59674607 contoh disain jembatan komposit

KONVERSI SATUAN

Quantity English System SI System

force 1 lb 4.45 N

mass 1 14.59 kg

length 1 ft 0.3 m1 ft 12 in1 in 0.0254 m2 in 2.54 cm

mass density 1 16.02

torque or moment 1 lb in 0.11 N m

acceleration 1 0.3

acceleration of gravity 32.2 9.81

386.22 9.81

spring constant 1 lb/in 175.1 N/mrotational spring constant 1 lb in/rad 0.11 N m/raddamping constant 1 lb sec/in 175.1 sec/m

moment of inertia 1 0.11

modulus of elasticity 1000000 ###

angle 1 degree 0.02 radian

lb sec2/ft

lb/ft3 kg/m3

ft/sec2 m/sec2

ft/sec2 m/sec2

in/sec2 m/sec2

lb in sec2 kg m2

lb/in2 N/m2

Perencanaan Jembatan Komposit

Gelagar Tengah - 2

PERENCANAAN GELAGAR JEMBATAN KOMPOSIT"GELAGAR TENGAH"

jembatan komposit dapat dilihat pada gambar di bawah ini.

DATA - DATA= 15.00 m= 7.60 m= 2.00 m= 0.80 m

Lebar trotoar = 1.00 m= 5 gelagar

= 20.0 cm= 5.0 cm

Tebal lapisan aspal = 5.0 cmTebal trotoar = 20.0 cmTebal genangan air hujan = 5.0 cm

Berat sendiri beton = 2.50

Berat sendiri aspal = 2.30

Berat sendiri trotoar = 2.20

Berat tiang + sandaran = 0.50 t/mBerat sendiri air = 1.00 t/m

= 25.0 MPa

= 400.0 MPa

= 200000.0 MPa

= 296.0 kg/m

= 40.0 cm

= 83.8 cm

= 380.0

= 10966.0

= 462017.0

Direncanakan suatu jembatan komposit dengan panjang bentang jembatan L = 15,00 m, dimana potongan melintang

Panjang jembatan (L)Lebar lantai kendaran (B)Jarak antar gelagar (s)Lebar kantilever (c')

Jumlah gelagar baja (Ng)

Tebal lantai beton (d)Tinggi voute (t)

t/m3

t/m3

t/m3

Kuat tekan beton (fc')

Tegangan leleh baja (fy)

Modulus elsatisitas baja (Es)

Profil Baja Wide Flange (W 838 × 296)ws

bs

hs

As cm2

Ws cm3

Is cm4

20,0 cm20,0 cm

1,0 m

0,8 m 0,8 m2,0 m 2,0 m2,0 m 2,0 m

1,0 m7,6 m


Gelagar Tengah - 3

ANALISA PEMBEBANAN

Berat sendiri lantai beton = 1.00000 t/m1.05625

= 0.05625 t/mBerat sendiri gelagar = 0.29600 t/m

0.35520Berat sendiri diafragma = 0.05920 t/m

= 1.41145 t/m

Berat sendiri aspal = 0.87400 t/mBerat sendiri trotoar = 0.88000 t/mBerat tiang + sandaran = 1.00000 t/mBerat air hujan = 0.38000 t/m

= 3.13400 t/m

Karena beban mati sekunder dipikul sama besar oleh setiap gelagar, maka besarnya beban mati sekunder

=

= 0.62680 t/m

q = 2.20 t/m

q' =dimana :

α = faktor distribusi= 1.0 (jika tidak ada gelagar melintang pada jembatan)

maka :q' =q' = 1.60 t/m

p = 12.00 ton

p' =dimana :

K = Koefisien kejut== 1.30769

maka :p' =p' = 11.41259 ton

BEBAN MATI (DEAD LOAD)

Beban Mati Primer (wDL

)

Berat sendiri voute

wDL

Beban Mati Sekunder (wSDL

)

w'SDL

(wSDL

) yang dipikul oleh gelagar tengah adalah :

wSDL

wSDL

/ Ng

wSDL

BEBAN HIDUP (LIVE LOAD)

Beban merata (q)

Beban merata (q) yang bekerja pada jembatan dengan panjang bentang 15,0 m adalah :(jika panjang bentang jembatan L < 30,0 m)

Besarnya beban merata (q) yang dipikul oleh setiap gelagar tengah adalah :(q / 2,75) × α × s

(q / 2,75) × α × s

Beban garis (p)Beban garis (p) yang bekerja pada jembatan adalah :

Besarnya beban garis (p) yang dipikul oleh setiap gelagar tengah adalah :(p / 2,75) × α × s × K

1 + [ 20 / (50 + L) ]

(p / 2,75) × α × s × K


Gelagar Tengah - 4

LEBAR EFEKTIF LANTAI BETON

Berdasarkan SPESIFIKASI AASHTO

dari nilai-nilai berikut :b = = 375.0 cmb = s = 200.0 cmb = = 240.0 cm

Jadi, lebar efektif lantai beton (b) untuk gelagar tengah berdasarkan spesifikasi AASHTO adalah :b = 200.0 cm

Berdasarkan PERATURAN BINA MARGA

b =dimana :

e = = 50.00 cma = = 75.00 cm

a/L = 0.05

= 1.0maka :

λ = a = 75.00 cm

Jadi, lebar efektif lantai beton (b) untuk gelagar tengah menurut peraturan BINA MARGA adalah :b =b = 200.0 cm

PERENCANAAN GELAGAR JEMBATAN SECARA ULTIMIT (ULTIMATE DESIGN)

a =a = 357.64706 mm

x =dimana :

= 0.85maka :

x =x = 420.76125 mm

berarti Asumsi Salah --> garis netral penampang komposit berada si daerah baja.

Lebar efektif lantai beton (b) untuk gelagar tengah berdasarkan spesifikasi AASHTO, adalah nilai terkecil

L / 4

12 d

Lebar efektif lantai beton (b) untuk gelagar tengah menurut peraturan BINA MARGA adalah :2 λ + e

bs + 2 t

(s - e) / 2

Untuk a / L = 0.05, dari tabel diperoleh :λ/a

2 λ + e

Perencanaan gelagar jembatan secara ultimit (ultimate design) bertujuan untuk mendapatkan momen kapasitasultimit (ultimate momen capacity, M

u) yang dapat dipikul oleh penampang gelagar jembatan.

Asumsikan Garis Netral Penampang Komposit berada di daerah Beton

l Tinggi Blok Tegangan Tekan Beton (a) :(A

s f

y) / (0,85 f

c' b)

l Lokasi garis netral penampang komposit (x) :a / β

1

β1

( jika kuat tekan beton, fc' < 30,0 MPa)

a / β1

Karena lokasi garis netral penampang (x = 420,76125 mm) lebih besar dari tebal lantai beton (d = 200,0 mm),


Gelagar Tengah - 5

x =dimana :

= 0

= = 0maka :

x =x = 652.80 mm

=

= 8500000.0 N

=

= 3350000.0 N

=dimana :

= = 820.40 mm

d" = = 444.00 mmmaka :

= 8460800000.00 N.mm

= 8460.80 kN.m

= 846.08 t.m

PERENCANAAN GELAGAR JEMBATAN SECARA ELASTIS (ELASTIC DESIGN)

PROPERTIES PENAMPANG

b = 200.0 cm

n =dimana :

= 200000.0 MPa

= = 23500.0 MPa

maka :

n =n = 8.510638 à n = 9

=

Untuk Garis Netral Penampang Komposit berada di daerah Baja

l Lokasi garis netral penampang komposit terhadap serat atas beton (x) :[ ε

cu / (ε

cu + ε

s) ] (d + h

s)

εcu

εs

fy / E

s

[ εcu

/ (εcu

+ εs) ] (d + t + h

s)

l Gaya Tekan Ultimit Beton (Cc) :

Cc

0,85 fc' b d

Cc

l Gaya Tarik Ultimit Baja (Cs) :

Cs

0.5 (As f

y - C

c)

Cs

l Momen Kapasitas Ultimit Penampang Komposit (Mu) :

Mu

Cc d' + C

s d"

d' 0.5 (hs + 3t + x)

0.5 (hs + t)

Mu

Mu

Mu

Perencanaan gelagar jembatan secara elastis (elastic design) bertujuan untuk mendapatkan teganganyang terjadi pada penampang komposit.

l Lebar efektif Lantai Beton (b) :

l Modulus Rasio (n) :E

s / E

c

Es

Ec

4700 (fc')0,5

Es / E

c

l Lokasi Garis Netral Penampang Komposit :

Asumsikan Garis Netral Penampang Kmposit berada di daerah Bajay

c[ A

c (d/2) + A

s (d + t + h

s/2) ] / [A

c + A

s]


Gelagar Tengah - 6

dimana :

= = 444.44maka :

=

= 36.22615 cm

Selanjutnya dapat dihitung :

= = 72.57385 cm

= = 26.22615 cm

= = 30.67385 cm

=dimana :

= 14814.81

= 305693.64913

= 462017.00000

= 357536.43173maka :

=

= 1140061.89568

TEGANGAN PADA PENAMPANG KOMPOSIT

1. Tanpa Tumpuan Sementara (Unshoring)

Sebelum Komposit :

= à = 39.69703 t.m

Setelah Komposit :

= à = 17.62875 t.m

= à = 87.79720 t.m

=dimana :

= 0.0 = 0.00000

= = 6.22404

= = 30.99787maka :

= <

= 37.22192 < 112.50 ……… OK !!

=

Ac (b/n) d cm2

yc

[ Ac (d/2) + A

s (d + t + h

s/2) ] / [A

c + A

s]

yc

Karena yc > d, berarti asumsi benar à garis netral penampang komposit berada di daerah baja

ys

(d + t + hs) - y

c

dc

yc - 0.5 d

ds

ys - 0.5 h

s

l Momen Inersia Total Penampang Komposit (It) :

It

Ic + A

c d

c2 + I

s + A

sd

s2

Ic 1/12 (b/n) d3 = cm4

Ac d

c2

cm4

Is cm4

Asd

s2

cm4

It

Ic + A

c d

c2 + I

s + A

sd

s2

It cm4

l Momen Maksimum pada Gelagar Jembatan (Momen di Tengah Bentang Jembatan)

M1

1/8 wDL

L2 M1

M2

1/8 wSDL

L2 M2

M3 1/8 q' L2 + 1/4 p' L M

3

l Tegangan pada Penampang Komposit di Tengah Bentang Jembatan

Tegangan pada serat atas beton (fc-a

) :

fc-a

fc-a1

+ fc-a2

+ fc-a3

fc-a1 kg/cm2

fc-a2 (M2 yc) / (n It) kg/cm2

fc-a3

(M3 y

c) / (n I

t) kg/cm2

fc-a

fc-a1

+ fc-a2

+ fc-a3

0.45 fc'

fc-a kg/cm2 kg/cm2

Tegangan pada serat bawah beton (fc-b

) :

fc-b

fc-b1

+ fc-b2

+ fc-b3


Gelagar Tengah - 7

dimana :

= 0.0 = 0.00000

= = 2.78783

= = 13.88434maka :

= <

= 16.67216 < 112.50 ……… OK !!

=dimana :

= = 362

= = 17.35896

= = 86.45357maka :

= <

= 465.81354 < 2666.67 ……… OK !!

=dimana :

= = 362

= = 112.22078

= = 558.89785maka :

= <

= 1033.11964 < 2666.67 ……… OK !!

2. Dengan Tumpuan Sementara Sepanjang Bentang (Full Shoring)

Sebelum Komposit :

= 0 à = 0.00000 t.m

Setelah Komposit :

= à = 39.69703 t.m

= à = 17.62875 t.m

= à = 87.79720 t.m

=dimana :

= 0.0 = 0.00000

= 14.01552

= = 6.22404

= = 30.99787

fc-b1 kg/cm2

fc-b2

[ M2 (y

c - d) ] / (n I

t) kg/cm2

fc-b3 [ M3 (yc - d) ] / (n It) kg/cm2

fc-b

fc-b1

+ fc-b2

+ fc-b3

0.45 fc'

fc-b kg/cm2 kg/cm2

Tegangan pada serat atas baja (fs-a

) :

fs-a

fs-a1

+ fs-a2

+ fs-a3

fs-a1

M1 / W

s kg/cm2

fs-a2

[ M2 (h

s - y

s) ] / I

t kg/cm2

fs-a3

[ M3 (h

s - y

s) ] / I

t kg/cm2

fs-a

fs-a1

+ fs-a2

+ fs-a3

fy / 1.5

fs-a kg/cm2 kg/cm2

Tegangan pada serat bawah baja (fs-b

) :

fs-b

fs-b1

+ fs-b2

+ fs-b3

fs-b1

M1 / W

s kg/cm2

fs-b2

(M2 y

s) / I

t kg/cm2

fs-b3

(M3 y

s) / I

t kg/cm2

fs-b

fs-b1

+ fs-b2

+ fs-b3

fy / 1.5

fs-b kg/cm2 kg/cm2


M1

M1

M2

1/8 wDL

L2 M2

M3

1/8 wSDL

L2 M3

M4 1/8 q' L2 + 1/4 p' L M

4



) :

fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

fc-a1 kg/cm2

fc-a2

(M2 y

c) / (n I

t) kg/cm2

fc-a3

(M3 y

c) / (n I

t) kg/cm2

fc-a4

(M4 y

c) / (n I

t) kg/cm2


Gelagar Tengah - 8

maka :

= <

= 51.23744 < 112.50 ……… OK !!

=dimana :

= 0.0 = 0.00000

= = 6.27773

= = 2.78783

= = 13.88434maka :

= <

= 22.94989 < 112.50 ……… OK !!

=dimana :

= 0.0 = 0.00000

= = 39.08951

= = 17.35896

= = 86.45357maka :

= <

= 142.90204 < 2666.67 ……… OK !!

=dimana :

= 0.0 = 0.00000

= = 252.70265

= = 112.22078

= = 558.89785maka :

= <

= 923.82127 < 2666.67 ……… OK !!

3. Dengan Satu Tumpuan Sementara di Tengah Bentang (Partial Shoring)

Sebelum Komposit :

= à = -9.92426 t.m

Setelah Komposit :

= à = 49.62129 t.m

= à = 17.62875 t.m

fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

0.45 fc'

fc-a kg/cm2 kg/cm2


) :

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

fc-b1 kg/cm2

fc-b2

[ M2 (y

c - d) ] / (n I

t) kg/cm2

fc-b3

[ M3 (y

c - d) ] / (n I

t) kg/cm2

fc-b4

[ M4 (y

c - d) ] / (n I

t) kg/cm2

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

0.45 fc'

fc-b kg/cm2 kg/cm2


) :

fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

fs-a1 kg/cm2

fs-a2 [ M2 (hs - ys) ] / It kg/cm2

fs-a3

[ M3 (h

s - y

s) ] / I

t kg/cm2

fs-a4

[ M4 (h

s - y

s) ] / I

t kg/cm2

fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

fy / 1.5

fs-a kg/cm2 kg/cm2


) :

fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

fs-b1 kg/cm2

fs-b2

(M2 y

s) / I

t kg/cm2

fs-b3

(M3 y

s) / I

t kg/cm2

fs-b4 (M4 ys) / It kg/cm2

fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

fy / 1.5

fs-b kg/cm2 kg/cm2


M1

1/32 wDL

L2 M1

M2

5/32 wDL

L2 M2

M3

1/8 wSDL

L2 M3


Gelagar Tengah - 9

= à = 87.79720 t.m

=dimana :

= 0.0 = 0.00000

= 17.51940

= = 6.22404

= = 30.99787maka :

= <

= 54.74132 < 112.50 ……… OK !!

=dimana :

= 0.0 = 0.00000

= = 7.84716

= = 2.78783

= = 13.88434maka :

= <

= 24.51932 < 112.50 ……… OK !!

=dimana :

= = -90.50025

= = 48.86189

= = 17.35896

= = 86.45357maka :

= <

= 62.17417 < 2666.67 ……… OK !!

=dimana :

= = -90.50025

= = 315.87831

= = 112.22078

= = 558.89785maka :

= <

= 896.49668 < 2666.67 ……… OK !!

M4 1/8 q' L2 + 1/4 p' L M

4



) :

fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

fc-a1 kg/cm2

fc-a2

(M2 y

c) / (n I

t) kg/cm2

fc-a3

(M3 y

c) / (n I

t) kg/cm2

fc-a4

(M4 y

c) / (n I

t) kg/cm2

fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

0.45 fc'

fc-a kg/cm2 kg/cm2


) :

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

fc-b1 kg/cm2

fc-b2

[ M2 (y

c - d) ] / (n I

t) kg/cm2

fc-b3

[ M3 (y

c - d) ] / (n I

t) kg/cm2

fc-b4

[ M4 (y

c - d) ] / (n I

t) kg/cm2

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

0.45 fc'

fc-b kg/cm2 kg/cm2


) :

fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

fs-a1

M1 / W

s kg/cm2


fs-a3

[ M3 (h

s - y

s) ] / I

t kg/cm2

fs-a4

[ M4 (h

s - y

s) ] / I

t kg/cm2

fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

fy / 1.5

fs-a kg/cm2 kg/cm2


) :

fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

fs-b1

M1 / W

s kg/cm2

fs-b2

(M2 y

s) / I

t kg/cm2


fs-b4

(M4 y

s) / I

t kg/cm2

fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

fy / 1.5

fs-b kg/cm2 kg/cm2


Gelagar Tengah - 10

4. Dengan Dua Tumpuan Sementara di titik D dan titik E (Partial Shoring)

Sebelum Komposit :

= à = -3.52863 t.m

Setelah Komposit :

= à = 38.81488 t.m

= à = 15.67000 t.m

= à = 68.53147 t.m

=dimana :

= 0.0 = 0.00000

= 13.70406

= = 5.53248

= = 24.19587maka :

= <

= 43.43242 < 112.50 ……… OK !!

=dimana :

= 0.0 = 0.00000

= = 6.13822

= = 2.47807

= = 10.83763maka :

= <

= 19.45393 < 112.50 ……… OK !!

=dimana :

= = -32.17787

= = 38.22086

= = 15.43019

= = 67.48267maka :

= <

= 88.95585 < 2666.67 ……… OK !!

Tegangan di titik D dan titik E

l Momen pada Gelagar Jembatan di titik D dan titik E

M1

1/90 wDL

L2 M1

M2

11/90 wDL

L2 M2

M3 1/9 wSDL L2 M3

M4 1/9 q' L2 + 1/6 p' L M

4



) :

fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

fc-a1 kg/cm2

fc-a2

(M2 y

c) / (n I

t) kg/cm2


fc-a4

(M4 y

c) / (n I

t) kg/cm2

fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

0.45 fc'

fc-a kg/cm2 kg/cm2


) :

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

fc-b1 kg/cm2

fc-b2

[ M2 (y

c - d) ] / (n I

t) kg/cm2

fc-b3

[ M3 (y

c - d) ] / (n I

t) kg/cm2

fc-b4

[ M4 (y

c - d) ] / (n I

t) kg/cm2

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

0.45 fc'

fc-b kg/cm2 kg/cm2


) :

fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

fs-a1

M1 / W

s kg/cm2

fs-a2

[ M2 (h

s - y

s) ] / I

t kg/cm2

fs-a3

[ M3 (h

s - y

s) ] / I

t kg/cm2

fs-a4

[ M4 (h

s - y

s) ] / I

t kg/cm2

fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

fy / 1.5

fs-a kg/cm2 kg/cm2


Gelagar Tengah - 11

=dimana :

= = -32.17787

= = 247.08703

= = 99.75180

= = 436.25639maka :

= <

= 750.91735 < 2666.67 ……… OK !!

Sebelum Komposit :

= à = 0.88216 t.m

Setelah Komposit :

= à = 38.81488 t.m

= à = 17.62875 t.m

= à = 87.79720 t.m

=dimana :

= 0.0 = 0.00000

= 13.70406

= = 6.22404

= = 30.99787maka :

= <

= 50.92598 < 112.50 ……… OK !!

=dimana :

= 0.0 = 0.00000

= = 6.13822

= = 2.78783

= = 13.88434maka :

= <

= 22.81039 < 112.50 ……… OK !!


) :

fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

fs-b1

M1 / W

s kg/cm2

fs-b2

(M2 y

s) / I

t kg/cm2

fs-b3

(M3 y

s) / I

t kg/cm2

fs-b4

(M4 y

s) / I

t kg/cm2

fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

fy / 1.5

fs-b kg/cm2 kg/cm2

Tegangan di titik C (titik di tengah bentang balok)


M1

1/360 wDL

L2 M1

M2 11/90 wDL L2 M2

M3

1/8 wSDL

L2 M3

M4 1/8 q' L2 + 1/4 p' L M

4



) :

fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

fc-a1 kg/cm2

fc-a2

(M2 y

c) / (n I

t) kg/cm2


fc-a4

(M4 y

c) / (n I

t) kg/cm2

fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

0.45 fc'

fc-a kg/cm2 kg/cm2


) :

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

fc-b1 kg/cm2

fc-b2

[ M2 (y

c - d) ] / (n I

t) kg/cm2

fc-b3

[ M3 (y

c - d) ] / (n I

t) kg/cm2

fc-b4 [ M4 (yc - d) ] / (n It) kg/cm2

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

0.45 fc'

fc-b kg/cm2 kg/cm2


) :


Gelagar Tengah - 12

=dimana :

= = 8.04447

= = 38.22086

= = 17.35896

= = 86.45357maka :

= <

= 150.07785 < 2666.67 ……… OK !!

=dimana :

= = 8.04447

= = 247.08703

= = 112.22078

= = 558.89785maka :

= <

= 926.25013 < 2666.67 ……… OK !!

5. Pengecoran Bertahap

Sebelum Komposit :

= à = 8.88000 t.m

= à = 13.20313 t.m

= à = 13.20313 t.m

Setelah Komposit :

= à = 15.67000 t.m

= à = 68.53147 t.m

=dimana :

= 0.00 = 0.00000

= 0.00 = 0.00000

= 0.00 = 0.00000

= = 5.53248

= = 24.19587maka :

= <

fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

fs-a1

M1 / W

s kg/cm2

fs-a2

[ M2 (h

s - y

s) ] / I

t kg/cm2

fs-a3

[ M3 (h

s - y

s) ] / I

t kg/cm2

fs-a4

[ M4 (h

s - y

s) ] / I

t kg/cm2

fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

fy / 1.5

fs-a kg/cm2 kg/cm2


) :

fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

fs-b1

M1 / W

s kg/cm2

fs-b2

(M2 y

s) / I

t kg/cm2

fs-b3

(M3 y

s) / I

t kg/cm2

fs-b4

(M4 y

s) / I

t kg/cm2

fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

fy / 1.5

fs-b kg/cm2 kg/cm2



M1

1/9 ws L2 M

1

M2

1/18 wc L2 M

2

M3

1/18 wc L2 M

3

M4 1/9 wSDL L2 M4

M5 1/9 q' L2 + 1/6 p' L M

5

l Tegangan pada Penampang Komposit pada titik D dan titik E


) :

fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

+ fc-a5

fc-a1 kg/cm2

fc-a2 kg/cm2

fc-a3 kg/cm2

fc-a4

(M3 y

c) / (n I

t) kg/cm2

fc-a5

(M4 y

c) / (n I

t) kg/cm2

fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

+ fc-a5

0.45 fc'


Gelagar Tengah - 13

= 29.72835 < 112.50 ……… OK !!

=dimana :

= 0.0 = 0.00000

= 0.0 = 0.00000

= 0.0 = 0.00000

= = 2.47807

= = 10.83763maka :

= <

= 13.31570 < 112.50 ……… OK !!

=dimana :

= = 80.97757

= = 120.40056

= = 120.40056

= = 15.43019

= = 67.48267maka :

= <

= 404.69154 < 2666.67 ……… OK !!

=dimana :

= = 80.97757

= = 120.40056

= = 120.40056

= = 99.75180

= = 436.25639maka :

= <

= 857.78687 < 2666.67 ……… OK !!

Sebelum Komposit :

= à = 9.99000 t.m

= à = 16.50391 t.m

Setelah Komposit :

= à = 13.20313 t.m

fc-a kg/cm2 kg/cm2


) :

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

+ fc-b5

fc-b1 kg/cm2

fc-b2 kg/cm2

fc-b3 kg/cm2

fc-b4

[ M4 (y

c - d) ] / (n I

t) kg/cm2

fc-b5

[ M5 (y

c - d) ] / (n I

t) kg/cm2

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

+ fc-b5

0.45 fc'

fc-b kg/cm2 kg/cm2


) :

fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

+ fs-a5

fs-a1

M1 / W

s kg/cm2

fs-a2

M2 / W

s kg/cm2

fs-a3

M3 / W

s kg/cm2

fs-a4

[ M4 (h

s - y

s) ] / I

t kg/cm2

fs-a5

[ M5 (h

s - y

s) ] / I

t kg/cm2

fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

+ fs-a5

fy / 1.5

fs-a kg/cm2 kg/cm2


) :

fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

+ fs-b5

fs-b1

M1 / W

s kg/cm2

fs-b2

M2 / W

s kg/cm2

fs-b3

M3 / W

s kg/cm2

fs-b4

(M4 y

s) / I

t kg/cm2

fs-b5

(M5 y

s) / I

t kg/cm2

fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

+ fs-b5

fy / 1.5

fs-b kg/cm2 kg/cm2



M1 1/8 ws L2 M1

M2

5/72 wc L2 M

2

M3

1/18 wDL

L2 M3


Gelagar Tengah - 14

= à = 17.62875 t.m

= à = 87.79720 t.m

=dimana :

= 0.0 = 0.00000

= 0.0 = 0.00000

= 4.66152

= = 6.22404

= = 30.99787maka :

= <

= 41.88344 < 112.50 ……… OK !!

=dimana :

= 0.0 = 0.00000

= 0.0 = 0.00000

= = 2.08796

= = 2.78783

= = 13.88434maka :

= <

= 18.76012 < 112.50 ……… OK !!

=dimana :

= = 91.09976

= = 150.50070

= = 13.00107

= = 17.35896

= = 86.45357maka :

= <

= 358.41405 < 2666.67 ……… OK !!

=dimana :

= = 91.09976

= = 150.50070

M4

1/8 wSDL

L2 M4

M5 1/8 q' L2 + 1/4 p' L M

5



) :

fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

+ fc-a5

fc-a1 kg/cm2

fc-a2 kg/cm2

fc-a3

(M2 y

c) / (n I

t) kg/cm2

fc-a4

(M3 y

c) / (n I

t) kg/cm2

fc-a5

(M4 y

c) / (n I

t) kg/cm2

fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

+ fc-a5

0.45 fc'

fc-a kg/cm2 kg/cm2


) :

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

+ fc-b5

fc-b1 kg/cm2

fc-b2 kg/cm2

fc-b3

[ M2 (y

c - d) ] / (n I

t) kg/cm2

fc-b4

[ M3 (y

c - d) ] / (n I

t) kg/cm2

fc-b5

[ M4 (y

c - d) ] / (n I

t) kg/cm2

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

+ fc-b5

0.45 fc'

fc-b kg/cm2 kg/cm2


) :

fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

+ fs-a5

fs-a1

M1 / W

s kg/cm2

fs-a2

M2 / W

s kg/cm2

fs-a3

[ M2 (h

s - y

s) ] / I

t kg/cm2

fs-a4

[ M3 (h

s - y

s) ] / I

t kg/cm2

fs-a5

[ M4 (h

s - y

s) ] / I

t kg/cm2

fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

+ fs-a5

fy / 1.5

fs-a kg/cm2 kg/cm2


) :

fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

+ fs-b5

fs-b1

M1 / W

s kg/cm2

fs-b2

M2 / W

s kg/cm2


Gelagar Tengah - 15

= = 84.04821

= = 112.22078

= = 558.89785maka :

= <

= 996.76730 < 2666.67 ……… OK !!

pada gelagar komposit, tergantung dari metoda pelaksanaan yang digunakan.

A. TANPA TUMPUAN SEMENTARA

Gaya Lintang akibat Beban MatiSebelum Komposit

= ; = 0.00 ton= 10.59 ton = 0.00 kg= 10585.875 kg

Setelah Komposit

= ; = 0.00 ton= 4.70100 ton = 0.00 kg= 4701.000 kg

Gaya Lintang akibat Beban Hidup

= == 23.41259 ton = 11.70629 ton= 23412.587 kg = 11706.294 kg

== 33406.52 kg

== 11706.29 kg

Diameter paku = 0.625 in = 1.58750 cmTinggi paku = 2.500 in = 6.35000 cm

Karena perbandingan tinggi paku dengan diameter paku :H/d = 4.0 < 5.5

Q == 1593.89 kg

fs-b3

(M2 y

s) / I

t kg/cm2

fs-b4

(M3 y

s) / I

t kg/cm2


fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

+ fs-b5

fy / 1.5

fs-b kg/cm2 kg/cm2

PERENCANAAN PENGHUBUNG GESER (SHEAR CONNECTOR)

Perhitungan gaya lintang maksimum di tumpuan (Dmax

) dan gaya lintang maksimum tengah bentang (Dmin

)

Gaya Lintang Maksimum di Tumpuan (Dmax

) dan di Tengah Bentang (Dmin

) pada Gelagar Jembatan

DA1

0,5 wDL

L DC1

DA2

0,5 wSDL

L DC2

DA3 0,5 q' L + p' D

C3 0,25 q' L + 0,5 p'



)

Dmax

50% DA1

+ 100% (DA2

+ DA3

)

Dmin

50% DC1

+ 100% (DC2

+ DC3

)

Kekuatan shear connector (Q)

Shear connector yang digunakan adalah Stud ⅝″ × 2½″, dimana :: d: H

maka, kekuatan sebuah Stud Connector (Q) adalah :

10 H d (fc')0.5


Gelagar Tengah - 16

Q == 4781.66 kg

q =

dimana :

=

= 11656.06469

= 1140061.89568

maka :

== 341.55042 kg/cm

== 119.68588 kg/cm

<< 13.99987 cm

Untuk interval 0,00 m - 2,00 m

= 10.0 cm ; = 2.00 m


= 15.0 cm ; = 2.00 mUntuk interval 4,00 m - 6,00 m


= 25.0 cm ; = 1.50 m


= 10.0 cm

= 2.0 m

= >

= 478.16603 kg/cm > 341.55042 …… OK !!


= 15.0 cm

= 2.0 m

== 27619.7966 kg

= >

Jika direncanakan dipasang tiga baris shear connector pada setiap penaapang gelagar, maka :3 Q

Longitudinal Shear (q) :(D × S

x) / I

t

Sx

[ (b/n) × d ] × dc

cm3

It cm4

Gaya geser longitudinal maksimum (qmax

) :

qmax

(Dmax

× Sx) / I

t

Gaya geser longitudinal minimum (qmin

) :

qmin

(Dmin

× Sx) / I

t

Jarak antara Shear Connector (l) :

Jarak antara Shear Connector di tumpuan (l1) :

l1

Q / qmax

Direncanakan jarak antara shear cinnector dibagi atas 4 (empat) interval jarak, dimana :

l1

L1

l2

L2

l3

L3

l4

L4

Kontrol jarak antar shear connector

l1

L1

q'1

Q / l1

qmax

l2

L2

D2

Dmin

+ [ [ (L/2 - L1) / (L/2) ] (D

max - D

min) ]

q'2

Q / l2

q2


Gelagar Tengah - 17

= 318.77735 kg/cm > 282.38654 …… OK !!


= 20.0 cm

= 2.0 m

== 21833.0683 kg

= >= 239.08301 kg/cm > 223.22267 …… OK !!


= 25.0 cm

= 1.5 m

== 16046.3399 kg

= >= 191.26641 kg/cm > 164.05879 …… OK !!

B. DENGAN TUMPUAN SEMENTARA SEPANJANG BENTANG

Gaya Lintang akibat Beban MatiSebelum Komposit

= 0.000 ton ; = 0.00 ton= 0.000 kg = 0.00 kg

Setelah Komposit

= ; = 0.00 ton= 15.286875 ton = 0.00 kg= 15286.875 kg

Gaya Lintang akibat Beban Hidup

= == 23.41259 ton = 11.70629 ton= 23412.587 kg = 11706.294 kg

== 38699.46 kg

== 11706.29 kg

Diameter paku = 0.625 in = 1.58750 cmTinggi paku = 2.500 in = 6.35000 cm

Karena perbandingan tinggi paku dengan diameter paku :H/d = 4.0 < 5.5

l3

L3

D3

Dmin

+ [ [ (L/2 - (L1 + L

2)) / (L/2) ] (D

max - D

min) ]

q'3

Q / l3

q3

l4

L4

D4

Dmin

+ [ [ (L/2 - (L1 + L

2 + L

3)) / (L/2) ] (D

max - D

min) ]

q'4

Q / l4

q4



) pada Gelagar Jembatan

DA1

DC1

DA2

0,5 (wDL

+ wSDL

) L DC2

DA3 0,5 q' L + p' D

C3 0,25 q' L + 0,5 p'



)

Dmax

50% DA1

+ 100% (DA2

+ DA3

)

Dmin

50% DC1

+ 100% (DC2

+ DC3

)

Kekuatan shear connector (Q)

Shear connector yang digunakan adalah Stud ⅝″ × 2½″, dimana :: d: H


Gelagar Tengah - 18

Q == 1593.89 kg

Q == 4781.66 kg

q =

dimana :

=

= 11656.06469

= 1140061.89568

maka :

== 395.66574 kg/cm

== 119.68588 kg/cm

<< 12.08510 cm


= 10.0 cm ; = 2.50 m



= 20.0 cm ; = 2.50 m


= 10.0 cm

= 2.5 m

= >

= 478.16603 kg/cm > 395.66574 …… OK !!


= 15.0 cm

= 2.5 m

=

maka, kekuatan sebuah Stud Connector (Q) adalah :

10 H d (fc')0.5

Jika direncanakan dipasang tiga baris shear connector pada setiap penaapang gelagar, maka :3 Q

Longitudinal Shear (q) :(D × S

x) / I

t

Sx

[ (b/n) × d ] × dc

cm3

It cm4

Gaya geser longitudinal maksimum (qmax

) :

qmax

(Dmax

× Sx) / I

t

Gaya geser longitudinal minimum (qmin

) :

qmin

(Dmin

× Sx) / I

t

Jarak antara Shear Connector (l) :

Jarak antara Shear Connector di tumpuan (l1) :

l1

Q / qmax

Direncanakan jarak antara shear cinnector dibagi atas 3 (tiga) interval jarak, dimana :

l1

L1

l2

L2

l3

L3

Kontrol jarak antar shear connector

l1

L1

q'1

Q / l1

qmax

l2

L2

D2

Dmin

+ [ [ (L/2 - L1) / (L/2) ] (D

max - D

min) ]


Gelagar Tengah - 19

= 29701.7395 kg

= >= 318.77735 kg/cm > 303.67246 …… OK !!


= 20.0 cm

= 2.5 m

== 20704.0166 kg

= >= 239.08301 kg/cm > 211.67917 …… OK !!

q'2

Q / l2

q2

l3

L3

D3

Dmin

+ [ [ (L/2 - (L1 + L

2)) / (L/2) ] (D

max - D

min) ]

q'3

Q / l3

q3


Gelagar Tengah - 20


Gelagar Tengah - 21


Gelagar Tengah - 22


Gelagar Tengah - 23


Gelagar Tengah - 24


Gelagar Tengah - 25


Gelagar Tengah - 26


Gelagar Tengah - 27


Gelagar Tengah - 28


Gelagar Tengah - 29


Gelagar Tengah - 30


Gelagar Tengah - 31


Gelagar Tengah - 32


Gelagar Tengah - 33


Gelagar Tengah - 34

33406.52

-2893.36

x

0.0 33406.52491

1.0 30513.160752.0 27619.79659

3.0 24726.43243

Di = ax + b

b = [ Dmax

]

a = [ {(Dmin

- Dmax

) / (L/2)} + Dmax

]

Di


Gelagar Tengah - 35

4.0 21833.068275.0 18939.704116.0 16046.33995

7.5 11706.29371


Gelagar Tengah - 36

38699.46

-3599.09

x

0.0 38699.46241

1.0 35100.37325

Di = ax + b

b = [ Dmax

]

a = [ (Dmin

- Dmax

) / (L/2) ]

Di


Gelagar Tengah - 37

2.0 31501.28409

2.5 29701.739513.0 27902.194934.0 24303.105775.0 20704.01661

6.0 17104.92745

7.5 11706.29371

Perencanaan Gelagar Tengah

38

RESUME HASIL PERHITUNGAN TEGANGANPADA PENAMPANG KOMPOSIT UNTUK GELAGAR TENGAH

ANALISA PEMBEBANAN= 1.41145 t/m ; q' = 1.60000 t/m

= 0.62680 t/m ; p' = 11.41259 ton

= 1.05625 t/m ; = 0.35520 t/m

PROPERTIES PENAMPANGn = 9 ;L = 15.00 m d = 20.0 cm

= 36.22615 cm ; = 83.8 cm

= 72.57385 cm ; = 10966.0

= 1140061.89568 ; = 462017.0


Sebelum Sesudah Komposit TEGANGAN Tegangan1 2 3 4 5 TOTAL Izin

Unshoring

MOMEN (t.m) 39.69703 17.62875 87.79720

TEGANGAN BetonAtas 0.00000 6.22404 30.99787 37.22192 112.50

Bawah 0.00000 2.78783 13.88434 16.67216 112.50

BajaAtas 362.00101 17.35896 86.45357 465.81354 2666.67

Bawah 362.00101 112.22078 558.89785 1033.11964 2666.67

Full Shoring

MOMEN (t.m) 0.00000 39.69703 17.62875 87.79720

TEGANGAN BetonAtas 0.00000 14.01552 6.22404 30.99787 51.23744 112.50

Bawah 0.00000 6.27773 2.78783 13.88434 22.94989 112.50

BajaAtas 0.00000 39.08951 17.35896 86.45357 142.90204 2666.67

Bawah 0.00000 252.70265 112.22078 558.89785 923.82127 2666.67

wDL

wSDL

wc

ws

yc hs

ys

Ws cm3

It cm4 I

s cm4

(kg/cm2)

(kg/cm2)


39

MOMEN (t.m) -9.92426 49.62129 17.62875 87.79720


Bawah 0.00000 7.84716 2.78783 13.88434 24.51932 112.50

BajaAtas -90.50025 48.86189 17.35896 86.45357 62.17417 2666.67

Bawah -90.50025 315.87831 112.22078 558.89785 896.49668 2666.67

MOMEN (t.m) -3.52863 38.81488 15.67000 68.53147


Bawah 0.00000 6.13822 2.47807 10.83763 19.45393 112.50

BajaAtas -32.17787 38.22086 15.43019 67.48267 88.95585 2666.67

Bawah -32.17787 247.08703 99.75180 436.25639 750.91735 2666.67

MOMEN (t.m) 0.88216 38.81488 17.62875 87.79720


Bawah 0.00000 6.13822 2.78783 13.88434 22.81039 112.50

BajaAtas 8.04447 38.22086 17.35896 86.45357 150.07785 2666.67

Bawah 8.04447 247.08703 112.22078 558.89785 926.25013 2666.67

Sebelum Komposit Sesudah Komposit1 2 3 4 5

MOMEN (t.m) 8.88000 13.20313 13.20313 15.67000 68.53147

TEGANGAN BetonAtas 0.00000 0.00000 0.00000 5.53248 24.19587 29.72835 112.50

Bawah 0.00000 0.00000 0.00000 2.47807 10.83763 13.31570 112.50

BajaAtas 80.97757 120.40056 120.40056 15.43019 67.48267 404.69154 2666.67

Bawah 80.97757 120.40056 120.40056 99.75180 436.25639 857.78687 2666.67

Satu Tumpuan Sementara

(kg/cm2)

Dua Tumpuan Sementara (titik D dan titik E) (kg/cm2)

Dua Tumpuan Sementara (titik C)

(kg/cm2)

Pengecoran Bertahap (titik D dan titik E)

(kg/cm2)


40


MOMEN (t.m) 9.99000 16.50391 13.20313 17.62875 87.79720


Bawah 0.00000 0.00000 2.08796 2.78783 13.88434 18.76012 112.50

BajaAtas 91.09976 150.50070 13.00107 17.35896 86.45357 358.41405 2666.67

Bawah 91.09976 150.50070 84.04821 112.22078 558.89785 996.76730 2666.67

Pengecoran Bertahap (titik C)

(kg/cm2)


41

………. OK !!………. OK !!

………. OK !!

………. OK !!

………. OK !!………. OK !!

………. OK !!

………. OK !!


42

………. OK !!………. OK !!

………. OK !!

………. OK !!

………. OK !!………. OK !!

………. OK !!

………. OK !!

………. OK !!………. OK !!

………. OK !!

………. OK !!

………. OK !!………. OK !!

………. OK !!

………. OK !!


43

………. OK !!………. OK !!

………. OK !!

………. OK !!


Gelagar Pinggir - 81

PERENCANAAN GELAGAR JEMBATAN KOMPOSIT"GELAGAR PINGGIR"


DATA - DATA= 15.00 m= 7.60 m= 2.00 m= 0.80 m= 1.00 m= 5 gelagar

= 20.0 cm= 5.0 cm

Tebal lapisan aspal = 5.0 cmTebal trotoar = 20.0 cmTebal genangan air hujan = 5.0 cm

Berat sendiri beton = 2.50

Berat sendiri aspal = 2.30

Berat sendiri trotoar = 2.20

Berat tiang + sandaran = 0.50 t/mBerat sendiri air = 1.00 t/m

= 25.0 MPa

= 400.0 MPa

= 200000.0 MPa

= 296.0 kg/m

= 40.0 cm

= 83.8 cm

= 380.0

= 10966.0

= 462017.0


Panjang jembatan (L)Lebar lantai kendaran (B)Jarak antar gelagar (s)Lebar kantilever (c') Lebar trotoar (c")Jumlah gelagar baja (N

g)

Tebal lantai beton (d)Tinggi voute (t)

t/m3

t/m3

t/m3




Profil Baja Wide Flange (W 838 × 296)ws

bs

hs

As cm2

Ws cm3

Is cm4

20,0 cm20,0 cm

1,0 m

0,8 m 0,8 m2,0 m 2,0 m2,0 m 2,0 m

1,0 m7,6 m



ANALISA PEMBEBANAN

Berat sendiri lantai beton = 0.98000 t/m= 0.05625 t/m

Berat sendiri gelagar = 0.29600 t/mBerat sendiri diafragma = 0.05920 t/m

= 1.39145 t/m

Berat sendiri aspal = 0.87400 t/mBerat sendiri trotoar = 0.88000 t/mBerat tiang + sandaran = 1.00000 t/mBerat air hujan = 0.38000 t/m

= 3.13400 t/m

Karena beban mati sekunder dipikul sama besar oleh setiap gelagar, maka besarnya beban mati sekunder

=

= 0.62680 t/m

q = 2.20 t/m

q' =dimana :

s' == 0.90 m

maka :q' =q' = 0.72000 t/m

p = 12.00 ton

p' =dimana :

K == 1.30769

maka :p' =p' = 5.13566 ton

BEBAN MATI (DEAD LOAD)


)

Berat sendiri voute

wDL


)

w'SDL

(wSDL

) yang dipikul oleh gelagar tengah adalah :

wSDL

wSDL

/ Ng

wSDL

BEBAN HIDUP (LIVE LOAD)

Beban merata (q)

Beban merata (q) yang bekerja pada jembatan dengan panjang bentang 15,0 m adalah :(jika panjang bentang jembatan L < 30,0 m)

Besarnya beban merata (q) yang dipikul oleh setiap gelagar tengah adalah :(q / 2,75) × s'

1/2 (s + c' - c")

(q / 2,75) × s'

Beban garis (p)

Beban garis (p) yang bekerja pada jembatan adalah :

Besarnya beban garis (p) yang dipikul oleh setiap gelagar tengah adalah :(p / 2,75) × s' × K

1 + [ 20 / (50 + L) ]

(p / 2,75) × s' × K



= 500.00 = 0.50

Beban hidup pada trotoar yang diperhitungkan dalam analisa pembebanan gelagar pinggir adalah : q'' =q'' = 0.34500 t/m

LEBAR EFEKTIF LANTAI BETON

Berdasarkan SPESIFIKASI AASHTO

dari nilai-nilai berikut :b = = 125.0 cm

b = = 120.0 cmb = = 120.0 cm

Nilai terkecil dari nilai - nilai di atas adalah :b = 120.0 cm

Pada gelagar pinggir terdapat kantilever dengan panjang (c) :

c =c = 55.00 cm

b =b = 180.0 cm

Berdasarkan PERATURAN BINA MARGA

b =dimana :

e = = 50.00 cma = = 75.00 cm

a/L = 0.05

= 1.0maka :

λ = a = 75.00 cm

Pada gelagar pinggir terdapat kantilever dengan panjang (c) :

c =c = 55.00 cm

menurut peraturan BINA MARGA adalah :b =b = 180.0 cm

Beban hidup trotoar (q")

Beban hidup pada trotoar (wtr) :

wtr kg/cm2 t/m2

(1/s) (0.6 wtr c") (s + c' - (c"/2))

Lebar efektif lantai beton (b) untuk gelagar tengah berdasarkan spesifikasi AASHTO, adalah nilai terkecil

L / 121/2 (s + b

s)

6 d

c' - (bs/2 + t)

Jadi, lebar efektif lantai beton (b) untuk gelagar tengah berdasarkan spesifikasi AASHTO adalah :b + c + t

Lebar efektif lantai beton (b) untuk gelagar tengah menurut peraturan BINA MARGA adalah :2 λ + e

bs + 2 t

(s - e) / 2

Untuk a / L = 0.05, dari tabel diperoleh :λ/a

c' - (bs/2 + t)

Karena nilai λ lebih besar dari penjang kantilever, maka lebar efektif lantai beton (b) untuk gelagar tengah

λ + e + c



PERENCANAAN GELAGAR JEMBATAN SECARA ULTIMIT (ULTIMATE DESIGN)

a =a = 397.39 mm

x =dimana :

= 0.85maka :

x =x = 467.51 mm

berarti Asumsi Salah --> garis netral penampang komposit berada si daerah baja.

x =dimana :

= 0

= = 0maka :

x =x = 652.80 mm

=

= 7650000.0 N

=

= 3775000.0 N

=dimana :

= = 820.40 mm

d" = = 444.00 mmmaka :

= 7952160000.00 N.mm

= 7952.16 kN.m

= 795.22 t.m

Perencanaan gelagar jembatan secara ultimit (ultimate design) bertujuan untuk mendapatkan momen kapasitasultimit (ultimate momen capacity, M

u) yang dapat dipikul oleh penampang gelagar jembatan.

Asumsikan Garis Netral Penampang Komposit berada di daerah Beton

l Tinggi Blok Tegangan Tekan Beton (a) :(A

s f

y) / (0,85 f

c' b)

l Lokasi garis netral penampang komposit (x) :a / β

1

β1

( jika kuat tekan beton, fc' < 30,0 MPa)

a / β1

Karena lokasi garis netral penampang (x = 420,76125 mm) lebih besar dari tebal lantai beton (d = 200,0 mm),

Untuk Garis Netral Penampang Komposit berada di daerah Baja

l Lokasi garis netral penampang komposit terhadap serat atas beton (x) :[ ε

cu / (ε

cu + ε

s) ] (d + h

s)

εcu

εs

fy / E

s

[ εcu

/ (εcu

+ εs) ] (d + t + h

s)

l Gaya Tekan Ultimit Beton (Cc) :

Cc

0,85 fc' b d

Cc

l Gaya Tarik Ultimit Baja (Cs) :

Cs

0.5 (As f

y - C

c)

Cs

l Momen Kapasitas Ultimit Penampang Komposit (Mu) :

Mu

Cc d' + C

s d"

d' 0.5 (hs + 3t + x)

0.5 (hs + t)

Mu

Mu

Mu



PERENCANAAN GELAGAR JEMBATAN SECARA ELASTIS (ELASTIC DESIGN)

PROPERTIES PENAMPANG

b = 180.0 cm

n =dimana :

= 200000.0 MPa

= = 23500.0 MPa

maka :

n =n = 8.510638 à n = 9

=

dimana :

= = 400.00

maka :

=

= 37.72051 cm

Selanjutnya dapat dihitung :

= = 71.07949 cm

= = 27.72051 cm

= = 29.17949 cm

=

dimana :

= 13333.33

= 307370.73241

= 462017.00000

= 323548.13938

maka :

=

= 1106269.20513

Perencanaan gelagar jembatan secara elastis (elastic design) bertujuan untuk mendapatkan teganganyang terjadi pada penampang komposit.

l Lebar efektif Lantai Beton (b) :

l Modulus Rasio (n) :E

s / E

c

Es

Ec

4700 (fc')0,5

Es / E

c

l Lokasi Garis Netral Penampang Komposit :

Asumsikan Garis Netral Penampang Kmposit berada di daerah Bajay

c[ A

c (d/2) + A

s (d + t + h

s/2) ] / [A

c + A

s]

Ac (b/n) d cm2

yc

[ Ac (d/2) + A

s (d + t + h

s/2) ] / [A

c + A

s]

yc

Karena yc > d, berarti asumsi benar à garis netral penampang komposit berada di daerah baja

ys

(d + t + hs) - y

c

dc

yc - 0.5 d

ds

ys - 0.5 h

s

l Momen Inersia Total Penampang Komposit (It) :

It Ic + Ac dc2 + Is + Asds

2

Ic 1/12 (b/n) d3 = cm4

Ac d

c2

cm4

Is cm4

Asd

s2

cm4

It

Ic + A

c d

c2 + I

s + A

sd

s2

It cm4




1. Tanpa Tumpuan Sementara (Unshoring)

Sebelum Komposit :

= à = 39.13453 t.m

Setelah Komposit :

= à = 17.62875 t.m

= = 49.21187 t.m

=

dimana :

= 0.0 = 0.00000

= = 6.67876

= = 18.64421

maka :

= <

= 25.32297 < 112.50 ……… OK !!

=

dimana :

= 0.0 = 0.00000

= = 3.13758

= = 8.75876

maka :

= <

= 11.89634 < 112.50 ……… OK !!

=

dimana :

= = 356.87

= = 20.27054

= = 56.58660

maka :

= <

= 433.72867 < 2666.67 ……… OK !!


M1 1/8 wDL L2 M1

M2

1/8 wSDL

L2 M2

M3 1/8 (q' + q") L2 + 1/4 p' L M

3



) :

fc-a

fc-a1

+ fc-a2

+ fc-a3

fc-a1 kg/cm2


fc-a3

(M3 y

c) / (n I

t) kg/cm2

fc-a

fc-a1

+ fc-a2

+ fc-a3

0.45 fc'

fc-a kg/cm2 kg/cm2


) :

fc-b

fc-b1

+ fc-b2

+ fc-b3

fc-b1 kg/cm2

fc-b2

[ M2 (y

c - d) ] / (n I

t) kg/cm2

fc-b3

[ M3 (y

c - d) ] / (n I

t) kg/cm2

fc-b

fc-b1

+ fc-b2

+ fc-b3

0.45 fc'

fc-b kg/cm2 kg/cm2


) :

fs-a

fs-a1

+ fs-a2

+ fs-a3

fs-a1

M1 / W

s kg/cm2

fs-a2

[ M2 (h

s - y

s) ] / I

t kg/cm2

fs-a3

[ M3 (h

s - y

s) ] / I

t kg/cm2

fs-a

fs-a1

+ fs-a2

+ fs-a3

fy / 1.5

fs-a kg/cm2 kg/cm2



=

dimana :

= = 356.87

= = 113.26741

= = 316.19376

maka :

= <

= 786.33270 < 2666.67 ……… OK !!

2. Dengan Tumpuan Sementara Sepanjang Bentang (Full Shoring)

Sebelum Komposit :

= 0 à = 0.00000 t.m

Setelah Komposit :

= à = 39.13453 t.m

= à = 17.62875 t.m

= = 49.21187 t.m

=dimana :

= 0.0 = 0.00000

= 14.82635

= = 6.67876

= = 18.64421

maka :

= <

= 40.14933 < 112.50 ……… OK !!

=dimana :

= 0.0 = 0.00000

= = 6.96519

= = 3.13758

= = 8.75876

maka :

= <

= 18.86153 < 112.50 ……… OK !!


) :

fs-b

fs-b1

+ fs-b2

+ fs-b3

fs-b1

M1 / W

s kg/cm2

fs-b2

(M2 y

s) / I

t kg/cm2

fs-b3

(M3 y

s) / I

t kg/cm2

fs-b

fs-b1

+ fs-b2

+ fs-b3

fy / 1.5

fs-b kg/cm2 kg/cm2


M1

M1

M2

1/8 wDL

L2 M2

M3

1/8 wSDL

L2 M3

M4 1/8 (q' + q") L2 + 1/4 p' L M

4



) :

fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

fc-a1 kg/cm2

fc-a2

(M2 y

c) / (n I

t) kg/cm2

fc-a3

(M3 y

c) / (n I

t) kg/cm2

fc-a4

(M4 y

c) / (n I

t) kg/cm2

fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

0.45 fc'

fc-a kg/cm2 kg/cm2


) :

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

fc-b1 kg/cm2

fc-b2

[ M2 (y

c - d) ] / (n I

t) kg/cm2

fc-b3

[ M3 (y

c - d) ] / (n I

t) kg/cm2

fc-b4

[ M4 (y

c - d) ] / (n I

t) kg/cm2

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

0.45 fc'

fc-b kg/cm2 kg/cm2



=dimana :

= 0.0 = 0.00000

= = 44.99911

= = 20.27054

= = 56.58660maka :

= <

= 121.85625 < 2666.67 ……… OK !!

=dimana :

= 0.0 = 0.00000

= = 251.44534

= = 113.26741

= = 316.19376maka :

= <

= 680.90652 < 2666.67 ……… OK !!

3. Dengan Satu Tumpuan Sementara di Tengah Bentang (Partial Shoring)

Sebelum Komposit :

= à = -9.78363 t.m

Setelah Komposit :

= à = 48.91816 t.m

= à = 17.62875 t.m

= = 49.21187 t.m

=dimana :

= 0.0 = 0.00000

= 18.53294

= = 6.67876

= = 18.64421

maka :

= <

= 43.85592 < 112.50 ……… OK !!


) :

fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

fs-a1 kg/cm2

fs-a2

[ M2 (h

s - y

s) ] / I

t kg/cm2

fs-a3

[ M3 (h

s - y

s) ] / I

t kg/cm2

fs-a4

[ M4 (h

s - y

s) ] / I

t kg/cm2

fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

fy / 1.5

fs-a kg/cm2 kg/cm2


) :

fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

fs-b1 kg/cm2

fs-b2

(M2 y

s) / I

t kg/cm2

fs-b3

(M3 y

s) / I

t kg/cm2

fs-b4

(M4 y

s) / I

t kg/cm2

fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

fy / 1.5

fs-b kg/cm2 kg/cm2


M1

1/32 wDL

L2 M1

M2

5/32 wDL

L2 M2

M3

1/8 wSDL

L2 M3

M4 1/8 (q' + q") L2 + 1/4 p' L M4



) :

fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

fc-a1 kg/cm2

fc-a2

(M2 y

c) / (n I

t) kg/cm2

fc-a3

(M3 y

c) / (n I

t) kg/cm2


fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

0.45 fc'

fc-a kg/cm2 kg/cm2



=dimana :

= 0.0 = 0.00000

= = 8.70649

= = 3.13758

= = 8.75876maka :

= <

= 20.60283 < 112.50 ……… OK !!

=dimana :

= = -89.21788

= = 56.24889

= = 20.27054

= = 56.58660

maka :

= <

= 43.88815 < 2666.67 ……… OK !!

=dimana :

= = -89.21788

= = 314.30668

= = 113.26741

= = 316.19376

maka :

= <

= 654.54997 < 2666.67 ……… OK !!

4. Dengan Dua Tumpuan Sementara di titik D dan titik E (Partial Shoring)

Sebelum Komposit :

= à = -3.47863 t.m

Setelah Komposit :

= à = 38.26488 t.m

= à = 15.67000 t.m

= = 39.46416 t.m


) :

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

fc-b1 kg/cm2

fc-b2

[ M2 (y

c - d) ] / (n I

t) kg/cm2

fc-b3

[ M3 (y

c - d) ] / (n I

t) kg/cm2

fc-b4

[ M4 (y

c - d) ] / (n I

t) kg/cm2

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

0.45 fc'

fc-b kg/cm2 kg/cm2


) :

fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

fs-a1 M1 / Ws kg/cm2

fs-a2

[ M2 (h

s - y

s) ] / I

t kg/cm2

fs-a3

[ M3 (h

s - y

s) ] / I

t kg/cm2

fs-a4

[ M4 (h

s - y

s) ] / I

t kg/cm2

fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

fy / 1.5

fs-a kg/cm2 kg/cm2


) :

fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

fs-b1

M1 / W

s kg/cm2

fs-b2

(M2 y

s) / I

t kg/cm2


fs-b4

(M4 y

s) / I

t kg/cm2

fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

fy / 1.5

fs-b kg/cm2 kg/cm2



M1

1/90 wDL

L2 M1

M2

11/90 wDL

L2 M2

M3

1/9 wSDL

L2 M3

M4 1/9 (q' + q") L2 + 1/6 p' L M

4



=dimana :

= 0.0 = 0.00000

= 14.49688

= = 5.93667

= = 14.95124

maka :

= <

= 35.38479 < 112.50 ……… OK !!

=dimana :

= 0.0 = 0.00000

= = 6.81041

= = 2.78896

= = 7.02386

maka :

= <

= 16.62323 < 112.50 ……… OK !!

=dimana :

= = -31.72191

= = 43.99913

= = 18.01826

= = 45.37814

maka :

= <

= 75.67361 < 2666.67 ……… OK !!

=dimana :

= = -31.72191

= = 245.85767

= = 100.68214

= = 253.56326

maka :

= <

= 568.38117 < 2666.67 ……… OK !!



) :

fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

fc-a1 kg/cm2

fc-a2

(M2 y

c) / (n I

t) kg/cm2

fc-a3

(M3 y

c) / (n I

t) kg/cm2

fc-a4

(M4 y

c) / (n I

t) kg/cm2

fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

0.45 fc'

fc-a kg/cm2 kg/cm2


) :

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

fc-b1 kg/cm2

fc-b2

[ M2 (y

c - d) ] / (n I

t) kg/cm2

fc-b3

[ M3 (y

c - d) ] / (n I

t) kg/cm2

fc-b4

[ M4 (y

c - d) ] / (n I

t) kg/cm2

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

0.45 fc'

fc-b kg/cm2 kg/cm2


) :

fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

fs-a1

M1 / W

s kg/cm2

fs-a2

[ M2 (h

s - y

s) ] / I

t kg/cm2


fs-a4

[ M4 (h

s - y

s) ] / I

t kg/cm2

fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

fy / 1.5

fs-a kg/cm2 kg/cm2


) :

fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

fs-b1

M1 / W

s kg/cm2

fs-b2

(M2 y

s) / I

t kg/cm2

fs-b3

(M3 y

s) / I

t kg/cm2

fs-b4

(M4 y

s) / I

t kg/cm2

fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

fy / 1.5

fs-b kg/cm2 kg/cm2



Sebelum Komposit :

= à = 0.86966 t.m

Setelah Komposit :

= à = 38.26488 t.m

= à = 17.62875 t.m

= = 49.21187 t.m

=dimana :

= 0.0 = 0.00000

= 14.49688

= = 6.67876

= = 18.64421maka :

= <

= 39.81985 < 112.50 ……… OK !!

=dimana :

= 0.0 = 0.00000

= = 6.81041

= = 3.13758

= = 8.75876maka :

= <

= 18.70675 < 112.50 ……… OK !!

=dimana :

= = 7.93048

= = 43.99913

= = 20.27054

= = 56.58660maka :

= <

= 128.78675 < 2666.67 ……… OK !!



M1 1/360 wDL L2 M1

M2

11/90 wDL

L2 M2

M3

1/8 wSDL

L2 M3

M4 1/8 (q' + q") L2 + 1/4 p' L M

4



) :

fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

fc-a1 kg/cm2

fc-a2

(M2 y

c) / (n I

t) kg/cm2

fc-a3

(M3 y

c) / (n I

t) kg/cm2

fc-a4

(M4 y

c) / (n I

t) kg/cm2

fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

0.45 fc'

fc-a kg/cm2 kg/cm2


) :

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

fc-b1 kg/cm2

fc-b2

[ M2 (y

c - d) ] / (n I

t) kg/cm2

fc-b3 [ M3 (yc - d) ] / (n It) kg/cm2

fc-b4

[ M4 (y

c - d) ] / (n I

t) kg/cm2

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

0.45 fc'

fc-b kg/cm2 kg/cm2


) :

fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

fs-a1

M1 / W

s kg/cm2

fs-a2

[ M2 (h

s - y

s) ] / I

t kg/cm2

fs-a3

[ M3 (h

s - y

s) ] / I

t kg/cm2


fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

fy / 1.5

fs-a kg/cm2 kg/cm2


) :



=dimana :

= = 7.93048

= = 245.85767

= = 113.26741

= = 316.19376

maka :

= <

= 683.24932 < 2666.67 ……… OK !!

5. Pengecoran Bertahap

Sebelum Komposit :

= à = 8.88000 t.m

= à = 12.95313 t.m

= à = 12.95313 t.m

Setelah Komposit :

= à = 15.67000 t.m

= = 39.46416 t.m

=dimana :

= 0.00 = 0.00000

= 0.00 = 0.00000

= 0.00 = 0.00000

= = 5.93667

= = 14.95124

maka :

= <

= 20.88791 < 112.50 ……… OK !!

=dimana :

= 0.0 = 0.00000

= 0.0 = 0.00000

= 0.0 = 0.00000

= = 2.78896

fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

fs-b1

M1 / W

s kg/cm2

fs-b2

(M2 y

s) / I

t kg/cm2

fs-b3

(M3 y

s) / I

t kg/cm2

fs-b4

(M4 y

s) / I

t kg/cm2

fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

fy / 1.5

fs-b kg/cm2 kg/cm2



M1

1/9 wc L2 M

1

M2 1/18 wc L2 M2

M3

1/18 wc L2 M

3

M4

1/9 wSDL

L2 M4

M5 1/9 (q' + q") L2 + 1/6 p' L M

5



) :

fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

+ fc-a5

fc-a1 kg/cm2

fc-a2 kg/cm2

fc-a3 kg/cm2

fc-a4

(M3 y

c) / (n I

t) kg/cm2

fc-a5

(M4 y

c) / (n I

t) kg/cm2

fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

+ fc-a5

0.45 fc'

fc-a kg/cm2 kg/cm2


) :

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

+ fc-b5

fc-b1 kg/cm2

fc-b2 kg/cm2

fc-b3 kg/cm2

fc-b4

[ M4 (y

c - d) ] / (n I

t) kg/cm2



= = 7.02386

maka :

= <

= 9.81282 < 112.50 ……… OK !!

=dimana :

= = 80.97757

= = 118.12078

= = 118.12078

= = 18.01826

= = 45.37814maka :

= <

= 380.61553 < 2666.67 ……… OK !!

=dimana :

= = 80.97757

= = 118.12078

= = 118.12078

= = 100.68214

= = 253.56326maka :

= <

= 671.46454 < 2666.67 ……… OK !!

Sebelum Komposit :

= à = 9.99000 t.m

= à = 16.19141 t.m

Setelah Komposit :

= à = 12.95313 t.m

= à = 17.62875 t.m

= = 49.21187 t.m

=dimana :

fc-b5

[ M5 (y

c - d) ] / (n I

t) kg/cm2

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

+ fc-b5

0.45 fc'

fc-b kg/cm2 kg/cm2


) :

fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

+ fs-a5

fs-a1

M1 / W

s kg/cm2

fs-a2

M2 / W

s kg/cm2

fs-a3

M3 / W

s kg/cm2

fs-a4

[ M4 (h

s - y

s) ] / I

t kg/cm2

fs-a5

[ M5 (h

s - y

s) ] / I

t kg/cm2

fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

+ fs-a5

fy / 1.5

fs-a kg/cm2 kg/cm2


) :

fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

+ fs-b5

fs-b1

M1 / W

s kg/cm2

fs-b2

M2 / W

s kg/cm2

fs-b3

M3 / W

s kg/cm2

fs-b4

(M4 y

s) / I

t kg/cm2

fs-b5

(M5 y

s) / I

t kg/cm2

fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

+ fs-b5

fy / 1.5

fs-b kg/cm2 kg/cm2



M1

1/8 ws L2 M

1

M2 5/72 wc L2 M2

M3

1/18 wDL

L2 M3

M4

1/8 wSDL

L2 M4

M5 1/8 (q' + q") L2 + 1/4 p' L M

5



) :

fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

+ fc-a5



= 0.0 = 0.00000

= 0.0 = 0.00000

= 4.90737

= = 6.67876

= = 18.64421maka :

= <

= 30.23034 < 112.50 ……… OK !!

=dimana :

= 0.0 = 0.00000

= 0.0 = 0.00000

= = 2.30541

= = 3.13758

= = 8.75876maka :

= <

= 14.20175 < 112.50 ……… OK !!

=dimana :

= = 91.09976

= = 147.65098

= = 14.89424

= = 20.27054

= = 56.58660maka :

= <

= 330.50213 < 2666.67 ……… OK !!

=dimana :

= = 91.09976

= = 147.65098

= = 83.22581

= = 113.26741

= = 316.19376maka :

= <

= 751.43772 < 2666.67 ……… OK !!

fc-a1 kg/cm2

fc-a2 kg/cm2


fc-a4

(M3 y

c) / (n I

t) kg/cm2

fc-a5

(M4 y

c) / (n I

t) kg/cm2

fc-a

fc-a1

+ fc-a2

+ fc-a3

+ fc-a4

+ fc-a5

0.45 fc'

fc-a kg/cm2 kg/cm2


) :

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

+ fc-b5

fc-b1 kg/cm2

fc-b2 kg/cm2

fc-b3

[ M2 (y

c - d) ] / (n I

t) kg/cm2

fc-b4 [ M3 (yc - d) ] / (n It) kg/cm2

fc-b5

[ M4 (y

c - d) ] / (n I

t) kg/cm2

fc-b

fc-b1

+ fc-b2

+ fc-b3

+ fc-b4

+ fc-b5

0.45 fc'

fc-b kg/cm2 kg/cm2


) :

fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

+ fs-a5

fs-a1

M1 / W

s kg/cm2

fs-a2

M2 / W

s kg/cm2

fs-a3

[ M2 (h

s - y

s) ] / I

t kg/cm2


fs-a5

[ M4 (h

s - y

s) ] / I

t kg/cm2

fs-a

fs-a1

+ fs-a2

+ fs-a3

+ fs-a4

+ fs-a5

fy / 1.5

fs-a kg/cm2 kg/cm2


) :

fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

+ fs-b5

fs-b1

M1 / W

s kg/cm2

fs-b2

M2 / W

s kg/cm2

fs-b3

(M2 y

s) / I

t kg/cm2

fs-b4

(M3 y

s) / I

t kg/cm2


fs-b

fs-b1

+ fs-b2

+ fs-b3

+ fs-b4

+ fs-b5

fy / 1.5

fs-b kg/cm2 kg/cm2


58

RESUME HASIL PERHITUNGAN TEGANGANPADA PENAMPANG KOMPOSIT UNTUK GELAGAR TENGAH

ANALISA PEMBEBANAN= 1.39145 t/m ; q' = 0.72000 t/m

= 0.62680 t/m ; p' = 5.13566 ton

= 1.03625 t/m ; = 0.35520 t/m

PROPERTIES PENAMPANGn = 9 ;L = 15.00 m d = 20.0 cm

= 37.72051 cm ; = 83.8 cm

= 71.07949 cm ; = 10966.0

= 1106269.20513 ; = 462017.0


Sebelum Sesudah Komposit TEGANGAN Tegangan1 2 3 4 5 TOTAL Izin

Unshoring

MOMEN (t.m) 39.13453 17.62875 39.50874

TEGANGAN BetonAtas 0.00000 6.67876 14.96813 21.64689 112.50

Bawah 0.00000 3.13758 7.03179 10.16937 112.50

BajaAtas 356.87152 20.27054 45.42940 422.57146 2666.67

Bawah 356.87152 113.26741 253.84970 723.98864 2666.67

Full Shoring

MOMEN (t.m) 0.00000 39.13453 17.62875 39.50874


Bawah 0.00000 6.96519 3.13758 7.03179 17.13456 112.50

BajaAtas 0.00000 44.99911 20.27054 45.42940 110.69905 2666.67

Bawah 0.00000 251.44534 113.26741 253.84970 618.56246 2666.67

wDL

wSDL

wc

ws

yc hs

ys

Ws cm3

It cm4 I

s cm4

(kg/cm2)

(kg/cm2)


59

MOMEN (t.m) -9.78363 48.91816 17.62875 39.50874


Bawah 0.00000 8.70649 3.13758 7.03179 18.87586 112.50

BajaAtas -89.21788 56.24889 20.27054 45.42940 32.73095 2666.67

Bawah -89.21788 314.30668 113.26741 253.84970 592.20591 2666.67

MOMEN (t.m) -3.47863 38.26488 15.67000 30.83916


Bawah 0.00000 6.81041 2.78896 5.48878 15.08814 112.50

BajaAtas -31.72191 43.99913 18.01826 35.46062 65.75610 2666.67

Bawah -31.72191 245.85767 100.68214 198.14632 512.96422 2666.67

MOMEN (t.m) 0.86966 38.26488 17.62875 39.50874


Bawah 0.00000 6.81041 3.13758 7.03179 16.97978 112.50

BajaAtas 7.93048 43.99913 20.27054 45.42940 117.62955 2666.67

Bawah 7.93048 245.85767 113.26741 253.84970 620.90526 2666.67


MOMEN (t.m) 8.88000 12.95313 12.95313 15.67000 30.83916


Bawah 0.00000 0.00000 0.00000 2.78896 5.48878 8.27773 112.50

BajaAtas 80.97757 118.12078 118.12078 18.01826 35.46062 370.69801 2666.67

Bawah 80.97757 118.12078 118.12078 100.68214 198.14632 616.04760 2666.67

Satu Tumpuan Sementara

(kg/cm2)

Dua Tumpuan Sementara (titik D dan titik E) (kg/cm2)

Dua Tumpuan Sementara (titik C)

(kg/cm2)

Pengecoran Bertahap (titik D dan titik E)

(kg/cm2)


60


MOMEN (t.m) 9.99000 16.19141 12.95313 17.62875 39.50874


Bawah 0.00000 0.00000 2.30541 3.13758 7.03179 12.47478 112.50

BajaAtas 91.09976 147.65098 14.89424 20.27054 45.42940 319.34492 2666.67

Bawah 91.09976 147.65098 83.22581 113.26741 253.84970 689.09366 2666.67

Pengecoran Bertahap (titik C)

(kg/cm2)


61

………. OK !!………. OK !!

………. OK !!

………. OK !!

………. OK !!………. OK !!

………. OK !!

………. OK !!


62

………. OK !!………. OK !!

………. OK !!

………. OK !!

………. OK !!………. OK !!

………. OK !!

………. OK !!

………. OK !!………. OK !!

………. OK !!

………. OK !!

………. OK !!………. OK !!

………. OK !!

………. OK !!


63

………. OK !!………. OK !!

………. OK !!

………. OK !!

PERENCANAAN GELAGAR JEMBATAN KOMPOSIT

"GELAGAR TENGAH"


Data - Data := 15.00 m= 7.60 m= 2.00 m= 0.80 m

Lebar trotoar = 1.00 mJumlah gelagar baja = 5 gelagar

Tebal lantai beton = 20.0 cm= 5.0 cm

Tebal lapisan aspal = 5.0 cmTebal trotoar = 20.0 cm

Berat sendiri beton = 2.50Berat sendiri aspal = 2.30Berat sendiri trotoar = 2.20Berat tiang + sandaran = 0.50 t/m

= 25.0 MPa= 400.0 MPa= 200000.0 MPa

Profil Baja= 296.0 kg/m

= 40.0 cm

= 83.8 cm

= 380.0

= 10966.0

= 462017.0


Panjang bentang jembatan (L)Lebar lantai kendaran (B)Jarak antar gelagar (s)Lebar kantilever (c)

Tinggi voute

t/m3

t/m3

t/m3




(W 838 × 296) ws

bs

hs

As cm2

Ws cm3

Is cm4

20,0 cm

20,0 cm

0,80 m 0,80 m

1,00 m1,00 m7,60 m

2,00 m 2,00 m 2,00 m 2,00 m

1. Analisa Pembebanan

Beban Mati


Berat sendiri gelagar = 0.29600 t/m Berat sendiri diafragma = 0.05920 t/m

= 1.41145 t/m

Berat sendiri aspal = 0.8740 t/m Berat sendiri trotoar = 0.8800 t/m Berat tiang + sandaran = 1.0000 t/m Berat air hujan = 0.3800 t/m

= 3.1340 t/m

Beban mati untuk setiap gelagar tengah= 1.41145 t/m= 0.62680 t/m

Beban Hidup

q = 2.200 t/mq' = 1.600 t/m

P = 12.000 tonP' = 8.727 ton

K = 1.30769P' = 11.41259 ton

b = = 375.00 cmb = s = 200.00 cmb = = 240.00 cm

Jadi :b = 200.00 cm

b =dimana :

e = = 50.00 cma = = 75.00 cm

a/L = 0.05000= 1.0 + [ (0.89 - 1.00) * (0.05 - 0.05) / (0.10 - 0.05) ]= 1.00000 cm

λ = 75.00 cm


)

Berat sendiri voute

wDL


)

wSDL

wDL

wSDL

Beban merata q

Beban garis P

Beban Kejut (K)

2. Lebar Efektif (b)

Menurut spesifikasi AASHTO, lebar efektif diambil sebagai nilai terkecil dari nilai-nilai berikut :L/4

12d

Menurut peraturan Bina Marga, lebar efektif adalah :2 λ + e

bs + 2 t

(s - e)/2

λ/a

Jadi :b = 200.00 cm

3. Disain Ultimit

a == 357.64706 mm= 35.76471 cm

Sebelum Komposit; = 39.697 ton-m

; = 10.586 ton

Setelah Komposit

; = 17.629 ton-m

; = 4.701 ton

Tinggi Blok Tegangan Beton (a)(A

s f

y) / (0,85 f

c' b)

2. Momen Maksimum (Mmax

) dan Lintang Maksimum (Dmax

)

Momen Maksimum akibat beban mati primer (M1) M

1

Lintang Maksimum akibat beban mati primer (D1) D

1

Momen Maksimum akibat beban mati sekunder (M2) M

2

Lintang Maksimum akibat beban mati sekunder (M2) D2

; = 87.797 ton-m

; = 23.413 ton

3. Tegangan Total pada Penampang Komposit

= 23500.0 MPa= 200000.0 MPa

n = = 8.5 ; n = 8Lebar efektif lantai beton

= 375.000 cm= 200.000 cm= 240.000 cm ; b = 200.0 cm

Lokasi Garis Netral (dari serat atas beton)b/n = 25.0 cm y = 30.765 cm

d = 20.0 cm = 30.765 cm

= 83.8 cm = 73.035 cm

= 500.0 = 15.383 cm

= 380.0 = 31.135 cm

Momen Inersia Penampang Komposit

= 60665.00

= 181994.99

= 462017.00

= 368362.7 = 1073039.68

Tegangan pada Penampang Komposit Sebelum Komposit

Serat atas beton = 0

Serat bawah beton fc1' = 0

Serat atas baja = -362.001

Serat bawah baja = 362.001

Setelah Komposit

Serat atas beton = -37.783


Serat bawah baja = -105.768


Setelah KompositSerat atas beton

= -37.783

= -3.778 MPa < 11.250 MPa………. OK !!!

Serat bawah beton

= -13.221

Momen Maksimum akibat beban hidup q' dan P' (M3) M

3

Lintang Maksimum akibat beban hidup q' dan P' (D3) D

3

Modulus Rasio (n)E

c

Es

Es/E

c

b = L/4b = sb = 12 d

yc

hs

ys

Ac cm2 d

c

As cm2 d

s

It = I

c’ + A

c’ d

c2 + I

s + A

s d

s2

Ic’ cm4

Ac’ d

c2 cm4

Is cm4

As d

s2 cm4 I

t cm4

; fc1

' = 0 fc1

'

; fc1

= 0

; fs1

' = - M

1/W

sfs1

' kg/cm2

; fs1

= M1/W

sfs1 kg/cm2

; fc2

' = - (M

2 + M

3) y

c / (n I

t) f

c2 kg/cm2

; fc2

= - (M2 + M

3) (y

c- d

c) / (n I

t) f

c2 kg/cm2

; fs2

'

= (M2 + M

3) (y

s - h) / I

tfs2 kg/cm2

; fs2

= (M2 + M

3) y

s / I

tfs2 kg/cm2

; fc' = fc1' + fc2'

kg/cm2

; fc

= fc1

+ fc2

kg/cm2

= -1.322 MPa

Serat atas baja

= -467.77

= -46.78 MPa

Serat bawah baja

= 1079.567

= 107.957 MPa < 266.667 MPa………. OK !!!

; fc' = f

c1' + f

c2'

kg/cm2

; fs

= fs1

+ fs2

kg/cm2

a/L0.05 1.000.10 0.890.15 0.780.20 0.680.25 0.580.30 0.50

λ/a

0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.110.8

0.9

1 f(x) = -2.20000x + 1.11000R² = 1.00000

0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.110.8

0.9

1 f(x) = -2.20000x + 1.11000R² = 1.00000

a = 50.0b = 760.0c = -23522.0

= 15.383

= = 31.135

= = 30.765

= = 73.035

2 (b/n) dc2 + 2A

s d

c – A

s (d + h/2) = 0

= 2 (b/n)= 2A

s

= – As (d + h/2)

dc = ( – b + ( b2 – 4ac)1/2 ) / 2ad

s(d + h/2) – 2 d

c

yc

2 dc

ys

(d + h) – 2 dc

PERENCANAAN GELAGAR JEMBATAN KOMPOSIT

"GELAGAR PINGGIR"


Data - Data := 15.00 m= 7.60 m= 2.00 m= 0.80 m

Lebar trotoar = 1.00 mJumlah gelagar baja = 5 gelagar

Tebal lantai beton = 20.0 cm= 5.0 cm

Tebal lapisan aspal = 5.0 cmTebal trotoar = 20.0 cm

Berat sendiri beton = 2.50Berat sendiri aspal = 2.30Berat sendiri trotoar = 2.20Berat tiang + sandaran = 0.50 t/m

= 25.0 MPa= 400.0 MPa= 200000.0 MPa

Profil Baja= 296.0 kg/m

= 40.0 cm

= 83.8 cm

= 380.0

= 10966.0

= 462017.0


Panjang bentang jembatan (L)Lebar lantai kendaran (B)Jarak antar gelagar (s)Lebar kantilever (c)

Tinggi voute

t/m3

t/m3

t/m3




(W 838 × 296) ws

bs

hs

As cm2

Ws cm3

Is cm4

20,0 cm

20,0 cm

0,80 m 0,80 m

1,00 m1,00 m7,60 m

2,00 m 2,00 m 2,00 m 2,00 m

1. Analisa Pembebanan

Beban Mati


Berat sendiri gelagar = 0.29600 t/m Berat sendiri diafragma = 0.05920 t/m

= 1.39145 t/m

Berat sendiri aspal = 0.8740 t/m Berat sendiri trotoar = 0.8800 t/m Berat tiang + sandaran = 1.0000 t/m Berat air hujan = 0.3800 t/m

= 3.1340 t/m

Beban mati untuk setiap gelagar tengah= 1.39145 t/m= 0.62680 t/m

Beban Hidup

q = 2.20000 t/mq' = 0.72000 t/m

P = 12.00000 tonP' = 3.92727 ton

K = 1.30769P' = 5.13566 ton

= 0.30000 t/mq" = 0.34500 t/m

b = = 125.00 cmb = = 120.00 cmb = = 120.00 cm

Nilai terkecil dari ketiga nilai di atas adalah :b = 120.00 cm

Karena pada gelagar pinggir terdapat kantilever dengan panjang :c =

= 55.0 cmMaka, lebar efektif gelagar pinggir ditambakan dengan panjang kantilrver :

b =


)

Berat sendiri voute

wDL


)

wSDL

wDL

wSDL

Beban merata q

Beban garis P

Beban Kejut (K)

Beban Hidup pada Trotoar (q")w

t

2. Lebar Efektif (b)

Menurut spesifikasi AASHTO, lebar efektif diambil sebagai nilai terkecil dari nilai-nilai berikut :L/12(s + b

s)/2

6d

80,0 - (0.5 bs + t)

b + c + t

b = 180.0 cm

b =dimana :

e = = 50.00 cma = = 75.00 cm

a/L = 0.05000= 1.0 + [ (0.89 - 1.00) * (0.05 - 0.05) / (0.10 - 0.05) ]= 1.00000 cm

λ = 75.00 cmPada gelagar pinggir terdapat kantilever dengan panjang :

c == 55.0 cm

Karena :λ > c

75.00 > 55.0maka, lebar efektif gelagar pinggir yang mempunyai kantilever adalah :

b =b = 180.00 cm

Sebelum Komposit; = 39.135 ton-m

; = 10.436 ton

Setelah Komposit; = 17.629 ton-m

; = 4.701 ton

; = 39.509 ton-m

; = 10.536 ton

3. Tegangan Total pada Penampang Komposit

= 23500.0 MPa

= 200000.0 MPa

n = = 8.5 ; n = 8Lebar efektif lantai beton

= 375.000 cm

= 200.000 cm

= 240.000 cm ; b = 200.0 cm

Lokasi Garis Netral (dari serat atas beton)b/n = 25.0 cm y = 30.765 cm

d = 20.0 cm = 30.765 cm

= 83.8 cm = 73.035 cm

= 500.0 = 15.383 cm

Menurut peraturan Bina Marga, lebar efektif adalah :2 λ + e

bs + 2 t

(s - e)/2

λ/a

80,0 - (0.5 bs + t)

λ + e + c

2. Momen Maksimum (Mmax

) dan Lintang Maksimum (Dmax

)

Momen Maksimum akibat beban mati primer (M1) M

1

Lintang Maksimum akibat beban mati primer (D1) D

1

Momen Maksimum akibat beban mati sekunder (M2) M

2

Lintang Maksimum akibat beban mati sekunder (M2) D

2

Momen Maksimum akibat beban hidup q' dan P' (M3) M

3

Lintang Maksimum akibat beban hidup q' dan P' (D3) D

3

Modulus Rasio (n)E

c

Es

Es/E

c

b = L/4

b = s

b = 12 d

yc

hs ys

Ac cm2 d

c

= 380.0 = 31.135 cm

Momen Inersia Penampang Komposit

= 60665.00

= 181994.99

= 462017.00

= 368362.7 = 1073039.68

Tegangan pada Penampang Komposit Sebelum Komposit

Serat atas beton = 0

Serat bawah beton fc1' = 0

Serat atas baja = -356.872


Setelah KompositSerat atas beton = -20.477


Serat bawah baja = -57.323


Setelah KompositSerat atas beton

= -20.477

= -2.048 MPa < 11.250 MPa………. OK !!!

Serat bawah beton

= -7.165

= -0.717 MPa

Serat atas baja

= -414.19

= -41.42 MPa

Serat bawah baja

= 745.769

= 74.577 MPa < 266.667 MPa………. OK !!!

As cm2 d

s

It = I

c’ + A

c’ d

c2 + I

s + A

s d

s2

Ic’ cm4

Ac’ dc2 cm4

Is cm4

As d

s2 cm4 I

t cm4

; fc1

' = 0 fc1

'

; fc1

= 0

; fs1' = - M1/Ws fs1' kg/cm2

; fs1

= M1/W

sfs1 kg/cm2

; fc2

' = - (M

2 + M

3) y

c / (n I

t) f

c2 kg/cm2

; fc2

= - (M2 + M

3) (y

c- d

c) / (n I

t) f

c2 kg/cm2

; fs2

'

= (M2 + M

3) (y

s - h) / I

tfs2 kg/cm2

; fs2

= (M2 + M

3) y

s / I

tfs2 kg/cm2

; fc' = f

c1' + f

c2'

kg/cm2

; fc

= fc1

+ fc2

kg/cm2

; fc' = fc1' + fc2'

kg/cm2

; fs = fs1 + fs2

kg/cm2

a/L0.05 1.000.10 0.890.15 0.780.20 0.680.25 0.580.30 0.50

λ/a

0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.110.8

0.9

1 f(x) = -2.20000x + 1.11000R² = 1.00000

a = 50.0b = 760.0c = -23522.0

= 15.383

= = 31.135

= = 30.765

2 (b/n) dc2 + 2As dc – As (d + h/2) = 0

= 2 (b/n)= 2A

s

= – As (d + h/2)

dc = ( – b + ( b2 – 4ac)1/2 ) / 2a

ds

(d + h/2) – 2 dc

yc

2 dc

0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.110.8

0.9

1 f(x) = -2.20000x + 1.11000R² = 1.00000

= = 73.035ys

(d + h) – 2 dc

59674607 contoh disain jembatan komposit

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