Questions: The floor of school corridor is made of 200 mm-thick cinder concrete and this floor fill was installed on the top of concrete having a specific weight of 23.6 kN / m^3 as shown figure below. If the floor and bottom concrete have a length 8 m and width of 1.25 m, what is the resultant force caused by the dead load and the live load?

Solution Steps

The thickness of the cinder concrete is given as 200 mm (0.2 m). The area of the floor is given by the length and width: \[ \text{Area} = 8 \, \text{m} \times 1.25 \, \text{m} = 10 \, \text{m}^2 \] The volume of the cinder concrete is: \[ \text{Volume} = \text{Area} \times \text{Thickness} = 10 \, \text{m}^2 \times 0.2 \, \text{m} = 2 \, \text{m}^3 \]

The specific weight of the cinder concrete is given as 23.6 kN/m³. The dead load is calculated by multiplying the volume by the specific weight: \[ \text{Dead Load} = \text{Volume} \times \text{Specific Weight} = 2 \, \text{m}^3 \times 23.6 \, \text{kN/m}^3 = 47.2 \, \text{kN} \]

From Table 1-4, the live load for corridors above the first floor is given as 3.83 kN/m². The area of the floor is 10 m²: \[ \text{Live Load} = \text{Area} \times \text{Live Load per Unit Area} = 10 \, \text{m}^2 \times 3.83 \, \text{kN/m}^2 = 38.3 \, \text{kN} \]

Final Answer