A rectangular footing of size 2.8 m $\times$ 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_\gamma = 0.0 $; Shape factors: $ s_c = 1.16 $, $ s_q = 1.0 $ and $ s_\gamma = 1.0 $; Depth factors: $ d_c = 1.1 $, $ d_q = 1.0 $ and $ d_\gamma = 1.0 $; and Inclination factors: $ i_c = 1.0 $, $ i_q = 1.0 $ and $ i_\gamma = 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 $\underline{\hspace{1cm}}$.

Question

A rectangular footing of size 2.8 m $\times$ 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_\gamma = 0.0 $; Shape factors: $ s_c = 1.16 $, $ s_q = 1.0 $ and $ s_\gamma = 1.0 $; Depth factors: $ d_c = 1.1 $, $ d_q = 1.0 $ and $ d_\gamma = 1.0 $; and Inclination factors: $ i_c = 1.0 $, $ i_q = 1.0 $ and $ i_\gamma = 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 $\underline{\hspace{1cm}}$.

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Meyerhoff's method for the ultimate bearing capacity of a footing under eccentric load is given by the formula: $$ q_u = c N_c s_c d_c i_c + \gamma q N_q s_q d_q i_q + \frac{1}{2} \gamma B N_\gamma s_\gamma d_\gamma i_\gamma $$ Where: - $ c = 40 \, \text{kN/m}^2 $ is the cohesion, - $ \gamma = 18.2 \, \text{kN/m}^3 $ is the unit weight of the soil,\ - $ B = 2.8 \, \text{m} $ is the width of the footing, - $ L = 3.5 \, \text{m} $ is the length of the footing, - $ e = 0.8 \, \text{m} $ is the eccentricity of the load. Step 1: Calculate the ultimate bearing capacity Substitute the values into the formula for $ q_u $: $$ q_u = 40 \times 5.14 \times 1.16 \times 1.1 + 18.2 \times 1 \times 1.0 \times 1.0 + \frac{1}{2} \times 18.2 \times 2.8 \times 0.0 $$ $$ q_u = 40 \times 5.14 \times 1.16 \times 1.1 + 18.2 \times 1 \times 1.0 \times 1.0 $$ $$ q_u = 40 \times 5.14 \times 1.16 \times 1.1 + 18.2 $$ $$ q_u = 40 \times 5.14 \times 1.16 \times 1.1 + 18.2 = 439.60 \, \text{kN/m}^2 $$ Step 2: Apply factor of safety Now, using a factor of safety of 2.5: $$ q = \frac{q_u}{\text{FoS}} = \frac{439.60}{2.5} \approx 175.84 \, \text{kN/m}^2 $$ Thus, the load that can be applied on the footing is: $$ \text{Load} = q \times \text{Area} = 175.84 \, \text{kN/m}^2 \times 2.8 \times 3.5 \, \text{m}^2 $$ $$ \text{Load} = 439.00 \, \text{kN} $$ Thus, the load that can be applied on the footing is $ \boxed{439.00} \, \text{kN} $.

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Correct Answer: 439

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