A rectangular footing of size 2.8 m × 3.5 m is embedded in a clay layer and a vertical load is placed with an eccentricity of 0.8 m as shown in the figure (not to scale). Take Bearing capacity factors: N_{c} = 5.14, N_{q} = 1.0, and N_{γ} = 0.0; Shape factors: s_{c} = 1.16, s_{q} = 1.0 and s_{γ} = 1.0; Depth factors: d_{c} = 1.1, d_{q} = 1.0 and d_{γ} = 1.0; and Inclination factors: i_{c} = 1.0 and i_{q }= 1.0 and i_{γ} = 1.0.

Using Meyerhoff's method, the load (in kN, round off to two decimal places) that can be applied on the footing with a factor of safety of 2.5 is _______

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GATE CE 2021 Official Paper: Shift 2

CT 1: Ratio and Proportion

2672

10 Questions
16 Marks
30 Mins

**Concept:**

**Meyerhoff's Method:**

(i) Mayerhoff's theory takes into account depths, shape, and inclination factors together in the bearing capacity equation.

(ii) It can be applied to shallow, deep and foundations to all shapes and loading conditions, except foundations on built-up shapes.

(iii) The ultimate bearing capacity is given by

**q _{u} = c N_{c} S_{c }d_{c} i_{c} + γD_{f} N_{q} S_{q} d_{q} i_{q} + 0.5Bγ N_{γ} S_{γ} d_{γ} i_{γ}**

Where,

S_{c}, S_{q}, S_{γ} are shape correction factor

d_{c}, d_{q}, d_{γ} are depth correction factor

i_{c}, i_{q}, i_{γ} are inclination correction factor

**Calculation:**

Given,

Soil is clay for which, γ = 18.2 kN/m^{3}, ϕ = 0°, c = 40 kN/m^{2}

Footing size = (2.8 × 3.5) m, D_{f} = 1.5 m

For ϕ = 0°, N_{c} = 5.14, N_{q} = 1 and N_{γ} = 0

Shape factors: S_{c} = 1.16, S_{q} = 1.0 and S_{γ} = 1.0

Depth factors: d_{c} = 1.1, d_{q} = 1.0 and d_{γ} = 1.0

Inclination factors: i_{c} = 1.0, i_{q} = 1.0, i_{γ} = 1.0

Given loading is eccentric for which, e = 0.8 m

B' = 2(B/2 - e) = (B - 2e) = 2.8 - 2 × 0.8 = 1.2 m

L' = L = 3.5 m

The ultimate bearing capacity is given by,

qu = c Nc Sc dc ic + γDf Nq Sq dq iq + 0.5B'γ Nγ Sγ dγ i_{γ}

∵ N_{γ} = 0, so

qu = c Nc Sc dc ic + γDf Nq Sq dq i_{q}

q_{u} = (40 × 5.14) × (1.16 × 1.1 × 1) + (18.2 × 1.5) × (1) × (1 × 1 × 1) = 289.64 kN/m^{2}

Net ultimate bearing capacity (q_{nu}) = q_{u} - γD_{f} = 289.64 - 18.2 × 1.5 = 262.35 kN/m^{2}

Net safe bearing capacity (q_{ns}) = (qu - γDf)/Fos = 262.35/2.5 = 104.94 kN/m^{2}

Net safe Load that can be applied on the footing is the (Net safe bearing capacity) × L'B'

So, Net safe load = 104.94 × 3.5 × 1.2 = 440.75 kN