CHURL JONG KIM by Churl Jong Kim

Undergraduate Work Portfolio CHURL JONG KIM

academic year : 2005.03 - 2011.12 date updated : 2021.11

Space as an architectural problem;

PROJECT LIST 01 Visual Training

2010 Fall - 2011 Spring

06 - 15

02 Space Problem Exercise

2010 Fall

16 - 23

2011 Summer

24 - 31

2010 Fall

34 - 47 48 - 75 76 - 107

03 The Farnsworth Barn 04 Space Problem

05 Extra Works

Space Problem 1 : Court house Space Problem 2 : Art Gallery Space Problem 3 : Low-rise Apartment

2011 Spring 2011 Fall

Freehand drawings

01 Visual Training Aesthetic expression as experience. Exercises in the study of from: proportion and rhythm, texture and color, mass and space. Exercise in visual perception and aesthetic judgement. Isolation and analysis; interdependence and integration of sensuous qualities. Aesthetic unity under restrictive condition.

Plate No. 1: Four White Rectangles

Plate No. 2: Curve

Plate No. 3: Falling Planes

Plate No. 4: Black and White Rectangles

Plate No. 5: Texture

Plate No. 6: Materials

Plate No. 7: Ink Drop

Aesthetic importance “The mullion functioned as support for the window frame and wind-bracing for the full height of the eight-foot, four-and-a-half-inch distance between floor and ceiling. However, on the column, the mullion had no such function.” One of the things that Mies kept in his mind was function. He had always driven by objective and a reason for everything he did. The spaces he provides are function oriented. What people need here and how people use those was the way he constructs one building. What we need has to be in it, and nothing else. In this manner, W-shapes in front of columns have no reason to be set on its facade. Those elements set in front of columns have no functions like support for the window frame and wind-bracing. It should not be there, if we only think Mies as a functionalist. Then why those I-beams are there? Did he change his mind? Or is this a simple exception? I do not believe it is a simple exception, since I think there is something more to talk about Mies’s philosophy in addition to the importance of function. It is aesthetics. One other magnitude on Mies’s architecture is in aesthetics. In other words, I would use the term ‘visual impression’. For instance, in 860’s case, probably it just does not look right without the additional mullion pieces. It would have destroyed the continuity of the envelope, if there were no W-shape in front of column. What Mies was looking for was aesthetical value. This reminds me the lesson I learnt from vertical space problem in 2011 spring semester. At the beginning of semester I was looking for the module that perfectly fit for golden ratio in vertically as well as horizontally. What I was looking for was the perfect module that helps me to control the whole. However, this kind of approach was not adequate. What I was actually looking for was mathematical balance, and that is not right at all. Even if I used the word ‘control’, it would end up with ‘dominate’. This was very arrogant way of thinking, since in the architecture, human-being who uses the space has to be in the center, not one architect. Architect works for people not for himself, and what Mies was trying to do was creating the very comfortable space for people. I assume that this is why professor Takeuchi was telling me “Use your module as a tool. Module is just a tool.” I think he omitted some sentences to give me chance to think. Now I would guess the blank like this “Module is just a tool to find aesthetically fine space. Use the tool only for the base. We use that just it helps us to find easier way to achieve aesthetical valuable space.” If we only care about how the elements are perfectly fit as a number, the thing will look very stiff and rigid. It cannot be looked free and natural. Even though one architect was able to make all the numbers he uses under his control, unfortunately there is no importance in domination. In order to have visual value, aesthetic decision has to be made and for this, we need to train our eyes. It could sound very vague, but it is literally like that. If we keep train our eyes, I guess we will see it and feel it. The class in IIT called visual training is the course that does this. During the course, students keep looking at the boards and pretend they know what they are doing. It sounds very non-educational, but this is training. Once you go through it, you will find better you, even if you do not notice that. When I was in the class, I misunderstood the class once, and ask questions such as reason for the beauty and logical order in the board. Now I think that is not what course is trying to teach student. Visual training gives students an opportunity to keep them looking at certain things and show how it can look better. At this point, I impudently state that Mies’s architecture can only be understood when a person experiences the same process that he went through, and it cannot be summarized by some academic sentences.

02 Space Problem Exercise Preliminary study of Space problem Exercise in visual perception in 3 dimensional space Placing partition walls or volume in various dimension, height, and material helps to train eyes to make sensuous spatial quality.

1 White Partition Wall

Length : 3 1/2 module Height : 10’

Length : 4 module Height : 8’

2 Perpendicular White Partition Walls Length / Height : 4 module / 7.5’ Length / Height : 2 1/2 module / 10’

2 Parellel White Partition Walls

Length / Height : 4 module / 7’ Length / Height : 2 module / 10’

Length / Height : 5 1/2 module / 6.5’ Length / Height : 1 1/2 module / 10’

Length / Height : 4 module / 7’ Length / Height : 2 1/2 module / 10’

2 Parellel White/Black Partition Walls

White Partition Length / Height : 4 module / 7’ Black Partition Length / Height : 2 module / 10’

White Partition Length / Height : 4 module / 7.5’ Black Partition Length / Height : 2 module / 10’

2 Parellel White/Black Partition Walls

2 Perpendicular White/Black Partition Walls

White Partition Length / Height : 3 1/2 module / 7.5’ Black Partition Length / Height : 1 1/2 module / 10’

White Partition Length / Height : 4 module / 7.5’ Black Partition Length / Height : 2 module / 10’

2 Parellel White/Black Partition Walls

Volume

1 module x 2 1/2 module at 10’ height

Volume

1 module x 3 module

at 10’ height

White Partition Length / Height : 2 1/2 module / 10’ Black Partition Length / Height : 4 1/2 module / 7’

Volume

1 module x 4 module

1 Copper partition wall at 8’ height

Length : 3 1/2 module Height : 10’

1 Copper partition wall Length : 2 module Height : 7’

1 = 1

Mathematical approach

1 = 2 - 1

Spatial approach

Making spatial balance does not always mean that it has exact same quantity on each side. What we are pursuing in architecture is not mathematical equilibrium. Mathematical equilibrium is somewhat too stiff and rigid. One can be bigger than the other, but it can be further at the back. One can be longer than the other in length, but it can be shorter in height. Spatial harmony may come beyond mathematical equality.

03 The Farnsworth Barn AN ADDITIONAL FACILITY FOR AN EXISTING VISITING CENTER The Farnsworth House, designed by Ludwig Mies van der Rohe in 1951, is one of the most important works of 20th century architecture. Originally designed as a country retreat for Dr. Edith Farnsworth, the house today belongs to the National Trust for Historic Preservation and is open to the pubric. More than 1000 visitors annually make the pilgraimage to this masterpiece of Modernism. The existing visitor center is in a desperate need of an additional office space, storage and an exhibition space for artifacts and original furniture of the Farnsworth House.

Drawing #1 Natural Flow from Visitor Center to Farnsworth House

Parking lot

Visitor Center

Lumen House

Drawing #2 Visitor’s various possible sequences

1 Parking lot 7

6 Visitor center 2

5 Lumenhouse

Farnsworth Barn 3

Farnsworth House

Parking lot - Visitor Center [ticketing- orientation] - Farnsworth Barn - Farnsworth House - Lumen House - Gift Shop - Parking lot

Parking lot - Gift Shop - Visitor Center [ticketing- orientation] Farnsworth Barn - Farnsworth House - Lumen House - Parking lot

1 7

Parking lot - Visitor Center [ticketing- orientation] - Gift Shop Farnsworth Barn - Farnsworth House - Lumen House - Parking lot

6 2

4 Farnsworth Barn

Parking lot - Visitor Center [ticketing- orientation] - Farnsworth House - Farnsworth Barn - Lumen House - Gift Shop - Parking lot

Parking lot - Gift Shop - Visitor Center [ticketing- orientation] Farnsworth House - Farnsworth Barn - Lumen House - Parking lot

Parking lot - Visitor Center [ticketing- orientation] - Gift Shop Farnsworth House - Farnsworth Barn - Lumen House - Parking lot

Drawing #3 Universal Space

Coat Closet

Kitchen

Mechanical Entrance

Exhibition Area

Drawing #4 Golden Ratio

1.618

Drawing #5 Floor Plan

Drawing #6 North Elevation

Drawing #7 North East Elevation 29

Drawing #8 Section 30

04 Space Problem Structure as an architectural factor; space as an architectural problem; proportion as a means of architectural expression; the expression value of materials; painting and sculpture in their relationship to architecture.

Space Problem 1 A PRIVATE RESIDENCE WITH ONE OR MORE COURTS A ‘professional’ couple desires a residence in the city with ample space in which to entertain friends and to display a collection of modern paintings and sculpture encompassing notable works from Cubism to “Pop Art”. In addition to the usual basic requirements provide a studio or study that, depending on the arrangement, could possibly double as an overnight quest facility. If not, provide a separate overnight guest facility. Provide off-street parking for at least one auto. Provide a suitable scheme in the ‘Open Plan’ mode.

STRUCTURAL CALCULATION

LOAD TABLE ROOFING INSULATION METAL DECK ESTM WT OF STL CEILING MISC

Fy = 50 ksi fb = 0.66 x Fy = 33 ksi fv = 0.40 x Fy = 20 ksi E = 29 x 106

2.0 2.0 2.5 6.5 5.0 2.5

TOTAL D.L. L.L.

8’

20.5 30.0

24’

6’ 6’ 6’ 6’ 8’

50.5

DESIGN B1

DESIGN B2

W = 6’ X 50.5 = 303 PSF WL m= = 8 2

m f

303 x (24 x 12) 8

TRY W8x21 S = 18.2 in3 I = 75.3 in4

202 x 242 x 122 8 x 12

m f

= 202 = 174528

= 5.288727273

TRY C12x20.7 S = 21.5 in3 I = 129 in4

CHK DEFLECTION

ACTUAL =

50.5 x 8 2

= 261792

261792 = 7.933 33000

: ALLOWABLE =

CHK DEFLECTION 24 x 12 = 1.2” 240

5 x 303 x 244 x 124 384 x 29 x 106 x 75.3 x 12

: ALLOWABLE = = 1.06” < 1.2”

ACTUAL =

24 x 12 = 0.8” 360

5 x 202 x 244 x 124 384 x 29 x 106 x 129 x 12

= 0.403” < 0.8”

DESIGN G1 P = 7272 W = 1212 Mc =

1212 242 x 122 82 x 122 ( ) = 581760 12 8 2

Mc = f

581760 33000

= 17.62909091

TRY PAIR OF W8x40 S = 35.5 in3 I = 146 in4 CHK DEFLECTION AT CENTER : ALLOWABLE = ACTUAL =

24 x 12 = 1.2” 240

1212 12 x 29 x 106 x (146 x 2)

(

5 x 244 x 124 384

(

8 x 124 8

82 x 122 x 242 x 122 ) = 0.4986” < 1.2” 16

CHK DEFLECTION AT TIP : ALLOWABLE = ACTUAL =

8 x 12 360

= 0.27”

1212 12 x 29 x 106 x (146 x 2)

83 x 123 x 24 x 12 4

8 x 12 x 243 x 123 24

) = - 0.25” < 0.27”

USE PAIR OF W8X40 FOR G1 THE ALTERNATEIVE IS TO USE A LIGHTER “W” SECTION SUCH AS PAIR OF W8X28 AND INCREASE CANTILEVER LENGTH FROM 8’ TO 8.5’

Drawing #1 Floor Plan

Drawing #2 Interior perspective Collage

Model Photo

Space Problem 2 A PRIVATE ART GALLERY A collector of contemporary art (Cubism, to Abstract Expressionism and Pop Art) needs a space to exhibit and store the collection. From time to time the owner plans to use the facility to entertain guests that at times may number only a handful and sometimes as many as 30 persons. The facility should also accomodate overnight guests in a comfortable manner. The facility will be a standalone annex to the main residence nearby, both suitated on a slightly slopping (11% slope NW to SE), lightly wooded typically mid-western site.

STRUCTURAL CALCULATION

B2 G1

(1) ROOF FRAMING

B1 G1

LOAD TABLE ROOFING INSULATION METAL DECK ESTM WT OF STL CEILING MISC TOTAL D.L. L.L.

6’

2.0 2.0 6.5 9.0 5.0 2.5

Fy = 50 ksi fb = 0.66 x Fy = 33 ksi fv = 0.40 x Fy = 20 ksi E = 29 x 106

27.0 30.0

B1 B1

B1 B1 B1

3@18’ = 54’

57.0

6’ 6’

18’

6’

DESIGN B1 W = 57 x 6 ÷ 2 = 171 PLF m = 0.0479 x W x L2 = 0.0479 x

171 x 542 x 122 = 286616 12

1) TRY PAIR OF 14CS 3.75x135 S = 11.22 in3 I = 78.57 in4 CHK DEFLECTION : ALLOWABLE = ACTUAL =

m 286616 = = 8.69 in3 f 33000

2) TRY PAIR OF 16CS 3.75x090 S = 9.233 in3 I = 73.87 in4 CHK DEFLECTION

54 x 12 = 2.7” 240

0.00406 x 171 x 544 x 124 29 x 106 x 12 x (78.57 x 2)

: ALLOWABLE = = 2.24” < 2.7”

ACTUAL =

54 x 12 = 2.7” 240

0.00406 x 171 x 544 x 124 29 x 106 x 12 x (73.87 x 2)

= 2.381” < 2.7”

14CS 3.75x135 HAS [PSSIBILITY TO HAVE NOT ENOUGH DEPTH FOR ALL THE MECHANICAL SYSTEM. THUS, USE 16CS 3.75x090 FOR B1

66’

DESIGN B2 B2 BEAMS ONLY CARRY HALF OF THE LOAD ON B1. HOWEVER, FOR EASIER CONSTRUCTION USE SAME CS AS B1. IT COSTS MORE TO KEEP TRACKING OF DIFFERENT BEAM SECTIONS.

DESIGN G1 P = 5643 m1 =

PL = 6

5643 x 18x 12 6

= 203148

m2 =

Wa2 = 2

171 x 62 x 122 2

= 443232

m = m1 + m2 =646375 S=

m f

646375 33000

= 19.587 in3

TRY PAIR OF 16CS 3.75x090 S = 9.233 in3 I = 73.87 in4 CHK DEFLECTION : ALLOWABLE = ACTUAL 1 = ACTUAL 2 =

18 x 12 240

= 0.90”

0.0095 x P x L3 EI

0.0095 x 5643 x 183 x 123 29 x 106 x (73.87 x 2)

W 5 a2 x L2 ( x L4 ) EI 384 16

= 0.126095276”

171 29 x 106 x (73.87 x 2) x 12

(

5 x 184 x 124 384

62 x 122 x 182 x 122 16

) = 0.04399257”

TOTAL ACTUAL DEFLECTION = ACTUAL 1 + ACTUAL 2 = 0.170087846” < 0.9” G1 CAN BE SMALLER AND LIGHTER, BUT FOR EASIER CONSTRUCTION, USE SAME CS AS BEAMS.

(2) MEZZANINE FRAMING LOAD TABLE METAL DECK ESTM WT OF STL CEILING PARTITION MISC

8.0 6.5 50.0 10.0 2.5

TOTAL D.L. L.L.

32.0 40.0

Fy = 50 ksi fb = 0.66 x Fy = 33 ksi fv = 0.40 x Fy = 20 ksi E = 29 x 106

6’

18’

6’

18’

30’

6’ 6’

48’

72.0 PSF

DESIGN B1 AND B2 W = 72 x 6 ÷ 2 = 216 PLF m = 0.0335 x S=

m f

216 12

ACTUAL AT CENTER =

x 182 x 122 = 28134

28134 = 0.8525 in3 33000

CHK DEFLECTION : ALLOWABLE AT CENTER = ALLOWABLE AT TIP =

(

5 a 2L 2 L4 ) 384 16

216 29 x 106 x (11.05 x 2) x 12

(

5 62 x 122 x 182 x 122 x 184 x 124 384 16

= 0.371486297” < 0.9”

TRY PAIR OF 9CS 3x060 S = 2.443 in3 I = 11.05 in4

W EI

18 x 12 240 = 0.9”

6 x 12 = 0.3” 240

ACTUAL AT TIP = =

W a4 a3 x L ( + EI 8 4

a x L2 ) 24

216 29 x 106 x (11.05 x 2) x 12 x(

64 x 124 63 x 123 x 18 x 12 + 8 4

= - 0.188691452” < 0.3”

6 x 12 x 183 x 123 ) 24

)

DESIGN G1 AND G2 FIND W mc = 11688 = W x (

L2 8

a2 ) 2

W = 519 PLF m=Wx( S=

m f

L2 8 =

a2 182 ) = 519 x ( 2 8

62 ) = 11677.5 in4 = 140130 in4 2

140130 = 4.246 in3 33000

TRY PAIR OF 9CS 3x090 S = 7.22 in3 I = 32.48 in4 CHK DEFLECTION : ALLOWABLE AT CENTER =

18 x 12 = 0.90” 240

6 x 12 = 0.30” 240

ALLOWABLE AT TIP = ACTUAL AT CENTER =

W EI

(

5 a 2L 2 L4 )= 384 16

519 29 x 106 x (16.24 x 2) x 12

(

5 x 184 x 124 384

62 x 122 x 182 x 122 16

)

= 0.371486297” < 0.90” ACTUAL AT TIP =

W a4 a3 x L ( + EI 8 4

a x L2 519 )= 24 29 x 106 x (16.24 x 2) x 12

64 x 124 63 x 123 x 18 x 12 + 8 4

6 x 12 x 183 x 123 24

)

= - 0.188691452” < 0.30”

(3) DESIGN COL 1

COL HT = 50’

4 6’

LOAD ON COL = 2 x 11060 = 22120 PIN-PIN CONNECTION KL = 50 x 1.0 = 50

REF CHAPTER E AISC DETERMINE Cc FROM TABLE 4

3@18’ = 54’ C

Cc = 126.1 TRY 8” IN DIAMETER XS PIPE A = 12.76 KL r Fa =

1.0 x 50 x 12 2.88

12 x π2 x E KL 23 x r

= 208.3 > 126.1 12 x π2 x 29 x 106 23 x 208.32

6’ 6’

= 3441.7

18’

6’

ALLOW LOAD = 12.76 x 3441.7 = 43916.092 > 22060

(4) DESIGN COL WITH WIND LOAD LOAD ON CORNER COL

GRAVITY LOAD = 22060 WIND LOAD = 19250 TOTAL = 41310

TRY 8” IN DIAMETER XS PIPE A = 12.76 KL r Fa =

1.0 x 50 x 12 2.88

12 x π2 x E KL 23 x r

= 208.3 > 126.1 12 x π2 x 29 x 106 23 x 208.32

= 3441.7

ALLOW LOAD = 12.76 x 3441.7 = 43916.092 > TOTAL LOAD = 41310

66’

(5) WIND BRACING

4 6’

WIND LOAD = 20 PSF AREA OF C.W. = 40 x 66 = 2640 PSF TOTAL = 20 x 2640 = 52800 WIND LOAD ON ROOF AND SIL =

52800 = 26400 2

3@18’ = 54’

66’

IF WIND BRACING @ EACH PERIMETER COL LINES LOAD ON EACH SYSTEM = 26400 ÷ 2 = 13200 tan-1 c=

50 = 54.25 36 b cos

13860 = 23725 0.5842

6’ 6’

18’

6’

FIND VERTICAL COMPONENT OF TENSION LOAD

a = c x sin = 23725 x sin

= 19255 40’

DESIGN TENSION LOAD = 2 x ACTUAL LOAD

50’

c a

DESIGN LOAD = 2 x 23725 = 47450 = 47.5 kips = 23.75 tons USE 5/8” IN DIAMETER STRUCTURAL CABLE : ALLOW LOAD = 24.0 tons > 23.75 tons

10’ b

South Elevation

East Elevation

North Elevation

West Elevation

Drawing #1 Lower Level Floor Plan

Main Entrance Above

Mezzane 3

Section line 2

Mezzane 1

Lower Level Mezzane 1 Mezzane 2 Mezzane 3

Mezzane 2

Section line 1

a. Work Room b. Kitchen c. Dining d. Pantry e. General Storage f. Coat Room g. Mechanical Room h. Art Storage i. Guest Room j. Gallery

Drawing #2 Interior perspective Collage

Drawing #3 Section1

Drawing #4 Section2

Model Photo

Space Problem 3 LOW-RISE APARTMENT the application of the Space Problem to a multi-story apartment building in the urban environment.

CODE ANALYSIS

APPLIED CODE INTERNATIONAL BUILDING CODE

CONSTRUCTION TYPE TYPE 1 B

OCCUPANCY TYPE RESIDENTIAL GROUP R, R-2, APARTMENT

OCCUPANCY CLASSIFICATION LOAD FACTOR : 200 GROSS ...... 1 (Refer to Plan)

BUILDING LIMITATION HEIGHT(S) : 12 STORY WITH SPRINKLER SYSTEM ...... 2 (Refer to Plan) AREA(A) : UNLIMITED

FIRE RESISTANCE RATING BUILDING ELEMENT STRUCTURAL FRAME EXTERIOR NON-BEARING WALLS INTERIOR NON-BEARING WALLS TYPICAL DWELLING UNIT SEPARARION RESIDENTIAL CORRIDORS VERTICAL SHAFT ENCLOSURES EGREES STAIRS ELEVATORS TRASH CHUTE EXHAUST DUCT RISERS FLOOR CONSTRUCTION ROOF CONSTRUCTION OPENING PROTECTION 0HR - EXTERIOR WALL 2HR - SHAFT WALL 1HR - PARTITION 1/2HR - PARTITION

EXIT ARRANGEMENT TYPE 1B RATING 2HR 0HR 0HR 1HR 1/2HR 2HR 2HR 2HR 2HR 2HR 1HR UNLIMITED 1 1/2HR 45MIN 20MIN

RESIDENTIAL AT LESS THAN 500 OCCUPANCY LOAD MINIMUM # OF EMERGENCY EXIT : 2 ...... 3 (Refer to Plan)

EXIT TRAVEL DISTANCE GROUP R-2 MAXIMUM COMMON PATH OF EGREES TRAVEL DISTANCE : 250FT WITH SPRINKLER SYSTEM ...... 4 (Refer to Plan)

EXIT DISTANCE NOT LESS THAN 1/3 OF THE LENGTH OF THE MAX DIAGONAL DIMENSION ...... 5 (Refer to Plan)

VENTILATION REQUIREMENT 4% OF THE FLOOR AREA ...... 6 (Refer to Plan)

FLOOR AREA : 4 BAY UNIT 3820 sf 6 BAY UNIT 5767 sf

OCCUPANCY ...... 1 : 4 BAY UNIT 19.10 (3820 sf/200) 6 BAY UNIT 28.84 (5767 sf/200)

TOTAL STOREY :12 STOREY ...... 2

OPERABLE WINDOW ...... 6 : 1 AT EACH CENTER OF BAY 7’ BY 2’-4” OPERABLE AREA 424.67 sf 4% OF FLOOR AREA : 420.8 sf

EMERGENCY EXIT

133’ 133’/3 = 44.3’...5 2

45’-1/16”...5 MAIN ENTRANCE

MAIN ENTRANCE

EMERGENCY EXIT

EMERGENCY EXIT 3

134’-6”

85’-6” 4

Drawing #1 Typical Floor Plan without improvement

STRUCTURAL CALCULATION

DESIGN COL C1 AND C2 FOR GRAVITY LOADS

LOAD TABLE FIN FLOOR CONC DECK ESTN WT OF STL CEILING MISC

12.0 PSF 26.0 PSF 10.0 PSF 5.0 PSF 3.0 PSF

TOTAL D.L. L.L.

56.0 PSF 40.0 PSF 96.0 PSF

Fy = 50 ksi fb = 0.66 x Fy = 33 ksi fv = 0.40 x Fy = 20 ksi E = 29 x 106

LOAD ON C1 = 21 x 21 x 96 = 42.336 kips/FLR C2 = 10.5 x 21 x 96 = 21.168 kips/FLR

MAX LOAD ON TIER4 TIER3 TIER2 TIER1 COL SIZE

DESIGN B1 W = 7 x 96 = 672 PLF 2 2 2 m = WL = 672 x 21 x 12 8 8 x 12

= 444528

KL = 12’ KL = 12’ KL = 12’ KL = 12’

TIER4 TIER3 TIER2 TIER1

C1 127K 254K 381K 508K

C2 63.5K 127K 190.5K 254K

C1 W10x33 W10x45 W10x60 W12x87

C2 W10x33 W10x33 W10x33 W12x87

444528 S= = 13.47 in3 33000 TRY W8x28 S = 24.3 in3 I = 98.0 in4 CHK DEFLECTION SPAN 21 x 12 : ALLOWABLE = = = 1.05” 240 240 ACTUAL =

5WL4 5 x 672 x 214 x 124 = = 1.03” < 1.05” 384EI 384 x 29 x 106 x 98 x 12

DESIGN G1 EQUIV W = 1.7778P = 1.7778 x 14112 = 1194.68 PLF L 21 Mc = a x P x L = 0.1111 x 14112 x 21 x 12 = 395096 395096 Sc = = 11.9726 in3 33000 Ms = b x P x L = 0.2222 x 14112 x 21 x 12 = 790193 790193 Ss = = 23.95 in3 33000 TRY W12x35 S = 45.6 in

I = 285.0 in

CHK DEFLECTION SPAN 21 x 12 : ALLOWABLE = = = 1.05” 240 240 e x P x L3 0.0077 x 14112 x 213 x 123 ACTUAL = = = 1.03” < 1.05” EI 29 x 106 x 285

RIGID FRAMING CHK 1. SHEAR CHK WIND FORCE = 21 x 146 x 22 = 67452 AREA OF W12x87 = 25.6 in2, 25.6 x 3(# of COL) = 76.8 in2 76.8 x fy = 76.8 x 20 = 1536000 > 67452 2. MOMENT CHK MOMENT DUE TO WIND 146 146 x WIND FORCE = x 67452 = 4923996 lb*ft 2 2

V1 = 21 x 22PSF x (10+1’-10”) = 2.73K 2 x 1000 V2 = V1 + 21 x 22PSF x 11.83 = 8.20K 1000 V3 = V2 + 5.47K = 13.67K V4 = V3 + 5.47K = 19.14K V5 = V4 + 5.47K = 24.61K V6 = V5 + 5.47K = 30.08K V7 = V6 + 5.47K = 35.55K V8 = V7 + 5.47K = 41.02K V9 = V8 + 5.47K = 46.49K V10 = V9 + 5.47K = 51.96K V11 = V10 + 5.47K = 57.43K V12 = V11 + 21 x 22PSF x ( 11.83 + 18.83 ) / 1000 = 64.51K 2 2 INCREASE TIER SIZE FOR WIND FORCE TIER4 = W10x60 I = 341in4 = 0.0164ft4 TIER3 = W12x87 I = 740in4 = 0.0357ft4 TIER2 = W14x120 I = 1380in4 = 0.0667ft4 TIER1 = W14x211 I = 2660in4 = 0.1283ft4 USE COMPOSITE W-SHAPE GIRDER FOR WIND FORCE FROM 8TH TO ROOF W12x30 I = 900in4 = 0.0434ft4 FROM 3RD TO 7TH W12x72 I = 2266in4 = 0.1094ft4 2ND FLOOR W18x76 I = 5054in4 = 0.2437ft4

TOTAL DEFLECTION

MOMENT AT COLUMN 25.6 x 33000 x 42 = 35481600 > 4923996

3. DEFLECTION CHK Vh2 ( 12

∑EI

) <

1 x TOTAL HEIGHT 500 +

DEFLECTION 1 = 0.0441in DEFLECTION 2 = 0.1326in DEFLECTION 3 = 0.2210in DEFLECTION 4 = 0.2260in DEFLECTION 5 = 0.2906in DEFLECTION 6 = 0.3552in DEFLECTION 7 = 0.3585in DEFLECTION 8 = 0.4139in DEFLECTION 9 = 0.4690in DEFLECTION 10 = 0.4748in DEFLECTION 11 = 0.5248in DEFLECTION 12 = 1.3343in

TOTAL DEFLECTION = 2.6987in

FINAL STRUCTURAL MEMBER DETERMINATION B1 AND B2 : W8x28 G1 : 2ND FLOOR FROM 3RD to 7TH FROM 8TH to ROOF C : TIER4 W10x60 TIER3 W12x87 TIER2 W14x120 TIER1 W14x211

21’-0”

W18x76 W12x72 W12x30

A 1

7’-0” 21’-0”

B1 G1

7’-0” 7’-0” 2

7’-0” 21’-0”

21’-0” B

B1 B1

7’-0” 3

21’-0” D

B2 C2

B2 B1

7’-0”

21’-0” C

B1 B1

B1 G1

B1 B1

L B2

B1 G1

B1 B1

B2 C2

21’-0” K

B2 C2

21’-0” J

B2 C2

21’-0” I

B2 C2

21’-0” H

B2 C2

21’-0” G

B2 C2

21’-0” F

B2 B1

21’-0” E

B1 B1

B1 G1

B1 B2

Drawing #2 Typical framing plan

V2 11’-10”

10’-0”

10’-0” 1’-10”

11’-10” V5

10’-0” 1’-10”

TIER3 FROM 7TH FLOOR TO 9TH FLOOR

11’-10”

10’-0” 1’-10”

11’-10” 148’-6”

10’-0” 1’-10”

11’-10” V8

10’-0” 1’-10”

TIER2 FROM 4TH FLOOR TO 6TH FLOOR

11’-10”

10’-0” 1’-10”

11’-10” V10

10’-0” 1’-10”

11’-10” V11

V12

1’-10” 11’-10”

10’-0” 1’-10”

TIER4 FROM 10TH FLOOR TO 12TH FLOOR

1’-10” 11’-10”

10’-0” 1’-10”

TIER1 FROM 1ST FLOOR TO 3RD FLOOR

Drawing #3 Transverse Section

11’-10”

10’-0”

18’-6”

16’-0”

2’-6”

Drawing #4 Longitudinal Section

MECHANICAL SYSTEM ANALYSIS

Heating and Cooling Radient floor and ceiling panel (diagram only shows two floors)

Bathroom and Kitchen Exhaust 10” by 6” steel exhaust, waste goes to main mechanical room in the basement (diagram only shows one typical example)

Supply water pipes 1” in diameter pipe each for hot water and cold water from main core supplies water from main mecanical room in the basement

Sprinkler pipes two main branch from center core, 2” in diameter with less than 1/8” slope supplies water from main mechanical room in the basement (diagram only shows one typical example)

Waste pipes 4” in diameter free standing riser and 1/8” slope at 2nd floor waste goes to main mechanical room in the basement

Roof drainage 1 above each unit and 1 above main core

Drawing #5 Vertical Section at Column Line1 and 3

Drawing #6 Typical unit floor plan

Drawing #7 Ground floor plan

Drawing #8 Typical unit Elevation

SLAB THICKNESS 1’-10” FLOOR TO CEILING HEIGHT 11’-10”

10’-0”

1’-10” FLOOR TO CEILING HEIGHT 11’-10”

10’-0”

1’-10”

11’-10”

10’-0”

1’-10”

10’-0”

11’-10”

1’-10”

11’-10”

10’-0”

1’-10” TOTAL BUILDING HEIGHT 148’-6”

11’-10”

10’-0”

1’-10”

11’-10”

10’-0”

1’-10”

10’-0”

11’-10”

1’-10”

11’-10”

10’-0”

1’-10”

11’-10”

10’-0”

1’-10”

11’-10”

10’-0”

2’-6”

FLOOR TO FLOOR HEIGHT AT GROUND FLOOR LEVEL 18’-6”

16’-0” FLOOR TO CEILING HEIGHT AT GROUND FLOOR LEVEL

Drawing #9 and #10 Elevations

Drawing #11 Interior perspective Collage

Drawing #12 Interior perspective Collage

Model Photo of 6 bay unit

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101

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Model Photo of 6 bay unit

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Model Photo of 6 bay unit

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105

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Model Photo of 6 bay unit

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05 Extra works

Free hand drawing

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Free hand drawing #1

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Guggenheim Museum

Free hand drawing #2

Schroeder House

Free hand drawing #3

La chapelle de Ronchamp

Free hand drawing #4

Seagram Building

Free hand drawing #5

Robie House

Free hand drawing #6

House of Parliament London

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Free hand drawing #7

Venice

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Free hand drawing #8

Water color painting

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Contact info. T. +852 2908 4050 / +852 2528 3031 C. +852 9787 2229 E. churl-jong.kim@arup.com / kimcj2190@gmail.com

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To discover more publications of Churl jong Kim Visit

1. Work Samples https://issuu.com/churljongkim/docs/worksamples-1 https://issuu.com/churljongkim/docs/worksamples-2 2. Undergraduate work portfolio https://issuu.com/churljongkim/docs/portfolio-undergraduate 3. Graduate work portfolio https://issuu.com/churljongkim/docs/portfolio-graduate 4. Master Thesis Full text http://share.iit.edu/handle/10560/3439

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