tugas pak zardan ojul

8/19/2019 Tugas Pak Zardan Ojul

1/34

MERENCANAKAN JEMBATAN KAYU

A. DATA STRUKTUR JEMBATAN KAYU

Data Jembatan

- Panjang bentang (L) = 16 m- Lebar jalan (b) = 3.5 m

- Lebar trotoar = 0.75 m- Lebar jembatan = 3.5 + 2 (0.75)= 5 m

- Tinggi (H) = 5 m- Beban ter!"at = 5 ton- Beban merata = 2.5 ton#m$

- Te%anan angin (&) = 100

Data Kayu

- 'eni" %a! = a! %er!ing

- Berat jeni" %a! = 0.6- ela" %a! = ela" **

- o,!l!" ela"ti"ita" () = 100000

Konstruksi Tak Terlindun

- = 5#6

Muatan Teta!

- / = 1

Teanan "#inela" !at

* ** *** *

150 100 75 50

130 5 60 5

0 25 15 10 20 12 5

%g#m2

%g#4m2

'eni"Tegangan

lt

tr##

t%


2/34

- = 5#6 (150) = 125

- = 5#6 (130) = 10.333

- = 5#6 (0) = 33.333

- = 5#6 (20) = 16.667

B. $EMBEBANAN

1 Trotoir

= 100 %g

= 1# (100) (75)= 175 %g.4m

'i%a lebar aan trotoir ,iambil = 20 4m

& =

= 3 1#3

lt %g#4m2

tr## %g#4m2

t% %g#4m2

%g#4m2

Porang

ma

1#6 b2


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4/34


5/34

MERENCANAKAN JEMBATAN KAYU

A. DATA STRUKTUR JEMBATAN KAYU

Data Jembatan

- 'embatan %ela" **- Panjang bentang (L) = 10.5 m

- Lebar jalan (b) = 3.5 m- Lebar trotoar = 0.75 m

- Lebar jembatan = 3.5 + 2 (0.75)= 5 m

- θ = 35

Data Kayu

- 'eni" %a! = a! jati

- Berat jeni" %a! = 0.8

- ela" %a! = ela" *- o,!l!" ela"ti"ita" () = 125000

Konstruksi Tak Terlindun

- = 5#6

Muatan Teta!

- / = 1

Teanan "#in

'eni" Tegangan ela" !at* ** *** *

150 100 75 50

130 5 60 5

0 25 15 10

20 12 5

o

t#m3

%g#4m2

lt

tr##

t%

E

35°

A CB

D


6/34

Tegangan i9in !nt!% %a! %ela" *

- = 5#6 (150) = 125

- = 5#6 (130) = 10.333

- = 5#6 (0) = 33.333

- = 5#6 (20) = 16.667

B. $EMBEBANAN

% Trotoir

= 100 %g

= 0.1 ton:%!ran aan ; b = 20 4m

; = 5 4m

!atan berat "en,iri ; < = 0.2 (0.05) (0.80)= 0.008 t#m

omen ma%"im!m - >%ibat P = 1# Pl

= 1# (0.1) (0.75)= 0.01 t.m

- >%ibat berat "en,iri =

=

= 0.00063 t.m

a%a momen ma%"im!m = 0.01 + 0.00063= 0.018 t.m

& =

== 0.0000

= #?

= 0.01830.0000

= 232.58375

= 23.258375 @ 125

lt

tr##

t%

Porang

1#


7/34

& 'antai Kendaraan

- Lebar jalan (b) = 3.5 m

'ara% gelagar memanjang ,an melintang"arat jara% gelagar memanjang

0. m A C 0.6 m

,iambil = 0.5 m

0.5 =3.5n-1

0.5 n - 0.5 = 3.50.5 n = 3.85

n = .77777 ≈ 8

- '!mla balo% gelagar = 8

Dierit!ng%an a"al + %eri%il "etebal 10 4m b, = 2. t#m

10 4m

Dierit!ng%an !nt!% ren4ana m!atan tr!4% ata! "emi trailer.

- 'embatan %ela" **P = 5000 %g= 5 ton

:%!ran lantai ; b = 20 4m; = 15 4m

!atan - < er%era"an = 0.20 (0.10) (2.0)

= 0.05 t#m$

=

=

= 0.0012 t.m

- < = 0.25 (0.1) (0.80)= 0.03 t#m$

berat "en,iri =

=

er%era"an

=

ma

1#


8/34

= 0.0007 t.m

- !atan bergera% Te%anan gan,ar = 5 tonTe%anan 1 ro,a (P) = 1#2 te%anan gan,ar

= 2.5 ton= 2500 %g

>ng%a "ent!

E = 1 +0

100 + 0.5= 1.38201

P "ent! = 1.38 2500= 385.5 %g

=

= 1# 385.5 = 3832.6 %g.4m

er%era"an = 0.0012 t.m= 121.500 %g.4m

b" = 0.0007= 6.3 %g.4m

=

= 3832.6 + 121.50 += 3851. %g.4m

:%!ran lantai ; b = 25 4m; = 15 4m

& =

=

= 837.500

= #?

3851. %g.4m

ma

1# P" l

ma

total ma

+ er%era"an + b"

1#6 b2

1#6 (25) (15)2

4m3

lantai

P = 2500 %g


9/34

837.500 4m3

= 2.18

= 5#6 (1;1) (150)

= 137.5

( Balok )elaar ABC

'ara% gelagar = 0.5 m= 5 4m

Panjang bentangan = 1050 4m

- >%ibat beban "en,iriita a%ai balo% ,engan !%!ran

b = 20 4m

= 30 4m

= 0.20 (0.30) (0.80)

= 0.05 t#m$

- (b, = 2.

= F0.10 (0.80) + (0.1) (2.)G (0.5)

= (0.080+0.2) (0.5)= 0.168 t#m$

= 0.05 + 0.168

= 0.223 t#m$

=

=

= 3.068 t.m

- !atan tr!4%#trailer ----- te%anan 1 o, = 2.5 ton" = 1.38

= 5'ara% gelagar = 0.5 m

P$ =.85

2.5 0.55

= 1.11 ton

%g#4m2

IltJ

%g#4m2

,imana %on"tr!%"i terlin,!ngi; %arena lantai @ IltJ

< b"

arena aan lantai (20 15 4m2) + er%era"an 10 4m

<lantai

< b"

+ <lantai

ma 1#


10/34

= 2;5 (P$) (;85)(")

10.5= 1.35

=

= 1;35(;85) - (1#2) (1;11) (3) (1;38)= 6.78 tm

-=

(1#2)P$(;25) + P$(5;25) + P$(2;25)

10.5

=11.625

P$10.5

= 1.107 1.11 = 1.72 ton

= 1;72(5;25) - (1#2) (1;11) (3) (1;38)

= 6.716 ton

:nt!% m!atan tr!4%#"emi trailer

6.78 tm ait! 0;3 m ,ari %iri#%anan tenga-tenga bentang >

- arena m!atan me"in elin,a"

'embatan %ela" ** P = 5 ton; ,itaan ole 3 gelagar

>

D

>(;85) - 1#2 P$ (5;25) (")

>

>

B

ma

P

2.2500

4.9500

P

2.2500

10.5000

12 P

2,5 t

P

212 P

112 P

10.5000

1.500 1.000

1

2 P

3 . 0 0 0

P

2 . 2 5 0 0

3 . 0 0 0

P

2 . 2 5 0 0

1 0 . 5 0 0 0


11/34

P$ = 5 = 1.666666667 ton3

= 2;5 (P$) 3 =10.5

= 1.25 1.666666667= 2.033333333

= 2.033333333 5.25 =

10.83 tm ,i tenga-tenga bentang

omen a%ibat berat "en,iri + aan lantai + er%era"an

= 3.068 tm ,engan < = 0.223

= 3.068 + 10.8375= 1.0073 tm

'embatan %ela" ** P = 5 ton ,itaan ole 3 gelagar

P$ =5

=1.667 ton

3

= 2;5 P$ ;7510.5

= 1.131 1.667= 1.5 ton

= = .853373016

omen ma ang menent!%an (,iambil)

K

B

:nt!% me"in elin,a" ma

=

ma

total

K

K ;75

212 P

10.5000

P 112 P

212 P

10.5000

1.500 1.000


12/34

= 3.068 + 10.8375 =

= 3.068 + .853373016 =

ang ,iambil = 1.0073 tm

ang menimb!l%an

=

=

1.0073 =

= 1.016007837 t#m

Dimen"i balo% gelagar >B ,an BK

l = 5.25 m< = 1.016 t#m

ma =

= 3.502 t#m

= (5#6) 150

= 125 %g#4m

?n =

= 3000

B

!atan berat "en,iri + aan lantai - er%era"an - me"in elin,!ng ,iganti ,engan m!atan terbagi

ma

tot

(1#) < l2

(1#) (


13/34

?

=35011.3583 %g 4m

= 116.733000 4m3

:%!ran balo% ,aat ,ia%ai 20 30 4m

Tinja!an "tati" ti,a% tertent!

= 8 1.016

123= 0.380 tm @ 3.502

Perit!ngan "elanj!tna ,engan ma = 3.5011,imana %eamanan terjamin.

= 2 (1#2) < l + 2 (#l)

= 5.336 + 1.33

= 6.670 ton

* Balok Melintan Mena+an )elaar

lt

= 116;73 %g#4m2 @ I lt J = 125 %g#4m2

ma

B

B

5.255.25

> B K

0.45P = 6,670 t

0.450.75 0.45


14/34

i"al%an !%!ran balo% = 20 304m< = 0;20 0;3 0;80

= 0.05 t#m

< =

= 1# 0;05 1= 0.00735

< = 1# P l= 1# 6;670 1= 1.66

tot = < + = 0.00735 + 1.66= 1.675 tm

= 0; (1#6) (20) (30)

= 200

=167.25520

= 68.77200

M = 20 20 = 00

= P = 6670.13020

M 00

= 16.675

= (5#6) 1;1 130

= 118.166666667

16.675 %g#4m2 = 118.167 %g#4m2

, Tian $enyokon

1# < l2

? N

lt

Tinja!an a,a lt O antara gelagar 20#30 ,engan balo% melintang 20#30

lt##

%g#4m2

Ilt##

J

t%##

= @ It%##

J

0.45P = 6,670 t

0.450.75 0.45

K2

C

0.90 0.90

K1

0.90 0.90

B

0.90 0.900.90 0.900.90

A

ED


15/34

P = 6.670 ton

= 6;670 + (2 (6;670)#(2))

= 6;670 + 2 (0.07)= 13.30260167 + 0.1176= 13.5

=(1#2) (13;5)

=6.7280.57

= 11.732 ton

Diambil balo% !%!ran b = 20 4m = 20 4m

M = 20 20

= 00

= 0.28

= 0.28 (20)= 5.7 4m

l% =5.25

= 6.08 4m

=l%

=

6.08

5.7= 1.108

&0 1

---- %.%-

1 1.01 & %.-%%

? P

ton + berat"eanjang 2

1 =

2 "in 35o

4m2

imin

4o" 35o

imin

K2

C

0.90 0.90

K1

0.90 0.90

B

0.90 0.900.90 0.900.90

A

ED


16/34

M

=1.011(6;670)

00

= 16.60

/ Sandaran

- Qelagar "an,aran'ara% tiang "an,aran = 1 mDia%ai !%!ran tiang "an,aran

b = 10 4m = 12 4m

!atan ori9ontal = 75 %g#m

=

=

= 8.375 %g.m = 837.5

=

=

= 182

=?

=837.5182

= .3

= 5#6 (1.1) (150)

= 137.5

t%##

%g#4m2

ma 1# (m!atan ori9ontal) (jara% tiang "an,aran)2

1# (75) (1)2

?n 0. (1#6) b 2

0. (1#6) (10) (12)2

4m2

lt

%g#4m2

IltJ

%g#4m2


17/34

--------- >man

- Tiang "an,aranDiambil !%!ran balo%

b = 12 4m

= 12 4m!atan ori9ontal = 75

< = 12#12 (75)= 75 %g#m$

=

=

= 8.375 + .675= 1.0625

=

=

= 230.

=?

=1.0625

230.

= 0.061

= 5#6 (1.1) (150)

= 36.6666666667

---------- >man

- Bo!t-bo!t

=150 (1) (75)

25= 50 %g

= 2 (50)5

= 10 %g

Diambil ba!t ; R =

0 $er+itunan sambunan

t%##

= ;3 %g#4m2 @ It%##

J = 137;5 %g#4m2

%g#4m2

B$

"an,aran +

tiang

1# 75 12 + 1#2 1# 75 12

?n 0. (1#6) b 2

0. (1#6) (12) (12)2

4m2

lt

%g#4m2

IltJ

%g#4m2

t%##

@ It%##

J

P1

P2


18/34

= ton

= ,an K

Eamb!ngan ,i,a"ar%an ata" be"arna momen ang ,aat ,itaan.Eamb!ngan ,engan ba!t ,an "amb!ngan "aming ,engan lat enamb!ng

= 2 10#25 ang ,itaan balo%.

= ==

= 1666.67

= 5#6 (1;1) (150)

= 137.5

=

?

= = 1666.66666667 137.5

= 228166.666667 %g 4m

Diambil ba!t R 2 4m

P = 0.65 30

= 0.65 30 = 111 %g

P = 0. 100 = 0. 100 = 1600 %g

P = 0. 200 = 0. 200 = 1600 %g

Diambil P = 111 %g; ,iambil 2 %elomo% ma"ing-ma"ing %elomo% b!a

B

>

>= >

?n . lt 0; (1#6) b 2

0; (1#6) 20 252

t%

t%

?n .

lt


19/34

---------- on"tr!%"i ta% terlin,!ng

SSS 5#6 0.8 e l1 = 228167ll = 25


20/34

E

C


21/34

>man

%g#4m2

%g#4m2

%g#4m2

%g#4m2

%g#4m2

3.50 0.75

0.75


22/34


23/34

5

6.3375


24/34

t#m)

=3,28/..


25/34

1.38

BK


26/34

10.8375 tm

t#m (


27/34

1.0073 tm

12.02316 tm

rata (=<e<

)

C


28/34

5.25

%g#4m2

= 6,670 t


29/34

aman

= 6,670 t


30/34

alo% melintang00 m


31/34

%g.4m


32/34


33/34

m ,20 2

l ,10 2

,2


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