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Selasa, 06 Maret 2012

STRUKTUR JEMBATAN KAYU


Blok ini memang saya tulis khusut Untuk Teman-teman yang lagi kulyah, yang mengambil jurusan Teknik sipil
Semoga bermanpat dan berguna untu Kita semua sesama Warga Teknik Sipil.

RENCANA KONSTRUKSI JEMBATAN






 



                           Trotoir                                                       it aspal

                                                                                                                                                                                      4



 


                                                                                   Gelagar melintang
                        1,2                                                      4                                                               1,2



 










                                                                KEPALA JEMBATAN

Lantai Jembatan
Akibat Beban Mati ( Dead Load )
-          Berat jenis aspalt = 2100 kg/cm 2
-          Kayu kelas I mutu B
Lt = 112,5 kg/cm 2 ; bj = 700 kg/cm 2
                                                L. atas 8/25                       7 cm
                                                                                                                                 7
                                                                                                                                 8
                                                                                                                                 10
                                                                  L. bawah 10/25  

Pembebanan pada selembar papan lapis bawah
-          Bj aspalt                                               = 0,07 . 21 . 21                    = 30,87
-          Berat sendiri lantai atas                 = 0,08 . 0,20 . 7                   = 0,112
-          Berat sendiri lantai bawah            = 0,10 . 0,25 . 7                   = 0,175    +          
                                                                                                      Q       = 31,157               t/mI
                                                                                Tot log 10 %        = 3,1157               t/mI  +
                                                                                                Q tot      = 34,2727             t/mI 
                                                                                                 ~            = 34                        t/mI 
Factor  distribusi                         
                                                                                  Q =34 t/mI 



 

E
 
A
 
B
 
C
 
D
 
                                                                                    

Ø  MBA                      = MBC   =  :  = 3 : 4                                MBA      = O,43                   MBC = O,57
Ø  MCB      = MCD  =  :  = 4 : 4                                MCB      = O,5                     MCD = O,5
Ø  MCD      = MDE   =  :  = 4 : 3                                MDC      = O,57                   MDE = O,43

Momen Primer
Ø  Mo BA   = Mo DE                =  . Q. L2 =  . (34) . (1)2  = 4,25 kg.m
Ø  Mo BC    = Mo Dc                =  . Q. L2 =  . (34) . (1)2  = 2,83 kg.m
BA
BC
CB
CD
DC
DE
7,08
5,66
7,08
0,43
0,57
0,5
0,5
0,57
0,43
4,25
2,83
2,83
2,83
2,83
4,25
- 3,044
- 4,036
2,83
2,83
- 4,036
- 3,044
-
1,415
- 2, 018
- 2, 018
1,415
-
- 0,608
- 0,806
1,009
1,009
- 0,806
- 0,608
-
0,504
- 0,403
- 0,403
0,504
-
- 0,217
- 0,287
0,201
0,201
- 0,287
- 0,217
-
0,100
- 0,143
- 0,143
0,100
-
- 0,043
- 0,057
0,071
0,071
- 0,057
- 0,043
-
0,035
- 0,028
- 0,028
0,035
-
0,338
- 0,302
1, 311
1, 311
- 0,302
0,338
                               
Reaksi perletakan dan momen maksimum                                                  x
VA        =  =  kg
Mx       = VA . X – ½ . Q . X2                                                                                                           VA                                    1                                 VB
Mmax =  = 0
VA – QX = 0                                                                                                             y
              X =      =  0,490 m
Momen lapangan            =  . 0,490 – ½ . 34 . 0,490 = -0,166
Momen tumpuan            = 0,347
                                             50                                                                                                   20
                                                              = 50+2+7+8+10=79                                                                                                                 =20+2+7+8+10=49










 



v  Tekanan gandar      = 20 ton
v  Tekanan roda          = ½ . 6,47 = 3,235
v  Faktor kejut             = k = 1 + 20/50 + L = 1  + 20/50+21 = 1,282
v  Jembatan kelas II   = 50% T Loading k   = 5 . 1,282 = 6,41 ton
Penyebaran gaya arah melintang papan ( ukuran 10/25 )
P           =   . 3,235           = 1,650
Penyebaran gaya arah memanjang papan
Q           =   = 2,089 t/m = 1089 kg/cm
Perataan Momen
a=0,05      b=0,05   c=0,05               Q                             Q
 

A             1                  B         1              C             1        D                   1                 E          

Pada kegiatan I
Momen Primer
Ø  Mo BA   =  
               =  
               = - 1,044 ( 0,375 – 0,234 )
               = - 0,147 t/m = -147 kg/cm
Dengan :
Ø  Mo BC   =  
               =         
               = 0,130 . 1.020 = 0,133 t/m = 133 kg/cm
Ø  Mo CB   =         
               =     
               = - 0,076 . (0,0755) = -0,048 t/m = 48 kg/cm



Kesimpulan momen
Ø  Mo BA   = -147 kg/cm
Ø  Mo BC    = +133kg/cm
Ø  Mo CB    = -48 kg/cm
Ø  Mo CD   = +48 kg/cm
Ø  Mo DC   = -133 kg/cm
Ø  Mo DE    = +147 kg/cm
Distribusi momen ( metode cross)
MK
-14
0
14
Batang
BA
BC
CB
CD
DC
DE
M
0,43
0,57
0,5
0,5
0,57
0,43
Mo
- 147
133
- 48
58
133
- 147
M
6,02
7,98
0
0
7,98
6,02


0
3,99
- 3,99
0


- 140,98
140,98
- 44,01
44,01
140,98
169,26

Momen lapangan dan reaksi perletakan

                                                                                                                    M = 140,98
                                                               0,5













VA
 

VB
 


 


+
 
                                                                                               -      = - 140,98
                                                                                          X   
                                                                          0,5

v RA            = 0,5 . 2,089 . 0,21 – (          = -140,761 kg
v RB             = 0,5 . 2,089 . 0,21 + (          = 141,199 kg
Mmak : Dmak = 0     X =   =  = 0,067                  y = 1 – 0,067 = 0,933
     

My                                 = -140,761 . 0,933 + ½ . 2089 . (0,933)2
                        = -131,330 + 909,226 = 777,809
Momen tumpuan = 140,98

Pada keadaan II
                                                    2 m
EB
 
DB
 
CB
 
AB
 
BB
 
     
                                 1                        1                       1                          1
                                                                                4 m
Momen primer
v Mo BA     = -1/8 . Q . P2 = -1/8 . 2089 . (1)2                 = -261,125           kg/cm
v Mo CD     = 1/12 . Q . P2 = 1/12 . 2089 . (1)2                = 174,083             kg/cm
v Mo DC     = -1/12 . Q . P2 = -1/12 . 2089 . (1)2             = -174,083           kg/cm

Distribusi momen ( Methoda Cross )
-261,125
174,083
174,083
BA
BC
CB
CD
DC
DE
0,43
0,57
0,5
0,5
0,57
0,43
-261,125
-
-
174,083
174,083
-
112,284
148,841
-87,041
-87,041
-99,27
-74,856
-
-43,520
74,420
-49,613
-43,520
-
18,714
24,806
-37,21
-24,806
24,806
-18,714
-
-18,605
12,403
12,403
-12,403
-
8,000
10,605
-6,201
-6,201
7,070
5,333
-
-3,100
5,302
3,535
-3,100
-
-122,127
119,027
-45,607
71,972
-275,651
-50,809





X
 
Reaksi perletakan dan momen
A
 
M = 119,027
 
B
 
RA          =                                                                                                             Q = 2089
=  925,473 kg    ( ↑ )
RB           =  
                =  1163,527 kg    ( ↑ )

Mmak : Dmak = 0            
RA . X = -1/2 . Q . X2 = 0
RA – Q . X = 0
              X   =  =  = 0,443 
Mx       = 925,473. 0,443 - ½ . 2089 . (0,433)2
              = 400,730 – 195,832 = 204,898 kg/m
M tumpuan = 119,027
Akibat beban mati
Akibat beban hidup
M total
Mtp
Mlp
Keadaan I
Keadaan II
Mtp
Mlp
-
-
Mtp
Mlp
Mtp
Mlp
-
-
Kg
Kg
Kg.m
Kg.m
Kg.m
Kg.m
Kg.m
Kg.m
-0,338
-0,166
140,98
777,896
119,027
204,898
141,368
118,861

Control tegangan ( Mlp = 69,919 kg.m )
v    lt       = 5/6 .   lt = 5/6 . 112,5 = 93,75 kg/cm2
v    Wbr = 1/6 b.h2 = 1/6 . 0,75 . (4)2 = 2 = 200  kg/cm2
v    Wn   = 0,8 . 200 = 160  cm2
T      =  =  = 0,43699 kg.m = 43,699 kg/cm2
                                43,699 kg/cm2 112,5 kg/cm2     ( oke )



Perencanaan Trotoar :












 



                                                                                                                                           4/20

3 komentar:

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    BalasHapus
  2. Minta file.y dong sulkifliarrahman@gmail.com

    BalasHapus
  3. minta filenya dong kak.. pas bangat ni dengan tugas ku yuliusamirinoburdam@gmail.com

    BalasHapus