DD Beban Eksentris
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Transcript of DD Beban Eksentris
![Page 1: DD Beban Eksentris](https://reader034.fdocument.pub/reader034/viewer/2022042522/55cf992e550346d0339c0b24/html5/thumbnails/1.jpg)
Yulvi Zaika
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Sub Pokok Bahasan Daya dukung pondasi dangkal untuk beban eksentris
satu arah Daya dukung pondasi dangkal untuk beban eksentris
dua arah Tugas Analisa daya dukung pondasi pondasi dangkal
eksentris (cantilever retaining wall).
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Pendahuluan Ketika pondasi dangkal dibebani oleh beban eksentris
maka diasumsikan tegangan kontak akan menurunsecara linear dari ujung ke tumit. Walaupunsebenarnya tegangan tersebut tidak linear. Meyerhof memperkenalkan konseep lebar efektif.
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BEBAN EKSENTRIS PADA PONDASI DANGKAL
Bila pondasi telapak tidak saja menahan bebanvertikal tetapi juga menahan momen guling makaresultan tegangan tanah tidak terletak pada titik pusatpondasi Bila kolom tidak terletak di pusat masa pondasi
PM
P
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DISTRIBUSI TEGANGAN TANAH AKIBAT TEG. VERTIKAL DANMOMEN
P
M
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Tegangan kontak akibat beban vertikal dan momen
P
M
P
Me
e = B/6qmaxqmin
e < B/6qmaxqmin
e > B/6qmaxqmin
B
eR
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P
e = B/6qmaxqmin
e < B/6qmaxqmin
e > B/6qmaxqmin
e
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Pondasi Lajur
e
B
B-2e
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Pondasi telapak
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TEGANGAN KONTAK
Untuk e = B/6 atau e < B/6
Dimana qmax dan qmin merupakan tegangan tekan
L
e
BL
Pq
L
e
BL
Pq
61
61
min
max
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TEGANGAN KONTAK
Untuk e > B/6
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EKSENTRISITAS BEBAN e > B/6 satu arah
B’ = B- 2e
B X L
B
P M
B X L
B
e
2e B’
L
=
e < B/6
qmax
e > B/6
qmax
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Contoh 1
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Contoh2P=1000kN, Mx = 500kNm, My= 400kN mBila L = 2.5m dan B = 2.5 m maka tentukan tegangan
kontak yang terjadi ?
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TUGAS
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Dua arah eksentrisitas, ex dan ey B’= B- 2e L’ = L – 2e
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Kasus 1
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KASUS 2
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KASUS 3
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KASUS 4