struktur kayu

MERENCANAKAN JEMBATAN KAYU

WINDA SEPTIAN

03061001067

MERENCANAKAN JEMBATAN KAYU

6 meter

4 meter

Ditentukan :

1. L = 18 m

2. H1 = 7 mH2 = 3 m

3. Kayu Kelas I Mutu A

4. Jembatan Kayu untuk jalan raya Kelas III

5. Beban Terpusat = 7 Ton

6. Beban Merata = 1,25 Ton

7. Peraturan Kayu PPKI

8. Lebar Jembatan = 5 m

9. Tekanan Angin (W) = 80 kg/m2 Batang Bawah ( B )

B1 = B2 = B3 = B4 = B5 = B6 = B7 = B8 = = 2.25 meter Batang Tegak ( T )

f = ( 7-3 ) meter = 4 meter

Rumus Parabola :

T1 = T7 = 3 meter

T2 = T6 = 3 + = 6 meter

T3 = T5 = 3 + = 6,75 meter

T4 = 3 + = 7 meter

Batang Diagonal ( D )

D1 = D2 = D7 = D8 = = 3,75 m

D3 = D4 = D5 = D6 =

= 7,115 m

Batang Atas ( A )

A1 = A6 = = 3,75 m

A2 = A5 = = 2,3717 m

A3 = A4 = = 2,2638 m

TABEL REKAPITULASI PANJANG BATANG

BatangPanjang (m)BatangPanjang (m)

A13,75T26

A22,3717T36,75

A32,2638T47

A42,2638T56,75

A52,3717T66

A63,75T73

B12,25D13,75

B22,25D23,75

B32,25D37,115

B42,25D47,115

B52,25D57,115

B62,25D67,115

B72,25D73,75

B82,25D83,75

T13

PERHITUNGAN LANTAI KENDARAAN Lantai Kendaraan yang digunakan adalah papan double

Papan atas = 5/25

Papan bawah = 15/25 Lapisan aspal dengan tebal minimum 5 cm

Berat Jenis Aspal: 2500 kg/m3Berat Jenis Kayu: 900 kg/m3Berat Jenis Air: 1000 kg/m3 Tegangan yang diperkenankan untuk kayu kelas I mutu A dari buku PPK adalah :

= 150 kg/cm2

= 130 kg/cm2

= 40 kg/cm2

= 20 kg/cm2 Menurut buku PPKI untuk konstruksi yang tidak terlindung, Tetapi kayu dapat mengerng dengan cepat ,aka tegangan-tegangan yang diizinkan dikalikan dengan faktor reduksi 5/6

= 150 kg/cm2 x 5/6 = 125kg/cm2

= 130 kg/cm2 x 5/6 = 108,33 kg/cm2

= 40 kg/cm2 x 5/6 = 33,33kg/cm2

= 20 kg/cm2 x 5/6 = 16,67 kg/cm2 Lebar Jalan= Lebar jembatan- lebar trotoar

= 5 m 2 m

= 3 m

Tebal lapisan rata-rata : X =

Y = x + 5

= 3 + 5

= 8 cm

Tebal aspal rata- rata :

= 6,5 cm

= 0,065 m

PEMBEBANAN

1. Akibat Beban Sendiri

- Beban sendiri aspal= tebal aspal x aspal x lebar papan

= 0,065 m x 2500 kg/m3 x 0,25 m

= 40,625 kg/m

Bs. Papan atas= tebal papan x kayu x lebar kayu

= 0,05 m x 900 kg/m3 x 0,25 m= 11,25 kg/m

Bs. Papan bawah= tebal papan x kayu x lebar kayu

= 0,15 m x 900 kg/m x 0,25 m= 33,75 kg/m

Berat air hujan= tebal air x air x lebar papan

= 0,05 m x 1000 kg/m3 x 0,25 m= 12,5 kg/m

Q total

=40,625 kg/m + 11,25 kg/m + 33,75 kg/m + 12,5 kg/m

= 98,125 kg/m2

Toeslag 10 %= 10% x 98,125

= 9,8125 kg/m2

TOTAL= 98,125 kg/m2 + 9,8125 kg/m2

= 107,9375 kg/m2 Faktor Kejut= dimana L = 18 m

= = 0,294 m

Faktor Pembebanan= 1 + 0,294

qtotal

= ( 1 + 0,294 ) x 107,9375

= 139,671 kg/m2. Akibat Beban T. Loading

Menurut peraturan muatan untuk jembatan jalan raya No 12/1970

P untuk dua roda=20 ton

P untuk satu roda=10 ton

Untuk jalan raya kelas III diambil beban 50 %PERHITUNGAN BEBAN RODA TERHADAP ANGIN

Y = 0,5 x t

, L = 1,75 (jarak antar As Roda)

Y = 0,5 x 200 cm

= 100 cm

= 1 m

PA = Tekanan Angin = 80 kg/m2

= 80 kg/m2 x 2m x 7,5 m

= 1200 Kg

*Beban angin merupakan beban merata pada sisi selebar 2m sepanjang 7,5m (PPJR Pasal 2)

MA = 0 RB . 1,75 PA . 1 = 0

RB =

=

= 685,714 kg

= 0,685 Ton

Faktor Pembebanan = 1,294

Po = 50 % x 10 Ton

= 5 Ton

P = ( Po + P ) x Faktor Pembebanan

= ( 5 + 0,685 ) x 1,294

= 7,356 Ton(PENYEBARAN GAYA)

a = (6,5 + 5 + 5) cm

b = 50 + (16,5) x 2

= 16,5 cm

= 83 cm

b = 20 + 2 a

= 20 + (2x16,5)

= 53 cm

Jadi untuk peyebaran daerah lebar papan :

P = P (kotak) x 25/53

= 7,356 x 25/53

= 3,469 Ton

= 3469 KgFaktor Pembebanan = 1,294

Q = 1.294 x 3469 Kg/m

= 4488,886 Kg/m

Untuk penyebaran daerah panjang papan :

q = = = 57,159 kg/cm

q =

= 41,79 Kg/cm

= 4179 kg/m

MOMEN YANG TERJADI

Jarak antara gelagar memanjang = 0,5 m Mmaks= 1/8.ql2 ( Momen akibar beban sendiri)

= 1/8 . 4179 kg/m . (0,5 m)2

= 130,59 kg.m

= 1/8.ql2 ( Momen akibar beban bergerak)

= 1/8 x 4488,886 kg/m x ( 0,5 m )2

= 140,27 kg.mSehungga didapatkan bahwa :

Mtotal= 130,59 kg.m + 140,27 kg.m= 270,86 kg.mKONTROL TEGANGAN

Kontrol Tegangan Lentur

= 1/6.b.h2

= 1/6 . 25 . 152

= 937,5 cm3 = = = 28,89 kg/cm2 Syarat Aman

<

28,89 kg/cm2< 100 kg/cm2 .................... AMAN!!!Kontrol Tegangan Geser

Q beban sendiri= 5/8 . q beban sendiri . L

= 5/8 41,79 kg/cm) . 50 cm

= 1305,94 kg

Q beban bergerak= 5/8 . q beban bergerak . L

= 5/8 (44,886 kg/cm) . 50 cm

= 1402,78 kg Jadi Qtotal = Q1 + Q2

= 1305,94 kg + 1402,78 kg= 2708,72 kg =

dimana : b= lebar papan = 25 cm

=

h= lebar aspal + papan

= 8,126 kg/cm2

= 5 + 15 = 20 cm

Syarat Aman

<

8,126 kg/cm2 < 10,00 kg/cm2 .......................AMAN!!!PERHITUNGAN GELAGAR MEMANJANG Menentukan gelagar memanjang

Jembatan Lalu lintas dengan lebar 5 meter

Syarat jarak gelagar memanjang 0,4 0,6

Rumus

:

Dimana :

n=Jumlah gelagar memanjang ( harus ganjil )

B=lebar jembatan = 5 m

=Jarak gelagar memanjang

Untukn = 12

Untukn = 11

Untukn = 10

Diambil jarak gelagar memanjang ( ) = 0,5 m

Jumlah gelagar memanjang ( n )

= 11 m

Menentukan Gelagar Melintang

Rumus yang digunakan :

dimana :

n=jumalah titik bahul batang ( 9 titik )

L=Panjang jembatan = 18 m

=Jarak gelagar melintang

(

Jadi diambil jarak gelagar melintang = 2,25 mPERHITUNGAN GELAGAR MEMANJANG

Gelagar memanjang dicoba dengan kayu 20/25

Akibat beban sendiri / beban tetap

a) Berat sendiri aspal= tebal aspal x x BJ aspal= 0,065 x 0,5 x 2500

= 81,25 kg/m

b) Berat sendiri papan atas

= 0,05 x 0,5 x 900= 45 kg/m

c) Berat sendiri papan bawah

= 0,15 x 0,5 x 900 = 67,5 kg / m

d) Berat sendiri gelagar memanjang

= 0,2 x 0,5 x 900

= 90 kg/m

e) Berat air

= 0,02 x 0,5 x 1000 kg/m3= 10 kg/m

Q total= Qa + Qb + Qc + Qd + Qe= 81,25 kg/m + 45 kg/m + 67,5 kg / m + 90 kg/m + 10 kg/m

= 271,25 kg/mPerhitungan M dan Q akibat gelagar memanjang maka kontruksi gelagar melintang dianggap Perletakan Sendiri

2,25 m

GAMBAR PERLETAKANMmax=

=1/8 . (271,25 kg/m) ( 2,25)2

=171,65 kg.m

Qmax=

=

=305,156 kg

Akibat D. LoadingUntuk jembatan kelas II=50%

Muatan garis P

=7 ton

Muatan merata Q

=1,25 t/m

Untuk L 30 m

( berdasarkan pedoman perencanaan pembebanan jembatan jalan raya DPU)Maka : D loading untuk :

P = 50 % . 7 ton= 3,5 ton

Q = 50 % . 1,25 t/m= 0,625 t/m

Lebar jalan 3 meter

7 t/m

1,25 t/m2

GAMBAR PELAT

Faktor pembebanan

Jadi, pembebanan yang bekerja pada suatu gelagar memanjang, adalah :

PEMBEBANAN

1. Akibat muatan garis P

Ra 2,25 m Rb Gambar Perletakan

Mmax = . P . L

= . 1,509 . 2,25

= 0,848 tm

= 848 kg.m

Q max= Vb = Va = P

= .1,509 ton

= 0,7545 T = 754,5 kg2. Akibat Muatan terbagi rata

2,25

GAMBAR PERLETAKAN

Mmax= 1/8 . 0,269 . 2,252= 0,170 tm

= 170 kg.m

Q max = . 0,269 . 2,25

= 0,302 ton.m= 302 kg.mTotal Muatan D

Mmax= 848 kg.m + 170 kg.m

= 1018 kg.mQmax= 754,5 kg + 302 kg.m

= 1056,5 kg.m

Total Pembebanan

Mmax= Berat sendiri + D Loading

= 171,65 kg.m + 1018 kg.m= 1189,65 kg.m

Qmax= Berat sendiri + D Loading

= 305,156 kg.m + 1056,5 kg.m

= 1361,656 kg.m

Kontrol terhadap Gelagar Memanjang

W= 1/6. b . h2 25

=1/6. 20. 252

=2083,33 cm3

Wn=0,8 x 2083,33 cm3

20

=1666,64 cm3

A.Kontrol tegangan lentur

= kg/cm2Syarat aman = lt lt kg/cm2 125 kg/cm2.............................. amanB.Kontrol tegangan geser

=

Syarat aman = II =

4,08 kg/cm2 16,67 kg/cm2.................. aman

C. Kontrol Lendutan

=

= 0,45 cm

E = 100.000 kgcm3I =

=

= 26041,67 cm4

=

= 0,00722 cm

Syarat :

<

0,0072 cm < 0,375 cm......................AMAN!!!

Perhitungan Trotoar

Papan Lantai Trotoar

Menurut peraturan muatan umum untuk jembatan jalan raya no.12 tahun 1970 konstruksi trotoar harus diperhitungkan terhadap muatan hidup sebesar 500 kg.

Lebar rencana trotoar : 100 cm

Muatan trotoar : 500 kg/m

Papan lantai kelas 1 dengan mutu A, berat jenis 900 kg/cm

Papan lantai yang direncanakan = 5/20

PembebananBerat sendiri papan lantai:0,05 . 0,2 . 900 = 9,0 kg/m

Berat sendiri beban bergerak: 0,2 . 500 = 100 kg/m

= 109 kg/m

Toeslag 10 % = 10,9 kg/m

q = 119,9 kg/m

1m

Mmax= . q L2

= . 119,9 kg/m . 1 m2

= 14,9875 kg.m

Qmax= . q . L

= . 119,9 kg/m . 1 m

= 59,95 kg

Kontrol Tegangan LenturW = 0,8 . . b . h2

= 0,8 . . 5 cm . (20 cm)2 = 266,67 cm3

Syarat,

5,620 kg/cm2 125 kg/cm2 ..AMANKontrol Tegangan Geser

Syarat,

0,89 kg/cm2 16,67 kg/cm2 AMANKontrol lendutan

I =

cm

=

Syarat,

0,00374 cm 0,25 cm ..AMANKesimpulan : Trotoar menggunakan papan lantai dengan ukuran 5/20PERHITUNGAN GELAGAR MEMANJANG

TROTOAR

Jumlah gelagar melintang= 9 buahJumlah gelagar memanjang= 2 buah ( direncanakan 10/15 )

Jarak gelagar melintang= 2,25 cm

Jarak gelagar memanjang= 80 cm

Menurut PMUJJR, konstruksi trotoar harus diperhitungkan muatan sebesar 60 % terhadap beban hidup di atas trotoar.

Pembebanan

Berat sendiri papan lantai= 0,05 x 0,2 x 900= 9 kg/m

Berat sendiri gelagar melintang= 0,1 x 0,15 x 900 = 13,5 kg/m

Beban hidup

= 60% x 0,5 x 900 = 270,00 kg/m+

= 292,5 kg/m

Toeslag 10%= 29,25 kg/m +

q= 321,75 kg/m

Mmax = 1/8 . qbs . L2

= 1/8 . 321,75. 2,252

= 40,218 kgm

Qmax = 1/2 . qbs . L

= 1/2 . 321,75. 2,25

= 361,968 kgKontrol Tegangan LenturW = 0,8 . . b . h2

= 0,8 . 1/6 . 10 cm . (15 cm)2 = 300 cm3

= = = 13,406 kg/cm2Syarat Aman

<

13,406 kg/cm2 < 125 kg/cm2 .................... AMAN!!!

Kontrol Tegangan Geser

=

= = 3,619 kg/cm2

Syarat Aman

<

3,619 kg/cm2 < 16,67 kg/cm2.....................AMAN!!!

Kontrol Lendutan

I= = . 10 . 153

= 2812,5 cm4

=

= 0,305 cm

=

= 0,5625 cm

Syarat :

<

0,305 cm < 0,5625 cm......................AMAN!!!

PERHITUNGAN BALOK SANDARAN TROTOAR

Menurut peraturan Bina Marga / PU No.12 / 1970. Balok-balok dan tiang sandaran harus diperhitungkan terhadap tekanan horzontal 100 kg/cm yang bekerja setinggi 90 cm dari lantai trotoar.

Pembebanan

- Ukuran balok yang digunakan 10/15

- Berat sendiri balok sandaran= 0,1 x 0,15 x 900=13,50 kg/m

Toeslag 10%= 1,35 kg/m +

qtotal=14,85 kg/m

- Beban hidup ( P )

P ( sb y

q ( sb x

a. q yang bekerja ( sumbu x

>> Mx max=1/8 . q . L2

=1/8 . 14,85 . (2,25)2

=9,39 kgm

>> Qx max= . q . L

= . 14,85 . 2,25

=16,70 kg

b. P yang bekerja ( sumbu y

>> My max=1/8 . q . L2

=1/8 . 100 . (2,25)2

=63,28 kgm

>> Qy max= . q . L

= . 100 . 2,25

=112,5 kg

Kontrol Tegangan Lentur

# Wx bruto = 0,8 . 1/6 . b . h2 = 0,8 . 1/6 . 15. 102 = 200 cm3# Wy bruto = 0,8 . 1/6 . b . h2 = 0,8 . 1/6 . 10. 152 = 300 cm3

# = =

= 34,77 kg/cm2Syarat Aman

<

34,77kg/cm2 < 125 kg/cm2 .................... AMAN!!!

Kontrol Tegangan GeserQmax = Qx max + Qy max

= 16,7 kg + 112,5 kg = 129,2 kg

=

=

= 1,292 kg/cm2Syarat aman :

1,292 kg/cm2 16,67 kg/cm2 .................... AMAN!!!

Perhitungan Tiang Sandaran

Menurut PU Bina Marga No. 12 / 1970, balok-balok dan tiang sandaran harus diperhitungkan terhadap tekanan horizontal sebesar 100 kg/m yang bekerja setinggi 90 cm dari lantai kendaraan.

Pembebanan

direncanakan tiang sandaran balok ukuran 2 x 20/25

beban horizontal (tiap tiang sandaran)

- P = 100 . 2 = 200 kg

-M = 200 kg . 0,9 m = 180 kg m

Qmax = P = 200 kg

y

20 x

25

Kontrol Tegangan Lentur Ix=

=

= 26041,67 cm4

=

=

= 1666,67 cm3

=

=

= 10,79 kg/cm2Syarat Aman :

10,79 kg/cm2 125 kg/cm2 .AMAN

Kontrol Tegangan Geser

Syarat Aman :

0,6 kg/cm2 16,67 kg/cm2 .AMANPERHITUNGAN GELAGAR MELINTANG

Diketahui:Jarak gelagar memanjang

=0,5 m

Jumlah gelagar memanjang=11 buah

Jarak gelagar melintang

=2,25 m

Lebar Jembatan

=5 m

PEMBEBANAN

1. Akibat Lantai kendaraan

Beban P1Berat gelagar memanjang trotoar(10/15) = 0,1 x 0,15 x 900 = 13,5 kg/m

Berat papan Lantai trotoar (5/20)

= 0.05 x 0,5 x 900 = 22,5 kg/m

Berat Sandaran (10/15)

= 3(0,10 x 0,15 x 900) = 40,5 kg/m

Bergerak

= 0,5 x500

= 250 kg/m

Berat air Hujan

=0,05 x 0,5 x 1000 = 25 kg/m

( q = 351,5 kg/m

Toeslag 10%= 35,15 kg/m

Q total= 386,65 kg/m

Dengan jarak gelagar melintang adalah 2,25 m, maka didapat nilai P1P1=q1 x L

=386,65 kg/m x 2,25 m

=869,9625 kg

Beban P2Berat gelagar memanjang trotoar(10/15) = 0,1 x 0,15 x 900= 13,5 kg/m

Berat papan Lantai trotoar (5/20)

= 0.05 x 0,5 x 900 = 22,5 kg/m

Berat gelagar memanjang (20/25)

= 0,2 x 0,25 x 900= 45 kg/m

Berat papan Lantai kendaraan

Papan atas (5/25)

= 0.05 x 0,25 x 900 = 11,25 kg/m

Papan bawah (15/25)

= 0,15 x 0,25 x 900 = 33,75 kg/m

Berat air Hujan

=0,05 x 0,5 x 1000 = 25 kg/m

Berat beban bergerak

= 0,5 x 500

= 100 kg/m

( q = 251 kg/m


Q total= 276,1 kg/m


=276,1 kg/m x 2,25 m

=621,225 kgBeban P3

Berat gelagar memanjang (20/25)

= 0,2 x 0,25 x 900= 45 kg/m

Berat papan Lantai kendaraan

Papan atas (5/25)

= 0.05 x 0,50 x 900 = 22,5 kg/m

Papan bawah (15/25)

= 0,15 x 0,50 x 900 = 67,5 kg/m

Berat Aspal = 0,065 x 0,5 x 2500 = 81,25 kg/m

Berat air Hujan

=0,05 x 0,5 x 1000 = 25 kg/m

( q = 241,25 kg/m


Q total= 265,375 kg/m


=265,375 x 2,25

=597,09375 kg

REAKSI PERLETAKAN

MA= P2 x1 + P3 x1,5 + P3 x 2 + P3 x 2,5 + P3 x3 + P3 x 3,5 + P2 x 1 + P1 x 5 RB x 5

= P1 .5 + P2 ( 1+ 4 ) + P3 (1,5+2+2,5+3+3,5) 5RB

= (869,9625 x 5) +( 621,225 x5) + (597,09375 x 12,5) 5RB 5RB = 14919,60938

RB =

= 2983,921875 kg

RA = RB = 2983,921875 kg

Momen Maksimum yang terjadi ditengah bentang

Mmax= RA ( 2,5 ) P1 ( 2,5 ) P2 (1,5) - P3 ( 1 + 0,5 + 0 )

= 2983,921875 (2,5) 869,9625 (2,5) 621,225 (1,5)- 597,09375 (1,5)

= 7459,804688 kgm 2174,90625 kgm 931,8375 kgm-895,640625kgm

= 3457,4203 kgm

Qmax= RA = RB = 2983,921875 kg

2. Akibat Muatan Loading

Untuk Jembatan kelas III diambil 50 %

Muatan garis P

=7 ton x 50% = 3,5 ton

Muatan merata q=1,25 ton x 50% = 0,625 ton

Jarak gelagar melintang adalah 2,25

maka faktor pembebanan = 1 +

= 1,383

Jarak gelagar melintang = 0,5 m maka beban yang bekerja pada tiap gelagar adalah :

P = x 0,5 x 1,383

= 1,210 t

= 1210 kg

q = x 0,5 x 1,383

= 0,216 t/m

= 216 kg/m

Qmax = q . L

= . 216 kg/m . 2,25 m

= 243 kg

Maka besarnya gaya :

P2 = P + Q

= 1210 kg + 243 kg

= 1453 kg

P1 = . P2

= . 1453 kg

= 726,5 kg

REAKSI PERLETAKAN

RA = RB= P1 + 2,5 P2

= 726,5 kg + 2,5 (1453 kg)

= 4359 kg

M max2= RA (2,5 m) P1 (1,5 m) P2 (1 + 0,5 + 0)

= 4359 kg (2,5 m) 726,5 kg (1,5 m) 1453 kg (1,5m)

= 10897,5 kgm 1089,75 kgm 2179,5 kgm

= 7628,25 kgm

Q max2= RA

= 4359 kg

3. Gelagar melintang direncanakan 30/80

q= 0,3 x 0,8 x 900

= 216kg/m

Mmax= 1/8 . q . L

= 1/8 . 180 . 52

= 562,5 kgm

Qmax= . q . L

= . 216 . 5

= 540 kg

Mmax total= (3457,4203 kgm + 7628,25 kgm + 562,5 kgm )

= 11648,1703 kgm

Qmax total= (2983,921875 kg + 4359 kg + 540 kg)

= 7882,9218 kg

w

= 1/6 . b . h2

= 1/6 . 30 cm . (80 cm)2

= 32000 cm31. Kontrol Tekuk

= = 36,40 kg/cm2syarat AMAN

<

36,40 kg/cm2 < 108,33 kg/cm2 .................... AMAN!!!

2. Kontrol Tegangan Geser

=

= = 4,93 kg/cm2

Syarat Aman

<

4,93 kg/cm2 < 16,67 kg/cm2.....................AMAN!!!3. Kontrol Lendutan

q= 216 kg/m

E= 125.103 kg/cm2 ( kayu kelas I )

I= = . 30 . 803

= 1280000 cm4Lendutan maksimum terjadi di tengah bentang Akibat P1 di titik 1

P= 869,9625 kg a= 0 meter

b= 5 m 0 m = 5 meter

=

= 0 cm Akibat P2 dan P1 di titik 2 P= 621,225 kg + 726,5 kg = 1347,725 kg a= 1 meter


=

= 0,0089 cm Akibat P3 dan P2 di titik 3 P= 597,09375 kg + 1453 kg = 2050,09375 kg a= 1,5 meter

b= 5 m 1,5 m = 3,5 meter

=

= 0,023 cm Akibat P3 dan P2 di titik 4 P= 597,09375 kg + 1453 kg = 2050,09375 kg a= 2 meter


=

= 0,030 cm Akibat P3 dan P2 di titik 5 P= 597,09375 kg + 1453 kg = 2050,09375 kg a= 2,5 meter

b= 5 m 2,5 m = 2,5 meter

=

= 0,033 cm Akibat q

=

= 0,0109 cm

Sehingga lendutan maksimum adalah :

(max =2(1 + 2(2 + 2(3 +2 (4 + (5 + (q= (2(0) + 2(0,0089) + 2(0,023) + 2(0,030) + 0,033+ 0,0109) cm

= 0,1677 cm

= = 1,25 cm

Syarat aman :

<

0,1677 cm < 1,25 cm......................AMAN!!!

PERHITUNGAN PASAK GELAGAR MELINTANG

Untuk gelagar melintang digunakan balok 30/80 disusun tersusun 3 lapis :

Bidang Lintang gelagar tersebut :

Diketahui :

P1= P1 = 869,9625 kg

P2= P2 + P1 = 621,225 kg + 726,5 kg = 1347,725 kg

P3= P3 + P2 = 597,09375 kg + 1453 kg = 2050,09375 kg

q= 216 kg/m

RA = RB

= . q . L + P1 + P2 + 2,5 P3

=.216 kg/m .5m + 869,9625 kg + 1347,725 kg + (2,5)2050,09375 kg

= 7882,921875 kg

Karena batang simetris maka ditinjau setengah saja....

QA

= 7882,921875 kg

QA kanan=7882,921875 kg - 869,9625 kg = 7012,959375 kg QB kiri = QA kanan - ( 0,5 . 216)

= 6904,959375 kg

QB kanan = QB kiri 1347,725 kg

= 5557,234375 kg

QC kiri = QB kanan - ( 0,5 . 216) = 5549,234375kg

QC kanan = QC kiri 2050,09375

= 3399,140625 kg

QD kiri = QC kanan - ( 0,5 . 216 ) = 3291,140625 kg

QD kanan = QD kiri 2050,09375 = 1241,046875 kg

QE kiri = QD kanan - ( 0,5 . 216 ) = 1133,046875 kg

QE kanan = QE kiri . 2050,09375 = 108 kg QF kiri = QE kanan - ( 0,5 . 216 ) = 0 kg

GAYA GESER PER SATUAN PANJANG

Dengan catatan I dapat diambil tanpa reduksi ( felix yap hal 71 ). ( max dapat dicari dengan menggunakan sifat-sifat titik parabola.

maka :

Gaya geser yang terjadi untuk satu balok :

= 147,805 kg/cm

= 131,493 kg/cm

= 129,467 kg/cm

= 104,198 kg/cm

= 104,048 kg/cm

= 63,734 kg/cm

= 61,708 kg/cm

= 23,269 kg/cm

= 21,244 kg/cm

= 2,025kg/cm

0 kg/cm

Kekuatan Pasak

kayu kelas I mutu A = 20,00 kg/cm2dipakai pasak kayu dengan syarat :

u > 5 t

t > 1,5 cm

u > 15 cm

Dipakai pasak :

- u = 15 cm

- t = 3 cm

- b = 30 cm

Maka kekuatan pasak : S1 = u . b . pasak= 15 cm x 30 cm x 10,00 kg/cm2 = 4500 kg

S2 = b . t . balok= 30 cm x 3 cm x 20,83 kg/cm2 = 1874,7 kg

Diambil nilai S minimum yaitu S = 1874,7 kg Gaya Geser yang Harus Dipikul pasak

LAB = x 100 cm

= x 100 cm = 5708,55 kg

LBC = x 50 cm

= x 50 cm = 2277,5 kg

LCD = x 50 cm

= x 50 cm = 1371,75 kg LDE = x 50 cm

= x 50 cm= 486,75 kg

LEF = x 50 cm

= x 50 cm= 88 kg LTotal = 9932,55 kg

Jumlah Pasak untuk Setengah Bentang

= = 5,29 ( 5 buah

Jumlah Pasak yang Diperlukan Per Satu Baris

(AB= = = 3,04 ( 3 pasak

(BC= = = 1,21 ( 1 pasak

(CD= =

= 0,73 ( 1 pasak

(DE= =

= 0,25 ( 0 pasak

(EF= =

= 0,046 ( 0 pasak

PERHITUNGAN IKATAN ANGIN

Dalam memperhitungkan jumlah bagian-bagian jembatan pada setiap sisi dapat digunakan ketentuan menurut PMUJR Bina Marga No.12 / 1970 : Jembatan rangka diambil 30 % terhadap luas bidang sisi jembatan yang bersangkutan. Pengaruh tekanan angin sebesar 90 kg/cm2 pada jembatan ditinjau berdasarkan bekerjanya angin horizontal terbagi rata pada bidang vertikal jembatan dalam arah tegak lurus sumbu memanjang jembatan Jumlah luas bidang vertikal jembatan yang dianggap terkena angin, ditetapkan sebesar 1,5 kali jumlah luas bagian sisi jembatan. Bila ada muatan hidup di jembatan, maka luas tersebut ditambah dengan luas bidang vertikal muatan hidup yang tidak terlindungi oleh bagian sisi jembatan. Bidang vertikal muatan hidup tersebut ditetapkan sebagai suatu permukaan bidang vertikal yang mempunyai tinggi terus menerus sebesar 2m di lantai kendaraan.

Perhitungan Luas Bidang Sisi Jembatan

Luas ABCD = = 47,25 m2 Luas CDE= 27 m2 Luas total= 47,25 m2 + 27 m2 = 74,25 m2 Jumlah luas bagan vertikal jembatan yang dianggap terkena angin adalah

1,5 x Luas total = 1,5 x 74,25 m2 = 113,375m2 Untuk jembatan rangka luasnya 30 %

L = 0,3 x 113,375m2 = 33,4125 m2Tekanan angin ( w ) = 80 kg/m2 Besar gaya angin pada rangkanya :

= 33,4125 m2 x 80 kg/m2= 2673kg

Gaya angin pada muatan hidup setinggi 2 m

= 2. L .W

= 2 m x 18 m x 80 kg/m2 = 2880 kg

Total gaya akibat angin pada jembatan rangka

= 2673kg + 2880 kg

= 5553 kg Jumlah medan pada jembatan rangka = 8tiap simpul (P) = P/2 =

A = RbRA = P/2 + 3,5 P

= 347,0625 kg + (3,5) 694,125 kg

= 2776,5TABEL GAYA BATANG

Nama BatangGaya Batang (kg)

Tarik (+)Tekan (-)

A1 = A8--

A2 = A7-1092,067

A3 = A6-1874,554

A4 = A5-2343,123

B1 = B8--

B2 = B71092,067-

B3 = B61874,554-

B4 = B52343,123-

T1 = T16-347,0625

T2 = T17-2776,5

T3 = T14521,947-

T4 = T15-1214,719

T5 = T12173,323-

T6 = T13-867,656

T7 = T10-173,531

T8 = T11-520,594

T9-347,0625

D1 = D15-1634,2479

D2 = D161634,2479-

D3 = D13-1169,427

D4 = D141169,427-

D5 = D11-726,228

D6 = D12726,228-

D7 = D9-233,468

D8 = D10233,468-

DIMENSI BATANG IKATAN ANGIN

Batang atas dan batang bawah merupakan gelagar memanjang.

Batang tegak merupakan gelagar melintang jembatan.

Jadi, hanya mendimensi batang diagonal ( D ) pada ikatan angin

Perhitungan Hubungan Gelagar Melintang dengan Ikatan Angin. Ikatan Angin Atas Akibat Gaya Tarik

Dicoba ukuran balok 15/20

Pmax batang D = 2343,123 kg Panjang batang maksimum

Lk =

= 3,363 m

= 336,3 cm

Ix= 1/12 x b x h3

= 1/12 x 15 cm x (20 cm)3

= 10000 cm4 Fbr= 15 cm x 20 cm = 300 cm 2 Fr= 0.8 x Fbr = 0,8 x 300 cm2

= 240 cm2(tr= = = 9,76 kg/cm2Syarat Aman :

(tr <

9,76kg/cm2 < 108,33 kg/cm2.........................AMAN!!! Akibat Gaya Tekan

Dicoba ukuran balok 15/20

Pmax batang D = 2343,123 kg

Panjang batang maksimum

Lk =

= 3,363 m

= 336,3 cm

Ix= 1/12 x b x h3

= 1/12 x 15 cm x (20 cm)3

= 10000 cm4 Fbr= 15 cm x 20 cm = 300 cm 2 Fr= 0.8 x Fbr = 0,8 x 300 cm2

= 240 cm2 imax= = = 5,774 cm

(=

= = 58,24 ( 59Syarat Aman :

( < 150

59 < 150........................AMAN!!!

Dari tabel tekuk ( = 59 didapat w = 1,65 ( PPKI, hal 12 )

(tk= = = 12,89 kg/cm2Syarat Aman :

(tk <

12,89 kg/cm2 < 33,33 kg/cm2.........................AMAN!!!

PERHITUNGAN GELAGAR INDUK Pembebanan

Akibat Beban Mati

1. Jalur Utama

Aspal

= 0,065 x 18 x 3 x 2500 = 8775 kg

Papan Lantai Kendaraan = 0,20 x 3 x 18 x 900 = 9720kg

Gelagar Memanjang

= 0,20 x 0,25 x 18 x 900 x 7= 5670kg

Gelagar Melintang

= 0,3 x 0,8 x 900 x 9 x 5 = 9720 kg +

= 33885 kg

Toeslag 10% = 3388,5 kg +

q1

= 37273,5 kg

2. Trotoar

Papan Lantai Kendaraan = 0,05 x 1 x 18 x 900

= 810 kg

Gelagar Memanjang= 0,1 x 0,15 x 18 x 900 x 2= 486 kg

Balok Sandaran

= 0,1 x 0,15 x 18 x 900 x 3= 729 kg

Tiang Sandaran

= 0,2 x 0,25 x 0,9 x 900 x 9 = 364,5 kg +

= 2389,5 kg

Toeslag 10%

= 238,95kg +

q2

= 2628,45 kg3. Ikatan Angin

Batang Diagonal = 0,15 x 0,2 x 3,363 x 900 x 16 = 1452,816 kg

Toeslag 10%

=145,2816kg +

q3

= 1598,0976 kg4. Berat Sendiri Gelagar Induk

Batang Atas 2 x 20/30

Batang Bawah 2 x 20/30

Batang Tegak 2 x 20/30 ( kanan-kiri )

Batang Diagonal 2 x 20/30 ( kanan-kiri )

Perhitungan Beban :

Batang Atas

= 2 x 0,2 m x 0,3 m x 16,771 m x 900 kg/m3= 1811,268 kg

Batang Bawah

= 2 x 0,2 m x 0,3 m x 18 m x 900 kg/m3

= 1944 kg

Batang Tegak

= 2 x 0,2 m x 0,3 m x38,5m x 900 kg/m3= 4158 kg Batang Diagonal

= 2 x 0,2 m x 0,3 m x 43,46 m x 900 kg/m3

= 4693,68 kg

q4Total = q + Toeslag 10 %= 12606,948 kg + 12606,6948 kg

= 13867,6428 kg

Berat Gelagar Induk + beban mati

Q= q1 + q2 + q3 + q4 = (37273,5 kg)+ 2628,45 kg+(1598,0976 kg)+ 13867,6428 kg

= 35931,8916 kg Tiap Simpul Menerima :

P = =

= 4491,486 kg

P/2 = 2245,743 kgRA = RB = 8/2 P = 4P

= 4 (4491,486 kg )

= 17965,944 kg Akibat Beban Angin pada Jembatan

Gaya angin akibat muatan setinggi 2 m

= Tekanan angin x 2 m x panjang jembatan

= 80 kg/m2 x 2 m x 18 m

= 2880 kg

Besar gaya akibat angin pada rangka

Perhitungan luas bidang sisi jembatan = 74,25 m2 Jumlah luas bagian vertikal jembatan yang dianggap terkena angin adalah = 1,5 x Luas total = 1,5 x 74,25 m2= 111,375 m2 Untuk jembatan rangka, Luasnya diambil 30%

L = 0,3 x 111,375 m2 = 33,4125 m2 Tekanan angin ( w ) = 80 kg/m2 Besarnya gaya angin pada rangka = 33,4125 m2 x 80 kg/m2 = 2673 kg

Tiap simpul menerima gaya = = 334,125 kg

Tekanan Vertikal ke bawah pada Gelagar Utama Akibat Angin

K x 5= 2880 kg ( 2 + 0.07 + 0.2 + 0,2 + 0,8 )

K= = 1883,52 kg

RA = RB = = 941,76 kg

Untuk tiap titik simpul menerima gaya = = 117,72 kg

Jadi, P = 117,72 kg

P/2 = 58,86 kg Perbandingan RA beban angin dan beban RA beban mati

= = 0,0523 Tambahan gaya untuk tiap gelagar induk

= 0,0523 x berat sendiri gelagar utama= 0,0523 x 35931,8916 kg= 1879,238 kg Akibat Beban Bergerak T Loading

Muatan garis P

=7 ton x 50% = 3,5 ton

Muatan merata q=1,25 ton x 50% = 0,625 ton

Panjang jembatan = 18 m

Faktor kejut

=

K = 1 + = 1,294Lebar jembatan = 5 m, lebar minimum jalur = 2,75 m P = x 5 m x 1,294 = 8,234 t

Q = x 5 m x 1,294

= 1,47 t/m Gaya ini akan ditahan oleh 2 gelagar induk

maka tiap gelagar induk menerima gaya :

P = = 4,117 t = 4117 kg q = = 0,735 t/m = 735 kg/m

PERHITUNGAN GARIS PENGARUH

Garis Pengaruh

( MB = 0

RA . L - P ( L-x ) = 0

RA =

GP. RA = ( MA = 0

- RB . L + P . x = 0

RB =

GP. RB =

Potongan I I

tgn ( = = 1,33

(= 53,06o Batang B1 (tinjau kiri)

( MJ = 0

RA x 2,25 - B1 x 3 + D1 x 0 = 0

3 x B1 =

B1 =

Batang B1 (tinjau kanan)

( MJ = 0

-RB x 15,75 + B1 x 3 - A1 x 0 = 0

3 B1 = x 15,75

B1 =

Batang D1 (tinjau kiri)

( MC = 0

RA x 2,25 + D1 x 1,352 = 0

1,352 D1 = -

D1 = -

Batang D1 (tinjau kanan)

( MC = 0

-RB x 15,75 - D1 x 1,352 = 0

1,352 D1 = -x 15,75

D1 = -

Potongan II II

Batang B2 (tinjau kiri)

( MJ = 0

RA x 2,25 - B2 x 3 = 0

3 B2 =

B2 =


( MJ = 0

-RB x 15,75 + B2 x 3 = 0

3 B2 =

B2 =

Batang T1 (tinjau kiri)

( MD = 0

x04,518

Y T1-2-1,50

RA x 4,5 + T1 x 2,25 = 0

2,25 T1 = -

T1 = -

Batang T1 (tinjau kanan)

( MD = 0

-RB x 13,5 - T1 x 2,25 = 0

2,25 T1 = -x 13,5

T1 = -

Potongan III III

Batang A1 (tinjau kiri)

( MD = 0

RA x 4,5 + A1 x 3,75 = 0

3,75 A1 = - x 4,5

A1 = -

Batang A1 (tinjau kanan)

( MD = 0

-RB x 13,5 - A1 x 3,75 = 0

3,75 A1 = -x 13,5

A1 = -


( MK = 0

RA x 4,5 D2 x 3,6 = 0

3,6 D2 = x 4,5

D2 =


( MK = 0

-RB x 13,5 + D2 x 3,6 = 0

3,6 D2 = x 13,5

D2 =

Potongan IV IV


( MD = 0

RA x 4,5 + A2 x 5,77 = 0

5,77 A2 = -x 4,5

A2 = -


( MD = 0

- RB x 13,5 - A2 x 5,77 = 0

5,77 A2 = -x 13,5

A2 = -


( MJ = 0

RA x 2,25 A2 x 2,25 - T2 x 2,25 B2 x 3 = 0

2,25 T2 = x 2,25 - x 2,25 - x 3

T2 = -


( MJ = 0

-RB x 15,75 + T2 x 2,25 + B2 x 3 + A2 x 2,25 = 0

2,25 T2 =x 15,75 - x 3 - x 2,25

T2 = -

Potongan V - V


( ML = 0

RA x 6,75 - B3 x 6,75 = 0

6,75 B3 = x 6,75

B3 =


( ML = 0

-RB x 11,25 + B3 x 6,75 = 0

6,75 B3 = x 11,25

B3 =


( MK = 0

RA x 6,75 - D3 x 2,107 - B3 x 6 = 0

2,107 D3 = x 6,75 - x 6

D3 =


( MK = 0

-RB x 11,25 + D3 x 2,107 + B3 x 6= 0

2,107 D3 = x 11,25 - x 6

D3 = -

=

=

Potongan VI VI Batang T3 (tinjau kiri)

( MD = 0

RA x 4,5 - T3 x 2,25 - A2 x 3,75 = 0

2,25 T3 = x 4,5 + x 3,75

T3 = + Batang T3 (tinjau kanan)

( MF = 0

-RB x 13,5 + T3 x 2,25 + A2 x 3,75 = 0

2,25 T3 = x 13,5 + x 3,75

T3 = + 1,67 .


( ML = 0

RA x 6,75 - B4 x 6,75 = 0

6,75 B4 = x 6,75

B4 =


( ML = 0

-RB x 11,25 + B4 x 6,75 = 0

6,75 B4 = x 11,25

B4 =

Potongan VII VII


( MF = 0

RA x 9 + A3 x 6,96 = 0

6,96 A3 = - x 9

A3 = -


( MF = 0

-RB x 9 - A3 x 6,96 = 0

6,96 A3 = - x 9

A3 = -


( MM = 0

RA x 9 - D4 x 2,214 B4 x 7 = 0

2,214 D4 = x 9 - x 7

D4 = 1,107


( MM = 0

-RB x 9 + D4 x 2,214 + B4 x 7 = 0

2,214 D4 = x 9 - x 7

D4 = -

Potongan VIII VIII

X011,2518

y T4 -2,79-1,0460


( MN = 0

RA x 11,25 - T4 x 2,25 D4 x 4,793 B4 x 6,75 = 0

2,25 T4 = x 11,25 - 1,107 x 4,793 - x 6,75

T4 = - 2,79

x011,2518

y T400,5710,913


( ML = 0

-RB x 5,75 + T4 x 2,25 - D4 x 4,793 + B4 x 6,75 = 0

2,25 T4 = x 5,75 + x 4,793 - x 6,75

T4 =

Gambar Garis Pengaruh Batang Atas

Gambar Garis Pengaruh Batang Bawah

Gambar Garis Pengaruh Batang Tegak

Gambar Garis Pengaruh Batang Diagonal

PERHITUNGAN GAYA BATANG

Akibat Beban Bergerak

P = 8234 kg

q = 1470 kg/m

contoh perhitungan = P. A1 = P. ( y . A1

q . A1 = q . A ( A = . L . MBatangPanjang ( ( y )Beban Bergerak (P)

+-+-

A1 = A6-3,577-29453,018

A2 = A5-2,319-19094,646

A3 = A4-2,583-21268,422

B1 = B82,625-21614,25-

B2 = B72,625-21614,25-

B3 = B62,533-20856,722-

B4 = B52,533-20856,722-

T1 = T7-6-49404

T2 = T6-2,726-22445,884

T3 = T59,993-82282,362-

T42,284-18806,456-

D1 = D8-5,819-47913,646

D2 = D73,834-31569,156-

D3 = D60,888-7311,792-

D4 = D51,3860,90111412,3247418,834

BatangLuas GP (A)Beban Bergerak (q)

A1 = A67,3210760,4

A2 = A55,227673,4

A3 = A46,048878,8

B1 = B85,888643,6

B2 = B75,888643,6

B3 = B65,638276,1

B4 = B55,638276,1

T1 = T713,519845

T2 = T66,149025,8

T3 = T522,2932766,3

T45,147555,8

D1 = D813,1019257

D2 = D78,43712402,39

D3 = D61,992925,3

D4 = D54,596747,3

Akibat beban sendiri

P = 1879,238 kg BatangPanjang ( ( y )qbs . ( y = P . ( y

+-+-

A1 = A6-3,577-6722,034

A2 = A5-2,319-4357,953

A3 = A4-2,583-4854,187

B1 = B82,625-4933,118-

B2 = B72,625-4933,118-

B3 = B62,533-4760,224-

B4 = B52,533-4760,224-

T1 = T7-6-11275,698

T2 = T6-2,726-5122,925

T3 = T59,993-18779,67-

T42,284-4292,179-

D1 = D8-5,819-10935,54

D2 = D73,834-7204,99-

D3 = D60,888-1668,763

D4 = D51,3860,9012604,6861693,193

Akibat beban Angin

P = 694,195 kg

BatangPanjang ( ( y )qbs . ( y = P . ( y

+-+-

A1 = A6-3,577-2843,136

A2 = A5-2,319-1609,838

A3 = A4-2,583-1793,106

B1 = B82,625-1822,261-

B2 = B72,625--

B3 = B62,533-1758,396-

B4 = B52,533--

T1 = T7-6-4165,17

T2 = T6-2,726-1892,375

T3 = T59,993-6937,09-

T42,284-1585,541-

D1 = D8-5,819-4039,521

D2 = D73,834-2661,54

D3 = D60,888-616,445-

D4 = D51,3860,901962,154625,469

REKAPITULASI GAYA BATANG

BatangKombinasi (1+2+3)Kombinasi (1+2+4)

+-+-

A1 = A6-39018,188-20325,57

A2 = A5-25062,432-13641,191

A3 = A4-27915,528-15526,093

B1 = B828369,629-15398,979-

B2 = B728369,629-15398,979-

B3 = B627375,342-14794,72-

B4 = B527375,342-14794,72-

T1 = T7-64844,868-35285,868

T2 = T6-29461,184-16041,1

T3 = T5107999,122-58493,06-

T424684,176-13433,52-

D1 = D8-62888,707-64232,061

D2 = D741435,686-22268,92-

D3 = D69597-5210,508-

D4 = D514979,1389737,49610314,149065,962

DIMENSI GELAGAR UTAMA

Batang Atas

Gaya batang maksimum = 39018,188 kg Ukuran balok ditaksir= 2 x 20/30

Panjang batang

= 375 cm

Gaya batang maksimum diambil dari kombinasi ( 1+2+3 ) atau ( 1+2+4 ) Fbr= 2 x 20 x 30 = 1200 cm2 Ix= 2 . . bh3

= 2 . . (20) . (30)3

= 90000 cm4 ix= = = 8,66 cm

it= 2 . . b3 h + 2 { b.h (b+H) }

= 2 . . (20)3 30 + 2 { 20.30 (20+30) }

= 70000 cm4 Iq= . (b+b)3 . h

= . (20+20)3 . 30

= 160000 cm4 Iy= . (It + 3Iq)

= . (70000 + 3(160000))

= 137500 cm4 iy= = = 10,70 cm

= = = 35,04 = 35 , w = 1,3 = = = 52,83 kg/cm2 Syarat Aman :

EMBED Equation.3

52,83 kg/cm2 < 108,33 kg/cm2 AMAN!!! Kontrol terhadap Tegangan tekan sumbu x

imin= 0,289 . b

= 0,289 . ( 20 cm )

= 5,78 cm

= = 64,87 ( 65Dari tabel didapat w = 1,76 untuk ( = 65 ( PPKI haL.11 )

= = 57,23 kg/cm2Syarat Aman :

57,23 kg/cm2< 108,33 kg/cm2................................AMAN !!! Batang Tegak


Panjang batang

= 700 cm

Ix= . bh3

= . (20) . (30)3

= 45000 cm4 ix= = = 8.66 cm

it= . b3 h + b.h (b+h)

= . 203 30 + 20.30 (20+30)

= 35000 cm4 Iq= . (b+b)3 . h

= . (20+20)3 . 30

= 160000 cm4 Iy= . (It + 3Iq)

= . (35000 + 3(160000))

= 128750 cm4

iy= = = 14,64 cm

= = = 47,81 = 48, w = 1.47 = = = 104,67 kg/cm2Syarat Aman :

103,12 kg/cm2 < 108,33 kg/cm2................................AMAN !!! Batang Diagonal


Panjang batang

= 711,5 cm

Kontrol terhadap Tegangan tekan sumbu x

Fb = 2 x 0,8 x b x h

= 2 x 0,8 x 20 x 30

= 960 cm2 = = 65,51 kg/cm2Syarat Aman :

65,51 kg/cm2 < 108,33 kg/cm2................................AMAN !!! Kontrol Tegangan tekan terhadap sumbu y

It = momen lembam teoritis

Iq = momen lembam geser

It= 2 ( 1/12 x b x h3 ) + ( 2 b h ) ( a + b2 )

= 2 ( 1/12 x 20 x (30)3 ) + ( 2 x 20 x 30 ) ( 30 + (20) 2 )

= 90000 + 516000

= 606000 cm4 Iq= 1/12 x ( 2xb )3 x 30

= 1/12 x ( 2 x 20 )3 x 30

= 160000 cm4 Iy= x ( It + 3Iq )

= x (606000 + 3 x 160000 )

= 271500 cm4 iy= = = 15,04 (= = 47,31 ( 47Dari tabel didapat w = 1,46 untuk ( = 47 ( PPKI haL. 10 )


76,51 kg/cm2 < 108,33 kg/cm2................................AMAN !!!

Batang Bawah


Panjang batang

= 225 cm

Kontrol terhadap Tegangan tekan sumbu x

Fb = 0,8 . 2 . b . h

= 0,8 . 2 . 20 . 30 = 960 cm2


29,55 kg/cm2 < 108,33 kg/cm2................................AMAN !!! Kontrol Profil Tunggal

P = x 28369,629 kg

= 14184,8145 kg

Fb= 0,8 . 20 . 30

= 480 cm2 = = 29,55 kg/cm2Syarat Aman :

29,55 kg/cm2 < 108,33 kg/cm2................................AMAN !!!

PERHITUNGAN SAMBUNGAN

Dipakai kokot bulldog persegi 5 x 5

Diameter baut = 1

P = 2 ton = 2000 kg

= = 3600 kg

P = ( 1 0,25 sin ( )

= 3600 ( 1 0,25 sin ( )

Titik Simpul I

D1

I B1 Sambungan batang B1 dan batang D1 ( = 53,130 P = 3600 ( 1 0,25 sin 56,310 )

= 2880 kg

P = 2 x 2880 kg = 5760 kg

= 10,92 baut

= 11 baut Titik Simpul II

A1

D1D2

T1

Tan (4 = = 1,33 (4 = 53,130 Sambungan batang T1 dan batang D1 ( = 36,86 P = 3600 ( 1 0,25 sin 36,86)

= 3060,12 kg

P = 2 x 3060,12 kg = 6120,24 kg

= 10,59 baut

= 11 baut Sambungan batang T1 dan batang D2 ( = 36,86

P = 3600 ( 1 0,25 sin 36,86)

= 3060,12 kg

P = 2 x 3060,12 kg = 6120,24 kg

= 10,59 baut

= 11 baut Sambungan batang D2 dan batang A1 ( = (3 + (4 = 53,130 + 53,130 = 106,260 P = 3600 ( 1 0,25 sin 106,260)

= 2725,99 kg

P = 2 x 2725,99 kg = 5471,99 kg

= 7,57 baut

= 8 baut

Titik Simpul III

B1 = 28369,629 kg

T1

B2 = 28369,629 kg

T1 = 64844,868 kg

B1 III B2

Sambungan batang B1 dan batang T1 ( = 900 P = 3600 ( 1 0,25 sin 900 )

= 2700 kg

P = 2 x 2700 kg = 5400 kg

= 12,01 baut

= 12 baut Sambungan batang B2 dan batang T1 ( = 900 P = 3600 ( 1 0,25 sin 900 )

= 2700 kg

P = 2 x 2700 kg = 5400 kg

= 12,01 baut

= 12 baut Titik Simpul IV

IVA2

A1

T2

Sambungan batang A1 dan batang T2 ( = 20,560 P = 3600 ( 1 0,25 sin 20,560)

= 3283,93 kg

P = 2 x 3283,93 kg = 6567,86 kg

= 5,94 baut

= 6 baut Sambungan batang T2 dan batang A2 ( = 900 + (2 = 900 + 18,430= 108,430 P = 3600 ( 1 0,25 sin 108,430)

= 2746,16 kg

P = 2 x 2746,16 kg = 5492,32 kg

= 5,36 baut

= 5 baut Titik Simpul V

Tan (3 = = 0,375

(3 = 20,560

Sambungan batang B2 dan batang D2 ( = 53,130P = 3600 ( 1 0,25 sin 53,130)

= 2880 kg

P = 2 x 2880 kg = 5760 kg

= 7,19 baut

= 7 baut Sambungan batang D2 dan batang T2 ( = 20,560 P = P = 3600 ( 1 0,25 sin 20,560)

= 3283,93 kg

P = 2 x 3283,93 kg = 6567,86 kg

= 6,3 baut

= 6 baut

Sambungan batang D3 dan batang T2 ( = 20,560 P = P = 3600 ( 1 0,25 sin 20,560)

= 3283,93 kg

P = 2 x 3283,93 kg = 6567,86 kg

= 4,48 baut

= 5 baut Sambungan batang D3 dan batang B3 ( = 71,50 P = 3600 ( 1 0,25 sin 71,50)

= 2746,51 kg

P = 2 x 2746,51 kg = 5493,02 kg

= 4,98 baut 5 baut Titik Simpul VI

Tan (4 = = 2,937

(4 = 71,20

Sambungan batang A2 dan batang D3 ( = 60,530 P = 3800 ( 1 0,25 sin 60,530)

= 2972,92 kg

P = 2 x 2972,92 kg = 5945,83 kg

= 11,35 baut

= 12 baut Sambungan batang D3 dan batang T3 ( = 18,790 P = 3800 ( 1 0,25 sin 18,790)

= 3494 kg

P = 2 x 3494 kg = 6988 kg

= 9,66 baut

= 10 baut Sambungan batang D4 dan batang T3 ( = 18,790 P = 3800 ( 1 0,25 sin 18,790)

= 3494 kg

P = 2 x 3494 kg = 6988 kg

= 2,31 baut

= 3 baut Sambungan batang D4 dan batang A3 ( = (3 + (4 = 71,20 + 3,580 = 74,780 P = 3800 ( 1 0,25 sin 74,780)

= 2883,32 kg

P = 2 x 2883,32 kg = 5766,64 kg

= 2,79 baut

= 3 baut

Titik Simpul VII

B3 = 30984,71 kg

T3

B4 = 30984,71 kg

T3 = 4540,28 kg

B3 VII B4

Sambungan batang B1 dan batang T1 ( = 900 P = 3800 ( 1 0,25 sin 900 )

= 2850 kg

P = 2 x 2850 kg = 5700 kg

= 5,43 baut

= 6 baut Sambungan batang B1 dan batang T1 ( = 900 P = 3800 ( 1 0,25 sin 900 )

= 2850 kg

P = 2 x 2850 kg = 5700 kg

= 5,43 baut

= 6 baut Titik Simpul VIII

A3A4

T4

Sambungan batang A3 dan batang T4 ( = 86,420 P = 3800 ( 1 0,25 sin 86,420)

= 2851,85 kg

P = 2 x 2851,85 kg = 5703,7 kg

= 1,65 baut

= 2 baut Sambungan batang A4 dan batang T4 ( = 86,420 P = 3800 ( 1 0,25 sin 86,420)

= 2851,85 kg

P = 2 x 2851,85 kg = 5703,7 kg

= 1,65 baut

= 2 baut Titik Simpul IX

T4

D4D5

B4 IX

B5 Sambungan batang D4 dan batang B4 ( = 71,20 P = 3800 ( 1 0,25 sin 71,20)

= 2900,68 kg

P = 2 x 2900,68 kg = 5801,36 kg

= 5,34 baut

= 6 baut Sambungan batang D4 dan batang T4 ( = 18,80 P = 3800( 1 0,25 sin 18,80)

= 3491,84 kg

P = 2 x 3491,84 kg = 6986,68 kg

= 2,31 baut

= 3 baut Sambungan batang D5 dan batang T4 ( = 18,80 P = 3800( 1 0,25 sin 18,80)

= 3491,84 kg

P = 2 x 3491,84 kg = 6986,68 kg

= 2,31 baut

= 3 baut Sambungan batang D5 dan batang B5 ( = 71,20 P = 3800 ( 1 0,25 sin 71,20)

= 2900,68 kg

P = 2 x 2900,68 kg = 5801,36 kg

= 5,34 baut

= 6 bautPERHITUNGAN LANDASAN

Pembebanan

Reaksi akibat beban mati ( RA1 ) = 16495,4 kg Reaksi akibat angin ( RA2 )

= 372,094 kg Reaksi akibat beban bergerak ( RA3 ) = (4975 + 829) kg +

Rtot= 19769,494 kg

Dimensionering landasan perletakan direncanakan menggunakan kayu kelas I.

Pembebanan terhadap rol dan sendi :

R = . Rtotal

= . (19769,494 kg)

= 9884,747 kg

Luas yang dibutuhkan :

( RoL

= 164,745 cm2 Direncanakan L = 25 cm ; b = 40 cm

Harga ( ditetapkan

( = bentang

( = 1600 cm

( = 1,6 cm

F = b . L

= 25 cm . 40 cm = 1000 cm2 w = 1/6 . b . h2

= 1/6 . 25 cm . (40 cm)2

= 6666,67 cm3 Rm= R x (

= 9884,747 kg x 1,6 cm

= 15815,595 kgcm

=

= 70,035 kg/cm2

M= x x L x b x ( . L/2 )

= x 70,035 kg/cm2 x 25 cm x 40 cm ( . 12,5 cm )

= 218859,375 kgcm =

t2 = =

t2= 468,75

t= 21,65 cm 22 cm

= tegangan kontak 6500 kg/cm2 , s = 1/d

(6500)2=

d1= 4,39 cm = 5 cm

d2= d1 + t

= 5 cm + 22 cm = 27 cm

Sendi

M= 12,5 cm x 70,035 kg/cm2 x 25 cm ( . 12,5 cm )

= 136787,109 kgcm

=

t2 = =

t2= 292,969

t= 17,12 cm = 18 cm

t1 = t = x 18 cm = 9 cm h= 3 t1 = 3 x 9 cm = 27 cm t3= 1/6 h = 1/6 x 27 cm = 4,5 cm Diameter rolr = . d1 (

d1 (

d1 ( 6,33 cm

d1 ( 7 cm

d2= d1

= x 7 cm = 1,75 cm

d3= d1 + 2 d2 = 6 cm + 2(1,75 cm) = 9,5 cm

5 cm

Balok

Sandaran

S

T

R

K

T

U

R

K

A

Y

U

x0918y RA10,50

EMBED AutoCAD.Drawing.17

30

cm


225 cm

diambil t > 3 cm

sehingga u = 5t =5x3 = 15 cm

u

t

5

P

q

30

cm

15

cm

10

cm

30

cm

225 cm

Balok

Sandaran

10 cm

x02,2518y B100,6565,25

x02,2518y B10,750,6560


x0918y RB00,51


+

+

+

+

+

+

x02,2518y D1-1,664-1,4560

x02,2518y D10-1,456-11,650


x02,2518y B20,750,6560

x02,2518y B200,6565,25

x04,518y T10-1,5-6


x04,518y A1-1,2-0,90

x04,518y A10-0,9-3,6

x04,518y D21,250,93750

x04,518y D200,93753,75



x02,2518y T20-0,097

-2,339

x02,2518y T2-0,78-0,682

0

x04,518y A20-0,58-0,78

x04,518y A2-0,78-0,580

x06,7518y B300,6250

x06,7518y B300,6251,667

x06,7518y D30,3560,2220

x06,7518y D300,2220,593


x04,518y T332,4770

x04,518y T302,4779,907

x06,7518y B410,6250

x06,7518y B400,6251,667


x0918y A3-1,293-0,6460

x0918y A30-0,646-1,293

x0918y D41,1070,5540

X0918y D40-0,602-1,204


















Luas penampang bruto

D1 = 62888,707 kg

A1 = 39018,188 kg

T1 = 64844,868 kg

D2 = 41435,686 kg

Tan (1 = EMBED Equation.3 = 0,75

(2 = (1 = 36,86 0

Tan ( = EMBED Equation.3 = 1,33

(= 53,130

D1 = 62888,707 kg

B1 = 28369,629 kg

(4

(3

(2

(1


(4 = 53,130

A1 = 39018,188 kg

A2 = 25062,432 kg

T2 = 29461,184 kg


(1 = 20,560

900

(2

(1


(2 = 18,430

B2 = 28369,629 kg

D2 = 41435,686 kg

T2 = 29461,184 kg

D3 = 9597 kg

B3 = 27375,342 kg

Tan ( = EMBED Equation.3 = 1,33

(= 53,130


(2

(3

(4

(1

Tan (4 = EMBED Equation.3 = 3

(4 = 71,50


(2 = 20,560


A3 = 27915,528 kg

A4 = 8497,83 kg

D3 = 67491,57 kg

D4 = 16126,68 kg

T3 = 4540,28 kg


(1 = 60,530

(5

(1

(4

(3

(2


(3 = (2 = 18,790


(5 = 3,580

VIII

A3 = A4 = 8497,83

T4 = 9428 kg

(1 = (2 = 900 - 3,580

= 86,420

(1

(2

B4 = B5 = 30984,71 kg

D4 = D5 = 16126,68 kg

T4 = 9428 kg


(4 = (1 = 71,20

(3

(2


(3 = (2 = 18,80

(1

(4

Balok perletakan direncanakan mengunaka mutu beton K175 dengan EMBED Equation.3 = 60 kg/cm2

syarat ( < 1600 kg/cm2

diambil ( = 100 kg/cm2

Civil Engineering of Sriwijaya University

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