4.2 Analisa Profil Pada Rangka Baja Tipe B 4.2.1 Bentang 20 m
Direncanakan Profil IWF 200 x 100 x 5,5 x 8 untuk kolom
Data –data profil : F = 27,16 cm
2
; q = 21,3 kgm h = 200 mm ; b = 100 mm
tw = 5,5 mm ; tf = 8 mm ix = 8,24 cm ; iy = 2,22 cm
Ix = 1840 cm
4
; Iy = 134 cm
4
Wx = 184 cm
3
; Wy = 26,8 cm
3
Panjang batang L = 9 m ; P maks = 3688,500 Tekuk terhadap sumbu x-x
24 ,
8 900
= =
x x
x
i lk
λ = 109,223
Kelangsingan batas
l g
x E
σ π
λ
7 ,
=
=
2400 7
, 10
1 ,
2
6
x x
π = 111,015
Faktor tekuk ω = 2,071 dari hasil interpolasi Tabel Baja
Perhhitungan gaya geser gan gaya normal D = 0,02 P
= 0,02 x 3688,500 = 73,77 kg
Universitas Sumatera Utara
kontrol gaya geser gan gaya desak P geser = 2.
baut d
. .
85 ,
4 1
σ π
⋅ ⋅
=
2400 .
85 ,
, 1
14 ,
3 4
1 ⋅
⋅
= 1092,72 kg P desak = d. t. 2. profil
σ = 1,0.0,50.2.1600
= 1600 kg Tegangan yang terjadi
σ =
A P
ω
.
= 16
, 27
071 ,
2 .
500 ,
3688
= 281,254 kgcm
2
Tegangan geser yang terjadi
tw Ix
Sx D
⋅ ⋅
= max
τ
= 55
, 1840
184 77
, 73
⋅ ⋅
= 13,412 kgcm
2
Tegangan yang diijinkan
2 2
3
τ σ
σ
+ =
n i
2 2
412 ,
13 3
254 ,
281 +
= = 282,211 kgcm
2
≤
σ = 1600 kgcm
2
………..Aman
Universitas Sumatera Utara
Direncanakan Profil 2L 50 x 50 x 5 untuk Balok Batang A1 sd A8
Data –data profil : F = 4,80cm
2
; q = 3,77 kgm b = 50 mm = 5 cm
t = 5 mm = 0,5 cm ix = iy = 1,51 cm
Ix = Iy = 11,0 cm
4
Wx = Wy = 3,05 cm
3
In = 4,59 cm
4
; e = 1,40 Panjang batang Lk = 1,33 m ; P maks = 6019,300 kg
Kontrol kelangsingan =
= =
51 ,
1 133
x x
x
i lk
λ 88,079
Kelangsingan batas
l g
x E
σ π
λ
7 ,
=
=
2400 7
, 10
1 ,
2
6
x x
π = 111,015
Faktor tekuk ω = 2,071 dari hasil interpolasi Tabel Baja
Perhhitungan gaya geser gan gaya normal a = e +
2 1
= 1,40 +
2 1
= 1,90 cm Iy. gab = 2 Iy + F.a
2
= 2 11,0 + 4,80. 1,90
2
= 56,656 cm
4
D = 0,02 P = 0,02 x 6019,300 = 120,386 kg Sy = F. a = 2 x 4,80 . 1,90 = 18,24 cm
3
Universitas Sumatera Utara
L = gab
Iy L
Sy D
. .
.
1
= 656
, 56
5 ,
66 .
24 ,
18 .
386 ,
120 = 2577,368 kg
N =
u a
L. .
2
= ,
3 90
, 1
. 368
, 2577
. 2
= 3264,667 kg kontrol gaya geser gan gaya desak
P geser = 2.
baut d
. .
85 ,
4 1
σ π
⋅ ⋅
=
2400 .
85 ,
, 1
14 ,
3 4
1 ⋅
⋅
= 1092,72 kg P desak = d. t. 2.
profil
σ = 1,0.0,50.2.1600 = 1600 kg
Tegangan yang terjadi σ =
A P
ω .
= 80
, 4
2 071
, 2
. 300
, 6019
⋅ = 1298,538 kgcm
2
Tegangan geser yang terjadi τ
2 3
=
.
F L
2 3
=
. 80
, 4
2 368
, 2577
⋅ = 402,713 kgcm
2
Tegangan yang diijinkan
2 2
3
τ σ
σ
+ =
n i
2 2
713 ,
402 3
538 ,
1298 +
= = 1474,019 kgcm
2
≤
σ = 1600 kgcm
2
………..Aman
Universitas Sumatera Utara
Direncanakan Profil 2L 50 x 50 x 5 untuk Balok Batang B1 sd B8
Data –data profil : F = 4,80cm
2
; q = 3,77 kgm b = 50 mm = 5 cm
t = 5 mm = 0,5 cm ix = iy = 1,51 cm
Ix = Iy = 11,0 cm
4
Wx = Wy = 3,05 cm
3
In = 4,59 cm
4
; e = 1,40 Panjang batang Lk = 1,33 m ; P maks = 5914,700
Kontrol kelangsingan =
= =
51 ,
1 133
x x
x
i lk
λ 88,079
Kelangsingan batas
l g
x E
σ π
λ
7 ,
=
=
2400 7
, 10
1 ,
2
6
x x
π = 111,015
Faktor tekuk ω = 2,071 dari hasil interpolasi Tabel Baja
Perhhitungan gaya geser gan gaya normal a = e +
2 1
= 1,40 +
2 1
= 1,90 cm Iy. gab = 2 Iy + F.a
2
= 2 11,0 + 4,80. 1,90
2
= 56,656 cm
4
D = 0,02 P = 0,02 x 5914,700 = 118,294 kg Sy = F. a = 2 x 4,80 . 1,90 = 18,24 cm
Universitas Sumatera Utara
L = gab
Iy L
Sy D
. .
.
1
= 656
, 56
5 ,
66 .
24 ,
18 .
294 ,
118 = 2546,068 kg
N =
u a
L. .
2
= ,
3 90
, 1
. 068
, 2546
. 2
= 3255,019 kg kontrol gaya geser gan gaya desak
P geser = 2.
baut d
. .
85 ,
4 1
σ π
⋅ ⋅
=
2400 .
85 ,
5 ,
1 14
, 3
4 1
⋅ ⋅
= 2402,100 kg P desak = d. t. 2.
profil
σ = 1,5.0,50.2.1600 = 2400 kg
Tegangan yang terjadi σ =
A P
ω .
= 80
, 4
2 071
, 2
. 700
, 5914
⋅ = 1275,973 kgcm
2
Tegangan geser yang terjadi τ
2 3
=
.
F L
2 3
=
. 80
, 4
2 068
, 2546
⋅ = 397,823 kgcm
2
Tegangan yang diijinkan
2 2
3
τ σ
σ
+ =
n i
2 2
823 ,
397 3
973 ,
1275 +
= = 1450,136 kgcm
2
≤
σ = 1600 kgcm
2
………..Aman
Universitas Sumatera Utara
Direncanakan Profil 2L 50 x 50 x 5 untuk Balok Batang D1 sd D8
Data –data profil : F = 4,80cm
2
; q = 3,77 kgm b = 50 mm = 5 cm
t = 5 mm = 0,5 cm ix = iy = 1,51 cm
Ix = Iy = 11,0 cm
4
Wx = Wy = 3,05 cm
3
In = 4,59 cm
4
; e = 1,40 Panjang batang Lk = 1,36 m ; P maks = 5525,547 kg
Kontrol kelangsingan =
= =
51 ,
1 136
x x
x
i lk
λ 90,066
Kelangsingan batas
l g
x E
σ π
λ
7 ,
=
=
2400 7
, 10
1 ,
2
6
x x
π = 111,015
Faktor tekuk ω = 2,071 dari hasil interpolasi Tabel Baja
Perhhitungan gaya geser gan gaya normal a = e +
2 1
= 1,40 +
2 1
= 1,90 cm Iy. gab = 2 Iy + F.a
2
= 2 11,0 + 4,80. 1,90
2
= 56,656 cm
4
D = 0,02 P = 0,02 x 5525,547 = 110,510 kg Sy = F. a = 2 x 4,80 . 1,90 = 18,24 cm
Universitas Sumatera Utara
L = gab
Iy L
Sy D
. .
.
1
= 656
, 56
68 .
24 ,
18 .
510 ,
110 = 2419,298 kg
N =
u a
L. .
2
= ,
3 90
, 1
. 298
, 2419
. 2
= 3064,445 kg kontrol gaya geser gan gaya desak
P geser = 2.
baut d
. .
85 ,
4 1
σ π
⋅ ⋅
=
2400 .
85 ,
5 ,
1 14
, 3
4 1
⋅ ⋅
= 2402,100 kg P desak = d. t. 2.
profil
σ = 1,5.0,50.2.1600 = 2400 kg
Tegangan yang terjadi σ =
A P
ω .
= 80
, 4
2 071
, 2
. 547
, 5525
⋅ = 1192,021 kgcm
2
Tegangan geser yang terjadi τ
2 3
=
.
F L
2 3
=
. 80
, 4
2 298
, 2419
⋅ = 378,015 kgcm
2
Tegangan yang diijinkan
2 2
3
τ σ
σ
+ =
n i
2 2
015 ,
378 3
021 ,
1192 +
= = 1360 kgcm
2
≤
σ = 1600 kgcm
2
………..Aman
Universitas Sumatera Utara
Direncanakan Profil 2L 50 x 50 x 5 untuk Balok Batang V1 sd V8
Data –data profil : F = 4,80cm
2
; q = 3,77 kgm b = 50 mm = 5 cm
t = 5 mm = 0,5 cm ix = iy = 1,51 cm
Ix = Iy = 11,0 cm
4
Wx = Wy = 3,05 cm
3
In = 4,59 cm
4
; e = 1,40 Panjang batang Lk = 1,00 m ; P maks = 3688,488 kg
Kontrol kelangsingan =
= =
51 ,
1 100
x x
x
i lk
λ 66,225
Kelangsingan batas
l g
x E
σ π
λ
7 ,
=
=
2400 7
, 10
1 ,
2
6
x x
π = 111,015
Faktor tekuk ω = 2,071 dari hasil interpolasi Tabel Baja
Perhhitungan gaya geser gan gaya normal a = e +
2 1
= 1,40 +
2 1
= 1,90 cm Iy. gab = 2 Iy + F.a
2
= 2 11,0 + 4,80. 1,90
2
= 56,656 cm
4
D = 0,02 P = 0,02 x 3688,488= 73,769 kg Sy = F. a = 2 x 4,80 . 1,90 = 18,24 cm
Universitas Sumatera Utara
L = gab
Iy L
Sy D
. .
.
1
= 656
, 56
50 .
24 ,
18 .
769 ,
73 = 1187,470 kg
N =
u a
L. .
2
= ,
3 90
, 1
. 470
, 1187
. 2
= 1504,128 kg kontrol gaya geser gan gaya desak
P geser = 2.
baut d
. .
85 ,
4 1
σ π
⋅ ⋅
=
2400 .
85 ,
5 ,
1 14
, 3
4 1
⋅ ⋅
= 2402,100 kg P desak = d. t. 2.
profil
σ = 1,5.0,50.2.1600 = 2400 kg
Tegangan yang terjadi σ =
A P
ω .
= 80
, 4
2 071
, 2
. 488
, 3688
⋅ = 795,714 kgcm
2
Tegangan geser yang terjadi τ
2 3
=
.
F L
2 3
=
. 80
, 4
2 470
, 1187
⋅ = 185,542 kgcm
2
Tegangan yang diijinkan
2 2
3
τ σ
σ
+ =
n i
2 2
542 ,
185 3
714 ,
795 +
= = 858,159 kgcm
2
≤
σ = 1600 kgcm
2
………..Aman
Universitas Sumatera Utara
4.2.2 Bentang 25 m
Direncanakan Profil IWF 250 x 125 x 6 x 9 untuk kolom
Data –data profil : F = 37,66 cm
2
; q = 29,6 kgm h = 250 mm ; b = 125 mm
tw = 6 mm ; tf = 9 mm ix = 10,4 cm ; iy = 2,79 cm
Ix = 4050 cm
4
; Iy = 294 cm
4
Wx = 324 cm
3
; Wy = 47,0 cm
3
Panjang batang L = 9 m ; P maks = 6620,700 kg Tekuk terhadap sumbu x-x
4 ,
10 900
= =
x x
x
i lk
λ = 86,538
Kelangsingan batas
l g
x E
σ π
λ
7 ,
=
=
2400 7
, 10
1 ,
2
6
x x
π = 111,015
Faktor tekuk ω = 2,071 dari hasil interpolasi Tabel Baja
Perhhitungan gaya geser gan gaya normal D = 0,02 P
= 0,02 x 6620,700 = 132,414 kg
Universitas Sumatera Utara
kontrol gaya geser gan gaya desak P geser = 2.
baut d
. .
85 ,
4 1
σ π
⋅ ⋅
=
2400 .
85 ,
, 1
14 ,
3 4
1 ⋅
⋅
= 1092,72 kg P desak = d. t. 2. profil
σ = 1,0.0,50.2.1600
= 1600 kg Tegangan yang terjadi
σ =
A P
ω
.
= 66
, 37
071 ,
2 .
700 ,
6620
= 364,085 kgcm
2
Tegangan geser yang terjadi
tw Ix
Sx D
⋅ ⋅
= max
τ
= 60
, 4050
324 414
, 132
⋅ ⋅
= 17,652 kgcm
2
Tegangan yang diijinkan
2 2
3
τ σ
σ
+ =
n i
2 2
652 ,
17 3
085 ,
364 +
= = 364,157 kgcm
2
≤
σ = 1600 kgcm
2
………..Aman
Universitas Sumatera Utara
Direncanakan Profil 2L 60 x 60 x 6 untuk Balok Batang A1 sd A8
Data –data profil : F = 6,91cm
2
; q = 5,42 kgm b = 60 mm = 6 cm
t = 6 mm = 0,6 cm ix = iy = 1,82 cm
Ix = Iy = 22,8 cm
4
Wx = Wy = 5,29 cm
3
In = 9,43 cm
4
; e = 1,69 Panjang batang Lk = 1,66 m ; P maks = 7843,790 kg
Kontrol kelangsingan =
= =
82 ,
1 166
x x
x
i lk
λ 91,208
Kelangsingan batas
l g
x E
σ π
λ
7 ,
=
=
2400 7
, 10
1 ,
2
6
x x
π = 111,015
Faktor tekuk ω = 2,071 dari hasil interpolasi Tabel Baja
Perhhitungan gaya geser gan gaya normal a = e +
2 1
= 1,69 +
2 1
= 2,19 cm Iy. gab = 2 Iy + F.a
2
= 2 22,8 + 6,91. 2,19
2
= 111,882 cm
4
D = 0,02 P = 0,02 x 7843,790 = 156,875 kg Sy = F. a = 6,91 x 2,19 = 15,132 cm
Universitas Sumatera Utara
L = gab
Iy L
Sy D
. .
.
1
= 882
, 111
83 .
132 ,
15 .
875 ,
156 = 1761,034 kg
N =
u a
L. .
2
= ,
3 19
, 2
. 034
, 1761
. 2
= 2571,109 kg kontrol gaya geser gan gaya desak
P geser = 2.
baut d
. .
85 ,
4 1
σ π
⋅ ⋅
=
2400 .
85 ,
, 1
14 ,
3 4
1 ⋅
⋅
= 1092,72 kg P desak = d. t. 2.
profil
σ = 1,0.0,50.2.1600 = 1600 kg
Tegangan yang terjadi σ =
A P
ω .
= 91
, 6
2 071
, 2
. 790
, 7843
⋅ = 1175,433 kgcm
2
Tegangan geser yang terjadi τ
2 3
=
.
F L
2 3
=
. 91
, 6
2 034
, 1761
⋅ = 191,139 kgcm
2
Tegangan yang diijinkan
2 2
3
τ σ
σ
+ =
n i
2 2
139 ,
191 3
433 ,
1175 +
= = 1221,165 kgcm
2
≤
σ = 1600 kgcm
2
………..Aman
Universitas Sumatera Utara
Direncanakan Profil 2L 60 x 60 x 6 untuk Balok Batang B1 sd B8
Data –data profil : F = 6,91cm
2
; q = 5,42 kgm b = 60 mm = 6 cm
t = 6 mm = 0,6 cm ix = iy = 1,82 cm
Ix = Iy = 22,8 cm
4
Wx = Wy = 5,29 cm
3
In = 9,43 cm
4
; e = 1,69 Panjang batang Lk = 1,66 m ; P maks = 7839,484 kg
Kontrol kelangsingan =
= =
82 ,
1 166
x x
x
i lk
λ 91,208
Kelangsingan batas
l g
x E
σ π
λ
7 ,
=
=
2400 7
, 10
1 ,
2
6
x x
π = 111,015
Faktor tekuk ω = 2,071 dari hasil interpolasi Tabel Baja
Perhhitungan gaya geser gan gaya normal a = e +
2 1
= 1,69 +
2 1
= 2,19 cm Iy. gab = 2 Iy + F.a
2
= 2 22,8 + 6,91. 2,19
2
= 111,882 cm
4
D = 0,02 P = 0,02 x 7839,484 = 156,789 kg Sy = F. a = 6,91 x 2,19 = 15,132 cm
Universitas Sumatera Utara
L = gab
Iy L
Sy D
. .
.
1
= 882
, 111
83 .
132 ,
15 .
789 ,
156 = 1760,069 kg
N =
u a
L. .
2
= ,
3 19
, 2
. 069
, 1760
. 2
= 2569,700 kg kontrol gaya geser gan gaya desak
P geser = 2.
baut d
. .
85 ,
4 1
σ π
⋅ ⋅
=
2400 .
85 ,
, 1
14 ,
3 4
1 ⋅
⋅
= 1092,72 kg P desak = d. t. 2.
profil
σ = 1,0.0,50.2.1600 = 1600 kg
Tegangan yang terjadi σ =
A P
ω .
= 91
, 6
2 071
, 2
. 484
, 7839
⋅ = 1174,788 kgcm
2
Tegangan geser yang terjadi τ
2 3
=
.
F L
2 3
=
. 91
, 6
2 069
, 1760
⋅ = 191,034 kgcm
2
Tegangan yang diijinkan
2 2
3
τ σ
σ
+ =
n i
2 2
034 ,
191 3
788 ,
1174 +
= = 1175,031 kgcm
2
≤
σ = 1600 kgcm
2
………..Aman
Universitas Sumatera Utara
Direncanakan Profil 2L 60 x 60 x 6 untuk Balok Batang D1 sd D8
Data –data profil : F = 6,91cm
2
; q = 5,42 kgm b = 60 mm = 6 cm
t = 6 mm = 0,6 cm ix = iy = 1,82 cm
Ix = Iy = 22,8 cm
4
Wx = Wy = 5,29 cm
3
In = 9,43 cm
4
; e = 1,69 Panjang batang Lk = 1,62 m ; P maks = 9230,398 kg
Kontrol kelangsingan =
= =
82 ,
1 162
x x
x
i lk
λ 89,010
Kelangsingan batas
l g
x E
σ π
λ
7 ,
=
=
2400 7
, 10
1 ,
2
6
x x
π = 111,015
Faktor tekuk ω = 2,071 dari hasil interpolasi Tabel Baja
Perhhitungan gaya geser gan gaya normal a = e +
2 1
= 1,69 +
2 1
= 2,19 cm Iy. gab = 2 Iy + F.a
2
= 2 22,8 + 6,91. 2,19
2
= 111,882 cm
4
D = 0,02 P = 0,02 x 9230,398 = 184,607 kg Sy = F. a = 6,91 x 2,19 = 15,132 cm
Universitas Sumatera Utara
L = gab
Iy L
Sy D
. .
.
1
= 882
, 111
81 .
132 ,
15 .
607 ,
184 = 2022,410 kg
N =
u a
L. .
2
= ,
3 19
, 2
. 410
, 2022
. 2
= 2952,718 kg kontrol gaya geser gan gaya desak
P geser = 2.
baut d
. .
85 ,
4 1
σ π
⋅ ⋅
=
2400 .
85 ,
, 1
14 ,
3 4
1 ⋅
⋅
= 1092,72 kg P desak = d. t. 2.
profil
σ = 1,0.0,50.2.1600 = 1600 kg
Tegangan yang terjadi σ =
A P
ω .
= 91
, 6
2 071
, 2
. 398
, 9230
⋅ = 1383,223 kgcm
2
Tegangan geser yang terjadi τ
2 3
=
.
F L
2 3
=
. 91
, 6
2 410
, 2022
⋅ = 219,509 kgcm
2
Tegangan yang diijinkan
2 2
3
τ σ
σ
+ =
n i
2 2
509 ,
219 3
223 ,
1383 +
= = 1434,523 kgcm
2
≤
σ = 1600 kgcm
2
………..Aman
Universitas Sumatera Utara
Direncanakan Profil 2L 60 x 60 x 6 untuk Balok Batang V1 sd V8
Data –data profil : F = 6,91cm
2
; q = 5,42 kgm b = 60 mm = 6 cm
t = 6 mm = 0,6 cm ix = iy = 1,82 cm
Ix = Iy = 22,8 cm
4
Wx = Wy = 5,29 cm
3
In = 9,43 cm
4
; e = 1,69 Panjang batang Lk = 1,00 m ; P maks = 6620,605 kg
Kontrol kelangsingan =
= =
82 ,
1 100
x x
x
i lk
λ 54,945
Kelangsingan batas
l g
x E
σ π
λ
7 ,
=
=
2400 7
, 10
1 ,
2
6
x x
π = 111,015
Faktor tekuk ω = 2,071 dari hasil interpolasi Tabel Baja
Perhhitungan gaya geser gan gaya normal a = e +
2 1
= 1,69 +
2 1
= 2,19 cm Iy. gab = 2 Iy + F.a
2
= 2 22,8 + 6,91. 2,19
2
= 111,882 cm
4
D = 0,02 P = 0,02 x 6620,605= 132,412 kg Sy = F. a = 6,91 x 2,19 = 15,132 cm
Universitas Sumatera Utara
L = gab
Iy L
Sy D
. .
.
1
= 882
, 111
50 .
132 ,
15 .
412 ,
132 = 895,434 kg
N =
u a
L. .
2
= ,
3 19
, 2
. 434
, 895
. 2
= 1307,334 kg kontrol gaya geser gan gaya desak
P geser = 2.
baut d
. .
85 ,
4 1
σ π
⋅ ⋅
=
2400 .
85 ,
, 1
14 ,
3 4
1 ⋅
⋅
= 1092,72 kg P desak = d. t. 2.
profil
σ = 1,0.0,50.2.1600 = 1600 kg
Tegangan yang terjadi σ =
A P
ω .
= 91
, 6
2 071
, 2
. 605
, 6620
⋅ = 992,132 kgcm
2
Tegangan geser yang terjadi τ
2 3
=
.
F L
2 3
=
. 91
, 6
2 434
, 895
⋅ = 97,188 kgcm
2
Tegangan yang diijinkan
2 2
3
τ σ
σ
+ =
n i
2 2
188 ,
97 3
132 ,
992 +
= = 1006,312 kgcm
2
≤
σ = 1600 kgcm
2
………..Aman
Universitas Sumatera Utara
4.2.3 Bentang 30 m
Direncanakan Profil IWF 300 x 150x 6,5 x 9 untuk kolom
Data –data profil : A = 46,78 cm
2
; q = 36,7 kgm h = 300 mm ; b = 150 mm
tw = 6,5 mm ; tf = 9 mm ix = 12,4 cm ; iy = 3,29 cm
Ix = 7210 cm
4
; Iy = 508 cm
4
Wx = 481 cm
3
; Wy = 67,7 cm
3
Panjang batang L = 9 m ; P maks = 5608,702 kg Tekuk terhadap sumbu x-x
4 ,
12 900
= =
x x
x
i lk
λ = 72,580
Kelangsingan batas
l g
x E
σ π
λ
7 ,
=
=
2400 7
, 10
1 ,
2
6
x x
π = 111,015
Faktor tekuk ω = 2,071 dari hasil interpolasi Tabel Baja
Perhhitungan gaya geser gan gaya normal D = 0,02 P
= 0,02 x 5608,702 = 112,174 kg
Universitas Sumatera Utara
kontrol gaya geser gan gaya desak P geser = 2.
baut d
. .
85 ,
4 1
σ π
⋅ ⋅
=
2400 .
85 ,
, 1
14 ,
3 4
1 ⋅
⋅
= 1092,72 kg P desak = d. t. 2. profil
σ = 1,0.0,50.2.1600
= 1600 kg Tegangan yang terjadi
σ =
A P
ω
.
= 78
, 46
071 ,
2 .
702 ,
5608
= 248,303 kgcm
2
Tegangan geser yang terjadi
tw Ix
Sx D
⋅ ⋅
= max
τ
= 65
, 7210
481 174
, 112
⋅ ⋅
= 11,511 kgcm
2
Tegangan yang diijinkan
2 2
3
τ σ
σ
+ =
n i
2 2
511 ,
11 3
303 ,
248 +
= = 249,102 kgcm
2
≤
σ = 1600 kgcm
2
………..Aman
Universitas Sumatera Utara
Direncanakan Profil 2L 70 x 70 x 7 untuk Balok Batang A1 sd A8
Data –data profil : F = 9,40 cm
2
; q = 7,38 kgm b = 70 mm = 7 cm
t = 7 mm = 0,7 cm ix = iy = 2,12 cm
Ix = Iy = 42,4 cm
4
Wx = Wy = 8,43 cm
3
In = 17,6 cm
4
; e = 1,97 Panjang batang Lk = 2,00 m ; P maks = 8737,356 kg
Kontrol kelangsingan =
= =
12 ,
2 200
x x
x
i lk
λ 94,339
Kelangsingan batas
l g
x E
σ π
λ
7 ,
=
=
2400 7
, 10
1 ,
2
6
x x
π = 111,015
Faktor tekuk ω = 2,071 dari hasil interpolasi Tabel Baja
Perhhitungan gaya geser gan gaya normal a = e +
2 1
= 1,97 +
2 1
= 2,47 cm Iy. gab = 2 Iy + F.a
2
= 2 42,2 + 9,40. 2,47
2
= 101,548 cm
4
D = 0,02 P = 0,02 x 8737,356 = 174,747 kg Sy = F. a = 9,40 x 2,47 = 23,218 cm
Universitas Sumatera Utara
L = gab
Iy L
Sy D
. .
.
1
= 548
, 101
100 .
218 ,
23 .
747 ,
174 = 3995,426 kg
N =
u a
L. .
2
= ,
3 47
, 2
. 426
, 3995
. 2
= 6579,134 kg kontrol gaya geser gan gaya desak
P geser = 2.
baut d
. .
85 ,
4 1
σ π
⋅ ⋅
=
2400 .
85 ,
, 1
14 ,
3 4
1 ⋅
⋅
= 1092,72 kg P desak = d. t. 2.
profil
σ = 1,0.0,50.2.1600 = 1600 kg
Tegangan yang terjadi σ =
A P
ω .
= 40
, 9
2 071
, 2
. 356
, 8737
⋅ = 962,503 kgcm
2
Tegangan geser yang terjadi τ
2 3
=
.
F L
2 3
=
. 40
, 9
2 426
, 3995
⋅ = 318,783 kgcm
2
Tegangan yang diijinkan
2 2
3
τ σ
σ
+ =
n i
2 2
783 ,
318 3
503 ,
962 +
= = 1109,630 kgcm
2
≤
σ = 1600 kgcm
2
………..Aman
Universitas Sumatera Utara
Direncanakan Profil 2L 70 x 70 x 7 untuk Balok Batang B1 sd B8
Data –data profil : F = 9,40 cm
2
; q = 7,38 kgm b = 70 mm = 7 cm
t = 7 mm = 0,7 cm ix = iy = 2,12 cm
Ix = Iy = 42,4 cm
4
Wx = Wy = 8,43 cm
3
In = 17,6 cm
4
; e = 1,97 Panjang batang Lk = 2,00 m ; P maks = 8744,528 kg
Kontrol kelangsingan =
= =
12 ,
2 200
x x
x
i lk
λ 94,339
Kelangsingan batas
l g
x E
σ π
λ
7 ,
=
=
2400 7
, 10
1 ,
2
6
x x
π = 111,015
Faktor tekuk ω = 2,071 dari hasil interpolasi Tabel Baja
Perhhitungan gaya geser gan gaya normal a = e +
2 1
= 1,97 +
2 1
= 2,47 cm Iy. gab = 2 Iy + F.a
2
= 2 42,2 + 9,40. 2,47
2
= 101,548 cm
4
D = 0,02 P = 0,02 x 8744,528= 174,890 kg Sy = F. a = 9,40 x 2,47 = 23,218 cm
Universitas Sumatera Utara
L = gab
Iy L
Sy D
. .
.
1
= 548
, 101
100 .
218 ,
23 .
890 ,
174 = 3998,696 kg
N =
u a
L. .
2
= ,
3 47
, 2
. 696
, 3998
. 2
= 6584,519 kg kontrol gaya geser gan gaya desak
P geser = 2.
baut d
. .
85 ,
4 1
σ π
⋅ ⋅
=
2400 .
85 ,
, 1
14 ,
3 4
1 ⋅
⋅
= 1092,72 kg P desak = d. t. 2.
profil
σ = 1,0.0,50.2.1600 = 1600 kg
Tegangan yang terjadi σ =
A P
ω .
= 40
, 9
2 071
, 2
. 528
, 8744
⋅ = 963,293 kgcm
2
Tegangan geser yang terjadi τ
2 3
=
.
F L
2 3
=
. 40
, 9
2 696
, 3998
⋅ = 319,044 kgcm
2
Tegangan yang diijinkan
2 2
3
τ σ
σ
+ =
n i
2 2
044 ,
319 3
293 ,
963 +
= = 1110,540 kgcm
2
≤
σ = 1600 kgcm
2
………..Aman
Universitas Sumatera Utara
Direncanakan Profil 2L 70 x 70 x 7 untuk Balok Batang D1 sd D8
Data –data profil : F = 9,40 cm
2
; q = 7,38 kgm b = 70 mm = 7 cm
t = 7 mm = 0,7 cm ix = iy = 2,12 cm
Ix = Iy = 42,4 cm
4
Wx = Wy = 8,43 cm
3
In = 17,6 cm
4
; e = 1,97 Panjang batang Lk = 1,90 m ; P maks = 8743,473 kg
Kontrol kelangsingan =
= =
12 ,
2 190
x x
x
i lk
λ 89,622
Kelangsingan batas
l g
x E
σ π
λ
7 ,
=
=
2400 7
, 10
1 ,
2
6
x x
π = 111,015
Faktor tekuk ω = 2,071 dari hasil interpolasi Tabel Baja
Perhhitungan gaya geser gan gaya normal a = e +
2 1
= 1,97 +
2 1
= 2,47 cm Iy. gab = 2 Iy + F.a
2
= 2 42,2 + 9,40. 2,47
2
= 101,548 cm
4
D = 0,02 P = 0,02 x 8743,473 = 174,869 kg Sy = F. a = 9,40 x 2,47 = 23,218 cm
Universitas Sumatera Utara
L = gab
Iy L
Sy D
. .
.
1
= 548
, 101
95 .
218 ,
23 .
869 ,
174 = 3798,305 kg
N =
u a
L. .
2
= ,
3 47
, 2
. 305
, 3789
. 2
= 6239,722 kg kontrol gaya geser gan gaya desak
P geser = 2.
baut d
. .
85 ,
4 1
σ π
⋅ ⋅
=
2400 .
85 ,
, 1
14 ,
3 4
1 ⋅
⋅
= 1092,72 kg P desak = d. t. 2.
profil
σ = 1,0.0,50.2.1600 = 1600 kg
Tegangan yang terjadi σ =
A P
ω .
= 40
, 9
2 071
, 2
. 473
, 8743
⋅ = 963,177 kgcm
2
Tegangan geser yang terjadi τ
2 3
=
.
F L
2 3
=
. 40
, 9
2 305
, 3798
⋅ = 303,056 kgcm
2
Tegangan yang diijinkan
2 2
3
τ σ
σ
+ =
n i
2 2
056 ,
303 3
177 ,
963 +
= = 1096,923 kgcm
2
≤
σ = 1600 kgcm
2
………..Aman
Universitas Sumatera Utara
Direncanakan Profil 2L 70 x 70 x 7 untuk Balok Batang V1 sd V8
Data –data profil : F = 9,40 cm
2
; q = 7,38 kgm b = 70 mm = 7 cm
t = 7 mm = 0,7 cm ix = iy = 2,12 cm
Ix = Iy = 42,4 cm
4
Wx = Wy = 8,43 cm
3
In = 17,6 cm
4
; e = 1,97 Panjang batang Lk = 1,00 m ; P maks = 5608,702 kg
Kontrol kelangsingan =
= =
12 ,
2 100
x x
x
i lk
λ 47,169
Kelangsingan batas
l g
x E
σ π
λ
7 ,
=
=
2400 7
, 10
1 ,
2
6
x x
π = 111,015
Faktor tekuk ω = 2,071 dari hasil interpolasi Tabel Baja
Perhhitungan gaya geser gan gaya normal a = e +
2 1
= 1,97 +
2 1
= 2,47 cm Iy. gab = 2 Iy + F.a
2
= 2 42,2 + 9,40. 2,47
2
= 101,548 cm
4
D = 0,02 P = 0,02 x 5608,702 = 112,174 kg Sy = F. a = 9,40 x 2,47 = 23,218 cm
Universitas Sumatera Utara
L = gab
Iy L
Sy D
. .
.
1
= 548
, 101
50 .
218 ,
23 .
174 ,
112 = 1282,376 kg
N =
u a
L. .
2
= ,
3 47
, 2
. 376
, 1282
. 2
= 2111,645 kg kontrol gaya geser gan gaya desak
P geser = 2.
baut d
. .
85 ,
4 1
σ π
⋅ ⋅
=
2400 .
85 ,
, 1
14 ,
3 4
1 ⋅
⋅
= 1092,72 kg P desak = d. t. 2.
profil
σ = 1,0.0,50.2.1600 = 1600 kg
Tegangan yang terjadi σ =
A P
ω .
= 40
, 9
2 071
, 2
. 702
, 5608
⋅ = 617,852 kgcm
2
Tegangan geser yang terjadi τ
2 3
=
.
F L
2 3
=
. 40
, 9
2 376
, 1282
⋅ = 102,317 kgcm
2
Tegangan yang diijinkan
2 2
3
τ σ
σ
+ =
n i
2 2
317 ,
102 3
852 ,
617 +
= = 642,765 kgcm
2
≤
σ = 1600 kgcm
2
………..Aman
Universitas Sumatera Utara
4.3 Perbandingan Profil Tunggal Dan Profil Majemuk
