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RANKINE’S THEORY OF BEARING
CAPACITY
PREPARED
BY
ER. SANJEEV SINGH
SNJV@432@GMAIL.COM
RANKINE’S THEORY OF BEARING CAPACITY
 Rankine 1885 considered the plastic equilibrium of two
elements, one immediately beneath the footing and the other
just beyond the edge of the footing.
 For the element I beneath the footing, the vertical stress in
the major principal stress and the lateral stress is the minor
principal stress.
 However, for the element II, the lateral stress becomes the
major stress, and the vertical stress becomes the minor
principal stress.
 When the footing pressure approaches the ultimate bearing capacity (𝑞 𝑢),
the element I attains a state of plastic equilibrium.
However, the element I can fails only when adjacent element II also fails.
For the active case
For element I, substituting
For element II,
Therefore
σ3 = 𝑡𝑎𝑛2
(450
−
′
2
)σ1
σ3=γ𝐷𝑓
σ1=
σ3
𝑡𝑎𝑛2(450−
′
2
)
=
γ 𝐷 𝑓
𝑡𝑎𝑛2(450−
′
2
)
………….. (b)
 As σ3 of element I is equal to σ1 of element II, from eqs, (a) and(b)
 The approximate value of the ultimate bearing capacity of the soil.
Where q is the intensity of loading at base.
𝑡𝑎𝑛2(450 −
′
2
)𝑞 𝑢 =
γ 𝐷 𝑓
𝑡𝑎𝑛2(450−
′
2
)
Or 𝑞 𝑢= γ𝐷𝑓×
1
𝑡𝑎𝑛4(450−
′
2
)
Or 𝑞 𝑢= γ𝐷𝑓 𝑡𝑎𝑛4
(450
+
′
2
)
Or 𝑞 𝑢= γ𝐷𝑓
1−𝑠𝑖𝑛 ′
1+𝑠𝑖𝑛 ′
2
(𝐷 𝑓) min =
𝑞
γ
1−𝑠𝑖𝑛 ′
1+𝑠𝑖𝑛 ′
2
Rankine’s Theory of Bearing Capacity,   Rankine’s ANALYSIS FOR BEARING CAPACITY

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Rankine’s Theory of Bearing Capacity, Rankine’s ANALYSIS FOR BEARING CAPACITY

  • 1. RANKINE’S THEORY OF BEARING CAPACITY PREPARED BY ER. SANJEEV SINGH SNJV@[email protected]
  • 2. RANKINE’S THEORY OF BEARING CAPACITY  Rankine 1885 considered the plastic equilibrium of two elements, one immediately beneath the footing and the other just beyond the edge of the footing.  For the element I beneath the footing, the vertical stress in the major principal stress and the lateral stress is the minor principal stress.  However, for the element II, the lateral stress becomes the major stress, and the vertical stress becomes the minor principal stress.
  • 3.  When the footing pressure approaches the ultimate bearing capacity (𝑞 𝑢), the element I attains a state of plastic equilibrium. However, the element I can fails only when adjacent element II also fails. For the active case For element I, substituting For element II, Therefore σ3 = 𝑡𝑎𝑛2 (450 − ′ 2 )σ1 σ3=γ𝐷𝑓 σ1= σ3 𝑡𝑎𝑛2(450− ′ 2 ) = γ 𝐷 𝑓 𝑡𝑎𝑛2(450− ′ 2 ) ………….. (b)
  • 4.  As σ3 of element I is equal to σ1 of element II, from eqs, (a) and(b)  The approximate value of the ultimate bearing capacity of the soil. Where q is the intensity of loading at base. 𝑡𝑎𝑛2(450 − ′ 2 )𝑞 𝑢 = γ 𝐷 𝑓 𝑡𝑎𝑛2(450− ′ 2 ) Or 𝑞 𝑢= γ𝐷𝑓× 1 𝑡𝑎𝑛4(450− ′ 2 ) Or 𝑞 𝑢= γ𝐷𝑓 𝑡𝑎𝑛4 (450 + ′ 2 ) Or 𝑞 𝑢= γ𝐷𝑓 1−𝑠𝑖𝑛 ′ 1+𝑠𝑖𝑛 ′ 2 (𝐷 𝑓) min = 𝑞 γ 1−𝑠𝑖𝑛 ′ 1+𝑠𝑖𝑛 ′ 2