Certainly! Let's walk through the detailed solution step-by-step:
1. Uniform Live Load: The first given load is a uniform live load of [tex]\( 2.8 \, \text{kN/m} \)[/tex]. This means that for every meter length of the floor beam, there is a continuous load of [tex]\( 2.8 \, \text{kN} \)[/tex].
2. Concentrated Live Load: The second load is a single concentrated live force of [tex]\( 20 \, \text{kN} \)[/tex]. This type of load acts at a specific point on the beam.
3. Uniform Dead Load: The third load is a uniform dead load given as [tex]\( 700 \, \text{N/m} \)[/tex]. To convert this from Newtons per meter to kilonewtons per meter, we divide by 1000 (since [tex]\( 1 \, \text{kN} = 1000 \, \text{N} \)[/tex]):
[tex]\[
\frac{700 \, \text{N/m}}{1000} = 0.7 \, \text{kN/m}
\][/tex]
So, summarizing the given loads:
- The uniform live load is [tex]\( 2.8 \, \text{kN/m} \)[/tex]
- The concentrated live load is [tex]\( 20 \, \text{kN} \)[/tex]
- The uniform dead load is [tex]\( 0.7 \, \text{kN/m} \)[/tex]
There you have it. The loads on the floor beams are:
- A uniform live load of [tex]\( 2.8 \, \text{kN/m} \)[/tex],
- A single concentrated live force of [tex]\( 20 \, \text{kN} \)[/tex],
- And a uniform dead load of [tex]\( 0.7 \, \text{kN/m} \)[/tex].
These values help in the design and analysis of the floor beams to ensure they can support the applied loads safely and efficiently.
