dbt - pondasi sondir

8/16/2019 Dbt - Pondasi Sondir

1/15

perencanaan struktur pondasi

Df

B

a. data perencanaan

Fc = 30 Mpa

Fy = 400 Mpa

b. menentukan beban berfaktor

qu = 1227.12 !

= 122712 k"

Mu = 300.4#$ !m

= 3004#.%%& k"m

c. daya dukun" tana'

dicoba dimensi

B = 1.# m

( = 2.2# m


2/15

Df = 2.4 m

Df)B = 1.& * 1.# +adi termasuk +enis pondasi dan"ka,

qc = 14-1%-14-1&-14-1#-1&-1%-20

$

= 1&.111 k")cm2

fr rata2 = 12.%&-&.&7-4.2$-&.2#-7.14-7.33-3.7#-7.7%-#.00

$= &.7%& / 4 maka termasuk tana' o'esif

q u,t = #-0.34 qc

= 10.47% k")cm2

qa,, = q u,t)F

= &.$%# k")cm2

= &$%#1.%#2 k")m2

q u,t = q a,,.B.(

= 23#7#0.000 k" / qu

k coy

d. ksentrisitas

e = Mu)qu

= 0.244%4%7$$4 m

e. e"an"an netto berfaktor


3/15

= 71$&$.04$$24%2#1 k")m2

= 574$.1722$73$71k")m2

karna q a,, / q ma6&$%#1.%#2 / 71$&$.04$$24%2#1

ok

+adi pondasi yan" direncanakan den"an keda,aman 2.4 m pan+an" 2.2# dan ,ebar 1.#

mampu mena'an beban sebesar %$%1& k"

f. penu,an"an pondasi

1. kontro, kekuatan "eser

teba, pondasi '8 = #00 mm

diameter tu,an"an = 22 mm

penutup beton = 70 mmtin""i efektf 9

db = '5p5D50.#D

= 3$7 mm

d, = '5p50.#D

= 41$ mm

untuk aksi 2 ara'9

te"an"an netto berfaktor = :u)B.(8

= 3&3#$.111111 k")m2 3&3.#$11111

;u = te"an"an netto 6 ,uas beban "eser

= te"an"an netto 6


4/15

= 247.0&01& !

"aya "eser norma,

= 1)3 bC.d fc>8?0.# @

= ड.#$%%

.337#$%%7 !

teba, p,at mencukupi untuk memiku, "aya "eser tanpa memer,ukan tu, "eser

Momen ,entur akibat beban berfaktor

a8 ara' pendek

Mu = 1)2 A te"an"an nettoA,?2AB , = B5a8)2

= %2.%30& !m = 1.#50.&8)2

= 0.4#

a8 ara' pan+an"

Mu = 1)2 A te"an"an nettoA,?2AB , = B5a8)2

= 1%#.&01$ !m = 2.2#50.&8)2= 0.%2#

er'itun"an ,uas tu,an"an ,entur

a) Tulangan arah panjang

En =

= 0.%%0$$43##2 Mpa

m = Fy) 0.%#Afc8

= 1#.&%&274#0$%

ρ = 0.002241$0&&

= 1.4)fy

= 0.003#

s =

= 21$$.7# mm2

,uas tu,an"an per meter ,ebar = 14&&.# mm2

b) Tulangan arah pendek

En =

= 0.2$1$&$3&72 Mpa

m = Fy) 0.%#Afc8

= 1#.&%&274#0$%

φ ;c

Mu)φ b d,

ρ min

ρ min / ρ maka pakai ρ min

ρ b.d,

Mu)φ b d,


5/15

ρ = 0.0007341#07

= 1.4)fy

= 0.003#

s == 312&.37# mm2

,uas tu,an"an per meter ,ebar = 13%$.# mm2

c) Luas tulangan jalur tengah (1.5 m)

= 2#01.1 mm2

,uas tu,an"an per meter ,ebar = 1&&7.4 mm2

dipi,i' tu,an"an % 5 D1&

d) luas tulangan jalur tepi (0.75) m

s tepi = s ara' pendek 5 s ten"a'

= &2#.27# mm2

dipi,i' tu,an"an D12 5 1%

ρ min

ρ min / ρ maka pakai ρ min

ρ b.db


6/15

0.%#


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!)m2

eser


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m

m


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dbt - pondasi sondir

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