P2 cos α = 0
0,91 S3 + 0,97 S16 = ..Pers. 2
Eliminasi Pers. 1 dan Pers. 2
= x
= x
=
= -
-0.57 S16 =
S16 = kg
Subtitusi ke Pers. 1
=
= kg
◊ Joint F
Σ H = 0
- S3 cos α + S4 cos α + P4 cos α = 0
S4 = kg
Σ V = 0
- S3 sin α - P4 sin α- S17 - S4 sin α = 0
S17 = kg
1.686
-2.316
-18.249
0, 41 S3 - 0,25 S16 -2.316 0.93
0,37 S3 - 0,23 S16 -2.147
0,37 S3 + 0,39 S16 -6.836
S4 =
-13.260
8.249
cos α (S3 - P4)
cos α
0, 91 S3 + 0,97 S16 -18.249 0.37
4.689
-8.242
0, 41 S3 - 0,25 S16 -2.316
S3 -11.01019
E
δo
αo
P3
S2
S3
S15
S16
F
P4
S3 S4
S17
αo αo
◊ Joint G
Σ V = 0
S16 sin δ + S17 + S18 sin δ - S10 sin β - S9 sin β = 0
- 0,21 S9 + 0,25 S18 = ..Pers. 1
Σ H = 0
- S16 cos δ + S18 cos δ - S10 cos β + S9 cos β = 0
0,98 S9 + 0,97 S18 = ..Pers. 2
Eliminasi Pers. 1 dan Pers. 2
= x
= x
=
= -
0.40 S18 =
S18 = kg
Subtitusi ke Pers. 1
=
= kg
◊ Joint H
Σ H = 0
- S4 cos α + S5 cos α - S18 cos δ = 0
Σ V = 0
S4 sin α - S5 sin α - S18 sin δ - S19 = 0
S19 = S4 sin α - S5 sin α - S18 sin δ
S19 = kg
◊ Joint I
Σ V = 0
S20 sin γ + S19 + S9 sin β - S8 sin β = 0
- 0,21 S8 + 0,02 S20 = ..Pers. 1
Σ H = 0
S20 cos γ - S9 cos β + S8 cos β = 0
0,98 S8 + 0,998 S20 = ..Pers. 2
-2.582
12.528
-13.512 kg
0.147
S4 cos α + S18 cos δ
cos α= =
-0.19
S9 12.762
0, 98 S9 + 0,97 S18 12.295
- 0,20 S9 + 0,24 S18 -2.442
- 0,20 S9 - 0,20 S18
- 0, 21 S9 + 0,25 S18 -2.488
-2.346
-2.488
12.295
0.982
S5
-0.096
-0.239
- 0, 21 S9 + 0,25 S18 -2.488
G S10 S9
S17
βo βo
δo δo
S16 S18
H
δo
αo
S5
S4
S19
S18
I
S9
S8
S20
S19
βo γo
Eliminasi Pers. 1 dan Pers. 2
= x
= x
=
= -
0.21 S20 =
S20 = kg
Subtitusi ke Pers. 1
=
= kg
◊ Joint J
Σ H = 0
- S5 cos α + S6 cos α - S20 cos γ = 0
Σ V = 0
S5 sin α - S6 sin α - S20 sin γ - S21 = 0
S21 = S5 sin α - S6 sin α - S20 sin γ
S21 = kg
◊ Joint K
Σ H = 0
- S8 cos β + S7 cos β = 0
Kontrol Σ V
Σ V = 0
S8 sin β - S7 sin β + S21 = 0
S7 =sin β
S8 sin β + S21= 15.014 kg
= -14.267 kgcos α
0.294
S6 =S5 cos α + S20 cos γ
S7 =S8 cos β
= 13.476 kgcos β
- 0,20 S8 - 0,21 S20 -2.390
-0.144
-0.700
- 0, 21 S8 + 0,02 S20 -2.582
S8 13.476
12.528
- 0, 21 S8 + 0,02 S20 -2.582 0.98
0, 98 S8 + 0,998 S20 -0.19
- 0,20 S8 + 0,02 S20 -2.535
.....OK
J γo
αo S6
S5
S20
S21
K
βo S8
S7
S21
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
S11
S12
S13
S6
S7
S8
S9
S4
S5
1
S1
S2
S3
Kode Batang
2.615
2.858
2.610
2.858
2.610
2.610
2.465
2.465
2.700
2.700
2.470
2.470
0.508
2.425
1.017
2.716
1.570
2.716
S14
S15
S16
S17
S10
1.686
-8.242
8.249
-0.239
0.147
-0.700
0.294
S18
S19
S20
S21
Tarik
2 3
Panjang Batang Gaya Batang
Tekan
Tarik
1.015
2.420
0.507
-19.355
-15.183
-11.010
-13.260
-13.512
-14.267
13.476
13.476
0.147
20.716
28.907
28.907
0.000
-8.042
Tabel Rekapitulasi Gaya - Gaya Batang Akibat Beban Angin Tekan
Keterangan
4
Tekan
Tekan
Tekan
Tekan
Tekan
Tekan
Tarik
Tarik
Tarik
Tarik
Tarik
Tarik
-
Tekan
Tarik
Tekan
Tarik
Tekan
• Panjang Jarak dan Beban
L1 = 2.425 m H1 = 0.980 m P1 = kg
L2 = 2.425 m H2 = 0.980 m P2 = kg
L3 = 2.650 m H3 = 1.071 m P3 = kg
L4 = 2.650 m H4 = 0.470 m P4 = kg
L5 = 2.420 m H5 = 0.470 m
L6 = 2.420 m H6 = 0.515 m
H7 = 1.570 m
L7 = L1 + L2 + L3 + L4 + L5 + L6 = m
L8 = L2 + L3 + L4 + L5 + L6 = m
L9 = L3 + L4 + L5 + L6 = m
L10 = L4 + L5 + L6 = m
L11 = L5 + L6 = m
L12 = L1 + L2 = m
L13 = L1 + L2 + L3 = m
L14 = L1 + L2 + L3 + L4 = m
L15 = L1 + L2 + L3 + L4 + L5 = m
• Sudut - sudut :
*) αo = 22.0 o *) γo = 0.875 o
tan αo = 0.40 tan γo = 0.015
sin αo = 0.37 sin γo = 0.015
cos αo = 0.93 cos γo = 0.9999
*) βo = 11.0 o *) δo = 12.676 o
tan βo = 0.19 tan δo = 0.225
sin βo = 0.19 sin δo = 0.219
cos βo = 0.98 cos δo = 0.976
22.500
45.000
45.000
22.500
12.570
PERHITUNGAN GAYA-GAYA BATANG AKIBAT BEBAN ANGIN HISAP
14.990
12.565
10.140
7.490
4.840
4.850
7.500
10.150
RAH A
B
C D
E
F
G
H
I
J
K
L
δo
γo
αo βo
P4
P3
P
P1
S1
S2
S3 S4
S5
S6
S7
S8
S9 S10
S11
S12
S1
S14
S15
S16
S17
S18 S19
S20
S21
L1 L2 L3 L4 L5 L6
H3
H2
H1
H7
H6
H4
H5
RA RL
• Reaksi Perletakan
Σ H = 0
= cos αo ( P1 + P2 + P3 + P4 )
= kg
Σ ML = 0
Σ MA = 0
Kontrol Σ V = 0
= 0.000
= 0.000 ............ Ok !
• Gaya - Gaya Batang
◊ Joint L
Σ V = 0
- RLV + P4 sin α + S6 sin α + S7 sin β = 0
0, 41 S6 + 0,21 S7 = ..Pers. 1
Σ H = 0
P4 cos α - S6 cos α - S7 cos β = 0
0, 91 S6 + 0,98 S7 = ..Pers. 2
Eliminasi Pers. 1 dan Pers. 2
= x
= x
=
= -
-0.19 S7 =
S7 = kg
Subtitusi ke Pers. 1
=
= kg
kg
RLV =- P3 cos αo (H1) - P2 cos αo (H1 + H2) - P1 cos αo (H1 + H2 + H3)
L7
RAH
RAH
P1 sin αo (L10) + P2 sin αo (L11) + P3 sin αo (L6)= 24.774 kg
RAV =+ P3 cos αo (H1) + P2 cos αo (H1 + H2) + P1 cos αo (H1 + H2 + H3)
125.170
- RAV - RLV + sin αo (P1 + P2 + P3 + P4)
0.000
17.369
20.862
0, 41 S6 + 0,21 S7 17.369 0.93
L7
P4 sin αo (L7) + P3 sin αo (L15) + P2 sin αo (L14) + P1 sin αo (L13)= 25.798
8.289
-43.443
0, 41 S6 + 0,21 S7 17.369
S6 68.494
0, 91 S6 + 0,98 S7 20.862 0.37
0,37 S6 + 0,19 S7 16.104
0,37 S6 + 0,40 S7 7.815
L
αo βo
P4
S6
S7
RLV
◊ Joint K
Σ H = 0
S7 cos β - S8 cos β = 0
S8 = kg
Σ V = 0
- S7 sin β + S21 + S8 sin β = 0
S21 = S7 sin β - S8 sin β
S21 = 0.000 kg
◊ Joint J
Σ V = 0
- S6 sin α - S21 + P3 sin α + S5 sin α - S20 sin γ = 0
0,41 S5 - 0,02 S20 = ..Pers. 1
Σ H = 0
S6 cos α - S5 cos α - S20 cos γ + P3 cos α = 0
0,91 S5 + 0,9998 S20 = ..Pers. 2
Eliminasi Pers. 1 dan Pers. 2
= x
= x
=
= -
-0.39 S20 =
S20 = kg
Subtitusi ke Pers. 1
=
= kg
◊ Joint I
Σ H = 0
S8 cos β + S20 cos γ - S9 cos β = 0
S9 = kg
105.230
0, 41 S5 - 0,02 S20 8.801 0.93
0, 91 S5 + 0,9998 S20 105.230 0.37
S8 =S7 cos β
cos β
-43.443
8.801
0, 41 S5 - 0,02 S20 8.801
S5 26.771
S9 =S8 cos β + S20 cos γ
cos β
0,37 S5 - 0,02 S20 8.160
0,37 S5 + 0,41 S20 39.420
-31.260
80.417
38.470
K
βo
S8
S7
S21
I
S9
S8
S20
S19
βo γo
J γo
αo
P3
S6
S5
S20
S21
Σ V = 0
- S8 sin β + S20 sin γ + S9 sin β + S19 = 0
S19 = S8 sin β - S20 sin γ - S9 sin β
S19 = kg
◊ Joint H
Σ V = 0
- S5 sin α - S19 + P2 sin α + S4 sin α - S18 sin δ = 0
0,41 S4 - 0,25 S18 = ..Pers. 1
Σ H = 0
S5 cos α - S4 cos α - S18 cos δ + P2 cos α = 0
0,91 S4 + 0,97 S18 = ..Pers. 2
Eliminasi Pers. 1 dan Pers. 2
= x
= x
=
= -
-0.57 S18 =
S18 = kg
Subtitusi ke Pers. 1
=
= kg
◊ Joint F
Σ H = 0
- S3 cos α + S4 cos α + P1 cos α = 0
S3 = kg
Σ V = 0
- S3 sin α + P1 sin α- S17 - S4 sin α = 0
S17 = kg
0, 91 S4 + 0,97 S18 66.545 0.37
0,37 S4 - 0,22 S18 -21.961
0,37 S4 + 0,39 S18 24.928
-16.857
-23.686
66.545
0, 41 S4 - 0,25 S18 -23.686 0.93
S3 =cos α (S4 + P1)
cos α
7.548
11.202
-46.890
82.417
0, 41 S4 - 0,25 S18 -23.686
S4 -14.9522
H
δo
αo
P2
S5
S4
S19
S18
F
P1
S3 S4
S17
αo αo
◊ Joint G
Σ V = 0
S16 sin δ + S17 + S18 sin δ - S10 sin β - S9 sin β = 0
- 0,21 S10 + 0,25 S16 = ..Pers. 1
Σ H = 0
- S16 cos δ + S18 cos δ - S10 cos β + S9 cos β = 0
0,98 S10 + 0,97 S16 = ..Pers. 2
Eliminasi Pers. 1 dan Pers. 2
= x
= x
=
= -
0.40 S16 =
S16 = kg
Subtitusi ke Pers. 1
=
= kg
◊ Joint E
Σ H = 0
+ S3 cos α - S2 cos α + S16 cos δ = 0
Σ V = 0
S3 sin α - S2 sin α - S16 sin δ - S15 = 0
S15 = S3 sin α - S2 sin α - S16 sin δ
S15 = kg
◊ Joint D
Σ V = 0
S14 sin γ + S15 + S10 sin β - S11 sin β = 0
- 0,21 S11 + 0,02 S14 = ..Pers. 1
Σ H = 0
- S14 cos γ + S10 cos β - S11 cos β = 0
0,98 S11 + 0,998 S14 = ..Pers. 2
- 0,20 S10 + 0,24 S16 -21.544
- 0,20 S10 - 0,20 S16 -22.548
1.005
2.502
-21.947
118.172
- 0, 21 S10 + 0,25 S16 -21.947 0.98
0, 98 S10 + 0,97 S16 118.172 -0.19
kgcos α
-1.535
-20.961
- 0, 21 S10 + 0,25 S16 -21.947
S10 117.897
S2 =S3 cos α + S16 cos δ
115.731
= 10.180
G S10 S9
S17
βo βo
δo δo
S16 S18
E
δo
αo
S2
S3
S15
S16
D
S10
S11
S14
S15
βo γo
Eliminasi Pers. 1 dan Pers. 2
= x
= x
=
= -
0.21 S14 =
S14 = kg
Subtitusi ke Pers. 1
=
= kg
◊ Joint C
Σ H = 0
- S1 cos α + S2 cos α + S14 cos γ = 0
Σ V = 0
S2 sin α - S1 sin α - S14 sin γ - S13 = 0
S13 = S2 sin α - S1 sin α - S14 sin γ
S13 = kg
◊ Joint B
Σ H = 0
- S12 cos β + S11 cos β = 0
Kontrol Σ V
Σ V = 0
S8 sin β - S7 sin β + S13 = 0
- 0, 21 S11 + 0,02 S14 -20.961 0.98
0, 98 S11 + 0,998 S14 115.731 -0.19
- 0, 21 S11 + 0,02 S14 -20.961
S11 109.852
S1 =S2 cos α + S14 cos γ
- 0,20 S11 + 0,02 S14 -20.576
- 0,20 S11 - 0,21 S14 -22.082
1.507
0.000
= 10.180 kgcos α
0.000
S12 =S11 cos β
= 109.852 kgcos β
S12 =S11 sin β + S13
= 109.85 kg .....OKsin β
C
γo
αo
S1
S2
S14
S13
B
βo
S11
S12
S13
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
m kg
Panjang BatangKode Batang Gaya Batang Keterangan
2 3
Tabel Rekapitulasi Gaya - Gaya Batang Akibat Beban Angin Hisap
1
S5
S6
S7
S8
S1
S2
S3
S4
S20 2.420
S13
S14
S15
S16
S9
S10
S11
S12
2.716
1.015
S17
S18
S21
2.615
2.858
2.610
2.858
2.610
2.610
2.465
2.465
2.700
2.700
2.470
2.470
0.508
2.425
1.017
2.716
1.570
S19
0.507
10.180
10.180
7.548
-14.952
26.771
68.494
-43.443
-43.443
38.470
117.897
109.852
109.852
0.000
0.000
-1.535
2.502
0.000
11.202
82.417
-16.857
80.417
4
Tarik
Tarik
Tarik
Tekan
Tarik
Tarik
Tekan
Tekan
Tarik
Tarik
Tarik
Tarik
Tarik
Tarik
Tekan
-
Tarik
-
-
Tekan
Tarik
Kode Beban Beban Beban B. Angin B. Angin Kombinasi I Kombinasi II Kombinasi III Kombinasi IV Kombinasi V Beban
Batang Mati Hidup Air hujan tekan hisap 1,4 D 1,2 D + 1,6 L 1,2 D + 1,6 H 1,2 D + 1,3 W1 1,2 D - 1,3 W2 Maksimum
(D) (L) (H) (W1) (W2) + 0,5 H + 0,5 L + 0,5 L + 0,5 H + 0,5 L + 0,5 H
1 2 3 4 5 6 7 8 9 10 11 14
S1 -822.04 -1285.45 -578.45 -19.35 10.18 -1150.86 -3332.40 -2554.70 -1943.57 -1893.24 -3332.40
S2 -657.55 -1028.23 -462.70 -15.18 10.18 -920.57 -2665.57 -2043.49 -1554.26 -1514.78 -2665.57
S3 -493.05 -771.00 -346.95 -11.01 7.55 -690.27 -1998.73 -1532.28 -1164.95 -1136.32 -1998.73
S4 -493.05 -771.00 -346.95 -13.26 -14.95 -690.27 -1998.73 -1532.28 -1167.87 -1133.40 -1998.73
S5 -657.55 -1028.23 -462.70 -13.51 26.77 -920.57 -2665.57 -2043.49 -1552.09 -1516.96 -2665.57
S6 -822.04 -1285.45 -578.45 -14.27 68.49 -1150.86 -3332.40 -2554.70 -1936.95 -1899.86 -3332.40
S7 776.45 1214.16 546.37 13.48 -43.44 1087.03 3147.58 2413.01 1829.52 1794.49 3147.58
S8 776.45 1214.16 546.37 13.48 -43.44 1087.03 3147.58 2413.01 1829.52 1794.49 3147.58
S9 621.08 971.20 437.04 0.15 38.47 869.51 2517.73 1930.16 1449.60 1449.22 2517.73
S10 621.08 971.20 437.04 20.72 117.90 869.51 2517.73 1930.16 1476.34 1422.48 2517.73
S11 776.45 1214.16 546.37 28.91 109.85 1087.03 3147.58 2413.01 1849.58 1774.43 3147.58
S12 776.45 1214.16 546.37 28.91 109.85 1087.03 3147.58 2413.01 1849.58 1774.43 3147.58
S13 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
S14 -152.54 -238.52 -107.34 -8.04 0.00 -213.55 -618.35 -474.04 -366.43 -345.52 -618.35
S15 31.98 50.00 22.50 1.69 -1.54 44.77 129.62 99.37 76.81 72.43 129.62
S16 -156.33 -244.46 -110.01 -8.24 2.50 -218.86 -633.73 -485.83 -375.54 -354.11 -633.73
S17 305.45 477.64 214.94 8.25 11.20 427.63 1238.24 949.26 723.55 702.11 1238.24
S18 -156.33 -244.46 -110.01 -0.24 82.42 -218.86 -633.73 -485.83 -365.14 -364.51 -633.73
S19 31.98 50.00 22.50 0.15 -16.86 44.77 129.62 99.37 74.81 74.43 129.62
S20 -152.54 -238.52 -107.34 -0.70 80.42 -213.55 -618.35 -474.04 -356.88 -355.06 -618.35
S21 0.00 0.00 0.00 0.29 0.00 0.00 0.00 0.00 0.38 -0.38 -0.38
Kesimpulan:
- Batang tunggal yang menerima gaya batang tekan terbesar
S1 = -3332.40
- Batang tunggal yang menerima gaya batang tarik terbesar
S7 = 3147.58 kg
- Batang rangkap yang menerima gaya batang tarik terbesar
S15 = 129.62 kg
TABEL GAYA - GAYA BATANG AKIBAT BEBAN YANG BEKERJA
Beban tetap Beban Angin
KONTROL KEKUATAN BATANG
- Batang tunggal yang menerima gaya batang tekan terbesar
b = 6 cm
h = 12 cm
S1 = -3332.40 kg
Lk = 261.545 cm
ix = 0.289 x h
= 3.468 cm
iy = 0.289 x b
= 1.734 cm
Lk
imin
Maka :
ω = 2.73 ( Lihat tabel faktor tekuk kayu )
Ʈtk = 130 kg/cm2 ( Lihat tabel tegangan ijin kayu )
Kontrol Tegangan :
ω . S
b . h
= 126.354 kg/cm2 < 130 kg/cm2 .....OK
151=λ = 150.833 ≈
=Ʈtk
h
b
- Batang tunggal yang menerima gaya Batang tarik terbesar
b = 6 cm
h = 12 cm
S7 = 3147.58 kg
Fnetto = h . b - ( 20% . h . b )
= 57.6
S
Fnetto
= 54.6455 kg/cm2 < 130 kg/cm2 .....OK
- Batang rangkap yang menerima gaya Batang tarik terbesar
b = 3 cm
h = 12 cm
S15 = 129.62 kg
Fnetto = h . b - ( 20% . h . b )
= 28.8
S
Fnetto
= 2.25036 kg/cm2 < 130 kg/cm2 .....OK
Ʈtr =
=Ʈtr
h
b
h
b
h
b
PERENCANAAN SAMBUNGAN
□ JOINT A
Digunakan sambungan gigi rangkap.
- Batang S1 = kg
- Sudut θ = 11.00 o
- Mencari tegangan ijin
Gigi belakang σds = σds a = σds // - ( σds // - σds ┴ ) sin θo
=
Gigi muka σds = σds ½ a = σds // - ( σds // - σds ┴ ) sin ½ θo
=
- Mencari ukuran gigi belakang (tv2 dan tm2)
Besarnya S2 diambil = = 1666.2 kg
dipakai tinggi gigi belakang = 3.00 cm
Dengan demikian gigi kedua dapat mendukung gaya S2 = kg
- Mencari ukuran gigi muka (tv1 dan tm1)
Besarnya S1 = = 1284.47 kg
dipakai tinggi gigi muka = 2.00 cm, sehingga tv2 - tv1 = 1 cm
- Mencari panjang muka gigi (panjang penyaluran = l m )
Maka pakai : l m1 = 15.00 cm
Maka pakai : l m2 = 15.00 cm
½ x S
Tinggi gigi muka tv1 =S1 x cos2 ½ θ
=
111.29
120.59
2047.9
3332.40
cm
=tv2
cos θ
= = 2.39S2 x cos2 θ
=
Tinggi gigi belakang tv2
Kemiringan gigi belakang tm2
b x σds a
1.76
l m1 = 10.47 cmPanjang kayu muka, =
cmcos θ
= 3 cm
S1 x cos θ
τ// x b< 15.0 cm
b x σds ½ a
S - S2
cm
3.07 cm
Kemiringan gigi muka tm1 =
cm < 15.0 cmτ// x b
13.58Panjang kayu belakang, l m2 =S2 x cos θ
=
tv2= 2.04
58,3915
2,04
3
3,1
84°
84°
24°
12°
12°
2
S
S1
S2
h
□ JOINT B
Digunakan sambungan bertampang dua
- Batang S11 = kg
- Direncanakan dengan baut berdiameter Ø = 3/8 " = cm
- Gol. Kelas Kuat I dgn sambungan tampang dua, didapat kekuatan perbaut
S = 125 x d x b3 x (1 - 0.6 sin θ) = kg
S = 250 x d x b1 x (1 - 0.6 sin θ) = kg
S = 480 x d2 x (1 - 0.35 sin θ) = kg
diambil S yang terkecil = kg
- Baut yang dibutuhkan = 0 / 299.64 = 0 ≈ 1 buah
- Jarak minimum antar sumbu baut dan ujung kayu = 7 d = 6.9 cm ≈ 7 cm
- Jarak minimum antara sumbu baut dalam arah gaya = 5 d = 4.9 cm ≈ 5 cm
- Jarak minimum antara baut tegak lurus arah gaya = 3 d = 3 x 0.98 = 2.94 cm
- Jarak minimum antara sumbu baut dgn tepi kayu = 2 d = 2 x 0.98 = 1.96 cm
□ JOINT C
Digunakan sambungan gigi tunggal.
- Batang S14 = kg
- Sudut θ = 22.87 o
- Mencari tegangan ijin
σds = σds ½ a = σds // - ( σds // - σds ┴ ) sin ½ θo
=
Mencari ukuran gigi belakang (tv dan tm)
Syarat untuk θ ≤ 50o tm harus ≤ ¼ h = 3.0 cm
Mencari panjang muka gigi (panjang penyaluran = l m )
Maka pakai : l m = 15.00 cm
< ¼ x h
= 0.88 cm
0.00
299.64
b x σds ½ a
Tinggi gigi tv =
0.98
147.00
294.00
299.64
618.35
112.15
S x cos θPanjang kayu muka,
tvKemiringan gigi tm = = 0.96 cm
cos θ
l m
S x cos2 ½ θ-
-
15.0 cmτ// x b
= = 4.75 cm <
58,3915
2,04
3
3,1
84°
84°
24°
12°
12°
2
S
S1
S2
h
12°
S
12
90°
12
Digunakan sambungan bertampang dua.
Batang S11 = kg
Direncanakan dengan baut berdiameter Ø = 3/8 " = cm
Gol. Kelas Kuat I dgn sambungan tampang dua, didapat kekuatan perbaut
- S = 125 x d x b3 x (1 - 0.6 sin 90) = kg
S = 250 x d x b1 x (1 - 0.6 sin 90) = kg
S = 480 x d2 x (1 - 0.35 sin 90) = kg
diambil S yang terkecil = kg
Baut yang dibutuhkan = 0 / 294 = 0 ≈ 1 buah
- Jarak minimum antar sumbu baut dan ujung kayu = 7 d = 6.9 cm ≈ 7 cm
- Jarak minimum antara sumbu baut dalam arah gaya = 5 d = 4.9 cm ≈ 5 cm
- Jarak minimum antara baut tegak lurus arah gaya = 3 d = 3 x 0.98 = 2.94 cm
- Jarak minimum antara sumbu baut dgn tepi kayu = 2 d = 2 x 0.98 = 1.96 cm
□ JOINT D
Digunakan sambungan gigi tunggal.
- Batang S14 = kg
- Sudut θ = 11.87 o
- Mencari tegangan ijin
σds = σds ½ a = σds // - ( σds // - σds ┴ ) sin ½ θo
=
Mencari ukuran gigi belakang (tv dan tm)
Syarat untuk θ ≤ 50o tm harus ≤ ¼ h = 3.0 cm
Mencari panjang muka gigi (panjang penyaluran = l m )
Maka pakai : l m = 15.00 cm
0.00
0.98
294.00
618.35
120.69
- Tinggi gigi tv =S x cos2 ½ θ
294.00
299.64
294.00
b x σds ½ a
- Kemiringan gigi tm =tv
= 0.86 cm
= 0.84 cm
< ¼ x hcos θ
Panjang kayu muka, l m =S x cos θ
= 5.04 cm < 15.0 cmτ// x b
12
12
12
25°
S
77°
0,810,73
24°
Digunakan sambungan bertampang dua.
Batang S15 = kg
Direncanakan dengan baut berdiameter Ø = 3/8 " = cm
Gol. Kelas Kuat I dgn sambungan tampang dua, didapat kekuatan perbaut
- S = 125 x d x b3 x (1 - 0.6 sin 90) = kg
S = 250 x d x b1 x (1 - 0.6 sin 90) = kg
S = 480 x d2 x (1 - 0.35 sin 90) = kg
diambil S yang terkecil = kg
Baut yang dibutuhkan = 129.62 / 294 = 0.4 ≈ 1 buah
- Jarak minimum antar sumbu baut dan ujung kayu = 7 d = 6.9 cm ≈ 7 cm
- Jarak minimum antara sumbu baut dalam arah gaya = 5 d = 4.9 cm ≈ 5 cm
- Jarak minimum antara baut tegak lurus arah gaya = 3 d = 3 x 0.98 = 2.94 cm
- Jarak minimum antara sumbu baut dgn tepi kayu = 2 d = 2 x 0.98 = 1.96 cm
□ JOINT E
Digunakan sambungan gigi tunggal.
- Batang S16 = kg
- Sudut θ = 34.68 o
- Mencari tegangan ijin
σds = σds ½ a = σds // - ( σds // - σds ┴ ) sin ½ θo
=
Mencari ukuran gigi belakang (tv dan tm)
Syarat untuk θ ≤ 50o tm harus ≤ ¼ h = 3.0 cm
Mencari panjang muka gigi (panjang penyaluran = l m )
Maka pakai : l m = 15.00 cm
294.00
299.64
294.00
633.73
103.18
129.62
0.98
294.00
0.93 cm- Tinggi gigi tv =S x cos2 ½ θ
=b x σds ½ a
1.13 cmcos θ
- Kemiringan gigi tm =tv
= < ¼ x h
4.34 cm < 15.0 cmPanjang kayu muka, l m =S x cos θ
=τ// x b
12
12
13° 0,7
0,72
83°
S
12
S
12
Digunakan sambungan bertampang dua.
Batang S15 = kg
Direncanakan dengan baut berdiameter Ø = 3/8 " = cm
Gol. Kelas Kuat I dgn sambungan tampang dua, didapat kekuatan perbaut
- S = 125 x d x b3 x (1 - 0.6 sin 90) = kg
S = 250 x d x b1 x (1 - 0.6 sin 90) = kg
S = 480 x d2 x (1 - 0.35 sin 90) = kg
diambil S yang terkecil = kg
Baut yang dibutuhkan = 129.62 / 294 = 0.4 ≈ 1 buah
- Jarak minimum antar sumbu baut dan ujung kayu = 7 d = 6.9 cm ≈ 7 cm
- Jarak minimum antara sumbu baut dalam arah gaya = 5 d = 4.9 cm ≈ 5 cm
- Jarak minimum antara baut tegak lurus arah gaya = 3 d = 3 x 0.98 = 2.94 cm
- Jarak minimum antara sumbu baut dgn tepi kayu = 2 d = 2 x 0.98 = 1.96 cm
□ JOINT F
Digunakan sambungan gigi rangkap.
- Batang S3 = kg
- Sudut θ = 68.00 o
- Mencari tegangan ijin
Gigi belakang σds = σds a = σds // - ( σds // - σds ┴ ) sin θo
=
Gigi muka σds = σds ½ a = σds // - ( σds // - σds ┴ ) cos ½ θo
=
- Mencari ukuran gigi belakang (tv2 dan tm2)
Besarnya S2 diambil = = 999.366 kg
dipakai tinggi gigi belakang = 2.00 cm
Dengan demikian gigi kedua dapat mendukung gaya S2 = kg
294.00
½ x S
294.00
299.64
294.00
1998.73
129.62
0.98
46.55
50.91
Tinggi gigi belakang tv2 = =b x σds a
S2 x cos2 θ0.50 cm < 1/6 x h = 2 cm
5.34 cmtv2
Kemiringan gigi belakang tm2 = =cos θ
1491.3
0,990,77
71°
12
12
38°
12
12
S
S
- Mencari ukuran gigi muka (tv2 dan tm2)
Besarnya S1 = = 507.46 kg
dipakai tinggi gigi muka = 1.00 cm, sehingga tv2 - tv1 = 1 cm
- Mencari panjang muka gigi (panjang penyaluran = l m )
Maka pakai : l m1 = 15.00 cm
Maka pakai : l m2 = 15.00 cm
□ JOINT G
Digunakan sambungan gigi rangkap.
- Batang S10 = kg
- Sudut θ = 51.00 o
- Mencari tegangan ijin
Gigi belakang σds = σds a = σds // - ( σds // - σds ┴ ) cos θo
=
Gigi muka σds = σds ½ a = σds // - ( σds // - σds ┴ ) cos ½ θo
=
- Mencari ukuran gigi belakang (tv2 dan tm2)
Besarnya S2 diambil = = 1258.87 kg
dipakai tinggi gigi belakang = 2.00 cm
Dengan demikian gigi kedua dapat mendukung gaya S2 = kg1132.8
= 1.73 cm < 1/6 x h = 2 cmb x σds a
Kemiringan gigi belakang tm2 =tv2
= 2.57 cmsin θ
S - S2
Tinggi gigi muka tv1 =S1 x sin2 ½ θ
= 0.38 cmb x σds ½ a
cmsin θ
Panjang kayu muka, l m1 =S1 x sin θ
= 3.92 cm
Kemiringan gigi muka tm1 =tv2
= 1.19
< 15.0 cmτ// x b
Panjang kayu belakang, l m2 =S2 x cos θ
= 3.12 cm < 15.0 cmτ// x b
2517.73
73.36
48.77
½ x S
Tinggi gigi belakang tv2 =S2 x sin2 θ
57°
57°
4,92
2
1
15
21,4
1
12 12
12
SS66°
1,19
- Mencari ukuran gigi muka (tv2 dan tm2)
Besarnya S1 = = 1384.95 kg
dipakai tinggi gigi muka = 1.00 cm, sehingga tv2 - tv1 = 1 cm
- Mencari panjang muka gigi (panjang penyaluran = l m )
Maka pakai : l m1 = 15.00 cm
Maka pakai : l m2 = 15.00 cm
Digunakan sambungan gigi tunggal.
- Batang S16 = kg
- Sudut θ = 23.68 o
- Mencari tegangan ijin
σds = σds ½ a = σds // - ( σds // - σds ┴ ) sin ½ θo
=
Mencari ukuran gigi belakang (tv dan tm)
Syarat untuk θ ≤ 50o tm harus ≤ ¼ h = 3.0 cm
Mencari panjang muka gigi (panjang penyaluran = l m )
Maka pakai : l m = 15.00 cm
cmτ// x b
Panjang kayu belakang, l m2 =S2 x sin θ
= 8.15 cm < 15.0 cmτ// x b
Panjang kayu muka, l m1 =S1 x sin θ
= 8.97 cm < 15.0
b x σds ½ a
Kemiringan gigi muka tm1 =tv2
= 1.29 cmsin θ
S - S2
Tinggi gigi muka tv1 =S1 x sin2 ½ θ
= 0.88 cm
633.73
111.54
0.99 cm < ¼ x h
0.91 cm
= 4.84 cm < 15.0 cmPanjang kayu muka, l m
S x cos θ=
τ// x b
- Tinggi gigi tv =S x cos2 ½ θ
=b x σds ½ a
- Kemiringan gigi tm =tv
=cos θ
1,29
2,57
1
2
0,84
0,75
26°
102°
51°
51°
12
12
SS
S S
12
PERENCANAAN KANTILEVER
Sudut :
αo = 24 o
tan α= 0.44523
sin α= 0.40674
cos α= 0.91355
Ukuran
L1 = 0.75 m
L2 = 0.75 m
H1 = 0.33392 m
H2 = 0.33392 m
Dimensi Kayu
b = 6 cm
h = 12 cm
Panjang Total Batang = 3.32 m
Pembebanan
Beban Sendiri = b x h x P.total Batang x BJ Kayu = 19.3782 kg
Beban Hidup = 100 kg
Beban Atap = Jarak efektif gording x Berat BJLS = 15 kg
= 134.378 kg
Σ MD = 0
Σ H = 0
301.82 kg
Σ V = 0
RDV = P1 + P2
RDV = 268.756 kg
Cek dimensi
S = 268.756 kg
Lk = 66.7843 cm
ix = 0.289 x b = 1.734 cm
iy = 0.289 x h = 3.468 cm
REH = RDH =
P1 = P2
REH =P1 x ( L1 + L2 )
H1 + H2
= 301.82 kg
RDH
RDV
A B
C
D
E
αo
P1
L1 L2
H2
H1
REH
P2
Lk
imin
Maka :
ω = 1.35 ( Lihat tabel faktor tekuk kayu )
Ʈtk = 96 kg/cm2 ( Lihat tabel faktor tekuk kayu )
Kontrol Tegangan :
ω . S
b . h
= 5.03918 kg/cm2 < 96 kg/cm2 .....OK
≈ 39
Ʈtk =
λ = = 38.51459215
PERENCANAAN KOLOM
Direncanakan :
- Dimensi Kolom = b = 20 cm
h = 20 cm
L = 700 cm
b x h2
6
- i x = 0.289 x h = 5.78 cm
- i y = 0.289 x b = 5.78 cm
- Luas Penampang Bruto (Fbr) = b x h = 400 cm2
- Luas Penampang Bersih (Fnt) = Fbr / 1.25 = 320 cm2
- Tegangan Ijin Kayu Kelas Kuat I
• σlt = 150 kg/cm2
• σtk // = σtr // = 130 kg/cm2
• σtk ┴ = 40 kg/cm2
• τ// = 20 kg/cm2
• α = σtk // / σlt = 0.86667
- Pembebanan yang terjadi arah vertikal (RV) :
Beban Atap = 777.37305 kg
Beban Sendiri = 672 kg
Total = 1449.3731 kg
- Pembebanan yang terjadi arah horizontal (RH) :
Beban Atap = -16.27208 kg
- Momen Yang Terjadi Pada kolom :
Mu = RH x L = 11390.454 kg.cm
- Mencari Faktor Tekuk (ω)
= 700 cm
l k =
i min
Maka diperoleh ω = 4.73 (lihat tabel faktor tekuk)
- Kontrol Tegangan
α x M ω x P
W Fnt
σ = 28.827 kg/cm2 < 130 kg/cm2 .....OK !
Maka kolom dengan dimensi 20/20 cm dapat digunakan.
cm3
Panjang Tekuk (l k ) = L
≈
Momen Tahanan Batang (W)- = = 1333.33
122
σ = +
Angka Kelangsingan, λ = 121.107
30 cm
700 cm
30
PERENCANAAN SLOOF
→ Pembebanan yang terjadi :
- Berat sendiri Sloof : 0,20 m x 0,40 m x 2400 kg = Kg
- Berat sendiri Dinding : 4,00 m x 250 kg x 1,00 = Kg
= Kg/m'
→ = q x 1,2
= 1192 kg/m' x 1,2
= Kg/m' -------> Knm
→ Momen Maksimal
- Mlap = 1/11 Wu L²
= 1/11 x 14,304 Kn/m' x 4²
= 20.81 Knm
- Mtump = 1/10 Wu L²
= 1/10 x 14,304 Kn/m' x 4²
= 22.89 Knm
→ Perhitungan Sloof
- F'c = 25 Mpa
Fy = 240 Mpa
d = h - Sb - Ø senkong - ½Ø Tulangan utama
= 400 - 70 - 10 - ½ 16
= 312 mm
192.00
1000.00
1000.00q
1430.4
- =0,2 (0,312)²
20.81= 10.689 Kn/m² ----->ρ = 0,0053
14.304
Mu Lap
b.d²
1/2 P P P P P P P
1/10 1/24
1/11 1/16
4,00 4,00 4,00 4,00 4,00 4,00 20 cm
40 cm
= p x b x d
= 0,0053 x 200 x 312
= mm² ---------------> 2 Ø 16 = 405 mm²
= p x b x d
= 0,0061 x 200 x 312
= mm² ---------------> 2 Ø 16 = 402 mm²
→ Check Tulangan Geser
- Vu = ½ . Wu . L
= ½ x 14,304 x 4
= Kg -------> Kn
Fc'
6
25
6
Vu
Ø
Vu
Ø
Vu
Ø
Tidak memerlukan tulangan geser, maka digunakan tulangan geser praktis
dengan dmax = 312
2
(dipakai Ø 10 - 150 mm)
b.d² 0,2 (0,312)²
As Lap
330.72
-Mu tump
=22.9
28.608
= 11.762 Kn/m² ----->ρ = 0,0061
Vc-
As tump
380.64
28608
= x
b.d
1000x=
47,7 Kn
200 x 312
1000= 52 Kn
- =28.608
0.6=
- ≤ Vc + 2/3 Fc'b.d
1000
≤ 52 + 2/3 25200 x 12
100047.7
= 156 mm
Ukuran penampang memenuhi (ok)
- ≤ Vc ---> 47,7 Kn ≤ 52 Kn
47.7 ≤ 2008 Kn ------->
PERENCANAAN PONDASI TELAPAK
→ Pembebanan :
- Beban konstruksi atas (kuda-kuda) = Kg
- Beban Kolom (0,20 x 0,20 x 2400 x 7,00) = Kg
- Beban Sloof (0,20 x 0,40 x 2400 x 1,00) = Kg
- Beban Lantai Kerja (0,05 x 1,20 x 2400 x 1,20) = Kg
- Beban Pondasi (0,25 x 1,20 x 2400 x 1,20) = Kg
- Beban Tanah Urug (1,20 x 1,20 x 1,00) x (0,20 x 0,40 x 0,8) x = Kg
(1,20 x 1,20 x 0,25) x 1100
∑G = Kg
→ Momen yang terjadi :
M = (5550,4 - 352,94) x 7,00 = 36642,093 Kgm
2588
676.8
192
172.8
864
1152.8
5550.4
→ Tegangan :
∑G M
A W
= ±
σ1 = 0,398 Kg/cm²
σ2 = 0,372 Kg/cm² ≤ σtanah = 0,52 Kg/cm²
→ Perhitungan Plat Pondasi
Beban yang diterima plat pondasi :
- Beban kuda-kuda = Kg
- Beban kolom = Kg
- Beban sloof = Kg
ℓ = Kg
→ Beban terfaktor :
= Kg
→ Momen max
- Mu = ½ . q . L²
= ½ x 3734,2 x 0,6²
= Kg -------> Kn m
- d = 250 - 70 - ½ 16
=
- b = 1.00 mm
± = ±5550.4
120 x 120
36642.093
1,6 x 120³
0.385 0.0132
σ =
L
2588
676.8
192
3460.8
ℓ
b.L1.6=q
3734.2
672.2 6.58
172 mm
= 1.6 1.001,20 x 1,20
3460.8
1.4
fy
- As = ℓ min x b x d
= 0,0058 x 1000 x 172
= =====> 998 mm² (Ø 16 - 200 = 1005 mm²)
-Mu
=6.58
= Kg/m²b . d² 1,00 (0,172)²
=1.4
240=
222.400
0.0058- didapat ℓ = 0,0024 < ℓ = ………….. Ok
997.6
PERENCANAAN PONDASI BATU KALI
→ Pembebanan :
- Gaya normal akibat berat sendiri kolom praktis
(0,15 x 0,15) x 2400 x 4,00 = Kg
- B.S dinding ½ bata (0,15 x 250 x 7,00) = Kg
- B.S Sloof (0,20 x 0,40 x 2400 x 1,00) = Kg
- B.S Pondasi ½ (0,3 x 0,6) x 2200 x 1,00 = Kg
- B.S Tanah Urug (0,6 x 1,3 x 1,00) - (0,20 x 0,40 x 1,00) - = Kg
(½ x 0,3 x 0,6) 1,00 x 1100
∑G = Kg
→ Momen yang terjadi :
M = (5550,4 - 352,94) x 7,00 = 36642,093 Kgm
→ Tegangan :
∑G
A
= 0,485 Kg/cm² ≤ σtanah = 0,52 Kg/cm² …….. Ok
2331.5
σ =
216
262.5
192
990
671
=2331.5
60 x 80
DAFTAR ISI
1. PERHITUNGAN PEMBEBANAN PADA GORDING
2. DATA PERENCANAAN KUDA - KUDA
3. PERHITUNGAN GAYA-GAYA BATANG AKIBAT BEBAN MATI
4. PERHITUNGAN GAYA-GAYA BATANG AKIBAT BEBAN HIDUP
5. PERHITUNGAN GAYA-GAYA BATANG AKIBAT BEBAN AIR HUJAN
6. PERHITUNGAN GAYA-GAYA BATANG AKIBAT BEBAN ANGIN TEKAN
7. PERHITUNGAN GAYA-GAYA BATANG AKIBAT BEBAN ANGIN HISAP
8. KONTROL KEKUATAN BATANG
9. PERENCANAAN SAMBUNGAN
10. PERENCANAAN KANTILEVER
11. PERENCANAAN KOLOM DAN SLOOF
12. PERENCANAAN PONDASI TELAPAK
1,50
Tanah Urug
Tanah Asli
1,20
0,30
1/2 P P P P P P P
1/10 1/24
1/11 1/16
4,00 4,00 4,00 4,00 4,00 4,00
20 cm
40 cm
Kolom 30/30
- 0,05 Lantai - 0,15 Pasir
- 0,55 Sloof 20/40
- 1,40
- 1,70 Lantai Kerja - 1,65
- 0,85 Pasir Urug
P
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