responsi pondasi dangkal

7/26/2019 Responsi Pondasi Dangkal

http://slidepdf.com/reader/full/responsi-pondasi-dangkal 1/23

Kelompok 4 Jurusan Teknik Sipil

Universitas Gunadarma

Responsi Pondasi Dangkal

2. Perhitungan Distribusi Tegangan Vertikal

1. Lapisan 1

Kedalaman 0,00 m – 9,00 m

q1’ = q1 = γ1. h1 = 17. 9 = 153 t/m2

2. Lapisan 2


q2 = q1 + (γ2 – γ! . h2 = 153 + (12 – 1! . 2 = 175 t/m2

3. Lapisan 3


q3 = q2 + (γ3 – γ! . h3 = 175 + (12 – 1! . " = 2#1 t/m2

#. Lapisan #


q# = q3 + (γ# – γ! . h# = 2#1 + (1#,5 – 1! . 3 = 2$1,50 t/m2

5. Lapisan 5

Kedalaman 20,00 m – 2#,00 m

q5 = q# + (γ5 – γ! . h5= 2$1,5 + (1$ – 1! . # = 3#9,50 t/m2

". Lapisan "Kedalaman 2#,00 m – 27,00 m

q" = q5 + (γ" – γ! . h" = 3#9,5 + (1" – 1! . 3 = 39#,50 t/m2

7. Lapisan 7


q7 = q" + (γ7 – γ! . h7 = 39#,50 + (1#,5 – 1! . 2 = #21,50 t/m2

$. Lapisan $


q$ = q7 + (γ$ – γ! . h$= #21,50 + (1" – 1! . 2,5 = #59 t/m2

9. Lapisan 9

Kedalaman 31,50 m – 3#,50 m

q9 = q$ + (γ9 – γ! . h9 = #59 + (19 – 1! . 3 = 513 t/m2

10. Lapisan 10

Kedalaman 3#,50 m – 37,50 m

%10 = q9 + (γ10 – γ! . h10 = 513 + (19 – 1! . 3 = 5"7 t/m2

11. Lapisan 11

Kedalaman 37,50 m – #0,00 m

"






%11 = q10 + (γ11 – γ! . h11 = 5"7 + (1" – 1! . 2,5 = "0#,50 t/m2

12. Lapisan 12Kedalaman #0,00 m – #3,50 m

%12 = q11 + (γ2 – γ! . h12 = "0#,50 + (1" – 1! . 3,5 = "57 t/m2

13. Lapisan 13

Kedalaman #3,50 m – #5, 00 m

%13 = q12 + (γ13 – γ! . h13= "57 + (19 – 1! . 1,5 = "$# t/m2

1#. Lapisan 1#

Kedalaman #5,00 m – #7,00 m

%1# = q13 + (γ1# – γ! . h1# = "$# + (1" – 1! . 2 = 71# t/m2

15. Lapisan 15

Kedalaman #7,00 m – 50,00 m

%15 = q1# + (γ15 – γ! . h15 = 71# + (1# – 1! . 3 = 753 t/m2

7






$

&'ai) 2.1 *eanan

e'ti)al






3. Menghitung Kapasitas Daya Dukung Tanah ( Bearing Capacity)

3.1 -enhitn Kapasitas aa )n *anah (ea'in apait! -et4de

-ee'h4

-ee'h4 ( 19"3 ! dan 'inh ansen ( 1970 ! mem6e'i)an pe'samaan

daa d)n denan mempe'tim6an)an 6ent) p4ndasi , )emi'inan

6e6aan dan )at ese' tanah diatas p4ndasina, se6aai 6e'i)t

...(1!

enan q = aa d)n ltimit

8, 8q, 8γ = a)t4' daa d)n nt) p4ndasi meman:an

;, ;q , ;γ = a)t4' 6ent) p4ndasi

, dq, dγ = a)t4'<a)t4' )edalaman p4ndasi

i, iq, iγ = a)t4' )emi'inan 6e6an

= Le6a' p4ndasi ee)ti

>0 = ? γ =*e)anan overburden pada dasa' p4ndasi

= Kedalam p4ndasi

γ = 6e'at @4lme tanah

Grafik 3.1 Faktr!Faktr Daya Dukung Meyerhf

a)t4' < a)t4' 6ent) p4ndasi -ee'h4 ( 19"3 ! se6aai 6e'i)t

9






a)t4' < a)t4' )emi'inan 6e6an p4ndasi -ee'h4 ( 19"3 ! se6aai

6e'i)t

a)t4' )edalaman p4ndasi -ee'h4 ( 19"3 ! se6aai 6e'i)t

>e'hitnan

10






Ga"bar 3.1 Pn#asi Telapak

Ai)a p4ndasi 6e'ada di6aah )edalaman 1,5 m (! denan a)t4'

)eamanan 3. *ent)an )apasitas daa d)n tanah ( Bearing Capacit!

denan met4de -ee'h4 pada p4ndasi 6:' san)a'.

e'dasa')an 'ai) 3.1 dipe'4leh a)t4'<a)t4' daa d)n -ee'h4

se6aai 6e'i)t

>4ndasi :' ;an)a' B= 0C

8(6s! = ".1" D 8q(6s! = 1 D 8γ(6s! = 0

>4ndasi -eman:an

B= 0C

8(m! = 5.1# D 8q(m! = 1D 8γ(m! = 0

a)t4' < a)t4' 6ent) p4ndasi meman:an men't -ee'h4 se6aai

6e'i)t

19$.1!11#.5

1"."!((1!1

!(

!(!((1 =−+=−+=

B

B

m !c

bs !c

"

BSc

1!11

1!((1!1

!(

!(!((1 =−+=−+=

B

B

m !#

bs !#

"

BS#

1!10

0!((1!1

!(

!(!((1 =−+=−+=

B

B

m !

bs !

"

BS

γ

γ γ

at4' < a)t4' )emi'inan 6e6an p4ndasi -ee'h4 se6aai 6e'i)t

1!90

1(1 20

=−+= δ

ci

1== c# ii

1!1(1 20

=−+=ϕ

δ γ i

a)t4' )edalaman p4ndasi -ee'h4 se6aai 6e'i)t

Btg

Btg

B

D$ d

c

3.01#5

5.12.01!

2#5(2.01 00 +=+=++=

ϕ

11






1== γ d d #

aa d)n ltimit γ γ γ γ

γ ! Bid S ! Poid S c! id S #####ccccu

⋅⋅⋅⋅⋅+⋅⋅⋅⋅+⋅⋅⋅= =5.0

0!155.2111(E13.31"."!3.0

1(19$.1F +⋅⋅⋅⋅+⋅⋅⋅+⋅= B

#u

55.2!30".7

(35#.2# ++= B

#u

!30".7

(902.2" B

#u +=

aa d)n tanah ( = 3 !

γ γ

⋅∑+⋅−

= D$ %k

D$ ## u

desain

!(

55.23

!55.2!30".7

902.2"((

+−+

= B#desain

55.2$5.0!#35.2

(9"7.$ +++= B

#desain

!#35.2

(""7.10 B

#desain +=

!(desain s struktur ## =

B B

P #35.2""7.10

2 +=

B B P #35.2""7.10 2 +=

;ehina dipe'4leh pe'samaan mm nt) menent)an dimensi p4ndasi

se6aai 6e'i)t

0#35.2""7.10 2 =+ B B

12






$. Menghitung Di"ensi Pn#asi %ete"pat #an Tulangan Pn#asi

#.1 -enhitn imensi >4ndasi

&ntuk P ' 2 tn

0#35.2""7.10 2 =+ B B

>e'hitnan dila))an denan menna)an 6antan microso$t e&cel .

-a)a didapat nilai = 1.#5 G 1.5m

Aadi,nt) > = 25 t4n, di6th)an dimensi p4ndasi 1.5 m H 1.5 m

e) daa d)n p4ndasi

0#35.2""7.10 2 =+ B B

10.""7(1.5!2 + 2.#35(1.5! = 25

27."53 I 25 ......JK

&ntuk P ' tn

0#35.2""7.10 2 =+ B B


-a)a didapat nilai = 2.05G 2.1 m

Aadi,nt) > = 50 t4n, di6th)an dimensi p4ndasi 2.1 m H 2.1 m

13

e) Ke'nthan

>1 = 15 m H 15 m

a I 6

27."53 I 25 JK






e) daa d)n p4ndasi

0#35.2""7.10

2 =+ B B

10.""7(2.1!2 + 2.#35(2.1! = 50

52.15# I 50 ...... JK

&ntuk P ' * tn

0#35.2""7.10 2 =+ B B


-a)a didapat nilai = 2.55G2." m

Aadi,nt) > = 70 t4n, di6th)an dimensi p4ndasi 2." m H 2." m

0#35.2""7.10 2=+ B B

10.""7(2."!2 + 2.#35(2."! = 75

7$.#39 I 75 ...... JK

1#

e) Ke'nthan

>2 = 15 m H 20 m

a I 6

52.15# I 50 JK

e) Ke'nthan

>3 = 20 m H 20 m

a I 6

7$.#39 I 75 JK






Tabel $.1 +asil Perhitungan Di"ensi Pn#asi

>4ndasi e6an (t4n! (m! aa )n (t/m2!

>1

>2

>3

25

50

75

1.5

2.1

2."

27."53

52.15#

7$.#39

$.2 Desain Tulangan Pn#asi %ete"pat

>4ndasi<p4ndasi >3 (2." m H 2."m!

ata ;t')t' 1. )'an K4l4m >3 = 20m H 20m2. aa d)n tanah (qn! 27.153 t/m2

3. e6an pada )4l4m 75 t4n

nt) tanah an )p )at denan p4ndasi dan)al, se)at<)atna ha's

te'sedia te)anan iin se6esa' 10 t.m2

nt) me'enana)an tlanan, dipe'l)an 6e6e'apa nilai, a)ni 1. *ida) te':adi m4men pada p4ndasi an dia)i6at)an 4leh )4l4m, an

did)n adalah a)i6at da'i p4ndasi dan tanah.2. *e6al p4ndasi (h! 1/3; D dimana tida) 64leh )'an da'i 15 m (;K;8M

2002!, sedan)an ; adalah :a'a) sisi te'pan:an p4ndasi denan )4l4m,

ma)a

2.12

2,0".2=

−=S

-enent)an nilai - (-4men ltimate!

291.1"2

2

== S

' u δ t4n

3. nt) )eamanan te'hadap lent' denan Nt = 27.153 t/m2

-a)a h I 1/3 ;, h I 0,2 m sedan)an men't ;K;8M 2002 hmin = 0,15 m.

enan demi)ian h = 0,2 m#. Oenana mt 6et4n (’! = )225 = 1$,"75 -pa5. -t 6esi <2# denan teanan leleh #00 -pa". Aa'a) psat tlanan te'hadap sisi la' 6et4n = 75 mm7. a)t4' dist'6si teanan 6et4n = 0,$5

Men,ari -asi Tulangan

!"00(

"00.=

$5,0. $ $

$cb

+= β ρ =

!#00"00(

"00.

#00

"75,1$$5,0.$5,0

+=

b ρ

0202,0=b ρ

15






×

×××−=

=$5,0

75,02/11=..75,0maH

$c

$ $ R b

b

ρ ρ

×

×××−=

"75,1$$5,0

#000202,075,02/11#00.0202,0.75,0maH R

917,#maH = R

φ u&

n

' ' =

$,0

291.1"=n '

3"3.20=n ' t4n.m

!(10

2

"

.

d b ' R n

n×

=

!2252"00(

10.3"3.202

"

×=n R

15#,0=n R

0,15# P O maH (JK!

-asi Tulangan yang #iperlukan

×−−=

$c

R

$

c $ n

$5,0

211

=

=$5,0 ρ

00039$#,0= ρ

=

".2min

$= ρ

#00

".2min = ρ

00"5,0min = ρ

0,00039$# P Qmin (*ida) JK!

-a)a, dipa)ai 'asi4 tlanan se6esa' 0,00"5 (Qmin!

uas tulangan yang #iperlukan /Rs = Q.6.d = 0,00"5 . 2"00 . 225 = 3$02.5 mm2

enan menna)an tlanan 22<225

Amlah tlanan = 2

#

1d

(s

×∏×

= 21"1#,3

#

1

"",11$2

××

= 10.00$ G 11 6atan

1"






Rs tlanan nt) 1 = :mlah tlanan H S H T H d2

= 11 H S H 3,1# H 222

= #179.3# mm2

Rs pasana U Rs'enana

#179.3# U 3$02.5 (JK!

nt) p4ndasi >1 dan >2 menna)an tlanan an sama denan >3, hal

ini di)a'ena)an daa d)nna le6ih )eil.

17






. Menghitung Penurunan ( Settlement ) yang Ter0a#i kibat Pe"bebanan

>en'nan ;ee'a *anah 4m4en denan *e6al *a) *e'hina

>e'samaan pen'nan see'a ata pen'nan elastis da'i p4ndasi an

te'leta) dipe'm)aan tanah an h4m4en, elastis, s4t'4pis, pada media

semi ta) te'hina dinata)an 4leh

....................................... (13)

enan

;i = >en'nan see'aq = *e)anan pada dasa' p4ndasi

= Le6a' p4ndasi

V = Rn)a p4iss4n

Mp = a)t4' pena'h

W = -4dls Wlasti

Tabel .1 Perkiraan M#ulus lastis () (Bowles, 1977)

Ma,a" tanah ( kg,"2 )

Lempn

;anat Lna)

Lna)

;edan

Ke'as

e'pasi'

3 – 30

20 – #0

#5 – 90

70 – 200

1$






>asi'

e'lana

*ida) >adat

>adat

>asi' dan Ke'i)il

>adat

*ida) >adat

Lana

L4ess

;e'pih

300 – #25

50 – 200

100 – 250

500 – 1000

$00 – 2000

500 – 1#00

20 – 200

150 – "00

1#00 < 1#000

Tabel .2 Perkiraan ngka Pissn (4) (Bowles, 1977)

Ma,a" Tanah 4

Lempn Aenh

Lempn *a) Aenh

Lempn e'pasi'

Lana

>asi' >adat

>asi' Kasa' (e = 0.# – 0.7 !

>asi' als ( e = 0.# – 0.7 !

at ( aa) te'antn da'i tipena !

L4ess

0.# – 0.5

0.1 – 0.3

0.2 – 0.3

0.3 – 0.35

0.2 – 0.#

0.15

0.25

0.1 – 0.#

0.1 – 0.3

19






Tabel .3 Faktr Pengaruh Im (Lee,1962) #an Ip (Schleicher,1962) &ntuk

Pn#asi Kaku #an Faktr Pengaruh &ntuk Pn#asi Fleksibel

(Ter5aghi617$3) dari Bowles, 1977)

8antusFleksibel ( Ip ) Kaku

Pusat %u#ut -ata!rata Ip Im

Lin)a'an

:'san)a'

Wmpat pe'sei pan:an

" ) B * 1.5

2.0

5.0

10.0

100.0

1.00

1.12

1.3"

1.53

2.10

2.52

3.3$

0."#

0.3"

0."$

0.77

1.05

1.2"

1."9

0.$5

0.95

1.20

1.31

1.05

1.2"

1."9

0.$$

0.$2

1.0"

1.20

1.70

2.10

3.#0

3.70

#.12

#.3$

#.$2

#.93

5.0"

>e'hitnan

Ga"bar .1 Pn#asi Telapak

20






i)etahi p4ndasi )a) dan tanah h4m4en te6al ta) te'hina. Aenis tanah

lempn 6e'lana )e4)latan.

a'i *a6el 5.1, *a6el 5.2 dan *a6el 5.3 dipe'4leh data se6aai 6e'i)t

W = 300 t/m2

V = 0.3

Mp = 0,$2

.1 Menghitung Penurunan (Settlement) Met#e Meyerhf

a'i pe'hitnan se6elmna, denan menna)an met4de -ee'h4

dipe'4leh )'an p4ndasi se6aai 6e'i)t

Tabel .$ Di"ensi Pn#asi #engan Met#e Meyerhf

9 P ( tn ) 8 ( " )

1 25 1.5

2 50 2.1

3 75 2."

Penurunan akibat beban 2 tn pa#a pn#asi P1 ' 1. " : 1. "

*e)anan pada dasa' p4ndasi ( q !

2 B

P # =

25.1

25=#

111.11=# t/m2

>en'nan see'a ( ;i !

+p ,

B#S i

!1( 2 µ −⋅

=

$2.0!3.01(300

5.1111.11 2−⋅

=iS

0#1.0=iS m

#1=iS mm

>4ndasi )a), ma)a pen'nan see'a te')4'e)si ( ;i !

ii S S ⋅= 93.0=

21






#193.0= ⋅=iS

13.3$==iS mm

>4ndasi le)si6el, ma)a pen'nan see'a te')4'e)si ( ; i !

ii S S ⋅= $5.0=

#1$5.0= ⋅=iS

$5.3#==iS mm

Penurunan akibat beban tn pa#a pn#asi P2 ' 2.1 " : 2.1 "


2 B

P # =

21.2

50=#

337.11=# t/m2


+p ,

B#S i

!1( 2 µ −⋅

=

$2.0!3.01(300

1.2337.11 2−⋅

=iS

059.0=iS m

59=iS mm


ii S S ⋅= 93.0=

5993.0= ⋅=iS

$7.5#==iS mm


ii S S ⋅= $5.0=

59$5.0= ⋅=iS

5.50==iS mm

Penurunan akibat beban * tn pa#a pn#asi P3 ' 2.; " : 2.; "


2 B

P # =

2".2

75=#

09#.11=# t/m2


+p ,

B#S i

!1( 2 µ −⋅

=

$2.0!3.01(300

".209#.11 2−⋅

=iS

071.0=iS m

71=iS mm

22







ii S S ⋅= 93.0=

7193.0= ⋅=iS

03.""==iS mm


ii S S ⋅= $5.0=

71$5.0= ⋅=iS

35."0==iS mm

9P

(tn)

8

(")

(t"2)<p = (t"2)

%i

("")

%i>

("")

%i>

(,")

1 25 1.5 300 0,3 0.$2 11.111 #1 3$.13 3.$13

2 50 2.1 300 0,3 0.$2 11.337 59 5#.$7 5.#$7

3 75 2." 300 0,3 0.$2 11.0.9# 71 "".03 "."03

Tabel . Penurunan (Settlement) Meyerhf Pa#a Pn#asi Kaku

9P

(tn)

8

(")

(t"2)<p

=

(t"2)

%i

("")

%i>

("")

%i>

(,")

1 25 1.5 300 0,3 0.$2 11.111 #1 3#.$5 3.#$5

2 50 2.1 300 0,3 0.$2 11.337 59 50.15 5.015

23






3 75 2." 300 0,3 0.$2 11.09# 71 "0.35 ".035

Tabel .; Penurunan ( settlement ) Meyerhf Pa#a Pn#asi Fleksibel

2#






.2 Penurunan Knsli#asi Pri"er Met#e Mayerhf

>en'nan )4ns4lidasi p'ime' dihitn denan menna)an pe'samaan

imana

Xe >e'6ahan an)a p4'i a)i6at pem6e6anan

e4 Rn)a p4'i aal

e1 Rn)a p4'i saat 6e'ahi'na )4ns4lidasi

*e6al lapisan tanah an ditin:a

Ga"bar .2 apisan Tanah Pn#asi

25






Lempn te')4ns4lidasi 6e'le6ihan ( 4@e'4ns4lidated !.

> = 7,0 t/m2

= 0,59#

' = 0,03

?nth Perhitungan &ntuk Di"ensi 16 " : 16 "

>41 = 1,25 ×1,7 = 2,125

1 = L1 = 0,15/2 = 0,075

1/ = L1/ = 0,075/1,25 = 0, "

M = 0,103

X> = #Mq = # ? 0,103 ? 11,111 = #,577 t/m2

0,7

577,#125,2l4+59#,0

125,2

0,7l4+03,01

++=∆ e

1e∆ = 0,00#3

1,0"$"7,2#,01101 =×=×= Gse ω

- eo

Sc e ×+

∆=

1

1

11

0025,025,1

"$0,11

00#3,01 =×

+=Sc m

>42 = (2,5 × 1,7! + (1,25 × 0,7! = 5,125

1 = L1 = 0,15/2 = 0,075

1/ = L1/ = 0,075/3,75 = 0,2

M = 0,01$

X>2 = #Mq = # ? 0,01$ ? 11,111 = 0,799 t/m2

0,7

799,0125,5l4+59#,0

125,5

0,7l4+03,02

++=∆ e

2e∆ = 0,03$

!(!( 221102 GsGse ×+×= ω ω

"13,2"7!,2#,0("7!,2#,0(02 =×+×=e

- eo

Sc e ×+

∆=

2

2

21

#50,075,32,13"1

03$,02 =×

+=Sc m

2"






>43 = (2,5 × 1,7! + (2,5 × 0,7! + (1,25×0,7! = ",$75

1 = L1 = 0,15/2 = 0,075

1/ = L1/ = 0,075/",25 = 0, 12M = 0,021

X> = #Mq = # ? 0,021 ? 11,111 = 0,933 t/m2

0,7

933,0",$75l4+59#,0

",$75

0,7l4+03,01

++=∆e

1e∆ = 0,02$

20#,3"7!,2#,0("7!,2#,0("7!,2#,0(

!(!(!(

03

33221103

=×+×+×=

×+×+×=

e

GsGsGse ω ω ω

- eo

Sc e ×+

∆=

1

11

1

0#1,0",253,20#1

02$,01 =×

+=Sc m

; t4tal = 0,0025 + 0,0#5 + 0,0#1 = 0,0$$ m

Tabel .* Penurunan Knsli#asi Pri"er ( Meyerhf )

8 : apisan

P>81'1 5 <p

@P@e e

%, A%,

("2) (t"2) (t"2) (") (")

1,5 H 1,5

1 2,125 0,75 0," 0,103 #,577 0,00#3 1,0"$ 0,0025

0,0$$2 5,125 0,75 0,2 0,01$ 0,799 0,03$ 2,13" 0,0#1

3 ",$75 0,75 0,12 0,021 0,933 0,02$ 3,20# 0,0#5

2,1 H 2,1

1 2,125 1,05 0,$# 0,1#1 ",39# 0,0"" 1,0"$ 0,032

0,10"2 5,125 1,05 0,2$ 0,037 1,"7$ 0,057 2,13" 0,01$

3 ",$75 1,05 0,1"$ 0,251 11,3$ 0,239 3,20# 0,05"

2," H 2,"

1 2,125 1,3 1,0# 0,1"$ 7,#55 0,09" 1,0"$ 0,0#"

0,0"#2 5,125 1,3 0,3#" 0,05" 0,#17 0,05" 2,13" 0,013

3 ",$75 1,3 0,20$ 0,01$ 0,79$ 0,023 3,20# 0,005

27





2$

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