Structures and Soil Mechanics: Analysis of Brock Commons House

This article provides an analysis of the Brock Commons House, the tallest wood construction in North America, from a Structures and Soil Mechanics perspective. It covers load calculations, wind load path, earthquake resistance, geotechnical testing, and more. The article also discusses the use of GLT and PSL, concrete shear walls, and connections in the construction of the building.

SCDM71-316 STRUCTURES AND SOIL MECHANICS
[Author Name(s), First M. Last, Omit Titles and Degrees]
[Institutional Affiliation(s)]
Question 1

The Brock Commons house has already been considered the tallest wood construction in the North
America. The structure is 13-storey that is found in the Pointe-aux-Lièvre’s eco-neighborhood in
the city of Quebec. The tall solid timber structure has been constructed on the podium made of
concrete/.The structure is 134 feet in its height. The structure is unique by the fact that it is made up
of entirely the solid wood(Su 2018). The wooden structure is found in the staircases and also in the
shafts of the elevator. The exterior walls and the cross are also made of the wooden components.
The structure is a combination of the glulam beams and columns and also the CLT columns. The
chance that has been created by the new, larger and the taller building made of wood has been
recognized as an option that is viable in the country especially by the building and the design
community(Al-Mukhtar, Khattab and Alcover 2012). The process has been made possible by the
evolution of the products of the woods. The approach and the philosophy of the entire project was
basically to use the materials in the most recommended way that allowed for the achievement of the
hybrid system of concrete cores and timber.
Question 2
Dead loads on respective floors.

Figure 1:Loading of the structure (Willebrands 2017)
Figure 2:Dead loads on the structure from the roof made of Iron sheets(Willebrands 2017.).
L
1
L
L
2

Figure 3:Dead loads on the structure made of timber(Willebrand 2017)
Question 3
1st Storey:
3a)Dead Load:
Assumption: It is assumed that slab to be non-suspended slab
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
Overall = 121.4 kN
3b)Second Storey:

Dead Load:
Slab = 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 Primary Beam = 24 x
0.6m x 0.8m x (6/2 + 4/2) = 57.6 kN
Secondary Beam = 24 x 0.25 x 0.45 x (6/2 + 4/2) = 13.5 kN
Overall = 256.4 kN
Live Load:
LL = 4 KN//m2 (assumed) x (6/2 + 4/2) x (5.6/2 + 4.4/2) = 100 kN
2nd 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
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

Overall = 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
Overall = 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
Summation of Dead Load = 122.4 + 255.4 + 197.77 + 51.75 + 56.25 = 682.57 kN
Footing strip Size of 2m width (instead of 4m) x 4m (range Length) x .75mm thick
Weight of Ground Footing Foundation = 24.0 x 2 x 4 x .75 =145.0 KN
Soil Pressure force = 989.64 / (2.0 x 4.0) = 124.71 KN//m2
5. Dynamic load calculation
The provided spacing is ta the interval of 20feet
=20ft
=20x0.305
=6.1m

The height of the colum will be taken as 5.9 metres
The pitch of the roof will bwe taken as 27 degrees
H(roof)=5.9+6.63X TAN 27
=9.28M.
The average height will be calculated as
h=5.9+9.28-5.9/2=7.6m
The working life obtained becomes 50 years
The region becomes A4
The equivalent VR=45m/s for the ultimate loading. And also 37m/s for
serviceability.
.
4. Live load plan

Figure 5typical floor plan(Willebrands 2017)
Taking the prime equation
VS=VtxMdxMZxMt
Mt=1.0
The value of the velocity is taken as 45x0.95x1.0x1.0x1.0
V=42.75m/s
This equation will completely define the pressure of the wind

=37x0.97
=35.15m/assuming that the density of the air remain the same as 1.2kg/m3
and also taking the factor of the dynamic loading to be constant that is;
C=1.0
P=0.5x1.2x42.75x42.75x1/1000
=1.097(This is described as Ultimate limit)
The internal pressure will translates into the cross wind
K=0.8 for the cross wind.
Pressure becomes; 1.097x (-0.24) remember -0.24 was obtained from -
0.3x0.8
P=-1.61kN/m.
For upward slope
P=-.4x1.097
=-0.44kPa
6. Durability
Steel is normally affected by the process of the corrosion. This can be

controlled by the use of the galvanic corrosion protection. The concrete may
be affected by improper compaction and other ASR effects. The effects of
ASR(Alkali Silica Reaction) can be prevented using pozzolanic admixtures.
Timber is affected by exposure to water. This should be minimized as low
as possible.
7. Fire protection.
The building has employed the use of the timber with the thick layer. This
thick layer normally chars to provide natural protection against protection of
fire. The building also enjoys access to the fire extinguishers and
standpipes(Willebrands, O., 2017)..
8. Reasoning behind GLT and PSL
The GLT has excellent strength to weight ratio
It is durable
It poses proper flexibility to size and shape
It has high dimensional stability and strength besides being locally
available.

Parallel strand lumber is suited for use in the places where high bending
stress is needed.
It is also very attractive with proper surface finish.
9.Load path
Figure 5:Load distribution for a person on roof side view(Willebrands 2017)

Figure6: plan of load distribution
10. Wind Load Path

Figure 7:Wind load system
11. Building Stability
In order to ensure proper stability, the building has been supplied with a
frame work of properly fixed columns and also systems of beams with joints
that ere stiff and cable of taking moments.
12. Use of concrete shear wall
The shear walls are basically used in the provision of the lateral resistance
and stiffness. When wood is exposed to a lot of moisture, it decays up. This
is also true when it comes into the contact with the ground. Also use of the
nails weakens the strength of the wood. If shear walls made from the

concrete are used instead, the structural strength of the building will
improve. When the shear wall are not used, the structure will utilize other
alternatives that include the following;
(a) Moment resisting frames: These are girders and columns that are
fixed with the semi rigid connections.)
(b) Braced Walls: They have single diagonals. This can be used in lace
of the shear walls.
Use of the shear wall will give perfect straightness of the RC walls as
another advantage.
13. .Earthquake
Most of the engineers and the designers normally rely on the previously
published reports on the earthquake to improve their design work. The
reports have indicated that proper design will easily assist in the resistance
of the seismic loading. In order to improve the resistance, the follow b in
design criteria must be followed(Teshnizi, Pilon, Storey, Lopez and Froese
2018).Thorough foundation survey or research should be conducted in the

sand that may be loose, clay locations and other silty places. Much
consideration should be given to the seismic waves or flooding
The cantilevers like parapets and towers are known to be undumped
elements. These components should be given some high coefficient of the
seismic design. The accelerations that are inclined vertically should be given
much consideration.
14. Tributary in vertical elements.
Shown in shaded parts.
Given: Live Load, L=50 psf Dead Load, D= 70 psf 1. Girder BC 1-1
Tributary area, A T = 40 x 15 = 600 sq ft Influence area, AI = 2 A T = 2 x
600 = 1200 sq ft Live load reduction = 0.25 + [15 / √ (AI) ] = 0.25 + [15 / √

(1200)] = 0.25 + 0.433 =0.683 Amplified loads per linear foot: Dead Load =
1.2 (70 psf) (15 ft) = 1260 plf Live Load = 1.6 (0.683)(50 psf) (15 ft) =
819.6 plf Wu = 1.2 D + 1.6 L = 1260 + 819.6 = 2079.6 plf = 2.08 k/ft
The materials component, the angle of inclination, the period of treatment of
the concrete. The thickness of the slab especially suspended one will give
wider crack at the point of the connection.
The light slab though suspenede gives smaller crack during the testing period.
Figure :Illustration of the cracks

15. Connections
Simple logic within the system
.

Figure9:Wall connection with slab
Special requirement on the concrete finish will definitely influence the type
of the slab wall connection. There is lots of demand for high quality
formwork for the case of the fair-faced concrete.
Finally the involvements of other techniques for the construction will mean
inclusion or use of specific type of connection.
16. Geotechnical testing
Direct Shear Test
This is a laboratory test that is used basically by the geotechnical engineers in the
measurements of the shear strength of the soil. The specimen is placed in a shear box with
the two stack rings that hold the sample. The contact that exists between the two rings is
taken as the mid-height of the sample. The other soil characteristics that are determined
include the soil cohesion and the angle of the internal friction.
Proctor Compaction Test
This is a laboratory technique for measuring the maximum moisture content of the soil
type. These lab tests for the most part comprise of compacting soil at realized dampness
content into a tube shaped form of standard measurements utilizing a compactive exertion
of controlled size. The soil is generally compacted into the form to a specific measure of
equivalent layers, each accepting various blows from a standard weighted mallet at a
predetermined tallness. This procedure is then rehashed for different dampness substance
and the dry densities are resolved for each. The graphical relationship of the dry thickness

to dampness content is then plotted to set up the compaction bend. The most extreme dry
thickness is at last acquired from the pinnacle purpose of the compaction bend and its
relating dampness content, otherwise called the ideal dampness content(Taylor 2014).
Triaxial shear test.
This method is used to measure the mechanical properties of the soil. In this test, the soil
sample is subjected to stress. This is regularly accomplished by setting the example
between two parallel platens which apply worry in one (generally vertical) heading, and
applying liquid weight to the example to apply worry in the opposite ways. (Testing
mechanical assembly which permits use of various dimensions of worry in every one of
three symmetrical headings are talked about underneath, under "Genuine Triaxial test".)
R-Value testing
The R-Value test[1] measures the reaction of a compacted test of soil or total to a vertically
connected weight under particular conditions. This test is utilized by Caltrans for asphalt
configuration, supplanting the California bearing proportion test. Numerous different
organizations have received the California asphalt plan technique, and indicate R-Value
testing for subgrade soils and street totals.
Sieve Analysis
A sifter investigation (or degree test) is a training or strategy utilized (ordinarily utilized in
structural designing) to evaluate the molecule measure conveyance (additionally called
degree) of a granular material by enabling the material to go through a progression of
strainers of continuously littler work size and gauging the measure of material that is
ceased by each strainer as a small amount of the entire mass.
The size circulation is regularly of basic significance to the manner in which the material

performs being used. A sifter investigation can be performed on non-natural or natural
granular materials including sands, smashed shake, soil, stone, feldspars, coal, soil, an
extensive variety of produced powders, grain and seeds, down to a base size contingent
upon the correct strategy. Being such a straightforward method of molecule estimating, it
is likely the most well-known.
The test strategy states
17. Footing system
The footing system refers to the component that assists in the transfer of the load from the
building to the laid foundation. The building that is under the study has employed the use
of the shallow rooting system that is fastened using the screw and bolts on the ground.
Figure 10:Footing system(Willebrand 2017)

Question 18
Deflection
C1 AS1170.0) stipulates the limiting deflections under two heads as given below:
(a) The maximum deflection should not go beyond span/250 due to all loads
including the effects of temperatures, shrinkage and creep and measured from the as-cast
level of the supports of roof ,floor and all other horizontal members.
(b) The highest deflection should not normally exceed the lesser of span/350 or20
mm including the effects of temperature, creep and shrinkage occurring after erection of
partitions and the application of finishes.
It is a must that both the requirements are to be fulfilled for each structure.
Selections of initial dimensions
The checking of the two requirements is done after designing the members.
Despite the fact that, the revision of the structural design has to be confirmed on details
to ensure that it does not fails to meet any one of the two or both the requirements.By
avoiding this, C1 AS1170.0) the member initial dimensions are assumed as
recommended by guideline thus general satisfaction of the deflection limits. Illustration
in the Clause20 indicates various spans to effective depth ratios and cl. 24.5 suggests
limiting cross section of

beams, a relation of b and d of the members, hence achieving lateral
stability. They are indicated below:
(A) For the deflection requirements
Different basic span figures to effective ratios of depth for three various
support conditions are prescribed for spans up to 10 m, in which the modification
should be done under any or all of the four different situations:
(i) In the cases for spans that are beyond 10 m,
(ii) In the case of the consideration of the amount and the stress of
tension steel reinforcement,
(iii) when putting into consideration the amount of compression
reinforcement.
(iv) In the cases for flanged beams.
(B) For lateral stability
Any beam lateral stability depends upon the ratio of slenderness and the
support situations. Accordingly section of the code that stipulates the following:
(i) In case of simply supported and continuous beams, the clear distance
between the lateral restraints shall be beyond the lesser of 60b or 250b2/d, in
which d is the effective depth and b refers to compression breath face lateral

⎥
restraints between the midway.
(ii) In the case of dealing with cantilever beams, the clear distance from the
free end of the cantilever to the lateral restraint shall not go beyond the lesser of 25b
or 100b2/d.
C1 AS1170.0) prescribes computation steps of the short-term deflection. The usual
methods for elastic deflections using the short-term modulus of elasticity of concrete
Ec and effective moment of inertia Ieff given by the following equation as
recommended by the code:
In the continuous beams, therefore ,modification of Ir, Igr and Mr values are done
as in the equation below;
⎡ X 1 + X 2 ⎤
X e
=
q1
⎢
+ (1 - q1 ) X o
2 ⎦
where Xe = modified value of X,
X1, X2 = values of X at the supports,

Xo = refers to value of X at mid span,
q1 = coefficient shown in the table below and
X = value of Ir, Igr or Mr as appropriate.
coefficient table
k
1
0.5 or
less
0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4
k
2
0 0.03 0.08 0.16 0.30 0.50 0.73 0.91 0.97 1.0
Remember: k2 is given by (M1 +
M2)/(MF1 + MF2), i.e. M1 and M2 = support
moments, and MF1 and MF2 = fixed end moments.
Deflection due to Shrinkage
C1 AS1170.0) indicates the method of calculating the
deflection due to shrinkage α
cs
in the equation below
α
cs
= k3 ψ cs l
I.e. k3 is a constant which is 0.5 for cantilevers, 0.125 for simply supported members,
0.086 for members continuous at one end, and 0.063 for fully

(Fast and Jackson 2017)
continuous members; ψ cs is shrinkage curvature equal to k4 ε cs /D where ε cs is
the ultimate shrinkage strain of concrete. For ε
cs
, cl. 6.2.4.1 of IS 456
recommends an approximate value of 0.0003 in the absence of test data.
where pt = 100Ast/bd and pc = 100Asc/bd, D is the total depth of the section, and l
is the length of span.
Deflection Due to Creep
C1 AS1170.0) stipulates the following method of calculating deflection due to creep.

3 3 3
Solution 1:
I gr
= 300(429.48)
3
+ 2234(170.52)
3
- 1934(70.52)
3
= (11.384) (10) 9 mm 4
The calculation of this section can be done as shown below;
Here, bf /bw = 7.45, Df /D = 0.17. the generated figures from the values becomes
k1 = 2.10.
Igr = k1bw D3/12 = (2.10)(300)(600)3/12 = (11.384)(10)9 mm4
Characteristics of the section cracked

With Consideration compressive concrete moment and steel tensile about the neutral
axis, we have
bf x2/2 = m Ast (d – x) gives (2234)(x2/2) = (8.94)(1383)(550 – x)
or x2 + 11.07 x – 6087.92 = 0 . from the equation solution, we get x = 72.68 mm.
z = lever arm = d – x/3 = 525.77 mm
I r
=
2234(72.68)3
3 + 8.94(1383)(550 - 72.68) 2 = 3.106(10)9 mm 4

M = wl2/8 = (9.3 + 10.7)(8)(8)/8 = 160 kNm
=
1.2 - (82.96
160
I r
) ( 525.77 ) (1
-
550
72.68 ) (
550
300 )
2234
= 0.875 I r . But I r ≤ I
eff
≤ I gr
So, Ieff = Ir = 3.106(10)9 mm4.
Characteristics of cracked section

using compressive concrete moment and tensile steel about the neutral axis (distance is
assumed to be x from the bottom of the flange as illustrated below:
2234(100)(50 + x ) = (23.255)(1383)(450 - x ) or x = 12.92 mm which
gives x = 112.92 mm. Accordingly, z = lever arm = d – x/3 =
512.36 mm.
Ir = 2234(100)3/12 + 2234(100)(62.92)2 + 23.255(1383)(550 – 112.92)2
+ 300(12.92)3/3 = 7.214(109) mm4
Mr = 82.96 kNm (see Step 2)
M = wperm l2/8 = 9.3(8)(8)/8 = 74.4 kNm.
I eff =
(1.2) - (82.96
74.4
I r
) ( 512.36
550
) (1 - 112.92 ) (
550
300 )
2234
= 0.918 I
≤
Ief≤
Igr, Ieff should be equivalent to Igr. So, Ieff = Igr =
11.384(109mm
Confirmation of the requirements of C1 AS1170.0)

The two requirements in regard to the deflection control are given in parts. Therefore they
are confirmed as follows
Confirmation of the first requirement
Highest allowable deflection = 8000/250 = 32 mm
The actual final deflection due to all loads
= 15.358 + 2.64
Perfect..
+ 3.118 ` = 21.116 mm < 32 mm. Hence,

Confirmation of the second requirement
The highest allowable deflection is the lower of span/350 or 20 mm. Here, span/350 =
22.86 mm. Hence, the highest allowable deflection = 20 mm. The actual final deflection =
1.948 + 2.64 4 + 3.118 = 7.706 mm < 20 mm. thus, o.k.
Thus, both the requirements of C1 AS1170.0) as well as as given in section are
met.
QUESTION 19
Floor Vibration
The design of the building has incorporated several management strategies to control the floor
vibrations.
Reduction of effects
There have been alterations that lower the annoyance that are linked to the vibrations. The noise
level has been eliminated by the removal of the articles that are known to vibrate noticeably.
Relocation of the activities

The building has employed the techniques of relocation of the vibration sources. This has
included the relocation of the area of the aerobic exercise from the top floor to the ground floor.
The walking vibrations have been controlled by having the specific areas near the columns.
Stiffening
The vibration due to the walking and rhythmic activities in the building has been reduced
through increase of the stiffness of the floor. The structure has incorporated new columns
between the already existing columns from the affected floor to the foundation down. The
stiffening in the structure has also been achieved using transverse methods of blocking or
decking.
Figure 11:Stiffening system(Willebrands 2017)
Isolation
The floor vibration that results from the machinery components within the building has been
reduced by placing them on the soft springs.
QUESTION 20

The creep refers to the deformation on the structure as a result of the imposed stress. This may
results from the self-weight of the elememt.The building has sorted out the creep effects through
load reduction(Connolly, Loss, Iqbal and Tannert 2012)
The structural members of the used timber have had increases surface area hence reduced effects
vof the creep. The water cement ratio of the building especially in the foundation was made as
little as possible to reduce the creeping effect.
The heat flow within the building takes place in three, two or one dimension. In most of the real
situations the heat flow takes place through three dimensions. The transfer of the heat takes place
through each of the following means
Convection
Conduction
Radiation.
Conduction is the flow of heat via a material through direct molecular contact. This particular
contact will take place within two objects that are in contact(Wood 2014).
Convection is transfer of heat by the movement of the molecular particles of the liquid or gas.
Radiation is the heat transfer by the electromagnetic radiations.Since conduction is regarded as

the primary heat transfer means,the material has incorporated the use of low density insulation
materials including glass of 2500kg/m3.
Air Leakage
The warm air that is genetrated from within the building rises and is quickly replaced with the
cold air from outside.During the warm weather,the leaking air is replaced with hot air that must
be dehumified and cooled.The structural component of the building has incorporated mechanism
that allows for the proper exchange of air or ventilation(Calderon 2018).

Figure 12:Ventilation means of the building(Willebrands 2017)
The building has allowed for some gaps so as to mitigate the adverse effects of the axial
shortening.
QUESTION21
Concrete admixtures
Concrete admixtures are normally added to the batch of the concrete before mixing the concrete.
The admixtures of the concretes tend to improve the quality of the concrete, retardation and
manageability(Poirier, Moudgil, Fallahi, Staub and Tannert 2016).
1. Set retarding Admixtures

These are normally added to delay the chemical reactions that occur when settling process of the
concrete starts. It is commonly used in the reduction of the high temperature effects that can
possibly lead to the faster setting of the concrete.
Advantages
It reduces additional costs in the placement of the new concrete batch
They help in the reduction of the cracking that results from the form deflection.
Disadvantages
Some retarders normally play a major role in the water reduction and may keep some air as well
in the concrete.
2. Air-Entrainment
An entrained concrete normally increase the durability of the concrete in relation to the freeze-
thaw properties. This type of the admixture produces a concrete that is more workable.
Advantages
The degree of workability is high
The durability is high as well
The cycles of wetting and drying are very high.
Disadvantages
The ratio of the entrained air and compressed strength normally vary and is difficult to maintain.

3. Water reducing concrete admixtures
These are chemical products that when added to the concrete may create a desired slump at a
lower ratio of the water cement. They are normally used to obtain a specific strength.
Advantages
Helps in the improvement of the properties of the concrete especially in the decks of the bridge.
Disadvantages
Low cement content tend to compromise the strength of the structure made from such
compounds.
4. The Shrinkage Reducing
These are chemical substances that are added during the initial mixing of the concrete.
Advantages
It reduces the long term and early shrinkage
It is suitable in the cases where shrinkage cracking leads to the problems of durability
Disadvantages
It may sometimes lead to the reduction of the strength of the compound
5. Super plasticizers
The core reason for the use of super plasticizers is to ensure that there is production of a flowing
concrete that has proper slump

Advantages
It improves the workability of the concrete.
Disadvantages
It is associated with the lump loss.

REFERENCES
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