Question

The steel framework is used to support the reinforced stone concrete slab that is used for an office. The slab is 200 mm thick. Calculate the value of the distributed load on BE. Take a = 2 m, b = 5 m. Assume density for concrete slab = 23.6 kN/m3. Hint: Look at Example 2 from class notes.

Answer 1

Answer:

Total distributed load on BE = 5 m²

Explanation:

The first process is to get the value for the Dead load (DL) on the slab;

This is determined by using the formula:

DL = ρ × t

DL = 23.6 kK/m³ × 0.2 m

DL = 4.72 kN/m²

From table 1.4 which relates to the office buildings, we derive the value for the minimum live load (LL) = 2.40 kN/m³

Hence the total load TL = Dead Load DL) + Live loaf (LL)

DL = (4.72 + 2.40) kN/m³

DL = 7.12 kN/m³

Now; from the imaginative view of the information given; the member BE which get the load from half area of the panel BEDC & half the area of BEFA panel parallel to member BE; then the tributary area on member BE can be calculated as;

$A_{BE} = ( \dfrac{a}{2}+\dfrac{a}{2}) \times width$

$A_{BE} = ( \dfrac{2}{2}+\dfrac{2}{2}) \times 1$

$A_{BE} =2\times 1$

$A_{BE} =2 m^2/m$

The total distributed load acting on BE is:

$Total \ load = TL \times A_{BE}$

$Total \ load = 7.12 \ \dfrac{kN}{m^2 }\times (2 \times \dfrac{m^2}{m})$

Total load = 2.5 × 2

Total load = 5 m²