tugas analisa plat & cangkang jawaban 3
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7/23/2019 Tugas Analisa Plat & Cangkang Jawaban 3
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Nama : Muhammad Yusuf
NPM : 12.11.106.701101.0703
Jurusan : Teknik Sipil
Semester : VII (tujuh)
Kelas : C
Sign :
3 Diketahui : a. Dimensi
2 D 1 3
h = 350 mm
b = 260 mm
d' = 50 mm
35 4 D 13 d = 300 mm
26
b. Mutu
1 Mpa = 10.1971 Kg/cm²
1 Kg/cm² = 0.0980671 MPa
fc' = 200 Kg/cm² ≈ 20 MPafy = 4000 Kg/cm² ≈ 400 MPaβ1 = 0.85
Ø = 0.8
Ditanya : Berapa momen yang mampu dipikul penampang tersebut?
Penyelesaian :
a. Menghitu ng
As = 4 D 13
= 4 x 1/4 x π x D²
= 4 x 0.25 x 3.14 x 169
= 530.66 mm²
As' = 2 D 13
= 2 x 1/4 x π x D²
= 2 x 0.25 x 3.14 x 169
= 265.33 mm²
1.4
fy
1.4
4000
= 0.00035
As
b . d
260 x 300
= 0.0068 > 0.00035 → OK
As'
b . d
260 x 300
= 0.003402
b. memeriksa apakah tulangan tekan sudah leleh atau belum
ρ min =
=
ρ =
= 530.66
ρ' =
=265.33
7/23/2019 Tugas Analisa Plat & Cangkang Jawaban 3
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ρ - ρ' = 0.0068 - 0.003402
= 0.003398
0.85 f'c d' 600
fy d 600 fy
0.85 . 20 . 50 600
400 . 300 600 - 400
= 0.85 . 0.00708333 . 3= 0.018063
ρ - ρ' < k → Tulangan tekan belum leleh
c. Karena tulangan belum leleh, maka fs' < fy , menentukan fs' dan ρ max
0.85 . f'c . . d'
fy d
0.85 . 20 . 0.85 . 50
. 400 . 300
= 600 . -0.7718756
= -463.12537 Mpa
d. Cek Tulangan maksimum
0.85 f'c 600
fy 600 + fy
0.85 20 600
400 600 + 400
= 0.75 . 0.036125 . 0.6 + -0.0039
= 0.01231737
ρ = 0.0068 < ρ max = 0.01232 OK
e. Menghitung Mn dan Mu
a = 127.5
Mn = 212264 - -122881.05 . 236.25 + -122881 . 250
= 48457755.5 Nmm
Mu = 0.8 . Mn
= 0.8 . 48457755.5
= 38766204.4 Nmm
= 38.7662044 KNm
1= -k .
= 0.85 . -
0.003398
( ρ - ρ' )
= 600 . 1 -
fs' = . 1 -600
ρ max
0.75 -
-
=
0.75=