Answer:
a) A suspended floor is a ground floor with a void underneath the structure. The floor can be formed in various ways, using timber joists, precast concrete panels, block and beam system or cast in-situ with reinforced concrete. However, the floor structure is supported by external and internal walls.
b) Soil exploration consists of determining the profile of the natural soil deposits at the site, taking the soil samples and determining the engineering properties of soils using laboratory tests as well as in-situ testing methods
c) Bulking in sand Occurs When dry sand interacts with the atmospheric moisture. Presence of moisture content forms a thin layer around sand particles. This layer generates the force which makes particles to move aside to each other. This results in the increase of the volume of sand.
d) In a nutshell, bearing capacity is the capacity of soil to support the loads that are applied to the ground above. It depends primarily on the type of soil, its shear strength and its density. It also depends on the depth of embedment of the load – the deeper it is founded, the greater the bearing capacity.
Explanation:
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Answer:
Explanation:
Ohms Law I=E/R (resistive requires no power factor correction)
150/25= 6 amps
Answer:
The ability to read electrical schematics is a really useful skill to have. To start developing your schematic reading abilities, it's important to memorize the most common schematic symbols. ... You should also be able to get a rough idea of how the circuit works, just by looking at the schematic.
Explanation:
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Answer:
The duration of the consolidation process for the same clay is 32 min
Explanation:
for clay 1:
t1=0
H1=thickness=2 cm
for the clay 2:
t2=?
H2=2 cm
The time factor is equal to:
![T=(\frac{Cv}{d^{2} })t](https://tex.z-dn.net/?f=T%3D%28%5Cfrac%7BCv%7D%7Bd%5E%7B2%7D%20%7D%29t)
where Cv is the coefficient of consolidation
![(\frac{Cvt}{d^{2} })_{1}= (\frac{Cvt}{d^{2} })_{2}](https://tex.z-dn.net/?f=%28%5Cfrac%7BCvt%7D%7Bd%5E%7B2%7D%20%7D%29_%7B1%7D%3D%20%20%28%5Cfrac%7BCvt%7D%7Bd%5E%7B2%7D%20%7D%29_%7B2%7D)
if Cv is constant, we have:
![(\frac{t1}{(\frac{H1}{2}) ^{2} })_{1}=(\frac{t2}{H2^{2} })_{2}\\\frac{0}{(\frac{2}{2})^{2}) }=\frac{t2}{2^{2} }](https://tex.z-dn.net/?f=%28%5Cfrac%7Bt1%7D%7B%28%5Cfrac%7BH1%7D%7B2%7D%29%20%5E%7B2%7D%20%7D%29_%7B1%7D%3D%28%5Cfrac%7Bt2%7D%7BH2%5E%7B2%7D%20%7D%29_%7B2%7D%5C%5C%5Cfrac%7B0%7D%7B%28%5Cfrac%7B2%7D%7B2%7D%29%5E%7B2%7D%29%20%20%7D%3D%5Cfrac%7Bt2%7D%7B2%5E%7B2%7D%20%7D)
Clearing t2:
t2=32 min