Andrii_tugas Struktur Beton II ( Tugas Perorangan )

7/21/2019 Andrii_tugas Struktur Beton II ( Tugas Perorangan )

http://slidepdf.com/reader/full/andriitugas-struktur-beton-ii-tugas-perorangan- 1/7

Soal nomor 2.

Mutu beton

.20)'( Mpa fc =

Mutu baja

.400)( Mpa fy =

Dimensi kolom

.3030 cm=

Tebal selimut beton

.40 mmds =

(tebal selimut betonminimum).

.57

.125

kNm M

kN P

u

u

=

=

Rencanakan penulangan pada 2 sisi ?

Soal nomor 3.

Mutu beton

.20)'( Mpa fc =

Mutu baja

.400)( Mpa fy =

Dimensi kolom

.4040 cm=

Tebal selimut beton

.40 mmds =


.25 mm pokok tulangan =φ

.57

.125

kNm M

kN P

u

u

=

=

Rencanakan penulangan pada 4 sisi ?

Penyelesaian.



Soal nomor 2.

Data - data.

Mutu beton

.20)'( Mpa fc =

Mutu baja

.400)( Mpa fy =

Dimensi kolom

.3030 cm=

Tebal selimut beton

.40 mmds =


.57

.125

kNm M

kN P

u

u

=

=

Perhitungan tulangan longitudinal kolom.

Direncanakan penulangan pada kedua sisi kolom

'dshd −=

.260

40300

mm

mmmm

=

−=

itung nilai eksentrisitas

u

u

P

M e =

.456

10456,0

456,0

3

12557

mm

mm

m

kN kNm

=

⋅=

=

=

.13,0300

40

.52,1300

456

==

==

mm

mm

h

ds

mm

mm

h

e



.90000300300)( 22 mmmmmmhbmmbruto penampang luas Agr =⋅=⋅==

!enentuan nilai koe"esien reduksi

:)(ϕ

73,0

069,080,0

180

125.10,080,0

.10,080,0

,125180

10180000

180000

180000

180000

3003002010,0

'10,0

3

2

2

2

=

−=

−=

−=

=>=

⋅=

=

=

=

⋅⋅⋅=

⋅⋅⋅=

−

kN

kN

Pu

Pu

makakN PukN

kN

N

mm

mm Mpa

mmmm Mpa

hb fc

mm N

ϕ

ϕ

Pada sumbu vertikal:

Puφ.Agr .0,85 . fc

=125.103

0,73. 90000.0,85 .20=0,11

Pada sumbu horisontal

Pu

φ.Agr .0,85 . fc ×

e

h=0,11.1,52=0,17

Dari gra#k per$itungan tulangan longitudinal didapat

r% &.&&' % &*'+

ρ=r × β

% &*&&' , &*'+ % &*&&'

uas tulangan ( /st*u)

/st*u % ρ × Agr

%&*&&' , 0&&&& mm % 12 mm

3umla$ tulangan (n)



n= Ast

1

4. π . D ²

= 612mm²

1

4.3,14 .25mm ²

=¿

Soal nomor .

Mutu beton

.20)'( Mpa fc =

Mutu baja

.400)( Mpa fy =

Dimensi kolom

.4040 cm=

Tebal selimut beton

.40 mmds =


.25 mm pokok tulangan =φ

.57

.72

mton M

ton P

u

u

=

=

.720

.720107200072000

7200072

10001

101

1001

2

2

kN Pu

kN kN kg

kg ton

kg ton

kN kg

kg kN

=

=⋅=

=

=

=

=

−

−

Perhitungan tulangan longitudinal kolom.

Direncanakan penulangan pada kedua sisi kolom

'dshd −=



.360

40400

mm

mmmm

=

−=

itung nilai eksentrisitas

u

u

P

M e =

.347

.10347,0

347,0

3

7225

mm

mm

m

tonmton

=

⋅=

=

=

.1,0400

40

.87,0400

347

==

==

mm

mm

h

ds

mm

mm

h

e

.400400)( 2 mmmmhbmmbruto penampang luas Agr ⋅=⋅==

.160000

2

mm=

!enetuan nilai koe"esien reduksi

:)(ϕ

.65,0,720320

10320000

320000

320000

320000

4004002010,0

'10,0

3

2

2

2

==<=

⋅=

=

=

=

⋅⋅⋅=

⋅⋅⋅=

−

ϕ nilaimakakN PukN

kN

N

mm

mm Mpa

mmmm Mpa

hb fc

mm N

!ada sumbu 5ertikal 6

Mpamm

kN

fc Agr

Pu

2085,016000065,0

720

'85,0 2⋅⋅⋅

=

⋅⋅⋅ϕ



.41,0

41,0

1768000

10720

2

2

3

2

=

⋅

=

⋅=

mm

N

mm Mpa

N

mm N

36,087,041,0'85,0

=⋅=

⋅⋅⋅ h

e x

fc Agr

Pu

ϕ

Grafk perhtungan tulangan longitudinal kolom lantai 1.

85,0

05,0

=

=

β

r

0425,085,005,0 =⋅=⋅= β ρ r

uas tulangan

:)( ,u st A

Agr A u st ⋅= ρ ,

.6800

1600000425,0

2

2

mm

mm

=

⋅=

3umla$ tulangan (n) 6

2

41

,

D

A

n u st

⋅= π

.1486,13

625,490

6800

)25(14,3

6800

2

2

2

41

2

batng batng

mm

mm

mm

mm

⇒=

=

⋅⋅

=

3adi* digunakan tulangan total 6



).(680075,68682514 2

,

2 OK mm Amm A u st st ⇒=>== φ

3umla$ tulangan maksimum perbaris 6

12

+

−

⋅−=

Sn D

dshm

.692,5

14025

402400

btng btng

mmmm

mmmm

⇒=

+

+

⋅−=

7ambar $asil per$itungan tulangan longitudinal kolom.

400 mm

φ

400 mm

Andrii_tugas Struktur Beton II ( Tugas Perorangan )

Documents

Transcript of Andrii_tugas Struktur Beton II ( Tugas Perorangan )