**Question 1**

There are two types of foundations to be installed whenever there is construction: a deep foundation and a pad foundation. Deep foundations are mainly used when the soil bearing capacity at the upper stratum is very low and unlikely to support. In essence, the foundation is driven deep into the earth’s crust until it reaches the bedrock or a soil of remarkable bearing capacity. Some of the types of deep foundations include the pier and the caisson foundation. On the other hand, a pad foundation is designed when the bearing capacity of the topsoil strata is adequate enough to hold a structure. However, regardless of the type of foundation designed, the settlement is always the determining factor on the shape and size of the foundation (valley, 2009). Some foundations may be circular with square footing etc.

In determining the total amount of soil to be excavated for the installation of piers, the depth needs to be determined. As a rule of thumb, the excavation needs to go into the deeper soil stratum of adequate bearing capacity. However, all the stratum, regardless of the depth, need not be subject to seasonal variation Taking the depth of the piers given=2.5m, the total area to be excavated for one pile =0.1767m3.As a matter of fact, piers are of shallow foundation and are mainly installed when the soil stratum has adequate bearing capacity to hold the building and all the structural members.Moreso, the house is a residential one and unlikely to have immense pressure on the soil and the bearing capacity

Therefore, basing our assumption on the fact that 2.5m will be sufficient enough for the loading requirements of the site, the total amount of soil that will be excavated for 1 pile =0.1767m3.However, the amount excavated will be more and as such, we may assume =0.18m3The number of piles=number of piles at the intersections (considering the 1800mm maximum spacing) =16+ (1+3+2+1+2+2+2+2+2+3) =36piles needed

Amount to be excavated for one pile (including wastage=0.18m3).Therefore, the total amount of excavation basing our assumption on the fact that 36 piers will be sufficient=0.18*36

**Question 2**

The site excavation is generally done to ensure that the field and the site are ready for the earthworks. In this, the top surface of the soil is excavated in order to remove the plant and organic material. Basically, bulk excavation involves the removal and addition of soil materials to/from a particular region. In this regard, the site is made suitable for construction, both in height and level. There are two methods through which bulk earthworks can be done: excavation and cutting, the addition of soil material to an area. In excavation, a piece of land that may be assumed to be of a relatively higher level than the adjacent land is leveled through the removal of material while filling is the addition of materials to a relatively lower lying areas.

One method that may be used in the excavation is borrow pitting. In this, the site engineer divides the land into grids and each level is determined. As the works progress, the level of each grid will be determined and compared to the tabulated levels. Thereafter, the land can either be excavated or filled. The process of bulk excavation necessitates the use of heavy machinery such as graders, bulldozers etc.

Cost plays a major role in bulk excavation whereby the materials can be reused therefore preventing waste collection costs or the need for material delivery. As per our site, the difference in elevation=0.500m.On this, note, the area should be level so as to provide the necessary platform for excavation of the foundation and the ground slab.However, our site is unproportioned and as such, the total volume of soil=(0.5*40) /2=10*18=180m3. Because the land is unproportioned, there is an area that needs fill which may be obtained from the cut part.

The calculation of bulk excavation is based on a lot of assumptions and as such, the amount of cut on the upper part =amount of fill on the lower parts=0.5*0.25*40*18=90m3

**Question 3**

The slab that is in direct contact with the ground is basically exposed and likely to degrade because of the environmental conditions. The groundwater table, Leaching, alkalinity etc. are all factors to be considered when designing a ground floor slab. Therefore, the ground floor slab has to be designed with the proper mix design and site preparation. However, the design has to consider the placing of concrete as well as proper curing. According to research, it has been recommended that the ground floor slab should be about 4 inches in thickness. Extreme scenarios may call for thicknesses between 5 and 6(areas with loadings such as garages.) (Wanderwerf, n.d.).

Volume of concrete to be used for the patio slab= area of the slab= (1670*6350)-(590*350)-(590*350)-(240*350) = 87500mm2*450=39.4m3

Volume of concrete used for the garage=5.5*5.5*0.175m= 5.30m3

Volume of concrete used for the rest of the house= (11.8*18.58-(5.5*5.5))*0.45= 85.0473m3

The assumption in this is that the area that will be occupied by the slab is rectangular in shape with the walling and the partitioning of the ground floor done on top of the slab. Therefore, some areas may not be partitioned or within the housing elements but are still covered by the slab.

**Question 4**

In calculating the mesh required for reinforcement, we basically take the whole area to be covered by the slab. As in our case, the slab has been assumed to cover, a rectangular area. Therefore, the area to be reinforced=11.8*18.58= 219.244m2*2= 438.488m2 including the wastage

Area of one sheet considering overlap=5.8*2.2=12.76m2.Therefore, the number of mesh/sheet to be used=35 sheets

**Question 5**

There are two calculations to obtain the whole length of the mesh to be obtained:

Edge beam= (assuming a uniform edge beam thickness of 300mm)

Perimeter=11870+12830+6350+1540+5520+12830= 50940

Therefore, the total length of the edge beam=15282m

Trench mesh=70*11870=830900mm=830.9m

**Question 6**

There are a variety of steel grades to be used in the construction: 250UB, 310UB, 150UB and the T Bar. However, they have been used in different sections of the building and as such, the numbers differ

250UB=4080/25.7=159 bars*5030=798536mm

310UB=12830/32=401 bars*5520=2213520mm

150UB=11870/14=848 bars*1740=1475520mm

150UB=6350/14=33 bars*1670=54104mm

150Ub=1670/14=119mm*6350=767462mm

The total amount of steel likely to be used=5299144mm for the ground floor slab

Assuming that the first-floor slab will consume as much, then the total weight of streel=5299.14*14.6kg/m=77,367kg=77.4tonnes

**Question 7**

Basically, the functions of a plasterboard are varied but may be a necessity in most of the current building structure. However, the plasterboard is made out of gypsum and as such, the design of the house should consider the occupancy in terms of health effects.One major function of plasterboard is fireproofing. Plasterboards are very helpful in the case of a fire outbreak by preventing the rate of spread. Moreover, their insulation property is very helpful in cutting down the spread of noise and as such, a reliable means in rooms where there is a need for noise reduction.

To begin with, the height of the floor is 2450mm.Furthermore, the design has to consider the wet areas such as bathrooms because of the corrosive effects of water.

Total length to be plastered= (12830-850-850-810) +950+600+470+120+1740+540+710+150+4500+ (3340-1810) +710+3110+830=26290

Therefore, assuming that only the internal areas require plaster boarding, the total area =26290*2450=64m2.Furthermore, the number of sheets required (assuming that their height =wall height) = 64410500/1200=53675

**Question 8**

Prime skirting for the first floor rooms. Initially, we have to calculate the total size area of the rooms and as such:

Bed 1=5270*4050=21343500

Bed 2=3540*3490=12354600

Bed 3=3230*3540=11434200

Bed 4=3340*3291=10991940

Rumpus=4500*7260=32670000

Void=1610*1710=2753100

Stair area=3400*4160-(2570*2023) = 5199110

Therefore the total area to be covered=100492230mm2

Area of prime skirting=66*18.Therefore, the total prime required=100492230mm2/ (66*18) =84590 considering the 66*18 plasterboards

**Question 9**

Tiles are basically used to enhance the living environment through beautification. Tiles can be used in ceilings, rooftops, table tops etc. It is therefore important to consider the size and the shapes as well as the design of all these tiles. More so, tiles are made up of mortar and cast into the desired shapes, most commonly rectangular. In establishing the floor area, grout has to be used to fill the spaces left between the tiles.

There are a variety of tile sizes available in the market. As a matter of fact, the recent past has observed a remarkable change in the tile industry and consequently the use. Most people have been shifting from smaller tiles to larger ones with the squared tile mostly preferred.Most designers have observed that larger tiles tend to increase the aesthetics of the floor as well as reduce the usage of grout. The tiles vary in size from 10*10 to 20*20 all the way to rectangular ones such as 60*120, 20*120 etc. Therefore, the designers and buildings are provided with an array of options.

Ceramic floor tiles (taking 0.2m by 0.2m tiles)

The choice of the tiles depending on the floor areas to be covered and as such, the number is obtained by dividing the total floor area by the size of a single tile.

- Family room=4080*5030= 20522400=51306
- Meals room=4589*2170= 9958130=254896 tiles
- Kitchen – allow tile to go under the cupboards=4589*6110= 28038790=7097 tiles
- Foyer/stair area=(1710*7070)+(2570*1910)= 16883800mm2=42210 tiles

**References**

Valley, M. (2009). Foundation analysis and design.

Wanderwerf, P. (n.d.). Concrete floor slab concrete construction. Retrieved from www.concreteconstruction.net/how-to/construction/concrete-floor-slabws_o