Contoh Perhitungan Turap

PERKUATAN PERKUATAN LERENG ( Revetments) dan Turap Turap

Perkuatan Perkuatan lereng (revetments) adalah adalah bangun bangunan an yang yang ditemp ditempatk atkan an  pada permukaan suatu lereng guna melindungi suatu tebing aIur sungai atau  permukaan lereng tanggul dan secara keseluruhan berperan meningkatkan stabilitas alur sungai atau tubuh tanggul yang dilindunginya.

Perhitungan Profil 6 Parameter: γ'

=

1,11

tm!

γb

=

1,6"

tm!

γ#ater

=

1

tm!

= γbeton = γ pasangan batu &edalaman sungai

1$ ",% ","

o

() cohesi

",1 *,!

m

1,"!

meter  

ϴ

=

+inggi muka air =

kgm! kgm!

Rumus2 :

I.

Ka = Kp = Pa $ =

tg2 (45  !"2) tg2 (45 # !"2) 2%&%'a*5

Pa 2 =

+%'a%(,#2*$)

Pa - =

%5%ϒ.%h22%'a

Pa 4 =

ϒ/%h2%Ka%(,#2*$)

Pa 5 = Pa 0 = Pp$ = Pp2 = Pp- =

%5%ϒ/%Ka% (,#2*$)2 %5%ϒ1%(,#2*$)2 2%&%'p %5%ϒ/%'p%, %5%ϒ1ater%,

3 =

ƩPpƩpa

nalisis -aya yang bekera pada turap &a = *,/"$ &p = 1,$0% ol. beton ol. pas.batu ol.tanah 5 e7etments 5 tanah 5.sin8 5.cos8

Profl

3 4 & !

Beton

0.404 0.43# 0.40" 0.4#4

Pasangan Batu Volume 0.4! 0.$4% 0.!# 0.$%3

II.

 Tanah

0."#$ %.%" 0."!$ %.%0%

= = = = = = =

*.!2*.!2",3*04")) *.!2",3*0) ","*!" 4 !,"6*%) *,%"%2".%)4*,$1!2".") 1,6"2/,%6) ",$*/ ( sin %/) ",$*/ ( cos %/)

berat (t/m) Revetme nt #.!%% #.$$$ #.!& #.$0&

tanah %.&04 %."%0 %.&!$ %.$3

+ekanan +anah Pa1 = 9 ( &a ( *,!) 4 9 ( &a ( h air) 4 9 ( &a ( d)  = %,10* 4 ",3!0 d Pa" = γtnh ( &a ( *,! ( h air) 4 γtnh ( &a ( *,! ( d)

= *,%"% m ! = *,$1! m ! = 1,1*1 m ! = ",$*/ tm = 1,3$! tm = ",661 ton = *,$$3 ton

q sin #.4 #.40 #.&#% #.!!%

q cos 0.$#! 0."%3 0.$40 0.$$

 = *,!16 4 *,"/3 d Pa! = 1" ( γtnh ( *,!" ( &a  = *,/ ( 1,6" ( *,!" ( *,/"$  = *,*!$ Pa% = 1"2γsat2ka2h air")41"2γsat2ka2"2h air2d)41"2γsat2ka2d")  = 6,*6* 4 %,130 d 4 *,3"1 d "

Profl 3

Pa % ' (  *a  0+3) , (  *a  h air) , (  *a  -) !.#&% #.!% +

4

&.0"4

+

#.!"

-

&

&.0$

+

#.!$

-

!

4.%"0

+

#.3"

-

Profl

Pa # ' (tnh  *a  0+3  h air) , (tnh  *a  0+3  -)

3

0.

+

0.#&

-

4

0.3"&

+

0.#&

-

&

0.40$

+

0.#&

-

!

0.3%!

+

0.#&

-

Profl 3

Prof l

Pa 3 ' %/#  tnh  0+3#  *a 0.03$

4

0.03$

&

0.03$

!

0.03$

Pa 4 ' (%/#  sat  a  h air#),(%/#  sat  a  #  h air  -),(%/#  sat  a  -#)

3

#."""

+

4

%.0"

+

&

%.$##

+

!

%.0"0

+

#."4 0 #.#% " #.#" % %. 3

-

+

-

+

-

+

-

+

0.# % 0.# % 0.# % 0.# %

-# -# -# -#

III.

Persamaan omen ktif  ;Pa = /,6!% 4 %,36$ d 4 *,3"1 d " Profl

Pa -

3

".$%%

,

&.$!$

- ,

#

0.#%

4

.#3

,

&.#44

- ,

#

0.#%

&

.34!

,

&.#3&

- ,

#

0.#%

!

&.!34

,

4.!$

- ,

#

0.#%

a = ;Pa ( 1! ( h air 4 *,!) = ",$3! 4 ",%!" d 4 *,!63 d " Profl

Pa  %/3  (h air , 0+3)

3

.!&3

,

4.&

- ,

0.&!#

-#

4

4.43$

,

3.#%

- ,

0.44#

-#

&

4.!#$

,

3.#"$

- ,

0.4&4

-#

!

#.$3

,

#.43#

- ,

0.3!

-#

a = ;Pa ( 1! d = 1,$3$ 4 1,/$0 d " 4 *,"%* d ! Profl

Pa  %/3 -

3

3.#0

- ,

%."&!

-#

,

0.#40

-3

4

#.4%#

- ,

%.4$

-#

,

0.#40

-3

&

#.44"

- ,

%.4&

-#

,

0.#40

-3

!

%.$$

- ,

%.&$"

-#

,

0.#40

-3

a =  ; Pa ( 1! ( h air 4 *,!)) 4  ; Pa ( 1! d ) = ",$3! 4 %,!1* d 4 1,0/3 d " 4 *,"%* d ! Profl

1a -

3

.!&3

,

.$4

- ,

#.&%$

#

,

0.#40

4

4.43$

,

&.!#"

- ,

#.%"0

#

,

0.#40

&

4.!#$

,

&.4

- ,

#.%""

#

3

3

,

0.#40

3

!

#.$3

,

4.3%0

- ,

#

%."&

I.

Persamaan omen Pasif  Pp = d"" ( γsat ( &p  = *,/ ( ",3! ( 1,$0%  = ",/$6 p = Pp ( 1! d  = ",/$6 d" ( 1! d  = *,$6" d!

.

Perhitungan omen +otal omen total = omen ktif < omen Pasif 

,

3

0.#40

total = ",$3! 4 %,!1* d 4 1,0/3 d " 4 *,"%* d !) 4 *,$6" d !) = ",$3! 4 %,!1* d 4 1,0/3 d " 4 *,6"" d !) Profl

1a21P'0 -

3

.!&3

,

.$4

- ,

#.&%$

#

,

20.!##

4

4.43$

,

&.!#"

- ,

#.%"0

#

,

20.!##

&

4.!#$

,

&.4

- ,

#.%""

#

3

3

,

20.!##

3

-

-

#

3

! #.$3 , 4.3%0 - , %."& , 20.!## >ntuk mendapatkan nilai ? dilakukan +rial and @rror  ?1 = %,3$* meter  ?" = %,30* meter  ap)1 = ",$3! 4 %,!1* d 4 1,0/3 d " 4 *,6"" d !) = ",$3! 4 %,!1* %,3$)4 1,0/3 %,3$) " 4 *,6"" %,3$) !) = *,"3"

ap)" = ",$3! 4 %,!1* d 4 1,0/3 d " 4 *,6"" d !) = ",$3! 4 %,!1* %,30) 4 1,0/3 %,30) " 4 *,6"" %,30) !) = *,*3/  Trial ror Profl

3 4 & !

-% !.33 0 &.4# 0 &.4! 0 4.$ 0

-# !.34 0 &.43 0 &.4 0 4." 0

(1a 2 1)%

(1a 2 1)#

0.&0

0.%&!

0.#

%$0.0#&

0.34$

0.0$"

0.##

0.0&

!.34 4 &.43 % &.4 3 4." 4

Pa !.04%

P %04.0$ 

&!."%

!.#3

&.&$&

.4!"

4&.04$

&".4#!

56 %.3! " %.33 " %.34 & %.3% "

7  Tura $.!$4 .#% .3!3 !.3#4

 Ailai ? profil 6 ? = B?" ?1) ( ap)")  ap)1  ap)"))C 4 ?" = B*,*1 ( *,*3/)  *.103C 4 %,30* = %,30% meter  ;Pa = /,6!% 4 %,63$ ( %,30%) 4 *,3"1 ( %,30% ")) = %/,*%$ ;Pp = ",/$6 ( %,30% ") = /0,%"6 DE = ;Pp  ;Pa = /0,%"6  %/,*%$ = 1,!10 F 1," G +urap = ? 4 *,! 4 uka ir  G +urap = %,30% 4 *,! 4 1,"! = 6,!"% meter  ?ari persamaan diatas didapat nilai ? = %,30% meter. Hadi tinggi +urap yang dibutuhkan untuk profil 6 adalah %,30% meter