A Chair for M.C. Escher by Philippe Fournier

A CHAIR FOR M.C. ESCHER by Wayne Yan + Philippe Fournier

TABLE OF CONTENTS

CHAPTER 1: CHAPTER 2: CHAPTER 3: CHAPTER 4: CHAPTER 5:

ESCHER DESIGN BUILD CHAIR ANALYSIS CREDITS

4 12 26 54 66 112

TABLE OF CONTENTS

CHAPTER 1: ESCHER

CLIENT

Maurits Cornelis Escher (1898 - 1972) was a Dutch graphic artist and printmaker famous for drawing impossible shapes, optical illusions, tessellated patterns and other mathematically-inspired artwork. Over his lifetime he produced hundreds of lithographs and woodcut prints as well as thousands of drawings and sketches. Our chair for MC Escher, like his artwork, challenges the user's perceptions of what is "impossible" by repeating a simple geometric form, then manipulating it to create an illusion. The motifs of defying gravity and laws of perspective -- prevalent in Escher's work -- manifests in a chair that seems to magically suspend itself above the user's head. Another recurring motif in Escher's art is the dissolution of geometries into fractals and tessellations. Likewise, our client's chair begins as a solid monolith which deconstructs into three cubes as the user interacts with it, and which can potentially be completely separated from one another if desired.

CLIENT

MANIFESTO

What is impossible? Is it that which we do not see? Is it that which we do not feel? Or is it simply that which we do not yet understand? If so, maybe anything is possible...

MANIFESTO

ESCHERâ&#x20AC;&#x2122;S CONCEPT

1. GEOMETRY

2. REPEAT

3. METAMORPHOSIS

CHAIR CONCEPT

1. GEOMETRY

2. REPEAT

3. METAMORPHOSIS

CHAPTER 2: DESIGN

PRECEDENTS

The first step of every design project is researching precedents. In the spirit of our client’s work, we were interested in furniture that had some aspect of “impossibility”, whether by creating optical illusions , appearing unstable, or otherwise tricking the user’s senses of proportion and physics. We had two major conceptual directions we were exploring, each with very different implications for the final design; restrained (left) or radical (right). The “restrained” approach would be to create a small geometric chair that would be aesthetically referential to Escher’s work, either with simple optical tricks or subtle dissonant components. The “radical” approach would be to create a large, dynamic chair with significant moving components seemingly defying physics. We ultimately opted for the latter approach with the final chair. We were also inspired by furniture designs that involved geometric forms dissolving into smaller subcomponents. We researched tower-like shelving with rotating structural joints, especially hidden ones.

← RESTRAINED RADICAL → 14

PRECEDENTS

PLAYING WITH 3-DIMENSIONS

Starting with the â&#x20AC;&#x153;restrainedâ&#x20AC;? design approach, we began modeling ideas out of paper. The initial thrust of these design iterations was to create whole 3D geometric forms out of smaller 2D shapes. Playing with triangular cuts offered the potential to create a chair that would have different, conflicting appearances from different perspectives, and could be taken apart in a variety of configurations.

PLAYING WITH MIRRORS Building on these forms, we explored the possibility of using mirrored surfaces to create optical illusions. We digitally modeled our paper shapes in rhino, then rendered mirrored surfaces to create illusions of disappearing corners, floating surfaces and infinite patterns. Through experimentation, we realized that some of these effects would be impossible to replicate in real life, when the observerâ&#x20AC;&#x2122;s own reflections would obstruct the illusions.

SKETCHES After experimenting with and modeling cubes in 3D, we decided to go for a more radical overall design approach, using the cube as a base geometry which would be replicated and stacked in impossible ways. Working with sketch models of foam, we cut identical cubes and began stacking them precariously on each other, using toothpicks and dowels to skewer them around a rotational axis. We tried many configurations until settling on the one we found most compositionally and ergonomically satisfying. In the first iterations of our â&#x20AC;&#x153;boxâ&#x20AC;? design, the chair was composed of four boxes, joined together by a fixed internal vertical spine. The fourth box would be suspended directly above the user while fully opened for seating. This would have required giving the bottom box extra ballast, while also adding significant weight, height and material cost to the whole project. We also decided that, compositionally, three boxes were more proportionally harmonious to the body, and eventually removed the fourth box. We explored the possibility of turning the whole chair horizontally, to be used as communal seating. However, this would have required ensuring each box had extra structural strength -- and thus weight -- to hold a person on its own, creating additional ballasting issues when in vertical position. It would have also required the boxes be in fixed position relative to each other, without the ability to rotate them, to prevent boxes from swiveling around each other when positioned horizontally. 18

SKETCHES

PROTOTYPING

← FOAM SKETCH MODELS FIRST 1:2 CARDBOARD SKETCH MODEL → 20

PROTOTYPING

Having decided on the chairâ&#x20AC;&#x2122;s overall form and function, the next phase of prototyping was to figure out the mechanism to give the chair rotation and structural stability. We designed and laser-cut a 1:1 cardboard mock-up of a typical rotator joint connection. A single structural pole will span through the interior of all boxes at a corner, and braced at regular intervals. Where the faces between two boxes meet, the cuts around the pole are designed so that when rotated, dowels pierced through the pole itself will lock at 90 degrees.

PROTOTYPING

The next phase of prototyping was to construct our first 1:1 mock-up of the entire chair. In order to do so, we first had to resolve how to hold the faces together. We decided upon glued puzzle joints around the perimeters and tested laser-cutting teeth at different scales and patterns. We re-cut enough rotation joints and pole bracing for the entirety of the chair and used cardboard tubes as the pole itself. For the sake of the midterm presentation, we left some faces off to reveal the interior joints in action and also to understand how we would assemble the final chair in such a way that we could install the interior joints before sealing a box. Constructing this prototype at 1:1 allowed us to infer the critical areas of structural weakness and sag that would occur in the final chair. In particular, it prompted us to make the final executive decision to remove the fourth box and scale the remaining three boxes up to 550mm x 550mm each. We realized we could secure box joints in plywood with fewer teeth than we had modeled. We also decided that in order to properly ballast the chair, we would build the upper boxes with thinner plywood than the bottom box. 24

PROTOTYPING

CHAPTER 3: BUILD

BUILD

What’s in the box? The seemingly precarious cantilevers are achieved by a continuous 2” (OD) aluminum pole, which is attached to the inner face of the boxes through a series of plywood ‘sleeves’. There are three in each box, located at the top, middle, and bottom. This combination resists the overturning moment in the upper boxes, due to their self-weight. Furthermore, although the three boxes appear to be identical, the bottom box is actually composed of 1/2” plywood sheets while the upper two boxes are 1/4” plywood. This distinction reduces the overall weight as well as reducing the strain from the cantilever and the upper volume’s self weight. This difference helps to lower the chair’s center of gravity, preventing it from tipping over when the boxes are rotated.

BUILD

ROTATION

Using 1/4” machine screws, fastened through the pole, the sleeves limit the rotation of the boxes around the pole. The lower sleeve (A), doesn’t rotate. The middle sleeve (B), is limited to 90° clockwise rotation while the upper sleeve (C) can only move 90° counterclockwise. The sleeves are milled from 1/2” Baltic birch plywood, and laminated in pairs. The pole itself isn’t bolted to any of boxes - it simply sits in the sleeves which allows the entire assembly to come apart easily for storage.

By concealing the mechanism of rotation - the chair aspires to generate a sense of mystery that is often found in Escher’s drawings. A

POLE SECTION 30

1:5

BUILD

ROTATION

SLEEVE A

1:1

BOX 1 - FIXED 32

1:10

ROTATION

SLEEVE A DETAIL 33

ROTATION

SLEEVE B

1:1

ROTATE -90 34

1:10

ROTATION

SLEEVE B DETAIL 35

ROTATION

SLEEVE C

1:1

ROTATE 90 36

1:10

ROTATION

SLEEVE C DETAIL 37

ROTATION

JOINTS

1 6

The 1/2â&#x20AC;? panels in the bottom box have routed lap joints on the edges, 6mm deep. Since the edge condition of each panel is unique, we engraved a number on the inside face of each face to identify the panels during assembly. This numbering system, starting from the bottom panel, is consistent through the three boxes.

JOINTS 12

6 6

JOINT TYPE C

JOINT TYPE B 6

6 6

BOTTOM BOX JOINT DIAGRAM

JOINT TYPE A 41

JOINTS

The upper boxes have a puzzle joint -since routing out a lap joint on the 1/4â&#x20AC;? plywood would compromise itâ&#x20AC;&#x2122;s stability. The edges were created through laser cutting, with the identifying numbers engraved on the inside.

JOINTS

UPPER BOXES CUT FILES 43

FINISHING

Each face of the panels were sanded to 320 grit before we applied a coat of Varathane. The exterior faces received two coats while the interior faces received one. Continuity of the stain + varnish helped to prevent warping of the panels due to differential moisture. At the same time, it was important to protect the edges of the panels where glue would be applied from being varnished.

FINISHING

CLAMP + GLUE

The boxes were assembled by gluing the edges and then clamping the entire box. Prior to gluing the faces together, the sleeves were glued to the interior of the boxes. Since the pieces were already finished, it was crucial to protect the exterior faces during the assembly.

CLAMP + GLUE

Since the faces had to be clamped together before glue dried - we had to work quickly to create equal pressure all sides. It took approximately an hour for each to box - to dry-fit the assembly, apply glue, and then add the clamps. We then left the clamps on overnight before removing them.

CLAMP + GLUE

BOXES

ASSEMBLY

COMPONENTS UNITS

COST PER UNIT ($)

TOTAL COST ($) TOTAL WEIGHT (kg)

BALTIC BIRCH (12mm) 2438mm X 1219mm

54.75

15.236

BALTIC BIRCH (6mm)

1524mm X 1219mm

19.98

59.95

15.594

ALUMINUM POLE (50.8mm) OD

3.175mm X 1219mm

30.00

6.1

M4 1/4” X 2” MACHINE SCREWS

5.50

11.00

SANDING DISCS

150 / 220 / 320 GRIT

28.00

VARATHANE STAIN

946ml

22.47

44.94

PAINT THINNER

8.35

CNC TIME

33.90

9.04

36.16

307.05

36.93

ITEM

UNIT SIZE

LASER CUTTING TIME 1 HOUR

COMPONENTS

CHAPTER 4: CHAIR

OPENING THE CHAIR

MAXIMUM OPEN POSITION

ENDLESS CONFIGURATIONS

CHAPTER 5: ANALYSIS

CENTER OF GRAVITY

0.45m

ANALYSIS 1 : CENTRE OF GRAVITY

BOX 2

CLOCKWISE, FROM RIGHT: PLAN VIEW, FRONT ELEVATION, RIGHT ELEVATION

0.3875m

0.55m

LABEL

BOX 3

BOX 1

0.6125m

0.55m

0.45m 1m

0.3875m 0.55m

0.45m 1m

0.6882m

0.55m

0.55m 0

0.55m

0.6882m

1.65m

0.55m 0.55m

1.65m

0.55m

0.6125m 0.45m

0.55m 1m

ANALYSIS 1: CENTRE OF GRAVITY

CALCULATION ASSUMPTIONS:

CENTROIDS

1. Weight of pole & shear bracing at internal rotating joints is negligible; not factored into equations.

CBOX1

= (X1, Y1, Z1) = (0.725m, 0.275m, 0.275m)

CBOX2

= (X2, Y2, Z2) = (0.725m, 0.725m, 0.825m)

CBOX3

= (X3, Y3, Z3) = (0.275m, 0.275m, 1.375m)

2. All boxes made from Baltic birch plywood, with the corresponding weight properties: 6mm thick 12mm thick

= 4.296kg/m2 = 8.394kg/m2

CENTRE OF GRAVITY

3. Box 2 & Box 3 have the same dimensions and plywood thickness (6mm), thus the same volume and weight. Box 1 has a 12mm plywood thickness on all sides.

= V1X1 + V2X2 + V3X3 V1 + V2 + V3 = (0.02178m3 x 0.725m) + (0.01089m3 x 0.725m) + (0.01089m3 x 0.275m) (0.02178m3 + 0.01089m3 + 0.01089m3) = 0.6125m

4. Calculations do not account for overlap or cutting at lap joint and box joint edges; the same surface area is assumed for each face. (0.55m x 0.55m = 0.3025m2)

= V1Y1 + V2Y2 + V3Y3 V1 + V2 + V3 = (0.02178m3 x 0.275m) + (0.01089m3 x 0.725m) + (0.01089m3 x 0.275m) (0.02178m3 + 0.01089m3 + 0.01089m3) = 0.3875m

VOLUME CALCULATIONS:

= V1Z1 + V2Z2 + V3Z3 V1 + V2 + V3 = (0.02178m3 x 0.275m) + (0.01089m3 x 0.825m) + (0.01089m3 x 1.375m) (0.02178m3 + 0.01089m3 + 0.01089m3) = 0.6882m

VBOX1

= 6 (0.55m x 0.55m x 0.012m) = 0.02178m3

VBOX2

= 6 (0.55m x 0.55m x 0.006m) = 0.01089m3 (0.6125m, 0.3875m, 0.6882m)

REACTIONS

ANALYSIS 2: REACTIONS - POSITION 1

FBX = 302.35N

FRONT

RIGHT BOX 3

AL BR L2 L1

R2 R1

BOX 2

BOX 1 FBX L2 L1

62.5

487.5

R1 FRONT VIEW 74

AL 50

450

ANALYSIS 2: REACTIONS

REACTIONS DUE TO SELF-WEIGHT

POSITION 1 REACTIONS

Having identified the combined selfweight of the boxes, we can use the total load at its centroid to find the reactions at the bottom of the chair. To simplify the calculations, we will treat the bottom panel, where the chair touches the ground, as a series of perpendicular beams. Using this method, we will analyze the reactions at the base of the chair in 3 different positions.

∑MAL = FBX(12.5mm) + BR (450mm) = 0 -BR (450mm) = 302.35N(12.5mm) -BR = 3 779.375Nmm / 450mm BR = 8.3986N

For all the calculations, we will assume that the chair sits on four supports located at 50mm inset from the bottom of Box 1.

∑FY = AL + BR + FBX = 0 AL + 8.3986 = 302.35N BR = 293.9514N

F3 = 6 x (0.55m x 0.55m)(4.296kg/m2) F3 = 7.797kg (9.81m/s2) F3 = 76.465N

∑MR1 = BR(337.5mm) + R2 (450mm) = 0 -R2 (450mm) = 8.399N(337.5mm) -R2 = 2 834.66Nmm / 450mm R2 = 6.299N ∑FY = R1 + R2 + BR= 0 R1 + 6.299N = 8.399N R1 = 2.1N

FBX = 76.465N + 76.465N + 149.456N FBX = 302.386N

BR = 8.399N

450

62.5

487.5

0.0278FBX = BR = 8.399N

AL : FBX = 293.95N / 302.35N AL : FBX = 0.972 RIGHT SIDE REACTIONS

F1 = 6 x (0.55m x 0.55m)(8.394kg/m2) F1 = 15.235kg (9.81m/s2) F1 = 149.456N

AL = 293.951N

BR : FBX = 8.399N / 302.35N BR : FBX = 0.0278

FORCE DUE TO SELF-WEIGHT

F2 = 6 x (0.55m x 0.55m)(4.296kg/m2) F2 = 7.797kg (9.81m/s2) F2 = 76.465N

FBX = 302.35N

R1 = 2.1N

R2 = 6.299N

450

50 162.5

387.5 0.972FBX = AL = 293.95N

LEFT SIDE REACTIONS ∑ML1 = AL(112.5mm) + L2 (450mm) = 0 -L2 (450mm) = 293.95N(112.5mm) -L2 = 33 069.38Nmm / 450mm L2 = 73.488N ∑FY = L1 + L2 + AL= 0 L1 + 73.488N = 293.95N L1 = 220.462N

L1 = 220.462N 50

L2 = 73.488N 450

162.5

50 387.5 75

ANALYSIS 2: REACTIONS - POSITION 2

FBX = 302.35N

FRONT

RIGHT BOX 3

AL BR L2 L1

R2 R1

BOX 2

BOX 1 FBX L2 L1

275

R2 R1

FRONT VIEW 76

275 BR

AL 50

450

ANALYSIS 2: REACTIONS

FBX = 302.35N

POSITION 2 REACTIONS ∑MAL = FBX(225mm) + BR (450mm) = 0 -BR (450mm) = 302.35N(275mm) -BR = 68 028.75Nmm / 450mm BR = 151.175N ∑FY = AL + BR + FBX = 0 AL + 151.175N = 302.35N BR = 151.175N

AL = 151.175N

BR = 151.175N 450

275

BR : FBX = 151.175N / 302.35N BR : FBX = 0.5

0.5FBX = BR = 151.175N

AL : FBX = 151.175N / 302.35N AL : FBX = 0.5 RIGHT SIDE REACTIONS ∑MR1 = BR(225mm) + R2 (450mm) = 0 -R2 (450mm) = 151.175N(225mm) -R2 = 34 014.375Nmm / 450mm R2 = 75.5875N ∑FY = R1 + R2 + BR= 0 R1 + 75.5875N = 151.175N R1 = 75.5875N

R1 = 75.5875N 50

R2 = 75.5875N 450

50 275

275

0.5FBX = BR = 151.175N

LEFT SIDE REACTIONS ∑ML1 = AL(225mm) + L2 (450mm) = 0 -L2 (450mm) = 151.175N(225mm) -L2 = 34 014.375Nmm / 450mm L2 = 75.5875N ∑FY = L1 + L2 + AL= 0 L1 + 75.5875N = 151.175N L1 = 75.5875N

L1 = 75.5875N 50

L2 = 75.5875N 450

275

50 275 77

POSITIONING THE BODY

ASSIGNMENT 3 BODY WEIGHT

BOX 3

BOX 2

BOX 1 FBX FA

RIGHT ELEVATION 1 : 10 80

X RA

REACTION CALCULATIONS: POSITION 1

BODY WEIGHT DISTRIBUTION FA= Head&Neck + Upper Arms + Lower Arms + Lower Legs + Feet = 90 + 92 + 40 + 130 + 28 = 380N FB: Upper Legs = 200N

FC: Torso = 420N

FT: 420N + 200N = 620N

∑MRA= FT x 285.48 + RB x 450 = 0 -620N x 285.48 + RB x 450 = 0 RB x 450 = 177 000 RB = 393.33N ∑FY = RA + RB + FT = 0 393.33 + RA = 620N RA = 226.6N

C.O.G. OF BODY ON CHAIR

COMBINED REACTIONS

x (from RB)= FC x 180 + FB x 350 FC + FB

∑MRA = -FB(285.48mm) + (-FBX(337.5mm)) + RBY (450mm) = -620N(285.48mm) - 302N(337.5mm) + RBY (450mm)

= 420N (180) + 200N (350) 420 + 200 = 214.516mm

FT = 620N

BODY WEIGHT REACTIONS

FB = 200N

RA = 226.6N 50

RAY = 302.17N

RB = 427.465N 450

200

170

180

335.484

214.516

FT = 620N

RBY = - 176997.6Nmm + 101925Nmm 450mm = 619.83N ∑Fy = 0 0 = -620N - 302N + 619.83N + RAY

FC = 420N

FBX = 302.35N

RA = 302.17N 50

RB = 619.83N 450

387.5

335.484

162.5

214.516

REACTION CALCULATIONS: POSITION 2

Y BOX 3

BOX 2 FA FB

BOX 1 FBX

FC RIGHT ELEVATION 1 : 10 82

X RA

REACTION CALCULATIONS: POSITION 2

BODY WEIGHT DISTRIBUTION

BODY WEIGHT REACTIONS

FA: Head & Neck + Upper Arms + Torso 90 + 92 + 420 = 602N

∑MRA: (332.19-50)702 + RB(450) = 0 0 = 198 097.38 + RB(450) RB = 198 097.38 450 =440.216N

FB: 1/2 Upper Legs = 200N x (1/2) = 100N FC: Lower Arms + Lower Legs + Feet + 1/2 Upper Legs

∑F = RA + RB + FB = 0 440.216 + RA = 702N RA = 261.784N

FT = 702N FB = 100N

RA = 261.784N

x (from RB)= FA x 200mm + FB x 325mm FA + FB

∑MRA = FT(332.19 - 50) + FBX (387.5 - 50) + RB(450) = 0 -RB(450) = 198 097.38 + 102 043.125 -RB = 300 140.505 450 RB = 666.98N

= 602N (200) + 100N (325) 702N = 217.806mm

450

175 COMBINED REACTIONS

∑FY = RA + RB + FT + FBX = 0 -RA = 666.98N - 702N - 302.35 RA = 337.37N

RB = 440.216N

FT: 602N + 100N = 702N C.O.G. OF BODY ON CHAIR

FA = 602N

125

250

217.806

332.194

FT = 702N FBX = 302.35N

RA = 337.37N 50

RB = 666.98N 450

387.5

332.194

162.5

217.806

REACTION CALCULATIONS: POSITION 3

BOX 3

BOX 2

BOX 1 FBX

RIGHT ELEVATION 1 : 10 84

REACTION CALCULATIONS: POSITION 3

BODY WEIGHT DISTRIBUTION

COMBINED REACTIONS

FA: Head & Neck + Torso + Upper Arms + Lower Arms + (1/5) Upper Legs

BODY WEIGHT REACTIONS

∑MRA = FA(100-50) + FBX (387.5 - 50) + RB(450) = 0 -RB(450) = 34 100 + 102 043.125 -RB = 136 193.75 450 RB = 302.65N

∑MRA: (100-50)682 + RB(450) = 0 -RB(450)= 34 100 RB = 75.78N

∑FY = RA + RB + FT + FBX= 0 -RA = 302.65 - 682N - 302.35 RA = 681.7N

90 + 420 + 92 + 40 + 1/5(200) = 682N

FA = 682N

RA = 606.22N 50

RB = 75.78N 450

100

450

∑FY = RA + RB + FB = 0 75.78N + RA = 682N RA = 606.22N

FA = 682N

FBX = 302.35N

RA = 681.7N

RB = 302.65N

450

387.5

100

162.5

450

ANALYSIS 3: OVERTURNING MOMENT

550

BOX 3

550

BOX 2

550

BOX 1 FT

RIGHT ELEVATION 1 : 10 86

150

FBX

237.5

162.5

ANALYSIS 3: OVERTURNING MOMENT

Horizontal forces from body: Head&Neck + Torso + Upper Arms + Lower Arms = 90 + 420 + 92 + 40 = 642N cos60 = adj hyp y

cos60 = x 642N

60 x

STATIC CHAIR OVERTURNING MOMENT CALCULATION CHAIR WEIGHT: 6 faces @ 12mm thick 12 faces @ 6mm thick (6(0.55m x 0.55m)(8.394kg/m2)) + (12(0.55m x 0.55m)(4.296kg/m2)) = 30.83kg

MOMENT RESISTANCE:

y = 6422 - x2 = 555.988N

MRESIST = (302.35N)(0.1625m) = 0.04913kNm

HORIZONTAL OVERTURNING FORCE (P)1.1m = (0.3kN)(0.1625m) P = (0.3kN)(0.1625m) + (.556kN)(0.4m) 1.1m = 0.2465kN

MRESIST

Force due to gravity : 30.83 x 9.807 = 302.35N

x = cos60(642N) = 321N

This is the horizontal force exerted on Box 2 from the body. P = 321N or 0.321kN

P MO ARM = 1.65m

WORST CASE SCENARIO OVERTURNING MOMENT CALCULATION

HORIZONTAL OVERTURNING FORCE: (P)1.65m = (0.3kN)(0.1625m) P = (0.3kN)(0.1625m) 1.65m = 0.0295kN or 29.5N This is the force required to push over the chair, when there is no one sitting on it.

It would appear that if the one was to lean back while seated at this position, the chair would tip over. 87

LOAD TAKEDOWN

ANALYSIS 4: LOAD TAKEDOWN

A1 = 10 000mm2

A2 = 1 164mm2

F3 = 76.465N

F2 = 76.465N

A3 = 55 000mm

A4 = 8 712 mm2

F1 = 76.465N

A5 = 55 000mm2

LOAD TRANSFER AREAS 90

ANALYSIS 4: LOAD TAKEDOWN

VERTICAL LOAD TRANSFER We will analyze the areas of overlap between each box to determine the axial stresses.

We are also treating the box as ‘welded’ , and will look at them as a whole instead of a collection of panels.

We assume that the boxes rest directly on each other, and that the vertical load is borne primarily through the top of the vertical panels. A1 = 100mm x 100mm A1 = 10 000mm2

A5 = 550mm x 550mm A5 = 302 500mm2

(P/A1) = 76.465N / 10 000mm2 (P/A1) = 7.647kPa

(P/A1) = 76.465N + 76.465N + / 10 000mm2 (P/A1) = 7.647kPa

A2 = 6mm x 100mm x 6(100-6)mm A2 = 1 164mm2 (P/A2) = 76.465N / 1 164mm2 (P/A2) = 65.69kPa A3 = 550mm x 100mm A3 = 55 000mm2 (P/A3) = (76.465N + 76.465N) / 55000mm2 (P/A3) = 2.78kPa A4 = 550mm x 12mm + (100-12)12mm A4 = 8 712mm2 (P/A4) = (76.465N + 76.465N) / 8 172mm2 (P/A4) = 18.71kPa

ANALYSIS 5 BEAMS + COLUMNS BOX 3

550

F3 = 76.465N BOX 2

550

FT = 854.93N

F2 = 76.465N

R2 = 427.465N

R1 = 427.465N

275

175

275

*All units in mm 92

BOX 1

550

WORSE CASE LOADING

ANALYSIS 5 BEAMS + COLUMNS

WORST CASE SCENARIOS.

ASSUMPTIONS

Since the chair only has one sitting position, we will first identify the seating scenarios where individual members should be under the greatest possible stress. Looking at these diagrams Scenario 1 should generate the greatest possible bending force in a horizontal member. Scenario 2 should lead to the greatest axial force in a vertical member.

1. We assume that the horizontal panel of Box 1 only transfers the load to 2 of the 4 vertical panels - thus simplifying the axial forces to be equivalent to the reactions from the top ‘beam’ panel.

SCENARIO 1 - RIGHT

SCENARIO 2 - RIGHT

SCENARIO 3 - FRONT

The body is seated near the center of the box. The total load is concentrated at the center - which puts the top panel under maximum bending stress. The vertical panels either end of the chair would be equally likely to buckle.

The body is seated as close to the front edge as possible. The top panel is still most likely to bend, while the vertical panel closest to the front should be the most likely to buckle, as the reaction forces near the front will be greater than the back.

The body is seated as close to the left side as possible. In this case, the left vertical panel should be under the greatest axial stress. Once again, the top panel should be under the greatest bending stress.

2. We assume that the aluminum pole and plywood sleeves will not contribute significantly to the bending and axial forces in the panels - so for the purposes of the analysis, we will treat the boxes as simply ‘welded’ on top of one another. 3. All distance units are in millimeters.

ANALYSIS 5 BEAMS + COLUMNS BOX 3

550

F3 = 76.465N BOX 2

550

FT = 854.93N

F2 = 76.465N

R1 = 427.465N

R2 = 427.465N

275

175

275

*All units in mm 94

BOX 1

550

WORST CASE LOADING

ANALYSIS 5 BEAMS + COLUMNS F3 = 76.465N

FB = Head & Neck + Upper Arms + Torso + Upper Legs (1/2) = 90 + 92 + 420 + 200N = 702N

FB = 702N FT = 854.93N F2 = 76.465N

R1 = 427.465N

R2 = 427.465N

225.984

324.016

175

275

175

x = 275mm (from R1) Solve for the location of the body: xb x = Fb(xb) + F3(275mm) + F2(725mm) Fb+ F3 + F2 235 105.75Nmm =Fb(xb)+21027.875Nmm+55437.125Nmm 158640.75Nmm = 702N (xb) (xb) = 225.984mm MMAX = FL / 4 MMAX = 854.93N x 550mm / 4 MMAX = 117 552.875Nmm

V1 = 427.465N

V2 = -427.465N

Sx = bh2 / 6 Sx = 550(12)2 / 6 Sx = 79 200 / 6 = 13 200mm3 fbx = MMAX / Sx fbx = 117 552.875Nmm / 13 200mm3 fbx = 8.95N/mm2

MMAX = 117.553 Nm

This would be the greatest possible bending force experienced by the top panel of Box 1.

ANALYSIS 5 BEAMS + COLUMNS BOX 3

550

F3 = 76.465N BOX 2

550

FT = 854.93N

F2 = 76.465N

R1 = 764.562N

R2 = 90.368N

275

175

275

*All units in mm 96

BOX 1

550

WORST CASE LOADING

ANALYSIS 5 BEAMS + COLUMNS

Solve for the location of the total load (from R2) FB = 702N

x = Fb(550) + F3(275mm) + F2(-175mm)

FT = 854.93N F3 = 76.465N

R1 = 764.562N 58.14

F2 = 76.465N

R2 = 90.368N 491.86

275

Fb+ F3 + F2 = (386100Nmm + 21027.875Nmm + 13381.375Nmm) / 854.93N = 420 509.25Nmm / 854.93N x = 491.86mm

175

Find the reactions

175

∑MR1 = 854.93N (58.14mm) + R2(550)= 0 R2(550mm) = 49 702.25Nmm R2 = 90.368N ∑F = FT + R1 + R2 = 0 -R1 = -854.93N + 90.368N R1 = 764.562N

R1 = 764.562N

Using R1 as the axial force acting on our ‘column’ - like panel, we can solve for the compression stress in the member.

A = 6 600mm2 fa = P/A fa = 764.562N / (12x550mm) fa = 764.562N / (6 600mm) fa = 0.115843N/mm2 Next, we will check for buckling since the panel is quite thin.

hx = 12 bx = 550

(P/A)cr = [π2 E] / (L/r)2 r = √(IX / A) r = √[(bh3 / 12) / (bh)] r = √[(550(12)3 / 12) / (550 x 12)] r = √12

(P/A)cr = (π2 E) / (L/r)2 (P/A)cr = π2 (8 167N/mm2) / (550/√12)2 (P/A)cr = 80 605.06N/mm2 / (158.77)2 (P/A)cr = 80 605.06N/mm2 / (25 208.33) (P/A)cr = 3.198 N/mm2

Critical buckling stress is well above actual axial stress, the panel should be quite safe from buckling.

COMBINED LOADING

ANALYSIS 6 COMBINED LOADING

C 50

500

BOX 3

257

514

550

MOMENT ARM DIAGRAM (FRONT)

CL BOX 2

275

225 225

275 550

F2 BOX 1

550

*All units in mm 100

ANALYSIS 6 COMBINED LOADING

CALCULATION ASSUMPTIONS:

POLE ANALYSIS

1. Weight of pole & shear bracing at internal rotating joints is negligible; not factored into equations.

The goal is to understand how moment forces as well as axial forces affect a vertical panel in the second box. We will begin by looking at the moment forces caused by the overhanging upper boxes - and how they are resisted in the aluminum pole.

2. All boxes made from Baltic birch plywood, with the corresponding weight properties:

FORCE DUE TO GRAVITY

6mm thick 12mm thick

= 4.296kg/m2 = 8.394kg/m2

F3 = 6 x (0.55m x 0.55m)(4.296kg/m2)g F3 = 7.797kg (9.81m/s2) F3 = 76.465N

3. Box 2 & Box 3 have the same dimensions and plywood thickness (6mm), thus the same volume and weight. Box 1 has a 12mm plywood thickness on all sides.

BOX 3 - SUM OF THE MOMENTS ABOUT B ∑M B = (F3 x 225mm) + (P1 x 514mm) ∑M B = 17 191.125Nmm - (P1 x 514mm) P1 = 17 191.125Nmm*/ 514mm P1 = 33.446N

4. Calculations do not account for overlap or cutting at lap joint and box joint edges; the same surface area is assumed for each face. (0.55m x 0.55m = 0.3025m2)

The moment force is resisted at the base and top of the first box (B & C), where the pole is braced through the plywood sleeves. For Box 3 to be in equilibrium - the equal and opposite force must be present at C, 33.446N. These horizontal forces, in turn, are assumed to bear primarily on the panel of the box directly perpendicular to the moment force. Using this method, we can analyze the middle box (Box 2) the same way.

5. While each box contains three ‘sleeves’ that brace the pole, it is assumed that only the top and bottom sleeve factor into the reactions.

25.4mm

With this moment force, we can evaluate the maximum bending force in a section of pole. fb = Mc / I I = 135 280.28 mm3 c = 25.4mm fb = 17 191.125Nmm x 25.4mm / 135 280.28 mm3 fb = 3.228Nmm2

3.175mm 101

ANALYSIS 6 COMBINED LOADING

BOX 3 F3

500

P4 275

225 550

F3 = 76.465N P4 = 33.446N 257

MOMENT ARM DIAGRAM (RIGHT)

225

BOX 2

275

257

550

514

F3 P4

P3 = 33.446N

F2 = 76.465N BOX 1

F2 dx 550

*All units in mm 102

ANALYSIS 6 COMBINED LOADING

HORIZONTAL FORCES In Box 2, the magnitude of the moment forces due to its self weight are the same as in Box 3, since they weigh the same. However, the moment induced is in a different direction. Having identified the horizontal forces from the cantilevered position of the box, we will analyze the panel highlighted in red to evaluate the horizontal forces as well as vertical forces.

532

F3 = 76.465N F2 = 76.465N

Therefore:

P3 = 33.446N P4 = 33.446N

We can understand the horizontal forces acting on the panel by modeling the panel as a vertical â&#x20AC;&#x2DC;beamâ&#x20AC;&#x2122;, that is cantilevered from the bottom. The second box is restrained from rotation - at point E, so we can treat this joint as resisting moment.

MX = 33.446N x 532mm MX = 17 793.272Nmm

Sx = bh2 /6 Sx = 550mm(6mm)2 / 6 = 3 300mm3

fbx = Mx / Sx = 17 793.272Nmm / 3 300mm3 fbx = 5.209 N / mm2 We can also evaluate potential bending force in the other direction (fby) - which comes from the cantilevered position of Box 3.

M MX = 17 793.272Nmm hx = 6

Sy = bh2 /6 Sy = 6mm(550mm)2 / 6 = 302 500mm3 My = F3 x dy = F2 x dx = 17 793.272Nmm

bx = 550

fby = My / Sy = 17 793.272Nmm / 302 500mm3 fby = 0.0588 N / mm2

by = 6 hy = 550

103

ANALYSIS 6 COMBINED LOADING

BOX 3 50

500

275

225 550

F3 = 76.465N

257

P4 = 33.446N

225

BOX 2

275

257

550

514

P3 = 33.446N

F2 = 76.465N BOX 1

550

*All units in mm 104

ANALYSIS 6 COMBINED LOADING

F3 = 76.465N

AXIAL FORCES We will simplify the axial forces running through Box 2 by treating all the vertical panels as one column. Thus the weight of Box 3 (F3) will be transferred through the surface area of the tops of the vertical columns (AT). fA = F3 / AT AT = (550mm)2 - (544mm)2 AT = 302 500mm2 - 295 936mm2 = 6 564mm2 fA = 76.465 N/ 6 564mm2 fA = 0.011649N/mm2 (compression) Now we have the axial forces as well as horizontal forces in the panel. Since the panel is relatively thin (6mm), we should check to see if the panel is likely to buckle.

AT = 6 564mm2

fA= 0.011649MPa

r = √(IX / A) r = √[(bh3 / 12) / (bh)] r = √[(550(6)3 / 12) / (550 x 6)] r = √3 (P/A)cr = π2 E / (L/r)2 (P/A)cr = π2 (7 656N/mm2) / (550/√3)2 (P/A)cr = 75 561.69N/mm2 / (317.54)2 (P/A)cr = 75 561.69N/mm2 / (100 832) (P/A)cr = 0.749 N/mm2 fA = (P/A)A < (P/A)cr The vertical panel is quite safe from buckling, as the critical stress is much greater than the actual stress.

105

ACTUAL LOADING

106

107

ANALYSIS 7 ACTUAL LOADING BOX 3

550

F3 = 76.465N CL

COLUMN PANEL (6.5mm)

BOX 2

225

275 550

FT = 854.93N

BEAM PANEL (12mm)

BOX 1

P = 33.446N

550 275

F2 = 76.465N

225

50 550

*All units in mm 108

ANALYSIS 7 ACTUAL LOADING

CALCULATION ASSUMPTIONS:

PLYWOOD PROPERTIES

1. When determining whether to use the values for bending forces perpendicular or parallel to the face grain from the tables- we will always match the lowest given values with the highest predicted values. This assumes the worse case scenario. ie. fbx : FB |

A quick glance at the material properties of plywood reveals that it has a lower tolerance for axial forces then bending forces. However, from our previous analysis, the greatest stresses are generally due to bending. Therefore, we will examine a vertical panel from Box 2 as well as a horizontal panel from Box 1 to see how our predicted loads compared to the critical loads.

FB = Head & Neck + Upper Arms + Torso + Upper Legs (1/2) = 90 + 92 + 420 + 200N = 702N FC = F3 + F2 = 76.465N + 76.465N = 152.93N FT = FB + FC = 854.93N P = 33.446N fA (12MM BEAM PANEL) =P/A = 33.446N / (550mm x 12mm) = 0.0050675N/mm2

6.5MM BALTIC BIRCH BENDING (FB) || BENDING (FB) | COMPRESSION (FA) || COMPRESSION (FA) |

6.5MM ‘COLUMN’ PANEL 50.9 29.0 29.3 22.8

N/mm2 N/mm2 N/mm2 N/mm2

12MM BALTIC BIRCH BENDING (FB) || BENDING (FB) | COMPRESSION (FA) || COMPRESSION (FA) |

BENDING (fbx) || BENDING (fby) | AXIAL FORCE (fa)

5.209 N/mm2 0.0588 N/mm2 0.01165 N/mm2

12MM ‘BEAM’ PANEL 42.9 33.2 27.7 24.3

N/mm2 N/mm2 N/mm2 N/mm2

BENDING (fbx) BENDING (fby) AXIAL FORCE (fa)

8.6598 N/mm2 0.1934 N/mm2 0.0051 N/mm2

COLUMN PANEL fA / FA + fbx / FB | + fby / FB || = (0.01165 / 22.8) + (5.209 / 29.0) + (0.0588 / 50.9) = 0.00051096 + 0.1796 + 0.001155 = 0.1813

1.0

BEAM PANEL fA / FA + fbx / FB | + fby / FB || = (0.0050675 / 24.3) + (8.6598 / 33.2) + (0.1934 / 42.9) = 0.00020854 + 0.2608 + 0.0045 = 0.2655

109

110

CHEERS

111

CITATIONS

ON MC ESCHER Lemle, Natalie. “How M.C. Escher Transfixed the World with His Mind-Bending Works.” Artsy, 21 Mar. 2018, artsy.net/article/artsy-editorial-mc-escher-transfixed-mind-bending-works. Poole, Steven. “The Impossible World of MC Escher.” The Guardian, Guardian News and Media, 20 June 2015, theguardian.com/artanddesign/2015/jun/20/the-impossible-world-of-mc-escher. ON THE PROPERTIES OF BALTIC BIRCH PLYWOOD Finnish Forest Industries Association Handbook of Finnish Plywood, Finnish Forest Industries Association, Finland 2002 Available Online

PRECEDENT IMAGES All referenced images of existing chair designs were accessed on the Internet and are the property of their respective owners.

112

CITATIONS

A CHAIR FOR M.C. ESCHER by Wayne Yan + Philippe Fournier

Special thanks to Heinz & Michael in the woodshop for their assistance

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