Site preparation and investigation

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This report will highlight ground encountered risk and foundation issues in order to prepare the tendering phase of the proposed development, discuss about solution may provided with a brief stating recommendation for the proposed site. Calculation and detailed section drawing shall be provided as a final indication for the proposed development. An explanation to why this particular design has been used will be discussed. All data's used in this report regard ground soil profile obtained from "SOILTECHNICS" handout that was given in the lecture.

History Development of the Site:

Most of the Coventry cities developments have been destroyed through the Second World War include the site. According to the desk the study before the war it was a residential building. After the war in 1963 was the first record of the demolished Frank Whittle building which consisted of single storey building and Surface Park.

Ground encountered:

Through the exploratory excavation in the proposed area soil profile has been determined. Comprised made ground soil overlies naturally deposit Corley sandstone. Further investigation in site located a number of masonry walls were encountered to the east of the area in the made ground which indicate the presence of former underground building thus requires re-fill it with compacted soil.

According to the report the comprised made ground soil excess to 3.5 meter (Borehole WSC by MP Consulting Engineers). it generally comprised loose and medium dense red brown sand with varying concentrations of gravels of brick, sandstone and ash. The Corley sandstone has a typical encountered which has been mentioned in the report without a detail depth of it. As over all depth of this soil considered to be 9.5 m which its moderately strong medium, occasionally thinly, bedded arenaeceous sandstone. The profile of the Corley sandstone is showing in figure 1, 2 and 3

Stiff orange brown sandy clays

Sandstone cannot be penetrated

Stiff orange brown sandy clays becoming weak mudstones

Sandstone cannot be penetrated

Sandstone cannot be penetrated

Loose medium dense & reddish brown slightly silty, slightly gravelly sand

Figure 1 Figure 2 Figure 3

According to the report through redevelopment of site from the year 1960, there have been some earthworks which caused lowering in southern part of the site. A retaining of structure shall be the solution for height of 2.5 m

Ground water condition:

According to exploratory excavations at depth 0.88 m below ground seepage observed thus requires a pump to dewater all the water may be found below the ground to ensure no water near the foundation or the cellars. Settlement must occur due to draining the water off the soil where the ground level will go down in some area in the project. It is not enough information about groundwater level more investigations are required to be more accurate through the building stage.

In case of heave raining suds will ensure to collect the higher amount of rainfall around the project to make it safe, offsite water must be consider. Groundwater levels are also subject to variations caused by changes in the local drainage conditions and seasonal effects hence long term monitoring of levels in standpipes must be provided to measure of seasonal variations in groundwater levels.

Foundation Solution:

Foundation designed for the proposed area will be pad foundation as it is the best solution to the price and soil condition. Pad depth shouldn't exceed more than 1.5 meter below ground as made ground is 3.5 meter depth below ground. Its recommended that if the soil underneath the foundation is weak then more excavation should be adopted until locates denser soils. Worst case scenario pad foundation is designed to be a depth of 1.1 m.

Foundation cost the construction companies and insurance companies more than 250 million pound a year most of the greatest failure cost due the trees and sulphate attacks or could be defined as chemical attack. The failure of the foundation result to the fall of the whole project, the site isn't well investigated against all risks.

The influence of trees across is very important specially through bedding the foundation. As stated in the desk study report it have been determined that within the depth of 0.75 m will enter the zone of trees and vegetations therefore pad foundation is designed to go deeper to 1.1 m.

At the northern part of the site a number of trees and other vegetation have been observed, these trees can't be ignored or removed therefore a qualified Arboriculturist is recommended to identify all necessity such as what species of tree are present, how old are they, what is the soil condition and what are the weather condition.

Contaminations were identified on site towards the south side of the project. As stated in the report that it won't be a great amount of risk to harm human or water receptors. It should be noted that a number of industries located northward of the site such as Dye Works, Timber Yard, Cycle Works, Motor cycle Works. It is yet not enough information collected but as overall its low.

As a small basement decided to be constructed to fit with requirements, cantilever retaining wall has been designed for it which it's the best choice to avoid instability may be caused from the soil so it won't fail. Soil is Corley sandstone which basically sand, permeability is one of sand properties so it will drain all the water which is great since water could damage the basement. If a case water reaches the basement it may freeze and cause the basement to heave so dewatering is must to avoid any heave could occur.

As the overall soil is okay but more ground investigation is needed. It should be noted that Coventry was heavily bombed during the second world war there possibility of finding unexploded bombs. Soil requires dewatering to avoid any problems and set long term monitoring of levels in standpipes.

Calculations:

Pad Foundation design:

Given Data:

Dead Load (reaction) = 1993.05 kN,

Live Load (reaction) = 1275.75kN

Allowable SBC = 225kN / Sq.M @ 1.0M depth assumed.

Concrete Density = 24kN/Cub.m

fck = 40Mpa for concrete & fyk = 500Mpa for Rebars.

Soil Density = 170 kN/Cub.m

Try pad size as 4.2M X 4.2M X 1.0M Thick Concrete pad with column neck 0.4mx0.4m x0.15m high.

Hence, Self weight of foundation slab = 424kN

Check for maximum pressure below foundation base:

Maximum pressure under footing slab pmax = ( P / A )

Where; P = DL+LL = (1993.05 + 424) + 1275.75 = 3693

A = 4.2 X 4.2 = 17.64 Sq.m

Hence; pmax = 209.33 kN / Sq.M < 225kN / Sq.m (safe)

Design of Foundation Pad:

Critical Section for Bending is at the face of the column neck i.e (4.2-0.4) / 2 = 1.9m

Hence BM acting over foundation Slab due to Imposed Loads :

BMDL = {(1993.05+424)/ (4.2 x 4.2)} x (1.9)2 / 2 = 247.34 kN-m per m width of foundation

BMLL = {(1275.75) / (4.2 x 4.2)} (1.9)2 / 2 = 130.6 kN-m per m width of foundation

Ultimate (Mu) = 1.2 x BMDL + 1.6 BMLL = 506 kN-m

Therefore, (Mn) = (Mu) / 0.9 = 562 kN-m

Required Reinforcement (Ast) = (Mn) / fyk x d , (considering concrete cover = 75mm)

Hence (Ast) = (562) x 106 / (500 x 925) = 1215 sq.mm / m

But considering beam action for foundation pad, minimum steel required = 0.2% of BxD

= (0.2/100) x 1000 x 1000 = 2000 Sq.mm / m

Hence provide Dia 20mm Bars at every 150mm = 2093 sq. mm/m > 2000 Sq.mm / m

Provide Dia 20 Bars @ 150mm at bottom, both ways with alternate bars box shaped.

Check for shear :

Critical section for shear is at a distance d = 925mm from face of the column neck.

Hence maximum design shear acting on footing slab due to soil pressure ;

VDL = {(1993.05+424)/ (4.2 x 4.2)} x (1.9 - 0.925) = 133.6 kN

VLL = {(1275.75) / (4.2 x 4.2)} x (1.9 - 0.925) = 70.6 kN

Vu = 1.2 x VDL + 1.6 x VLL = 273.28 kN

Vn = Vu / 0.9 = 304 kN = 304000 N

Actual shear capacity of foundation section;

√ fck X b X d / 6 = 975036 N > 304000 N (Safe)

Design cantilever retaining wall:

Density of the soil is = 1700 kg/m^2

check the stability of the wall

Assume ka = 0.33 suitable for the mix soil

Pa = Ka x Density x g x h

Pa = 0.33 x 1700x10^-3 x 9.81 x 3.8 = 21 kN/m^

Allowing for the minimum required surcharge of the 10 kN/m^2

Ps = Ka x 10 = 3.3 kN/m^2 which act uniformly over the whole depth h

Therefore the horizontal force on 1 m length of wall is given by:

Hk(earth) = 0.5 x Pa x h = 0.5 x 21 x 3.8 = 40 kN from the active earth pressure

And Hk (surcharge) = Ps x h = 3.3 x 3.8 = 12.4 kN

Vertical loads

Permanent loads

Wall = 0.5 x (0.4 + 0.3) x 3.5 x 25 = 30.6

Base = 0.4 x 3 x 25 = 30

Earth = 2 x 3.5 x 1700 x10^-3 x 9.81 = 116.7

Total = 177.3 kN

Variable loads

Surcharge = 2 x 10 = 20 kN

Overturning: taking moments about point A at the edge of the toe, at the ultimate limit state.

Overturning moment (unfavourable)= (1.1 x 40 x 3.5/3 ) + (1.5 x 12.4 x 3.5/2) = 83.9 kNm

For the restraining (favourable) moment a factor of 0.9 is applied to the permanent loads and 0 to the variable surcharge load

Restraining moment = 0.9 x (30.6 x 1 + 30 x 1.5 + 2 x 116.7) = 309 kNm

Thus the criterion for overturning is satisfied.

Sliding: for the sliding (unfavourable) effect a factor of 1.35 is applied to the earth pressure and a factor of 1.5 to the surcharge pressure

Sliding force = 1.35 x 40 + 1.5 x 12.4 = 72.6 kN

For the restraining (favourable) effect a factor of 1 is applied to the permanent loads and 0 to the variable surcharge load. Assuming a value of coefficient of friction = 0.6

Frictional resisting force = 0.6 x 1 x 116.7 = 70.02 kN

Since the sliding force exceeded the frictional force, a passive earth pressure resistance must also provide against the heel beam

Hp = factor f x 0.5 x Kp x density x g x a^2

Where kp is coefficient of passive pressure which assumed is 3.5 for this soil and a is 0.5 m trench allowance.

Hp = 1 x 0.5 x 3.5 x 1700x10^-3 x 9.81 x 0.5^2 = 7.3 kN

Therefore total resisting force is 70.02 + 7.3 = 77.32

This exceeds the sliding force

0.4 Surcharge 10 kN/m^2

3.5 m Soil

0.3 m

0.6 m

3

Bearing pressure at ultimate limit state (STR & GEO)

P = N/D -+ 6M/D^2

Where M is the moment about the base centreline therefore

M = 1.35 (40 x 3.8/3) + 1.5 (12.4 x 3.8/2) + 1.35 (30.6 x (1.5-1)) - 1 x 116.7 x (2-1.5)

M = 66 kNm

Therefore, bearing pressure at toe and heel of wall

P1 = (1.35 x (30.6 + 30) + 1 x 116.7) / 3 +- 6 x 66 / 3^2

= 66.2 +- 44

= 110.2, 22.2 kN/m^2

P1 = 110.2 P3 = 80.9 P2 = 22.2

2m

3 m

Bending reinforcement

Wall

Horizontal force

1.35 x 0.5 x Ka x density x g x h^2 + 1.5 x Ps x h

1.35 x 0.5 x 0.33 x 1700x10-3 x 9.81 x 3.5^2 + 1.5 x 3.3 x 3.5

45.5 + 17.3 = 62.8 kN

Considering the effective span, the maximum moment is

Med = 45.5 x (0.2 + 3.5/3) + 17.3 x (0.2+3.5/2) = 44.4 kNm

Med/bd^2fck = 44.4x10^6 / 1000 x 330^2 x 30 = 0.014

Using the lever arm cure la can determined as 0.98

Fyk = 500 kN/mm^2

As = 44.4x10^6 / 0.98 x 330 x 0.87 x 500 = 315 mm^2/m

Provide H12 bars at 200 mm centres (As = 339 mm^2).

Base

The bearing pressures at the ultimate limit state are obtained from part (2) of these calculations. Using the figures from part (2):

Pressure P1 = 110.2

P2 = 22.2

Figure 2

P3 = 22.2 + (110.2 - 22.2) 2/3 = 80.9 kN/m^2

Heel: taking moments about the stem centreline for the vertical loads and bearing pressures

Med = 1.35 x 30 x (1.5-1) + 1 x 116.7 x 1.3 - 22.2 x 2 x 1.3 - (80.9 - 22.2) x 2/2 x (2/3 + 0.2)

= 63.4 kNm

Therefore

Med / bd^2 fck = 63.4x10^6 / 1000 x 330^2 x 30 = 0.019

Using the lever arm cure la can determined as 1

As = 63.4x10^6 / 1 x 330 x 0.87 x 500 = 442 mm^2/m

Provide H12 bars at 250 mm centres (As = 452 mm^2). Top steel

Toe: taking moments about the stem centreline

Med = 1.35 x 30 x 0.6 x 0.8/3 - 110.2 x 0.8 x 0.6 = -46.4 kNm

Therefore

Med/bd^2fck = 46.4x10^6 / 1000 x 330^2 x 30 = 0.014

Using the lever arm cure la can determined as 0.97

As = 46.4x10^6 / 0.98 x 330 x 0.87 x 500 = 330 mm^2/m

The minimum area for this

As, min = 0.15btd/100 = 0.0015 x 1000 x 330 = 495 mm^2

Thus, provide H12 bars at 200 mm centres (As = 566 mm^2/m)

bottom and distribution steel also same steel should be provided in the compression face to prevent cracking.

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