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-
-
′
K 0 =
σ h σ v
′
K 0 = 1 − sin φ ′ K 0 = (1 − sin φ ) OCR
(1 − sin φ )(OCR)sin φ K0 =
-
’
-
⎡ γ d ⎤ − 1⎥ × 5.5 K 0 = (1 − sin φ ) + ⎢ ⎢⎣ γ d ,min ⎥⎦
OCR =
σ c σ v
P0 =
1 2
K 0γ H 2
′ ′
′
σ h = K 0 ⋅ γ ⋅ z
′
σ h = K 0 [γ ⋅ H 1 + γ ′( z − H 1 )]
u = γ w ( z − H 1 )
′
σ h = σ h + u P0 =
1 2
K 0γ H 1 + K 0γ H 1 H 2 +
1 2
( K 0γ ′ + γ w ) H 22
’
z
σ v H
σ h
’
’
’
(σ v − σ a ) / 2 sin φ = ⎛ σ + σ a ⎞ c ⋅ cot φ + ⎜ v ⎟ 2 ⎝ ⎠