CVG 3140

Theory of Structures I

Building Design Project

Daniel Pereira

Fortuna Sophia Konan

Josiane Nku

Salma Nasr

Saloua Zoubir
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ABSTRACT

In order for a building to be safe, a careful and detailed structural design must be
developed. Therefore, it is essential to understand the properties of civil engineering
materials, the effects of loads in physical structures and the components used to build a
trustable structure.
This project aims to discuss how two buildings (“Desmarais” and “Advanced
Research Complex”), both localized at the University of Ottawa, were designed. Going
through a structural analysis, the importance and role of each structural component can be
determined.
Furthermore, analyze the precautions and considerations that were made to improve its
design, including a consult to the National Building Code of Canada.
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TABLE OF CONTENTS

1. Desmarais building (University of Ottawa): Reinforced concrete flat slab structure ............ 7
1.1. STRUCTURAL SYSTEM AND LOAD TRANSFER ................................................... 7
1.2. DEAD AND LIVE LOADS .......................................................................................... 11
1.3. SNOW LOADS ............................................................................................................. 12
1.4. EARTHQUAKE LOADINGS ...................................................................................... 18
1.5. SELECTED STRUCTURAL ANALYSIS UNDER GRAVITY LOADS .................. 21

2. Advanced Research Complex (University of Ottawa): Hybrid steel/concrete structure ..... 29

2.1. STRUCTURAL SYSTEM AND LOAD TRANSFER ................................................. 29
2.2.STRUCTURAL LAYOUTS ......................................................................................... 31
2.3. DEAD AND LIVE LOADS .......................................................................................... 33
2.4. SNOW LOADS ............................................................................................................. 36
2.5. SELECTED STRUCTURAL ANALYSIS UNDER GRAVITY LOADS .................. 39

3. REFERENCES ..................................................................................................................... 43
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LIST OF TABLES

Table 1: Summary of loads………………………….………………………………………..11

Table 2: Axial load rundown table……………………………………………………………22

Table 3: Axial load rundown table for ARC…………………………………………………40

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LIST OF FIGURES

Figure 1: Concrete shear walls with gravity frames………………………...…………………7

Figure 2: Gravity load’s path…………………………………………………………………..8
Figure 3: Columns inside Desmarais…………………………………………………………..8

Figure 4: Two-way flat slab with drop panels……………………………………………...….8

Figure 5: Lateral load’s path…………………………………………………………………...9
Figure 6: Steel reinforcements………………………………………………………...……….9

Figure 7: Structural drawing of Desmarais’ first floor ………………………….…...………10

Figure 8: Desmarais building…………………………………………………………………10

Figure 9: Snow calculations and sketches…………………………………………….………15

Figure 10: Comparison between theoretical and obtained values……………………….……16

Figure 11: Snow load distribution sketch…………………………………………………….17

Figure 12: Ottawa – Design Spectrum (Site Class C)………………………………………...20

Figure 13: Desmarais architectural drawing………………………………………………….22

Figure 14: Architectural drawing illustration ………………………………………………...22

Figure 15: Axial load diagram………………………………………………………………..23

Figure 16: South-east entrance canopy framing plan’s structural drawing…………….…….24

Figure 17: South-east entrance canopy framing plan’s illustration…………………..………24

Figure 18: Loading in beam AB, and the respective shear and moment diagrams…….……..25

Figure 19: Loading in beam CD, and the respective shear and moment diagrams…….……..25

Figure 20: Upper roof mechanical room architectural drawing…………………………..…26

Figure 21: Upper roof mechanical room structural components……………………………..26

Figure 22: Secondary beams’ load distribution, shear and moment diagrams…………….....27

Figure 23: Primary beams’ load distribution, shear and moment diagrams………………….27

Figure 24: Steel frames with concrete shear walls……………………………………………28

Figure 25: Advanced research complex………………………………………………………29
Figure 26: ARC during construction………………………………………………………….29
Figure 27: Steel structure with metal decks and concrete topping…………………………...30
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Figure 28: Structural representation..........................................................................................30

Figure 29: Beams and deck………………………………………………………..………….31
Figure 30: Beam AB, shear and moment diagrams…………………………………………..31
Figure 31: Sketches calculate snow load parameters……………………………..…………..37
Figure 32: Sketches to illustrate and calculate snow load parameters………………………..38
Figure 33: ARC’s Level 4 architectural drawing……………………………………………..39
Figure 34: Sketch using values of ARC’s Level 4 architectural drawing…………………….39
Figure 35: Axial load diagram (kPa) for Case 2…………………………………………...…40
Figure 36: Secondary beam’s distributed load, shear and moment diagrams……………...…41
Figure 37: Primary beam’s distributed load, shear and moment diagrams………………...…42
Figure 38: Beam selection table………………………………………………………………42
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1. Desmarais building (University of Ottawa): Reinforced concrete flat slab

structure

1.1. STRUCTURAL SYSTEM AND LOAD TRANSFER

The Desmarais building, located at University of Ottawa, has a concrete structural system.
This type of system can present concrete beams, columns and bearing walls.

A predominant characteristic of Desmarais’ structural design is the presence of concrete

shear walls; they provide lateral stiffness and strength to support lateral loads. The shear-wall
type of building, in this case, is the one which a column-supported framing system carries the
gravity loads (Fig. 1). (“Seminar 1”, 2013)

Figure 1: Concrete shear walls with gravity frames (“Seminar 1”, 2013)

The role of each structural component in this system is described below (“Seminar 1”,
2013):

- Diaphragms: diaphragms are horizontal-resistance members that transfer lateral

forces between vertical-resistance elements (ex: shear walls). They receive lateral
loads. Essentially, they are floors and the roof systems; in this case, the system
used was a “two-way flat slab with drop panels”. (Fig. 4)
- Shear walls: receive lateral loads from diaphragms and transmit them to the
ground.
- Columns: resist gravity loads

Thus, the structure resists gravity loads with columns and beams; dead, live and snow
loads are transmitted from the slabs to the beams, that transmit them to the columns and
finally to the foundation. Figure 2 illustrates this path.
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Figure 2: Gravity load’s path (“Seminar 1”, 2013)

Figure 3 shows columns, which resist gravity loads, in Desmarais building.

Figure 3: Columns inside Desmarais

A building has to be prepared to also support lateral loads, coming mostly from wind and
earthquake. First, the horizontal lateral force resistance systems, in this case floors with flat
slabs with drop panels (Fig. 4), receive the load. They transmit the load to vertical lateral
force resistance systems, in this case shear walls, which take to load to the foundations. Figure
5 illustrates this lateral load’s path through a structure.

Figure 4: Two-way flat slab with drop panels (CRSI, 2013)

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Figure 5: Lateral load’s path (“Seminar 1”, 2013)

Since a concrete flat slab is used in the floors, the diaphragm can be categorized as rigid,
what means that the horizontal forces distribution is proportional to the vertical elements
relative stiffness (“Seminar 1”, 2013).

Moreover, the shear walls were placed in a symmetric arrangement and in both directions
of the building, to avoid the center of mass to be different from the center of resistance. This
could have created torsional forces, which, to be avoided, would require a more detailed
analysis from the engineer in charge.

Concrete is a material that has high compressive and low tensile strength, and steel a
material that is strong in tension; because of these characteristics, steel might be combined
with concrete to provide a better structural performance, making the structure more ductile
and more resistant to bending. Desmarais contains steel reinforcements, in two directions, in
the concrete slabs; it can be observed in Fig.7.

Figure 6: Steel reinforcements

Overall, analyzing Desmarais’ structural drawings, it is possible to clearly identify some

of the components previously described. Figure 7 shows a shear wall, a column, a drop panel
and steel reinforcements in Desmarais’ first floor structural drawing.
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Figure 7: Structural drawing of Desmarais’ first floor

Desmarais has a mechanical room in its roof, which is made of a steel structure. It can be
observed in Fig. 8, where it is clear the difference between this room and the rest of the
building, made of concrete.

Figure 8: Desmarais building

Since not many details were provided, the structure format can be speculated. It could
be a steel light frame, single-story, structure. In this kind of structures, the roof and walls are
formed by lightweight metals and the lateral forces are resisted by transverse steel bents and
diagonal bracings in longitudinal direction (“Seminar 1”, 2013).
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1.2. DEAD AND LIVE LOADS

From the definition in our notes: dead load is a permanent load caused by the weight of
building components; live load is a variable load caused by intended use and occupancy.

1.2.1. Second floor

Dead load: 6.30 Kpa + drops

Live load: 4.80 Kpa

From table 4.1.5.3 in NBCC 2010 part 4, we can see that the minimum live load required
for office areas on floors above the first storey is 2.40Kpa. This load does not include
computer rooms.

1.2.2. Third floor

(a) Office

Dead load: 7.78 +drops

Live load: 3.60

From table 4.1.5.3 in NBCC 2010 part 4, we can see that the minimum live load required
for offices is 2.4Kpa.

(b) Corridors/assembly area

Dead load: 6.30+ drops

Live load: 4.80

From table 4.1.5.3 in NBCC 2010 part 4, we can see that the minimum live load required
for corridors is 4.80 Kpa.

(c) Total minus snow load

Dead load: 14.08 + drops

Live load: 8.4Kpa

1.2.3. Lower mechanical room

Dead load: 2.24

Live load: 9.48

From table 4.1.5.3 in NBCC 2010 part 4, we can see that the minimum live load required
for equipment area and service rooms is 3.6 Kpa. From table “design dead load of materials”
of dead load extra hand, we obtained a minimum dead load of 1.13Kpa.
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1.2.4. Calculation

Roofing= 0.32

Insulation= 0.07

Mech’L= 0.5

Deck=0.10

Joists=0.14

Total dead load= 1.13Kpa

Table 1: Summary of loads

Levels Live loads (Kpa) Dead loads (Kpa)

Parking plan 2.40 6.98+drops
First floor 16.5 19.5+drops
Second floor 4.80 6.30+drops
Third load 8.40 14.08+drops
Fourth floor 10.55 22.95+drops
Fifth floor 9.36 26.75+drops
Sixth floor 8.40 14.03+drops
Seventh to twelfth floors 7.60 14.05
Lower mechanical room 10.22 2.24
Mechanical penthouse floor plan 3.60 10.10
Service elevator (roof plan/roof canopy) 0 10.10
Service elevator (room floor/roof plan) 9.05 10.94
Information desk 1 0.60
South east entrance 5 0.60
North entrance 5.08 0.60
Total 101.96 159.82

1.3. SNOW LOADS

Depending on the snow load , a structure must be designed to support the maximum
weight of snow. Effectively, the amount of snow weight will vary with:

- Geographical location (SS, Sr)

- Site exposure (Cw)
- Shape and type of roof/ structure (Ca)

The specific snow load (S) acting on a roof is calculated using the following formula:

S= Is [ Ss[ CbCwCaCs] + Sr ]

Figure 11 shows the snow accumulation used to complete the calculations below.
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1.3.1. Ground snow load (Ss) and associated rain load ( Sr)

According to the table C-2 (Appendix C, NBCC), “Design climatic data for selected
location in Canada”, we can assume that Sr = 0.4 and Ss = 2.4 in Ottawa.

1.3.2. Basic snow load factor (Cb) (see Figure 9)

Snow accumulation on roofs is typically smaller than ground snow due to wind erosion. In
our case, the parameter Cb is taken as 0.8 small roof Ic < 70m .

Proof:

In case of a large roofs when Ic >70 m:



 = 2 −   And  = 1 –  
 

View of the upper roof (Figure 9)

W= 18 m and L= 60 m
()^

 = 2(18) −  
= 30.6 m < 70m small roof

Where: w= smaller plan dimension of roof (m)

l = Larger plan dimension of roof (m)

1.3.3. Wind exposure factor (Cw)

A several conditions have to be respected to consider a roof in an exposed location. In our

case, the Desmarais building is in a sheltered location, so Cw= 1.0

1.3.4. Roof slop factor Cs

The value of Cs is a function of the angle of the roof slope. For non- slippery roofs were
alpha is too small ( α <30 ° ). In this case, we have a flat roofs so we can take Cs= 1.0.

1.3.5. Importance factor for snow load Is

The value of Is depends on the degree of importance of the building. In our case, we analyzed
a faculty so we can consider the building as a school with a lot of accumulation snow. Thus,
Is= 1.15

1.3.6. Shape factor (Ca) :

The shape factor takes account of possible redistribution of snow load due to wind action.
If there is no shape, the Ca can be considered equal to 1. But, in this case, there is a lot of
snow accumulation assumed as a lower roof as a shape:

Assuming hp=0 and γ= 3 KN/m2, h= 88154-84750= 3404 mm= 3.4m

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(a) Step 1 : Getting F

Factor F = greater of :

I) F=2
II) F= 0.35 (γ* lc / Ss*6 (γ*hp/ Ss)2)0.5= 0.35 (3* 30.6m / 2.4)0.5 +0.8 = 2.96
(maximum value)

So F= 2.96 < 5.0 Ok

(b) Step 2: Getting Ca (0) see Figure 9

h= 52 – 14 = 38 m

Ca(0) = lesser of

I) Ca(0) = (γ*h)/ (Cb*Ss)= (3*38)/ (0.8*2.4) = 59.37 m

II) Ca(0)= F/ C b = 2.96/ 0.8 = 3.7 m

So Ca(0)= 3.7 m

(c) Step 3 : Getting xd

xd = lesser of


.&
I) xd= 5 ∗ ℎ −   ! " = 5 ∗ (38 – (0.8 ∗ (  ))= 186.8m

.&
II) xd= 5 ∗  !  (' − ) = 5 ∗    ∗ (2.96 − 0.8)= 8.64m

So, xd = 5.44m

(d) Step 4 : Getting the snow load

Beyond the step:

S= Is [Ss[ CbCwCaCs] + Sr ]= 1.15 [ 2.4[ 1*1*1*0.8] + Sr ]= 2.7 Kpa

At the step:

S=1.15 [2.4[ 1*3.7*1*0.8] + Sr ]= 8.64 Kpa

In conclusion, according to the theoretical values, the results for the distribution snow
loads founded from the calculations done for the Desmarais building are acceptable (see
Figure 10).
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Figure 9: Snow calculations and sketches

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Figure 10: Comparison between theoretical and obtained values

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Figure 11: Snow load distribution sketch

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1.4. EARTHQUAKE LOADS

Earthquake is the result of acceleration of a structure because of ground motion during a

seismic event. All the parameters calculate to find the seismic base shear are very important
according to the influence on the ground motion on a structure. Desmarais building has
structural walls which will helps to resist lateral loads as it has a rigid connection to the
foundation. All those earthquake loads prevents the building from collapsing, major failure
and causing loss of life.

The building is designed with clear load paths which will transfer the initial forces
generated in an earthquake to the supporting ground. And all the factors were calculated
according to the National Building Code.

1.4.1. Seismic base shear

The equation to calculate the seismic base shear is:

+(,- ) × /0 ×
1 × 2 56778869:;
*= = × 268Aℎ:
34 × 3 <5=> ?6>@:859 7=:5?;

1.4.2. Weight

The weight given on the drawings is W = 225862 KN.

1.4.3. Coefficients

Calculation of different coefficients:

(a) Spectral response acceleration +(,- )

Seismic data: Ottawa

- Sa (0.2) = 0.64

- Sa (0.5)= 0.31

- Sa (1.0) = 0.14

- Sa (2.0) = 0.046

- PGA = 0.32

According to Microzonation maps Ottawa, Desmarais belongs to soil of Site Class C. The
ground profile name is: Very dense soil and soft rock.

The following graph is the representation of the Period Vs the Response Spectral
acceleration (g) which is going to help finding Building period (Ta).
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Ottawa design Spectrum (Site Class C)

0.7
Response Spectral acceleration (g)
0.6

0.5

0.4

0.3
Série1
0.2

0.1

0
0 0.5 1 1.5 2 2.5
Period (s)

Figure 12: Ottawa – Design Spectrum (Site Class C)

(b) Period of vibration (Ta): Building Period

The types of structures used for Desmarais are Shear wall and other structures. The
building period will follow the equation:

,- = 0.05 × (ℎB )/&

According to the architectural drawings of the Desmarais building, the height of the
entire structure is 59.94 m.

Then the Building period is

,- = 0.05 × (59.94)/& = E. E F

The corresponding response spectral acceleration S (Ta) corresponding to the period

T=1.1 s is equal to 0.13 using the graph above.

S (Ta) = 0.13

(c) Factor to account for higher mode effect on base shear (Mv):

For T<=1s, Mv = 1.0

For T = 1.1s, Mv ≈ 1.0 (doing linear interpolation)

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(d) Importance factor for Earthquake IE:

Category: Normal ==> IE = 1.0

(e) Force factor modification (Rd) and Over strength related modification factor (R0)

System Description: Moderately ductile shear walls

According to CSA A23.3,

Rd=2.0

R0= 1.4

As all coefficients have been calculated, the seismic base shear V can be calculated:

0.13 × 1 × 1 × 225862
*= = 10486.45 KN
(2.0) × (1.4)

Comparison using factors on drawings and base shear found by engineers:

On drawings, values of the coefficients are:

S (Ta) = 0.13

Mv = 1

Rd = 2

R0 = 1.4

IE = 1

W = 225862 KN

The base shear is going to be the same since the values of the coefficients are the
same.

V = 10486.45 KN

The base shear found by engineers is:

V = 11163 KN
Building Design Project 21

Those two values are in the same range, there is not a big difference. The percentage
of error is 6%, not too high.

According to the values we found, the Equivalent Static Force Procedure (ESFP)
would be permitted?

Check some criteria:

- Regions of low / moderate seismicity (IEFaS(Ta) less than 0.35

For Soil Class C: Fa = 1.0

We calculated S (Ta) = 0.13 and IE = 1

1*1*0.13 = 0,13 < 0.35

Yes, ESFP would be permitted

- Regular Structures:

The total height found for Desmarais building is 59.94 m < 60 m

The Building period found was Ta = 1.1 s < 2s
Yes, ESFP would be permitted.

According to our data, ESFP would be required since the building has regions of
low/moderate seismicity and also has regular structures.

1.5. SELECTED STRUCTURAL ANALYSIS UNDER GRAVITY LOADS

1.5.1. Axial load rundown for selected column

After choosing column G1-7, shown in Fig. 13, it is possible to calculate its tributary area:

,I = 9000JJ × (4900JJ ⁄2) = 22.05J²

The values of dead (D), live (L) and snow (S) loads were found in the structural drawings.
Two combinations were considered:

Case 1: T=1.4D

Case 2: T=1.25D+1.25L+0.5S

Results are shown in Table 2.

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Figure 13: Desmarais architectural drawing

Figure 14: Architectural drawing illustration

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Table 2: Axial load rundown table

Level TA(m²) D(kN) L(kN) S(kN) Pd(kPa) Pdt(kPa) Pl(kPa) Plt(kPa) Ps(kPa) Pst(kPa) Case 1 Case 2
Roof 22.05 2.24 1.13 2.5 49.392 49.392 24.9165 24.9165 55.125 55.125 69.1488 126.6773
12 22.05 7.75 3.6 0 170.8875 170.8875 79.38 104.2965 0 90 239.2425 415.0541
11 22.05 7.75 3.6 0 170.8875 170.8875 79.38 183.6765 0 90 239.2425 534.1241
10 22.05 7.75 3.6 0 170.8875 170.8875 79.38 263.0565 0 90 239.2425 653.1941
9 22.05 7.75 3.6 0 170.8875 170.8875 79.38 342.4365 0 90 239.2425 772.2641
8 22.05 7.75 3.6 0 170.8875 170.8875 79.38 421.8165 0 90 239.2425 891.3341
7 22.05 7.75 3.6 0 170.8875 170.8875 79.38 501.1965 0 90 239.2425 1010.404
6 22.05 7.75 3.6 0 170.8875 170.8875 79.38 580.5765 0 90 239.2425 1129.474
5 22.05 6.5 4.8 0 143.325 143.325 105.84 686.4165 0 90 200.655 1253.781
4 22.05 6.5 4.8 0 143.325 143.325 105.84 792.2565 0 90 200.655 1412.541
3 22.05 7.75 3.6 0 170.8875 170.8875 79.38 871.6365 0 90 239.2425 1566.064
2 22.05 7.75 3.6 0 170.8875 170.8875 79.38 951.0165 0 90 239.2425 1685.134
1 22.05 9.9 4.8 0 218.295 218.295 105.84 1056.857 0 90 305.613 1903.154

Figure 15: Axial load diagram (kPa) for Case 2

1.5.2. Analysis of the south-east entrance canopy framing plan

Using Figure 16, it is possible to calculate the loading in the beams of the south-east
entrance canopy framing plan.
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Figure 16: South-east entrance canopy framing plan’s structural drawing

Figure 17: South-east entrance canopy framing plan’s illustration

The dead and snow loads are found in the drawing:

D=0.6kPa

S=5kPa

The tributary width for secondary beam AB is:

TW=300mm

The load combination can be considered:

q=1.4D+0.5S=1.4(0.6)+0.5(5)=3.34kPa
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Multiplying it by the TW:

Q=3.34(0.3)=1.002kN/m

Figure 18 shows the loading in beam AB, and the respective shear and moment diagrams.

Figure 18: Loading in beam AB, and the respective shear and moment diagrams

For primary beam CD, the loadings and respective shear and moment diagrams are shown
in Figure 19.

Figure 19: Loading in beam CD, and the respective shear and moment diagrams
Building Design Project 26

1.5.3. Analysis of interior joists/girders on the upper roof mechanical room

Figure 20: Upper roof mechanical room architectural drawing

Analyzing the panel shown in Figure 20, limited by KL, it is possible to notice the
components shown in Figure 21. The metal decks are in the shorter direction, as always. They
receive the gravity load, and transmit it to the secondary beams; these beams transmit the load
to the primary ones, that address them to the columns.

Figure 21: Upper roof mechanical room structural components

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From the structural drawings, the following values are found for dead and snow loads:

D=2.25kPa

S=2.16kPa

Using the combination used previously for other parts of the building, we have:

q=1.4D+0.5S

q=1.4(2.25)+0.5(2.16)=4.23kPa

Multiplying this value by the tributary width (TW), we have:

Q=4.23(4.5)=19.035kN/m

Therefore, Fig. 22 represents the load distribution for secondary beams in Fig. 20, that are
located between K and J, and the consequent shear and moment diagrams.

Figure 22: Secondary beams’ load distribution, shear and moment diagrams

For primary beams, the load distribution is shown in Figure 23.

Figure 23: Primary beams’ load distribution, shear and moment diagrams
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2. Advanced Research Complex (University of Ottawa): Hybrid steel/concrete

structure

2.1. STRUCTURAL SYSTEM AND LOAD TRANSFER

The two types of loads we can find in a building are:

- Gravity load path: The gravity load can be characterized by snow loads.

- Lateral load path: The lateral load can be characterized by wind and earthquake.

We have some structural systems in the building which can help resist those loads. As we
observe the Advanced Research Complex, we can notice the presence of shears walls, braced
frames and moment resistant frame. It is, therefore, a hybrid structure consisted of a steel
frame with concrete shear walls (Fig. 24).

Figure 24: Steel frames with concrete shear walls (“Seminar 1”, 2013)

The columns and beams will provide resistance for gravity loads; braced frames will
provide strength for lateral loads. We can classify the Lateral Force Resistant Systems in two:
horizontal, which is for the floor and the roof and vertical, for the shear walls.

The Shear walls (solid wall) are placed to take lateral forces from diaphragms and
transmit them to the ground. We can see in the building (Fig. 25) that it is started on the top
until to the foundation. The walls are provided in the direction of load, thus, in two
directions; loads can come from the left or from the bottom.
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Figure 25: Advanced research complex

Brace frames provide less resistance but more ductility; they take lateral load and give
stability to the building. In the building, the type of brace frame used is ‘eccentric brace
frame’. Moreover, there is a special connection between beams and columns which transfer
the moment. The moment can take gravity load and lateral loads. Also steel are used to
reinforce the resistance. Steel moment frame to take lateral load; some details too are added to
the connection between and columns.

Figure 26: ARC during construction

In steel structures, a metal deck can be used in the diaphragms. It should always span in
the short direction, as it is shown in Fig. 26 and in Fig. 27.
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Figure 27: Steel structure with metal decks and concrete topping

2.2.STRUCTURAL LAYOUT

Layout: Level 4 North:

W=4.11+0.892+2.319+1.388+3.379+1.914+1.189+2=17.2mm

L=0.577+4.083+2.830+0.476+7.320+6.710+6.710+6.710+6.710+6.710+6.710+6.710+6.710+
7.020+0.559+6.880+0.95+0.796+3.482+0.482+0.348=91mm

Figure 28: Structural representation

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Figure 29: Beams and deck

Beam AB:

Figure 30: Beam AB, shear and moment diagrams

qf=1.25D+1.5L

qf=1.25(5.65) + 1.5(24)

qf= 43.06 Kn/m

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wf =qf * tribwidth

wf = 43.06 * 1.75

wf =75.36 Kn/m

Mf = wf*L2/8

Mf = 75.36*82/8

Mf = 602.88 Kn.m

∆ = 5 wl * L4/ 384EI

= 5*75.36*84/384*200*82.6*10-6

=243.29m

2.3. DEAD AND LIVE LOADS

From table 4.1.5.3 in NBCC 2010 part 4, we can the live loads for the different rooms on
each level. We will estimate the values of dead loads too.

2.3.1. Level 1

(a) Calculation of live loads:

Mechanical room = 3.6 Kpa

Laboratories= 3.6 Kpa

Assembly area= 4.8 Kpa

Main lobby= 4.8 Kpa

Study room= 2.9 Kpa

Elevator=3.6 Kpa

Corridor= 4.8 Kpa

Electrical room= 3.6 Kpa

Total live load: 31.7 Kpa

(b) Calculation of dead loads

Partition load= 1 Kpa

Finishing=0.50 Kpa
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Suspended mechanical +electrical= 0.25 Kpa

Joists=0.1Kpa

Beams=0.2 Kpa

Girders=0.1 Kpa

Roof materials= 0.8

Roof deck= 0.1 Kpa

Floor finishing= 0.4 Kpa

Floor decking= 0.1 Kpa

Floor toping= 2.1 Kpa

Total dead load= 5.65 Kpa

2.3.2. Level 5

(a) Calculation of live loads:

Mechanical room = 3.6 Kpa

Laboratories= 3.6 Kpa

Main lobby= 4.8 Kpa

Elevator=3.6 Kpa

Corridor= 4.8 Kpa

Electrical room= 3.6 Kpa

Total live load: 24 Kpa

(b) Calculation of dead loads

Partition load= 1 Kpa

Finishing=0.50 Kpa

Suspended mechanical +electrical= 0.25 Kpa

Joists=0.1Kpa

Beams=0.2 Kpa

Girders=0.1 Kpa

Roof materials= 0.8

Building Design Project 34

Roof deck= 0.1 Kpa

Floor finishing= 0.4 Kpa

Floor decking= 0.1 Kpa

Floor toping= 2.1 Kpa

Total dead load= 5.65 Kpa

2.3.3. Level 6

(a) Calculation of live loads:

Mechanical room = 3.6 Kpa

Elevator=3.6 Kpa

Vestibul= 3.6 Kpa

Total live load: 10.8 Kpa

(b) Calculation of dead loads:

Partition load= 1 Kpa

Finishing=0.50 Kpa

Suspended mechanical +electrical= 0.25 Kpa

Joists=0.1Kpa

Beams=0.2 Kpa

Girders=0.1 Kpa

Roof materials= 0.8

Roof deck= 0.1 Kpa

Floor finishing= 0.4 Kpa

Floor decking= 0.1 Kpa

Floor toping= 2.1 Kpa

Total dead load= 5.65 Kpa

Building Design Project 35

2.4. SNOW LOADS

The assumptions and calculations done for Desmarais building in terms of the parameters
Is, Ss ,Sr , Cw, Cs and Cb have the same values . Effectively, the building Arc in and sheltered
condition so Cw= 1 , have non slippery roof so Cs= 1 . The Arc building can be classify in the
category with a high importance factor so Is = 1.15. According to the table C-2 (Appendix C,
NBCC), “Design climatic data for selected location in Canada” , we can assume that Sr = 0.4
and Ss = 2.4 in Ottawa .

2.4.1. Basic snow load factor (Cb)

Snow accumulation on roofs is typically smaller than ground snow due to wind erosion. In
our case , the parameter Cb is taken as 0.8 small roof Ic < 70m .

Proof:

- In case of a large roofs when Ic >70 m:



 = 2 −    And  = 1 –   

View of the upper roof (see Figure 32) :

w= 20,450m and L= 76,545 m

(
.&M)N

 = 2(20.45) −  O.M&M = 35.43 m < 70m small roof

So, Cb= 0.8

2.4.2. Shape factor (Ca) on parapet roof

Calculation of the snow load on a parapet roof (see Figure 32)

Assuming B=20 m and γ = 3KN/m3

I) Step 1: Check if drift occurs

∗ ∗
.&
B> = = 2.4 m since 20> 2.4 Drift occurs
γ 

h= 85200-84510= 690mm= 0.69 m (see Figure 32)

II) Step 2 : Getting Ca(0)

.O∗ !∗P .O∗ ∗.S
Ca(0)=   =   = 0.72 m
QR∗ .∗
.&

a) When Ca(0) < (0.8/ Cb= 1) : Ca(0) = 0.8 /Cb

b) When Ca(0) > (2/Cb= 2.5) : Ca(0)= 2/ 0.8

The result satisfy the first condition so Ca(0)= 1

Building Design Project 36

III) Step 3: Getting xd

Xd= 2h = 2*0.69= 1.38m

IV) Step 4 : Calculation of snow load

At obstruction : x=0

Cw= 1 and Ca(0)= 1 m

S= Is [ Ss[ CbCwCaCs] + Sr ]= 1.15 [ 2.4[ 1*1*1*0.8] + 0.4 ]= 2.7 Kpa

At xd :

S= 1.15 [ 2.4[ 1*1*1*0.8] + 0.4 ]= 2.7 Kpa .

We can see that since the height of the parapet is too small , the snow accumulation is
almost nonexistent . For the load distribution shape see Figure 32.

2.4.3. Shape factor (Ca) on lower roof

The shape factor takes account of possible redistribution of snow load due to wind action.
If there is no shape , the Ca can be considered equal to 1 . But, in this case, there is a lot of
snow accumulation assumed as a lower roof as a shape :

Assuming hp=0 and γ= 3 KN/m2 , h= 88154-84750= 3404 mm= 3.4m

I) Step 1 : Getting F

Factor F = greater of :

a) F=2
b) F= 0.35 (γ* lc / Ss*6 (γ*hp/ Ss)2)0.5= 0.35 (3* 35.43m / 2.4)0.5 +0.8 = 3.13 (maximum
value)

So F= 3.13 < 5.0 Ok

II) Step 2: Getting Ca (0) (see Figure 9)

h= 88154 – 84750 = 3.4 m

Ca(0) = lesser of

(a) Ca(0) = (γ*h)/ (Cb*Ss)= (3*3.4)/ (0.8*2.4) = 5.3125 m

(b) Ca(0)= F/ C b = 3.13/ 0.8 = 3.91 m

So, Ca(0)= 3.91 m

III) Step 3 : Getting xd

Building Design Project 37

xd = lesser of


.&
(a) xd= 5 ∗ ℎ −   " = 5 ∗ (3.4 – (0.8 ∗ ( ))= 13.8m
! 

.&
(b) xd= 5 ∗   (' − ) = 5 ∗   ∗ (3.13 − 0.8)= 9.32m
! 

So xd = 9.32m

IV) Step 4 : Getting the snow load :

Beyond the step:

S= Is [ Ss[ CbCwCaCs] + Sr ]= 1.15 [ 2.4[ 1*1*1*0.8] + Sr ]= 2.7 Kpa

At the step:

S=1.15 [ 2.4[ 1*3.91*1*0.8] + 0.4 ]= 9.09 Kpa

In conclusion, according to the theoretical values, the results for the distribution snow
loads founded from the calculations done for the Desmarais building are acceptable (see
Figure 10).

Figure 31: Sketch to calculate snow load parameters

Building Design Project 38

Figure 32: Sketches to illustrate and calculate snow load parameters

Building Design Project 39

2.5.SELECTED STRUCTURAL ANALYSIS UNDER GRAVITY LOADS

2.5.1. Axial load rundown

Figure 34 uses values and locations from Figure 33 in a sketch to simplify loading
calculations.

Figure 33: ARC’s Level 4 architectural drawing

Figure 34: Sketch using values of ARC’s Level 4 architectural drawing

 Building Design Project 40

In Figure 34 is the data necessary to calculate the tributary area (TA), considering the
column G1:

TA=[(4100/2)+(4500/2)]x(7490/2)=15.35m²

The values of dead (D), live (L) and snow (S) loads were calculated in the previous
sessions. Two combinations were considered:

Case 1: T=1.4D

Case 2: T=1.25D+1.25L+0.5S

Results are shown in Table 3 for the combined loadings.

Table 3: Axial load rundown table for ARC

Level TA(m²) D(kN) L(kN) S(kN) Pd(kPa) Pdt(kPa) Pl(kPa) Plt(kPa) Ps(kPa) Pst(kPa) Case 1 Case 2
Roof 15.35 5.65 1 9.09 86.7275 86.7275 15.35 15.35 139.5315 139.5315 121.4185 201.2001
6 15.35 5.65 24 0 86.7275 86.7275 368.4 383.75 0 90 121.4185 729.0344
5 15.35 5.65 24 0 86.7275 86.7275 368.4 752.15 0 90 121.4185 1281.634
4 15.35 5.65 24 0 86.7275 86.7275 368.4 1120.55 0 90 121.4185 1834.234
3 15.35 5.65 24 0 86.7275 86.7275 368.4 1488.95 0 90 121.4185 2386.834
2 15.35 5.65 24 0 86.7275 86.7275 368.4 1857.35 0 90 121.4185 2939.434
1 15.35 5.65 31.7 0 86.7275 86.7275 486.595 2343.945 0 90 121.4185 3669.327

Figure 35: Axial load diagram (kPa) for Case 2

Building Design Project 41

2.5.2. Analysis of beam/girders on level 4

Using Figure 34, it is possible to calculate the tributary width for the secondary beam in
the middle (G1):

TW=(4100/2)+(4500/2)=4300mm

Using the previous load combination Case II and the values shown in Table 3, we have the
combined loading acting in the structure:

T=1.25D+1.25L+0.5S

T=1.25(5.65)+1.25(24)=37kPa

The distributed load acting in this secondary beam is, therefore:

q=T(TW)=37(4.3)=159.1kN/m

Figure 36 shows the beam with the distributed load, and its respective shear and moment
diagrams.

Figure 36: Secondary beam’s (G1) distributed load, shear and moment diagrams

Regarding the primary beams, Figure 37 shows the distributed load, and its respective
shear and moment diagrams. The point load represent the loads that are transferred from the
secondary beams.
Building Design Project 42

Figure 37: Primary beam’s distributed load, shear and moment diagrams

2.5.3. Beam selection

Using the beam selection table (Fig. 38), based on the previous values for moments found
for the primary beam (Fig. 37), one can choose which type of beam should be used in the
construction.

Figure 38: Beam selection table

Building Design Project 43

Since the value of maximum moment found was 1278kNm (Fig. 37), one has to choose a
beam design with higher moment resistance. Looking at Figure 38, the Designation
W760x134 follows this requirement.

Using the following equation, it is possible to verify if the I-shaped designation will
support the deflection:
.
XY
5U0V <& 5( )(8.6J)&
Z
∆= = = 0.0245J
384W
384(200 × 10 [\=) × (1500 × 10] J& )

Then, the comparison with the value of deflection found with the value described in the
code must be done; the I-shape beam, chose using the beam selection table, must have a lower
value of deflection than the one specified in the code. In this case, this demand is
accomplished.

3. REFERENCES

CRSI – Concrete Reinforcing Steel Institute. (2013). Concrete structural floor systems.
Available at: http://www.crsi.org/index.cfm/engineering/floor

Seminar 1 (2013). Building Structural Systems. Theory of Structures I (CVG3140), University

of Ottawa, Ottawa, ON.

Hibbeler, R. C. (2011). Mechanics of Materials, 8th edition, Jurong, Singapore.