Analisis Optimasi Profil Rangka Baja Dalam Perencanaan Bangunan Industri
TUGAS AKHIR BENTANG 20 m
20 m
DIMENSIONING
BAB IIIGORDING
cos
sin
20
20
=
=
Gaya angin
z1
z2
=
=
0,940
0,342
=
=
1,5 m
20 m
H =
h =
7m
5,250 m
65 kg/cm2
=
1,5
10,0
a
L
/ cos
/ cos
=
=
=
=
=
6m
240 Mpa
160 Mpa
Mutu Baut
Mutu Las
=
=
1,596 m
10,642 m
Asumsikan jarak gording =
1,400 m
==>
Maka banyak gording untuk bentang z 1 + z2
Asumsikan berat sendiri atap
=
Berat sendiri gording
S b .Y
=
×
q cos
1,596
4,00 =
q=
q total = ( 5,600
+
10,642 ) / 1,400 =
8,7
5,600 kg/m
q kg/m
+ q ) kg/m
9
buah
+
MX1
= q total × cos × L2 / 8
5,600 + q ) × cos
20 ×
= (
23,6803 + 4,2286 q ) kg m
= (
6,0 2 / 8
MY1
= q total × sin × L2 / 8
5,600 + q ) × sin
20 ×
= (
8,6189
= (
+ 1,5391 q ) kg m
6,0 2 / 8
S b .X
q s in
n=(
4,00 kg/m2
Dimensi Gording
A. Dimensi Gording Variasi Tanpa Ada Track Stang
1. Muatan Tetap
Berat sendiri atap
1,4
=
q
Jarak Portal ( )
Mutu Profil
Mutu Profil
Page 1
360 Mpa
360 Mpa
TUGAS AKHIR BENTANG 20 m
2. Muatan Angin
2a.
Angin Datang
qa =
65
kg/cm2
Koefisien =
0.02 × × 0.4
20
= 0,02 ×
qa datang
Koefisien
=
× Jarak Gording
×
0,000
=
× 1,4 × 65
0,000
kg/m
=
q d a ta n g
qa
0,4
= 0,000
MX2a
= qa datang × L2 / 8
= 0,000
× 6,0
kg m
= 0,0000
MY2a
=
S b .X
2 /8
0
S b .Y
2b.
Angin Pergi
qa =
65
kg/cm2
Koefisien =
-0,4
qa datang
Koefisien
=
× Jarak Gording
-0,400
=
× 1,4 × 65
-36,400
kg/m
=
q p e rg i
×
qa
MX2b
= qa datang × L2 / 8
= -36,400
× 6,0 2 / 8
kg m
= -163,8000
MY2b
=
S b .X
0
S b .Y
3. Muatan Tak Terduga
P
P = 100 kg
Sb.Y
Sb.X
P sin
P cos
MX3
= P × cos × / 4
cos
= 100
×
= 140,954 kg m
20
×
6,0
/4
MY3
= P × sin × / 4
sin
= 100
×
= 51,3030 kg m
20
×
6,0
/4
Kombinasi Momen
Akibat Muatan 1 + Muatan 2a (Angin Datang)
MX Total
23,6803
4,2286
=(
+
MY Total
8,6189
1,5391
=(
+
Akibat Muatan 1 + Muatan 2b (Angin Pergi)
MX Total
23,6803
4,2286
=(
+
MY Total
8,6189
1,5391
=(
+
q ) + ( 0,0000 )
q) +( 0 )
=(
=(
+
+
4,2286 q ) kg m
1,5391 q ) kg m
q ) + ( -163,800 )
q) +( 0 )
= ( -140,120 +
= ( 8,6189 +
4,2286 q ) kg m
1,5391 q ) kg m
=(
=(
164,634
59,9219
+
+
4,2286 q ) kg m
1,5391 q ) kg m
=(
=(
164,634
59,9219
+
+
4,2286 q ) kg m
1,5391 q ) kg m
Akibat Muatan 1 + Muatan 3 (Muatan Tak Terduga)
MX Total
23,6803
4,2286 q ) + ( 140,954 )
=(
+
MY Total
8,6189
1,5391 q ) + ( 51,3030 )
=(
+
Maka momen maksimum yang menentukan :
Dicoba profil Channel Lips
Data:
q
=
Wx
=
Wy
=
q total
=(
7,51 kg/m
44,27 cm3
12,26 cm3
5,600
+
MX Total
MY Total
C 150×65×20×3.2
7,510 ) =
Ix
Iy
=
=
F
=
13,110 kg/m
23,6803
8,6189
332 cm4
53,8 cm4
9,567 cm2
ix
iy
E
=
=
=
5,89 cm
2,37139 cm
2100000 kg/cm2
Page 2
TUGAS AKHIR BENTANG 20 m
Kontrol Tegangan
MX
MY
=
+
Terjadi
WX
WX
( 16463,4
+
=
profil
422,86
44,27
Kontrol Lendutan
5
qTotal × cos × L4
FX
=
384
E × Ix
5
0,131
cos
×
=
384
2100000
= 0,905 cm
FY
=
=
=
F
5
qTotal × cos × L4
384
E × Iy
5
0,131
cos
×
384
2100000
2,032 cm
=
Fizin
FX2 + FY2
=
F <
L
180
600
180
=
Fizin
0,905
=
..
=
×
7,5
)
+
( 5992,19
+
153,91 ×
7,5
)
12,26
1026,659 kg/cm2
1 P × cos × L3
48
E × Ix
20
× 600 4
+
×
332
1
48
100
20 ×
× cos
2100000 × 332
600 3
1 P × cos × L3
48
E × Iy
20
× 600 4
+
×
53,8
1
48
100
20 ×
× cos
2100000 × 53,8
600 3
1600 kg/cm2
1600 kg/cm2
..
+
+
+ 2,032 2
2
=
2,224 cm
3,333 cm
OK
B. Dimensi Gording Variasi Dengan Track Stang
Direncanakan gading-gading kap memiliki 3 medan ekonomis
G a d in g K a p
2,00 m
Ly
S b .Y
S b .X
2,00 m
Lx
T ra c k S ta n g
2,00 m
G a d in g K a p
1. Muatan Tetap
q total
= ( 5,600
+ q ) kg/m
MX1
= q total × cos × LX2 / 8
5,600
+ q) ×
cos
= (
23,6803
4,2286
= (
+
20 ×
q ) kg m
6,0 2
/8
MY1
= q total × sin × LY2 / 8
5,600
+ q) ×
sin
= (
0,9577
0,1710
= (
+
20 × (
q ) kg m
6,0 /
3
2. Muatan Angin
2a.
Angin Datang
=
MX2a
=
MY2a
2b.
Angin Pergi
=
MX2b
=
MY2b
0,0000 kg m
0 kg m
-163,8000 kg m
0 kg m
3. Muatan Tak Terduga
MX3
= P × cos × X / 4
cos
= 100
×
= 140,954 kg m
MY3
)2 / 8
= P × sin × Y / 4
sin
= 100
×
= 17,1010 kg m
20
×
6,0
20
×(
6,0 /
/4
3 )/4
Page 3
OK
TUGAS AKHIR BENTANG 20 m
Kombinasi Momen
Akibat Muatan 1 + Muatan 2a (Angin Datang)
MX Total
23,6803
4,2286
=(
+
MY Total
0,9577
0,1710
=(
+
Akibat Muatan 1 + Muatan 2b (Angin Pergi)
MX Total
23,6803
4,2286
=(
+
MY Total
0,9577
0,1710
=(
+
q ) + ( 0,0000 )
q) +( 0 )
=(
=(
+
+
4,2286 q ) kg m
0,1710 q ) kg m
q ) + ( -163,800 )
q) +( 0 )
= ( -140,120 +
= ( 0,9577 +
4,2286 q ) kg m
0,1710 q ) kg m
=(
=(
164,634
18,0587
+
+
4,2286 q ) kg m
0,1710 q ) kg m
=(
=(
164,634
18,0587
+
+
4,2286 q ) kg m
0,1710 q ) kg m
Akibat Muatan 1 + Muatan 3 (Muatan Tak Terduga)
MX Total
23,6803
4,2286 q ) + ( 140,954 )
=(
+
MY Total
0,9577
0,1710 q ) + ( 17,1010 )
=(
+
Maka momen maksimum yang menentukan :
MX Total
MY Total
Dicoba profil Channel Lips
C 125×50×20×2.3
Data:
q
4,51 kg/m
Ix
=
=
Wx
21,92 cm3
Iy
=
=
Wy
6,22 cm3
F
=
=
q total
5,600
4,510 ) =
10,110 kg/m
=(
+
Kontrol Tegangan
MX
MY
=
+
Terjadi
WX
WX
( 16463,4
+
=
=
=
=
F
422,86
21,92
FX2 + FY2
=
F <
×
4,5
1
48
20
×
×
137
)
( 1805,87
+
P × cos × LX3
E × Ix
600 4
1
+
48
+
5
1
qTotal × cos × LY4
P × cos × LY3
+
384
E × Iy
48
E × Iy
5
0,101
cos
20 × ( 600 / 3 )4
×
384
2100000
×
20,6
0,148 cm
=
Fizin
137 cm4
20,6 cm4
5,747 cm2
L
180
Fizin
2,027
=
600
180
=
..
ix
iy
E
4,88 cm
1,89 cm
2100000 kg/cm2
=
=
=
profil
Kontrol Lendutan
5
qTotal × cos × LX4
FX
=
384
E × Ix
5
0,101
cos
×
=
384
2100000
= 2,027 cm
FY
23,6803
0,9577
=
2
+ 0,148 2
=
+
+
17,10
4,5
×
)
6,22
1140,803 kg/cm2
100
20 ×
× cos
2100000 × 137
1
48
100
1600 kg/cm2
1600 kg/cm2
..
OK
600 3
20 × ( 600 /
× cos
2100000 × 20,6
3 )3
2,032 cm
3,333 cm
OK
Kesimpulan : Profil gording yang di pakai adalah Channel Lips
Y
t
X
h
Cx
d
Cy
b
Jumlah profil Channel Lips C yang diperlukan :
h
b
d
t
F
q
Wx
Wy
Ix
Iy
C 125×50×20×2.3
=
=
=
=
=
=
=
=
=
=
125
50
20
2,3
5,747
4,51
21,92
6,22
137
20,6
n =
mm
mm
mm
mm
cm2
kg/m
cm3
cm3
cm4
cm4
( Jumlah Lapangan Gording ) + 1
=
9
+ 1 =
10
buah
Page 4
8,74
TUGAS AKHIR BENTANG 20 m
BAB II
DIMENSI BATANG TARIK (TRACK STANG)
2
m
2
m
2
m
b
a
a = 1,400
m
b = 1,038
m
Berat sendiri atap
Jarak batang tarik
Berat sendiri gording (q)
Beban tak terduga
a
=
=
=
=
4,00
2
4,51
100
Beban yang bekerja
Atap
A. Tipe 1 =
Gording
Beban Tak Terduga
A. Tipe 2 =
Atap
Gording
Beban Tak Terduga
Beban yang dipikul oleh track stang :
NTotal
8
= N1
+
Dibutuhkan
:
Terjadi
d
d
6
a
a
kg/m2
m
kg/m
kg
= 2,00
= 2,00
= 100
×
×
×
1,04
4,51
sin
×
×
= 2,00
= 2,00
= 100
×
×
×
1,4
4,51
sin
×
×
=
1600
kg/cm2
1600
kg/cm2
NTotal
1600 /4
369,069
1600
×
0,542
cm
d
d =
a
20
20
4,00
sin
× sin
20
20
N1
4,00
sin
× sin
20
20
N2
=
=
=
=
2,840
3,085
34,202
40,127
kg
kg
kg
kg
=
=
=
=
3,831
3,085
34,202
41,118
kg
kg
kg
kg
+
+
369,069 kg
profil
NTotal
A
NTotal
d2/4
Ambil :
N2
a
/
4
mm
Page 5
TUGAS AKHIR BENTANG 25 m
BAB III
DIMENSIONING
GORDING
Dimensioning Gording
cos
sin
20
20
=
=
Gaya angin
z1
z2
=
=
0,940
0,342
=
=
1,5 m
25 m
H =
h =
7m
5,250 m
65 kg/cm2
=
1,5
12,5
a
L
/ cos
/ cos
=
=
=
=
=
6m
240 Mpa
160 Mpa
Mutu Baut
Mutu Las
=
=
1,596 m
13,302 m
Asumsikan jarak gording =
1,400 m
==>
Maka banyak gording untuk bentang z 1 + z2
Asumsikan berat sendiri atap
=
Berat sendiri gording
S b .Y
=
×
q cos
1,596
4,00 =
q=
q total = ( 5,600
+
13,302 ) / 1,400 = 10,6
5,600 kg/m
q kg/m
+ q ) kg/m
11
buah
+
MX1
= q total × cos × L2 / 8
5,600 + q ) × cos
20 ×
= (
23,6803 + 4,2286 q ) kg m
= (
6,0 2 / 8
MY1
= q total × sin × L2 / 8
5,600 + q ) × sin
20 ×
= (
8,6189
= (
+ 1,5391 q ) kg m
6,0 2 / 8
S b .X
q s in
n=(
4,00 kg/m2
Dimensi Gording
A. Dimensi Gording Variasi Tanpa Ada Track Stang
1. Muatan Tetap
Berat sendiri atap
1,4
=
q
Jarak Portal ( )
Mutu Profil
Mutu Profil
Page 1
360 Mpa
360 Mpa
TUGAS AKHIR BENTANG 25 m
2. Muatan Angin
2a.
Angin Datang
qa =
65
kg/cm2
Koefisien =
0.02 × × 0.4
20
= 0,02 ×
qa datang
Koefisien
=
× Jarak Gording
×
0,000
=
× 1,4 × 65
0,000
kg/m
=
q d a ta n g
qa
0,4
= 0,000
MX2a
= qa datang × L2 / 8
= 0,000
× 6,0
kg m
= 0,0000
MY2a
=
S b .X
2 /8
0
S b .Y
2b.
Angin Pergi
qa =
65
kg/cm2
Koefisien =
-0,4
qa datang
Koefisien
=
× Jarak Gording
-0,400
=
× 1,4 × 65
-36,400
kg/m
=
q p e rg i
×
qa
MX2b
= qa datang × L2 / 8
= -36,400
× 6,0 2 / 8
kg m
= -163,8000
MY2b
=
S b .X
0
S b .Y
3. Muatan Tak Terduga
P
P = 100 kg
Sb.Y
Sb.X
P sin
P cos
MX3
= P × cos × / 4
cos
= 100
×
= 140,954 kg m
20
×
6,0
/4
MY3
= P × sin × / 4
sin
= 100
×
= 51,3030 kg m
20
×
6,0
/4
Kombinasi Momen
Akibat Muatan 1 + Muatan 2a (Angin Datang)
MX Total
23,6803
4,2286
=(
+
MY Total
8,6189
1,5391
=(
+
Akibat Muatan 1 + Muatan 2b (Angin Pergi)
MX Total
23,6803
4,2286
=(
+
MY Total
8,6189
1,5391
=(
+
q ) + ( 0,0000 )
q) +( 0 )
=(
=(
+
+
4,2286 q ) kg m
1,5391 q ) kg m
q ) + ( -163,800 )
q) +( 0 )
= ( -140,120 +
= ( 8,6189 +
4,2286 q ) kg m
1,5391 q ) kg m
=(
=(
164,634
59,9219
+
+
4,2286 q ) kg m
1,5391 q ) kg m
=(
=(
164,634
59,9219
+
+
4,2286 q ) kg m
1,5391 q ) kg m
Akibat Muatan 1 + Muatan 3 (Muatan Tak Terduga)
MX Total
23,6803
4,2286 q ) + ( 140,954 )
=(
+
MY Total
8,6189
1,5391 q ) + ( 51,3030 )
=(
+
Maka momen maksimum yang menentukan :
Dicoba profil Channel Lips
Data:
q
=
Wx
=
Wy
=
q total
=(
7,51 kg/m
44,27 cm3
12,26 cm3
5,600
+
MX Total
MY Total
C 150×65×20×3.2
7,510 ) =
Ix
Iy
=
=
F
=
13,110 kg/m
23,6803
8,6189
332 cm4
53,8 cm4
9,567 cm2
ix
iy
E
=
=
=
5,89 cm
2,37139 cm
2100000 kg/cm2
Page 2
TUGAS AKHIR BENTANG 25 m
Kontrol Tegangan
MX
MY
=
+
Terjadi
WX
WX
( 16463,4
+
=
profil
422,86
44,27
Kontrol Lendutan
5
qTotal × cos × L4
FX
=
384
E × Ix
5
0,131
cos
×
=
384
2100000
= 0,905 cm
FY
=
=
=
F
5
qTotal × cos × L4
384
E × Iy
5
0,131
cos
×
384
2100000
2,032 cm
=
Fizin
FX2 + FY2
=
F <
L
180
600
180
=
Fizin
0,905
=
..
=
×
7,5
)
+
( 5992,19
+
153,91 ×
7,5
)
12,26
1026,659 kg/cm2
1 P × cos × L3
48
E × Ix
20
× 600 4
+
×
332
1
48
100
20 ×
× cos
2100000 × 332
600 3
1 P × cos × L3
48
E × Iy
20
× 600 4
+
×
53,8
1
48
100
20 ×
× cos
2100000 × 53,8
600 3
1600 kg/cm2
1600 kg/cm2
..
+
+
+ 2,032 2
2
=
2,224 cm
3,333 cm
OK
B. Dimensi Gording Variasi Dengan Track Stang
Direncanakan gading-gading kap memiliki 3 medan ekonomis
G a d in g K a p
2,00 m
Ly
S b .Y
S b .X
2,00 m
Lx
T ra c k S ta n g
2,00 m
G a d in g K a p
1. Muatan Tetap
q total
= ( 5,600
+ q ) kg/m
MX1
= q total × cos × LX2 / 8
5,600
+ q) ×
cos
= (
23,6803
4,2286
= (
+
20 ×
q ) kg m
6,0 2
/8
MY1
= q total × sin × LY2 / 8
5,600
+ q) ×
sin
= (
0,9577
0,1710
= (
+
20 × (
q ) kg m
6,0 /
3
2. Muatan Angin
2a.
Angin Datang
=
MX2a
=
MY2a
2b.
Angin Pergi
=
MX2b
=
MY2b
0,0000 kg m
0 kg m
-163,8000 kg m
0 kg m
3. Muatan Tak Terduga
MX3
= P × cos × X / 4
cos
= 100
×
= 140,954 kg m
MY3
)2 / 8
= P × sin × Y / 4
sin
= 100
×
= 17,1010 kg m
20
×
6,0
20
×(
6,0 /
/4
3 )/4
Page 3
OK
TUGAS AKHIR BENTANG 25 m
Kombinasi Momen
Akibat Muatan 1 + Muatan 2a (Angin Datang)
MX Total
23,6803
4,2286
=(
+
MY Total
0,9577
0,1710
=(
+
Akibat Muatan 1 + Muatan 2b (Angin Pergi)
MX Total
23,6803
4,2286
=(
+
MY Total
0,9577
0,1710
=(
+
q ) + ( 0,0000 )
q) +( 0 )
=(
=(
+
+
4,2286 q ) kg m
0,1710 q ) kg m
q ) + ( -163,800 )
q) +( 0 )
= ( -140,120 +
= ( 0,9577 +
4,2286 q ) kg m
0,1710 q ) kg m
=(
=(
164,634
18,0587
+
+
4,2286 q ) kg m
0,1710 q ) kg m
=(
=(
164,634
18,0587
+
+
4,2286 q ) kg m
0,1710 q ) kg m
Akibat Muatan 1 + Muatan 3 (Muatan Tak Terduga)
MX Total
23,6803
4,2286 q ) + ( 140,954 )
=(
+
MY Total
0,9577
0,1710 q ) + ( 17,1010 )
=(
+
Maka momen maksimum yang menentukan :
MX Total
MY Total
Dicoba profil Channel Lips
C 125×50×20×2.3
Data:
q
4,51 kg/m
Ix
=
=
Wx
21,92 cm3
Iy
=
=
Wy
6,22 cm3
F
=
=
q total
5,600
4,510 ) =
10,110 kg/m
=(
+
Kontrol Tegangan
MX
MY
=
+
Terjadi
WX
WX
( 16463,4
+
=
=
=
=
F
422,86
21,92
FX2 + FY2
=
F <
×
4,5
1
48
20
×
×
137
)
( 1805,87
+
P × cos × LX3
E × Ix
600 4
1
+
48
+
5
1
qTotal × cos × LY4
P × cos × LY3
+
384
E × Iy
48
E × Iy
5
0,101
cos
20 × ( 600 / 3 )4
×
384
2100000
×
20,6
0,148 cm
=
Fizin
137 cm4
20,6 cm4
5,747 cm2
L
180
Fizin
2,027
=
600
180
=
..
ix
iy
E
4,88 cm
1,89 cm
2100000 kg/cm2
=
=
=
profil
Kontrol Lendutan
5
qTotal × cos × LX4
FX
=
384
E × Ix
5
0,101
cos
×
=
384
2100000
= 2,027 cm
FY
23,6803
0,9577
=
2
+ 0,148 2
=
+
+
17,10
4,5
×
)
6,22
1140,803 kg/cm2
100
20 ×
× cos
2100000 × 137
1
48
100
1600 kg/cm2
1600 kg/cm2
..
OK
600 3
20 × ( 600 /
× cos
2100000 × 20,6
3 )3
2,032 cm
3,333 cm
OK
Kesimpulan : Profil gording yang di pakai adalah Channel Lips
Y
t
X
h
Cx
d
Cy
b
Jumlah profil Channel Lips C yang diperlukan :
h
b
d
t
F
q
Wx
Wy
Ix
Iy
C 125×50×20×2.3
=
=
=
=
=
=
=
=
=
=
125
50
20
2,3
5,747
4,51
21,92
6,22
137
20,6
n =
mm
mm
mm
mm
cm2
kg/m
cm3
cm3
cm4
cm4
( Jumlah Lapangan Gording ) + 1
=
11
+ 1 =
12
buah
Page 4
10,6
TUGAS AKHIR BENTANG 25 m
BAB II
DIMENSI BATANG TARIK (TRACK STANG)
2
m
2
m
2
m
b
a
a = 1,400
m
b = 0,898
m
Berat sendiri atap
Jarak batang tarik
Berat sendiri gording (q)
Beban tak terduga
a
=
=
=
=
4,00
2
4,51
100
Beban yang bekerja
Atap
A. Tipe 1 =
Gording
Beban Tak Terduga
A. Tipe 2 =
Atap
Gording
Beban Tak Terduga
Beban yang dipikul oleh track stang :
NTotal
10
= N1
+
Dibutuhkan
:
Terjadi
d
d
6
a
a
kg/m2
m
kg/m
kg
= 2,00
= 2,00
= 100
×
×
×
0,9
4,51
sin
×
×
= 2,00
= 2,00
= 100
×
×
×
1,4
4,51
sin
×
×
=
1600
kg/cm2
1600
kg/cm2
NTotal
1600 /4
450,922
1600
×
0,599
cm
d
d =
a
20
20
4,00
sin
× sin
20
20
N1
4,00
sin
× sin
20
20
N2
=
=
=
=
2,458
3,085
34,202
39,745
kg
kg
kg
kg
=
=
=
=
3,831
3,085
34,202
41,118
kg
kg
kg
kg
+
+
450,922 kg
profil
NTotal
A
NTotal
d2/4
Ambil :
N2
a
/
4
mm
Page 5
TUGAS AKHIR BENTANG 30 m
BAB III
DIMENSIONING
30 GORDING
20
20
20
G
1
2
=
=
=
1,5
15,0
L
=
=
1,5
30
7
5,250
H =
=
J P ()
M P
65
2
=
=
=
0,364
0,940
0,342
/
=%
/
=%
1,596
15,963
A
& %
6
240 M
160
M B
M L
=
=
5,459554
=
1,400
==>
M $ $ 1 + 2
A
$
=
B
=
S b .Y
q cos
4,00 =
=
= (
1,596
+
15,963 ) / 1,400 = 12,5 "
5,600
5,600
+ )
13
$
+
M
=
2 / 8
5,600
+ )
20
= (
23,6803
= (
+ 4,2286 )
6,0 2 / 8
M
=
2 / 8
5,600
+ )
20
= (
8,6189
= (
+ 1,5391 )
6,0 2 / 8
S b .X
q s in
%(
4,00
2
D
A. D
! #
1. M
B
1,4
=
q
=
=
1
P'() *
360 M
360 M
TUGAS AKHIR BENTANG 30 m
2. M12324 Q4:E4
22L
A4:E4 R2324:
2
65
+, =
6:,?,@A
KC7DEFE74
J2826 GC89E4:
=
0,000
1,4 - 65
=
0,000
6:,?,@A - B2 / 8
= 0,000
- 6,0
6: ;
= 0,0000
M./J
=
S b .X
2 /8
0
S b .Y
2KL
A4:E4 M78:E
2
+, =
65
6:,?,@A
=
-0,400
1,4
=
-36,400
6:,?,@A - B2 / 8
= -36,400
= -163,8000
=
-
6,0 2 / 8
6: ;
0
S b .Y
3. M12324 526 57891:2
P
P = 100 kg
Sb.Y
Sb.X
P sin
P cos
M=l
= P - HCF N - m < P
HCF
= 100
140,954 6: ;
=
20
-
6,0
/4
M.l
= P - FE4 N - m < P
FE4
= 100
51,3030 6: ;
=
20
-
6,0
/4
KC;KE42FE iC;74
W A6EK23 i12324 j k i12324 g2 (A4:E4 R2324:)
Ma ^_?,`
23,6803
4,2286
=(
+
M] ^_?,`
8,6189
1,5391
=(
+
W A6EK23 i12324 j k i12324 gK (A4:E4 M78:E)
Ma ^_?,`
23,6803
4,2286
=(
+
M] ^_?,`
8,6189
1,5391
=(
+
0,0000
+) +(
+) +( 0 )
)
=(
=(
+
+
4,2286 + ) 6: ;
1,5391 + ) 6: ;
-163,800
+) +(
+) +( 0 )
)
= ( -140,120 +
= ( 8,6189 +
4,2286 + ) 6: ;
1,5391 + ) 6: ;
)
)
=(
=(
164,634
59,9219
+
+
4,2286 + ) 6: ;
1,5391 + ) 6: ;
=(
=(
164,634
59,9219
+
+
4,2286 + ) 6: ;
1,5391 + ) 6: ;
W A6EK23 i12324 j k i12324 h (M12324 526 57891:2)
Ma ^_?,`
23,6803
4,2286 + ) + (
=(
+
M] ^_?,`
8,6189
1,5391 + ) + (
=(
+
140,954
51,3030
M262 ;C;74 ;26FE;1; Z24: ;747431624 \
DEHCK2 b8CDEX cd2447X BEbF
+
D232\
=
Y[
YZ
+ 3C32X
=
=
=(
7,51 6:
20 m
DIMENSIONING
BAB IIIGORDING
cos
sin
20
20
=
=
Gaya angin
z1
z2
=
=
0,940
0,342
=
=
1,5 m
20 m
H =
h =
7m
5,250 m
65 kg/cm2
=
1,5
10,0
a
L
/ cos
/ cos
=
=
=
=
=
6m
240 Mpa
160 Mpa
Mutu Baut
Mutu Las
=
=
1,596 m
10,642 m
Asumsikan jarak gording =
1,400 m
==>
Maka banyak gording untuk bentang z 1 + z2
Asumsikan berat sendiri atap
=
Berat sendiri gording
S b .Y
=
×
q cos
1,596
4,00 =
q=
q total = ( 5,600
+
10,642 ) / 1,400 =
8,7
5,600 kg/m
q kg/m
+ q ) kg/m
9
buah
+
MX1
= q total × cos × L2 / 8
5,600 + q ) × cos
20 ×
= (
23,6803 + 4,2286 q ) kg m
= (
6,0 2 / 8
MY1
= q total × sin × L2 / 8
5,600 + q ) × sin
20 ×
= (
8,6189
= (
+ 1,5391 q ) kg m
6,0 2 / 8
S b .X
q s in
n=(
4,00 kg/m2
Dimensi Gording
A. Dimensi Gording Variasi Tanpa Ada Track Stang
1. Muatan Tetap
Berat sendiri atap
1,4
=
q
Jarak Portal ( )
Mutu Profil
Mutu Profil
Page 1
360 Mpa
360 Mpa
TUGAS AKHIR BENTANG 20 m
2. Muatan Angin
2a.
Angin Datang
qa =
65
kg/cm2
Koefisien =
0.02 × × 0.4
20
= 0,02 ×
qa datang
Koefisien
=
× Jarak Gording
×
0,000
=
× 1,4 × 65
0,000
kg/m
=
q d a ta n g
qa
0,4
= 0,000
MX2a
= qa datang × L2 / 8
= 0,000
× 6,0
kg m
= 0,0000
MY2a
=
S b .X
2 /8
0
S b .Y
2b.
Angin Pergi
qa =
65
kg/cm2
Koefisien =
-0,4
qa datang
Koefisien
=
× Jarak Gording
-0,400
=
× 1,4 × 65
-36,400
kg/m
=
q p e rg i
×
qa
MX2b
= qa datang × L2 / 8
= -36,400
× 6,0 2 / 8
kg m
= -163,8000
MY2b
=
S b .X
0
S b .Y
3. Muatan Tak Terduga
P
P = 100 kg
Sb.Y
Sb.X
P sin
P cos
MX3
= P × cos × / 4
cos
= 100
×
= 140,954 kg m
20
×
6,0
/4
MY3
= P × sin × / 4
sin
= 100
×
= 51,3030 kg m
20
×
6,0
/4
Kombinasi Momen
Akibat Muatan 1 + Muatan 2a (Angin Datang)
MX Total
23,6803
4,2286
=(
+
MY Total
8,6189
1,5391
=(
+
Akibat Muatan 1 + Muatan 2b (Angin Pergi)
MX Total
23,6803
4,2286
=(
+
MY Total
8,6189
1,5391
=(
+
q ) + ( 0,0000 )
q) +( 0 )
=(
=(
+
+
4,2286 q ) kg m
1,5391 q ) kg m
q ) + ( -163,800 )
q) +( 0 )
= ( -140,120 +
= ( 8,6189 +
4,2286 q ) kg m
1,5391 q ) kg m
=(
=(
164,634
59,9219
+
+
4,2286 q ) kg m
1,5391 q ) kg m
=(
=(
164,634
59,9219
+
+
4,2286 q ) kg m
1,5391 q ) kg m
Akibat Muatan 1 + Muatan 3 (Muatan Tak Terduga)
MX Total
23,6803
4,2286 q ) + ( 140,954 )
=(
+
MY Total
8,6189
1,5391 q ) + ( 51,3030 )
=(
+
Maka momen maksimum yang menentukan :
Dicoba profil Channel Lips
Data:
q
=
Wx
=
Wy
=
q total
=(
7,51 kg/m
44,27 cm3
12,26 cm3
5,600
+
MX Total
MY Total
C 150×65×20×3.2
7,510 ) =
Ix
Iy
=
=
F
=
13,110 kg/m
23,6803
8,6189
332 cm4
53,8 cm4
9,567 cm2
ix
iy
E
=
=
=
5,89 cm
2,37139 cm
2100000 kg/cm2
Page 2
TUGAS AKHIR BENTANG 20 m
Kontrol Tegangan
MX
MY
=
+
Terjadi
WX
WX
( 16463,4
+
=
profil
422,86
44,27
Kontrol Lendutan
5
qTotal × cos × L4
FX
=
384
E × Ix
5
0,131
cos
×
=
384
2100000
= 0,905 cm
FY
=
=
=
F
5
qTotal × cos × L4
384
E × Iy
5
0,131
cos
×
384
2100000
2,032 cm
=
Fizin
FX2 + FY2
=
F <
L
180
600
180
=
Fizin
0,905
=
..
=
×
7,5
)
+
( 5992,19
+
153,91 ×
7,5
)
12,26
1026,659 kg/cm2
1 P × cos × L3
48
E × Ix
20
× 600 4
+
×
332
1
48
100
20 ×
× cos
2100000 × 332
600 3
1 P × cos × L3
48
E × Iy
20
× 600 4
+
×
53,8
1
48
100
20 ×
× cos
2100000 × 53,8
600 3
1600 kg/cm2
1600 kg/cm2
..
+
+
+ 2,032 2
2
=
2,224 cm
3,333 cm
OK
B. Dimensi Gording Variasi Dengan Track Stang
Direncanakan gading-gading kap memiliki 3 medan ekonomis
G a d in g K a p
2,00 m
Ly
S b .Y
S b .X
2,00 m
Lx
T ra c k S ta n g
2,00 m
G a d in g K a p
1. Muatan Tetap
q total
= ( 5,600
+ q ) kg/m
MX1
= q total × cos × LX2 / 8
5,600
+ q) ×
cos
= (
23,6803
4,2286
= (
+
20 ×
q ) kg m
6,0 2
/8
MY1
= q total × sin × LY2 / 8
5,600
+ q) ×
sin
= (
0,9577
0,1710
= (
+
20 × (
q ) kg m
6,0 /
3
2. Muatan Angin
2a.
Angin Datang
=
MX2a
=
MY2a
2b.
Angin Pergi
=
MX2b
=
MY2b
0,0000 kg m
0 kg m
-163,8000 kg m
0 kg m
3. Muatan Tak Terduga
MX3
= P × cos × X / 4
cos
= 100
×
= 140,954 kg m
MY3
)2 / 8
= P × sin × Y / 4
sin
= 100
×
= 17,1010 kg m
20
×
6,0
20
×(
6,0 /
/4
3 )/4
Page 3
OK
TUGAS AKHIR BENTANG 20 m
Kombinasi Momen
Akibat Muatan 1 + Muatan 2a (Angin Datang)
MX Total
23,6803
4,2286
=(
+
MY Total
0,9577
0,1710
=(
+
Akibat Muatan 1 + Muatan 2b (Angin Pergi)
MX Total
23,6803
4,2286
=(
+
MY Total
0,9577
0,1710
=(
+
q ) + ( 0,0000 )
q) +( 0 )
=(
=(
+
+
4,2286 q ) kg m
0,1710 q ) kg m
q ) + ( -163,800 )
q) +( 0 )
= ( -140,120 +
= ( 0,9577 +
4,2286 q ) kg m
0,1710 q ) kg m
=(
=(
164,634
18,0587
+
+
4,2286 q ) kg m
0,1710 q ) kg m
=(
=(
164,634
18,0587
+
+
4,2286 q ) kg m
0,1710 q ) kg m
Akibat Muatan 1 + Muatan 3 (Muatan Tak Terduga)
MX Total
23,6803
4,2286 q ) + ( 140,954 )
=(
+
MY Total
0,9577
0,1710 q ) + ( 17,1010 )
=(
+
Maka momen maksimum yang menentukan :
MX Total
MY Total
Dicoba profil Channel Lips
C 125×50×20×2.3
Data:
q
4,51 kg/m
Ix
=
=
Wx
21,92 cm3
Iy
=
=
Wy
6,22 cm3
F
=
=
q total
5,600
4,510 ) =
10,110 kg/m
=(
+
Kontrol Tegangan
MX
MY
=
+
Terjadi
WX
WX
( 16463,4
+
=
=
=
=
F
422,86
21,92
FX2 + FY2
=
F <
×
4,5
1
48
20
×
×
137
)
( 1805,87
+
P × cos × LX3
E × Ix
600 4
1
+
48
+
5
1
qTotal × cos × LY4
P × cos × LY3
+
384
E × Iy
48
E × Iy
5
0,101
cos
20 × ( 600 / 3 )4
×
384
2100000
×
20,6
0,148 cm
=
Fizin
137 cm4
20,6 cm4
5,747 cm2
L
180
Fizin
2,027
=
600
180
=
..
ix
iy
E
4,88 cm
1,89 cm
2100000 kg/cm2
=
=
=
profil
Kontrol Lendutan
5
qTotal × cos × LX4
FX
=
384
E × Ix
5
0,101
cos
×
=
384
2100000
= 2,027 cm
FY
23,6803
0,9577
=
2
+ 0,148 2
=
+
+
17,10
4,5
×
)
6,22
1140,803 kg/cm2
100
20 ×
× cos
2100000 × 137
1
48
100
1600 kg/cm2
1600 kg/cm2
..
OK
600 3
20 × ( 600 /
× cos
2100000 × 20,6
3 )3
2,032 cm
3,333 cm
OK
Kesimpulan : Profil gording yang di pakai adalah Channel Lips
Y
t
X
h
Cx
d
Cy
b
Jumlah profil Channel Lips C yang diperlukan :
h
b
d
t
F
q
Wx
Wy
Ix
Iy
C 125×50×20×2.3
=
=
=
=
=
=
=
=
=
=
125
50
20
2,3
5,747
4,51
21,92
6,22
137
20,6
n =
mm
mm
mm
mm
cm2
kg/m
cm3
cm3
cm4
cm4
( Jumlah Lapangan Gording ) + 1
=
9
+ 1 =
10
buah
Page 4
8,74
TUGAS AKHIR BENTANG 20 m
BAB II
DIMENSI BATANG TARIK (TRACK STANG)
2
m
2
m
2
m
b
a
a = 1,400
m
b = 1,038
m
Berat sendiri atap
Jarak batang tarik
Berat sendiri gording (q)
Beban tak terduga
a
=
=
=
=
4,00
2
4,51
100
Beban yang bekerja
Atap
A. Tipe 1 =
Gording
Beban Tak Terduga
A. Tipe 2 =
Atap
Gording
Beban Tak Terduga
Beban yang dipikul oleh track stang :
NTotal
8
= N1
+
Dibutuhkan
:
Terjadi
d
d
6
a
a
kg/m2
m
kg/m
kg
= 2,00
= 2,00
= 100
×
×
×
1,04
4,51
sin
×
×
= 2,00
= 2,00
= 100
×
×
×
1,4
4,51
sin
×
×
=
1600
kg/cm2
1600
kg/cm2
NTotal
1600 /4
369,069
1600
×
0,542
cm
d
d =
a
20
20
4,00
sin
× sin
20
20
N1
4,00
sin
× sin
20
20
N2
=
=
=
=
2,840
3,085
34,202
40,127
kg
kg
kg
kg
=
=
=
=
3,831
3,085
34,202
41,118
kg
kg
kg
kg
+
+
369,069 kg
profil
NTotal
A
NTotal
d2/4
Ambil :
N2
a
/
4
mm
Page 5
TUGAS AKHIR BENTANG 25 m
BAB III
DIMENSIONING
GORDING
Dimensioning Gording
cos
sin
20
20
=
=
Gaya angin
z1
z2
=
=
0,940
0,342
=
=
1,5 m
25 m
H =
h =
7m
5,250 m
65 kg/cm2
=
1,5
12,5
a
L
/ cos
/ cos
=
=
=
=
=
6m
240 Mpa
160 Mpa
Mutu Baut
Mutu Las
=
=
1,596 m
13,302 m
Asumsikan jarak gording =
1,400 m
==>
Maka banyak gording untuk bentang z 1 + z2
Asumsikan berat sendiri atap
=
Berat sendiri gording
S b .Y
=
×
q cos
1,596
4,00 =
q=
q total = ( 5,600
+
13,302 ) / 1,400 = 10,6
5,600 kg/m
q kg/m
+ q ) kg/m
11
buah
+
MX1
= q total × cos × L2 / 8
5,600 + q ) × cos
20 ×
= (
23,6803 + 4,2286 q ) kg m
= (
6,0 2 / 8
MY1
= q total × sin × L2 / 8
5,600 + q ) × sin
20 ×
= (
8,6189
= (
+ 1,5391 q ) kg m
6,0 2 / 8
S b .X
q s in
n=(
4,00 kg/m2
Dimensi Gording
A. Dimensi Gording Variasi Tanpa Ada Track Stang
1. Muatan Tetap
Berat sendiri atap
1,4
=
q
Jarak Portal ( )
Mutu Profil
Mutu Profil
Page 1
360 Mpa
360 Mpa
TUGAS AKHIR BENTANG 25 m
2. Muatan Angin
2a.
Angin Datang
qa =
65
kg/cm2
Koefisien =
0.02 × × 0.4
20
= 0,02 ×
qa datang
Koefisien
=
× Jarak Gording
×
0,000
=
× 1,4 × 65
0,000
kg/m
=
q d a ta n g
qa
0,4
= 0,000
MX2a
= qa datang × L2 / 8
= 0,000
× 6,0
kg m
= 0,0000
MY2a
=
S b .X
2 /8
0
S b .Y
2b.
Angin Pergi
qa =
65
kg/cm2
Koefisien =
-0,4
qa datang
Koefisien
=
× Jarak Gording
-0,400
=
× 1,4 × 65
-36,400
kg/m
=
q p e rg i
×
qa
MX2b
= qa datang × L2 / 8
= -36,400
× 6,0 2 / 8
kg m
= -163,8000
MY2b
=
S b .X
0
S b .Y
3. Muatan Tak Terduga
P
P = 100 kg
Sb.Y
Sb.X
P sin
P cos
MX3
= P × cos × / 4
cos
= 100
×
= 140,954 kg m
20
×
6,0
/4
MY3
= P × sin × / 4
sin
= 100
×
= 51,3030 kg m
20
×
6,0
/4
Kombinasi Momen
Akibat Muatan 1 + Muatan 2a (Angin Datang)
MX Total
23,6803
4,2286
=(
+
MY Total
8,6189
1,5391
=(
+
Akibat Muatan 1 + Muatan 2b (Angin Pergi)
MX Total
23,6803
4,2286
=(
+
MY Total
8,6189
1,5391
=(
+
q ) + ( 0,0000 )
q) +( 0 )
=(
=(
+
+
4,2286 q ) kg m
1,5391 q ) kg m
q ) + ( -163,800 )
q) +( 0 )
= ( -140,120 +
= ( 8,6189 +
4,2286 q ) kg m
1,5391 q ) kg m
=(
=(
164,634
59,9219
+
+
4,2286 q ) kg m
1,5391 q ) kg m
=(
=(
164,634
59,9219
+
+
4,2286 q ) kg m
1,5391 q ) kg m
Akibat Muatan 1 + Muatan 3 (Muatan Tak Terduga)
MX Total
23,6803
4,2286 q ) + ( 140,954 )
=(
+
MY Total
8,6189
1,5391 q ) + ( 51,3030 )
=(
+
Maka momen maksimum yang menentukan :
Dicoba profil Channel Lips
Data:
q
=
Wx
=
Wy
=
q total
=(
7,51 kg/m
44,27 cm3
12,26 cm3
5,600
+
MX Total
MY Total
C 150×65×20×3.2
7,510 ) =
Ix
Iy
=
=
F
=
13,110 kg/m
23,6803
8,6189
332 cm4
53,8 cm4
9,567 cm2
ix
iy
E
=
=
=
5,89 cm
2,37139 cm
2100000 kg/cm2
Page 2
TUGAS AKHIR BENTANG 25 m
Kontrol Tegangan
MX
MY
=
+
Terjadi
WX
WX
( 16463,4
+
=
profil
422,86
44,27
Kontrol Lendutan
5
qTotal × cos × L4
FX
=
384
E × Ix
5
0,131
cos
×
=
384
2100000
= 0,905 cm
FY
=
=
=
F
5
qTotal × cos × L4
384
E × Iy
5
0,131
cos
×
384
2100000
2,032 cm
=
Fizin
FX2 + FY2
=
F <
L
180
600
180
=
Fizin
0,905
=
..
=
×
7,5
)
+
( 5992,19
+
153,91 ×
7,5
)
12,26
1026,659 kg/cm2
1 P × cos × L3
48
E × Ix
20
× 600 4
+
×
332
1
48
100
20 ×
× cos
2100000 × 332
600 3
1 P × cos × L3
48
E × Iy
20
× 600 4
+
×
53,8
1
48
100
20 ×
× cos
2100000 × 53,8
600 3
1600 kg/cm2
1600 kg/cm2
..
+
+
+ 2,032 2
2
=
2,224 cm
3,333 cm
OK
B. Dimensi Gording Variasi Dengan Track Stang
Direncanakan gading-gading kap memiliki 3 medan ekonomis
G a d in g K a p
2,00 m
Ly
S b .Y
S b .X
2,00 m
Lx
T ra c k S ta n g
2,00 m
G a d in g K a p
1. Muatan Tetap
q total
= ( 5,600
+ q ) kg/m
MX1
= q total × cos × LX2 / 8
5,600
+ q) ×
cos
= (
23,6803
4,2286
= (
+
20 ×
q ) kg m
6,0 2
/8
MY1
= q total × sin × LY2 / 8
5,600
+ q) ×
sin
= (
0,9577
0,1710
= (
+
20 × (
q ) kg m
6,0 /
3
2. Muatan Angin
2a.
Angin Datang
=
MX2a
=
MY2a
2b.
Angin Pergi
=
MX2b
=
MY2b
0,0000 kg m
0 kg m
-163,8000 kg m
0 kg m
3. Muatan Tak Terduga
MX3
= P × cos × X / 4
cos
= 100
×
= 140,954 kg m
MY3
)2 / 8
= P × sin × Y / 4
sin
= 100
×
= 17,1010 kg m
20
×
6,0
20
×(
6,0 /
/4
3 )/4
Page 3
OK
TUGAS AKHIR BENTANG 25 m
Kombinasi Momen
Akibat Muatan 1 + Muatan 2a (Angin Datang)
MX Total
23,6803
4,2286
=(
+
MY Total
0,9577
0,1710
=(
+
Akibat Muatan 1 + Muatan 2b (Angin Pergi)
MX Total
23,6803
4,2286
=(
+
MY Total
0,9577
0,1710
=(
+
q ) + ( 0,0000 )
q) +( 0 )
=(
=(
+
+
4,2286 q ) kg m
0,1710 q ) kg m
q ) + ( -163,800 )
q) +( 0 )
= ( -140,120 +
= ( 0,9577 +
4,2286 q ) kg m
0,1710 q ) kg m
=(
=(
164,634
18,0587
+
+
4,2286 q ) kg m
0,1710 q ) kg m
=(
=(
164,634
18,0587
+
+
4,2286 q ) kg m
0,1710 q ) kg m
Akibat Muatan 1 + Muatan 3 (Muatan Tak Terduga)
MX Total
23,6803
4,2286 q ) + ( 140,954 )
=(
+
MY Total
0,9577
0,1710 q ) + ( 17,1010 )
=(
+
Maka momen maksimum yang menentukan :
MX Total
MY Total
Dicoba profil Channel Lips
C 125×50×20×2.3
Data:
q
4,51 kg/m
Ix
=
=
Wx
21,92 cm3
Iy
=
=
Wy
6,22 cm3
F
=
=
q total
5,600
4,510 ) =
10,110 kg/m
=(
+
Kontrol Tegangan
MX
MY
=
+
Terjadi
WX
WX
( 16463,4
+
=
=
=
=
F
422,86
21,92
FX2 + FY2
=
F <
×
4,5
1
48
20
×
×
137
)
( 1805,87
+
P × cos × LX3
E × Ix
600 4
1
+
48
+
5
1
qTotal × cos × LY4
P × cos × LY3
+
384
E × Iy
48
E × Iy
5
0,101
cos
20 × ( 600 / 3 )4
×
384
2100000
×
20,6
0,148 cm
=
Fizin
137 cm4
20,6 cm4
5,747 cm2
L
180
Fizin
2,027
=
600
180
=
..
ix
iy
E
4,88 cm
1,89 cm
2100000 kg/cm2
=
=
=
profil
Kontrol Lendutan
5
qTotal × cos × LX4
FX
=
384
E × Ix
5
0,101
cos
×
=
384
2100000
= 2,027 cm
FY
23,6803
0,9577
=
2
+ 0,148 2
=
+
+
17,10
4,5
×
)
6,22
1140,803 kg/cm2
100
20 ×
× cos
2100000 × 137
1
48
100
1600 kg/cm2
1600 kg/cm2
..
OK
600 3
20 × ( 600 /
× cos
2100000 × 20,6
3 )3
2,032 cm
3,333 cm
OK
Kesimpulan : Profil gording yang di pakai adalah Channel Lips
Y
t
X
h
Cx
d
Cy
b
Jumlah profil Channel Lips C yang diperlukan :
h
b
d
t
F
q
Wx
Wy
Ix
Iy
C 125×50×20×2.3
=
=
=
=
=
=
=
=
=
=
125
50
20
2,3
5,747
4,51
21,92
6,22
137
20,6
n =
mm
mm
mm
mm
cm2
kg/m
cm3
cm3
cm4
cm4
( Jumlah Lapangan Gording ) + 1
=
11
+ 1 =
12
buah
Page 4
10,6
TUGAS AKHIR BENTANG 25 m
BAB II
DIMENSI BATANG TARIK (TRACK STANG)
2
m
2
m
2
m
b
a
a = 1,400
m
b = 0,898
m
Berat sendiri atap
Jarak batang tarik
Berat sendiri gording (q)
Beban tak terduga
a
=
=
=
=
4,00
2
4,51
100
Beban yang bekerja
Atap
A. Tipe 1 =
Gording
Beban Tak Terduga
A. Tipe 2 =
Atap
Gording
Beban Tak Terduga
Beban yang dipikul oleh track stang :
NTotal
10
= N1
+
Dibutuhkan
:
Terjadi
d
d
6
a
a
kg/m2
m
kg/m
kg
= 2,00
= 2,00
= 100
×
×
×
0,9
4,51
sin
×
×
= 2,00
= 2,00
= 100
×
×
×
1,4
4,51
sin
×
×
=
1600
kg/cm2
1600
kg/cm2
NTotal
1600 /4
450,922
1600
×
0,599
cm
d
d =
a
20
20
4,00
sin
× sin
20
20
N1
4,00
sin
× sin
20
20
N2
=
=
=
=
2,458
3,085
34,202
39,745
kg
kg
kg
kg
=
=
=
=
3,831
3,085
34,202
41,118
kg
kg
kg
kg
+
+
450,922 kg
profil
NTotal
A
NTotal
d2/4
Ambil :
N2
a
/
4
mm
Page 5
TUGAS AKHIR BENTANG 30 m
BAB III
DIMENSIONING
30 GORDING
20
20
20
G
1
2
=
=
=
1,5
15,0
L
=
=
1,5
30
7
5,250
H =
=
J P ()
M P
65
2
=
=
=
0,364
0,940
0,342
/
=%
/
=%
1,596
15,963
A
& %
6
240 M
160
M B
M L
=
=
5,459554
=
1,400
==>
M $ $ 1 + 2
A
$
=
B
=
S b .Y
q cos
4,00 =
=
= (
1,596
+
15,963 ) / 1,400 = 12,5 "
5,600
5,600
+ )
13
$
+
M
=
2 / 8
5,600
+ )
20
= (
23,6803
= (
+ 4,2286 )
6,0 2 / 8
M
=
2 / 8
5,600
+ )
20
= (
8,6189
= (
+ 1,5391 )
6,0 2 / 8
S b .X
q s in
%(
4,00
2
D
A. D
! #
1. M
B
1,4
=
q
=
=
1
P'() *
360 M
360 M
TUGAS AKHIR BENTANG 30 m
2. M12324 Q4:E4
22L
A4:E4 R2324:
2
65
+, =
6:,?,@A
KC7DEFE74
J2826 GC89E4:
=
0,000
1,4 - 65
=
0,000
6:,?,@A - B2 / 8
= 0,000
- 6,0
6: ;
= 0,0000
M./J
=
S b .X
2 /8
0
S b .Y
2KL
A4:E4 M78:E
2
+, =
65
6:,?,@A
=
-0,400
1,4
=
-36,400
6:,?,@A - B2 / 8
= -36,400
= -163,8000
=
-
6,0 2 / 8
6: ;
0
S b .Y
3. M12324 526 57891:2
P
P = 100 kg
Sb.Y
Sb.X
P sin
P cos
M=l
= P - HCF N - m < P
HCF
= 100
140,954 6: ;
=
20
-
6,0
/4
M.l
= P - FE4 N - m < P
FE4
= 100
51,3030 6: ;
=
20
-
6,0
/4
KC;KE42FE iC;74
W A6EK23 i12324 j k i12324 g2 (A4:E4 R2324:)
Ma ^_?,`
23,6803
4,2286
=(
+
M] ^_?,`
8,6189
1,5391
=(
+
W A6EK23 i12324 j k i12324 gK (A4:E4 M78:E)
Ma ^_?,`
23,6803
4,2286
=(
+
M] ^_?,`
8,6189
1,5391
=(
+
0,0000
+) +(
+) +( 0 )
)
=(
=(
+
+
4,2286 + ) 6: ;
1,5391 + ) 6: ;
-163,800
+) +(
+) +( 0 )
)
= ( -140,120 +
= ( 8,6189 +
4,2286 + ) 6: ;
1,5391 + ) 6: ;
)
)
=(
=(
164,634
59,9219
+
+
4,2286 + ) 6: ;
1,5391 + ) 6: ;
=(
=(
164,634
59,9219
+
+
4,2286 + ) 6: ;
1,5391 + ) 6: ;
W A6EK23 i12324 j k i12324 h (M12324 526 57891:2)
Ma ^_?,`
23,6803
4,2286 + ) + (
=(
+
M] ^_?,`
8,6189
1,5391 + ) + (
=(
+
140,954
51,3030
M262 ;C;74 ;26FE;1; Z24: ;747431624 \
DEHCK2 b8CDEX cd2447X BEbF
+
D232\
=
Y[
YZ
+ 3C32X
=
=
=(
7,51 6: