Capstone Final Report

Civil Engineering Capstone Progress Report

(06 - 87/93-400-01)

Design of New Windsor Fire Station

Submitted to the Department of Civil and Environmental Engineering in Fulfillment of the Requirements for the

Course 06-87-400 Capstone Design Project

Presented by:

Daniel LopezSabah Benou Sherif Shouman Tomide Olaniyi

Faculty Advisor: Dr. GhribIndustry Advisor: Dr. Tape

Department of Civil Engineering Faculty of Engineering

University of Windsor

NEW WINDSOR FIRE STATION 1

Windsor, ONAugust 2nd 2016

Executive Summary

The proposed project is a two bay fire station for the City of Windsor. The

facility will be approximately 12,000 square-feet and is a single storey

structure. The building will also house a communications Centre within it.

The building will be designed for post-disaster loading in accordance with the

Ontario building code. It will be located on a property with a seismic site

classification of D. Our group consists of four members and our group name

is Eccentric Solutions. We are tasked with the design of a new Windsor Fire

Station. The objective of our Capstone project is to apply engineering

knowledge to design a building that is economical, efficient and

environmentally friendly. The main construction materials that will be used in

this building are steel and masonry. This combination provides several

structural benefits that include; the inherent strength they both provide for

architectural and design flexibility, the recyclability of steel at the end of its

life and most importantly, the exceptional resistance to fire and corrosion

that this combination will provide. We will be taking a top-down approach to

design this building, which means that we will start with the design of the

upper components then proceed downwards to the foundation.

Acknowledgements

Special thanks to our Faculty advisor, Dr. Ghrib and our industry advisor, Dr. Tape for assisting us throughout the entire phase of the project.


Table of Contents1. Introduction.............................................................................52. Design criteria..........................................................................62.1. Environmental Loads...........................................................................................6

2.1.1. Snow Loads...................................................................................................72.1.3. Wind Loads...................................................................................................8

2.2. Roof deck............................................................................................................92.2.1. Justification of Roof Deck Selection...............................................................92.2.2. Roofing materials........................................................................................10

2.3. Joist selection....................................................................................................112.3.1. Methodology...............................................................................................11

2.4. Beam Selection..................................................................................................152.4.1. Methodology...............................................................................................15

2.5. Columns............................................................................................................172.4.1. Methodology...............................................................................................17

...................................................................................................182.4.2. Base Plate...................................................................................................18

2.6. Apparatus Bay Slab on Grade............................................................................192.7. Load Bearing Walls............................................................................................202.8. Shear Walls........................................................................................................21................................................................................................................................. 232.9. Lintels................................................................................................................232.10. Foundation......................................................................................................243. Cost Analysis...........................................................................274. Conclusion...............................................................................305. References..............................................................................31Appendix A..................................................................................32Snow Load Calculation.................................................................................................32

List of TablesTable 1: Wind Load Cases................................................................................8Table 2: Dead Load of Roofing Materials.......................................................10Table 3: Environmental Load Combinations..................................................11


Table 4: Joist Selection...................................................................................13Table 5: Bridging for Joists.............................................................................13Table 6: Beam Selection................................................................................15Table 7: Joist Girder Selection........................................................................16Table 8: Column Selection.............................................................................17Table 9: Base Plate Selection.........................................................................19Table 10: Slab on Grade................................................................................19Table 11: Load Bearing Wall Summary..........................................................20Table 12: Shear Walls Summary....................................................................22Table 13: Summary of Lintels........................................................................24Table 14: Summary of Spread foundation.....................................................25Table 15: Cost Analysis..................................................................................27

List of FiguresFigure 1: Fire Station Elevations......................................................................5Figure 2: Snow Load Distribution.....................................................................7Figure 3: Roof Deck.........................................................................................9Figure 4: Roofing materials assembly............................................................10Figure 5: Steel Joists and Bridging.................................................................12Figure 6: Roof Joist Layout.............................................................................14Figure 7: Beam and Joist Girder Layout.........................................................16Figure 8: Column Layout................................................................................18Figure 9: Base Plates.....................................................................................18Figure 10: Selected load bearing walls..........................................................21Figure 11: Shear walls...................................................................................22Figure 12: Spread Footing.............................................................................25Figure 13: Strip Footing.................................................................................26Figure 14: Cost Analysis Chart.......................................................................29


1. Introduction

The proposed project we are tasked with designing is a single-storey Fire

Station that is approximately 12,000 (Refer to Figure 1). The building also

houses a Communications Centre within it. The building has been designed

for post-disaster conditions in accordance with the Ontario building code

because a fire station is essential to the provision of fire-fighting services in


the event of a disaster. The property where the fire station sits has a soil

classification of D. A site classification of D means that the soil on the

property is considered to be “stiff soil”, and this is acceptable for a building

that is going to be designed for post disaster conditions.

The fire station was designed in accordance to the National Building Code of

Canada (NBCC 2015) and the Ontario Building Code (OBC 2015). The

Canadian Institute of Steel Construction (CISC) Design Handbook (10th

Edition) and the Cement Association of Canada (CAC) Design Handbook (3rd

Edition) are also extensively used throughout this project, due to the

combination of steel and masonry structure. The following textbooks;

Masonry Structures: Behavior and Design, and Reinforced Concrete Design:

A Practical Approach were also extensively used.

Using knowledge from course material, various resources, and code

documentation, Eccentric Solutions implemented many structural elements

for the new fire station, from masonry structural walls to beam and column

designs. Also, as a team and with help of professors and advisors, Eccentric

Solutions was able to resolve issues pertaining to complex design elements.

2. Results

2.1. Environmental LoadsThe first step was to first calculate the environmental loads that would act on

the structure. Given the building dimensions, we were able to calculate the

environmental loads following NBCC 2010 guidelines. Our results was cross


checked with the online tool (http://www.jabacus.com).

Environmental loads are loads that are imposed on structures directly or

indirectly by the environment. The environmental loads that are considered

for this project were the wind, snow and seismic loads. The snow load is the

load that acts on the roof due to the accumulation of snow. The wind load is

the load in pounds per square foot that is placed on the exterior and interior

of a structure by wind. The Seismic load is the lateral load an earthquake

might cause to act upon a structural system in any horizontal direction.

2.1.1. Snow Loads The average snow loads acting on the building was calculated to be 1.3 kPa.

Referring to figure 2 below, it can be observed that the snow load is


significantly higher on the edges on the building. This makes sense because

snow typically accumulates more around the edges of a building due to snow

drift and significant changes in the building roof elevation. The elevation of

the apparatus bay is significantly higher than the rest of the building which

results in a maximum snow load of 3.9 kPa. (Refer to Appendix A for detailed

Calculations)

Figure 2: Snow Load Distribution

2.1.3. Wind Loads The external wind pressure that acts on all sides of the building is shown in

Table 1. The maximum and minimum external wind pressure are 1.13 kPa

and 0.19 kPa respectively. It should be noted that the results were calculated

for both ultimate limit state(ULS) and service limit state(SLS). The internal

wind pressures for the leeward and windward sides were calculated to be

0.51 kPa and 0.34 kPa respectively under ULS conditions. The internal wind

pressures for the leeward and windward sides were calculated to be 0.31 kPa

and 0.20 kPa respectively under SLS conditions. These values are important

because they impact the fastener resistance required for the roof deck due

to uplift forces that will want to push the roof upwards. (Refer to Appendix A

for detailed Calculations)


2.2. Roof deck

2.2.1. Justification of Roof Deck Selection Steel deck is a cold formed corrugated steel sheet supported by steel joists

or beams. It is used to support concrete or insulating membrane of a roof.

We opted to use a steel deck for this building because it is the most

commonly used material therefore, making it the

most economical. Most of the sheet steel products used in Canada are

supplied by members of the Canadian Sheet Steel Building Institute (CSSBI).

NEW WINDSOR FIRE STATION

Figure 3: Roof Deck

Table 1: Wind Load Cases

9

ULS SLSSide Cg ∙C p Pe (kPa) Pe (kPa)1 0.75 0.42 0.251E 1.15 0.65 0.392 -1.3 -0.73 -0.442E -2 -1.13 -0.683 -0.7 -0.4 -0.243E -1 -0.57 -0.344 -0.55 -0.31 -0.194E -0.8 -0.45 -0.275 n/a n/a n/a5E n/a n/a n/a6 n/a n/a n/a6E n/a n/a n/a

Upon the suggestion of our advisors, we choose a steel deck that is produced

by Canam Group. Canam is a manufacturing company that specializes in the

design and fabrication of construction product and they are one of the top

structural steel construction companies in Canada. Canam produces P-3615,

P-3606, P-3623, and P-2432 steel deck profiles. To design the roof deck for

the entire building, we ran calculations for the apparatus bay as it had the

largest unsupported span. We selected profile P-3615 22 gauge (Refer to

Figure 3) because it weighs the least and it was very important to select a

steel deck profile that would transmit the least loads on the columns in case

of an earthquake.

2.2.2. Roofing materialsThe roof is going to be a flat roof and the roofing materials to be added on

the steel deck were selected from the CISC handbook page 7-57. Materials

chosen include a Gypsum wallboard, 3 Ply asphalt (no gravel) and a

Urethane rigid foam for insulation. Figure 4 shows the assembly of the

roofing materials on the roof deck. Table 2 shows the design dead load for

each of the selected materials. Additional dead load of 0.25 kPa was added

to the total dead load for electrical allowance.


Figure 3: Roof Deck

Figure 4: Roofing materials assembly

10

Table 2: Dead Load of Roofing Materials

Materials Dead Load (kPa)Gypsum wallboard per 100mm 0.08

3 Ply asphalt, no gravel 0.15Urethane foam 0.03

Ducts/pipes/wiring allowance 0.25

2.3. Joist selection

2.3.1. MethodologyDifferent types of steel joist designs are available based on loading and

mounting configurations. We choose most the commonly used type which is

the open-web steel joist. This design consists of two parallel members,

known as chords, with a repeating, triangular web structure located between

the chords. Since we know the weight of the roof deck and insulation, we


were able to calculate the load combinations (Table 3) and determine the

loads that the joist will need to resist.

From this knowledge we referred to the Canam Canadian joist catalogue to

select a joist that was able to withstand the loads from the roof.

Table 3: Environmental Load Combinations Case Loads

1 1.4DL = 26.88 psf2a 1.25DL + 1.5LL = 55.35 psf2b 1.25DL + 1.5LL +.5SL = 81.455

psf3a 1.25DL + 1.5SL = 102.15 psf3b 1.25DL + 1.5SL +.5LL = 112.6 psf

The seismic isn’t considered here because it doesn’t apply significant loads

onto the roof. It can be noted that case 3b governs as the factored load that

will be acting on the roof. In order to pick a joist spacing that was the most

economical, we considered two of the best alternatives which were spacing

the joist 6ft or 8ft apart.

Alternative 1: Joist spaced 6ft apart.

Factored Load = 112.6 lb./ft2 x 6ft = 675.6 lb./ft.

Service Load = DL + LL +.5SL = 66.21 lb./ft2 x 6ft = 397.23 lb./ft.

From the Canam Canadian Joist Catalogue, we selected a joist that could

withstand the calculated factored and service loads. Referring to the Canam

Catalogue, the selected joist needs withstand a factored load of 675.6 lb./ft.

and a service load of 397.23 lb./ft. The weight of the selected joist is 13.2

lb./ft. and the percentage of service load to produce a deflection of L/360 is


64%. This joist can withstand a factored load of 720 lb./ft. and a service load

of 480 lb./ft.

Alternative 2: Joist spaced at 8ft apart.

Factored Load = 112.6 lb./ft2 x 8ft = 900.8 lb./ft.

Service Load = DL + LL +.5SL = 66.21 lb./ft2 x 8ft = 529.6 lb./ft.

From the Canam Canadian Joist Catalogue, we selected a joist that could

withstand the calculated factored and service loads. Referring to the Canam

Catalogue, the selected joist needs withstand a factored load of 900.8 lb. /ft.

and a service load of 529.6 lb./ft. The weight of the selected joist is 19.2

lb./ft. and the percentage of service load to produce a deflection of L/360 is

64%. This joist can withstand a factored load of 930 lb./ft. and a service load

of 620 lb./ft.

By comparing the weight of steel needed for each option, we can see that

option 1 yields a smaller value therefore making it the cheaper alternative.

We continued with the calculations

considering only option 1 for the

apparatus bay. The same approach

was repeated for the other sections

of the building. We also designed

bridging members that will be

connected to the joist to brace it

from lateral movement (Refer

to Table 5).


Figure 5: Steel Joists and Bridging

13

Table 4: Joist SelectionSectio

nSpan(ft.)

Depth

(in.)

Spacing

(ft.)

Factored/Service Load Joist needs to

carry(lb./ft.)

Factored/Service Load Canam Joist

can carry (lb./ft.)

Self weight

of chosen

joist(lb./ft.)

Number of Joist require

d

A 10 18 6 675.6/397.23 720/480 7.2 7B 20 24 6 675.6/397.23 720/480 9.3 7C 36 24 6 675.6/397.23 720/480 11.7 7

D1 20 24 6 956/491 1035/690 10.6 3D2 20 24 6 451/322 510/340 7.9 2D3 20 24 6 932/483 1035/690 10.6 3E1 30 24 6 956/491 1035/690 13.1 3E2 30 24 6 451/322 510/340 9 2E3 30 24 6 932/483 1035/690 13.1 3F1 30 24 6 956/491 1035/690 13.1 3F2 30 24 6 451/322 510/340 9 3F3 30 24 6 932/483 1035/690 13.1 4G 33 22 5 376/269 405/270 8.6 11H 16 22 5 486/305 510/340 7.2 11I1 10 10 6 956/491 1035/690 5.6 3I2 10 10 6 451/322 510/340 5.6 13J1 20 16 6 956/491 1035/690 10.1 3J2 20 16 6 451/322 510/340 6.7 13K 26 24 6 932/483 1035/690 10.3 6

Due to differences in a snow loads on the roof, different joists had to be selected. The joists were separated into different sections (Figure 5) and an appropriate joist was selected for each section. The selected joists are summarized in the table above (Table 4). It can also be noted that a total of 107 steel open-web joists will be required for this building.

Table 5: Bridging for JoistsSpan(ft.)

Factored Load

(lb./ft.)

ServiceLoad

(lb./ft.)

Lines ofBridging

Type Spacing

(ft.)10 675.6 392.3 0 None 020 675.6 392.3 1 L 15/8

x .1187.92

36 675.6 392.3 2 L 15/8 x .118

7.92



Figure 6: Roof Joist Layout


2.4. Beam Selection

2.4.1. MethodologyRecognizing that we need 7 separate I beam, we realized that the critical

beam that carries the most loads is the beam running 47.5’ through the

center of the apparatus bay. This beam was the first to be analyzed.

Obtaining the factored and shear loads, we were able to select an I- beam

that could withstand the applied loads and deflection occurring. However,

considering this beam carries the largest load and with the given deflection

(3.2 inches), we noticed that it would be much more satisfactory if we

designed this beam as a joist girder which will cause a significant decrease in

deflection. A Joist Girder typically has an I-beam cross section composed of

two load-bearing flanges that are separated by a modified warren truss. The

girder was selected from the Canadian Canam Joist Girder catalogue based

on the selected depth (36in) and the given spans. The deflections were

calculated and the amount produced was much smaller than that of an I

section (0.17 inches). Therefore, we chose to implement the joist girder in

our design and followed through with calculations. Table 6 and 7 shows the

difference in deflection of the selected Beams and Girders (Refer to Appendix

A for detailed Calculations).


Table 6: Beam Selection

Table 7: Joist Girder Selection

Joist Girde

rType

Depth

(Inches)

Weight

(lb./ft.)

Span

(ft.)

Factored

Moment

(kip∙ft)

ActualDeflecti

on(inches)

AllowableDeflectio

n(Inches)

JG1ModifiedWarrenTruss

36 65 47.5 920.2 0.17 2

JG2 Modified WarrenTruss 36 40 47.5 548.8 0.13 2


Beam Section Span(ft.)

Factored

Moment (kip∙ft)

Resistive

Moment(kip∙ft)

Actual Deflectio

n(Inches)

Allowable

Deflection

(Inches)B1 W530 X 219 47.5 595.2 785.9 2.86 3.17B2 W360 X 122 47.5 163.2 269.5 2.79 3.17B3 W310 X 79 36 109.8 153.9 2.15 2.40B4 W310 X 79 36 109.8 153.9 2.15 2.40B5 W410 X 114 30 322.1 373.7 1.14 2.00B6 W410 X 114 30 322.1 373.7 1.14 2.00B7 W460 X 52 22 117.8 127.8 1.01 1.46

18

2.5. Columns

2.4.1. MethodologyOur selection of HSS sections were based on the sections benefits when

compared to our initial choice of I section columns. HSS section benefits

include its high strength to weight ratios, uniform strength, cost

effectiveness, recyclability and its aesthetic appeal. When we first designed

the columns as I sections, we were advised that due to bays relatively large

height, we needed to brace these sections in order to safely withstand the


Figure 7: Beam and Joist Girder Layout

19

lateral loads from the wind. However, in terms of cost, we had decided that it

would be cost beneficial if we implement HSS sections with I section bracing.

We braced each 8.83 m HSS column with an I section located at 5.5m from

the floor. This ensured our design was able to accompany for the wind load

and also be cost efficient. Table 8 shows the details of the selected columns

(Refer to Appendix A for detailed Calculations).

Column SectionHeigh

t(ft.)

FactoredAxial

Compressive Load(kip)

FactoredAxial

Compressive

Resistance (kip)

Factored

Moment (kip∙ft)

FactoredResistiveMoment(kip∙ft)

C1 HSS 152 X 76 X 6.4 29 37 39 14.4 29.1

C2 HSS 152 X 76 X 6.4 29 37 39 14.4 29.1

C3 HSS 152 X 76 X 6.4 29 35.1 39 14.4 29.1

C4 HSS 152 X 76 X 6.4 29 35.1 39 14.4 29.1

C5 HSS 152 X 76 X 9.5 29 39 51.7 14.4 39.8

C6 HSS 152 X 76 X 9.5 29 39 51.7 14.4 39.8


Table 8: Column Selection

2.4.2. Base PlateAlthough the base plate is among the last

components of the building to be designed,

they are typically one of the first components to be placed. The main

function of base plates to is to transfer the compressive axial loads from the


Figure 8: Column Layout

Figure 9: Base Plates21

columns to the foundation (Refer to Appendix A for detailed Calculations).

Table 9: Base Plate Selection

Base Plate Section Factored Load

(kip)Bearing Capacity

(kip)BP1 300 x 300 x 30 37.1 252BP2 300 x 300 x 30 37.1 252BP3 300 x 300 x 30 35.2 252BP4 300 x 300 x 30 35.2 252BP5 300 x 300 x 30 39.1 252BP6 300 x 300 x 30 39.1 252

2.6. Apparatus Bay Slab on Grade

Space

Slab Thicknes

s (In.)

Specified

Live Load (lb./ft2)

Joint Spacin

g(ft.)

F y

(MPa)

WeldedWireMesh

Section

Area Required(In.2/ft.)

ActualArea(In.2/

ft.)

Mass(lb./ft

Office 7 100 26.25 450 102 x 102-MW 11.1 X MW 11.1 0.49 0.56 0.38

Apparatus Bay 7 250 26.25 450 102 X 102-MW

25.8 X MW 25.8 1.22 1.29 0.88The floor in the apparatus bay needed to be designed such that it will be

capable to withstand the loads applied from a heavy load (fire truck). Slab-

on-grade are concrete slabs that serve as the foundation for the structure,

which is formed from a mold set into the ground. The concrete is then placed

into the mold, leaving no space between the ground and the structure. Low

areas are filled to grade using Granular ‘B’ placed in 250 mm thick lifts

uniformly compacted to 98 per cent of SPMDD. Final construction should

consist of at least 200 mm of Ontario Provincial Standard Specifications

(OPSS) Granular ‘A’ base material uniformly compacted to 100 per cent of


https://en.wikipedia.org/wiki/Concrete

SPMDD. The preferred thickness chosen for the bay concrete slab will be 7”

thick and be reinforced with welded wire mesh section of 102 X 102-MW 25.8

X MW 25.8. The office floor will have similar concrete slab but will be

reinforced with 102 x 102-MW 11.1 X MW 11.1 (Refer to Appendix A for

detailed Calculations). Table

10: Slab on Grade


2.7. Load Bearing Walls

The next step after completing the design of the apparatus bay was

structural or load bearing walls. A load bearing wall is a wall that transmit

loads from the roof to the foundation of a building. Deciding on which wall to

design as a load bearing wall was a critical step, as the load bearing walls

must be placed such that the joist layout is utilized optimally, but also

keeping in mind the foundation requirements. A wall carrying a significant

number of joists as well as a section of roof might require a deep and heavily

reinforced foundation, and in this case altering the joist layout and using two

structural walls may be significantly more cost effective. The design of each

load bearing wall was governed by the loads acting on the wall from the

joists and roof. The amount of reinforcement required was determined by the

amount of load the wall needed to resist. From Table 11, based on the

resistance moment, it can be observed that wall E had the most load acting

on it and this is mainly due to the snow load acting in that area of the

building (Refer to Appendix A for detailed Calculations).

Wall

Height of wall(m)

Length of wall(m)

#/Size of

rebar

Rebar Spacin

g(m)

Size of masonry block (mm)

Compressive

strength of

masonry unit (MPa)

f m'

(MPa)M r

(kN.m/m)

M f tot

(kN.m/m)

A 4.13 28.77 2-15M 0.8 240 30 17.5 21.8 15.5B 4.13 29.06 2-15M 0.39 190 30 17.5 32.3 15.86C 4.13 27.24 1-20M 0.162 190 30 17.5 59 15.2


Table 11: Load Bearing Wall

24

D 4.13 10.3 1-15M 0.162 190 30 17.5 40.6 21E 4.13 17.09 2-15M 0.162 240 30 17.5 107 73F 4.13 14.64 1-15M 0.162 190 30 17.5 42.6 34.9G 4.13 14.64 1-20M 0.8 240 30 17.5 17.4 16H 4.8768 17.72 2-15M 0.8 190 30 17.5 16 15.12I 4.8768 10.38 2-15M 0.8 240 30 17.5 27.3 23.8J 4.8768 18.34 1-15M 0.8 240 30 17.5 11.5 10.3

2.8. Shear WallsFor the buildings lateral stability, shear walls are implemented into the

diaphragm. Unlike the apparatus bay which has columns to resist both axial

and bending loads, masonry walls don’t perform well against shear forces, so

a select few walls must be selected to deal with the wind and seismic loads.


Figure 10: Selected load bearing walls

25

As seen in the environmental loading sections, the wind governs the design

of this buildings shear walls, as is the case in all of Windsor-Essex.

The shear walls in this building were designed to resist the maximum wind

load determined in the NBCC calculations for Windsor, Ontario. In accordance

with this projects environmental loads, and the NBCC, the maximum shear

force in the East-West direction is determined to be 44.51 kN (Refer to figure

12).

The Shear Force, V, is to be taken by 2 shear walls, and shall only resist the

force acting on the top half of the wall, because the top half of the

compression chord is the only area that contributes a notable amount to the

factored moment of the wall (Refer to Appendix A for detailed Calculations).

Wallhw

(mm)lw

(mm)Mf

(106

N.mm2)

Mr(106

N.mm2)

Vertical Rebar

(# , bar)

PositionSpacing of

grouted cells

A 4206 7588 294.42 685.4 2 10M100mm

from each end 48”

B 4206 9144 294.42 1677.4 2 10M100mm

from each end

Only at ends

C, D 4206 3700 235.5 255.8 2 15M100mm

from each end

Only at ends

E 4941 9144 553.4 988.5 2 15M100mm

from each end

Only at ends


Table 12: Shear Walls Summary

2.9. LintelsLintels are

structural

horizontal blocks

that spans the

opening

between two

vertical


Figure 12: Shear Forces acting on Foundation

Figure 11: Shear walls

27

supports. We had to design lintels for the low rise sections of the structure

that had gaps between them due to doors and windows. According to the

layout of the building, there were 8 lintels that had to be designed according

to the masonry behavior design.

The table below shows the results obtained for all of the lintels designed. 7 out

of 8 had to be designed as 2 course beam made of 200 mm masonry block.

The reason for this was because of the load acting on them due to transfer of

load from the joist to the masonry beam. Lintel #1 had two joists acting on the

beam with different spacing, resulting in two factored moments (M f= 45kN.m).

Therefore, stronger reinforcements were used (2-15M), also one extra course

was used and the result that the beam would be able to resist the moment.

However, for the rest of the lintels, 2 course beams and 15M rebars were

used to achieve of resisting moment (M r=15.6 kN.m) that was higher than all

the factored moments that ranged between 3-15 kN.m (Refer to Appendix A

for detailed Calculations).


Table 13: Summary of Lintels

2.10. Foundation Foundations are the base for any structure, it carries all the loads that are

transferred through the diaphragm. Starting from the environmental load

and working all the way to the bottom. Therefore, it is crucial to design an

appropriate foundation that is capable of carrying the loads acting on it. The

design of the foundation for the apparatus bay was chosen to be square

spread footing due to the following reason:

The loads exerted by the structure

The cost of foundations allotted

Nature of the structure


Lintels

length of beam (m)

# of course beams M f Max(kN.m) M r

(kN.m)Reba

r As(m m2)

1 2.84 3 45.947 47.55 2-15M 400

2 1.77 2 8.44 14.94 2-10M 200

3 2.38 2 14.5 15.6 1-15M 200

4 1.83 2 13.92 15.6 1-15M 200

5 1.82 2 11.67 14.94 2-10M 200

6 1.12 2 3.29 15 1-15M 200

7 1.12 2 6.3 15.6 1-15M 200

8 2.24 2 13.81 15.6 1-15M 200

29

The table 14 shows a summary that was made to display the 6 square

spread footings

designed for the

apparatus bay.

Foundations at

columns C1-C4 has

an area of 2.25m2

while columns C5&6

are 3 m2 and this is

due to the applied load on the foundation coming from the corresponding

columns. Shear reinforcements weren’t required as the calculated resistance

shear was more than the factored shear (V f ). However, for Flexural

Reinforcement Two types of rebar were used for the spread foundation: 4-

15M @ 430 mm spacing and 5-15M @ 400 spacing (Refer to Appendix A for

detailed Calculations).


Figure 12: Spread Footing

Table 14: Summary of Spread foundation

Column Type Length(m)

Width(m)

Thickness

(mm)Area(m2)

Column Base

(mm x mm)

Rebar Spacing(mm)

C1Square Spread Footin

g1.5 1.5 250 2.25 250x250 4-15M 430


g1.5 1.5 250 2.25 250x250 4-15M 430


g1.5 1.5 250 2.25 250x250 4-15M 430


g1.5 1.5 250 2.25 250x250 4-15M 430


g1.75 1.75 250 3 250x250 5-15M 400



g1.75 1.75 250 3 250x250 5-15M 400

For the rest of the building, a continuous strip foundation was selected to

take the linear loading pattern of the structural walls, and to easily resist

shifting, sliding or overturning for the shear walls. In all practicality, there

wasn’t a viable alternative to this considering the economic benefits, and the

continuous rigid system offered by a continuous strip shallow footing.

In accordance with OBC, NBCC, and the Concrete Handbook, the strip

foundation for all the load bearing walls was governed by wall E which had

the largest load acting on it.

Figure 13: Strip Footing


3. Cost Analysis

Table 15: Cost Analysis

Item Specific Item Unit Price unit Quantity Cost

A) Roof i) P-3615 Steel Deck 1.3 m21114.836

5 $1,449.29

ii) connection steel 1.5 Lb. 8890 $13,335.0

0

iii) steel shop drawings 10000 ea. 1 $10,000.0

0

B) Joisti) 10" 10.8 LF 160 $1,728.00ii)16" 12 LF 320 $3,840.00iii) 18" 14.5 LF 70 $1,015.00iv) 22" 15.8 LF 539 $8,516.20

v) 24" 18.5 LF 1218 $22,533.0

0vi) L 15/8 X 0.118 Steel

Bridging 3000 tn 0.375 $1,125.00

C) Joist Girder

ii) 36" 2500 Lb. 2.5 $6,250.00

D) Beams

i) W530 X 219 1.5 Lb.6990.157

523 $10,485.2

4

ii) W360 X 122 1.5 Lb.3894.060

355 $5,841.09

iii) W310 X 79 1.5 Lb.3822.159

672 $5,733.24

iv) W410 X 114 1.5 Lb.4596.267

96 $6,894.40

v) W460 X 52 1.5 Lb.768.7325

36 $1,153.10

E) Walls

i) 6" Masonry Block 10 ea. 2900 $29,000.0

0

ii) 8" Masonry Block 12 ea. 8845 $106,140.

00


iii) 30 MPA 8" Masonry Blocks 15 ea. 6010 $90,150.0

0

iv) 30 MPA 10" Masonry Block 18 ea. 2200 $39,600.0

0

v) 10" Masonry Block 14 ea. 5660 $79,240.0

0vi) 10M Structural

Reinforcement 1 Lb. 4068 $4,068.00vii) 15M Structural

Reinforcement 1 Lb. 1200 $1,200.00viii) 20M Structural

Reinforcement 1 Lb. 6000 $6,000.00x) 7.5 MPA Grout 4 cores 220 $880.00

xi) 17.5 MPA Grout 4 cores 1280 $5,120.00

F) Foundation

i) Forms and labour 43 m2 1620 $69,660.0

0

ii) 25 MPA concrete 210 m3 186 $39,060.0

0iii) 15M Structural

Reinforcement 1 Lb. 1124 $1,124.00iv) 20M Structural

Reinforcement 1 Lb. 984 $984.00

v) 175mm slab on grade 160 m2 320 $51,200.0

0vi) 102 X 102-MW25.8 X

MW25.8 Mesh Reinforcement 1.25 f t 2 3366 $4,207.50vii) 102 X 102-MW11.1 X

MW11.1 Mesh 1.25 f t 2 8700 $10,875.0

0

viii) 150mm slab floor 110 815 $89,650.0

0

G) Earth Works

i) Excavation 40 m3 1998 $79,920.0

0

ii) native backfill 20 m3 880 $35,200.0

0

H) Canopyi) 20 X 20 X2 SHS 2 Lb. 228 $456.00

ii) 3" X 2" x 0.25 HSS tubing 2 Lb. 5550 $11,100.0


0

I) Columni) HSS 152 X 76 X 9.5 2 Lb. 1110 $2,220.00ii) HSS 152 X 76 X 6.4 2 Lb. 1560 $3,120.00

iii) anchor bolts 15 Ea.. 48 $720.00iv) anchor bolts to base 10 ea. 24 $240.00

v) steel plate 2 Lb. 84 $168.00

vi) misc. steel 2 Lb. 6453 $12,906.0

0

Sum $874,107.

05

Figure 14: Cost Analysis Chart

The propose project is expected to cost approximately $4,500,000 in total.

However according to our industry advisor, the structural components will

cost approximately $900,000 which is relatively close to the amount we


calculated. Referring to figure 14, it can be seen that the walls will account

for 41% of the project cost, this is expected due to the high material cost of

masonry units and labour cost.

4. Conclusion

The main goal of our capstone project was to design a cost effective, modern

and safe post disaster structure to fit for a fire station, meaning that it should

be able to have specific necessities in terms of equipment, building design

and components. Some of those requirements were that it should be fire

resistive, composed of some steel components, able to resist seismic load

and be capable to safely withstand any fire drill practice that occurs within.

However our side goal was to achieve a building that is economical and

environmentally friendly. Considering our minimal utilization of materials

used while keeping safety as the highest factor, we believe that our goal has

been achieved through accounting for worst case scenario. Most components

we designed were governed by the most critical sections throughout the

whole building. The building was designed to satisfy ULS (Ultimate Limit

State) and SLS (Serviceable Limit State) loading conditions, this ensures that


our building will be a safe and sturdy. A major problem that we encountered

was finding a way to anchor the canopy on top of the roof to the building. We

resolved this issue by loading the canopy onto a specified Joist Girder which

then transferred the load to the Beams and the Walls, therefore stabilizing

the canopy. This project provided us with an extensive knowledge through

learning more about all the design procedures, different components of the

building and how to account for it. Also the various alternatives and whether

which one fits better in different situations. This projects has helped us

develop and broaden our knowledge about structural engineering.

5. References

Various Authors. Handbook of Steel Construction, Tenth Edition. Toronto:

Canadian Standards Association. 2009. Print

NBC 2010 Structural Commentaries. Ottawa: Canadian Commission on

Building and Fire Codes. 2010. Print

Ontario Building Code 2006. Toronto: Ontario Ministry of Municipal Affairs

and Housing. 2006. Print

Kulak, G.L., and G.Y. Grondin. Limit States Design in Structural Steel. 9.

Toronto: CISC-ICCA,2011. 403. Print.


Drysdale, R. G., Hamid, A. A., & Baker, L. R. (1994). Masonry structures:

Behavior and design. Englewood Cliffs, NJ: Prentice Hall.

Brzev, S., & Pao, J. (2006). Reinforced concrete design: A practical approach. Toronto: Pearson Prentice Hall.

Appendix A

Snow Load CalculationFor Apparatu Bay S=I s [S s (C b C w C s C a )+S r ] Location :Windsor , OntarioSs=0.8kPa

Sr=0.4 kPaImportance FactorsULS : I s=1.25SLS : I s=0.9C s=1C b=0.8

Density=γ=3 kNm3 Distribution w=15.09 m

L=15.73 m

l c=2w−w 2l


2 (15.09 m) – (15.09 m )215.73 m

15.7 m

Factor F=greater of :a¿ F=2b¿ F=0.35(γ∗l c /S s−6(γ∗h p /S s)2)0.5+C b

0.35( (3 Knm 3 ) (15.73 m)

.8 kpa– 6( (3 Kn

m3 ) ( .762 m)

.8 kpa )2)0.5+ .8

1.9

F=2

C a ( 0 )=lesser of :

a¿C a(0)=(γ∗h)/ (C b Ss)( 3 Knm3 )(3.81 m )

( .8 ) ( .8 kpa )17.859

b¿C a(0)=F /C b2.82.5C a ( 0 )=2.5

x d=lesser of :a¿ x d=5(h−C b S s/ γ )

5( (3.81 m ) – (.8 ) ( .8 kpa )3 Knm 3 )

17.983 mb¿ x d=5(S s/ γ)(F−C b)

5( .83 Knm3 ) (2−.8 )

1.6 mx d=1.6 m

h'=h−C bC w S sγ

(3.81 m ) – ( .8 ) (1 ) (.8 kpa )3 Knm3

3.597 mMin dist where (C w=1.0 )=10∗h'=35.97 m


Snow Load Summary( x=0 )C w=1C a ( 0 )=2.5

SULS=1.25 [0.8 (0.8∗1∗1∗2.5 )+0.4 ]=2.5 kPaS SLS=0.9 [0.8 (0.8∗1∗1∗2.5 )+0.4 ]=1.8 kPa( x=x d=1.6 m )C w=1C a ( x d )=1S ULS=1.25 [0.8 (0.8∗1∗1∗1 )+0.4 ]=1.3 kPaSSLS=0.9 [0.8 (0.8∗1∗1∗1 )+0.4 ]=0.936 kPa

x>( xd=1.6m )C w=1.0C a ( x )=1SULS=1.25 [0.8 (0.8∗1.0∗1∗1 )+0.4 ]=1.3 kPaSSLS=0.9 [0.8 (0.8∗1.0∗1∗1 )+0.4 ]=0.936 kPa

Lower adjacent RoofEoc CentreWind Load Calculation

……

Seismic AnalysisSection1 : Apparatus BayLocation :Windsor ,Ontario

Sa (0.2 )=0.15Sa (0.5 )=0.085Sa (1.0 )=0.045Sa (2.0 )=0.014

PGA=0.073Site class=DImportance Factor , I E=1.5Material=steel

System=momentSFRS=Ductile moment−resisting frames

weight=1712 kNhn=8.83 mF a=1.3F v=1.4


R d=5R o=1.5T acomputed=0.44 sM v=1

S (T a )=S (0.44 )=0.14V=S (T a )∗M v∗I E∗WR d∗R o

=46.35 kN S (2.0 )=F v∗S a (2.0 )=0.0196

V min=S (2.0 )∗M v∗I E∗WR d∗R o

=6.71kN S (0.2 )=F a∗S a (0.2 )=0.195

V max=( 23 )∗S (0.2 )∗I E∗W

R d∗R o=44.51 kNV >V max∧R d ≥ 1.5V Design=44.51 kN

Section2 : Lower Level

Location :Windsor ,Ontario

Sa (0.2 )=0.15Sa (0.5 )=0.085Sa (1.0 )=0.045Sa (2.0 )=0.014

PGA=0.073Site class=DImportance Factor , I E=1.5Material=masonry

System=shear wallSFRS=Conventional construction ( Moment−resisting frames )weight=520 kNhn=4.13 m

F a=1.3F v=1.4R d=1.5R o=1.5

T acomputed=0.1 s

M v=1


=65 kN

S (2.0 )=F v∗S a (2.0 )=0.0196V min=S (2.0 )∗M v∗I E∗WR d∗R o

=6.53 kN


S (0.2 )=F a∗S a (0.2 )=0.195V max=( 23 )∗S (0.2 )∗I E∗W

R d∗R o=43.33 kN

V >V max∧R d≥ 1.5

V Design=43.33 kN

Section3 : Higher Level

Location :Windsor ,Ontario

Sa (0.2 )=0.15Sa (0.5 )=0.085Sa (1.0 )=0.045Sa (2.0 )=0.014

PGA=0.073Site class=DImportance Factor , I E=1.5Material=masonry

System=shear wallSFRS=Conventional construction ( Moment−resisting frames )weight=325 kNhn=4.88 m

F a=1.3F v=1.4R d=2R o=1.5T acomputed=0.16 sM v=1


=31.69 kN

S (2.0 )=F v∗S a (2.0 )=0.0196V min=( S (2.0 )∗0.5 )∗M v∗I E∗WR d∗R o

=1.59 kN

S (0.2 )=F a∗S a (0.2 )=0.195V max=( 23 )∗S (0.2 )∗I E∗W

R d∗R o=21.13 kN

V >V max∧R d≥ 1.5

V Design=21.13 kN

BeamCalculation


Sample beam selection.Beam1 Analysis Insert A 1

∑ Fy=−675.6 lbft

x36 ft−13.2 (36 )+2 R

R=12.4 kips

Insert A 2

∑ Fy=−2 (12.16 kips )−7 (12.4 kips )+2 RR=61.3 kips

Insert A 3

Mf =595.2 kips∙ ft=806.98 kN ∙mFactored Live Load=122.6 kips=545.35 kN

Factor Dead Load=.7 kNm

x14.478 m=10.13 kN

Wf =545.35 kN+10.13 kN14.418 m

=38.37 kNm

Try W 530−219

Dead Load ( kNm )=12.15

Mass( kgm )=219

Area (mm 2 )=27900I x (106 mm 4 )=1510S x (10 3mm 3 )=5390r x (mm )=233Z x (10 3 mm3 )=6110I y (10 6mm 4 )=157S y (10 3 mm3 )=986r y (mm )=75Z y (103mm 3 )=1520J (103 mm4 )=6420C w (10 9mm6 )=11000Depth , d ( mm )=560


Flange width, b (mm )=318Flange thickness , t (mm )=29.2Web thickness , w (mm )=18.3F y ( MPa )=350

ˇlocal buckling class II

Flange Slenderness

b2t= 159 mm

29.2 mm=5.45

170√Fy

= 170√350 MPa

=9.1

5.45<9.1∴No Flangebuckling .

Web Slenderness

hw

= 50218.3

=27.43

1700√Fy

= 1700√350 MPa

=90.87

27.43<90.87∴No Web buckling .

ˇDeflection

∆= L180

=14478 mm180

=80.43 mm

∆=5w f L4

384 E I x=

(5 )(23.5 kNm ) (14478 mm )4

384 (200000 MPa ) (1510 x106 m m4 )=44.5 mm<80.43 mmOK

Design for laterally unsupported members


M u=π w2

L √ E I y GJ +( πEL )

2

I y Cw

w2=(4 M max)/√ (M max 2+4 M a2+7 M b

2+4 M c2)≤ 2.5

M max=M b=595.2 kip . ft .a=M c=297.6 kip . ft .W 2=1.26≤ 2.5

E I y GJ=1.55 x 1025

( π x 200000L ) x I y Cw=3.25 x1024

M u=π x 1.51

6705x√1.55 x1025+3.25 x 1024=1183.9 kN . m

M p=Zx F y=( 6110 x103 m m3 ) (350 MPa )=2138.5 kN ∙ m0.67 M p=.67 (2138.5 kN ⋅m )=1432.8 kN .m

∴M u ≤0.67 M p

M r=∅ M u=0.9 x1183.9 kN ⋅m=1065.1 kN .m>M f OK

Sample joist girder calculation

JoistGirder 1.

Insert A 4

∑ Fy=−675.6 lbft

x20 ft−9.3 (10 )+2R

R=6.8 kipsInsert A 5

∑ Fy=−.104 kipsft

x 47.5 ft−7 (10.2 )+2 R

R=38.17kips

Insert A 6

Mf =548.88 kips∙ ft=746.5 kN ∙mFactored Live Loa=545.35 kN


Factor Dead Load=10.13 kN

Wf =545.35 kN+10.13 kN14.418 m

=24.16 kNm

I=.132 M f D=.132 (548.88 kips⋅ ft ) ¿

∆=w f L4

154667 M f D=

( 77.4 kipft ) (47.5 ft )4

154667 (548.8 kips ⋅ ft ) ¿¿

Therefore using a Modified Warren Truss¿ the designated CanamJoist catalogue witha weight of 40 lbft

for this span .

¿∗TOMMY REFERENCE THE ABOVE FORUMALS ( I∧deflection ) ¿THECANAM CATALOGUE∗¿

Column Design

Insert A 7

Sample Analysis for Column Design

C f =3.4 kips+35.6 kips=39kips=173.5 kN

K y Ly=5500 mm (weak axis )K x Lx=8830 mm (strong axis )

¿CSA CISC table on page 4 – 46

HSS 152 X 76 X 9.5 :

C r=230 kN>173.5 kN OK

( r xr y )=1.74

M rx=53.9 kN ∙ mM ry=32.8 kN ∙ m

〖 ( K 〗¿¿ y Ly )( r xr y )=(5500 ) (1.74 )=9570>8830 mmOK ¿

Insert A 8


Insert A 9

∑ M o=Fbx (5.5 )−(8.18 ) (8.83 ) (4.415 )Fbx=57.98 kN

∑ F x=−F x−57.98+72.2¿14.22 kN

M f =19.55 kN ∙m<M r OK

Beam Analysis : (between joist G∧H )

G=376 lbft

;self −weight=8.6 lbft

; H=486 lbft

; self −weight=7.2 lbft

;Span of 1 st part=32.43 ft . ;span of 2nd part=14.93 ft .

Load=( span x factored load )+(self −weight x span )

¿(376 lbft

x32.43 ft .+8.6 lbft

x32.43 ft .)+(486 lbft

x14.93 ft .+7.2 lbft

x14.93 ft .)¿9.92kips∑ Fy=0

R 1=R 2=9.92 x52

=24.8 kips

M f =159.7 kN .mFactored≪¿9.92 x2=19.84 kips=88 kNFactored DL=0.7 x 6.7056 m=4.69 kN

w f =¿ 88+4.696.7056

=13.82 kNm

∆= L180

= 6.7180

=37.2mm

∆=5 x Wf x L4

384 x E x I= 5 x13.82 x6705.64

384 x 200000 x 212 x 106 =25.8<37.2 mm

Therefore nodeflection will occur :M max=M b=117.8kip . ft .a=M c=58.9 kip. ft .W 2=1.51≤ 2.5E I y GJ=4.08 x 1020

( π x 200000L ) x I y Cw=7.37 x1022

M u=π x 1.51

6705x√4.08 x1020+7.37 x 1022=192.6 kN .m

M p=Zx F y=1490 x 103 x 350=521.5 kN .m0.67 M p=0.6 x 521.5=349.4 kN .m∴M u ≤0.67 M p

M r=∅ M u=0.9 x192.6=173.3kN . m>M f

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−¿Beam Analysis : B 2


Factored load=112.6 lbft2 .

x6 ft .=675.6 lbft

∑ Fy=0 ; R 1=R 2=675.6 x362

=12.37 kips

∑ Fy=0 ; Rc 1=Rc 2=2 x18.9+7 x 19.72

=86 kips

M f =920 kN . mI=0.132 x920.22 x36 ft .=4372.89∈.

∆= w x L4

154667 M f x D= 172 x 47.54

154667 x 920.2 x 36=0.17<2

Geom etry=modified truss

Weight=65 lbft

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−¿

Beam Analysis : B 3For 20 ’ span ;for 10’ span;


x6 ft .=675.6 lbft

Factored resistive load=720 lbft

>675.6

SL=DL+¿+0.5 SLmass of joist=72 lbft

¿397.23 lbft

Factored resistive load=720 lbft

>675.6 lbft

Mass of joist=93 lbft

∑ Fy=0 ; R 1=R 2=R=675.6 x 10+7.2 x 102

=3.4 kips

∑ Fy=0 ; Rc 1=Rc 2=7 x 3.4+2 x3.32

=15.2 kips

∴M f =163.2 kip. ft=221.95 kN .mFactored≪¿27.1 kips=120.55 kNFactored DL=0.7 x14.478 m=10.13 kN

w f =9.03 kNm

V f =52.96 kN ¿Use W 360 x 1220.5 b

t=5.92<9.1√❑

hw

=24.6<90.87√❑

∆= L180

=6.714478180

=80.43 mm

∆ actual=70.77<80.43 mm√❑


M max=M b=163.2 kip. ft .a=M c=81.6 kip . ft .W 2=1.26≤ 2.5E I y GJ=1.998 x 1024

( π x 200000L ) x I y Cw=2.07 x 1023

M u=π x 1.51

6705x√1.998 x1024+2.07 x1023=406 kN . m

0.67 M p=0.6 x 532.3=349.4 kN .m∴M u ≤0.67 M p

M r=∅ M u=0.9 x 406=365.4 kN .m>M f

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−¿

Beam Analysis : B 7


x6 ft .=675.6 lbft

∑ Fy=0 ; R 1=R 2=R=675.6 x 20+9.3 x102

=6.8 kips

∑ Fy=0 ; R 1=R 2=R=10.2 x7+104 x 47.52

=38.17 kips

∴M f =548.88 kip . ft=746.5 kN .mFactored≪¿339.6 kNFactored DL=0.7 x14.478 m=10.13 kN

w f =24.16 kNm

I=0.132 x548.88 x36 ft .=2607.89 ¿4

∆= w x L4

154667 M f x D= 77.4 x47.54

154667 x 548.88 x 36=0.13<2

Geometry=modified Warrentruss

Weight=40 lbft

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−¿

Beam Analysis : B 6

∑ MC 2=0 ;−R c1 x30+675.6 x30 x 15+38.17 x10R c1=22.86 kips∑ Fy=0 ; R c1+R c2−675.6∗30R c2=35.6 kips∴M f =322.1 kip . ft=436.7 kN .mFactored≪¿260 kNFactored DL=0.7 x14.478 m=10.13 kN

w f =29.54 kNm

Use W 410 x114


0.5 bt

=6.76<9.1√❑

hw

=32.84<90.87√❑

∆= L180

=9144180

=50.8 mm

∆ actual=29 mm<50.8 mm√❑M max=M b=322.1 kip. ft .a=M c=214.7 kip . ft .W 2=1.18≤ 2.5E I y GJ=1.31 x1024

( π x 200000L ) x I y Cw=6.21 x1023

M u=π x 1.51

6705x√6.21x 1023+1.31x 1024=563 kN . m

0.67 M p=576.87 kN .m∴M u ≤0.67 M p

M r=∅ M u=0.9 x563=506.7 kN . m> M f

Base Plate

Sample analysis for Base plate

For Column 1Pf =165 kNAllowable Bearing Pressure=12 MPa

Bearingarea required=165 x 103 N12 MPa

=13750 mm2

300 mm

300 mm

Area = 300 mm x 300 mm = 90000 mm2 > 13750 mm2

Actual Bearing Stress=165 x103 N90000 mm

=1.83 MPa


M max=(1.83 Nmm ) (36 mm x36 mm )=2372 N ∙mm

Br=∅m .85 f m A=( .6 ) ( .85 ) (25 MPa ) (300 x300 )=1147.5kN>P f OK

Let Mr = Mmax

M max=∅ Fy t 2

4

t = 30mm

Floor Design

1 Truck = 80 kips

Traffic limit = 20 vehicles/day

Concrete Flexural Strength = 650 lb/in2

Modulus of subgrade reaction = k = 185 lb/in3

Based on 80 kips gross weight, we chose the Forklift Truck Category (IV)Design Index 8


from above figure. Chosen thickness (t) will 7 inchs.

A s=w( l

2 ) F

fs

w = 250 lb/ft2 = 11970 N/m2

F = (friction factor) = 1.5L = Joint Spacing = 8mfs= 2/3 FyFy = 450 MPa


A s=(11970 N

m2 )( 4 m )(1.5)

( 23)(450 MPa)

=240 mm2/m

∴Use 102 x102 MW 25.8 X MW 25.8

As = 253 > 240 mm2/m OK

For Office floor

w = 100 lb/ft2 = 4788 N/m2

F = (friction factor) = 1.5L = Joint Spacing = 8mfs= 2/3 FyFy = 450 MPa

A s=(4788 N

m2 )( 4m )(1.5)

( 23)(450 MPa)

=96 m m2/m

∴Use 102 x102 MW 11.1 X MW 11.1

As = 109 > 96 mm2/m OK


Using 20 x 20 x 2 HSS Height = 6ft = 1.83mMass =1.1kg/m x 1.83 = 2.013 kgLoad = 4.15 kNW= 104.3 lb/ft

The Canopy will be designed for 1 meter or 3.28ft length of the roof. For each section, there will be 2 20 x 20 x 2 HSS sections at a height of 6ft. These sections will the be braced with 3 horizontal 20 x 20 x 2 HSS sections which will allow panels or cladding to be bolted with. The same sections will span over the roof and behind the building for fastening and moment resistance.


Sample Calculations for Load bearing walls

For Wall A

Height of wall = 4.12m

R f=(956 lbft

×20 ft )+(10.2 lbft

×20 ft )=19,3222

lb=9661 lb

Pf =9661 lb×16 joist=154,576 lb

Pf =154,576 lb

94' 421 } ft} =1637.9 {lb} over {ft} =23.9 {kN} over {m¿¿

Trying reinforced concrete masonry with a reinforcing bar and grouted cell at 0.8 m spacing using 30 MPa, normal weight, 25cm hollow block in running bond with Type S mortar and F y= 400 MPa.

kht

=(1 ) ( 4130 )

240=17.2<30

Special provisions for slender walls are not required (CSA S304.1(10.7.4.6¿7.1 ¿

For reinforcement spaced at 0.8 m, from CSA S304.1 (10.6.1¿7.1for running bond,

beff=4 ( t )=4 (240 )0.8 m

=1200 mmm

Factored moment, M f =0.9 (23.9 )

2+ 1.4 (1 ) ( 4.13 )2

8=13.7 kN . m /m

Using an average entire effective mortared face shell thickness, t f of 35.65mm and f m' =17.5 MPa

for Type S mortar with a masonry unit compressive strength of 30 MPa

C=∅m 0.85 f m' X beff t f =(0.6 ) (0.85 ) (17.5 ) (1 ) (1200 ) (35.65 ×10−3 )=382 kN

M r=C( t2−

t f

2 )=382( 2402

−35.652 )×10−3=39 kN .m /m


M r>M f , so compression block is within the face shell and the use of f m' =17.5 MPawil be correct

For steel,

M r=∅ s A s f y (d−β1 c2

)

Setting M r=M f and solving,

A s (approximate )= 13.7 ×106

0.85(400)(2402

−10 ( estimated ))=366 m m2 /m∨293 m m2/0.8 m

Therefore, try 2-15M @ 0.8 m, giving A s=400 mm2

0.8 m=500 mm2

m> 366 mm2

m

A s>0.0013 beff t=(0.0013 ) (1200 ) (240 )=374.4 m m2

At midheight of the wall,

Pf =0.9(23.9+ 4.132 ( 0.54 (0.24 ) 1.0 (2,100 )+ 3

4(0.46 ) (0.24 ) (0.2 ) (2350 )

1000 )9.81)=27.2 kNm

[In the calculation of self-weight per meter, 2,100 kg/m3 density is used for the 54% solid content of hollow blocks plus ¾ of the 2,350 kg/m3 grout in the 1 cell out of the 4 that is grouted per 0.8m wall length. The grouted cell occupies 46% of the volume of the grouted length of the wall.]

e = M f

Pf=0.505 m≫3 ek

n=ES

Em= 200,000

(17.5)(850 )=13.4

Where Em=850 f m'

n A s (d−kd )=beff kd ( kd2 )

(13.4 ) (500 ) (120−kd )=1200( k d2

2 )600k d2+6700 kd−804000=0 , kd=31.4 mm

I cr=n A s (d−kd )2+ b (kd )3

3=13.4 (500 ) (120−31.4 )2+ (1200 ) (31.4 )3

3=65× 106 mm4 /m

Using the moment magnifier method


Pcr=π2∅ er ( EI )eff

(1+0.5 βd ) (kh )2=

π 2(0.75)(850)(17.5)(65× 106)(1+0.5(1.57)) (4130 )2

×10−3=235 kN /m

Where,

βd=M fd

M f=

(0.9 ) (23.9 )13.7

=1.57

( EI )eff=Em I cr

∅ er=0.75¿Cl10.7 .4 .2.2 Therefore,

M f tot=M fp( Cm

1−Pf

Pcr)=13.7( 1

1−27.2235 )=15.5 kN . m /m

Where Cm=0.6+0.4 (1 )=1

From equilibrium of internal and external axial forces per meter of wall length Pf =Cm−T=∅m 0.85 f m

' X beff β1 c−∅ s A s f y

27,200=(0.6 ) (1 ) (0.85 ) (17.5 ) (1200 ) β1 c−0.85 (500)(400)β1 c=18.4 mm<t f =35.65 mm( f ' m=17.5 MPa assumption is correct)

M r= (0.6 ) (0.85 ) (17.5 ) (1 ) (1200 ) (18.4 )(120−18.42 )×10−6=21.8 kN /m

M r=¿ M f tot ∴OK

∴Use 2−15 M rebar @ 0.8m spacingusing 30 MPa unit, normal weight, 25cm hollow block in running bond with Type S mortar

Bearing Plate for Wall A


R f required=9661 lb=42.9 kN

Br=∅m 0.85 f m' Abp=0.6 (0.85 ) (17.5 ) (100 )2=89.25 kNBr>R f required

−OK∴Use 100 mmBearing Plate

Wall B


R f=(956 lbft

×20 ft )+(10.1 lbft

×20 ft )+(956 lbft

× 10 ft )+(5.6 lbft

×10 ft )=28,9382

lb=14,469lb

Pf =14,469 lb×16 joist=231,504 lb

Pf =231,504 lb

94' 421 } ft} =2427.8 {lb} over {ft} =35.4 {kN} over {m¿¿


kht

=(1 ) ( 4130 )

190=21.7<30




beff=4 ( t )=4 (190 )0.39 m

=1948.72 mmm


2+

1.4 (1 ) (4.13 )2

8=18.9 kN . m /m



C=∅m 0.85 f m' X beff t f =(0.6 ) (0.85 ) (17.5 ) (1 ) (1948.72 ) (35.65× 10−3 )=620 kN

M r=C( t2−

t f

2 )=620(1902

−35.652 )×10−3=47.8 kN . m /m


For steel,


)



0.85(400)(1902

−10 ( estimated ))=653.98 m m2 /m∨255.05 m m2/0.39 m

Therefore, try 2-15M @ 0.39m, giving A s=400 mm2

0.39 m=1025.6 mm2

m> 653.98 mm2

m

A s>0.0013 beff t=(0.0013 ) (1948.72 ) (190 )=481 mm2


Pf =0.9(35.4+4.13

2 ( 0.54 (0.19 ) 1.0 (2,100 )+12

(0.46 ) (0.19 ) (0.2 ) (2350 )

1000 )9.81)=36.2 kNm

[In the calculation of self-weight per meter, 2,100 kg/m3 density is used for the 54% solid content of hollow blocks plus 1/2 of the 2,350 kg/m3 grout in the 1 cell out of the 2 that is grouted per 0.39m wall length. The grouted cell occupies 46% of the volume of the grouted length of the wall.]

e = M f

Pf=0.527 m≫3 ek


n=ES

Em= 200,000

(17.5)(850 )=13.4

Where Em=850 f m'


(13.4 ) (1025.6 ) (95−kd )=1948.72( k d2

2 )974.36k d2+13743 kd−1305589=0 , kd=30.2mm


3=13.4 (1025.6 ) (95−30.2 )2+ (1948.72 ) (30.2 )3

3=75.6× 106 mm4 /m



(1+0.5 βd ) (kh )2=

π 2(0.75)(850)(17.5)(75.6 × 106)(1+0.5(1.6857)) (4130 )2

×10−3=265 kN /m

Where,

βd=M fd

M f=

(0.9 ) (35.4 )13.7

=1.6857

( EI )eff=Em I cr

∅ er=0.75¿Cl10.7 .4 .2.2 Therefore,

M f tot=M fp( Cm

1−Pf

Pcr)=13.7( 1

1−36.2265 )=15.86 kN .m /m

Where Cm=0.6+0.4 (1 )=1



36,200= (0.6 ) (1 ) (0.85 ) (17.5 ) (1948.72 ) β1 c−0.85(1025.6)(400)β1 c=22.1mm<t f=35.65 mm (f ' m=17.5 MPa assumptionis correct )

M r= (0.6 ) (0.85 ) (17.5 ) (1 ) (1948.72 ) (22.1 )(95−22.12 )×10−6=32.3 kN /m

M r>¿ M f tot ∴OK

∴Use 2−15 M rebar @ 0.39 mspacingusing 30 MPa unit, normal weight, 20cm hollow block in running bond with Type S mortar


Bearing Plate for Wall BR f required

=14,469lb=64.4 kNBr=∅m 0.85 f m

' Abp=0.6 (0.85 ) (17.5 ) (100 )2=89.25 kNBr>R f required


Wall C


R f=(956 lbft

×10 ft )+(5.6 lbft

×10 ft)+(956 lbft

×30 ft )+(13.1 lbft

×30 ft )=38,6892

lb=19,344.5 lb

Pf =19,344.5 lb×16 joist=309,512lb


Pf =309,512 lb

89' 441 ft} =3464 {lb} over {ft} =50.5 {kN} over {m¿


kht

= (1 ) ( 4130 )190

=21.7<30



beff=4 ( t )= 4 (190 )0.162m

=4691.4 mmm


2+

1.4 (1 ) ( 4.13 )2

8=25.7 kN . m /m



C=∅m 0.85 f m' X beff t f =(0.6 ) (0.85 ) (17.5 ) (1 ) ( 4691.4 ) (35.65 ×10−3 )=1492 kN

M r=C( t2−

t f

2 )=1492( 1902

−35.652 )×10−3=115kN .m /m


For steel,


)



0.85(400)(1902

−10 ( estimated ))=889.3 m m2/m∨144 mm2/0.162m


0.162 m=1852 mm2

m> 889.3 m m2

m

A s>0.0013 beff t=(0.0013 ) (4691.4 ) (190 )=1159m m2



Pf =0.9(50.5+4.13

2 ( 0.54 (0.19 )1.0 (2,100 )+( 0.46 ) (0.19 ) (0.2 ) (2350 )1000 )9.81)=50.1 kN

m[In the calculation of self-weight per meter, 2,100 kg/m3 density is used for the 54% solid content of hollow blocks plus of the 2,350 kg/m3 grout since all cells are grouted. The grouted cell occupies 46% of the volume of the grouted length of the wall.]

e = M f

Pf=0.527 m ≫3 ek

n=ES

Em= 200,000

(17.5)(850 )=13.4

Where Em=850 f m'


(13.4 ) (1852 ) (95−kd )=4691.4( k d2

2 )2345.7k d2+24817kd−2357596=0 , kd=27 mm


3=13.4 (1852 ) (95−27 )2+ (4691.4 ) (27 )3

3=146 ×106m m4/m



(1+0.5 βd ) (kh )2=

π 2(0.75)(850)(17.5)(146 × 106)(1+0.5(1.768)) (4130 )2

×10−3=500 kN /m

Where,

βd=M fd

M f=

(0.9 ) (50.5 )13.7

=1.768

( EI )eff=Em I cr

∅ er=0.75¿Cl10.7 .4 .2.2 Therefore,

M f tot=M fp( Cm

1−Pf

Pcr)=13.7( 1

1−50.1500 )=15.2 kN . m /m

Where Cm=0.6+0.4 (1 )=1



50,100= (0.6 ) (1 ) (0.85 ) (17.5 ) (4691.4 ) β1 c−0.85(1852)(400)


β1 c=16.2mm< t f=35.65 mm( f ' m=17.5 MPa assumptionis correct)

M r= (0.6 ) (0.85 ) (17.5 ) (1 ) (4691.4 ) (16.2 )(95−16.22 )×10−6=59 kN /m



Bearing Plate for Wall CR f required

=19,344.5=86 kNBr=∅m 0.85 f m



Wall D



R f=(932 lbft

× 26 ft )+(10.3 lbft

×26 ft)=24,499.82

lb=12,249lb

Pf =12,249 lb× 6 joist=73,499.4 lb

Pf =73,499.4 lb

33' 1141 ft} =2165.7 {lb} over {ft} =31.6 {kN} over {m¿


kht

=(1 ) ( 4877 )

190=25.7<30



beff=4 ( t )= 4 (190 )0.162m

=4691.4 mmm


2+ 1.4 (1 ) (4.877 )2

8=18.4 kN .m /m



C=∅m 0.85 f m' X beff t f =(0.6 ) (0.85 ) (17.5 ) (1 ) ( 4691.4 ) (35.65 ×10−3 )=1492 kN

M r=C( t2−

t f

2 )=1492( 1902

−35.652 )×10−3=115kN .m /m



For steel,


)



0.85(400)(1902

−10 ( estimated ))=636.7 m m2/m∨103.15 m m2 /0.162 m


0.162 m=1234.6 m m2

m>636.7 mm2

m

A s>0.0013 beff t=(0.0013 ) (4691.4 ) (190 )=1159m m2


Pf =0.9(31.6+4.877

2 ( 0.54 (0.19 ) 1.0 (2,100 )+ (0.46 ) (0.19 ) (0.2 ) (2350 )1000 )9.81)=33.4 kN


e = M f

Pf=0.527 m≫3 ek

n=ES

Em= 200,000

(17.5)(850 )=13.4

Where Em=850 f m'


(13.4 ) (1234.6 ) (95−kd )=4691.4( k d2

2 )2345.7k d2+16,543.6 kd−1,571,645.8=0 , kd=22.6 mm


3=13.4 (1234.6 ) (95−22.6 )2+ (4691.4 ) (22.6 )3

3=104.8 ×106 mm4/m



(1+0.5 βd ) (kh )2=

π 2(0.75)(850)(17.5)(104.8 × 106)(1+0.5(1.5457)) (4130 )2

× 10−3=274 kN /m

Where,


βd=M fd

M f=

(0.9 ) (31.6 )18.4

=1.5457

( EI )eff=Em I cr

∅ er=0.75¿Cl10.7 .4 .2.2 Therefore,

M f tot=M fp( Cm

1−Pf

Pcr)=18.4( 1

1−33.4274 )=21kN . m/m

Where Cm=0.6+0.4 (1 )=1



33,400= (0.6 ) (1 ) (0.85 ) (17.5 ) (4691.4 ) β1 c−0.85(1234.6)(400)β1 c=10.8 mm<t f=35.65mm (f ' m=17.5 MPa assumptionis correct)

M r= (0.6 ) (0.85 ) (17.5 ) (1 ) (4691.4 ) (10.8 )(95−10.82 )×10−6=40.6 kN /m



Bearing Plate for Wall DR f required

=12,249lb=54 kNBr=∅m 0.85 f m




Wall E


R f=(956 lbft

×30 ft )+(13.1 lbft

×30 ft )+(956 lbft

× 30 ft )+(13.1 lbft

× 30 ft )=58,1462

lb=29,073 lb

Pf =29,073 lb×18 joist=523,314 lb

Pf =523,314 lb

55' 5 ft} =9443 {lb} over {ft} =138 {kN} over {m ¿


kht

=(1 ) ( 4130 )

240=17.2<30



beff=4 (t )= 4 (240 )0.162m

=5926 mmm

Factored moment, M f =0.9 (138 )

2+

1.4 (1 ) ( 4.13 )2

8=65 kN .m /m



C=∅m 0.85 f m' X beff t f =(0.6 ) (0.85 ) (17.5 ) (1 ) (5926 ) (35.65 ×10−3 )=1886 kN


M r=C( t2−

t f

2 )=1886 ( 2402

−35.652 )× 10−3=193 kN . m /m


For steel,


)


A s (approximate )= 65 ×106

0.85(400)(2402

−10 ( estimated ))=1738 m m2/m∨282 m m2 /0.162 m


0.162 m=2469 mm2

m> 1738 mm2

m

A s>0.0013 beff t=(0.0013 ) (5926 ) (240 )=1849 m m2


Pf =0.9(138+4.13

2 ( 0.54 (0.24 ) 1.0 (2,100 )+(0.46 ) (0.24 ) (0.2 ) (2350 )1000 )9.81)=130 kN


e = M f

Pf=0.5 m≫3ek

n=ES

Em= 200,000

(17.5)(850 )=13.4

Where Em=850 f m'


(13.4 ) (2469 ) (120−kd )=5926( k d2

2 )2963k d2+33,084.5 kd−3970152=0 , kd=31.4 mm


3=13.4 (2963 ) (120−31.4 )2+ (4691.4 ) (31.4 )3

3=360× 106 mm4 /m




(1+0.5 βd ) (kh )2=

π 2(0.75)(850)(17.5)(360 × 106)(1+0.5(1.91)) (4130 )2

×10−3=1189kN /m

Where,

βd=M fd

M f=

(0.9 ) (138 )65

=1.91

( EI )eff=Em I cr

∅ er=0.75¿Cl10.7 .4 .2.2 Therefore,

M f tot=M fp( Cm

1−Pf

Pcr)=65( 1

1− 1301189 )=73 kN .m /m

Where Cm=0.6+0.4 (1 )=1



130,000= (0.6 ) (1 ) (0.85 ) (17.5 ) (5926 ) β1 c−0.85 (2469)(400)β1 c=18.3 mm<t f=35.65 mm (f ' m=17.5 MPa assumptionis correct)

M r= (0.6 ) (0.85 ) (17.5 ) (1 ) (5926 ) (18.3 )(120−18.32 )×10−6=107 kN /m



Bearing Plate for Wall ER f required

=29,073 lb=129 kN


Br=∅m 0.85 f m' Abp=0.6 (0.85 ) (17.5 ) (200 )2=357 kNBr>R f required

−OK∴Use 200 mm Bearing Plate

Calculations for flexure wall around apparatus bay

Height of wall = 25.01ft = 7.623mWind Load = 1 kPaUsing 240mm Concrete masonry unit, 20 MPa unit

t f =38 mm

Pf =0.9( 7.6232 )( 2.81 kN

m2 (self weight of concrete ))=9.639 kNm

M f =(1.4 ) (7.6232 ) (1 )

8=10.2 kN . m

Sx, required=M f

∅ f t

10.2× 106

0.6 ×0.4=42.5 ×106 m m3

m

Se=beff d2

6=2462× 2402

6=23.63× 106 m m3

m

Since Sx ,required>Se−Provide reinforcementTrying 15M rebar @ 0.39m spacing , f ' m=13, 20 MPa unit, Type S mortar

A s=200 mm2

0.39 m=512.8m m2/m

Assuming C=T

∅m 0.85 f m' X beff a=∅ s A s f y

beff=4 ts

=4 × 2400.39

=2461.5 mm/0.39 m

a=β1c=7.94 mm

M r=T (d−β1 c2 )= (0.85 ) (512.8 ) ( 400 )( 240

2−7.94

2 )×10−6=20.2 kN .m/0.39 m

∴M r>M f

Check to see if steel yields

cd=

10.680.8120

=0.11125≤ 600600+400

=0.6

Summary: 15M rebar @ 0.39m spacing, f ' m=13 MPa ,20MPa masonry unit, Type S mortar


Shear Wall Calculations

Shear Wall A: Sample Calculation

From environmental loads:W =1.13 kPaV=Awall× W¿ (164 ) (1.13 )¿ 186 kN

The Shear Force, V, is to be taken by 2 shear walls, and shall only resist the force acting on the top half of the wall, so,

V=1864

= 50 kN (rounded up)

Using load factors for masonry walls, and wall loading table:

Pf =(0.9 ) (70028 )=63025And,

Mf =(50000 ) (4.206 ) (1.4 )=294.42× 106

Setting Mr=Mf ,

Mf =[ Pf +(∅ s ) ( ys ) AS

2 ]×[ lw

2−

( β1 ) (c )2 ]+(∅ ¿¿ s)( y¿¿ s)( A ¿¿ s)

2[d−

lw

2]¿¿¿

294.42 ×106=[63025+(0.9 ) (400 ) AS

2 ]×[75882

−(0.8 ) (500 )

2 ]+ (0.9)(400) A s

2[4200−7588

2]

∴ A s=94 mm2

∴Use 210 M bars , A s=200 mm2

Balancing tension and compression forces, we find:


P=C−∑ TP=(∅m ) ( x ) (0.85 f ' m ) ( t ) ( β1) (c )−(∅ s)( A s)( f y )

63025=(0.6 ) (1 ) (0.85 ) (7.5 ) (190 ) (0.8 ) c−(0.9)(100)(400)c=170.3 mm

Mr=(∅ ¿¿m)( x ) (0.85 f 'm ) ( t ) ( β1 ) (c ) ¿¿

∴Mr=685.4 × 106

∴Mr ≫ Mf

Checking for shear:

Vf =1.4 (50 )=70 kN

Vr=∅m [ (vm ) ( bw ) (dv )+0.25 ( Pd ) ] γ g+[( 0.6∅ s ) ( Av ) ( f y )( dv

s )]70000=0.85 [( 0.209 ) (190 ) (0.8 ×7588 )+0.25 (63025 ) ]1.0+[ (0.6×0.85 ) (400 )( 0.8 × 7588

1200 )] Av

A v=¿0∴No horizontal reinforcement isrequired

Shear Wall B: Sample Calculation

V=1864



Pf =(0.9 ) (349000 )=314100And,

Mf =(50000 ) (4.206 ) (1.4 )=294.42× 106

Setting Mr=Mf ,

Mf =[ Pf +(∅ s ) ( ys ) AS

2 ]×[ lw

2−

( β1 ) (c )2 ]+(∅ ¿¿ s)( y¿¿ s)( A ¿¿ s)

2[d−

lw

2]¿¿¿

294.42 ×106=[63025+(0.9 ) (400 ) AS

2 ]×[91442

−(0.8 ) (500 )

2 ]+(0.9)(400)A s

2[4200−9144

2]


∴ A s=(−ve )number∴Still use2 10M bars , A s=200 mm2


P=C−∑ TP=(∅m ) ( x ) (0.85 f ' m ) ( t ) ( β1) (c )−(∅ s)( A s)( f y )

314100= (0.6 ) (1 ) (0.85 ) (7.5 ) (190 ) (0.8 ) c−(0.9)(100)(400)c=602mm

Mr=(∅ ¿¿m)( x ) (0.85 f 'm ) ( t ) ( β1 ) (c ) ¿¿

∴Mr=1677.4 ×106

∴Mr ≫ Mf

Checking for shear:

Vf =1.4 (50 )=70 kN


s )]70000=0.85 [( 0.209 ) (190 ) (0.8 × 9144 )+0.25 (63025 ) ] 1.0+[ (0.6 × 0.85 ) (400 )( 0.8 ×9144

1200 )] A v


Shear Wall C & D: Sample Calculation

V=1704



Pf =0And,

Mf =(50000 ) (4.206 ) (1.4 )=294.42× 106

Setting Mr=Mf ,


Mf =[ Pf +(∅ s ) ( ys ) AS

2 ]×[ lw

2−

( β1 ) (c )2 ]+(∅ ¿¿ s)( y¿¿ s)( A ¿¿ s)

2[d−

lw

2]¿¿¿

294.42 ×106=[0+(0.9 ) (400 ) AS

2 ]×[ 37002

−(0.8 ) (200 )

2 ]+(0.9)(400) As

2[4200−3700

2]

∴ A s=318.7 mm2



P=C−∑ TP=(∅m ) ( x ) (0.85 f ' m ) (t ) ( β1) (c )−(∅ s)( A s)( f y )

63025=(0.6 ) (1 ) (0.85 ) (7.5 ) (190 ) (0.8 ) c−(0.9)(200)(400)c=124.0 mm

Mr=(∅ ¿¿m)( x ) (0.85 f 'm ) ( t ) ( β1 ) (c ) ¿¿

∴Mr=255.8 ×106

∴Mr ≫ Mf

Checking for shear:

Vf =1.4 (50 )=70 k N


s )]70000=0.85 [( 0.209 ) (190 ) (0.8 ×3700 )+0.25 (63025 ) ]1.0+[ (0.6 ×0.85 ) (400 )( 0.8 × 3700

1200 )]Av


Shear Wall E: Sample Calculation

V=80 kN


Pf =84684 × 0.9=76216 NAnd,

Mf =(80000 ) (4.941 ) (1.4 )=533.4 ×106 kN . m

Setting Mr=Mf ,


Mf =[ Pf +(∅ s ) ( ys ) AS

2 ]×[ lw

2−

( β1 ) (c )2 ]+(∅ ¿¿ s)( y¿¿ s)( A ¿¿ s)

2[d−

lw

2]¿¿¿

294.42 ×106=[0+(0.9 ) (400 ) AS

2 ]×[ 91442

−(0.8 ) (200 )

2 ]+ (0.9)(400) A s

2[4200−9144

2]

∴ A s=298 mm2



P=C−∑ TP=(∅m ) ( x ) (0.85 f ' m ) (t ) ( β1) (c )−(∅ s)( A s)( f y )

76216=(0.6 ) (1 ) (0.85 ) (7.5 ) (190 ) (0.8 ) c−(0.9)(200)(400)c=255.0mm

Mr=(∅ ¿¿m)( x ) (0.85 f 'm ) ( t ) ( β1 ) (c ) ¿¿

∴Mr=988.5 ×106

∴Mr ≫ Mf

Checking for shear:

Vf =1.4 ( 80 )=112kN


s )]70000=0.85 [( 0.209 ) (190 ) (0.8 × 9144 )+0.25 (76216 ) ] 1.0+[ (0.6 × 0.85 ) (400 )( 0.8 ×9144

1200 )] A v


LINTEL BEAM DESIGN


In figure X it shows the building layout for the low rise of the structure. It is

found that there are 8 lintels needed based on the joist layout and Load

bearing walls.

The following part shows the method used for calculating the masonry

beams used as lintels fir the highlighted sections

Lintel #1 & Lintel #2 were shown how to calculate for different #no. of

courses for the masonry beam

Lintel #1: it lies between joist J1 and I2 I1 956 lb/ft ; self weight 5.6 lb/ft

∴Pl=956∗6+5.6∗6=5.76 kips

Reaction R 1=R 2=5.762

=2.88 kips

J1 956 lb/ft ; self-weight 6.7 lb/ft

∴Pl=956∗18+6.7∗18=17.32 kips


=8.66 kips

Therefore total load = R2+R3= 11.54 kips = 51.33 kN


Figure X: highlighted sections for lintels

R2= 2.88 R1=2.88

5.76

5.76

R4= 8.66 R4= 8.66

77

It is shown that M f =33.048 kN .m@ x=0.53 mM f =45.947 kN .m @ x=1.7 m

Masonry Beam design:(Assuming 3 course Beam) using 200 mm blocks h= 200*3-10= 590 d= h-100=490

( CD )max=0.6

Cbalanced=0.6∗490=294 mma=β∗c=294∗0.8=235.2mmFm=∅ m∗X∗(0.85∗f ' m )b∗a=0.6∗0.5∗0.85∗10∗190∗235.2=113.95 kNFm=F s

123000=0.85∗Asbal∗400∴ Asbal=335 mmUse 2−15 M=2∗200=400 mm2


Figure X: Shear diagram for lintel #1 Figure X: moment diagram for Lintel #1

78

F s=0.85∗400∗400=136 kN136000=0.6∗0.5∗0.85∗10∗190∗a∴a=280.7 mm

M r=136 kN (d−a2 )¿136∗(490−280.7

2 )=47.55 kN . m>M f =45.947 kN . m

Lintel #2: it lies between joist J2 and I2:

I2 451 lb/ft ; self-weight 5.6 lb/ft∴Pl=451∗6+5.6∗6=2.7396 kips


=1.369 kips

J1 956 lb/ft ; self-weight 6.7 lb/ft

∴Pl=451∗18+6.7∗18=8.23 kips


=4.12 kips

Therefore total load = R2+R3= 5.489 kips = 24.41 kN


R4= 24.41

R= 17.907 R= 6.5032

Figure X: Shear diagram for lintel #2 Figure X: moment diagram for Lintel #2

79

It is shown that M f =8.44 kN .m @ x=1.298 m

Masonry Beam design:(Assuming 2 course Beam) using 200 mm blocks h= 200*2-10= 390 d= h-100=290

( CD )max=0.6

Cbalanced=0.6∗290=174 mma=β∗c=174∗0.8=139.2 mmFm=∅ m∗X∗(0.85∗f ' m )b∗a=0.6∗0.5∗0.85∗10∗190∗139.2=67.4 kNFm=F s

67442.4 kN=0.85∗Asbal∗400

∴ Asbal=198 mm2

Use 2−10 M=2∗100=200mm2

F s=0.85∗200∗400=68kN68000=0.6∗0.5∗0.85∗10∗190∗a∴a=140.35 mm

M r=136 kN (d−a2 )¿136∗(290−140.35

2 )=14.94 kN .m>M f =8.44 kN .m

Square Spread Foundation DesignThe following figure shows the location for columns that are to be supported by Square Spread foundations.Columns 1-6 are shown how to calculate for reinforcement, development length and shear

Column C3 & C4:Width = 600mmSemax ≤25 mmPf =156 kN+1.79 kN=157.79 kNqall=100 kpa

A ≥( P f

qall)=157.79

100=1.57 m2

b=√1.57=1.25 m>1mA=b2=1.52=2.25 m2>1.57 m2


q f =P f

A=157.79

2.25=70 kN

m2

Check for Two Way shear:Vf =P f=157.79 kN1)vc=0 .38∗ℷ∗∅ c∗√ f ' c=0.38∗1∗0.65∗5=1.24 MPaSet Vc=VfVc=vc∗bo∗d

bo∗d=Vcvc

=157.791240

=0.13 m 2

bo=4∗( t+d )using t=0.25 mbo∗d=4 (t +d )∗d

0.13=4 ( 0.25∗d+d 2 )d=0.12m

Therefore , bo=4 ( t +d )=4∗(0.37 )=1.48 m2¿vc=(1+2/Bc )∗0.19∗λ∗φ∗√ f ' c¿ (1+2 )∗0.19∗1∗0.65∗5¿1.85 MPa3¿vc=(αs∗d /bo+0.19)∗λ∗φ∗√ f ' c

¿( 2∗0.121.48

+0.19)∗1∗0.65∗5

¿1.4 MPaTherefore , vc=1.24 MPa governsCover=75 mmd b=16 mm (15−M rebar )

h=d+d b+cover=120 mm+(162 )+75=203 mm=250 mm

d=250−(162 )−75=167 mm=150 mm

Check for one-way shearVf =qf∗b∗( b−t

2−d )

¿70∗1.5∗(1.5−0.252

−0.15)=50 kN

0.9*d = 0.9*150= 135 mmdv =

0.72*h= 0.72*250 = 180 mmTherefore , d v=180 mm

β= 2301000

+d v= 2301000

+180=0.19

bw=b=1500 mmVc=λ∗φ∗β∗√ f ' c∗bw∗dv¿1∗0.65∗0.19∗5∗1500∗180=166.7 kN


Since, Vc=166.7 ≥157.79 kN***Shear reinforcement not requiredCheck for flexural reinforcement:Mf =qf ∗(b− t

2 )∗(b− t4 )∗b

¿70∗(1.5−0.252 )∗(1.5−0.25

4 )∗1.5=20.5 kN . m

Mr ≥ MfSet Mr=Mf=20.5 kN .m

A s=0.0015∗f ’ c∗b∗(d−√d 2−3.85∗Mrf ' c∗b )

¿0.0015∗25∗1500∗¿¿404 mm2Ag=h∗b=250∗1500=375000 mm2Asmin=0.002∗Ag=750 mm2Since, As ≤ As min use As minuse 4−15 M rebarAs=4∗200=800 mm2SpacingS=1500−2∗100

3=433 mm=430 mm

Smax=500Since S ≤ Smax

l d=0.45 k 1∗k 2∗k 3∗k 4∗fs√ f ' c

∗db

¿ 0.45∗1∗1∗1∗0.8∗3505

∗16=403mm

L= b-t/2= 1500-250/2 -100 = 525 mm

Column C1 & C2:Width = 600mmSemax ≤25 mmPf =165 kN+(20.7∗8.89)kN=167 kNqall=100 kpa

A ≥( P f

qall)=175

100=1.67m2

b=√1.67=1.35 m>1mA=b2=1.52=2.25 m2>1.67 m2


q f =P f

A= 167

2.25=75 kN

m2

Check for Two Way shear:Vf =P f=167 kN1)vc=0 .38∗ℷ∗∅ c∗√ f ' c=0.38∗1∗0.65∗5=1.24 MPaSet Vc=VfVc=vc∗bo∗d

bo∗d=Vcvc

= 1671240

=0.13 m 2

bo=4∗( t+d )using t=0.25 mbo∗d=4 (t +d )∗d

0.13=4 ( 0.25∗d+d 2 )d=0.116m

Therefore , bo=4 ( t +d )=4∗(0.37 )=1.464 m2¿vc=(1+2/Bc )∗0.19∗λ∗φ∗√ f ' c¿ (1+2 )∗0.19∗1∗0.65∗5¿1.85 MPa3¿vc=(αs∗d /bo+0.19)∗λ∗φ∗√ f ' c

¿( 2∗0.1161.464

+0.19)∗1∗0.65∗5

¿1.13 MPaTherefore , vc=1.13 MPa governsCover=75 mmd b=16 mm (15−M rebar )

h=d+d b+cover=120 mm+(162 )+75=203 mm=250 mm

d=250−(162 )−75=167 mm=150 mm

Check for one-way shearVf =qf∗b∗( b−t

2−d )

¿75∗1.5∗(1.5−0.252

−0.15)=53.4kN

0.9*d = 0.9*150= 135 mmdv =

0.72*h= 0.72*250 = 180 mmTherefore , d v=180 mm

β= 2301000

+d v= 2301000

+180=0.19

bw=b=1500 mmVc=λ∗φ∗β∗√ f ' c∗bw∗dv¿1∗0.65∗0.19∗5∗1500∗180=166.7=167 kN


Since, Vc=167=Vf =167 kN***Shear reinforcement not requiredCheck for flexural reinforcement:Mf =qf ∗(b− t

2 )∗(b− t4 )∗b

¿75∗(1.5−0.252 )∗(1.5−0.25

4 )∗1.5=22 kN .m

Mr ≥ MfSet Mr=Mf=22 kN . m

A s=0.0015∗f ’ c∗b∗(d−√d 2−3.85∗Mrf ' c∗b )

¿0.0015∗25∗1500∗¿¿434.8 mm2Ag=h∗b=250∗1500=375000 mm2Asmin=0.002∗Ag=750 mm2Since, As ≤ As min use As minuse 4−15 M rebarAs=4∗200=800 mm2SpacingS=1500−2∗100

3=433 mm=430 mm

Smax=500Since S ≤ Smax

l d=0.45 k 1∗k 2∗k 3∗k 4∗fs√ f ' c

∗db

¿ 0.45∗1∗1∗1∗0.8∗3505

∗16=403mm

L= b-t/2= 1500-250/2 -100 = 525 mm > 403 mm

Strip Foundation Design

For the masonry assembly of the building, a continuous strip foundation is selected to take the linear loading pattern of the structural walls, and to easily resist shifting, sliding or overturning for the shear walls. In all practicality, there is no viable alternative to this considering the economic benefits, and the continuous rigid system offered by a continuous strip shallow footing.In accordance with OBC, NBCC, and the Concrete Handbook, using the largest room with tributary width of 9.2m (30ft), and the loading from the structural load bearing walls, the following parameters are obtained:

PD=51 kNm

qall=100 kPa


PL= (4.8 kPa ) × (9.2 m )= 45 kNm

Pf =1.5 PD+1.25PL

¿132.75 kNm

Ps=PD+PL

¿96.0 kNm

To determine the length of the foundation, the serviceable loading is divided by the allowable soil pressure to determine the necessary length of foundation to distribute the soil pressure within the allowable range:

l ≥Ps

b ×qall, l=0.96 m

Also, since the wall is 200mm thick,

t=200 mm

The applied load and the shear reinforcement required for the foundation can be determined by first calculating the area, and taking the divisor of the factored load over the area. Using the shear reinforcement equation for shallow continuous strip footing foundation as well Table A23.3 in the Concrete Design Handbook, we arrive at the following for a 1m section of the foundation using based on the calculated effective depth:

A=1.82m2

q f =P f

A=138.3kPa

…Try using15 M bars

d=h−cover−db

2=165mm

V f =q f × b×( l−t2

−d )=95 kNm

dv={ 0.9d0.72 h

=180 mm

From table A23.3 of Concrete Design Handbook,

β=0.21 Since h < 350


V c=(∅ c ) ( β ) (√ f 'c ) (bw ) ( dv )=128.8 kN

m

V f <V c

∴No shear reinforcemnet required

Now, the factored moment of the foundation must be calculated, and set equal to the resisting moment to determine the area of steel required to resist this moment. Also, it must be ensured that the minimum steel used for design purposes is greater than the gross minimum steel calculation stated in the handbook.

M f =(q f ) ( l−t2 )( l−t

4 ) (b )=15.0 kNm

M r=M f ,

A s=(0.0015)( f 'c)(b)(d)−√d2−

(3.85 ) ( M f )( f '

c) (b )=588.9 mm2

Ag=h × b=250 ×103mm2

A smin=0.002 Ag=500 mm2

∴ A s=588.9 mm2 satisfies

The spacing is simply determined by taking the area of 1 bar, and multiplying it by the ratio of 103 by the total area of minimum steel:

s≤ Ab× 1000A s

=≈ 330 mm , smax=500 mm

∴Use 15 M bars at 330 mm spacing


The final step is determine the required longitudinal continuous reinforcement. This is done by a gross area calculation as follows:

Ag=h ×l=240 ×103

ASmin=0.002 Ag=480mm2

∴Use 315 M bars , A s=3× 200 mm2=600 mm2

starting at 100 mm fro m theedges∧spaced 380 mm

For the 10” course block (t = 250), the design will remain the same, because a greater thickness of wall decreases the factored shear and moment forces. The values are as follows:

V f =(139 ) (1 )( 0.96−0.252

−0.165)=90 kNm

90<93(origional V f )

M f =(139 )(0.96−0.252 )( 0.96−0.25

4 ) (1 )=13.9 kN .m

9.2<10.0(origional M f )

∴SinceV f ∧M f are smaller thanthe origionalvalues ,the above foundation is valid

for 10 course blocks



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