soil mechanics Assignment PDF

SOIL MECHANICS
[Author Name(s), First M. Last, Omit Titles and Degrees]
[Institutional Affiliation(s)]

QUESTION ONE
The Brock Commons house has just been viewed as the tallest wood building in the North
America. The structure is 13-story that is found in the Pointe-aux-Lièvre's eco-neighborhood in the
city of Quebec. The tall strong timber structure has been developed on the platform made of
cement. The structure is 134 feet in its stature. The structure is special by the way that it is
comprised of totally the strong wood. The wooden structure is found in the staircases and
furthermore in the poles of the lift.
The outside walls and the cross are likewise made of the wooden segments. The structure is a mix
of the glulam bars and sections and furthermore the CLT segments. The shot which has been made
by the new, bigger and the taller building made of wood has been perceived as a choice which is
feasible in the nation particularly by the building and the construction network(Poirier et al 2016).
The procedure has been made conceivable by the advancement of the results of the forested areas.
The methodology and the reasoning of the whole project was essentially to utilize the materials in
the most prescribed way which took into account accomplishment of the cross breed arrangement
of solid characteristics and timber.
QUESTION TWO
The various loads on the floors selected for the study
Timber
Factors Solution
Temperature
Timber tends to become much stiffer and
brittle when exposed to very high temperature.
This normally results into the twisting effects.
The construction of the Brock building has
used technique of allowance provision to
compensate for the effects of the twisting.
Moisture
The quality of timber is lowered when it is
excessively exposed to the moist environment.
The timber will actually rot.
The construction has used water proof material
at the foundation to limit the access of water to
the building.

C
Concrete
Water/cement ratio
The construction of the foundation of the
building has been done using concrete whose
performance is normally affected by the water/
cement ratio.
The construction of the structure has employed
the right ratio of water /cement so as to
improve on the durability of the building.
Moisture exposure
The concrete is normally affected in terms of
the performance when it is exposed to too
much moisture content.
Use of plastic material that is water proof to
limit the moisture exposure
Steel
Moisture
Steel is capable of undergoing rusting when it
is excessively exposed to lots of moisture.
The steel has been reinforced within the
concrete to limit the exposure to moisture.
Expansion and contraction
The metal will need some allowance due to
temperature change for expansion and
contraction to take place
The construction process has allowed room for
the variation of length of the steel due to
changes in the temperature.

Figure 1: The dead loads are actually acting downwards (Zhuo 2016).
Figure 2: Dead loads acting from the roof system (Zhuo 2016).
Figure 3: Dead loads acting on the floors downwards (Zhuo 2016).

QUESTION THREE
1st Store y :
Dead Load:
What is assumed: The calculation that has been done is on the assumption that the floor is
suspended.
Floor system = 0.75 kPa (assumed) x (6/2 + 4/2) x (5.6/2 + 4.4/2) = 18.75 kN Column
= 0.45 x 0.4 x 4.75 x 24 = 20.52 kN
Wall unit = 20 x 0.175 x (6/2 + 4/2) x 4.75 = 83.13 kN
Total = 121.4 kN
Second Store y :
Dead Load:
Floor = 0.15m (assumed) x 24 x (6/2 + 4/2) x (5.6/2 + 4.4/2) = 90 kN Floor
system = 0.75 kPa (assumed) x (6/2 + 4/2) x (5.6/2 + 4.4/2) = 18.75 kN Column =
0.45 x 0.4 x 3.46 x 24 = 15 kN wall system = 20
x 0.175 x (6/2 + 4/2) x (5.6/2 + 4.4/2) = 60.55 main Beam = 24 x 0.6m
x 0.8m x (6/2 + 4/2) = 57.6 kN
Supporting beam = 24 x 0.25 x 0.45 x (6/2 + 4/2) = 13.5 kN
Total = 256.4 kN
Live Load:

LL = 4 KN//m2 (assumed) x (6/2 + 4/2) x (5.6/2 + 4.4/2) = 100 kN
Second upper or Third storey
(i) Dead Load:
System Slab = 0.15m (assumed) x 24 x (6/2 + 4/2) x (5.6/2) = 51.4 kN
Beam = 24 x 0.6m x 0.8m x 6 = 68.1 KN
Floor system = 0.75 kPa (assumed) x (6/2 + 4/2) x (5.6/2) = 10.5 kN
Vertical Column = 0.45 x 0.4 x 3.06 x 24 = 12.22 kN
wall system = 20 x 0.175 x (6/2 + 4/2) x 3.06 = 54.55 kN
Total = 197.77 kN
(ii) Live Load: (2nd Upper Storey Rooms)
LL = 4 KN/m2 (assumed) x (6/2 + 4/2) x (5.6/2) = 50 kN
(2d) Roof Level:
Wall unit = 20 x 0.175 x 1.81 x (6/2 + 4/2) =32 kN Dead Load service
= 0.75 kPa (assumed) x (5.6/2 + 4.4/2) x (6/2 + 4/2) = 18.75 kN
Total = 53.75 kN

Dead Loads of roof = 0.75KN/m2 x 5 x 5 = 14.17
Roof Materials kN = 1.5KN//m2 x 5 x 5 = 57.25 KN
Addition of Dead Load = 122.4 + 255.4 + 197.77 + 51.75 + 56.25 = 682.57 kN
Base strip Size of 2m width (instead of 4m) x 4m (range Length) x .75mm thick
Loads on the Ground Footing Foundation = 24.0 x 2 x 4 x .75 =145.0 KN
Pressure force on the soil = 989.64 / (2.0 x 4.0) = 124.71 KN//m2
4 Plan for live loads

5. Dynamic load calculation
Taking into account the height of the building that has been assumed to be 4.4m ,the
multiplication thus becomes 4.4*maximum number of floors
A=W*L
4.4*0.5*1/12
=0.125ft2 but P=0.00256*V2
=70*70*0.00256;12.5
The value of the coefficient is equivalent to 0.8; therefore F=0.125*12.5*0.8=1.25.
6 Durability
Steel is typically influenced by the procedure of the corrosion. This can be
controlled by the utilization of the galvanic erosion insurance. The solid might be