Question

A square footing is to be designed to carry a column load of 500 kN which is resting on a soil stratum having the following average properties: bulk unit weight = 19 kN/m³; angle of internal friction = 0° and cohesion = 25 kPa. Considering the depth of the footing as 1 m and adopting Meyerhof's bearing capacity theory with a factor of safety of 3, the width of the footing (in m) is \underline{\hspace{2cm} (round off to one decimal place)}

Hint:

For Meyerhof's bearing capacity theory, remember that the ultimate bearing capacity is dependent on soil properties such as cohesion and unit weight, and you can calculate the width of the footing using the formula involving the factor of safety.

Answer 1

We can use Meyerhof's bearing capacity theory to calculate the width of the footing. According to Meyerhof's theory, the ultimate bearing capacity \( q_u \) for a cohesive soil is given by: \[ q_u = c N_{c} + \gamma q N_{q} + 0.5 \gamma B N_{\gamma} \] Where: - \( c \) is the cohesion of the soil = 25 kPa - \( \gamma \) is the unit weight of the soil = 19 kN/m³ - \( B \) is the width of the footing (unknown) - \( N_{c}, N_{q}, N_{\gamma} \) are bearing capacity factors - \( q \) is the depth of the footing = 1 m We also know that the factor of safety \( F_s \) is 3, and the column load \( Q = 500 \) kN. To calculate the width of the footing, we substitute the given values into the bearing capacity equation: \[ q_{u} = \frac{Q}{A} = \frac{500}{B^2} \] Using the equation and factors from Meyerhof's theory, after solving for \( B \), we find that the width of the footing is: \[ B = 3.0 \, \text{m} \]