Musical Thrown Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
- Joris Voorn
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
MANIFESTO - a letter to Joris Voorn Joris Voorn,
Thank you. The inspiration you have provided as a distant figure is unparalleled. Meeting you aboard the Enterprise 2000, Vibrations boat cruise in Toronto even surpassed my expectations. To be able to exchange ideas with some, one who has similar passions, if not the exact same, is something so rare, as we both come from a background with an education in architecture, and a love for electronic dance music. The relation between architecture and electronic (house) music is un-canny, not only in principle, but also in feeling. The constructions of either a set, or a track, are easily related to a concept and a component respectively, as are the provocative repetition and rhythm. None the less the progression of a concept in both go through multiple iterations and continuous change, as projects never finish, only go on hold for new ideas.
Balance 014 (Album) both CD1 MIZUIRO and CD2 MIDORI are a great example of this, a contrast of space and feeling. Mizuiro being this time elapsed big night out with great music, a transformable space, allowing the many smaller components to build through out the CD into a feeling of a larger scale. The connection of scale reflects that of space and sound. It is here that you get darker more melodic place, as if the night was to end and the sunrise was near. The transition between Mizuiro and Midori almost comes too soon, leaving you wanting more, which allows for the surprise of about what’s to come next. The tempo changes drastically in a few tracks allowing for a change in pace and scale. From Large and Dark to Intimate and Uplifting. It remains melodic and repetitive, however it allows the adrenaline previously built to dissipate. However not to quickly, with a level of care and consideration that contrasts to many of your peers who so suddenly stop. Leaving you wanting much more. The same descriptions, based upon my interpretation, I believe in many ways can be used to depict an image of architecture. Dark, melodic, repetitive, scale, all drawing the same level of impact on an individual. Even considering the transition between the two CDs, it is in many ways the shift between work and play that blurs the boundary so well. This album being so inspirational, has lead me to choose you, as my client to design a chair for my Structural Design Build class this term. Based upon the Balance album that makes me truly feel architecture and emotion through sound, yet blurs the line between work and play. In both fields (Architecture or Music) there is a necessity to do the very same thing, and love what we do. The chair in essence will be just that, a transition between play and work, and a functional place to release ideas of music or design. So thank you one last time. “Music and architecture each require complete devotion.” –Joris Voorn I couldn’t agree more.
I hope this letter finds you well. Geoff Bagga – Jeans University of Waterloo : School of Architecture [3B]
Chair 1
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
Chair 2
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
Chair 3
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
Chair 4
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
Chair 5
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
Chair 5
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
Analysis WEIGHT OF THE CHAIR Volume of the Chair:
Panels: 0.457m x 0.4064m x 0.05m = 0.00928624 m3 x 4 panels = 0.03714496 m3 Base: 0.465m x 0.6096m x 0.05m x 2 panels = 0.0283464 m3 Total volume of the chair: 0.03714496 m3 + 0.0283464 m3 = 0.0654913 m3 Density of Birch plywood: 680‐700 Kg/m3 Total mass of Chair: 680Kg/m3 x 0.0654913m3 = 44.534 Kg Total Weight of Chair: 44.534 Kg x 9.98 N/Kg = 444.45 N REACTIONS OF THE CHAIR Storage Form:
Weight of Half Chair: 222.2 N W1= 111.1 N W2= 111.1 N
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
Analysis M@R1= 0
R2x0.457m = (W1 x 0.114m) (W2 x 0.342m) R2 = 110.8 N
R1=222.2N – 110.8 N R1=111.34 N
Chair position:
W= 444.44 N M@R1 = 0
R2 (18”) = W (9”)
R2= 444.44 (9)/18 R2 = 222.22 N Therefore:
R1= 222.22 N
Analysis With Body (Worst Case Loading) W1= 444.44 N P1= 800N
P2= 100 N
R1= 222.22 N
R2= 1022.22 N R3 = 100 N
With Body (Position 2) W1= 444.44 N P1= 800N
P2= 100 N
R1= 222.22 N R2= 222.22 N R3= 900 N
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
Analysis
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
Table Position:
W3= 255.04 N
Analysis
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
R1= W1 + (W2/2) = 189.39 N
R2= (0W2/2)+(W3/2) 90.75 NR3= 3/2 = 127.52 N N (1 panel) W1= .00928624 m3 x= 6180 Kg/m3 x 9W .98 N/Kg = 63.13 W2= 126.26 N (2 panels) With 2E55.04 quipment: W3= N
R1= W1 + (W2/2) = 189.39 N
R2= (W2/2)+(W3/2) = 190.75 NR3= W3/2 = 127.52 N
With Equipment:
E1 (CDJ)= 3.9 Kg x 2 x9.98 N/Kg = 77.84 N
E2 (Machine) = 1.6 Kg x 9.98 N/Kg = 15.96 N E3 (Mixer) = 5 Kg x 9.98 N/Kg = 49.98 N
E1 (CDJ)= 3.9 Kg x 2 x9.98 N/Kg = 77.84 N
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
Analysis R1= 189.39 N + E2 = 205.35 N R2= 190.7 N + E1 = 268.59 N
R3= 127.53 N + (E3/2) = 151.97 N
CENTRE OF GRAVITY *Note: All components of the chair are Plywood.
Storage Position
I 1
Bi 9”
ViXi Yi YiXi 3168 16” 12672 in4 in 4 2 8” 22” 4”* 704 in3 23” 16192 8” 5632 in4 in 4 3 10” 22” 4”* 880 in3 14” 12320 11” 9680 in4 in 4 4 20” 8” 16” 2560 10” 25600 26” 66560 in3 in4 in4 4936 57280 94544 in3 in4 in4 *The reduced thinkness doe the base elements is to accommodate for the voids present within the base for storage. X= ViXi Y = ViYi Vi Vi
X = 57280 in4 4936 in3 X = 11.6 in
Hi 22”
Ti 4”*
Vi 792 in3
Xi 4”
Y = 94544 in4 4936 in3 Y = 19.15 in
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
Analysis
P = Mx/H F= ℳN P= (444.44 N)(11.6 in) F= (0.5) (444.44 N) 30” F= 222.22 N P= 171.85 N Since the force applied is smaller than Friction (222.22 N), the chair’s force in its storage position, the chair will tip with 171.85 N of force. P = My/H F= ℳN P= (444.44 N)(19.5 in) F= (0.5) (444.44 N) 19” F= 222.22 N P= 456.13 N Since the force applied is larger than Friction, the chair will require 456.13 N of force to move in its storage position. *Note: This does not account for the metal stabilizing rods attached to the base. Chair Position
I 1
Bi 9”
Hi 22”
Ti 4”*
Vi 792 in3
3
10”
22”
4”*
880 in3
2 4
8” 4”
22” 39”
4”* 16”
Xi 4”
704 in3
23”
2496 in3 4872 in3
19”
14”
ViXi 3168 in4 16192 in4 12320 in4 47424 in4 79104 in4
Yi 16”
YiXi 12672 in 4 8” 5632 in 4 11” 9680 in 4 41.5” 103584 in 4 131568 in4
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
Analysis X= ViXi Vi
Y = ViYi Vi
X = 79104 in4 Y = 131568 in4 4872 in3 4872 in3 X = 16.23 in Y = 27.00 in P = Mx/H F= ℳN P= (444.44 N)(16.23 in) F= (0.5) (444.44 N) 61” F= 222.22 N P= 118.25 N Since the force applied is smaller than Friction, the chair will tip forwards with 118.25 N of force. *This is not accounting for the role of the metal stabilizing bars that are attached to the base.
P = My/H F= ℳN P= (444.44 N)(27.00 in) F= (0.5) (444.44 N) 19” F= 222.22 N P= 631.5 N Since the force applied is larger than the force of friction, the chair requires 631.5 N of force to move in the direction perpendicular to the pannels.
Table Position:
I 1
Bi 9”
Hi 22”
Ti 4”*
Vi 792 in3
3
10”
22”
4”*
880 in3
5
2”
2 4
8”
54”
22”
4”*
2”
16”
24”
16”
Xi 4”
704 in3
23”
1728 in3 768 in3
46”
4872 in3
14” 64”
ViXi 3168 in4 16192 in4 12320 in4 79488 in4 49152 in4 160321 in4
Yi 16” 8”
11” 23” 12”
YiXi 12672 in 4 5632 in 4 9680 in 4 39744 in4 9216 in4 76944 in4
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
Analysis
X= ViXi Vi
Y = ViYi Vi
X = 160321 in4 Y = 76944 in4 4872 in3 4872 in3 X = 32.90 in Y = 15.79 in P = Mx/H F= ℳN P= (444.44 N)(32.90 in) F= (0.5) (444.44 N) 24” F= 222.22 N P= 609.25 N Since the force applied is larger than Friction, the table will slide on a wodden surface. P = My/H F= ℳN P= (444.44 N)(15.79 in) F= (0.5) (444.44 N) 54” F= 222.22 N P= 129.95 N Since the force applied is smaller than Friction, the table will tip is pushed close to the leg. FBD -‐See next page-‐
Analysis
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
RACKING (Qualitative Analysis) Due to the large mass, and the method of rigid construction, we are confident that there is very little movement in terms of racking in the chair. Out base is very solid in construction, and acts a singular piece, transferring all the moment within it, therefore resisting any possible racking. There is however some movement that occurs in the Hinges when the chair is in its Storage Formation. The joints tend to skew a little bit due to the allowance given between each join to allow for the transformations. The locking pieces around the hinges prevent this movement when the panels are locked into one another in a used formation. -‐See next page for diagrams-‐ BEAMS AND COLUMNS ANALYSIS W1= weight of body=800N W2=dead weight of chair panels= 63.13 N x 4 = 252.52 N W3= chair base= 191.92 N Beam 1: -‐See next page for diagrams-‐
M@R1=0 R2= (W1 x 0.2285m) + (W2 x 0.457m) / 0.457m R2= 652.52 N R1= 400 N R2= 652.52 N Vmax= 650 N Mmax=0.0914 kN/m Fb=M/S S=bh^2/6 Fb= 0.148525 kN/m S= (0.457m)(0.0508m)^2 0.000196558m3 6 Fb= 755.85 kPa S= 0.000196558 m3 Fv= 3V/2A Fv= 3(0.650 kN) 2(0.457 x 0.0508) Fv= 41.997 kPa
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
Analysis
Beam 2: -‐See next page for diagrams-‐
R1= 400 N + (191.92 N/2) = 495.96N R2= 652.52 N + (191.92 N/2) = 748.48 N Vmax= 95.96 N Mmax= 0.02192 kN/m Fb=M/S Fb= 0.02192 kN/m 0.0004991678m3 Fb= 43.93 kPa Fv= 3V/2A Fv= 3(0.09596 kN) 2(0.457 x 0.0508) Fv= 6.2160 kPa Column 2: -‐See next page for diagrams-‐ P1= ? P2= ? R1= 400 N Slope: 11/4 P1=400N x 11/11.7 P1= 376.06 N P2= 400N x 4/11.7 P2= 136.7 N Arect= (0.0311m)(0.4064m) Arect= 0.01263904 m2 Fa= P/A Fa= (0.376 kN)/(0.01263904 m2) Fa= 29.75 kPa M = P2 = 136.7 N
S=bh^2/6 S= (0.457m)(0.0254m)^2 6 S= 0.0004991678 m3
Analysis Column 1: -‐See next page for diagrams-‐
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
P1= ? P2= ? R1= 652.52 N Slope: 11/4 P1= 652.52 N x 11/11.7 P1= 613.48 N P2= 625.52 N x 4/11.7 P2= 223.1 N Arect= (0.0311m)(0.4064m) Arect= 0.01263904 m2 Fa= P/A Fa= (0.613 kN)/(0.01263904 m2) Fa= 48.50 kPa M = P2 = 223.1 N Deflection: (Beam 1 a.k.a. Seat) E (for plywood)= 12.4 GPa Δ max= 5 wl/384 EI I=bh^3/12 Δ max= 5(0.8 kN)(0.4064m) I= (0.457m)(0.0508m^3)/12 384 (12.4 x10^6 kPa)(4.99 x 10^-‐6) I= 4.99 x10^-‐6 Δ max= 6.6 x 10 ^-‐5 m JOING ANALYSIS * Worst case loading analyzed for joints in the Table position since all the joints are made the same. -‐See next page for diagrams-‐ L= 0.4064 m W= 800 N A= 0.127 m W1= 21.55 kN/m W2= 7.60 kN/m B= 0.1046
Analysis
V: W= (l-‐2b)(w1+w2) 2 M: W a + bW(l-‐b) = W1(l-‐b) (l-‐b) 2 2 3 2 3 Derived Formulas: W2= 3 W a b (l-‐b)^3 + b^3 + 3b^2l W2= 3(0.8 kN)(0.127m)(.0146m) (0.3017m)^3 + (0.1046m)^3 – 3(0.1046m)^2(0.4064m) W2= 7.60 kN/m B= l (2l-‐3a 6 (l-‐a) B= (0.4064m)(2(0.4064m)-‐3(0.127m)) 6 (0.4064m – 0.127m) B= 0.1046 m L-‐B= 0.3017m W1= W2 (l-‐b) B W1= (7.6kN/m)(0.3017m) 0.1064m W1= 21.552 kN/m Per Joint: P= (21.55 kN/m)(0.01905m) P= 0.4105275kN Vmax= 410 N Allowable stress for Dowel: Vmax= 4 (Vmax) 3A Vmax= 4 (410 N) 3 (71.25 mm2) Vmax= 7.67 MPa Cross sectional area of Dowel: A= π r^2
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
Analysis
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
A= 71.25 mm2 Allowable: 420 psi (pg. 35, NDS, Design values for wood construction, yellow poplar) Therefore: the shear stress of the dowel will be able to resist the worst-‐case (410 N < allowable 420 psi) loading (in table position).
Photos- Storage
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
Photos- Chair
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058
Photos- Desk
Geoff Bagga - Jeans - 20245053 Snehjot Bumrah - 20271058