A chair for Emily Carr and her dog Koko

(wo)MAN’S BEST FRIENDA CHAIR FOR EMILY CARR AND HER DOG KOKO

David Donnelly - 20411299Natalie Krakovsky - 20427441

4MANIFESTO/ CLIENT RESEARCH/ DESIGN PARAMETERS

8PRECEDENT IMAGES

12DESIGN DEVELOPMENTDRAWINGS

18FINAL DRAWINGS/ WORKING DRAWINGS

22PERSPECTIVE DRAWINGS/ ASSEMBLY PROCESS

NESTING DIAGRAM/ COMPONENT DRAWINGS

PHOTOGRAPHED/ DOCUMENTED FINAL CONSTRUCTION

CONSTRUCTION PROCESS/ PROTOTYPING

MATERIAL LIST/ COMPONENT MANIFESTO

FINAL CALCULATIONS

REFLECTION

CONTENTS

48

28NESTING DIAGRAM/ COMPONENT DRAWINGS

32PHOTOGRAPHED/ DOCUMENTED FINAL CONSTRUCTION

36CONSTRUCTION PROCESS/ PROTOTYPING

44MATERIAL LIST/ COMPONENT MANIFESTO

46FINAL CALCULATIONS

60REFLECTION

MANIFESTO/CLIENT RESEARCH/DESIGN PARAMETERS

4

Emily is a great Canadian artist who loved her pets, particularly her dogs. Emily Carr painted Canadian landscapes in a modern and poetic way. Our chair design will aim to capture Emily Carr’s essence, not only through her incredible artistic talent, but through her love for her dogs, particularly Billy. We want to design and build a chair that Emily Carr would be able to read, write and paint along side her most loyal companion, her dog Billy. This would be especially beneficial since she worked predominately in oil and paper giving her freedom to work outside. The chair will be built to be set within the great Canadian wilderness at a cottage or an artists post. Our design will look to incorporate natural materials to create a relaxing escape to think deeply for both the Emily and Billy.

MANIFESTO/

5

CLIENT RESEARCH/

6

DESIGN PARAMETERS

1 We designed a patio chair. Therefore, comfort was a major concernfor both the owner and his/her dog. Using fabric in the palces where the dog and the person sits was the solution.

Since the chair is design not only for a person, but for a dog aswell, there were a few restrictions in how the chair could be designed.

2 The dog portion needed to be elevated off the ground. As a dog ages and health problems arise, cement and other cold/hard surfaces can irritate the dog. Therefore,we elevated the plateform slightly off the ground.

3 The portion for a dog needed to have a slight bounce to it to increase comfrot further. Therefore, the form work of the wood structure bends in a way that fores the fabric to becomealmost bubble like.

4 We wanted to limit the amount of wood members that penetrate the dogs sitting area because a dog needs space to site upright, lie down and lift his/her head. Therefore, the form of the chair involved a solid member that transferred the weight of the seat to either end of the chair leaving the space for the dog free and open. For further information on this visit the final calculation section.

5 We wanted to limit the amount of wood members that penetrate the dogs sitting area because a dog needs space to site upright, lie down and lift his/her head. Therefore, the form of the chair involved a solid member that transferred the weight of the seat to either end of the chair leaving the space for the dog free and open. For further information on this visit the final calculation section.

6 Sun need to penetrate trough too the dogs seat, but there still should be the ability to have the area shaded at certain times of the day. Using Upholstery instead of wood was a solutions to this.

7 Emily Carr loved literature and art. Therefore, we wanted to create a portion that could be used to rest a book or a sketch pad enabeling her to draw/write/read while in her chair.

8 There are many structural restrictions withing the chair that were solved by making stronger solid sections that could withstand tension.

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PRECEDENT IMAGES8

PRECEDENT IMAGES

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1 2

9

5

4

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These precedents inspired the simple form we designed; a seat on top for a person to sit and an area underneath that can be a dog bed.

1 https://www.pinterest.com/J0hnMoore/dog-houses/2 http://www.remals.com/attractive-pet-house-design-ideas/3 http://www.citylab.com/design/2012/10/behold-dog-house-sofa/3640/4 http://www.citylab.com/design/2012/04/rocking-chair-your-cat-or-dog/1843/5 http://www.designboom.com/design/cat-shelters-architects-for-animals-fixnation-los-angeles- 09-16-2014/?utm_campaign=daily&utm_medium=e-mail&utm_source=subscribers6 http://www.dezeen.com/2014/06/03/paul-loebach-launches-peg-chair-during-nyc-design-week/

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6 6

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DESIGNDEVELOPMENT DRAWINGS

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PHASE 1Early in the design process we envisioned our form of seating to be in the style of a bench. Large enough to seat two people and have a pull out compartment for the dog. The compartment would give the dog the ability to lay out in the sun beside his owner or underneath the bench in the shade, with the language of a dog house.

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PHASE 2The second phase shared a similar principle to the first, but was a simpler design. The idea was still for a bench, but part of the bench transformed (Figure1) to allow a stronger connection between the dog and the owner.

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1 Two person bench

2 Remove seat

FIGURE 1

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3 Slide piece into bench

4 A seat for you and your dog

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PHASE 3For our final phase we took inspiration and ideas from the first iterations, but reimagined how the chair would be used. We felt the bench was too bulky so we narrowed it down to simply a chair for man and his best friend. It became less “solid” to allow for light to penetrate beneath the chair yet it still allows for shade during the hotter portions of the day. This eliminated the need to have removable parts. A writing desk was also added to include Emily Carr’s love for literature and art.

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FINAL DRAWINGS/WORKING DRAWINGS

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FINAL DRAWINGS

1 FRONT ELEVATION

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2 SIDE ELEVATION

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3 SECTION

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116232

41843

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PERSPECTIVEDRAWINGS/ASSEMBLY PROCESS

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PERSPECTIVEDRAWINGS/

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The chair is not solid wood we designed it so that it could be upholstered. Cre-ating a more comfortable chair for relaxing. These are axonometric drawings in its bare form.

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ASSEMBLY PROCESS

The chair was designed so that we could use a CNC machine. Each piece of the chair is made of 11 mm plywood. The three pieces are glued together and then left to dry using clamps. The male joint was made slightly larger then the female joint so we would be able to shave it down using the router and the band saw, after the glue has dried. This gives us control over how tight the joint is. The pieces then are hammered into one another with a rubber mallet. The joints used throughout the chair are mortise and tenon.

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NESTING DIAGRAM/COMPONENT DRAWINGS

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NESTING DIAGRAM/1 SEAT PIECE

2 BASE PIECE

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COMPONENTDRAWINGS

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PIECE Hx 4

PIECE Ix 2

PIECE Lx 1

PIECE Mx 2

PIECE Nx 1

PIECE Ox 2

PIECE Jx 4

PIECE Kx 2

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PHOTOGRAPHED/DOCUMENTED FINAL CONSTRUCTION

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CONSTRUCTIONPROCESS/PROTOTYPING

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CONSTRUCTIONPROCESS/

1 DESIGN CHAIR / PROTOTYPE / CREATE AUTOCAD FILES FOR CNC MACHINE

2 CNC PIECES FOR CHAIR

3 SAND ROUGH EDGES / GLUE AND CLAMP SETS OF THREE TOGETHER

4 ROUTER THE EDGES TO ACHIEVE A TIGHT MORTISE AND TENON CONNECTION

5 HAMMER THE JOINTS TOGETHER

6 APPLY UPHOLSTERY WITH STAPLE GUN TO THE FRAME OF THE CHAIR

7 COMPLETED CHAIR

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PROTOTYPINGPHASE 1- 1:5

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PHASE 2- 1:5

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PHASE 3- 1:5

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PHASE 4- 1:2

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MATERIAL LIST/COMPONENTMANIFESTO

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MATERIAL LIST1 PLYWOOD

TYPE: 1/2 inches 4x8 Sanded Fir PlywoodAMOUNT: 4 sheets (2 were used)PRICE: $115

64% of final chair cost

2 CARPENTER’S GLUETYPE: LePageAMOUNT: 1 bottlePRICE: $16


3 UPHOLSTERYTYPE: 2 inches seat belt webbingAMOUNT: 28 yardsPRICE: $32


STORE PURCHASED: Home Depot

STORE PURCHASED: Home Depot

STORE PURCHASED: Lens Mill Store

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FINAL ANALYSISCALCULATIONS

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LEGENDSTANDARD JOINTWELDED PINSTRUCTURAL ELEMENTSNOT CONSIDERED ELEMENTSPINROLLERLOADSREACTIONS

A

B

C

ED

G H

F

V1

H1

0.5 kN

1 kN

V2

A

B

C

ED

G H

F

V1

H1

0.5 kN

1 kN

V2

1 kN

COMMON LOADS/REACTIONS

47

CRITICAL SECTIONThe critical section is around joint A. This joint carries the highest moment load and is crucial for the structure of the chair. We determined this point to be the critical section over joint E or joint H because it is here that both the vertical and lateral forces act together in the highest magnitude, creating more stress on this joint than any other. V1 is multiplied by 480mm (horizontal distance away form joint) and H1 is multiplied by 208mm (vertical distance away from joint). We also considered this to be the critical section because of our design in the construction of the chair. By building the main “beam” member of the chair, (member ACDE) from a solid piece of wood, we reduced the stress and moment experienced at points C and D considerable, however, this stress is transferred more to joint A than E.

We analyzed a number of joints to be sure joint D was the critical section. This becomes the critical section over joint C because the applied horizontal force is in line with joint C and therefore is negligible in the moment calculation; in addition the applied vertical force is only multiplied by 276mm (distance away from joint) in comparison to the 480mm in the case of joint D. Joint H experiences a smaller moment than joint D because the larger vertical force is located closer horizontally to the joint and therefore creates a smaller moment.

Analyzing the chair from the east elevation we see a number of areas that experience stress though none that could be considered the critical section because this chair section is shorter and built with fewer joints. However an interesting area we wanted to analyze was the vertical force experienced on joint A. The force from a person leaning back could cause uplift, and stress at this joint. We analyzed this section with a 1 KN force acting vertically and a 0.5 KN force acting horizontally applied to the middle section of the chairs back. What we found was there was an uplift force but only of 0.083 KN which told us that it was something to be considered in the design but was very manageable.

48

NARRATIVEThe structure chosen is completely reflective of the design intent. That is, a chair for a person, a chair for a dog, and a flat surface for writing/art. The structure bends and forms in the most direct way to accommodate these purposes. A place for a dog to rest just inches from the ground which allows for easy access of the dog and a comfortable place above the cold ground which causes great discomfort for older dogs with arthritis. The seat itself is reclined five degrees and uses a tensile fabric to hold the user while they wright, draw or read from the surface positioned beside the chair. The chair is intended to be used outdoors, on a patio, a place Emily Carr was fond of. The fabric easily dries when wet, provides a comfortable seat for both users and provides ample shading for the dog. Wood was chosen for its cost, workability, aesthetics and connection to nature, which was the focus of much of Emily Carr’s work. The structure is composed of three basic parts, which are the two sides, which transfer the vertical loads to the ground, the lateral bracing pieces, which, connect the two sides, and finally the seat, which carries the person and rests on the lateral members. The structural design started with many straight pieces with complex joinery and eventually developed into edited and simplified pieces to render the most efficient use of material, and the strongest structure possible. The most distinctive and unconventional part of the structure is the lateral bracing which forms two right angles, allowing space below for a dog sitting tall, and space above as a writing surface. Multiple iterations of this member held together with joinery resulted in unsecured and failed structures. This needed to resist the largest forces acting on the chair and the most structural and simplest of solutions was to build this piece solid.The joints used throughout the chair are mortise and tenon. These joints were used because of their strength and because they are easily hidden. The construction of the chair involves 3 pieces of 11mm plywood glued together with the central piece containing the male joints. These pieces glued together give the chair its cross sectional strength.

49

*Dog excluded from calculation

1. Sum of the horizontals

ΣFx = 0H1 - 0.5 kN = 0H1 = 0.5 kN

2. Moment around B

ΣMB = 0(0.5 kN)(0.432 m) + (V1)(0.61 m) - (1 kN)(0.409 m) = 00.216 kNm + V1(0.61 m) = 0.409 kNmV1(0.61 m) = 0.193 kNmV1 = 0.32 kN

3. Sum of the verticles

ΣFy = 0V1 + V2 = 0V2 = 1 kN - V1V2 = 1 kN - 0.32 kNV2 = 0.68 kN

V1 = 0.32 kNV2 = 0.68 kNH1 = 0.5 kN

Therefore the base reaction calculations are 0.32 kN for V1, 0.68 kN for V2 and 0.5 kN for H1.

BASE REACTION


A

B

C

ED

G H

F

V1 = 0.32 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.68 kN

A

B

C

ED

G H

F

V1 = 0.36 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.73 kN

0.045 kN 0.045 kN

50

BASE REACTIONS*Dog included in calculation

1. Sum of the horizontals

ΣFx = 0H1 - 0.5 kN = 0H1 = 0.5 kN

2. Moment around B

ΣMB = 0(0.5 kN)(0.432 m) + (V1)(0.61 m) - (1 kN)(0.409 m) - (0.045 kN)(0.2 m) - (0.045 kN)(0.41m) = 00.216 kNm + V1(0.61 m) - (0.409 m) - (0.009 m) - (0.018 m) = 0V1(0.61 m) = 0.22 kNm V1 = 0.36 kNm

V1 = 0.36 kN


ΣFy = 0V1 + V2 - 1kN - 0. 045 - 0. 045 = 0V1 + V2 - 1.09 kN = 0V2 = 1.09 kN - V1V2 = 1.09 kN - 0.36 kNV2 = 0.73 kN

V1 = 0.36 kNV2 = 0.73 kNH1 = 0.5 kN

Therefore the base reaction calculations are 0.36 kN for V1, 0.73 kN for V2 and 0.5 kN for H1. The dog is so light it is almost negligible.


A

B

C

ED

G H

F

V1 = 0.32 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.68 kN

A

B

C

ED

G H

F

V1 = 0.36 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.73 kN

0.045 kN 0.045 kN

51

FREE BODY DIAGRAMLEGENDSTANDARD JOINTWELDED PINSTRUCTURAL ELEMENTSNOT CONSIDERED ELEMENTSPINROLLERLOADSREACTIONS

A

B

C

ED

G H

F

V1 = 0.32 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.68 kN

A

B

C

ED

G H

F

V1 = 0.36 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.73 kN

0.045 kN 0.045 kN

MEMBER AB*Dog excluded in calculation


ΣFy = 0VA - V2 = 0VA = V2VA = 0.68 kN

2. Moment around A

ΣMA = 0(0.5 kN)(0.615 m) - (HG)(0.515 m) = 00.3075 kNm - (HG)(0.515 m) = 00.3075 kNm = (HG)(0.515 m) HG = 0.6 kN

3. Sum of the horizontalsΣFx = 0HG - HA - 0.5 kN = 0HA = HG - 0.5 kNHA = 0.6 kN - 0.5 kNHA = 0.1 kN

VA = 0.68 kNHG = 0.6 kNHA = 0.1 kN

Therefore the reactions are 0.68 kN for VA, 0.6 kN for HG and 0.1 kN for HA.


A

B

C

ED

G H

F

V1 = 0.32 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.68 kN

A

B

C

ED

G H

F

V1 = 0.36 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.73 kN

0.045 kN 0.045 kN

A

B

G

H1 = 0.5 kN

V2 = 0.68 kN

HG = 0.6 kN

HA = 0.1 kN

VA = 0.73 kN

AC

HA = 0.1 kN

VA = 0.68 kN

HA = 0.1 kN

VA = 0.68 kN

HC = 0.1 kNVC = 0.68 kN

C

D

HC = 0.1 kNVC = 0.68 kN

VC = 0.68 kN

VD = 0.68 kN

HC = 0.1 kN

HD = 0.1 kN

D

VD = 0.68 kN

HD = 0.1 kN

HD = 0.1 kNVD = 0.68 kN

1 kN

HE = 0.1 kN

VE = 0.32 kN

HE = 0.1 kN

VE = 0.32 kN

VE = 0.32 kN

HE = 0.1 kN

HH = 0.6 kN

E

E

V1 = 0.32 kN

0.5 kN

H

F

HH = 0.6 kN HGHG = 0.6 kN

HH = 0.6 kN


A

B

C

ED

G H

F

V1 = 0.32 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.68 kN

A

B

C

ED

G H

F

V1 = 0.36 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.73 kN

0.045 kN 0.045 kN

A

B

G

H1 = 0.5 kN

V2 = 0.68 kN

HG = 0.6 kN

HA = 0.1 kN

VA = 0.73 kN

AC

HA = 0.1 kN

VA = 0.68 kN

HA = 0.1 kN

VA = 0.68 kN

HC = 0.1 kNVC = 0.68 kN

C

D

HC = 0.1 kNVC = 0.68 kN

VC = 0.68 kN

VD = 0.68 kN

HC = 0.1 kN

HD = 0.1 kN

D

VD = 0.68 kN

HD = 0.1 kN

HD = 0.1 kNVD = 0.68 kN

1 kN

HE = 0.1 kN

VE = 0.32 kN

HE = 0.1 kN

VE = 0.32 kN

VE = 0.32 kN

HE = 0.1 kN

HH = 0.6 kN

E

E

V1 = 0.32 kN

0.5 kN

H

F


HH = 0.6 kN

52

*Dog excluded in calculation

VA = 0.68 kNHA = 0.1 kN

1. Sum of the verticlesΣFy = 0VA - VC = 0VC = VA VC = 0.68 kN

2. Sum of the horizontalsΣFx = 0HA - HC = 0HC = HA HC = 0.1 kN

VC = 0.68 kNHC = 0.1 kN

Therefore the reactions are 0.68 kN for VC and 0.1 kN for HC.

MEMBER AC


A

B

C

ED

G H

F

V1 = 0.32 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.68 kN

A

B

C

ED

G H

F

V1 = 0.36 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.73 kN

0.045 kN 0.045 kN

A

B

G

H1 = 0.5 kN

V2 = 0.68 kN

HG = 0.6 kN

HA = 0.1 kN

VA = 0.73 kN

AC

HA = 0.1 kN

VA = 0.68 kN

HA = 0.1 kN

VA = 0.68 kN

HC = 0.1 kNVC = 0.68 kN

C

D

HC = 0.1 kNVC = 0.68 kN

VC = 0.68 kN

VD = 0.68 kN

HC = 0.1 kN

HD = 0.1 kN

D

VD = 0.68 kN

HD = 0.1 kN

HD = 0.1 kNVD = 0.68 kN

1 kN

HE = 0.1 kN

VE = 0.32 kN

HE = 0.1 kN

VE = 0.32 kN

VE = 0.32 kN

HE = 0.1 kN

HH = 0.6 kN

E

E

V1 = 0.32 kN

0.5 kN

H

F


HH = 0.6 kN


A

B

C

ED

G H

F

V1 = 0.32 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.68 kN

A

B

C

ED

G H

F

V1 = 0.36 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.73 kN

0.045 kN 0.045 kN

A

B

G

H1 = 0.5 kN

V2 = 0.68 kN

HG = 0.6 kN

HA = 0.1 kN

VA = 0.73 kN

AC

HA = 0.1 kN

VA = 0.68 kN

HA = 0.1 kN

VA = 0.68 kN

HC = 0.1 kNVC = 0.68 kN

C

D

HC = 0.1 kNVC = 0.68 kN

VC = 0.68 kN

VD = 0.68 kN

HC = 0.1 kN

HD = 0.1 kN

D

VD = 0.68 kN

HD = 0.1 kN

HD = 0.1 kNVD = 0.68 kN

1 kN

HE = 0.1 kN

VE = 0.32 kN

HE = 0.1 kN

VE = 0.32 kN

VE = 0.32 kN

HE = 0.1 kN

HH = 0.6 kN

E

E

V1 = 0.32 kN

0.5 kN

H

F


HH = 0.6 kN

53

MEMBER CD*Dog excluded in calculation

VC = 0.68 kNHC = 0.1 kN

1. Sum of the verticlesΣFy = 0VC - VD = 0VD = VC VD = 0.68 kN

2. Sum of the horizontalsΣFx = 0HC - HD = 0HD = HC HD = 0.1 kN

VD = 0.68 kNHD = 0.1 kN

Therefore the reactions are 0.68 kN for VD and 0.1 kN for HD.


A

B

C

ED

G H

F

V1 = 0.32 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.68 kN

A

B

C

ED

G H

F

V1 = 0.36 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.73 kN

0.045 kN 0.045 kN

A

B

G

H1 = 0.5 kN

V2 = 0.68 kN

HG = 0.6 kN

HA = 0.1 kN

VA = 0.73 kN

AC

HA = 0.1 kN

VA = 0.68 kN

HA = 0.1 kN

VA = 0.68 kN

HC = 0.1 kNVC = 0.68 kN

C

D

HC = 0.1 kNVC = 0.68 kN

VC = 0.68 kN

VD = 0.68 kN

HC = 0.1 kN

HD = 0.1 kN

D

VD = 0.68 kN

HD = 0.1 kN

HD = 0.1 kNVD = 0.68 kN

1 kN

HE = 0.1 kN

VE = 0.32 kN

HE = 0.1 kN

VE = 0.32 kN

VE = 0.32 kN

HE = 0.1 kN

HH = 0.6 kN

E

E

V1 = 0.32 kN

0.5 kN

H

F


HH = 0.6 kN


A

B

C

ED

G H

F

V1 = 0.32 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.68 kN

A

B

C

ED

G H

F

V1 = 0.36 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.73 kN

0.045 kN 0.045 kN

A

B

G

H1 = 0.5 kN

V2 = 0.68 kN

HG = 0.6 kN

HA = 0.1 kN

VA = 0.73 kN

AC

HA = 0.1 kN

VA = 0.68 kN

HA = 0.1 kN

VA = 0.68 kN

HC = 0.1 kNVC = 0.68 kN

C

D

HC = 0.1 kNVC = 0.68 kN

VC = 0.68 kN

VD = 0.68 kN

HC = 0.1 kN

HD = 0.1 kN

D

VD = 0.68 kN

HD = 0.1 kN

HD = 0.1 kNVD = 0.68 kN

1 kN

HE = 0.1 kN

VE = 0.32 kN

HE = 0.1 kN

VE = 0.32 kN

VE = 0.32 kN

HE = 0.1 kN

HH = 0.6 kN

E

E

V1 = 0.32 kN

0.5 kN

H

F


HH = 0.6 kN

54

MEMBER DE*Dog excluded in calculation

VD = 0.68 kNHD = 0.1 kN

1. Sum of the verticlesΣFy = 01 kN - VD + VE = 0VE = 1 kN - VD VE = 0.32 kN

2. Sum of the horizontalsΣFx = 0HD - HE = 0HD = HE HE = 0.1 kN

VE = 0.32 kNHE = 0.1 kN

Therefore the reactions are 0.32 kN for VE and 0.1 kN for HE.


A

B

C

ED

G H

F

V1 = 0.32 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.68 kN

A

B

C

ED

G H

F

V1 = 0.36 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.73 kN

0.045 kN 0.045 kN

A

B

G

H1 = 0.5 kN

V2 = 0.68 kN

HG = 0.6 kN

HA = 0.1 kN

VA = 0.73 kN

AC

HA = 0.1 kN

VA = 0.68 kN

HA = 0.1 kN

VA = 0.68 kN

HC = 0.1 kNVC = 0.68 kN

C

D

HC = 0.1 kNVC = 0.68 kN

VC = 0.68 kN

VD = 0.68 kN

HC = 0.1 kN

HD = 0.1 kN

D

VD = 0.68 kN

HD = 0.1 kN

HD = 0.1 kNVD = 0.68 kN

1 kN

HE = 0.1 kN

VE = 0.32 kN

HE = 0.1 kN

VE = 0.32 kN

VE = 0.32 kN

HE = 0.1 kN

HH = 0.6 kN

E

E

V1 = 0.32 kN

0.5 kN

H

F


HH = 0.6 kN


A

B

C

ED

G H

F

V1 = 0.32 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.68 kN

A

B

C

ED

G H

F

V1 = 0.36 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.73 kN

0.045 kN 0.045 kN

A

B

G

H1 = 0.5 kN

V2 = 0.68 kN

HG = 0.6 kN

HA = 0.1 kN

VA = 0.73 kN

AC

HA = 0.1 kN

VA = 0.68 kN

HA = 0.1 kN

VA = 0.68 kN

HC = 0.1 kNVC = 0.68 kN

C

D

HC = 0.1 kNVC = 0.68 kN

VC = 0.68 kN

VD = 0.68 kN

HC = 0.1 kN

HD = 0.1 kN

D

VD = 0.68 kN

HD = 0.1 kN

HD = 0.1 kNVD = 0.68 kN

1 kN

HE = 0.1 kN

VE = 0.32 kN

HE = 0.1 kN

VE = 0.32 kN

VE = 0.32 kN

HE = 0.1 kN

HH = 0.6 kN

E

E

V1 = 0.32 kN

0.5 kN

H

F


HH = 0.6 kN


A

B

C

ED

G H

F

V1 = 0.32 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.68 kN

A

B

C

ED

G H

F

V1 = 0.36 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.73 kN

0.045 kN 0.045 kN

A

B

G

H1 = 0.5 kN

V2 = 0.68 kN

HG = 0.6 kN

HA = 0.1 kN

VA = 0.73 kN

AC

HA = 0.1 kN

VA = 0.68 kN

HA = 0.1 kN

VA = 0.68 kN

HC = 0.1 kNVC = 0.68 kN

C

D

HC = 0.1 kNVC = 0.68 kN

VC = 0.68 kN

VD = 0.68 kN

HC = 0.1 kN

HD = 0.1 kN

D

VD = 0.68 kN

HD = 0.1 kN

HD = 0.1 kNVD = 0.68 kN

1 kN

HE = 0.1 kN

VE = 0.32 kN

HE = 0.1 kN

VE = 0.32 kN

VE = 0.32 kN

HE = 0.1 kN

HH = 0.6 kN

E

E

V1 = 0.32 kN

0.5 kN

H

F


HH = 0.6 kN

55

MEMBER EF*Dog excluded in calculation

VE = 0.32 kNHE = 0.1 kN

1. Sum of the verticlesΣFy = 0VE + V1 = 0V1 = VE V1 = 0.32 kN

2. Sum of the horizontalsΣFx = 0HH - HE - 0.5 = 0HH = 0.5 + HE HH = 0.6 kN

V1 = 0.32 kNHH = 0.6 kN

Therefore the reactions are 0.32 kN for V1 and 0.6 kN for HH. Therefore the system is in equilibrium.


A

B

C

ED

G H

F

V1 = 0.32 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.68 kN

A

B

C

ED

G H

F

V1 = 0.36 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.73 kN

0.045 kN 0.045 kN

A

B

G

H1 = 0.5 kN

V2 = 0.68 kN

HG = 0.6 kN

HA = 0.1 kN

VA = 0.73 kN

AC

HA = 0.1 kN

VA = 0.68 kN

HA = 0.1 kN

VA = 0.68 kN

HC = 0.1 kNVC = 0.68 kN

C

D

HC = 0.1 kNVC = 0.68 kN

VC = 0.68 kN

VD = 0.68 kN

HC = 0.1 kN

HD = 0.1 kN

D

VD = 0.68 kN

HD = 0.1 kN

HD = 0.1 kNVD = 0.68 kN

1 kN

HE = 0.1 kN

VE = 0.32 kN

HE = 0.1 kN

VE = 0.32 kN

VE = 0.32 kN

HE = 0.1 kN

HH = 0.6 kN

E

E

V1 = 0.32 kN

0.5 kN

H

F


HH = 0.6 kN


A

B

C

ED

G H

F

V1 = 0.32 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.68 kN

A

B

C

ED

G H

F

V1 = 0.36 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.73 kN

0.045 kN 0.045 kN

A

B

G

H1 = 0.5 kN

V2 = 0.68 kN

HG = 0.6 kN

HA = 0.1 kN

VA = 0.73 kN

AC

HA = 0.1 kN

VA = 0.68 kN

HA = 0.1 kN

VA = 0.68 kN

HC = 0.1 kNVC = 0.68 kN

C

D

HC = 0.1 kNVC = 0.68 kN

VC = 0.68 kN

VD = 0.68 kN

HC = 0.1 kN

HD = 0.1 kN

D

VD = 0.68 kN

HD = 0.1 kN

HD = 0.1 kNVD = 0.68 kN

1 kN

HE = 0.1 kN

VE = 0.32 kN

HE = 0.1 kN

VE = 0.32 kN

VE = 0.32 kN

HE = 0.1 kN

HH = 0.6 kN

E

E

V1 = 0.32 kN

0.5 kN

H

F


HH = 0.6 kN

56


HH = 0.1 kN

1. Sum of the horizontalsΣFx = 0HG - HH = 0HG = HH HG = 0.6 kN

Therefore the base reaction calculations are 0.6 kN for HG. Therefore the chair is in equilibrium.

MEMBER HG


A

B

C

ED

G H

F

V1 = 0.32 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.68 kN

A

B

C

ED

G H

F

V1 = 0.36 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.73 kN

0.045 kN 0.045 kN

A

B

G

H1 = 0.5 kN

V2 = 0.68 kN

HG = 0.6 kN

HA = 0.1 kN

VA = 0.73 kN

AC

HA = 0.1 kN

VA = 0.68 kN

HA = 0.1 kN

VA = 0.68 kN

HC = 0.1 kNVC = 0.68 kN

C

D

HC = 0.1 kNVC = 0.68 kN

VC = 0.68 kN

VD = 0.68 kN

HC = 0.1 kN

HD = 0.1 kN

D

VD = 0.68 kN

HD = 0.1 kN

HD = 0.1 kNVD = 0.68 kN

1 kN

HE = 0.1 kN

VE = 0.32 kN

HE = 0.1 kN

VE = 0.32 kN

VE = 0.32 kN

HE = 0.1 kN

HH = 0.6 kN

E

E

V1 = 0.32 kN

0.5 kN

H

F


HH = 0.6 kN


A

B

C

ED

G H

F

V1 = 0.32 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.68 kN

A

B

C

ED

G H

F

V1 = 0.36 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.73 kN

0.045 kN 0.045 kN

A

B

G

H1 = 0.5 kN

V2 = 0.68 kN

HG = 0.6 kN

HA = 0.1 kN

VA = 0.73 kN

AC

HA = 0.1 kN

VA = 0.68 kN

HA = 0.1 kN

VA = 0.68 kN

HC = 0.1 kNVC = 0.68 kN

C

D

HC = 0.1 kNVC = 0.68 kN

VC = 0.68 kN

VD = 0.68 kN

HC = 0.1 kN

HD = 0.1 kN

D

VD = 0.68 kN

HD = 0.1 kN

HD = 0.1 kNVD = 0.68 kN

1 kN

HE = 0.1 kN

VE = 0.32 kN

HE = 0.1 kN

VE = 0.32 kN

VE = 0.32 kN

HE = 0.1 kN

HH = 0.6 kN

E

E

V1 = 0.32 kN

0.5 kN

H

F


HH = 0.6 kN

57

SHEAR AND BENDINGMOMENT DIAGRAMS


A

B

C

ED

G H

F

V1 = 0.32 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.68 kN

A

B

C

ED

G H

F

V1 = 0.36 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.73 kN

0.045 kN 0.045 kN

A

B

G

H1 = 0.5 kN

V2 = 0.68 kN

HG = 0.6 kN

HA = 0.1 kN

VA = 0.73 kN

AC

HA = 0.1 kN

VA = 0.68 kN

HA = 0.1 kN

VA = 0.68 kN

HC = 0.1 kNVC = 0.68 kN

C

D

HC = 0.1 kNVC = 0.68 kN

VC = 0.68 kN

VD = 0.68 kN

HC = 0.1 kN

HD = 0.1 kN

D

VD = 0.68 kN

HD = 0.1 kN

HD = 0.1 kNVD = 0.68 kN

1 kN

HE = 0.1 kN

VE = 0.32 kN

HE = 0.1 kN

VE = 0.32 kN

VE = 0.32 kN

HE = 0.1 kN

HH = 0.6 kN

E

E

V1 = 0.32 kN

0.5 kN

H

F


HH = 0.6 kN

HA = 0.1 kN

E

AV A = 0.68 kN HE = 0.1 kN

1 kN

0.5 kN

VE = 0.32 kN

0.68 kN 0.68 kN

0.32 kN 0.32 kN

0.0698 kN

0

0


A

B

C

ED

G H

F

V1 = 0.32 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.68 kN

A

B

C

ED

G H

F

V1 = 0.36 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.73 kN

0.045 kN 0.045 kN

A

B

G

H1 = 0.5 kN

V2 = 0.68 kN

HG = 0.6 kN

HA = 0.1 kN

VA = 0.73 kN

AC

HA = 0.1 kN

VA = 0.68 kN

HA = 0.1 kN

VA = 0.68 kN

HC = 0.1 kNVC = 0.68 kN

C

D

HC = 0.1 kNVC = 0.68 kN

VC = 0.68 kN

VD = 0.68 kN

HC = 0.1 kN

HD = 0.1 kN

D

VD = 0.68 kN

HD = 0.1 kN

HD = 0.1 kNVD = 0.68 kN

1 kN

HE = 0.1 kN

VE = 0.32 kN

HE = 0.1 kN

VE = 0.32 kN

VE = 0.32 kN

HE = 0.1 kN

HH = 0.6 kN

E

E

V1 = 0.32 kN

0.5 kN

H

F


HH = 0.6 kN

HA = 0.1 kN

E

AV A = 0.68 kN HE = 0.1 kN

1 kN

0.5 kN

VE = 0.32 kN


A

B

C

ED

G H

F

V1 = 0.32 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.68 kN

A

B

C

ED

G H

F

V1 = 0.36 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.73 kN

0.045 kN 0.045 kN

A

B

G

H1 = 0.5 kN

V2 = 0.68 kN

HG = 0.6 kN

HA = 0.1 kN

VA = 0.73 kN

AC

HA = 0.1 kN

VA = 0.68 kN

HA = 0.1 kN

VA = 0.68 kN

HC = 0.1 kNVC = 0.68 kN

C

D

HC = 0.1 kNVC = 0.68 kN

VC = 0.68 kN

VD = 0.68 kN

HC = 0.1 kN

HD = 0.1 kN

D

VD = 0.68 kN

HD = 0.1 kN

HD = 0.1 kNVD = 0.68 kN

1 kN

HE = 0.1 kN

VE = 0.32 kN

HE = 0.1 kN

VE = 0.32 kN

VE = 0.32 kN

HE = 0.1 kN

HH = 0.6 kN

E

E

V1 = 0.32 kN

0.5 kN

H

F


HH = 0.6 kN

HA = 0.1 kN

E

AV A = 0.68 kN HE = 0.1 kN

1 kN

0.5 kN

VE = 0.32 kN

0.68 kN 0.68 kN

0.32 kN 0.32 kN

0.0698 kN

0

0

SHEAR FORCE DIAGRAM

BENDING MOMENT DIAGRAM

58

MAXIMUM STRESS


Section of Interest

1. Moment around A

ΣMA = 0(0.1 kN + 0.5 kN)(0.15 m) - (0.32 kN)(0.61 m) + (1 kN)(0.175) - MA = 00.09 kNm - 0.1952 kNm + 0.175 - MA = 0MA = 0.265 kNm - 0.1952 kNmMA = 0.0698 kNm

Therefore the moment is 0.0698 kNm for MA.

Maximum Bending Stress Calculation at Critical Section

M = 0.0698/2 kNmM = 0.0349 kNm

The element is 36 mm x 36 mm

b = 36 mmd = 36 mm

S = bd^2/6S = (36)(36^2)/6S = 7776 mm^3

σ = M/Sσ = 34900 Nmm / (7776mm^3)σ = 4.5 MPa< σallow = 5 MPaThis section is adequate!


A

B

C

ED

G H

F

V1 = 0.32 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.68 kN

A

B

C

ED

G H

F

V1 = 0.36 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.73 kN

0.045 kN 0.045 kN

A

B

G

H1 = 0.5 kN

V2 = 0.68 kN

HG = 0.6 kN

HA = 0.1 kN

VA = 0.73 kN

AC

HA = 0.1 kN

VA = 0.68 kN

HA = 0.1 kN

VA = 0.68 kN

HC = 0.1 kNVC = 0.68 kN

C

D

HC = 0.1 kNVC = 0.68 kN

VC = 0.68 kN

VD = 0.68 kN

HC = 0.1 kN

HD = 0.1 kN

D

VD = 0.68 kN

HD = 0.1 kN

HD = 0.1 kNVD = 0.68 kN

1 kN

HE = 0.1 kN

VE = 0.32 kN

HE = 0.1 kN

VE = 0.32 kN

VE = 0.32 kN

HE = 0.1 kN

HH = 0.6 kN

E

E

V1 = 0.32 kN

0.5 kN

H

F


HH = 0.6 kN


A

B

C

ED

G H

F

V1 = 0.32 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.68 kN

A

B

C

ED

G H

F

V1 = 0.36 kN

H1 = 0.5 kN

0.5 kN

1 kN

V2 = 0.73 kN

0.045 kN 0.045 kN

A

B

G

H1 = 0.5 kN

V2 = 0.68 kN

HG = 0.6 kN

HA = 0.1 kN

VA = 0.73 kN

AC

HA = 0.1 kN

VA = 0.68 kN

HA = 0.1 kN

VA = 0.68 kN

HC = 0.1 kNVC = 0.68 kN

C

D

HC = 0.1 kNVC = 0.68 kN

VC = 0.68 kN

VD = 0.68 kN

HC = 0.1 kN

HD = 0.1 kN

D

VD = 0.68 kN

HD = 0.1 kN

HD = 0.1 kNVD = 0.68 kN

1 kN

HE = 0.1 kN

VE = 0.32 kN

HE = 0.1 kN

VE = 0.32 kN

VE = 0.32 kN

HE = 0.1 kN

HH = 0.6 kN

E

E

V1 = 0.32 kN

0.5 kN

H

F


HH = 0.6 kN

HA = 0.1 kN

E

AV A = 0.68 kN HE = 0.1 kN

1 kN

0.5 kN

VE = 0.32 kN

*Note that we have written M in the units of Nmm in this equation, rather than kNm, so we had to multiply it by 1,000,000. It is critical that all terms in an equation use the same system of units.

59

REFLECTION60

61

Our chair developed from countless sketches, digital models, two 1:5 scale models and one 1:2 scale model. This process of sketching, digital modeling and physical modeling allowed us to anticipate design problems and solve many issues before final construction. This process was extremely valuable to the final product however the final product was the only full scale, fully constructed iteration and so having completed it we have found a number of improvements and solvable design problems we faced with the final product.

Iteration 1: Iteration 1 was a 1:2 cardboard model of early design sketches. It was deemed unsatisfactory to our needs and proved a need for further design development.

Iteration 2: the second iteration was a 1:5 scale bass wood, laser cut model. We anticipated mortensen tenon joints would be the best solution for our chair and so designed this scale model with these joints. We found the joints worked well, though our major structural piece, the lateral member that supports the chair, which also happened to be our critical section, was not able to made from moment joints. It was far too weak and an unnecessary risk to keep this member jointed. We also found we could reduce the number of joints over all and increase stability. We would solve this in further iterations.

Iteration 3: The third iteration was a 1:5 scale bass wood, laser cut model. The third iteration solved the critical section by creating the member from one solid piece. And having the chair rest on it. In this iteration we also increased the thickness of the members and altered the design of the base. We would find a happy medium between member thicknesses and alter the base further in future iterations. We also decreased the number of jointed by flattening the chair section.

REFLECTION

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Iteration 4: The fourth iteration was a 1:2 cardboard laser cut model. This Iteration combined all design solutions and was prepared as if it was the final product with all current design issues solved. Only minor differences appear in the final product.

Final product: The final product was made from CNC cut plywood sections glued together to achieve the desired cross section. Mortensen Tenon joints which were meticulously shaved down to achieve the perfect fit, needing to be hammered into place, neglecting the use of glue. Fabric was purchased and stapled to the chair for the upholstery.

Future iterations: having finally completed the finished product we discussed further developments and potential iterations. In future iterations we discussed a larger angle for the back of the chair, probably around 85 degrees. We also discussed alterations and additions to the writing table, making it more versatile and more aesthetic.

In final, the finished product was a success. A structural, comfortable chair for both Emily Car and her many dogs to enjoy. We are very proud of the final product.

A chair for Emily Carr and her dog Koko

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Transcript of A chair for Emily Carr and her dog Koko