jörg schlaich
TRANSCRIPT
Jörg SchlaichSchlaich Bergermann und Partner, Stuttgart
The challenge and Joy of STrucTural engineeringthurSday, OctOBer 4, 2012
inaugural edward and Mary allenlecTure in STrucTural deSign
SchOOl Of architecture + Planning
2
edward and Mary allen lecTure in STrucTural deSign
Established in 2012, the annual Edward and Mary Allen Lecture aims to bring
the world’s leading structural designers to MIT to speak about engineering
within the realm of architecture, design, and creativity and to interact closely
with students in their design work. The program includes a public lecture
in the Architecture Lecture Series as well as a workshop reviewing graduate
student structural design projects.
For the inaugural Allen Lecture on October 4, 2012, the eminent German
structural engineer, educator, and designer Jorg Schlaich visited MIT to
speak about his decades of inspired design work, including long-span roofs,
innovative bridges, and multi-story facades, as well as his more recent
efforts in solar energy. Along with Edward Allen, Schlaich also worked with
graduate students in architecture and engineering on their designs for an
enclosed skybridge over Vassar Street on the MIT campus, a project which
he had also developed conceptual designs for several years prior.
This booklet commemorates Schlaich’s visit with photographs of his lecture
and from his review of student projects, and also contains a survey of the
student structural design work.
3
Jörg Schlaich
Prof. Dr. Ing. Drs. h.c. Jorg Schlaich is a German structural engineer. He
studied architecture and civil engineering from 1953 to 1955 at Stuttgart
University before completing his studies at the Technical University of
Berlin in 1959 (Dr. Ing.). He spent 1959-60 at the Case Western Reserve
University in Cleveland, Ohio, USA (MSc.).
Jorg Schlaich was made a partner of the structural engineers, Leonhardt und
Andra and was responsible for the Olympic Stadium Roof, Munich in 1968-
72.
From 1974 to 2000 he was full professor and director of the Institute for
Concrete Structures (Institut fur Massivbau), later called the Institute for
Structural Design (Konstruktion und Entwurf) at the University of Stuttgart.
In 1980 he and his partner, Dr. Rudolph Bergermann, founded their own
firm, Schlaich Bergermann und Partner in Stuttgart.
Most of his projects, as well of that of his company with offices in Stuttgart,
Berlin, New York, Sao Paulo and Shanghai, are documented on their website
(www.sbp.de). The work of Schlaich Bergermann und Partner focuses on the
three main themes of building, surveying and solar energy.
Jorg Schlaich is also the developer of the Solar Updraft Tower for large
scale solar energy generation. He is credited with advancing the strut and
tie model for reinforced concrete with his seminal 1987 paper, “Toward a
Consistent Design of Structural Concrete” in PCI Journal.
Jörg Schlaich with Caitlin Mueller, Edward Allen, and
John Ochsendorf
4
5
ABOVE: Jörg Schlaich with Edward AllenLEFT: Jörg Schlaich presenting during his
public lecture
6
ABOVE: Jörg Schlaich talking to students about bridge designRIGHT: Jörg Schlaich with John Ochsendorf
7
8
9
ABOVE: Jörg Schlaich with graduate students in architecture
and engineeringLEFT: Jörg Schlaich with John
Ochsendorf and Edward Allen
10
ABOVE AND RIGHT: Jörg Schlaich reviewing student projects with Edward Allen and John Ochsendorf
11
12
13
14
ABOVE: Jörg Schlaich with Edward Allen and John Ochsendorf at the end of the student project reviewRIGHT: Two skybridge designs proposed by Schlaich Bergermann und Partner for the MIT campus
15
STudenT deSign worK
The following pages showcase the projects reviewed by Jorg Schlaich during his visit to MIT. The student design work comes from a second-year structural design course in the Masters of Architecture program at MIT, taught by John Ochsendorf and Andrea Love with assistance from Caitlin Mueller. Each presentation proposes a conceptual structural design developed over the course of four weeks for an enclosed skybridge connecting two existing buildings on the MIT campus across Vassar Street.
Before the review, Jorg Schlaich presented the students with two of his own designs for the same project developed several years earlier.
16
In devising a scheme for a skybridge between buildings 36 and 46 our main architectural objective was to design a visually minimal bridge that did not compete with the surrounding buildings but was still interesting in its own right. The bridge bends in plan to allow for views of the Stata Center and the envelope is cladded in folded plates of glass. After several iterations, we decided upon a cable stayed structural system anchored in both buildings.
ant bridgeSuk Lee & Susanna Pho
46.80 k208.18 kn
23.29 k103.60 kN
19.35 k86.07 kn
42.64 k189.67 kn
109.20 k485.75 kn
10.28 k45.73 kn
89.94 k400.07 kn
21.43 k95.33 kn
20.79 k92.48 kn
19.76 k87.90 kn
17.68 k78.64 kn
74 ft
h=varies
b= 8in
7.4 ft
89.94 k
a) Maximum Tension force on Tension Rods: 12.28 k or 54.62 kn Diameter = 0.5 in, Stainless Steel Rod with allowable force: 12.56k > 12.28k
b) Arequired
= F/σ = 12.28k / 15ksi = 0.82 in2 >> Diameter = 1.02 in or 25.91 mm
Stainless Steel Rods with 1.25in or 31.75 mm Diameter will be used to carry tension force>> >>
Stainless Steel Rod is used instead of tension cable to reduce the amount of possible sag
1.25 in
M NLKJIHGFEDCBAA’
1’ 1O 2 3 4 5 6 7 8 9 10 11 13 142’4’5’6’
B’C’D’E’F’
123’
B
B’
A’
O
C
C’
D’
E’
F’
D
E
F
G
H
I
J
K
L
M
N
A 1
2’
1’
3’
4’
5’
6’
2
3
4
5
6
7
8
9
10
11
12
13
14
1
Maximum Tension Member B
Maximum Tension Member A
O
3’
4’
: 7.8 k
Axial l
oad conditions
forc
e polyg
on
load condit
ions
GIRDER SIZING
cable SIZING
Bridge B
Bridge A
Material Properties:concrete deck: 40 lbs/ft2
structural glass: 35 lbs/ft2
tributary area: 5 ft or 1.52 m
Uniform Live Load: 9ft x (80 lbs/ft2 + 30 lbs/ft2) =990 lbs/ftStructure Dead Load: (1041 lbs/ft x 5ft) + ( 1090 lbs/ft x 5)
DEFLECTION ANALYSISIncreased overall depth of beam to prevent deflection.
AXIAL LOAD ANALYSISINFORMED REQUIRED DEPTH OF BEAM
Crushing:A = F/σ = 89.94 k / 3 ksi = 29.97 in2
h = 3.74 in or 95 mm
Buckling:I = ((kL)2 x P
cr )/ ( π2E)
= (((74ft x 12in))2 x 89.94 k)/(π2 x 3,000 ksi) = 2395.29 in4
bhv/12 = 2395.29 in4 (b=width=8 in)h = 15.32 in or 389.19 mm
Moment:M= wL2/16 = (3.12 k/ft)(7.4 ft)2/16 = 10.68 k-ft = 128.16 k-inS = M/σ = 128.16 k-in / 3 ksi = 42.72 in3
S = I/y = 6/h3
h = 0.52 in 13.20 mm
Point Load: (990 lbs/ft x 5 ft) + (2131 lbs/ft x 5 ft) = 15.6 k
Point Load On One Side of Structure:15.6 k / 2 = 7.8 k or 34.7 kn
DEFLEC
TION
CROSS SECTION
CABLE-DECK CONNECTION DETAIL
EXPOSED EMBEDDED STEEL PLATECustomized to act as cable
anchoring system
STEEL REINFORCEMENT
17
In devising a scheme for a skybridge between buildings 36 and 46 our main architectural objective was to design a visually minimal bridge that did not compete with the surrounding buildings but was still interesting in its own right. The bridge bends in plan to allow for views of the Stata Center and the envelope is cladded in folded plates of glass. After several iterations, we decided upon a cable stayed structural system anchored in both buildings.
ant bridgeSuk Lee & Susanna Pho
46.80 k208.18 kn
23.29 k103.60 kN
19.35 k86.07 kn
42.64 k189.67 kn
109.20 k485.75 kn
10.28 k45.73 kn
89.94 k400.07 kn
21.43 k95.33 kn
20.79 k92.48 kn
19.76 k87.90 kn
17.68 k78.64 kn
74 ft
h=varies
b= 8in
7.4 ft
89.94 k
a) Maximum Tension force on Tension Rods: 12.28 k or 54.62 kn Diameter = 0.5 in, Stainless Steel Rod with allowable force: 12.56k > 12.28k
b) Arequired
= F/σ = 12.28k / 15ksi = 0.82 in2 >> Diameter = 1.02 in or 25.91 mm
Stainless Steel Rods with 1.25in or 31.75 mm Diameter will be used to carry tension force>> >>
Stainless Steel Rod is used instead of tension cable to reduce the amount of possible sag
1.25 in
M NLKJIHGFEDCBAA’
1’ 1O 2 3 4 5 6 7 8 9 10 11 13 142’4’5’6’
B’C’D’E’F’
123’
B
B’
A’
O
C
C’
D’
E’
F’
D
E
F
G
H
I
J
K
L
M
N
A 1
2’
1’
3’
4’
5’
6’
2
3
4
5
6
7
8
9
10
11
12
13
14
1
Maximum Tension Member B
Maximum Tension Member A
O
3’
4’
: 7.8 k
Axial l
oad conditions
forc
e polyg
on
load condit
ions
GIRDER SIZING
cable SIZING
Bridge B
Bridge A
Material Properties:concrete deck: 40 lbs/ft2
structural glass: 35 lbs/ft2
tributary area: 5 ft or 1.52 m
Uniform Live Load: 9ft x (80 lbs/ft2 + 30 lbs/ft2) =990 lbs/ftStructure Dead Load: (1041 lbs/ft x 5ft) + ( 1090 lbs/ft x 5)
DEFLECTION ANALYSISIncreased overall depth of beam to prevent deflection.
AXIAL LOAD ANALYSISINFORMED REQUIRED DEPTH OF BEAM
Crushing:A = F/σ = 89.94 k / 3 ksi = 29.97 in2
h = 3.74 in or 95 mm
Buckling:I = ((kL)2 x P
cr )/ ( π2E)
= (((74ft x 12in))2 x 89.94 k)/(π2 x 3,000 ksi) = 2395.29 in4
bhv/12 = 2395.29 in4 (b=width=8 in)h = 15.32 in or 389.19 mm
Moment:M= wL2/16 = (3.12 k/ft)(7.4 ft)2/16 = 10.68 k-ft = 128.16 k-inS = M/σ = 128.16 k-in / 3 ksi = 42.72 in3
S = I/y = 6/h3
h = 0.52 in 13.20 mm
Point Load: (990 lbs/ft x 5 ft) + (2131 lbs/ft x 5 ft) = 15.6 k
Point Load On One Side of Structure:15.6 k / 2 = 7.8 k or 34.7 kn
DEFLEC
TION
CROSS SECTION
CABLE-DECK CONNECTION DETAIL
EXPOSED EMBEDDED STEEL PLATECustomized to act as cable
anchoring system
STEEL REINFORCEMENT
18
SECTION PROPERTIES / SELF WEIGHT
M =
180 kips [800.69 kN]
1.8 kips [0.026 kN/m]
force polygon
from the cable sizing table:
COMBINED STRESS
BUCKLING
1-7/8” [47.625 mm] diameter,class “A” coating throughout(breaking strength: 216.0 tons)
SIZING THE CABLE
UNIFORM LOADING
SIZING THE GIRDER/STRUT
free body diagram
187.8 kips [835.38 kN]
187.8 kips [835.38 kN]
floor dead load: 0.244 kip/ft [0.004 kN/m]floor live load: 0.800 kip/ft [0.012 kN/m]roof dead load: 0.329 kip/ft [0.005 kN/m]
roof live load: 0.300 kip/ft [0.004 kN/m]cladding: 0.124 kip/ft [0.002 kN/m]TOTAL: 1.800 kip/ft [0.026 kN/m]
LOADS
LOAD CASE 1
110 ft [33.53 m]
187. kips = 93.9 tons93.9 tons2.2 s.fx
M = C x dC = 1.89 kipsCmax = 181.89 kips
A = = 12.19 in2 [0.008 m2]
d = .75 ft
σsteel = 15 ksi
= 1.42 k·ft
distributed load on girder = (0.329 + 0.3) / 2 = 0.4645 kip/ft
assume for full unbraced length
Pcr =
7 ft [2.133 m]
0.46 kip/ft
3.25 kip/ft3.25 kip/ft
7 ft [2.133 m]
180 kips
0.46 kip/ft
.75 ft[0.23 m]
1.89 kips
assume 18 inch strut depth10 ft [3.048 m]
2 ft [0.61 m]
180 kips
wL216
Fσ
π2E(I)(k L)2
I = 1096 in4 [45,618.97cm4]
180 kips = π2(29,000 ksi)(I)(1·110 ft·12 in)2
47.6 mmfrom the hollow steel section table:
ASYMMETRICAL LOADING
1.8 kips [0.026 kN/m]
SIZING THE GIRDER
DEFLECTION
free body diagram
LOAD CASE 2 DETAIL
assume cable carries no stiffnessassume vertical struts are infinitely stiff/transfer deformations from deck to top girder
Mmax =
Mmax = 340 kip·ft [460978.1 Nm]
=wL264
(1.8 kip/ft)(1102)64
5wL4
384E∆max
Ireq ≥ 34,000 in4
Ireq ≥
5(0.09 k/in)4
384(29000 k/in2)(3.7in)≥
Length = 1320 inWeight = 1100 lb/ft = 0.09 k/inEsteel = 2900 k/in2
check via multiframe:
I = ∑Aidi2
= (2·30in2·(52in)2)+(2·2.76in2·(0in)2)+(2·20in2·(78in)2)I = 405,600 in4 ≥ 34,000 in4 :) [ 0.17 m4 ≥ 0.01 m4 ]
σ =Ms
340 kip·fts
12 in1 ft
S = 136 in3 [2000 cm3]
20 in x 8 in x 5/8 in[508 mm x 203.2 mm x 15.9 mm]
A = 30.3 in2
[0.02 m2]I = 1440 in4
[0.0006 m4]S = 144 in3
[2000 cm3]
∆max = 3.7 in
∆ = 3.46 in ≤ 3.7 in [0.087 m ≤ 0.093 m]
508
mm
203.2 mm
15.9 mm
y=0
1.8 kips [0.026 kN/m]
= x 12
x
from the hollow steel section table:
SIZING THE SECONDARY STRUT
BUCKLING
CRUSHING
A = = 0.48 m2Fσ
Pcr =π2E(I)
(k L)27,150 lbs =
I = 0.230 in4 [9.57 cm4]
π2(29 x 107 psi)(I)
(96 in)28 ft
7.15 kips
1½ in x 1½ in x 3/16 in[38.1 mm x 38.1 mm x 4.76 mm]
A = 0.84 in2
[5.42 cm2]I = 0.235 in4
[9.78 cm4]
38.1
mm
38.1 mm
4.76 mm
3550
mm
3050mm 32500 mm
30 in2
.02 m2
52 in2
.03 m2
206.58 tons [1837 kN]
20 in2
.02 m2
2.76 in2
17.81 cm2
6.35
50.8
0
6.35
203.
20
R152.40
12.5
152.40
12.50
50.8
0
6.35
38.1
0
152.40
4.76
38.10
28.54
508.
00
254.00
15.8
8
ROOF SYSTEM
FACADE SYSTEM
FLOOR SYSTEM
Component BreakdownCable Connection
Elevation
Section
1/4” Glass Roof
1/4” Glass Floor
2” Steel Diagrid Mesh
2” Steel Diagrid Mesh
20 x 8 x 5/8 HSS
1.5 x 1.5 x 3/16 HSS
1/4” Glass Cladding
1 7/8” Steel Cable
34 lb/ft51 kg/m
150 lb/ft223 kg/m
160 lb/ft238 kg/m
110 lb/ft164 kg/m
34 lb/ft51 kg/m
7.39 lb/ft11 kg/m
34 lb/ft51 kg/m
FLOATING FOOT BRIDGE RENA YANG & YAN-PING WANG
19
SECTION PROPERTIES / SELF WEIGHT
M =
180 kips [800.69 kN]
1.8 kips [0.026 kN/m]
force polygon
from the cable sizing table:
COMBINED STRESS
BUCKLING
1-7/8” [47.625 mm] diameter,class “A” coating throughout(breaking strength: 216.0 tons)
SIZING THE CABLE
UNIFORM LOADING
SIZING THE GIRDER/STRUT
free body diagram
187.8 kips [835.38 kN]
187.8 kips [835.38 kN]
floor dead load: 0.244 kip/ft [0.004 kN/m]floor live load: 0.800 kip/ft [0.012 kN/m]roof dead load: 0.329 kip/ft [0.005 kN/m]
roof live load: 0.300 kip/ft [0.004 kN/m]cladding: 0.124 kip/ft [0.002 kN/m]TOTAL: 1.800 kip/ft [0.026 kN/m]
LOADS
LOAD CASE 1
110 ft [33.53 m]
187. kips = 93.9 tons93.9 tons2.2 s.fx
M = C x dC = 1.89 kipsCmax = 181.89 kips
A = = 12.19 in2 [0.008 m2]
d = .75 ft
σsteel = 15 ksi
= 1.42 k·ft
distributed load on girder = (0.329 + 0.3) / 2 = 0.4645 kip/ft
assume for full unbraced length
Pcr =
7 ft [2.133 m]
0.46 kip/ft
3.25 kip/ft3.25 kip/ft
7 ft [2.133 m]
180 kips
0.46 kip/ft
.75 ft[0.23 m]
1.89 kips
assume 18 inch strut depth10 ft [3.048 m]
2 ft [0.61 m]
180 kips
wL216
Fσ
π2E(I)(k L)2
I = 1096 in4 [45,618.97cm4]
180 kips = π2(29,000 ksi)(I)(1·110 ft·12 in)2
47.6 mmfrom the hollow steel section table:
ASYMMETRICAL LOADING
1.8 kips [0.026 kN/m]
SIZING THE GIRDER
DEFLECTION
free body diagram
LOAD CASE 2 DETAIL
assume cable carries no stiffnessassume vertical struts are infinitely stiff/transfer deformations from deck to top girder
Mmax =
Mmax = 340 kip·ft [460978.1 Nm]
=wL264
(1.8 kip/ft)(1102)64
5wL4
384E∆max
Ireq ≥ 34,000 in4
Ireq ≥
5(0.09 k/in)4
384(29000 k/in2)(3.7in)≥
Length = 1320 inWeight = 1100 lb/ft = 0.09 k/inEsteel = 2900 k/in2
check via multiframe:
I = ∑Aidi2
= (2·30in2·(52in)2)+(2·2.76in2·(0in)2)+(2·20in2·(78in)2)I = 405,600 in4 ≥ 34,000 in4 :) [ 0.17 m4 ≥ 0.01 m4 ]
σ =Ms
340 kip·fts
12 in1 ft
S = 136 in3 [2000 cm3]
20 in x 8 in x 5/8 in[508 mm x 203.2 mm x 15.9 mm]
A = 30.3 in2
[0.02 m2]I = 1440 in4
[0.0006 m4]S = 144 in3
[2000 cm3]
∆max = 3.7 in
∆ = 3.46 in ≤ 3.7 in [0.087 m ≤ 0.093 m]
508
mm
203.2 mm
15.9 mm
y=0
1.8 kips [0.026 kN/m]
= x 12
x
from the hollow steel section table:
SIZING THE SECONDARY STRUT
BUCKLING
CRUSHING
A = = 0.48 m2Fσ
Pcr =π2E(I)
(k L)27,150 lbs =
I = 0.230 in4 [9.57 cm4]
π2(29 x 107 psi)(I)
(96 in)28 ft
7.15 kips
1½ in x 1½ in x 3/16 in[38.1 mm x 38.1 mm x 4.76 mm]
A = 0.84 in2
[5.42 cm2]I = 0.235 in4
[9.78 cm4]
38.1
mm
38.1 mm
4.76 mm
3550
mm
3050mm 32500 mm
30 in2
.02 m2
52 in2
.03 m2
206.58 tons [1837 kN]
20 in2
.02 m2
2.76 in2
17.81 cm2
6.35
50.8
0
6.35
203.
20
R152.40
12.5
152.40
12.50
50.8
0
6.35
38.1
0
152.40
4.76
38.10
28.54
508.
00
254.00
15.8
8
ROOF SYSTEM
FACADE SYSTEM
FLOOR SYSTEM
Component BreakdownCable Connection
Elevation
Section
1/4” Glass Roof
1/4” Glass Floor
2” Steel Diagrid Mesh
2” Steel Diagrid Mesh
20 x 8 x 5/8 HSS
1.5 x 1.5 x 3/16 HSS
1/4” Glass Cladding
1 7/8” Steel Cable
34 lb/ft51 kg/m
150 lb/ft223 kg/m
160 lb/ft238 kg/m
110 lb/ft164 kg/m
34 lb/ft51 kg/m
7.39 lb/ft11 kg/m
34 lb/ft51 kg/m
FLOATING FOOT BRIDGE RENA YANG & YAN-PING WANG
20
GLOBAL SHAPE: PRIMARY MEMBER SIZING:DESIGN APPROACH:: ASYMMETRICAL LOADING CASE: using force polygon
Final Force Polygon
Previous Iterations
(d)
(a)
(b)
(c)
(d)
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
s
(a)
(a)
(b)(c)
(d)
(b)
(c)
107’ = 32.1 m
edge of verticalsupport zone edge of sidewalk
10 KIPS=44.5 KN
AB
CDEFGHI
JK
L
MN
O
O’
P
Q
R
S
T
20’45’
30’=9m
horizontal force = 56.3 KIPS = 250.5 kN
10 K
IPS=
44.5
kN
1’ = 1 KIP
Fmax = 128 KIPS= 569.6 kN
113.8
KIP
S
o’
o’
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
s
t
LOADS:
UNIIFORM LOADING CASE (w):
people
snow
80 lbs/ft2
110 lbs/ft2
30 lbs/ft2
Live Load
steel
concrete
glass
226 lbs/ft2
336 lbs/ft2
=16.6 kN/m2
(density of steel= 490 lbs/ft3)assuming that we use 2” of steel (1/6 of a foot), » 490 lbs/ft3 • 1/6 ft = 82 lbs/ft2
(density of concrete= 150 lbs/ft3)assuming that we use 6” of cocnrete (1/2 of a foot), » 150 lbs/ft3 • 1/2 ft = 75 lbs/ft2
(density of glass= 162 lbs/ft3)assuming that we use 2 of 2” of steel (1/3 of a foot), » 162 lbs/ft3 • 1/3 ft = 54 lbs/ft2
extra
Dead Load
15 lbs/ft2
σ = P/A , where steel allowable stress = 15ksi» 15 ksi = 128 kips/A » A=8.53 in2 » b=h= 3 in = 25.4 mm
due to axial force
Live Load+ Dead Load = 336 lbs/ft2
336 lbs/ft2 • 10ft = 3360 lbs/ft = 3.36 kips/ft
3.36 kips/ft for two arches = 1.68 kips/ft for 1 arch
1.68 kips/ft • 107 ft = 179.76 kips
» 179.76 kips/19 segments = 9.46 kips/seg
Dead Load = 226 lbs/ft2
» 226 lbs/ft2 • 10ft = 2260 lbs/ft » 2.26 kips/ft
Live Load = 110 lbs/ft2
» 110 lbs/ft2 • 10ft • 107ft = 117.7 kips » 117.7 kips / 92 ft= 1.28 kips /ft
arch-in-motion Maya Taketani / Jasmine Kwak
4.463 Building Systems II
simplified diagram of design proposal
15 feet
30 feet
10 feet
92 feet
steel columns
structural glass
107 feet
concrete deck as a horizontal tie
steel cables
steel arch SECONDARY MEMBER SIZING:
MAXMIMUM MOMENT DUE TO ASYMMETRICAL LOADING:
DEFLECTION DUE TO ASYMMETRICAL LOADING
σ= F/A» 15 ksi = 10 kips / A » A=0.66 in2
» A=0.66 in2=π(r)2
» r= 0.46 in » approx. r= 1/2 in = 12.7 mm MMAX= (wLL• L
2)/32 » where wLL= 0.11 kips/ft2 •10 ft /2 arch= 0.55 k/ft » 0.55k/ft • (92ft)2/32 » MMAX= 145.5 kip•ft = 194 kN•m
T=C=M/d » T= 145.5 kip•ft•(12in/ft) / 6 in » 291 kips
σ= F/A » 15 ksi = 291 kips / A » A= 19.4 in2
» d= 4 in = 101.6 mm
T=C=M/d » T= 145.5 kip•ft•(12in/ft) / 18 in » 97 kips
Pbuckling=(π2EI)/(KL)2
» 97 kips•3= (π2•29000 k/in2•Ireq) (112 ft•12ft/in)2
» Ireq= 1836.5 in4
σ= F/A » 15 ksi = 97 kips / A » A= 6.5 in2
Steel Cables in Tension
σ= F/A» 15 ksi = 56.3 kips / A » A=3.75 in2
» I-beam S6x17.25
I-beam decking in Concrete Decking in Tension
bridge
10 feet
336 lb/ft2
336 lb/ft
bridge
336 lb/ft 168 lb/ft
10 kips / ea
due to crushing
due to buckling
Steel Column in Compression
σ= F/A» 15 ksi = 10 kips / A » A=0.67 in2 » b=0.82 in
Pcr= (π2EI)/(kL)2
» 3 •10 kips = [(π2)(29000kips/in2)(Irequired)]/ [(24.25 ft)(12in/1ft)]2 » Irequired = 8.88 in4, where I= bh3/12» for a square section column, b= 4.74 in » b=4 in. or 101.6mm and h=3 in. 76.2 mm
I= bh3/12» I= 108/12= 9 in4 Pcr= (π2EI)/(kL)2
» 3 •10 kips = [(π2)(29000kips/in2)(9 in4)]/ [L in]2
» L < 293 inches = 24 ft = 7.3 m
226 lb/ft
bridge
92 feet
110 lb/ft
bridge
226 lb/ft
110 lb/ft
1”
3”
3”
4”
6”
18”
5/8”
20”
18”
5/8”
20”
20”
5/8”
22”
3-5/8”
6”
7/16”
5”
16.8 kips each
0.9560.6050.458
0.3650.293
0.23
0.172
0.115
0.058
2.6441.449
0.9810.508
1.961
3.987 3.5762.3730.859
0.5430.411
0.3280.263
0.207
0.154
0.103
0.052
1.761.3
0.8810.456
OR
When we used HSS20X18X5/8 as sectional propertiesin Multiframe software to calculate, the highest distance of deflection was 3.987 inches as shown in the graphon the left. Therefore, we had to increase the sectional dimension in order to accomodate ∆max≤ 3.7 in, given ∆max≤ L/360 = 110 ft/360. The new max deflection is 3.576 inches as shown on the right graphand new section is HSS22X20X5/8
street perspective elevation and material
plan
peoplesnow
steel concrete
glass extra
Vassar Street is one of MIT's prominent streets located on the north edge of MIT campus. Nearby the site is Frank Gehry's Stata Center and Charles Correa's Brain and Cognitive Sciences Building. Despite many well known buildings, Vassar Street lacks a visual key urban element that brings people to this street. With these observations, this proposal attempts to achieve the following criteria: • Create a visual focal point on Vassar St. using an archway as an indicator of a grand passageway; • Maximize transparency to maintain the openness that Vassar Street currently has; • Create a comfortable and visually engaging experience for the skybridge users;
C
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GLOBAL SHAPE: PRIMARY MEMBER SIZING:DESIGN APPROACH:: ASYMMETRICAL LOADING CASE: using force polygon
Final Force Polygon
Previous Iterations
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(a)
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107’ = 32.1 m
edge of verticalsupport zone edge of sidewalk
10 KIPS=44.5 KN
AB
CDEFGHI
JK
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MN
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O’
P
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20’45’
30’=9m
horizontal force = 56.3 KIPS = 250.5 kN
10 K
IPS=
44.5
kN
1’ = 1 KIP
Fmax = 128 KIPS= 569.6 kN
113.8
KIP
S
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LOADS:
UNIIFORM LOADING CASE (w):
people
snow
80 lbs/ft2
110 lbs/ft2
30 lbs/ft2
Live Load
steel
concrete
glass
226 lbs/ft2
336 lbs/ft2
=16.6 kN/m2
(density of steel= 490 lbs/ft3)assuming that we use 2” of steel (1/6 of a foot), » 490 lbs/ft3 • 1/6 ft = 82 lbs/ft2
(density of concrete= 150 lbs/ft3)assuming that we use 6” of cocnrete (1/2 of a foot), » 150 lbs/ft3 • 1/2 ft = 75 lbs/ft2
(density of glass= 162 lbs/ft3)assuming that we use 2 of 2” of steel (1/3 of a foot), » 162 lbs/ft3 • 1/3 ft = 54 lbs/ft2
extra
Dead Load
15 lbs/ft2
σ = P/A , where steel allowable stress = 15ksi» 15 ksi = 128 kips/A » A=8.53 in2 » b=h= 3 in = 25.4 mm
due to axial force
Live Load+ Dead Load = 336 lbs/ft2
336 lbs/ft2 • 10ft = 3360 lbs/ft = 3.36 kips/ft
3.36 kips/ft for two arches = 1.68 kips/ft for 1 arch
1.68 kips/ft • 107 ft = 179.76 kips
» 179.76 kips/19 segments = 9.46 kips/seg
Dead Load = 226 lbs/ft2
» 226 lbs/ft2 • 10ft = 2260 lbs/ft » 2.26 kips/ft
Live Load = 110 lbs/ft2
» 110 lbs/ft2 • 10ft • 107ft = 117.7 kips » 117.7 kips / 92 ft= 1.28 kips /ft
arch-in-motion Maya Taketani / Jasmine Kwak
4.463 Building Systems II
simplified diagram of design proposal
15 feet
30 feet
10 feet
92 feet
steel columns
structural glass
107 feet
concrete deck as a horizontal tie
steel cables
steel arch SECONDARY MEMBER SIZING:
MAXMIMUM MOMENT DUE TO ASYMMETRICAL LOADING:
DEFLECTION DUE TO ASYMMETRICAL LOADING
σ= F/A» 15 ksi = 10 kips / A » A=0.66 in2
» A=0.66 in2=π(r)2
» r= 0.46 in » approx. r= 1/2 in = 12.7 mm MMAX= (wLL• L
2)/32 » where wLL= 0.11 kips/ft2 •10 ft /2 arch= 0.55 k/ft » 0.55k/ft • (92ft)2/32 » MMAX= 145.5 kip•ft = 194 kN•m
T=C=M/d » T= 145.5 kip•ft•(12in/ft) / 6 in » 291 kips
σ= F/A » 15 ksi = 291 kips / A » A= 19.4 in2
» d= 4 in = 101.6 mm
T=C=M/d » T= 145.5 kip•ft•(12in/ft) / 18 in » 97 kips
Pbuckling=(π2EI)/(KL)2
» 97 kips•3= (π2•29000 k/in2•Ireq) (112 ft•12ft/in)2
» Ireq= 1836.5 in4
σ= F/A » 15 ksi = 97 kips / A » A= 6.5 in2
Steel Cables in Tension
σ= F/A» 15 ksi = 56.3 kips / A » A=3.75 in2
» I-beam S6x17.25
I-beam decking in Concrete Decking in Tension
bridge
10 feet
336 lb/ft2
336 lb/ft
bridge
336 lb/ft 168 lb/ft
10 kips / ea
due to crushing
due to buckling
Steel Column in Compression
σ= F/A» 15 ksi = 10 kips / A » A=0.67 in2 » b=0.82 in
Pcr= (π2EI)/(kL)2
» 3 •10 kips = [(π2)(29000kips/in2)(Irequired)]/ [(24.25 ft)(12in/1ft)]2 » Irequired = 8.88 in4, where I= bh3/12» for a square section column, b= 4.74 in » b=4 in. or 101.6mm and h=3 in. 76.2 mm
I= bh3/12» I= 108/12= 9 in4 Pcr= (π2EI)/(kL)2
» 3 •10 kips = [(π2)(29000kips/in2)(9 in4)]/ [L in]2
» L < 293 inches = 24 ft = 7.3 m
226 lb/ft
bridge
92 feet
110 lb/ft
bridge
226 lb/ft
110 lb/ft
1”
3”
3”
4”
6”
18”
5/8”20
”
18”
5/8”
20”
20”
5/8”
22”
3-5/8”
6”
7/16”
5”
16.8 kips each
0.9560.6050.458
0.3650.293
0.23
0.172
0.115
0.058
2.6441.449
0.9810.508
1.961
3.987 3.5762.3730.859
0.5430.411
0.3280.263
0.207
0.154
0.103
0.052
1.761.3
0.8810.456
OR
When we used HSS20X18X5/8 as sectional propertiesin Multiframe software to calculate, the highest distance of deflection was 3.987 inches as shown in the graphon the left. Therefore, we had to increase the sectional dimension in order to accomodate ∆max≤ 3.7 in, given ∆max≤ L/360 = 110 ft/360. The new max deflection is 3.576 inches as shown on the right graphand new section is HSS22X20X5/8
street perspective elevation and material
plan
peoplesnow
steel concrete
glass extra
Vassar Street is one of MIT's prominent streets located on the north edge of MIT campus. Nearby the site is Frank Gehry's Stata Center and Charles Correa's Brain and Cognitive Sciences Building. Despite many well known buildings, Vassar Street lacks a visual key urban element that brings people to this street. With these observations, this proposal attempts to achieve the following criteria: • Create a visual focal point on Vassar St. using an archway as an indicator of a grand passageway; • Maximize transparency to maintain the openness that Vassar Street currently has; • Create a comfortable and visually engaging experience for the skybridge users;
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ENVELOPE
Kukje Art Center by SO-IL
Primary material: steel mesh
Desired effects: varying opacity, openness, and
ENVELOPE
Kukje Art Center by SO-IL
Primary material: steel mesh
Desired effects: varying opacity, openness, and
4.463 Skybridge Concept DesignEvelyn Ting, Jie Zhang
09/21/2012
4.463 Skybridge Concept DesignEvelyn Ting, Jie Zhang
09/21/2012
70.00
35.09
13.22
13.0
0
90.00
28.18
35.09
55.00
13.0
0
5.00
70.00
90.00
15.26
8.00
25.00
55.00
5.715.71
16.83
54.7 kips
Tributary Area = 50sf Tributary Area = 300sf
8.6 kips
5 kips
5 kips
3.1 kips
0.3 kips/ft
17.5 kips
max. moment = 1770 kip-ft
8.75
10 kips
8.75
8.758.75
17.5 kips
16.83
18.97
0.85 kips/ft
10 kips
16.83
0.55 kips/ft
18.97
16.83
65.8 kips
6m4.5m3m
UNDER UNIFORM LOAD
TRUSSES FOR LOOKOUT POINTS
For Truss 1Truss self weight = 111.09 ft x 22.37 lbs/ft + 91.45ft x 5.41lbs/ft = 2.98 kipsTributary area = 103.75sf + 103.75sf = 207 sfTotal Load = 207 sf x (45 + 110) lbs/sf + 2.98 kips = 35.1 kips Reaction force (vertical) = 35.1 kips/2 = 17.5 kips
Max. axial tension force = 18.97 kipsReq. cross sectional area = 1.26 in^2< 1.51in^2
Max. axial compression force = 16.83 kips over 35.09 ftReq. I = 10.44 in^4 < 21.7 inch^4
Structure for truss 1 is sound.
For Truss 2Truss self weight = 0.55 kips Tributary area = 61.25 sfTotal load = 61.25 sf x (45 + 110) lbs/sf + 0.55 kips = 10 kipsReaction force (vertical) = 10 kips ; (horizontal) = 3.1 kips
Member sizes are su�cient for Truss 2 with signi�cantly smaller loads
UNDER ASYMMETRICAL LOAD MULTIFRAME ANALYSIS
The trusses are braced by diagonal �oor and ceiling beams and act as rigid tubes under asymmetrial load.
MULTIFRAME ANALYSISAxial stress < 15 ksi = allowable stressThe truss structure is sound under asymmetrical load
TRUSSES FOR MAIN BRIDGE
Self-weight of one truss = 37.69 lbs/ft x 110 ft x 2 +22.37 lbs/ft x 297.14 ft + 48.86 lbs/ft x13ft x 2 = 15.23 kipsUniform Load = 15.23 kips/110 ft +0.45 kips(other systems) + 1.1 kips/ft (live loads) = 1.7 kips/ftUniform Load on each truss = Uniform load/2 = 0.85 kips/ft
BEAM ANALOG METHOD FOR VERIFYING CORD MEMBER SIZING
Max.moment = 1770 kips x ftTension/compression force in cords = 1770kipsxft /13 ft= 136.1 kips
Crushing: σ= P/AReq. area =136.1 kips / 15ksi = 9.07 in^2 < 10.4 in^2
Buckling: P = �^2E(I) / (kL)^2 , L = 10 ftI = 6.85 in^4 < 99.6 inch ^4
MULTIFRAME METHOD FOR VERIFYING DIAGONAL MEMBER SIZING
Maximum axial stress in cords = 136.24 kips (tension) / 134.56 kips (compression)Maximum axial stress in webs = 78.96 kips (tension) / 67.63 kips (compression) Maximum moment on diagonal bracing= 93.978 k x ft; required sectional modulus = 6.27 < 8.67 in^3
Multiframe results are consistent with hand calculations;Steel sections will not crush or buckle under given load; The structure is sound.
BASIC PARAMETERSTotal length = 110 ftTotal height = 13 ftTotal width = 10 ftLive load = 80lbs/sfSnow load = 30lbs/sf
CONCEPT
Allowable Stress (steel) = 15 ksiModulus of Elasticity (steel) = 29,000Allowable Stress (concrete) = 1.5 ksiModulus of Elasticity (glass) = 3,000k (�xity) = 1
This design was conceived as an intersection of two bridges. Two lookouts branch o� the bridge, sloping up and down on either side to provide views out onto the surroundings. They create speci�c views and invite visual dialogue with street tra�c, making the bridge both a circulation path and a gathering space. However, the use of glass and steel mesh attempts to bring a sense of lightness and transparency to the site.
2-in-1EVELYN TING, JIE ZHANG
SECONDARY STRUCTURE ANALYSIS
FLOOR SYSTEM (steel beam)Tributary Area A= 300 sfTotal Load = 20 lbs/sf x 300sf +80 lbs/sf x 300 sf = 30 kips
w = Total Load/ Length = 30 kips / 35 ft = 0.86 k/ftM = (wL^2)/8 = (0.77 k/ft x 35 ft x 35 ft) / 8= 131.7.9 k*ftReq. Section Modulus of beam A= M/∂ = 8.78 in^3 < 13.1 in^3
Tributary Area B= 50 sf
Req. Section Modulus of beam B= 0.42 in^3 < 0.486 in^3
DEFLECTION ANALYSIS MULTIFRAME ANALYSISMaximum de�ection = 3.687 in <3.7 in (allowed de�ection)Hand calculations are inaccurate as the application of the equation assumes a well braced truss whereas in actuality, diagonal bracings are compromised at the openings to allow circulation into lookout spaces.
FLOOR SYSTEM (composite slab)
The composite slab is rested between the main cords, and the tertiary beams e�ectively shorten the span of the slab, allowing the thickness of the slab to be reduced.
Concrete allowable stress = 1.5 ksiM = (wL^2)/8 = (0.8 k/ft x 10 ft x 10 ft) / 8= 10 k*ftT=C=m/d= 53.3 kips
A=C/1.5ksi = 35.5 sq in< actual area 180 sq in. The �oor system is sound.
Total load = Truss self weight + other structural systems dead load + Live load
Floor deck consists of 16 short diagonal beams and 1 long diagonal beam (counted under Truss self weight )
For the main bridge area of 1100 sf:Total Load (�oor beam) = 3.05 lbs/ft x 10 ft x 16 = 0.488 kips; distributed = 4.4 lbs/ftTotal Load (concrete decking) = 1.5 inch x (1/12) x 10ft x 150 lbs/cf = 187.5 lbs/ftTotal Load (metal decking) = 10 lbs/ft
Total Load (�oor) = 202 lbs/ft, or 20 lbs/sfThe ceiling system is estimated to weigh half of the �oor system = 101 lbs/ft, or 10 lbs/sfApproximate facade weight = 15 lbs/sf
Truss self-weight = 0.147 kips/ftOther systems’ self-weight = Floor + Ceiling + Facades = 45 lbs/sf
Total load = Dead Load + Live Load = 0.147 kips/ft + 0.45 kips/ft +1.1kips/ft ~1.7 kips/ft
LOAD ANALYSIS1. Main truss cord member Size = 8 x 8 x 3/8 inchWeight = 37.69 lbs/ftCross Sectional Area = 10.4 in^2S = 24.9 in^3, I = 99.6 in^4
2. Main truss web / secondary truss compression member
Size = 12 x 12 x 5/16 inchWeight = 48.86 lb/ftCross Section Area = 13.4 in^2S = 50.7 in^3, I = 304 in^4
Size = 5 x 5 x 3/8 inchWeight = 22.37 lbs/ftCross Sectional Area = 6.18 in^2S= 8.67 inch^3, I = 21.7 in^4
3. Secondary truss tension member Size = 2 x 2 x 1/4 inchWeight = 5.41 lbs/ftCross sectional area = 1.51 inch^2
4. Tertiary beamSize = 2 x 2 x 1/8 inchWeight = 3.05 lbs/ftS = 0.486 in^3
5. Floor systemComposite �oor system of 2.25 inch thickness ( max. thickness of concrete = 1.5 in, max. corrugation thickness of metal decking = 1.5in) Density of Concrete = 150 lbs/cubic feetWeight of Metal Decking = 1lbs/sf
6. Cladding material Glass for thermal insulationMetal mesh for architectural expression
PLAN CONNECTION DETAIL 1- FLOOR
ELEVATION
CONNECTION DETAIL 2- JOINT
BUILDING SYSTEMS
steel mesh envelope
glass cladding
secondary
1” �nishing0.75” concrete1.5” corrugated metal decktertiary beam bracing
welded steel connection
steel mesh5/8” gypsum board
primary truss
tertiary truss
30.00°
70.00°
10.00
8.00
25.00
10.00
110.
00
0.125
2.00
0.25
2.00
0.375
5.00
8.00
0.375
12.0
0
0.3125
0.75
” 1.00
”1.
50”
2.00
”
8.00
”
elevation 1/16” scale
23
ENVELOPE
Kukje Art Center by SO-IL
Primary material: steel mesh
Desired effects: varying opacity, openness, and
ENVELOPE
Kukje Art Center by SO-IL
Primary material: steel mesh
Desired effects: varying opacity, openness, and
4.463 Skybridge Concept DesignEvelyn Ting, Jie Zhang
09/21/2012
4.463 Skybridge Concept DesignEvelyn Ting, Jie Zhang
09/21/2012
70.00
35.09
13.22
13.0
0
90.00
28.18
35.09
55.00
13.0
0
5.00
70.00
90.00
15.26
8.00
25.00
55.00
5.715.71
16.83
54.7 kips
Tributary Area = 50sf Tributary Area = 300sf
8.6 kips
5 kips
5 kips
3.1 kips
0.3 kips/ft
17.5 kips
max. moment = 1770 kip-ft
8.75
10 kips
8.75
8.758.75
17.5 kips
16.83
18.97
0.85 kips/ft
10 kips
16.83
0.55 kips/ft
18.97
16.83
65.8 kips
6m4.5m3m
UNDER UNIFORM LOAD
TRUSSES FOR LOOKOUT POINTS
For Truss 1Truss self weight = 111.09 ft x 22.37 lbs/ft + 91.45ft x 5.41lbs/ft = 2.98 kipsTributary area = 103.75sf + 103.75sf = 207 sfTotal Load = 207 sf x (45 + 110) lbs/sf + 2.98 kips = 35.1 kips Reaction force (vertical) = 35.1 kips/2 = 17.5 kips
Max. axial tension force = 18.97 kipsReq. cross sectional area = 1.26 in^2< 1.51in^2
Max. axial compression force = 16.83 kips over 35.09 ftReq. I = 10.44 in^4 < 21.7 inch^4
Structure for truss 1 is sound.
For Truss 2Truss self weight = 0.55 kips Tributary area = 61.25 sfTotal load = 61.25 sf x (45 + 110) lbs/sf + 0.55 kips = 10 kipsReaction force (vertical) = 10 kips ; (horizontal) = 3.1 kips
Member sizes are su�cient for Truss 2 with signi�cantly smaller loads
UNDER ASYMMETRICAL LOAD MULTIFRAME ANALYSIS
The trusses are braced by diagonal �oor and ceiling beams and act as rigid tubes under asymmetrial load.
MULTIFRAME ANALYSISAxial stress < 15 ksi = allowable stressThe truss structure is sound under asymmetrical load
TRUSSES FOR MAIN BRIDGE
Self-weight of one truss = 37.69 lbs/ft x 110 ft x 2 +22.37 lbs/ft x 297.14 ft + 48.86 lbs/ft x13ft x 2 = 15.23 kipsUniform Load = 15.23 kips/110 ft +0.45 kips(other systems) + 1.1 kips/ft (live loads) = 1.7 kips/ftUniform Load on each truss = Uniform load/2 = 0.85 kips/ft
BEAM ANALOG METHOD FOR VERIFYING CORD MEMBER SIZING
Max.moment = 1770 kips x ftTension/compression force in cords = 1770kipsxft /13 ft= 136.1 kips
Crushing: σ= P/AReq. area =136.1 kips / 15ksi = 9.07 in^2 < 10.4 in^2
Buckling: P = �^2E(I) / (kL)^2 , L = 10 ftI = 6.85 in^4 < 99.6 inch ^4
MULTIFRAME METHOD FOR VERIFYING DIAGONAL MEMBER SIZING
Maximum axial stress in cords = 136.24 kips (tension) / 134.56 kips (compression)Maximum axial stress in webs = 78.96 kips (tension) / 67.63 kips (compression) Maximum moment on diagonal bracing= 93.978 k x ft; required sectional modulus = 6.27 < 8.67 in^3
Multiframe results are consistent with hand calculations;Steel sections will not crush or buckle under given load; The structure is sound.
BASIC PARAMETERSTotal length = 110 ftTotal height = 13 ftTotal width = 10 ftLive load = 80lbs/sfSnow load = 30lbs/sf
CONCEPT
Allowable Stress (steel) = 15 ksiModulus of Elasticity (steel) = 29,000Allowable Stress (concrete) = 1.5 ksiModulus of Elasticity (glass) = 3,000k (�xity) = 1
This design was conceived as an intersection of two bridges. Two lookouts branch o� the bridge, sloping up and down on either side to provide views out onto the surroundings. They create speci�c views and invite visual dialogue with street tra�c, making the bridge both a circulation path and a gathering space. However, the use of glass and steel mesh attempts to bring a sense of lightness and transparency to the site.
2-in-1EVELYN TING, JIE ZHANG
SECONDARY STRUCTURE ANALYSIS
FLOOR SYSTEM (steel beam)Tributary Area A= 300 sfTotal Load = 20 lbs/sf x 300sf +80 lbs/sf x 300 sf = 30 kips
w = Total Load/ Length = 30 kips / 35 ft = 0.86 k/ftM = (wL^2)/8 = (0.77 k/ft x 35 ft x 35 ft) / 8= 131.7.9 k*ftReq. Section Modulus of beam A= M/∂ = 8.78 in^3 < 13.1 in^3
Tributary Area B= 50 sf
Req. Section Modulus of beam B= 0.42 in^3 < 0.486 in^3
DEFLECTION ANALYSIS MULTIFRAME ANALYSISMaximum de�ection = 3.687 in <3.7 in (allowed de�ection)Hand calculations are inaccurate as the application of the equation assumes a well braced truss whereas in actuality, diagonal bracings are compromised at the openings to allow circulation into lookout spaces.
FLOOR SYSTEM (composite slab)
The composite slab is rested between the main cords, and the tertiary beams e�ectively shorten the span of the slab, allowing the thickness of the slab to be reduced.
Concrete allowable stress = 1.5 ksiM = (wL^2)/8 = (0.8 k/ft x 10 ft x 10 ft) / 8= 10 k*ftT=C=m/d= 53.3 kips
A=C/1.5ksi = 35.5 sq in< actual area 180 sq in. The �oor system is sound.
Total load = Truss self weight + other structural systems dead load + Live load
Floor deck consists of 16 short diagonal beams and 1 long diagonal beam (counted under Truss self weight )
For the main bridge area of 1100 sf:Total Load (�oor beam) = 3.05 lbs/ft x 10 ft x 16 = 0.488 kips; distributed = 4.4 lbs/ftTotal Load (concrete decking) = 1.5 inch x (1/12) x 10ft x 150 lbs/cf = 187.5 lbs/ftTotal Load (metal decking) = 10 lbs/ft
Total Load (�oor) = 202 lbs/ft, or 20 lbs/sfThe ceiling system is estimated to weigh half of the �oor system = 101 lbs/ft, or 10 lbs/sfApproximate facade weight = 15 lbs/sf
Truss self-weight = 0.147 kips/ftOther systems’ self-weight = Floor + Ceiling + Facades = 45 lbs/sf
Total load = Dead Load + Live Load = 0.147 kips/ft + 0.45 kips/ft +1.1kips/ft ~1.7 kips/ft
LOAD ANALYSIS1. Main truss cord member Size = 8 x 8 x 3/8 inchWeight = 37.69 lbs/ftCross Sectional Area = 10.4 in^2S = 24.9 in^3, I = 99.6 in^4
2. Main truss web / secondary truss compression member
Size = 12 x 12 x 5/16 inchWeight = 48.86 lb/ftCross Section Area = 13.4 in^2S = 50.7 in^3, I = 304 in^4
Size = 5 x 5 x 3/8 inchWeight = 22.37 lbs/ftCross Sectional Area = 6.18 in^2S= 8.67 inch^3, I = 21.7 in^4
3. Secondary truss tension member Size = 2 x 2 x 1/4 inchWeight = 5.41 lbs/ftCross sectional area = 1.51 inch^2
4. Tertiary beamSize = 2 x 2 x 1/8 inchWeight = 3.05 lbs/ftS = 0.486 in^3
5. Floor systemComposite �oor system of 2.25 inch thickness ( max. thickness of concrete = 1.5 in, max. corrugation thickness of metal decking = 1.5in) Density of Concrete = 150 lbs/cubic feetWeight of Metal Decking = 1lbs/sf
6. Cladding material Glass for thermal insulationMetal mesh for architectural expression
PLAN CONNECTION DETAIL 1- FLOOR
ELEVATION
CONNECTION DETAIL 2- JOINT
BUILDING SYSTEMS
steel mesh envelope
glass cladding
secondary
1” �nishing0.75” concrete1.5” corrugated metal decktertiary beam bracing
welded steel connection
steel mesh5/8” gypsum board
primary truss
tertiary truss
30.00°
70.00°
10.00
8.00
25.00
10.00
110.
00
0.125
2.00
0.25
2.00
0.375
5.00
8.00
0.375
12.0
0
0.3125
0.75
” 1.00
”1.
50”
2.00
”
8.00
”
elevation 1/16” scale
24
1/4"=1'-0"CROSS SECTION
HSS 18x12x5/8
HSS 6x6x1/4
2-1/8” DIAMETERSTEEL CABLE
1-1/2” DIAMETER STEEL ROD
6” CONCRETE OF 3” METAL DECK
F.F.L.
+ 50'-3" A.S.L.F.F.L.
+ 49'-4" A.S.L.
+ 40'-3" A.S.L.B.O. STRUCTURE
T.O. SIDEWALK+ 20'3" A.S.L.
LONGITUDINAL SECTION1/8"=1'-0"
HSS 18x12x5/8
HSS 6x6x1/4
2-1/8” STEEL CABLE
LOCATION OF SECTION FOR AVERAGE GLOBAL MOMENT OF INERTIACALCULATION
6” CONCRETE OF 3” METAL DECK
Maximum Moment= (wL^2)/8=10153 kip-feet
w=2.5kips / linear foot
MOMENT 0
SHEAR 0
143 kips 143 kips
David Miranowski David Moses
LOADING
DEAD LOADstructural steel HSS 6x6x1/4 4122.0 lbs
HSS 6x2x1/4 7424.0 lbsHSS 6x9x1/4 5306.0 lbs
HSS 18x12x5/8 13019.0 lbs
cable length 125.0 feetweight per foot 9.5 pounds per foot
weight 1186.3 pounds
lightweight concrete volume 565.0 cubic feetweight by volume 100.0 pounds per cubic foot
weight 56500.0 pounds
skylight glass volume 174.0 cubic feetweight by volume 150.0 pounds per cubic foot
weight 26100.0 pounds
cladding allowance weight 52200.0 pounds
total distributed dead load 165.9 kips 737.8
DISTRIBUTED LIVE LOADcarried at roof weight per unit area 30.0 pounds per square foot
area 1391.0 square feetload allowance 41730.0 pounds
carried at floor weight per unit area 80.0 pounds per square footarea 1391.0 square feet
load allowance 111280.0 pounds
load allowance 111.3 kips 495.0
imperial units metric unitstotal span = 113.0 feet 34.4 meters
linear load = 2.5 kips/feet 35.8load segments = 11.0 segments
total load = 277.1 kips 1232.8load per segment = 25.2 kips 112.1
segment length = 10.3 feet 3.1 meters
kN
kN
kN/meters
kNkN
LOADING
PLAN1:192
VASSAR STREET
BRAIN AND COGNITIVE SCIENCES BUILDING BUILDING 36
ASYMETRIC LIVE LOADtotal carried by floor weight per unit area 110.0 pounds per square foot
area 565.0 square feetload allowance 62.2 kips
load length 56.5 feet ½ the total span
linear load = 1.1 kips/feet 4.9
imperial units metric unitstotal span = 113.0 feet 34.4 meters
PRIMARY STRUCTURE (LONGITUDINAL SECTION)
MAXIMUM MOMENTimperial units metric units
linear load (w) = 1.1 kips/feetload length (L) = 56.5 ft
109.7 kip*ft
T=C=M/d FORCE COUPLEimperial units metric units
Moment(M)= 109.7 kip*ftDepth(d) = 1.0 feet guess
Axial Load(T,C)= 109.7 kips
AXIAL STRESSimperial units metric units
maximum axial load(F) = 109.7 kips 488.1 (from force polygon)15.0 kips/square inch (safety factor included)
number of compression blocks = 1.0 membersrequired cross sectional area(A) = 7.3 square inches 4719.7 square mm in compression and tension blocks
actual area of tube wall = 10.1 square inches
kN
Mmax=(wL^2)/32
Maximum Moment (Mmax) =
σ=F/A
kNsteel allowable stress(σ) =
A=F/σ
DEFLECTIONCALCULATING CENTROID
3.13.1
22.087.3 in Distance between Column Center and Cable Center67.9 in location of centroid
REQUIRED MOMENT OF INERTIA
applied live load (w) = 0.1 K/inLength (L) = 1320.0 in
29000.03.7 in
required moment of inertia (I)= 33,459
AVERAGE GLOBAL MOMENT OF INERTIA
3.13.1
22.067.9 in67.9 in19.4 in
global moment of inertia (I)= 37,201
MAXIMUM ALLOWED DEFLECTION
Length (L)= 1320.0 in3.7 in
y=A3Y3/(A1+A2+A3) Area A1= in2
Area A2= in2
Area A3= in2
Y3=
y=
Ireq.d≥(5wL^4)/(384Edeltamax)
modulus of elasticitysteel (E) = K/in2
Max Deflection (∆max ) =
in4
IGLOBAL=∑Aidi2 area A1= in2
area A2= in2
area A3= in2
distance from centroid (d1)=distance from centroid (d2)=distance from centroid (d3)=
in4
∆max≤L/360Max Deflection (∆max ) =
ASYMETRICAL LOADING DEFLECTION
SECONDARY STRUCTURE - CROSS SECTION
PRIMARY STRUCTURE (LONGITUDINAL SECTION)
CABLE- AXIAL STRESSimperial units metric units
maximum axial stress = 490.0 kips 2179.6 (from force polygon)number of cables = 2.0 cables
working load per cable = 245.0 kips 1089.8cable safety factor = 2.2
cable breaking strength = 269.5 tons 2397.6 Class B cable diameter 2 1/ 8 inches FORM and FORCES p. 4754.0 mm
imperial units metric unitstotal max axial load(F) = 424.0 kips 1886.0 (from force polygon)
15.0 kips/square inch (safety factor included)number of members = 1.0 members
required cross sectional area(A) = 28.3 square inches 18236.5 square mmarea per member = 28.3 square inches 18236.5 square mm HSS Steel Tube 18 x 12 x 5/8 Inch STI HSS Properties Table
457 x 305 x 16 mm
imperial units metric unitstotal max axial load(P) = 424.0 kips 1886.0 (from force polygon)
modulus of elasticity(E) = 29000.0 kips/ square inchbuckling length(L) = 123.3 inches 3131.1 mm
fixity (k) = 1.0 fixed/fixedbuckling safety factor = 3.0
moment of inertia(I) = 212.2 inches^4 88309640.3 mm^4
SECONDARY STRUCTURE (CROSS SECTION)
imperial units metric unitsmaximum axial load(F) = 22.7 kips 101.0 (from force polygon)
15.0 kips/square inch (safety factor included)number of members = 1.0 members
required cross sectional area(A) = 1.5 square inches 976.3 square mmarea per member = 1.5 square inches 976.3 square mm HSS Steel Tube 6 x 6 x 1/4 Inch STI HSS Properties Table
152 x 152 x 6 mm
imperial units metric unitstotal max axial load(P) = 22.7 kips 101.0 (from force polygon)
modulus of elasticity(E) = 29000.0 kips/ square inchbuckling length(L) = 123.0 inches 3124.2 mm
fixity (k) = 1.5 fixed/pinbuckling safety factor = 3.0
moment of inertia(I) = 25.4 inches^4 10590753.2 Mm^4
imperial units metric unitsmaximum axial load(F) = 19.5 kips 86.7 (from force polygon)
15.0 kips/square inch (safety factor included)number of rods = 1.0 element
cross sectional area(A)= 1.3 square inches 838.7 square mmrod diameter = 1.3 inches 32.7 mm Steel rod diameter 1 1/ 2 Inch
38.1 mm
σ=F/A
kN
kN
kN
σ=F/ACOMPRESSION – AXIAL
STRESS
kNsteel allowable stress(σ) =
A=F/σ
P=πEI/(kL)^2COMPRESSION –
BUCKLING
kN
I=(P(kL)^2/πE)*3
σ=F/ACOMPRESSION STRUTS-
AXIAL STRESS
kNsteel allowable stress(σ) =
A=F/σ
P=πEI/(kL)^2COMPRESSION –
BUCKLING
kN
I=(P(kL)^2/πE)*3
σ=F/ATENSION ROD- AXIAL
STRESS
kNsteel allowable stress(σ) =
A=F/σD=2√(A/π)
PRIMARY STRUCTURE (LONGITUDINAL SECTION)
CABLE- AXIAL STRESSimperial units metric units
maximum axial stress = 490.0 kips 2179.6 (from force polygon)number of cables = 2.0 cables
working load per cable = 245.0 kips 1089.8cable safety factor = 2.2
cable breaking strength = 269.5 tons 2397.6 Class B cable diameter 2 1/ 8 inches FORM and FORCES p. 4754.0 mm
imperial units metric unitstotal max axial load(F) = 424.0 kips 1886.0 (from force polygon)
15.0 kips/square inch (safety factor included)number of members = 1.0 members
required cross sectional area(A) = 28.3 square inches 18236.5 square mmarea per member = 28.3 square inches 18236.5 square mm HSS Steel Tube 18 x 12 x 5/8 Inch STI HSS Properties Table
457 x 305 x 16 mm
imperial units metric unitstotal max axial load(P) = 424.0 kips 1886.0 (from force polygon)
modulus of elasticity(E) = 29000.0 kips/ square inchbuckling length(L) = 123.3 inches 3131.1 mm
fixity (k) = 1.0 fixed/fixedbuckling safety factor = 3.0
moment of inertia(I) = 212.2 inches^4 88309640.3 mm^4
SECONDARY STRUCTURE (CROSS SECTION)
imperial units metric unitsmaximum axial load(F) = 22.7 kips 101.0 (from force polygon)
15.0 kips/square inch (safety factor included)number of members = 1.0 members
required cross sectional area(A) = 1.5 square inches 976.3 square mmarea per member = 1.5 square inches 976.3 square mm HSS Steel Tube 6 x 6 x 1/4 Inch STI HSS Properties Table
152 x 152 x 6 mm
imperial units metric unitstotal max axial load(P) = 22.7 kips 101.0 (from force polygon)
modulus of elasticity(E) = 29000.0 kips/ square inchbuckling length(L) = 123.0 inches 3124.2 mm
fixity (k) = 1.5 fixed/pinbuckling safety factor = 3.0
moment of inertia(I) = 25.4 inches^4 10590753.2 Mm^4
imperial units metric unitsmaximum axial load(F) = 19.5 kips 86.7 (from force polygon)
15.0 kips/square inch (safety factor included)number of rods = 1.0 element
cross sectional area(A)= 1.3 square inches 838.7 square mmrod diameter = 1.3 inches 32.7 mm Steel rod diameter 1 1/ 2 Inch
38.1 mm
σ=F/A
kN
kN
kN
σ=F/ACOMPRESSION – AXIAL
STRESS
kNsteel allowable stress(σ) =
A=F/σ
P=πEI/(kL)^2COMPRESSION –
BUCKLING
kN
I=(P(kL)^2/πE)*3
σ=F/ACOMPRESSION STRUTS-
AXIAL STRESS
kNsteel allowable stress(σ) =
A=F/σ
P=πEI/(kL)^2COMPRESSION –
BUCKLING
kN
I=(P(kL)^2/πE)*3
σ=F/ATENSION ROD- AXIAL
STRESS
kNsteel allowable stress(σ) =
A=F/σD=2√(A/π)
PRIMARY STRUCTURE - LONGITUDINAL SECTION
EXPLODED AXONOMETRIC
w=2.5kips / linear foot
11 feet of bridge lengthsuppored by each segment
27.5 kips applied loadA B
1"=24 KIPS
b
c
a
1
22.7 KIPS (T)
26.6 KIPS (C)26.6 KIPS (C)
13.75 KIPS 13.75 KIPS
27.5 KIPS
1
C
1/4"=1'-0"CROSS SECTION
oblique plane22.7 kips per load
o
245 KIPS
k
j
i
h
g
f
e
d
c
b
1"=48 KIPS
a
212 KIPS
KJIHGFECBA D
FAT BELLY
INTERIOR VIEW
UNDERBELLY
CROSS SECTION1:48
LONGITUDINAL SECTION1:96
M=(wL^2)/32MOMENT DIAGRAM
w=1.1 kips / foot
67.8
7 "
19.4
1 "
87.2
8 "
A1
A1, A2 3.1 sq.in
A2
=
=
A3
A3
d1, d2
d3
y
22.0 sq.in
arbitrary x axis, for finding centroid
arbitrary y axis, for finding centroid
25
1/4"=1'-0"CROSS SECTION
HSS 18x12x5/8
HSS 6x6x1/4
2-1/8” DIAMETERSTEEL CABLE
1-1/2” DIAMETER STEEL ROD
6” CONCRETE OF 3” METAL DECK
F.F.L.
+ 50'-3" A.S.L.F.F.L.
+ 49'-4" A.S.L.
+ 40'-3" A.S.L.B.O. STRUCTURE
T.O. SIDEWALK+ 20'3" A.S.L.
LONGITUDINAL SECTION1/8"=1'-0"
HSS 18x12x5/8
HSS 6x6x1/4
2-1/8” STEEL CABLE
LOCATION OF SECTION FOR AVERAGE GLOBAL MOMENT OF INERTIACALCULATION
6” CONCRETE OF 3” METAL DECK
Maximum Moment= (wL^2)/8=10153 kip-feet
w=2.5kips / linear foot
MOMENT 0
SHEAR 0
143 kips 143 kips
David Miranowski David Moses
LOADING
DEAD LOADstructural steel HSS 6x6x1/4 4122.0 lbs
HSS 6x2x1/4 7424.0 lbsHSS 6x9x1/4 5306.0 lbs
HSS 18x12x5/8 13019.0 lbs
cable length 125.0 feetweight per foot 9.5 pounds per foot
weight 1186.3 pounds
lightweight concrete volume 565.0 cubic feetweight by volume 100.0 pounds per cubic foot
weight 56500.0 pounds
skylight glass volume 174.0 cubic feetweight by volume 150.0 pounds per cubic foot
weight 26100.0 pounds
cladding allowance weight 52200.0 pounds
total distributed dead load 165.9 kips 737.8
DISTRIBUTED LIVE LOADcarried at roof weight per unit area 30.0 pounds per square foot
area 1391.0 square feetload allowance 41730.0 pounds
carried at floor weight per unit area 80.0 pounds per square footarea 1391.0 square feet
load allowance 111280.0 pounds
load allowance 111.3 kips 495.0
imperial units metric unitstotal span = 113.0 feet 34.4 meters
linear load = 2.5 kips/feet 35.8load segments = 11.0 segments
total load = 277.1 kips 1232.8load per segment = 25.2 kips 112.1
segment length = 10.3 feet 3.1 meters
kN
kN
kN/meters
kNkN
LOADING
PLAN1:192
VASSAR STREET
BRAIN AND COGNITIVE SCIENCES BUILDING BUILDING 36
ASYMETRIC LIVE LOADtotal carried by floor weight per unit area 110.0 pounds per square foot
area 565.0 square feetload allowance 62.2 kips
load length 56.5 feet ½ the total span
linear load = 1.1 kips/feet 4.9
imperial units metric unitstotal span = 113.0 feet 34.4 meters
PRIMARY STRUCTURE (LONGITUDINAL SECTION)
MAXIMUM MOMENTimperial units metric units
linear load (w) = 1.1 kips/feetload length (L) = 56.5 ft
109.7 kip*ft
T=C=M/d FORCE COUPLEimperial units metric units
Moment(M)= 109.7 kip*ftDepth(d) = 1.0 feet guess
Axial Load(T,C)= 109.7 kips
AXIAL STRESSimperial units metric units
maximum axial load(F) = 109.7 kips 488.1 (from force polygon)15.0 kips/square inch (safety factor included)
number of compression blocks = 1.0 membersrequired cross sectional area(A) = 7.3 square inches 4719.7 square mm in compression and tension blocks
actual area of tube wall = 10.1 square inches
kN
Mmax=(wL^2)/32
Maximum Moment (Mmax) =
σ=F/A
kNsteel allowable stress(σ) =
A=F/σ
DEFLECTIONCALCULATING CENTROID
3.13.1
22.087.3 in Distance between Column Center and Cable Center67.9 in location of centroid
REQUIRED MOMENT OF INERTIA
applied live load (w) = 0.1 K/inLength (L) = 1320.0 in
29000.03.7 in
required moment of inertia (I)= 33,459
AVERAGE GLOBAL MOMENT OF INERTIA
3.13.1
22.067.9 in67.9 in19.4 in
global moment of inertia (I)= 37,201
MAXIMUM ALLOWED DEFLECTION
Length (L)= 1320.0 in3.7 in
y=A3Y3/(A1+A2+A3) Area A1= in2
Area A2= in2
Area A3= in2
Y3=
y=
Ireq.d≥(5wL^4)/(384Edeltamax)
modulus of elasticitysteel (E) = K/in2
Max Deflection (∆max ) =
in4
IGLOBAL=∑Aidi2 area A1= in2
area A2= in2
area A3= in2
distance from centroid (d1)=distance from centroid (d2)=distance from centroid (d3)=
in4
∆max≤L/360Max Deflection (∆max ) =
ASYMETRICAL LOADING DEFLECTION
SECONDARY STRUCTURE - CROSS SECTION
PRIMARY STRUCTURE (LONGITUDINAL SECTION)
CABLE- AXIAL STRESSimperial units metric units
maximum axial stress = 490.0 kips 2179.6 (from force polygon)number of cables = 2.0 cables
working load per cable = 245.0 kips 1089.8cable safety factor = 2.2
cable breaking strength = 269.5 tons 2397.6 Class B cable diameter 2 1/ 8 inches FORM and FORCES p. 4754.0 mm
imperial units metric unitstotal max axial load(F) = 424.0 kips 1886.0 (from force polygon)
15.0 kips/square inch (safety factor included)number of members = 1.0 members
required cross sectional area(A) = 28.3 square inches 18236.5 square mmarea per member = 28.3 square inches 18236.5 square mm HSS Steel Tube 18 x 12 x 5/8 Inch STI HSS Properties Table
457 x 305 x 16 mm
imperial units metric unitstotal max axial load(P) = 424.0 kips 1886.0 (from force polygon)
modulus of elasticity(E) = 29000.0 kips/ square inchbuckling length(L) = 123.3 inches 3131.1 mm
fixity (k) = 1.0 fixed/fixedbuckling safety factor = 3.0
moment of inertia(I) = 212.2 inches^4 88309640.3 mm^4
SECONDARY STRUCTURE (CROSS SECTION)
imperial units metric unitsmaximum axial load(F) = 22.7 kips 101.0 (from force polygon)
15.0 kips/square inch (safety factor included)number of members = 1.0 members
required cross sectional area(A) = 1.5 square inches 976.3 square mmarea per member = 1.5 square inches 976.3 square mm HSS Steel Tube 6 x 6 x 1/4 Inch STI HSS Properties Table
152 x 152 x 6 mm
imperial units metric unitstotal max axial load(P) = 22.7 kips 101.0 (from force polygon)
modulus of elasticity(E) = 29000.0 kips/ square inchbuckling length(L) = 123.0 inches 3124.2 mm
fixity (k) = 1.5 fixed/pinbuckling safety factor = 3.0
moment of inertia(I) = 25.4 inches^4 10590753.2 Mm^4
imperial units metric unitsmaximum axial load(F) = 19.5 kips 86.7 (from force polygon)
15.0 kips/square inch (safety factor included)number of rods = 1.0 element
cross sectional area(A)= 1.3 square inches 838.7 square mmrod diameter = 1.3 inches 32.7 mm Steel rod diameter 1 1/ 2 Inch
38.1 mm
σ=F/A
kN
kN
kN
σ=F/ACOMPRESSION – AXIAL
STRESS
kNsteel allowable stress(σ) =
A=F/σ
P=πEI/(kL)^2COMPRESSION –
BUCKLING
kN
I=(P(kL)^2/πE)*3
σ=F/ACOMPRESSION STRUTS-
AXIAL STRESS
kNsteel allowable stress(σ) =
A=F/σ
P=πEI/(kL)^2COMPRESSION –
BUCKLING
kN
I=(P(kL)^2/πE)*3
σ=F/ATENSION ROD- AXIAL
STRESS
kNsteel allowable stress(σ) =
A=F/σD=2√(A/π)
PRIMARY STRUCTURE (LONGITUDINAL SECTION)
CABLE- AXIAL STRESSimperial units metric units
maximum axial stress = 490.0 kips 2179.6 (from force polygon)number of cables = 2.0 cables
working load per cable = 245.0 kips 1089.8cable safety factor = 2.2
cable breaking strength = 269.5 tons 2397.6 Class B cable diameter 2 1/ 8 inches FORM and FORCES p. 4754.0 mm
imperial units metric unitstotal max axial load(F) = 424.0 kips 1886.0 (from force polygon)
15.0 kips/square inch (safety factor included)number of members = 1.0 members
required cross sectional area(A) = 28.3 square inches 18236.5 square mmarea per member = 28.3 square inches 18236.5 square mm HSS Steel Tube 18 x 12 x 5/8 Inch STI HSS Properties Table
457 x 305 x 16 mm
imperial units metric unitstotal max axial load(P) = 424.0 kips 1886.0 (from force polygon)
modulus of elasticity(E) = 29000.0 kips/ square inchbuckling length(L) = 123.3 inches 3131.1 mm
fixity (k) = 1.0 fixed/fixedbuckling safety factor = 3.0
moment of inertia(I) = 212.2 inches^4 88309640.3 mm^4
SECONDARY STRUCTURE (CROSS SECTION)
imperial units metric unitsmaximum axial load(F) = 22.7 kips 101.0 (from force polygon)
15.0 kips/square inch (safety factor included)number of members = 1.0 members
required cross sectional area(A) = 1.5 square inches 976.3 square mmarea per member = 1.5 square inches 976.3 square mm HSS Steel Tube 6 x 6 x 1/4 Inch STI HSS Properties Table
152 x 152 x 6 mm
imperial units metric unitstotal max axial load(P) = 22.7 kips 101.0 (from force polygon)
modulus of elasticity(E) = 29000.0 kips/ square inchbuckling length(L) = 123.0 inches 3124.2 mm
fixity (k) = 1.5 fixed/pinbuckling safety factor = 3.0
moment of inertia(I) = 25.4 inches^4 10590753.2 Mm^4
imperial units metric unitsmaximum axial load(F) = 19.5 kips 86.7 (from force polygon)
15.0 kips/square inch (safety factor included)number of rods = 1.0 element
cross sectional area(A)= 1.3 square inches 838.7 square mmrod diameter = 1.3 inches 32.7 mm Steel rod diameter 1 1/ 2 Inch
38.1 mm
σ=F/A
kN
kN
kN
σ=F/ACOMPRESSION – AXIAL
STRESS
kNsteel allowable stress(σ) =
A=F/σ
P=πEI/(kL)^2COMPRESSION –
BUCKLING
kN
I=(P(kL)^2/πE)*3
σ=F/ACOMPRESSION STRUTS-
AXIAL STRESS
kNsteel allowable stress(σ) =
A=F/σ
P=πEI/(kL)^2COMPRESSION –
BUCKLING
kN
I=(P(kL)^2/πE)*3
σ=F/ATENSION ROD- AXIAL
STRESS
kNsteel allowable stress(σ) =
A=F/σD=2√(A/π)
PRIMARY STRUCTURE - LONGITUDINAL SECTION
EXPLODED AXONOMETRIC
w=2.5kips / linear foot
11 feet of bridge lengthsuppored by each segment
27.5 kips applied loadA B
1"=24 KIPS
b
c
a
1
22.7 KIPS (T)
26.6 KIPS (C)26.6 KIPS (C)
13.75 KIPS 13.75 KIPS
27.5 KIPS
1
C
1/4"=1'-0"CROSS SECTION
oblique plane22.7 kips per load
o
245 KIPS
k
j
i
h
g
f
e
d
c
b
1"=48 KIPS
a
212 KIPS
KJIHGFECBA D
FAT BELLY
INTERIOR VIEW
UNDERBELLY
CROSS SECTION1:48
LONGITUDINAL SECTION1:96
M=(wL^2)/32MOMENT DIAGRAM
w=1.1 kips / foot
67.8
7 "
19.4
1 "
87.2
8 "
A1
A1, A2 3.1 sq.in
A2
=
=
A3
A3
d1, d2
d3
y
22.0 sq.in
arbitrary x axis, for finding centroid
arbitrary y axis, for finding centroid
26
BEOMKI LEE / SHIYU WEI4.463 Building Structural Systems II (Fall 2012)Professor: J.A. Ochsendorf / TA: Caitlin Mueller
Sky Illusion
Snow Load 30 lb/ft
80 lb/ft
42 lb/ft
20 lb/ft
2 lb/ft
58 lb/ft
232 lb/ft
0.4 kN/m
1.1 kN/m
0.6 kN/m
0.3 kN/m
0.03 kN/m
0.8 kN/m
3.4 kN/m
249 K 1106 kN
People Load
Steel Arch (x4)
Steel Struts (x10)
3/4” Glass Cladding
Total Distributed Load
Total Load BRIDGE AREA = 107 ft * 10 ft
12.5 K 55.6 kNPoint Loads
Deck
LOADS_
FORCE POLYGON_
MEMBER SIZING_
Shear Diagram
Maximum Moment = 1818 K*ft
12.5K55.6 kN
55.6 kN
55.6 kN
12.5K 12.5K12.5K 12.5K12.5K12.5K12.5K12.5K
606 K*ft
-606 K*ft
485 K*ft
-485 K*ft
364 K*ft
-364 K*ft
242 K*ft
-242 K*ft
121K*ft0 K*ft
-121 K*ft9.7 ft
32 m
278 kN
278 kN
107 ft K 0
62.5 K 62.5 K
Moment Diagram
T = C = M/d; Greater depth of the beam implies smaller forces on the top and bottom member, suggesting a beam shape that is thick in the middle
CONCEPT_ TransparencyTransparencyThe gray, monolithic concrete buildings on Vassar street dominate the view of the people passing through. Therefore, we strived for a design that had minimal visual impact on the site. Also, since the Stata Center already has a very strong presence on the street, a simple bridge would not interfere with the view so highly sought after by tourists and also would not clutter up an already packed street.
1/4 inch = 1 kipForce in vertical strut = 9.7K = 46kN
= Checking for Uneven Loading Conditions
ω = 30lb/ft + 80 lb/ft = 110 lb/ft = 0.11 k/ftM = �L2 / 32M = [(0.11k/ft)(107.2ft)2]/32 = 474 k*inσ�= 15 ksi = M/S σ = 15ksi = 474 k*in/SS = 31.6 in3 = 517 cm3
= Check for Buckling over L/2
L/2 = 53.6 ftE = 29,000 ksiP = � 2EI/(kL)k = 1 (pin-pin condition)158K = � 2(29,000ksi)I / (1(53.6ft))Ireq
= 228in4 = 9490 cm4 --> 684in4 = 28470 cm4 (safety factor)
= Checking for Max Allowable Stress
Largest force = 158KStress = force/area15 ksi = 158K / areaArea
req = 158K / 15ksi =10.5in2 = 67.7 cm2
Choose 16”x12”x3/8” Rectangular HSS 40.6cm x 30.5cm x 0.9cmS
x = 87.7in3 S
y = 75.3in3
Ix = 702 in4 I
y = 452 in4
12.5K
9.7 ft
3 m
1
62.5K 62.5K
2 3 4 5 6 7 8 109 11
KA B C D E F I J
L
G H
12.5K 12.5K 12.5K
12.5K
12.5K
12.5K
12.5K
12.5K
12.5K
12.5K12.5K
12.5K
12.5K
12.5K12.5K12.5K 12.5K 12.5K12.5K
= Check for Max Allowable Stress
Largest force = 9.7 K15 ksi = 9.7 K/Area
req
Areareq
= 9.7 K/15ksi = 0.65 in2
Choose member 4”x3”x3/16” 10.2cm x 7.6cmx 0.5cmIx = 4.9 in4
Iy
= 3.2 in4
= Check for Buckling
E = 29,000 ksiP = � 2EI/(kL)2
9.7 K = � 2(29,000ksi)I/(1(144in))2
Ireq
= 0.97 in4 = 40.4 cm4 --> 2.91in4 = 121 cm4 (safety factor)
= Checking for Max Allowable Stress
Largest Force = 164.3KArea
req = 164.3K / 15ksi = 10.9 in2
Choose 7”x7”x1/2” Square HSS 17.8cm x 17.8cm x 1.3cm
J
K
G
H
I
F
E
D
C
B
A158K703K
164K 730K
12
3
4
56
789
10
11
PRIMARY COMPRESSION MEMBER SIZING
PRIMARY TENSION MEMBER SIZING
SECONDARY COMPRESSION MEMBER SIZING
Exploded Axon
Render ViewInterior Render View Render View
Render View
9 ft
3/4” Glass Cladding
Steel Structure for Cladding
Concrete Deck
Tension Cable
Primary Tension Member
Secondary Compression Member
Primary Compression Member
Cross Section
Elevation0m 6m
0m 1.5m
Site Plan
Building 46
Building 36 Stata Center
Vassar Street
Connection Detail
PARALLEL AXIS THEOREM_
I = I + A d 2 I = 2*[702 in4 + 18.7 in2 *(23.1 in)2] + 2*[80.5 in4 + 11.6 in2 *(33.6 in)2 ] I = 47713.9 in4
i
i
i
DEFLECTION OF STRUCTURE_
allowed
= L/360 < 3.7 in
max = (5ωL4) / 384*EI
max
= (5*(0.09 K)(1320in)4) / (384*(29000 K/in2)(47713.9 in4))
max = 2.57 in
A1= 18.7 in2
I1
= 702 in4
d1= 23.1 in
A2= 18.7 in2
I2
= 702 in4
d2= 23.1 in
A3= 11.6 in2
I3
= 80.5 in4
d3= 33.6 in
A4= 11.6 in2
I4
= 80.5 in4
d4= 33.6 in
23.1 in
33.6 in
Axis
27
BEOMKI LEE / SHIYU WEI4.463 Building Structural Systems II (Fall 2012)Professor: J.A. Ochsendorf / TA: Caitlin Mueller
Sky Illusion
Snow Load 30 lb/ft
80 lb/ft
42 lb/ft
20 lb/ft
2 lb/ft
58 lb/ft
232 lb/ft
0.4 kN/m
1.1 kN/m
0.6 kN/m
0.3 kN/m
0.03 kN/m
0.8 kN/m
3.4 kN/m
249 K 1106 kN
People Load
Steel Arch (x4)
Steel Struts (x10)
3/4” Glass Cladding
Total Distributed Load
Total Load BRIDGE AREA = 107 ft * 10 ft
12.5 K 55.6 kNPoint Loads
Deck
LOADS_
FORCE POLYGON_
MEMBER SIZING_
Shear Diagram
Maximum Moment = 1818 K*ft
12.5K55.6 kN
55.6 kN
55.6 kN
12.5K 12.5K12.5K 12.5K12.5K12.5K12.5K12.5K
606 K*ft
-606 K*ft
485 K*ft
-485 K*ft
364 K*ft
-364 K*ft
242 K*ft
-242 K*ft
121K*ft0 K*ft
-121 K*ft9.7 ft
32 m
278 kN
278 kN
107 ft K 0
62.5 K 62.5 K
Moment Diagram
T = C = M/d; Greater depth of the beam implies smaller forces on the top and bottom member, suggesting a beam shape that is thick in the middle
CONCEPT_ TransparencyTransparencyThe gray, monolithic concrete buildings on Vassar street dominate the view of the people passing through. Therefore, we strived for a design that had minimal visual impact on the site. Also, since the Stata Center already has a very strong presence on the street, a simple bridge would not interfere with the view so highly sought after by tourists and also would not clutter up an already packed street.
1/4 inch = 1 kipForce in vertical strut = 9.7K = 46kN
= Checking for Uneven Loading Conditions
ω = 30lb/ft + 80 lb/ft = 110 lb/ft = 0.11 k/ftM = �L2 / 32M = [(0.11k/ft)(107.2ft)2]/32 = 474 k*inσ�= 15 ksi = M/S σ = 15ksi = 474 k*in/SS = 31.6 in3 = 517 cm3
= Check for Buckling over L/2
L/2 = 53.6 ftE = 29,000 ksiP = � 2EI/(kL)k = 1 (pin-pin condition)158K = � 2(29,000ksi)I / (1(53.6ft))Ireq
= 228in4 = 9490 cm4 --> 684in4 = 28470 cm4 (safety factor)
= Checking for Max Allowable Stress
Largest force = 158KStress = force/area15 ksi = 158K / areaArea
req = 158K / 15ksi =10.5in2 = 67.7 cm2
Choose 16”x12”x3/8” Rectangular HSS 40.6cm x 30.5cm x 0.9cmS
x = 87.7in3 S
y = 75.3in3
Ix = 702 in4 I
y = 452 in4
12.5K
9.7 ft
3 m
1
62.5K 62.5K
2 3 4 5 6 7 8 109 11
KA B C D E F I J
L
G H
12.5K 12.5K 12.5K
12.5K
12.5K
12.5K
12.5K
12.5K
12.5K
12.5K12.5K
12.5K
12.5K
12.5K12.5K12.5K 12.5K 12.5K12.5K
= Check for Max Allowable Stress
Largest force = 9.7 K15 ksi = 9.7 K/Area
req
Areareq
= 9.7 K/15ksi = 0.65 in2
Choose member 4”x3”x3/16” 10.2cm x 7.6cmx 0.5cmIx = 4.9 in4
Iy
= 3.2 in4
= Check for Buckling
E = 29,000 ksiP = � 2EI/(kL)2
9.7 K = � 2(29,000ksi)I/(1(144in))2
Ireq
= 0.97 in4 = 40.4 cm4 --> 2.91in4 = 121 cm4 (safety factor)
= Checking for Max Allowable Stress
Largest Force = 164.3KArea
req = 164.3K / 15ksi = 10.9 in2
Choose 7”x7”x1/2” Square HSS 17.8cm x 17.8cm x 1.3cm
J
K
G
H
I
F
E
D
C
B
A158K703K
164K 730K
12
3
4
56
789
10
11
PRIMARY COMPRESSION MEMBER SIZING
PRIMARY TENSION MEMBER SIZING
SECONDARY COMPRESSION MEMBER SIZING
Exploded Axon
Render ViewInterior Render View Render View
Render View
9 ft
3/4” Glass Cladding
Steel Structure for Cladding
Concrete Deck
Tension Cable
Primary Tension Member
Secondary Compression Member
Primary Compression Member
Cross Section
Elevation0m 6m
0m 1.5m
Site Plan
Building 46
Building 36 Stata Center
Vassar Street
Connection Detail
PARALLEL AXIS THEOREM_
I = I + A d 2 I = 2*[702 in4 + 18.7 in2 *(23.1 in)2] + 2*[80.5 in4 + 11.6 in2 *(33.6 in)2 ] I = 47713.9 in4
i
i
i
DEFLECTION OF STRUCTURE_
allowed
= L/360 < 3.7 in
max = (5ωL4) / 384*EI
max
= (5*(0.09 K)(1320in)4) / (384*(29000 K/in2)(47713.9 in4))
max = 2.57 in
A1= 18.7 in2
I1
= 702 in4
d1= 23.1 in
A2= 18.7 in2
I2
= 702 in4
d2= 23.1 in
A3= 11.6 in2
I3
= 80.5 in4
d3= 33.6 in
A4= 11.6 in2
I4
= 80.5 in4
d4= 33.6 in
23.1 in
33.6 in
Axis
28
H
11.79k
Z
E
68.95k
A
68.95k
11.79k
B
Mmax = wL^2 / 8 = (2k/ft)(15ft)^2 / 8 = 56.25 k-ft = 675 k-inσ = My/I = M/SSreq’d = 45 in^3
If solid and rectangular, then bh^2/6 = 45;Can use W12x35 (flange 12.5" deep / 6.56" wide)
G
B
Q
Q
15.0
1
68.95k23.58k
11.79k
11.79k
3.66k
6k
Bottom tied arches support deck and transfersload to bottom chord of truss.
Max internal force 7.46k so 0.5 sq in cross section neededHorizontal force 1.4k need tie of 0.1 sq in steel
Asymmetrical loading:Moment in arch: Mmax = wL^2 / 32= (80 psf)(18.33ft)(15ft)^2 / 32 = 10311 lb-ft = 123.7k-in15ksi = M/SSreq’d = 8.25 in^3
Max internal force 13.4k so 0.9 sq inHorizontal force 6k need tie of 0.4 sq in steel
Asymmetrical loading:Moment in arch: Mmax = wL^2 / 32= (130psf)(18.33ft)(15ft)^2 / 32 = 16755 lb-ft = 201k-inσ = My/I = M/SSreq’d = 13.4 in^3
3.66k
30.64k30.64k
U
F
116.07k
a
3.66k
Dominant Loads:
Top Chord C-T / D-U: 301.94k C 20 sq in or 10 sq in each sideBottom Chord K-V: 281.47k T 18.75 sq in or 9.4 sq each sideVertical G-Z: 173.85k C 11.6 sq in or 5.8 sq in each sideDiagonal Y-Z: 208.41k T 13.9 sq in or 6.9 sq in each side
Buckling check:Top chord C-T: Pcr = π^2 EI / (kL)^2 safety factor 3 * 301.94k = π^2 (30000 ksi) I / (0.5 * 220.33in)^2 Ireq = 37.1 in^4
Vertical G-Z: Pcr = π^2 EI / (kL)^2 safety factor 3 * 173.85k = π^2 (30000 ksi) I / (0.5 * 180in)^2 Ireq = 14.3 in^4
Top Chord C-T / D-U: 296.23k C 19.75 sq in or 10 sq in each sideBottom Chord K-V: 281.47k T 18.75 sq in or 9.4 sq each sideVertical G-Z: 173.85k C 11.6 sq in or 5.8 sq in each sideDiagonal Y-Z: 208.41k T 13.9 sq in or 6.9 sq in each side
Top Chord C-T / D-U: 301.94k C 20 sq in or 10 sq in each sideBottom Chord K-V: 268.39k T 17.9 sq in or 9 sq each sideVertical O-I,/ G-Z: 155.94k C 10.4 sq in or 5.2 sq in each sideDiagonal O-P / Y-Z: 194.79k T 13 sq in or 6.5 sq in each side
6k6k
X
23.58k
44.79k
24.16k11.79k
R1 + R2 = 346.5k
Moment equilibirum around a: (18.33*23.58k) + (18.33*2*23.58) + ... + (18.33*6*68.95k) = (18.33*6*R2)R2 = 218.64k, R1 = 127.86k
E
I
YR
C DE
2k/ft
F
S V
T UW
N
D
110.0018.33
J
TY
173.85k
R1 = 127.86k
b
AD
O
C
C
O
DB EA
WR
R2 = 218.64k
24.0
1
6k
23.58k
44.79k44.79k
6k
V
24.16k24.16k
6k Dead load distributed evenly, live load on half.
Dead load and live load distributed evenly.
Z
ML
P
O KJ
H
B
GS
I
For 18.33' segment (distance between truss joints), supporting roof load on top and pedestrian load on bottom.
Distributed load for 18' span
A
Approaching the center, the arch supports become shallow like beams.
18.85k
18.85k18.85k 18.85k 18.85k
18.85k
18.85k
30.64k30.64k
X
9.00
30.64k
123.76k 123.76k
NM
L
3.66k
P
C
O
3.66k
K
68.95k
24.16k
3.66k
D
E
1k = 25"
A
WUT
S
Q
P
F
EM
D
I
BA
N, O
K
J
G
H, ZX
Y
C
LE
F
D
A
B
H, Z
G
CK
M
I
N, O
B
J
L
A
U
X
C
Q
S
V
W
E
P, YR
R
T
B
C
VD
Assumed dead load:20 psf steel deck and horizontal supports25 psf 2" concrete topping20 psf steel truss35 psf cladding100 psf total
Assumed live load:80 psf pedestrian30 psf snow/rain
Total distributed load 210 psf
...split so that top chords take 80 psf and bottom 130 psf
x
y
4.75"
Sample Section: Truss element C-T in compression
1.5"
1"
Aa = 9 sq in Ya = 3.25" AaYa = 29.25in^3Ab = 8 sq in Yb = 6.75" AbYb = 54in^3 total 83.25in^3 / 17in^2 = 4.90" to centroid
Ia = (1.5)(6)^3 / 12 = 27 d = 3.25 - 4.90 = -1.65 Ad = (9)(-1.65)^2 = 24.5Ib = (8)(1)^3 / 12 = 0.67 d = 6.75 - 4.90 = 1.75 Ad = (8)(1.85)^2 = 27.38Iy = 79.55 in^4
Ix = (6)(1.5)^3 / 12 + (1)(8)^3 / 12 = 44.35 in^4
6"
8"
Estimate deflection: Allowed 3.7" deflection minimum3.7" = 5wL^4 / (384*E*I)LL = 110lb/sqft; 0.09k/inI req'd > 5(0.09k/in)(1320in)^4 / 384(30000k/sqin)(3.7in)I req'd > 32000 in^4
Taking just the abstracted four chords of the truss (two top, two bottom):
I = Σ Aidi^2 = 4 * 10 sqin * (90 in)^2 = 324000 in^4, an order of magnitude above what is necessary
90 in
10 sq in ea
29
H
11.79k
Z
E
68.95k
A
68.95k
11.79k
B
Mmax = wL^2 / 8 = (2k/ft)(15ft)^2 / 8 = 56.25 k-ft = 675 k-inσ = My/I = M/SSreq’d = 45 in^3
If solid and rectangular, then bh^2/6 = 45;Can use W12x35 (flange 12.5" deep / 6.56" wide)
G
B
Q
Q
15.0
1
68.95k23.58k
11.79k
11.79k
3.66k
6k
Bottom tied arches support deck and transfersload to bottom chord of truss.
Max internal force 7.46k so 0.5 sq in cross section neededHorizontal force 1.4k need tie of 0.1 sq in steel
Asymmetrical loading:Moment in arch: Mmax = wL^2 / 32= (80 psf)(18.33ft)(15ft)^2 / 32 = 10311 lb-ft = 123.7k-in15ksi = M/SSreq’d = 8.25 in^3
Max internal force 13.4k so 0.9 sq inHorizontal force 6k need tie of 0.4 sq in steel
Asymmetrical loading:Moment in arch: Mmax = wL^2 / 32= (130psf)(18.33ft)(15ft)^2 / 32 = 16755 lb-ft = 201k-inσ = My/I = M/SSreq’d = 13.4 in^3
3.66k
30.64k30.64k
U
F
116.07k
a
3.66k
Dominant Loads:
Top Chord C-T / D-U: 301.94k C 20 sq in or 10 sq in each sideBottom Chord K-V: 281.47k T 18.75 sq in or 9.4 sq each sideVertical G-Z: 173.85k C 11.6 sq in or 5.8 sq in each sideDiagonal Y-Z: 208.41k T 13.9 sq in or 6.9 sq in each side
Buckling check:Top chord C-T: Pcr = π^2 EI / (kL)^2 safety factor 3 * 301.94k = π^2 (30000 ksi) I / (0.5 * 220.33in)^2 Ireq = 37.1 in^4
Vertical G-Z: Pcr = π^2 EI / (kL)^2 safety factor 3 * 173.85k = π^2 (30000 ksi) I / (0.5 * 180in)^2 Ireq = 14.3 in^4
Top Chord C-T / D-U: 296.23k C 19.75 sq in or 10 sq in each sideBottom Chord K-V: 281.47k T 18.75 sq in or 9.4 sq each sideVertical G-Z: 173.85k C 11.6 sq in or 5.8 sq in each sideDiagonal Y-Z: 208.41k T 13.9 sq in or 6.9 sq in each side
Top Chord C-T / D-U: 301.94k C 20 sq in or 10 sq in each sideBottom Chord K-V: 268.39k T 17.9 sq in or 9 sq each sideVertical O-I,/ G-Z: 155.94k C 10.4 sq in or 5.2 sq in each sideDiagonal O-P / Y-Z: 194.79k T 13 sq in or 6.5 sq in each side
6k6k
X
23.58k
44.79k
24.16k11.79k
R1 + R2 = 346.5k
Moment equilibirum around a: (18.33*23.58k) + (18.33*2*23.58) + ... + (18.33*6*68.95k) = (18.33*6*R2)R2 = 218.64k, R1 = 127.86k
E
I
YR
C DE
2k/ft
F
S V
T UW
N
D
110.0018.33
J
TY
173.85k
R1 = 127.86k
b
AD
O
C
C
O
DB EA
WR
R2 = 218.64k
24.0
1
6k
23.58k
44.79k44.79k
6k
V
24.16k24.16k
6k Dead load distributed evenly, live load on half.
Dead load and live load distributed evenly.
Z
ML
P
O KJ
H
B
GS
I
For 18.33' segment (distance between truss joints), supporting roof load on top and pedestrian load on bottom.
Distributed load for 18' span
A
Approaching the center, the arch supports become shallow like beams.
18.85k
18.85k18.85k 18.85k 18.85k
18.85k
18.85k
30.64k30.64k
X
9.00
30.64k
123.76k 123.76k
NM
L
3.66k
P
C
O
3.66k
K
68.95k
24.16k
3.66k
D
E
1k = 25"
A
WUT
S
Q
P
F
EM
D
I
BA
N, O
K
J
G
H, ZX
Y
C
LE
F
D
A
B
H, Z
G
CK
M
I
N, O
B
J
L
A
U
X
C
Q
S
V
W
E
P, YR
R
T
B
C
VD
Assumed dead load:20 psf steel deck and horizontal supports25 psf 2" concrete topping20 psf steel truss35 psf cladding100 psf total
Assumed live load:80 psf pedestrian30 psf snow/rain
Total distributed load 210 psf
...split so that top chords take 80 psf and bottom 130 psf
x
y
4.75"
Sample Section: Truss element C-T in compression
1.5"
1"
Aa = 9 sq in Ya = 3.25" AaYa = 29.25in^3Ab = 8 sq in Yb = 6.75" AbYb = 54in^3 total 83.25in^3 / 17in^2 = 4.90" to centroid
Ia = (1.5)(6)^3 / 12 = 27 d = 3.25 - 4.90 = -1.65 Ad = (9)(-1.65)^2 = 24.5Ib = (8)(1)^3 / 12 = 0.67 d = 6.75 - 4.90 = 1.75 Ad = (8)(1.85)^2 = 27.38Iy = 79.55 in^4
Ix = (6)(1.5)^3 / 12 + (1)(8)^3 / 12 = 44.35 in^4
6"
8"
Estimate deflection: Allowed 3.7" deflection minimum3.7" = 5wL^4 / (384*E*I)LL = 110lb/sqft; 0.09k/inI req'd > 5(0.09k/in)(1320in)^4 / 384(30000k/sqin)(3.7in)I req'd > 32000 in^4
Taking just the abstracted four chords of the truss (two top, two bottom):
I = Σ Aidi^2 = 4 * 10 sqin * (90 in)^2 = 324000 in^4, an order of magnitude above what is necessary
90 in
10 sq in ea
30
60 cm2 Hollow Steel Arch Tension Tie
5 cm Thick Reinforced Concrete Deck
22 mm Diameter Steel Cable
25.4 cm Diameter 22mm Thick Hollow Steel Arch
20 cm deep I-BeamsCompression Flange: 18 cm2
Tension Flagne: 20 cm2
Duoble-Pane Low-E Glass
Insulated Panels
NETWORK NODE Precast Concrete Deck (Local Bending)
Deck Hangers (Tension)
Lattice Members (Bending)
Arch Sized by Evenly Distributed Loads (Compression)
Local Moment
Load in tributary area Compression/Tension Concrete Thickness
WL2
8
(45872N +18933N) / 3.929m = 16494 N/m
T=C=MD
(3.9m)2(16494 N/m)
31.8 kN-m 637 kN=5 cm
= 31.8 kN-m8
σFA
= AFσ
= = = =637 kN
.06 m2 .06 m2
3.05 m2.24 cmx100
9308 kN/m2
Steel thickness
σFA
= AFσ
= = = =637 kN
.006 m2 .006 m2
3.05 m 2 mm steel bottom
x1000103421 kN/m2
Tension in CableCross sectional cable area
AFσ
= = =39 kN
x10000 cm2/m2 = 103421 kN/m2
11.9 m2 (46 kN +19 kN)
= 38.7 kN .0003 m2 3.75 cm2 = 1.1cmradius
Tributary vertical load on deck
resultant angular forcecos(33.23o)
2
Loads Moment as a �xed beam Force in tension and compression of �anges
Cross sectional area of tension and compression �anges
WL2
16
= .0017 m2 186 kN103421 kN/m2
= 37.2 kN-m(2.55 kN/m) (15.3 m)2
1637.2 kN-m
= 186 kN20 cm
Lattice member max tributary area (measured with Rhino model) = 17.8 m2
Length of lattice member= 15.3 m17.8 m2 (1.4kN + 0.7 kN) = 39 kN39 kN / 15.3m = 2.55 kN/m
Total tributary load of member:Linear load of Lattice member:
AFσ
= = = x10000 = 68.67 cm2 cross section710 kN
.007 m2103421 kN/m2
Total deck load in plane of one arch: (Tension hanger load) x 7
Total vertical roof load and arch self weigth on one arch: ((32.1 kN x 14 roof ) + (17.6kN x 14 arch) / 2 = 348 kNAngle of Vertical load to plane of arch: 33.23o
(38.7 kN)(7) = 271 kN= = 417 kNcos(33.23o) resultant angular
force
Total vertical loads
Total Planar loads =
Force along each point = 688 kN / 7 = 98.3 kN
271kN + 417 kN = 688 kN
348 kN98 kN
98 kN
98 kN
98 kN
98 kN
98 kN
710 kN
Required Cross sectional Area
98 kN98 kN98 kN98 kN98 kN98 kN98 kN
CALCULATIONS
Roof Load(32.1 kN)
Lattice Reaction(32.6 kN)
Arch Reaction(98.3 kN)
Arch Self Weight
(17.6 kN)
Deck Load(38.7 kN)
5 cm concrete top
< 2x
A = π r2 3.75 cm2 = π r2 r
T = C =MD
98 kN
x10000 =
75.6 kN103421 kN/m2
Total Vertical Load on Individual Arch: 348 kNPoint loads on arch (at lattice points):Lattice Component of vertical force:
Additonal Tension Resulting from Oblique Angles in Plan:
Area to be added to Tension Flange:
348 kN / 7 = 49.8 kN
cos(64.4o)32.6 kN
tan(33.2o) 49.8kN = 32.6 kN
= 75.6 kN
= 0.0003 m2 = 3.15 cm2
Addiontal area for tension �ange (from arch interaction)
Total Tension Flange area: 21.15 cm2 cross sectiontension �ange
20cm
8cm
22.5 mm
26.5 mm
Arch Tension Tie
σFA
= AFσ
= = =621kN
103421kN/m2
Global moment under full asymmetrical loading(derived in Multiframe and compared with hand calculations - not shown)
Multiframe ResultsMoment of Inertia
Under even loading conditions Under asymmetrical loading conditions
Neutral Axis Finding
98 kN
621 kN
98 kN
98 kN
98 kN
98 kN
98 kN
98 kN
710 kN
Arch Sized by Assymetrical Load (Global Bending)
Arch/Total Bridge Deflection (Compression + Bending)
Horizontal component of forces = Cross sectional Area
Member sizing based on sectional Modulus Formula for hollow tube
Desired Outer radius:
Wall Thickness: .022 m
.011 m 11 mm Max De�ection: 59mm
Max Moment: 399 kN-m
621 kN
.006 m2
= 60 cm2
20 cm
20 cm
1 cm
π πr1 = =
4Sr2+πr24 4(399kN-m) 0.25m +π (0.25m)4
4
S =π(r2
4-r14)
4r2
4
25.4 cm
23.2 cminner radius
σMS
= σM
S
I = ΣI1 + ΣA1d12 (.001m4)+(4.14E-05m4)+(2(.066m2(.532))+(2(.013m2(2.642)) = .228m4
Max De�ection
max = = =5wL4
384EI
= =
Required Sectional modulus399 kN-m
.003 m3
103421 kN/m2241 kN-m
352 kN-m
304 kN-m 98.5 kN-M
265 kN-M399 kN-M
300 kN-M
-39 mm
-59 mm
-50 mm -17 mm 21 mm
42 mm
32 mm
5(38495)314
384(1.99E+11).228
3.17
8 m
0.53 mTop Axis
Bottom Axis
Neutral Axis
22 mm
25.4 cm
18 cm2
cross sectioncompression �ange
31
60 cm2 Hollow Steel Arch Tension Tie
5 cm Thick Reinforced Concrete Deck
22 mm Diameter Steel Cable
25.4 cm Diameter 22mm Thick Hollow Steel Arch
20 cm deep I-BeamsCompression Flange: 18 cm2
Tension Flagne: 20 cm2
Duoble-Pane Low-E Glass
Insulated Panels
NETWORK NODE Precast Concrete Deck (Local Bending)
Deck Hangers (Tension)
Lattice Members (Bending)
Arch Sized by Evenly Distributed Loads (Compression)
Local Moment
Load in tributary area Compression/Tension Concrete Thickness
WL2
8
(45872N +18933N) / 3.929m = 16494 N/m
T=C=MD
(3.9m)2(16494 N/m)
31.8 kN-m 637 kN=5 cm
= 31.8 kN-m8
σFA
= AFσ
= = = =637 kN
.06 m2 .06 m2
3.05 m2.24 cmx100
9308 kN/m2
Steel thickness
σFA
= AFσ
= = = =637 kN
.006 m2 .006 m2
3.05 m 2 mm steel bottom
x1000103421 kN/m2
Tension in CableCross sectional cable area
AFσ
= = =39 kN
x10000 cm2/m2 = 103421 kN/m2
11.9 m2 (46 kN +19 kN)
= 38.7 kN .0003 m2 3.75 cm2 = 1.1cmradius
Tributary vertical load on deck
resultant angular forcecos(33.23o)
2
Loads Moment as a �xed beam Force in tension and compression of �anges
Cross sectional area of tension and compression �anges
WL2
16
= .0017 m2 186 kN103421 kN/m2
= 37.2 kN-m(2.55 kN/m) (15.3 m)2
1637.2 kN-m
= 186 kN20 cm
Lattice member max tributary area (measured with Rhino model) = 17.8 m2
Length of lattice member= 15.3 m17.8 m2 (1.4kN + 0.7 kN) = 39 kN39 kN / 15.3m = 2.55 kN/m
Total tributary load of member:Linear load of Lattice member:
AFσ
= = = x10000 = 68.67 cm2 cross section710 kN
.007 m2103421 kN/m2
Total deck load in plane of one arch: (Tension hanger load) x 7
Total vertical roof load and arch self weigth on one arch: ((32.1 kN x 14 roof ) + (17.6kN x 14 arch) / 2 = 348 kNAngle of Vertical load to plane of arch: 33.23o
(38.7 kN)(7) = 271 kN= = 417 kNcos(33.23o) resultant angular
force
Total vertical loads
Total Planar loads =
Force along each point = 688 kN / 7 = 98.3 kN
271kN + 417 kN = 688 kN
348 kN98 kN
98 kN
98 kN
98 kN
98 kN
98 kN
710 kN
Required Cross sectional Area
98 kN98 kN98 kN98 kN98 kN98 kN98 kN
CALCULATIONS
Roof Load(32.1 kN)
Lattice Reaction(32.6 kN)
Arch Reaction(98.3 kN)
Arch Self Weight
(17.6 kN)
Deck Load(38.7 kN)
5 cm concrete top
< 2x
A = π r2 3.75 cm2 = π r2 r
T = C =MD
98 kN
x10000 =
75.6 kN103421 kN/m2
Total Vertical Load on Individual Arch: 348 kNPoint loads on arch (at lattice points):Lattice Component of vertical force:
Additonal Tension Resulting from Oblique Angles in Plan:
Area to be added to Tension Flange:
348 kN / 7 = 49.8 kN
cos(64.4o)32.6 kN
tan(33.2o) 49.8kN = 32.6 kN
= 75.6 kN
= 0.0003 m2 = 3.15 cm2
Addiontal area for tension �ange (from arch interaction)
Total Tension Flange area: 21.15 cm2 cross sectiontension �ange
20cm
8cm
22.5 mm
26.5 mm
Arch Tension Tie
σFA
= AFσ
= = =621kN
103421kN/m2
Global moment under full asymmetrical loading(derived in Multiframe and compared with hand calculations - not shown)
Multiframe ResultsMoment of Inertia
Under even loading conditions Under asymmetrical loading conditions
Neutral Axis Finding
98 kN
621 kN
98 kN
98 kN
98 kN
98 kN
98 kN
98 kN
710 kN
Arch Sized by Assymetrical Load (Global Bending)
Arch/Total Bridge Deflection (Compression + Bending)
Horizontal component of forces = Cross sectional Area
Member sizing based on sectional Modulus Formula for hollow tube
Desired Outer radius:
Wall Thickness: .022 m
.011 m 11 mm Max De�ection: 59mm
Max Moment: 399 kN-m
621 kN
.006 m2
= 60 cm2
20 cm
20 cm
1 cm
π πr1 = =
4Sr2+πr24 4(399kN-m) 0.25m +π (0.25m)4
4
S =π(r2
4-r14)
4r2
4
25.4 cm
23.2 cminner radius
σMS
= σM
S
I = ΣI1 + ΣA1d12 (.001m4)+(4.14E-05m4)+(2(.066m2(.532))+(2(.013m2(2.642)) = .228m4
Max De�ection
max = = =5wL4
384EI
= =
Required Sectional modulus399 kN-m
.003 m3
103421 kN/m2241 kN-m
352 kN-m
304 kN-m 98.5 kN-M
265 kN-M399 kN-M
300 kN-M
-39 mm
-59 mm
-50 mm -17 mm 21 mm
42 mm
32 mm
5(38495)314
384(1.99E+11).228
3.17
8 m
0.53 mTop Axis
Bottom Axis
Neutral Axis
22 mm
25.4 cm
18 cm2
cross sectioncompression �ange
32
SECONDARY
A required = 4 in2 Buckling
DC
3 x 1 1/2 x 1/4
A required = 7.3 in2
L = 5' (fixed to deck)
26.6 kips 63.42 kips
SIZING MEMBERS
ARCH AND FIN SYSTEM
I required = 21.13 in4
87.0 kips
25.67 kips
80.9 kips20.1 kips
21.2 kips
A required = 5 in2
56.9 kips
Crushing
Mullion / FinsCOMPRESSION STRUT
L = 37'Buckling I required = 150 in4
12.5 x 0.250
B
63.42 kips
56.9 kips
A
50'40' 25'
Crushing
G
F
37.5 kips37.5 kips
75 kips
E
D
PRIMARY COMP. MEMBER (C1)
80.6 kips
C
A
B
D
E
O
N
M
L
K
J
F
G
H
4 k4 k4 k4 k4 k4 k4 k4 k4 k4 k4 k4 k4 k4 k
C
B
39.6 kips41 kips
A required = 5.5 in2
TENSION CABLE (T1)
A
26.6 kips
B C D
12.5 x 0.625
6.0 x 0.312
A required = 5.34 in2
40’12192mm
29’37’
78’
8840mm
23780mm
11280mm
10’3050mm
25’7620mm
50’15240mm
L = 50.7'Buckling I required = 377.7 in4
A required = 7.26 in2
PRIMARY TENSION MEMBER (T2)
MAXIMUM DEFLECTION
COMPRESSION STRUT (C2)
A E F G H I J K L M
87.0 kips21.3 kips
N O
21.2 kips
I
Buckling
A required = 1.71 in2
DL + Live Load
DL + Snow Load
A required = 5.8 in2
CABLE 2
M
J
A
Vd
K
L
RIGID MOMENT FRAME
S required = 8.3 in3
S required = 27.1 in3
M max = 125 kips.ft
Point load = 5ft x 1.6 kips/ft / 2 = 4 kips
M max = 406.3 kips.ft
DL + Snow Load = 0.4 kips/ft
DL + Live Load = 1.3 kips/ft
H
1.6 kips/ft
Snow Load = 300lb/ft = 0.3 kips/ft
DETAIL A
W6x8.5 Beam and girdersCast steel tube Y-column
1.5” Double Glazed Frosted Glass1”x9” Steel strap
1.5” Double Glazed Clear Vision Glass
DETAIL A
DETAIL B
DETAIL B
CABLE 1SIZING MEMBERS
CABLE AND FIN SYSTEM
REACTION ON BUILDING 46 Top Level
25.7 kips
20.1 kips
C
REACTION ON BUILDING 46 3rd level
60.8 kips
DA E F G H
87.0 kips
87.0 kips
80.9 kips
H
G
87.5 kips
87.0 kips
25.7 kips
39.6 kips
F
J
K
L
87.0 kips
21.3 kips
41 kips
M
N
O
I
21.3 kips
63.4 kips
82.9 kips
37.5 kips
63.4 kips
21.3 kips
J K L M
4 k 4 k 4 k 4 k 4 k
Vc
Hc
Vb
4 k 4 k 4 k 4 k 4 k 4 k 4 k
T1
T1
C1
C1
T2
C2
C2108.9 kips
T2
SIZING MEMBERS
ARCH AND CABLE
A required = 4.23 in2
Hb
ARCH 1
F
CABLE 2
A required = 5.8 in2
FRAME 1
PRIMARY
PRIMARY MEMBERS
DEFLECTION ANALYSIS
S required = 27.1 in3W10X12W10X22S required = 8.3 in3
FRAME 2
C
A
B
C
D
E
F
G
80.9 kips
63.4 kips
61.1 kips
37.5 kips
26.6 kips
37.5 kips
61.1 kips
G H I J K L M N
56.9 kips
63.42 kips
4 k 4 k 4 k 4 k 4 k 4 k 4 k 4 k
O
Va
LOADS
26.6 kips63.42 kips
49.4 kips
61 kips
DL = .445 kips/ft = 0.5 kips/ft
Total Weight of Glass = 0.325kips/ft
79.2 kips
A required = 2.38 in2
TENSION CABLE
REACTION ON BUILDING 36
ARCH 1
B C D
Total Weight of Steel = 0.12 kips/ft
BC
A
ARCH 2
35.7 kips
A required = 4.23 in2
A E F G H
4 k 4 k 4 k 4 k 4 k 4 k 4 k 4 k 4 k 4 k
87.0 kips25.67 kips
B
B
I required = 1.46 in4
TRIBUTARY AREA
Width of Deck = 10ft
THE HANDSHAKEBARRY BEAGEN + TRYGVE WASTVEDT
LL = 800lb/ft = 0.8kips/ft
A required = 1.77 in2
Total Load = 1.6kips/ft10’
387 kN
359.9 kN89.4 kN
94.3 kN
118.3 kN282.1 kN
253.1 kN
550.9 kN.m
169.5 kN.m282.1 kN
253.1 kN
387 kN
359.9 kN
182.4 kN387 kN
94.7 kN
484.4 kN
368.8 kN94.7 kN
282 kN
166.8 kN
34.5 cm2
182.4 kN 176.1 kN
358.5 kN
176.1 kN
389.2 kN 114.3 kN
387 kN
118.3 kN
282 kN 166.8 kN
271.8 kN
15.4 cm227.3 cm2 11.4 cm2
31.8 cm x 0.6 cm
35.5 cm2
31.8 cm x 1.6 cm
152 mm x 7.9 mm
114.2 kN
37.4 cm227.3 cm2
37.4 cm211 cm2
270.5kN
158.8 kN
23.4 kN/m4.4 kN/m11.7 kN/m
3.1 m
1.8 kN/m4.7 kN/m
7.3 kN/m
17.9 kN
Column Side = -0.86 in
Cable Side = -3.3 in
33
SECONDARY
A required = 4 in2 Buckling
DC
3 x 1 1/2 x 1/4
A required = 7.3 in2
L = 5' (fixed to deck)
26.6 kips 63.42 kips
SIZING MEMBERS
ARCH AND FIN SYSTEM
I required = 21.13 in4
87.0 kips
25.67 kips
80.9 kips20.1 kips
21.2 kips
A required = 5 in2
56.9 kips
Crushing
Mullion / FinsCOMPRESSION STRUT
L = 37'Buckling I required = 150 in4
12.5 x 0.250
B
63.42 kips
56.9 kips
A
50'40' 25'
Crushing
G
F
37.5 kips37.5 kips
75 kips
E
D
PRIMARY COMP. MEMBER (C1)
80.6 kips
C
A
B
D
E
O
N
M
L
K
J
F
G
H
4 k4 k4 k4 k4 k4 k4 k4 k4 k4 k4 k4 k4 k4 k
C
B
39.6 kips41 kips
A required = 5.5 in2
TENSION CABLE (T1)
A
26.6 kips
B C D
12.5 x 0.625
6.0 x 0.312
A required = 5.34 in2
40’12192mm
29’37’
78’
8840mm
23780mm
11280mm
10’3050mm
25’7620mm
50’15240mm
L = 50.7'Buckling I required = 377.7 in4
A required = 7.26 in2
PRIMARY TENSION MEMBER (T2)
MAXIMUM DEFLECTION
COMPRESSION STRUT (C2)
A E F G H I J K L M
87.0 kips21.3 kips
N O
21.2 kips
I
Buckling
A required = 1.71 in2
DL + Live Load
DL + Snow Load
A required = 5.8 in2
CABLE 2
M
J
A
Vd
K
L
RIGID MOMENT FRAME
S required = 8.3 in3
S required = 27.1 in3
M max = 125 kips.ft
Point load = 5ft x 1.6 kips/ft / 2 = 4 kips
M max = 406.3 kips.ft
DL + Snow Load = 0.4 kips/ft
DL + Live Load = 1.3 kips/ft
H
1.6 kips/ft
Snow Load = 300lb/ft = 0.3 kips/ft
DETAIL A
W6x8.5 Beam and girdersCast steel tube Y-column
1.5” Double Glazed Frosted Glass1”x9” Steel strap
1.5” Double Glazed Clear Vision Glass
DETAIL A
DETAIL B
DETAIL B
CABLE 1SIZING MEMBERS
CABLE AND FIN SYSTEM
REACTION ON BUILDING 46 Top Level
25.7 kips
20.1 kips
C
REACTION ON BUILDING 46 3rd level
60.8 kips
DA E F G H
87.0 kips
87.0 kips
80.9 kips
H
G
87.5 kips
87.0 kips
25.7 kips
39.6 kips
F
J
K
L
87.0 kips
21.3 kips
41 kips
M
N
O
I
21.3 kips
63.4 kips
82.9 kips
37.5 kips
63.4 kips
21.3 kips
J K L M
4 k 4 k 4 k 4 k 4 k
Vc
Hc
Vb
4 k 4 k 4 k 4 k 4 k 4 k 4 k
T1
T1
C1
C1
T2
C2
C2108.9 kips
T2
SIZING MEMBERS
ARCH AND CABLE
A required = 4.23 in2
Hb
ARCH 1
F
CABLE 2
A required = 5.8 in2
FRAME 1
PRIMARY
PRIMARY MEMBERS
DEFLECTION ANALYSIS
S required = 27.1 in3W10X12W10X22S required = 8.3 in3
FRAME 2
C
A
B
C
D
E
F
G
80.9 kips
63.4 kips
61.1 kips
37.5 kips
26.6 kips
37.5 kips
61.1 kips
G H I J K L M N
56.9 kips
63.42 kips
4 k 4 k 4 k 4 k 4 k 4 k 4 k 4 k
O
Va
LOADS
26.6 kips63.42 kips
49.4 kips
61 kips
DL = .445 kips/ft = 0.5 kips/ft
Total Weight of Glass = 0.325kips/ft
79.2 kips
A required = 2.38 in2
TENSION CABLE
REACTION ON BUILDING 36
ARCH 1
B C D
Total Weight of Steel = 0.12 kips/ft
BC
A
ARCH 2
35.7 kips
A required = 4.23 in2
A E F G H
4 k 4 k 4 k 4 k 4 k 4 k 4 k 4 k 4 k 4 k
87.0 kips25.67 kips
B
B
I required = 1.46 in4
TRIBUTARY AREA
Width of Deck = 10ft
THE HANDSHAKEBARRY BEAGEN + TRYGVE WASTVEDT
LL = 800lb/ft = 0.8kips/ft
A required = 1.77 in2
Total Load = 1.6kips/ft10’
387 kN
359.9 kN89.4 kN
94.3 kN
118.3 kN282.1 kN
253.1 kN
550.9 kN.m
169.5 kN.m282.1 kN
253.1 kN
387 kN
359.9 kN
182.4 kN387 kN
94.7 kN
484.4 kN
368.8 kN94.7 kN
282 kN
166.8 kN
34.5 cm2
182.4 kN 176.1 kN
358.5 kN
176.1 kN
389.2 kN 114.3 kN
387 kN
118.3 kN
282 kN 166.8 kN
271.8 kN
15.4 cm227.3 cm2 11.4 cm2
31.8 cm x 0.6 cm
35.5 cm2
31.8 cm x 1.6 cm
152 mm x 7.9 mm
114.2 kN
37.4 cm227.3 cm2
37.4 cm211 cm2
270.5kN
158.8 kN
23.4 kN/m4.4 kN/m11.7 kN/m
3.1 m
1.8 kN/m4.7 kN/m
7.3 kN/m
17.9 kN
Column Side = -0.86 in
Cable Side = -3.3 in
34
Frame development
offset determined by need to reconcile center of gravity in relation to the possible torsion of the connec-tion between the arch and the deck
Fmax = 165 kips
o = F/A
o = 15 ksiF= 340 kips
o = 165 kips/A
A= 165 kips/ 15ksi
A= 11 in^2
rectangular section with A > 22.7 :14” x 12” x 1/2”
14”
12”
1/2”
x
y
I x = 678 in^4Iy = 536 in ^4
Sx = 96.9Sy = 89.3
Rectangular Arch Beam sized as follows
Checking for Buckling
since the arch is not being completely stabilized by the connection to the deck it is necessary for us to check for buckling at two areas.The most conservative span is halfway along the arch above the deck. Because the force is highest on the left side of the arch we assume that buckling will occur over the first half. A less con-serative span is the distance between the plates that connect the arch to the deck.
conservative estimate check Pcr = [pi^2(EI)]/kL^2
E = 29000ksiI = Ix = 678 in^4
L = 75’
Pcr = [pi^2(29000ksi*678 in^4)]/75ft(12inch/ft)^2
Pcr = 239.6 kips
less conservative estimate check
working backward we get the actual buckling length
Because this length occurs in between our two estimates it is safe to assume that the member size chosen will be adequate to prevent buckling along the arch.
Pcr = [pi^2(EI)]/kL^2
E = 29000ksiI = Ix = 678 in^4
L = 11’
Pcr = [pi^2(29000ksi*678 in^4)]/11ft(12inch/ft)^2
Pcr = 11137 kips
L = sqrt[ [pi^2(EI)]/(Pcr * 3)]L = 436 in or about 36 ft
Calculating Dead load
9’
15’
8x10.5
assuming 11 identical panels @ 5.1ft^2
Panel size:2 @ 110ft x 8ft x .25in1 @ 110ft x 10ft x .25in
Floor size:110ft x 8ft x .5ft
.1’
Dead Load steel = 35519lbs
Dead Load glass = 10335lbs
Frame Load
2k2k2k2k2k2k2k4k
load greater on arch because deck not supported on the left side10k
Bending and asymmetrical loadBy finding the maximum bending moment we can increase the size of the arch in order to stiffen it for asymmetrical loading
DeformationUsing a computer model of our arch and loads mentioned above in the deformation was calculated.The input for the beam size was the minimum dimension of 12” x 14“
Mmax =wl^2/32
w = live load (estimate to be about 110 psf or 1.1 ksf)
Mmax = [1.1 ksf(90ft)^2]/32Mmax = 278 kf
o = Mmax/Sxo = 15 ksi
90ft
Sx = Mmax/ oSx = 278kf(12in/ft)/15 ksiSx = 222.4 in^3
The rectangular section that has a section modulus of at least 223 and keeps the width and thickness of the current beam is 12” x 20” x 1/2”
b
B
hH
S = [BH^2/6] - (bh^3)/(6H)S = [12in*24in^2/6] - (11in*23in^3/(24in*6)S= 222.5in^3
The beam section is large enough because the two Section modulus values are equal
14”
24”
Steel = .28 lbs/in^3
20,000 lbs = 20 kips
20 kips/11 = 1.8 kips
Floor Dead load: 66147lbs/ (880) = 75psf80ft^2 x 75psf = 6000lbs
Roof Dead load: 2851lbs/(880) = 3.25psf80ft^2 x 3.25ps =260 lbs
Wall Dead Load: 3742lbs/(880) = 4.25psf100ft^2 x (4.25psf +4.25psf) = 850lbs
Steel Dead load 35519lbs/(165ft^2) = 215 psf15ft^2 x 215psf = 3229lbs
Live load Roof = 30psf
Live load Floor 80psf80ft^2 x (30psf + 80psf) = 8,800lbs
Total Load = 20,000 lbs
Wall and Roof Load
Floor Load
Calculating Point Loads
Concrete = .087 lbs/in^3Dead load Concrete = 66147 lbs
Glass = .09 lbs/in^3
157.9kN
45.9kN
294.3kN
3.59kN/m^2
.15kN/m^2
.20kN/m^2
10.29kN/m^2
3.83kN/m^2
1.43kN/m^2
89kN
8kN
734kN
279mm
.35mm x .3m x .025m
1065kN
49539kN
10.9m
5.6m^3
5.6m^3
110ft
28ft
52ft
Iguana Bridge
frame attachment to concrete deck
10’
1’
6”
Stata center
Building 46
DMAX = 1.442”
DMAX = 1.368”
DMAX = 0.930”
DMAX = 0.313”
DMAX = 0.988”
DMAX = 1.397”
Dmax = 1.483”
DMAX = 1.144”
Our arch is properly sized because 1.483” is lower than the allowable 3.5” for deformation of this arch
Chris Martin & Robert White
35
Frame development
offset determined by need to reconcile center of gravity in relation to the possible torsion of the connec-tion between the arch and the deck
Fmax = 165 kips
o = F/A
o = 15 ksiF= 340 kips
o = 165 kips/A
A= 165 kips/ 15ksi
A= 11 in^2
rectangular section with A > 22.7 :14” x 12” x 1/2”
14”
12”
1/2”
x
y
I x = 678 in^4Iy = 536 in ^4
Sx = 96.9Sy = 89.3
Rectangular Arch Beam sized as follows
Checking for Buckling
since the arch is not being completely stabilized by the connection to the deck it is necessary for us to check for buckling at two areas.The most conservative span is halfway along the arch above the deck. Because the force is highest on the left side of the arch we assume that buckling will occur over the first half. A less con-serative span is the distance between the plates that connect the arch to the deck.
conservative estimate check Pcr = [pi^2(EI)]/kL^2
E = 29000ksiI = Ix = 678 in^4
L = 75’
Pcr = [pi^2(29000ksi*678 in^4)]/75ft(12inch/ft)^2
Pcr = 239.6 kips
less conservative estimate check
working backward we get the actual buckling length
Because this length occurs in between our two estimates it is safe to assume that the member size chosen will be adequate to prevent buckling along the arch.
Pcr = [pi^2(EI)]/kL^2
E = 29000ksiI = Ix = 678 in^4
L = 11’
Pcr = [pi^2(29000ksi*678 in^4)]/11ft(12inch/ft)^2
Pcr = 11137 kips
L = sqrt[ [pi^2(EI)]/(Pcr * 3)]L = 436 in or about 36 ft
Calculating Dead load
9’
15’
8x10.5
assuming 11 identical panels @ 5.1ft^2
Panel size:2 @ 110ft x 8ft x .25in1 @ 110ft x 10ft x .25in
Floor size:110ft x 8ft x .5ft
.1’
Dead Load steel = 35519lbs
Dead Load glass = 10335lbs
Frame Load
2k2k2k2k2k2k2k4k
load greater on arch because deck not supported on the left side10k
Bending and asymmetrical loadBy finding the maximum bending moment we can increase the size of the arch in order to stiffen it for asymmetrical loading
DeformationUsing a computer model of our arch and loads mentioned above in the deformation was calculated.The input for the beam size was the minimum dimension of 12” x 14“
Mmax =wl^2/32
w = live load (estimate to be about 110 psf or 1.1 ksf)
Mmax = [1.1 ksf(90ft)^2]/32Mmax = 278 kf
o = Mmax/Sxo = 15 ksi
90ft
Sx = Mmax/ oSx = 278kf(12in/ft)/15 ksiSx = 222.4 in^3
The rectangular section that has a section modulus of at least 223 and keeps the width and thickness of the current beam is 12” x 20” x 1/2”
b
B
hH
S = [BH^2/6] - (bh^3)/(6H)S = [12in*24in^2/6] - (11in*23in^3/(24in*6)S= 222.5in^3
The beam section is large enough because the two Section modulus values are equal
14”
24”
Steel = .28 lbs/in^3
20,000 lbs = 20 kips
20 kips/11 = 1.8 kips
Floor Dead load: 66147lbs/ (880) = 75psf80ft^2 x 75psf = 6000lbs
Roof Dead load: 2851lbs/(880) = 3.25psf80ft^2 x 3.25ps =260 lbs
Wall Dead Load: 3742lbs/(880) = 4.25psf100ft^2 x (4.25psf +4.25psf) = 850lbs
Steel Dead load 35519lbs/(165ft^2) = 215 psf15ft^2 x 215psf = 3229lbs
Live load Roof = 30psf
Live load Floor 80psf80ft^2 x (30psf + 80psf) = 8,800lbs
Total Load = 20,000 lbs
Wall and Roof Load
Floor Load
Calculating Point Loads
Concrete = .087 lbs/in^3Dead load Concrete = 66147 lbs
Glass = .09 lbs/in^3
157.9kN
45.9kN
294.3kN
3.59kN/m^2
.15kN/m^2
.20kN/m^2
10.29kN/m^2
3.83kN/m^2
1.43kN/m^2
89kN
8kN
734kN
279mm
.35mm x .3m x .025m
1065kN
49539kN
10.9m
5.6m^3
5.6m^3
110ft
28ft
52ft
Iguana Bridge
frame attachment to concrete deck
10’
1’
6”
Stata center
Building 46
DMAX = 1.442”
DMAX = 1.368”
DMAX = 0.930”
DMAX = 0.313”
DMAX = 0.988”
DMAX = 1.397”
Dmax = 1.483”
DMAX = 1.144”
Our arch is properly sized because 1.483” is lower than the allowable 3.5” for deformation of this arch
Chris Martin & Robert White
36
custom glazing (planar slices)
top �ange (steel)
web (carbon composite)
bottom �ange (steel/aluminum)
Steel and insulated glass web; steel is perforated
Depth and frequency of curvature varies along elevation
BOTTOM FLANGE (STEEL/ALUMINUM CLADDING)
WEB (CARBON COMPOSITE SHELLS WITH FOAM CORE)
TOP FLANGE (STEEL)*This revision will allow us to add thermal performance to the envelope, lighten structure by using less steel, and make the previously uninhabitable spaces accessible. NOTE: Carbon Composite is not used in our calculations as of yet.
GLAZING (PLANAR WEB SLICES)
Buckling in Top FlangeI = (bh3/12)outer - (bh3/12)innerbo = 1.25ft, bi = 1.19ft, ho = 3.5ft, hi = 3.44ftI = ((3.5*12)(1.25*12)3/12) - ((3.44*12)(1.19*12)3/12)I = 1748in4
Pcr = π2EI/(kL2)Pcr = π2(29,000ksi)(1748i4)/(1*118ft(12)2)Pcr = 249kPsafety = 166k (Safety factor increased due to support from web)T=163k OKShear: For this calculation, both beams are considered simultaneously since the corrugations are asymmetrical. Divide bridge into four chunks of 29.5 ft or 354 inches. For the outermost part (with most shear:Stress = V/Av = must be under 15ksiAv = (5/6)(teq)(h)h = 112inTeq = (length of corrugation)(width)/length of chunkTeq = (117.75in)(.125in)/(354in)Teq= .50inAv = (5/6)(.50in)(112in)V from shear diagram – average shear over length measuredV = 84kStress = 84/(5/6)(.50in)(112in)Stress = 1.80ksi OK
Asymmetrical Loading:Assume a live load of 3.85k distributed over one half of the beam or L/2.Assume a dead load of 565k distributed over the entire beam.
Reactions:∑F = 0∑M = 0W1 = Live Load over ½ beamW2 = Dead Load over entire beamVB + VA = W1(L) + W2(L/2) VB(L) =( (W1*L)/2)(L/4) + (W2*L)(L/2) VB= (.39)(118)/(8) + (.56)(118)/2VB = 5.75 + 33.04VB = 39kVA= (.56)(118) + (.39)(59) - 39VA= 50kHA= 0k
Area under shear diagram: 1317k*fMax. M = 1317k*f
THE CORRIGATOR TYLER CRAIN & CLAUDIA BODECROSS SECTION 1/2” = 1’
1:24
PERFORATED WEB
1:96PLAN 1/8” = 1’
NEED FROM RHINO:- FINAL RENDERS(2)- PLAN(MAKE2D)- CROSS SECTION(MAKE2D)- EXPLODED AXON(MAKE2D OR RENDER)-LAYERS- QUICK AXON RENDERS FOR DIAGRAMS- ELEVATION(MAKE2D/RENDER) FOR STRUCTURAL DIAG
The inspiration for this bridge is a Guy Nordenson pedestrian bridge in New Haven, CT which utilizes a curved, perforated steel sheet as support for the handrail
Aluminum Framing
Glass Roof
Top Flange (Box Girder)
Interior Flooring
Corrugated Aluminum Decking
Tributary Steel Support Ribs (spanning ribs)
Bottom Flange (steel)
Aluminum Cladding and Insulation Cavity
EXPLODED ISOMETRIC 1/16” = 1’1:192
1’-6
”
0.13
0.10
DESIGN REVISIONS FOR FUTURE
Buckling in Top FlangeI = (bh3/12)outer - (bh3/12)innerbo = 1.25ft, bi = 1.19ft, ho = 3.5ft, hi = 3.44ftI = ((3.5*12)(1.25*12)3/12) - ((3.44*12)(1.19*12)3/12)I = 1748in4
Pcr = π2EI/(kL2)Pcr = π2(29,000ksi)(1748i4)/(1*118ft(12)2)Pcr = 249kPsafety = 166k (Safety factor increased due to support from web)T=163k OKShear: For this calculation, both beams are considered simultaneously since the corrugations are asymmetrical. Divide bridge into four chunks of 29.5 ft or 354 inches. For the outermost part (with most shear:Stress = V/Av = must be under 15ksiAv = (5/6)(teq)(h)h = 112inTeq = (length of corrugation)(width)/length of chunkTeq = (117.75in)(.125in)/(354in)Teq= .50inAv = (5/6)(.50in)(112in)V from shear diagram – average shear over length measuredV = 84kStress = 84/(5/6)(.50in)(112in)Stress = 1.80ksi OK
Asymmetrical Loading:Assume a live load of 3.85k distributed over one half of the beam or L/2.Assume a dead load of 565k distributed over the entire beam.
Reactions:∑F = 0∑M = 0W1 = Live Load over ½ beamW2 = Dead Load over entire beamVB + VA = W1(L) + W2(L/2) VB(L) =( (W1*L)/2)(L/4) + (W2*L)(L/2) VB= (.39)(118)/(8) + (.56)(118)/2VB = 5.75 + 33.04VB = 39kVA= (.56)(118) + (.39)(59) - 39VA= 50kHA= 0k
Area under shear diagram: 1317k*fMax. M = 1317k*f
ADDITIONAL CALCULATIONS
ASSUMPTIONS AND LOADS
MAXIMUM MOMENT
ASYMMETRICAL LOADING
FLANGE FORCES
COMPRESSION CHORD BUCKLING
Average width = 9ftWidth of center glass roof = 4ft Allowable stress of steel = 15ksiWeight of steel = 490pcfE steel = 29,000k/128,998 kNL = Length = 118ft/36md = depth = 9.875ftdepth of web = 9ft
Live Load = 110psf/537kg/m^2Dead Load = 166psf Top Flange = 51psf or 54420lbs Bottom Flange = 17psf or 18727lbs Steel Web = 74psf or 78553lbs Cladding/Glass = 23psf Glass Facade = 12psf or 12744lbs Glass Roof = 1.33psf or 1416lbs Flange Cladding = 10psfW = Total Load per linear ft = 2.49k/ft Reactions = Fy,A = 147k, Fy,A = 147k, Fx,A = 0
Find the maximum moment of the beam using WL2/8:M=WL2/8M = (2.49(118)2)/8M= 4326k*ft
Assume a live load of 58k (= 1k/linear ft)distributed over one half of the beam or L/2. Assume a dead load of 177k (= 1.5 k/linear ft) distributed over the entire beam.
Reactions:∑F = 0, ∑M = 0W1 = Live Load over ½ beamW2 = Dead Load over entire beamVB + VA = W1(L) + W2(L/2) VB(L) =( (W1*L)/2)(L/4) + (W2*L)(L/2) VB= (1)(118)/(8) + (1.5)(118)/2VB = 14.75 + 88.5VB = 103kVA= (1.5)(118) + (1)(118/2) - 103VA= 133kHA= 0k
Area under shear diagram: 3537.8k*fMax. M = 3537.8k*f
Find the stresses in the flanges by using T=C=M/d:T=C=M/dT=C=4326/9.875T=C=438k
Verify that the compression chord will not crush under the flange forces:The top flange is made of 2 connected boxes each with a depth of 1.5ft, average width of 2.5 ft, and a wall thickness of 0.85in. For the purposes of calculating axial stress, only the continuous cross-sectional area is used.A = cross-sectional area = 54 sq/inStress = Force/AreaStress = 438/54Stress = 8.1ksi OK
Verify that the tension chord can carry the necessary tension:The bottom flange is a steel box. It has an average depth of 8” and a wall thickness of .4”.A = cross-sectional area = 47 sq/inStress = Force/AreaStress = 438/47Stress = 9.4ksi OK
Buckling in Top FlangeFind the moment of inertia of a box by subtracting the I(inside wall) from I(outside wall) of each box:I = (bh3/12)outer - (bh3/12)innerb(outer) = 1.5ft, b(inner) = 1.43ft, h(outer) = 2.5ft, h(inner)= 2.43ftI = ((2.5*12)(1.5*12)3/12) - ((2.43*12)(1.43*12)3/12)I = 2327in4 for each sideI total= 4654in4
Use the buckling equation to find the point of failure in buckling:Pcr = π2EI/(kL2)Pcr = π2(29,000ksi)(4654i4)/(1*118ft(12)2)Pcr = 664k
Reduce by a safety factor. Because this chord is being supported on one side by the web, the safety factor is increased from 1/3 the critical value to 2/3 the critical value.Pcr = 664kPsafety = 442k, C=438k OK
SHEAR CALCULATION FOR WEBSDivide bridge into four chunks of 29.5 ft or 354 inches. For the outermost part (with most shear):
Stress = V/Av = must be under 15ksiAv = (5/6)(teq)(h)h = 108in (height of web)Teq = (length of corrugation)(width)/length of chunkTeq = (720in)(.125in)/(354in)Teq= .25in (effective cross sectional area)Av = (5/6)(.25in)(108in)V from shear diagram – max. shear over length measuredV = 147kStress = 147/(5/6)(.25in)(108in)Stress = 6.42ksi OK
DEFLECTION CALCULATIONAmax = (5wL^4)/(384EI) --> I = (5wl^4)/(384E*Amax)w= 110psf = .08k/in, L = 118’ or , Ama x = 118/360 = .33 or 4”
Imin = [(5)(.08)(1416^4)]/[(384)(29,000)(4)]Imin= 37861 in^4
To calculat e I of entire structure: use parallel axis theorem for simply supported beam: I = ∑I + ∑Ad^2
I = ∑I + ∑Ad^2I(1) = I(2) = (54in^2)(34in)^2I(3) = (47in^2)(85in)^2I(total) = 395571 in^4Deflection for actual I = 0.38 inch
COMPRESSION = UPPER DECK
TENSION = LOWER DECK
V
SYMMETRICAL LOADING
d2d1
A1 A2
A3d3
CALCULATIONS
STEEL SUPPORT FOR GLASS ROOF (BEYOND) @ 48” O.C
INSULATED GLASS ROOF DRAINS TO GUTTER
SLOPED RIGID INSULATION
VARIES4’-0”1.22mVARIES
1’-6
”46
cm8” 23
cm8” 23
cm9’
-0”
2.74
cm
18” DEEP BOX FLANGE W/ WALL THICK-NESS OF 0.85IN, VARIABLE WIDTH
PERFORATED, CORRUGATED STEEL WEB
(2) LAYERS INSULATED GLASS
CLADDING WITH SPACE FOR INSULATION AND ELECTRICAL WIRING
8” DEEP STL BOX WITH INTER-NAL CROSS SUPPORTS
37
custom glazing (planar slices)
top �ange (steel)
web (carbon composite)
bottom �ange (steel/aluminum)
Steel and insulated glass web; steel is perforated
Depth and frequency of curvature varies along elevation
BOTTOM FLANGE (STEEL/ALUMINUM CLADDING)
WEB (CARBON COMPOSITE SHELLS WITH FOAM CORE)
TOP FLANGE (STEEL)*This revision will allow us to add thermal performance to the envelope, lighten structure by using less steel, and make the previously uninhabitable spaces accessible. NOTE: Carbon Composite is not used in our calculations as of yet.
GLAZING (PLANAR WEB SLICES)
Buckling in Top FlangeI = (bh3/12)outer - (bh3/12)innerbo = 1.25ft, bi = 1.19ft, ho = 3.5ft, hi = 3.44ftI = ((3.5*12)(1.25*12)3/12) - ((3.44*12)(1.19*12)3/12)I = 1748in4
Pcr = π2EI/(kL2)Pcr = π2(29,000ksi)(1748i4)/(1*118ft(12)2)Pcr = 249kPsafety = 166k (Safety factor increased due to support from web)T=163k OKShear: For this calculation, both beams are considered simultaneously since the corrugations are asymmetrical. Divide bridge into four chunks of 29.5 ft or 354 inches. For the outermost part (with most shear:Stress = V/Av = must be under 15ksiAv = (5/6)(teq)(h)h = 112inTeq = (length of corrugation)(width)/length of chunkTeq = (117.75in)(.125in)/(354in)Teq= .50inAv = (5/6)(.50in)(112in)V from shear diagram – average shear over length measuredV = 84kStress = 84/(5/6)(.50in)(112in)Stress = 1.80ksi OK
Asymmetrical Loading:Assume a live load of 3.85k distributed over one half of the beam or L/2.Assume a dead load of 565k distributed over the entire beam.
Reactions:∑F = 0∑M = 0W1 = Live Load over ½ beamW2 = Dead Load over entire beamVB + VA = W1(L) + W2(L/2) VB(L) =( (W1*L)/2)(L/4) + (W2*L)(L/2) VB= (.39)(118)/(8) + (.56)(118)/2VB = 5.75 + 33.04VB = 39kVA= (.56)(118) + (.39)(59) - 39VA= 50kHA= 0k
Area under shear diagram: 1317k*fMax. M = 1317k*f
THE CORRIGATOR TYLER CRAIN & CLAUDIA BODECROSS SECTION 1/2” = 1’
1:24
PERFORATED WEB
1:96PLAN 1/8” = 1’
NEED FROM RHINO:- FINAL RENDERS(2)- PLAN(MAKE2D)- CROSS SECTION(MAKE2D)- EXPLODED AXON(MAKE2D OR RENDER)-LAYERS- QUICK AXON RENDERS FOR DIAGRAMS- ELEVATION(MAKE2D/RENDER) FOR STRUCTURAL DIAG
The inspiration for this bridge is a Guy Nordenson pedestrian bridge in New Haven, CT which utilizes a curved, perforated steel sheet as support for the handrail
Aluminum Framing
Glass Roof
Top Flange (Box Girder)
Interior Flooring
Corrugated Aluminum Decking
Tributary Steel Support Ribs (spanning ribs)
Bottom Flange (steel)
Aluminum Cladding and Insulation Cavity
EXPLODED ISOMETRIC 1/16” = 1’1:192
1’-6
”
0.13
0.10
DESIGN REVISIONS FOR FUTURE
Buckling in Top FlangeI = (bh3/12)outer - (bh3/12)innerbo = 1.25ft, bi = 1.19ft, ho = 3.5ft, hi = 3.44ftI = ((3.5*12)(1.25*12)3/12) - ((3.44*12)(1.19*12)3/12)I = 1748in4
Pcr = π2EI/(kL2)Pcr = π2(29,000ksi)(1748i4)/(1*118ft(12)2)Pcr = 249kPsafety = 166k (Safety factor increased due to support from web)T=163k OKShear: For this calculation, both beams are considered simultaneously since the corrugations are asymmetrical. Divide bridge into four chunks of 29.5 ft or 354 inches. For the outermost part (with most shear:Stress = V/Av = must be under 15ksiAv = (5/6)(teq)(h)h = 112inTeq = (length of corrugation)(width)/length of chunkTeq = (117.75in)(.125in)/(354in)Teq= .50inAv = (5/6)(.50in)(112in)V from shear diagram – average shear over length measuredV = 84kStress = 84/(5/6)(.50in)(112in)Stress = 1.80ksi OK
Asymmetrical Loading:Assume a live load of 3.85k distributed over one half of the beam or L/2.Assume a dead load of 565k distributed over the entire beam.
Reactions:∑F = 0∑M = 0W1 = Live Load over ½ beamW2 = Dead Load over entire beamVB + VA = W1(L) + W2(L/2) VB(L) =( (W1*L)/2)(L/4) + (W2*L)(L/2) VB= (.39)(118)/(8) + (.56)(118)/2VB = 5.75 + 33.04VB = 39kVA= (.56)(118) + (.39)(59) - 39VA= 50kHA= 0k
Area under shear diagram: 1317k*fMax. M = 1317k*f
ADDITIONAL CALCULATIONS
ASSUMPTIONS AND LOADS
MAXIMUM MOMENT
ASYMMETRICAL LOADING
FLANGE FORCES
COMPRESSION CHORD BUCKLING
Average width = 9ftWidth of center glass roof = 4ft Allowable stress of steel = 15ksiWeight of steel = 490pcfE steel = 29,000k/128,998 kNL = Length = 118ft/36md = depth = 9.875ftdepth of web = 9ft
Live Load = 110psf/537kg/m^2Dead Load = 166psf Top Flange = 51psf or 54420lbs Bottom Flange = 17psf or 18727lbs Steel Web = 74psf or 78553lbs Cladding/Glass = 23psf Glass Facade = 12psf or 12744lbs Glass Roof = 1.33psf or 1416lbs Flange Cladding = 10psfW = Total Load per linear ft = 2.49k/ft Reactions = Fy,A = 147k, Fy,A = 147k, Fx,A = 0
Find the maximum moment of the beam using WL2/8:M=WL2/8M = (2.49(118)2)/8M= 4326k*ft
Assume a live load of 58k (= 1k/linear ft)distributed over one half of the beam or L/2. Assume a dead load of 177k (= 1.5 k/linear ft) distributed over the entire beam.
Reactions:∑F = 0, ∑M = 0W1 = Live Load over ½ beamW2 = Dead Load over entire beamVB + VA = W1(L) + W2(L/2) VB(L) =( (W1*L)/2)(L/4) + (W2*L)(L/2) VB= (1)(118)/(8) + (1.5)(118)/2VB = 14.75 + 88.5VB = 103kVA= (1.5)(118) + (1)(118/2) - 103VA= 133kHA= 0k
Area under shear diagram: 3537.8k*fMax. M = 3537.8k*f
Find the stresses in the flanges by using T=C=M/d:T=C=M/dT=C=4326/9.875T=C=438k
Verify that the compression chord will not crush under the flange forces:The top flange is made of 2 connected boxes each with a depth of 1.5ft, average width of 2.5 ft, and a wall thickness of 0.85in. For the purposes of calculating axial stress, only the continuous cross-sectional area is used.A = cross-sectional area = 54 sq/inStress = Force/AreaStress = 438/54Stress = 8.1ksi OK
Verify that the tension chord can carry the necessary tension:The bottom flange is a steel box. It has an average depth of 8” and a wall thickness of .4”.A = cross-sectional area = 47 sq/inStress = Force/AreaStress = 438/47Stress = 9.4ksi OK
Buckling in Top FlangeFind the moment of inertia of a box by subtracting the I(inside wall) from I(outside wall) of each box:I = (bh3/12)outer - (bh3/12)innerb(outer) = 1.5ft, b(inner) = 1.43ft, h(outer) = 2.5ft, h(inner)= 2.43ftI = ((2.5*12)(1.5*12)3/12) - ((2.43*12)(1.43*12)3/12)I = 2327in4 for each sideI total= 4654in4
Use the buckling equation to find the point of failure in buckling:Pcr = π2EI/(kL2)Pcr = π2(29,000ksi)(4654i4)/(1*118ft(12)2)Pcr = 664k
Reduce by a safety factor. Because this chord is being supported on one side by the web, the safety factor is increased from 1/3 the critical value to 2/3 the critical value.Pcr = 664kPsafety = 442k, C=438k OK
SHEAR CALCULATION FOR WEBSDivide bridge into four chunks of 29.5 ft or 354 inches. For the outermost part (with most shear):
Stress = V/Av = must be under 15ksiAv = (5/6)(teq)(h)h = 108in (height of web)Teq = (length of corrugation)(width)/length of chunkTeq = (720in)(.125in)/(354in)Teq= .25in (effective cross sectional area)Av = (5/6)(.25in)(108in)V from shear diagram – max. shear over length measuredV = 147kStress = 147/(5/6)(.25in)(108in)Stress = 6.42ksi OK
DEFLECTION CALCULATIONAmax = (5wL^4)/(384EI) --> I = (5wl^4)/(384E*Amax)w= 110psf = .08k/in, L = 118’ or , Ama x = 118/360 = .33 or 4”
Imin = [(5)(.08)(1416^4)]/[(384)(29,000)(4)]Imin= 37861 in^4
To calculat e I of entire structure: use parallel axis theorem for simply supported beam: I = ∑I + ∑Ad^2
I = ∑I + ∑Ad^2I(1) = I(2) = (54in^2)(34in)^2I(3) = (47in^2)(85in)^2I(total) = 395571 in^4Deflection for actual I = 0.38 inch
COMPRESSION = UPPER DECK
TENSION = LOWER DECK
V
SYMMETRICAL LOADING
d2d1
A1 A2
A3d3
CALCULATIONS
STEEL SUPPORT FOR GLASS ROOF (BEYOND) @ 48” O.C
INSULATED GLASS ROOF DRAINS TO GUTTER
SLOPED RIGID INSULATION
VARIES4’-0”1.22mVARIES
1’-6
”46
cm8” 23
cm8” 23
cm9’
-0”
2.74
cm
18” DEEP BOX FLANGE W/ WALL THICK-NESS OF 0.85IN, VARIABLE WIDTH
PERFORATED, CORRUGATED STEEL WEB
(2) LAYERS INSULATED GLASS
CLADDING WITH SPACE FOR INSULATION AND ELECTRICAL WIRING
8” DEEP STL BOX WITH INTER-NAL CROSS SUPPORTS
38
C
B
J
L
I
E
L
B
F
D
G
K
G
D
B
C
K
A
L
I
E
H
J
F
E
G
H
D JH IE F
F G K
L
C I
C
B JD
KA
A
A
H 60.9 kN
Floor Tributary Area
w = 300 lbs/ft
3
Snow Load: 30 psf -> 300 lbs/ft
60.9 kN
59.2 kN69.8 kN
60.9 kN
69.8 kN 69.8 kN69.8 kN
69.8 kN
365 kN
69.8 kN
Bottom
69.8 kN69.8 kN 69.8 kN
DL: Concretew = 750 lbs/ft
Total: w = 1550 lbs/ft
LL: Peoplew = 800 lbs/ft
Tributary Area = 3m x 3m
445 kN 303 kN
Assymetric Maximum Stress: 13.6 ksi < 15 ksi OK
Roof Uniform Load
60.9 kN
2
24.9 kN
69.8 kN
69.8 kN
Additional Stress = 4.4 ksi
Uniform Load Stress: 9.36 ksi
60.9 kN Floor Beam SizingMmax = 232.5 kips * in
σ,steel = 15 ksi
Section Modulus = 15.5 in
W10X17 S = 16.2 in.
3
3
274 kN
Floor Uniform Load
35.1 kN
No.6 (US) BarsAs = 1700 mmSteel Ratio = 0.006
69.8 kN 69.8 kN 69.8 kN
Assymetric Maximum Stress: 18.3 ksi > 15 ksi too much
Top
60.9 kN
Assymetric Maximum Stress: 13.8 ksi < 15 ksi OK
60.9 kN
30.8 kN
60.9 kN
69.8 kN
30.8 kN
274 kN
60.9 kN60.9 kN 60.9 kN
Column Sizing
Concrete Strength: 3 ksiSteel Rebar Strength: 40 ksi
No.6 (US) BarsAs = 1700 mmSteel Ratio = 0.013
Load from Roof = 365 kNLoad from Floor = 445 kN
810 kN Through Column
2
Force Polygon
60.9 kN60.9 kN51.6 kN
Roof Tributary Area
60.9 kN
287 kN
21.4 kN
60.9 kN
30.8 kN
60.9 kN60.9 kN
Force Polygon
69.8 kN
314 kN
Mmax = 76 kips/ft
69.8 kN
35.1 kN
Mmax = 203 kips/ft
Maximum Force Based On Force Polygon
Floor: 406 kNRoof: 406 kN
Uniform Load Stress: 6.52 ksi
w = 800 lbs/ft
Assymetric Maximum Stress: 29.5 ksi > 15 ksi too much
Surface Area: 5000 mm
Primary Member Sizing
314 kN
k = 0.5 fixed-fixedE = 29,000 ksi
Pcr = 5700 kN >> 406 kNNo chance of buckling
Buckling Check
152mm x 102mm x 13mm
Buckling Check
2Surface Area: 5000 mm
2
3
152mm x 102mm x 13mm
New Primary Member Sizing
Primary Member Sizing
Additional Stress = 7.12 ksi
People Load: 80 psf -> 800 lbs/ft
Uniform Load Stress: 11.6 ksi
Assymetric Loading Floor
k = 0.5 fixed-fixedE = 29,000 ksi
Pcr = 5700 kN >> 406 kNNo chance of buckling
Current Member: 152mm x 102mm x 13mmSx = 185000 mm
Maximum Force Based On Force Polygon
Floor: 406 kNRoof: 406 kN
New Primary Member Sizing
Uniform Load Stress: 11.6 ksi
35.1 kN
2
3
Try 203mm x 152mm x 16mm
Surface Area = 9000 mm Sx = 467000 mm
Current Member: 152mm x 102mm x 13mmSx = 185000 mm
Assymetric Loading Roof
Mmax = 474 kips * in
Section Modulus = 31.6 in
33Use 230mm x 178mm x 13mmSx = 33 in > 31.6 in OK
Each strut carries 35 kN
3
Worst case scenario: 35kN
Equal horizontal forces allowforces to negate each other at their connection points
DL: Concrete, Glassw = 500 lbs/ft + 540 lbs/ft
Total: w = 1340 lbs/ft
LL: Snoww = 300 lbs/ft
Tributary Area = 3m x 3m
Roof Beam SizingMmax = 201 kips * in
σ,steel = 15 ksi
Section Modulus = 13.4 in
W10X15 S = 13.8 in.
3
3
Floor and Roof Strut Sizing
Secondary Member Sizing
Try 178mm x 127mm x 13mm
Surface Area = 6700 mm Sx = 284000 mm
2
3
Additional Stress = 6.73 ksi
Additional Stress = 17.9 ksi
A
A
B
B
PROCESS DIAGRAMSPROCESS DIAGRAMS SECTION DIAGRAM CALCULATIONS
DL: Concrete = 50 lbs/ft Beam = 150 lbs Glass = 536 lbs/ftLL: Snow = 30 lbs/ft
Total Weight (every 3.0m x 3.0m area)= 50*100 + 150 + 536*10 + 30*100= 60.9 kN
178 mm
102 mm
178
mm
13 m
m
127 mm
325 kN
610 mm
406 kN
400 kN
14 m
3.0 m2.1 m
3.0 m
3.0 m2.1 m
3.0 m
406 mm
152 mm
152
mm
230
mm
13 m
m
203
mm
152
mm
13 m
m
102 mm
13 m
m16
mm
102mm
254m
m27
5 m
m
102mm
2
2
2
2
ReducedTributaryArea
ReducedTributaryAreaReduced
TributaryArea
ReducedTributaryArea
ReducedTributaryArea
ReducedTributaryArea
DL: Concrete = 75 lbs/ft Beam = 170 lbsLL: People = 80 lbs/ft
Total Weight (every 3.0m x 3.0m area)= 75*100 + 170 + 80*100= 69.8 kN
SECTION B-B
7.5
m5.
4 m
3.5
m
0.4
m0.
4 m
1.2
m
SECTION A-A
7.0
m6.
0 m
0.4
m3.
5 m
0.4
m1.
8 m
Dual-Restrained Arch Skybridge Andrew Sang Sean T. Tang
39
C
B
J
L
I
E
L
B
F
D
G
K
G
D
B
C
K
A
L
I
E
H
J
F
E
G
H
D JH IE F
F G K
L
C I
C
B JD
KA
A
A
H 60.9 kN
Floor Tributary Area
w = 300 lbs/ft
3
Snow Load: 30 psf -> 300 lbs/ft
60.9 kN
59.2 kN69.8 kN
60.9 kN
69.8 kN 69.8 kN69.8 kN
69.8 kN
365 kN
69.8 kN
Bottom
69.8 kN69.8 kN 69.8 kN
DL: Concretew = 750 lbs/ft
Total: w = 1550 lbs/ft
LL: Peoplew = 800 lbs/ft
Tributary Area = 3m x 3m
445 kN 303 kN
Assymetric Maximum Stress: 13.6 ksi < 15 ksi OK
Roof Uniform Load
60.9 kN
2
24.9 kN
69.8 kN
69.8 kN
Additional Stress = 4.4 ksi
Uniform Load Stress: 9.36 ksi
60.9 kN Floor Beam SizingMmax = 232.5 kips * in
σ,steel = 15 ksi
Section Modulus = 15.5 in
W10X17 S = 16.2 in.
3
3
274 kN
Floor Uniform Load
35.1 kN
No.6 (US) BarsAs = 1700 mmSteel Ratio = 0.006
69.8 kN 69.8 kN 69.8 kN
Assymetric Maximum Stress: 18.3 ksi > 15 ksi too much
Top
60.9 kN
Assymetric Maximum Stress: 13.8 ksi < 15 ksi OK
60.9 kN
30.8 kN
60.9 kN
69.8 kN
30.8 kN
274 kN
60.9 kN60.9 kN 60.9 kN
Column Sizing
Concrete Strength: 3 ksiSteel Rebar Strength: 40 ksi
No.6 (US) BarsAs = 1700 mmSteel Ratio = 0.013
Load from Roof = 365 kNLoad from Floor = 445 kN
810 kN Through Column
2
Force Polygon
60.9 kN60.9 kN51.6 kN
Roof Tributary Area
60.9 kN
287 kN
21.4 kN
60.9 kN
30.8 kN
60.9 kN60.9 kN
Force Polygon
69.8 kN
314 kN
Mmax = 76 kips/ft
69.8 kN
35.1 kN
Mmax = 203 kips/ft
Maximum Force Based On Force Polygon
Floor: 406 kNRoof: 406 kN
Uniform Load Stress: 6.52 ksi
w = 800 lbs/ft
Assymetric Maximum Stress: 29.5 ksi > 15 ksi too much
Surface Area: 5000 mm
Primary Member Sizing
314 kN
k = 0.5 fixed-fixedE = 29,000 ksi
Pcr = 5700 kN >> 406 kNNo chance of buckling
Buckling Check
152mm x 102mm x 13mm
Buckling Check
2Surface Area: 5000 mm
2
3
152mm x 102mm x 13mm
New Primary Member Sizing
Primary Member Sizing
Additional Stress = 7.12 ksi
People Load: 80 psf -> 800 lbs/ft
Uniform Load Stress: 11.6 ksi
Assymetric Loading Floor
k = 0.5 fixed-fixedE = 29,000 ksi
Pcr = 5700 kN >> 406 kNNo chance of buckling
Current Member: 152mm x 102mm x 13mmSx = 185000 mm
Maximum Force Based On Force Polygon
Floor: 406 kNRoof: 406 kN
New Primary Member Sizing
Uniform Load Stress: 11.6 ksi
35.1 kN
2
3
Try 203mm x 152mm x 16mm
Surface Area = 9000 mm Sx = 467000 mm
Current Member: 152mm x 102mm x 13mmSx = 185000 mm
Assymetric Loading Roof
Mmax = 474 kips * in
Section Modulus = 31.6 in
33Use 230mm x 178mm x 13mmSx = 33 in > 31.6 in OK
Each strut carries 35 kN
3
Worst case scenario: 35kN
Equal horizontal forces allowforces to negate each other at their connection points
DL: Concrete, Glassw = 500 lbs/ft + 540 lbs/ft
Total: w = 1340 lbs/ft
LL: Snoww = 300 lbs/ft
Tributary Area = 3m x 3m
Roof Beam SizingMmax = 201 kips * in
σ,steel = 15 ksi
Section Modulus = 13.4 in
W10X15 S = 13.8 in.
3
3
Floor and Roof Strut Sizing
Secondary Member Sizing
Try 178mm x 127mm x 13mm
Surface Area = 6700 mm Sx = 284000 mm
2
3
Additional Stress = 6.73 ksi
Additional Stress = 17.9 ksi
A
A
B
B
PROCESS DIAGRAMSPROCESS DIAGRAMS SECTION DIAGRAM CALCULATIONS
DL: Concrete = 50 lbs/ft Beam = 150 lbs Glass = 536 lbs/ftLL: Snow = 30 lbs/ft
Total Weight (every 3.0m x 3.0m area)= 50*100 + 150 + 536*10 + 30*100= 60.9 kN
178 mm
102 mm
178
mm
13 m
m
127 mm
325 kN
610 mm
406 kN
400 kN
14 m
3.0 m2.1 m
3.0 m
3.0 m2.1 m
3.0 m
406 mm
152 mm
152
mm
230
mm
13 m
m
203
mm
152
mm
13 m
m
102 mm
13 m
m16
mm
102mm
254m
m27
5 m
m 102mm
2
2
2
2
ReducedTributaryArea
ReducedTributaryAreaReduced
TributaryArea
ReducedTributaryArea
ReducedTributaryArea
ReducedTributaryArea
DL: Concrete = 75 lbs/ft Beam = 170 lbsLL: People = 80 lbs/ft
Total Weight (every 3.0m x 3.0m area)= 75*100 + 170 + 80*100= 69.8 kN
SECTION B-B
7.5
m5.
4 m
3.5
m
0.4
m0.
4 m
1.2
m
SECTION A-A
7.0
m6.
0 m
0.4
m3.
5 m
0.4
m1.
8 m
Dual-Restrained Arch Skybridge Andrew Sang Sean T. Tang
40
virendeel bridge dicle uzunyayla & sayjel patel
DL + LL = 3.2 kip/ft
84 kip84 kip
Mmax = 2200 kip*ft
BEAMANALYSIS
MULTIFRAME ANALYSIS
PARAMETRICLOGIC
Moment (z) kN*m
Shear (V) kN
Axial Loads (P) kN
De�ection (mm)
CONCEPT
Fabrication
13731212 3732540912984 1681 693551 2088873
ASYMMETRICAL LOADING
66 C
2 C
66 T 63 T
63 C
12 C 9 C 4 C
62 T
62 C
25 T
1 C
9 C
9 T 3 T
3 C
2 C
25 C
17 T
3 C
17 C
13 C
40 C23 C52 C 44 C54 C
23 T9 T40 T
9 C
52 T
34 C
54 T
2 C1 C2 C 8 C 4 C 3 C
34 T44 T
2 C
3 3333 33
1. Loadsa. Dead Loads
A Cross Sectional Area of Top Beam 20 Ft 2 from geometryB Cross Sectional Area of Bottom Beam 16 Ft 2 "C Bridge Length 105 Ft "D Total Beam Volume 33,600 Ft 3 D = A*B*CE Percentage Steel 30.00% estimateF Volume Steel 10080 Ft 3 F = D*EG Density of Steel 22 lb/ft3 givenH Total Mass of Steel (Total Deadload) 222 Kips H = F * G / 1000I W DL 2.11 Kips / ft J = I*C
b. Live LoadsJ Live Load Typical Pedestrian Bridge 110 lbs/ft2 gIvenK Bridge Width 10 Ft from geometryL LL Total 116 Kips L = (K * C)*J / 1000M W LL 1.10 Kips / Ft M = L / C
1.49 KN / m // Metricc. Find W
N W 3.21 Kips / ft N = M + IO W/2 (split between right and left side) 1.605 Kips / ft O = N / 2
d. Find P 2.18 KN / m // MetricP Uniform Load (W)' 1605.0 lbs/ft P = O*1000Q Total Load 168.5 Kips P = O*CR Number of Segments 15 from geometryS P / NumSegments 11.2 Kips S = Q / R
50 KN // Metric2. Primary Elementsa. Find Max Allowable Bending Stress
T Reaction Force 84.3 Kips T = Q / 2U Max Bending Moment 2,211,891 Lb*ft U = P *C^2/8 (w L^2/8)V Allowable Bending Stress 19800 ơ
b. Find Section Modulus RequiredW Section Modulus (S req'd) 112 in^3 W = U /V Mmax / I (steel
c.Find moment of Inertia required usingdeflection formula
X Max Allowable De�ection 3.5 inches T = Q / 360 * 12 L / 360Y Modulus of Elasticity Steel 30,000,000 psi E constant Mod for SteelZ I Total (Required) 41,805 in^4 I = 5wL^4/384ED
d. Select Steel Section, check moment ofinertia and STop HSS a Height 20 in
b Width 12 inc Thickness 1/2 ind Section Modulus 117 in^3 > S req'de Iy 705 in^4
Bottom HSS d Height 20 ine Width 12 inf Thickness 1/2 ing Section Modulus 117 in^3 > S req'dh Iy 705 in^4
Check I req'dg Atop 240 in^2 g = a*bh Abottom 240 in^2 h = d*ei Height of Section 120 inj Dtop 60 ink Dbottom 60 inl Itotal 1,729,410 in^2 l = (g + h) + g*j^2+ h*k^2
Check A req'd Axial Force 55.0 KN (M.F)13.75 KIP
Allowable Bending Stress 15 KSIArea Required 0.9 in^2 p = m / o (A = P / ơ)
3. Secondary Elements - (Multi Framefind area Req'd
m Axial Force (vertical elements) 10.0 KN multi framen " 2.25 KIP // metricso Allowable Bending Stress 15 KSIp Area Required 0.2 in^2 p = m / o (A = P / ơ)
Choose Sectionq Height 8.0 inr Width 8.0 ins Thickness 0.625 in
(t > a) t Area 18 in^2u I 125 in^4v s 31 in^3
Check Buckling (x > n)w member length 120.0 inx Max Axial Force 2.6 KIPs p = m / o (F = pi^2 * E* I^2 / (KL)^2) buckling governs
4. Asymetrical Loading
Secondary Elements - check max axial forcefind area Req'd
m Axial Force (vertical elements) KNn " KIPo Allowable Bending Stress 15 KSIp Area Required in^2
Choose Sectionq Height inr Width ins Thickness 0.625 in
(t > a) t Area in^2u I in^4v s in^3
Check Buckling (x > n)w member length 120.0 inx Max Axial Force KIPs
In most extreme case, member ( Fc = 23 kN)requires 16 " x 16" HSS, governed bybuckling)
13.03.25
0.2
9.09.0
2118240
3.7
Massing
32.0 m
3.2 m
prefabricated vertical elements are built with welded steel plates
Custom plate steel sections serve as a deck and ceiling
Secondary ribs
ceiling and deck are perforated toreduce the weight of the structure
30 cm
30 cm
2
4
4
4
0
3
0 2
0
28
15
13
8
8
4
8
5
5
15
159
8
8
5
6
7
8 9
10
6
5
6
8
4
8
9 0
0
15
15
13
12
5
2
31
1
2 1
6 9 5
4
6
2
9
11 8
8 5 7
30
1115 6
8 5
7 0
09 8
55 C
0 C
55 T 58 T
58 C
8 C 5 C 5 C5 C
48 T
48 C
9 C
28 C16 C 38 C
16 T6 T 28 T
6 C
38 T
1 C1 C
852.4
41
virendeel bridge dicle uzunyayla & sayjel patel
DL + LL = 3.2 kip/ft
84 kip84 kip
Mmax = 2200 kip*ft
BEAMANALYSIS
MULTIFRAME ANALYSIS
PARAMETRICLOGIC
Moment (z) kN*m
Shear (V) kN
Axial Loads (P) kN
De�ection (mm)
CONCEPT
Fabrication
13731212 3732540912984 1681 693551 2088873
ASYMMETRICAL LOADING
66 C
2 C
66 T 63 T
63 C
12 C 9 C 4 C
62 T
62 C
25 T
1 C
9 C
9 T 3 T
3 C
2 C
25 C
17 T
3 C
17 C
13 C
40 C23 C52 C 44 C54 C
23 T9 T40 T
9 C
52 T
34 C
54 T
2 C1 C2 C 8 C 4 C 3 C
34 T44 T
2 C
3 3333 33
1. Loadsa. Dead Loads
A Cross Sectional Area of Top Beam 20 Ft 2 from geometryB Cross Sectional Area of Bottom Beam 16 Ft 2 "C Bridge Length 105 Ft "D Total Beam Volume 33,600 Ft 3 D = A*B*CE Percentage Steel 30.00% estimateF Volume Steel 10080 Ft 3 F = D*EG Density of Steel 22 lb/ft3 givenH Total Mass of Steel (Total Deadload) 222 Kips H = F * G / 1000I W DL 2.11 Kips / ft J = I*C
b. Live LoadsJ Live Load Typical Pedestrian Bridge 110 lbs/ft2 gIvenK Bridge Width 10 Ft from geometryL LL Total 116 Kips L = (K * C)*J / 1000M W LL 1.10 Kips / Ft M = L / C
1.49 KN / m // Metricc. Find W
N W 3.21 Kips / ft N = M + IO W/2 (split between right and left side) 1.605 Kips / ft O = N / 2
d. Find P 2.18 KN / m // MetricP Uniform Load (W)' 1605.0 lbs/ft P = O*1000Q Total Load 168.5 Kips P = O*CR Number of Segments 15 from geometryS P / NumSegments 11.2 Kips S = Q / R
50 KN // Metric2. Primary Elementsa. Find Max Allowable Bending Stress
T Reaction Force 84.3 Kips T = Q / 2U Max Bending Moment 2,211,891 Lb*ft U = P *C^2/8 (w L^2/8)V Allowable Bending Stress 19800 ơ
b. Find Section Modulus RequiredW Section Modulus (S req'd) 112 in^3 W = U /V Mmax / I (steel
c.Find moment of Inertia required usingdeflection formula
X Max Allowable De�ection 3.5 inches T = Q / 360 * 12 L / 360Y Modulus of Elasticity Steel 30,000,000 psi E constant Mod for SteelZ I Total (Required) 41,805 in^4 I = 5wL^4/384ED
d. Select Steel Section, check moment ofinertia and STop HSS a Height 20 in
b Width 12 inc Thickness 1/2 ind Section Modulus 117 in^3 > S req'de Iy 705 in^4
Bottom HSS d Height 20 ine Width 12 inf Thickness 1/2 ing Section Modulus 117 in^3 > S req'dh Iy 705 in^4
Check I req'dg Atop 240 in^2 g = a*bh Abottom 240 in^2 h = d*ei Height of Section 120 inj Dtop 60 ink Dbottom 60 inl Itotal 1,729,410 in^2 l = (g + h) + g*j^2+ h*k^2
Check A req'd Axial Force 55.0 KN (M.F)13.75 KIP
Allowable Bending Stress 15 KSIArea Required 0.9 in^2 p = m / o (A = P / ơ)
3. Secondary Elements - (Multi Framefind area Req'd
m Axial Force (vertical elements) 10.0 KN multi framen " 2.25 KIP // metricso Allowable Bending Stress 15 KSIp Area Required 0.2 in^2 p = m / o (A = P / ơ)
Choose Sectionq Height 8.0 inr Width 8.0 ins Thickness 0.625 in
(t > a) t Area 18 in^2u I 125 in^4v s 31 in^3
Check Buckling (x > n)w member length 120.0 inx Max Axial Force 2.6 KIPs p = m / o (F = pi^2 * E* I^2 / (KL)^2) buckling governs
4. Asymetrical Loading
Secondary Elements - check max axial forcefind area Req'd
m Axial Force (vertical elements) KNn " KIPo Allowable Bending Stress 15 KSIp Area Required in^2
Choose Sectionq Height inr Width ins Thickness 0.625 in
(t > a) t Area in^2u I in^4v s in^3
Check Buckling (x > n)w member length 120.0 inx Max Axial Force KIPs
In most extreme case, member ( Fc = 23 kN)requires 16 " x 16" HSS, governed bybuckling)
13.03.25
0.2
9.09.0
2118240
3.7
Massing
32.0 m
3.2 m
prefabricated vertical elements are built with welded steel plates
Custom plate steel sections serve as a deck and ceiling
Secondary ribs
ceiling and deck are perforated toreduce the weight of the structure
30 cm
30 cm
2
4
4
4
0
3
0 2
0
28
15
13
8
8
4
8
5
5
15
159
8
8
5
6
7
8 9
10
6
5
6
8
4
8
9 0
0
15
15
13
12
5
2
31
1
2 1
6 9 5
4
6
2
9
11 8
8 5 7
30
1115 6
8 5
7 0
09 8
55 C
0 C
55 T 58 T
58 C
8 C 5 C 5 C5 C
48 T
48 C
9 C
28 C16 C 38 C
16 T6 T 28 T
6 C
38 T
1 C1 C
852.4
42
Bldg 36
cable anchored to �oor platecable cable
roof
20*
deck3’
10’
15’
cable anchored to �oor plate
Bldg 46
Deck
Roof
Cable
Closing line
(30’)
A B C D E F G H I J K L
a
b
c
d
e
f
g
h
i
j
k
l
29.4 k
29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k
point load
245.7 kMaximum Force
Force Polygon
9.7ft
o
2.96 m
130.86 kN
1092.93 kN
9.14 m
θ=20o
ah
θ=20o
PPh
Bldg 36
cable anchored to �oor platecable cable
roof
20*
deck3’
10’
15’
cable anchored to �oor plate
Bldg 46
Deck
Roof
Cable
Closing line
(30’)
A B C D E F G H I J K L
a
b
c
d
e
f
g
h
i
j
k
l
29.4 k
29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k
point load
245.7 kMaximum Force
Force Polygon
9.7ft
o
2.96 m
130.86 kN
1092.93 kN
9.14 m
ESTIMATED LOADSFloor Dead Load: 3.11 kN/m2 (65 psf)Floor Live Load: 3.83 kN/m2 (80 psf)Roof Dead Load: 1.44 kN/m2 (30 psf)Roof Live Load: 2.15 kN/m2 (45 psf)Cladding Load: 0.48 kN/m2 (10 psf)
CONCEPTThe Vassar Skybridge will allow for visibility down the street and by its slanted geometry it will allow for people on the bridge to lean over and see the street life below.
DESIGN ASSUMPTIONSBridge length: 32.54 m (106.75 ft)Bridge Width: 3.05 m (10 ft)Bridge Height: 4.57 m (15 ft)
ALLOWABLE STRESSES (σ)Steel: 103 N/mm2 (15 ksi)
POINT LOADS
Tributary Area - Floor(65 psf+80 psf )(10 ft)(9.7 ft)=14.065 kips
Tributary Area - Roof(30 psf+45 psf )(10 ft)(9.7 ft)=10.67 kips
Tributary Area - Cladding(10 psf)(15 ft)(9.7 ft)(2)=2.91 kips
Vertical Point Loads (P)27.645 kips =122.97 kN
Point Loads for Cable (Ph)Ph =P/cos(20o)=29.419 kips=130.86 kN
Maximum Force in Cable (Pm)Pm =245.7 kips=1092.93 kN
Gross Required Area for 2 Cables (Ag)σ=Pm/AgAg= Pm/ σ=245.7 kips/15 ksi=16.4 in2
Required Area for 1 Cable (A)A= 1/2 Ag=8.2 in2=5290.31 mm2
A= πr2
r=1.41 in=35.81 mm
SIZING THE CABLES
Required Area for Secondary Cables (As)As= Ph/ σ= πrs
2
rs=0.79 in= 20.06 mm
DISTRIBUTED LOADS
w=(80 psf+45 psf)(10 ft) =1250 plf=21161.155 N/m
SIZING THE TRUSSES
Maximum Moment in Truss(Mmax)Mmax=wL2/32Mmax=(1250 plf)(106.75 ft)2/32=516.36 klbft=1110.50 kNm
Maximum Tension and Compression Forces in Truss(T/C)T=C=Mmax/dT=C=516.36 klbft/3 ft=172.12 kips=765.63 kN
Gross Required Area for 2 Trusses (Ag)σ=T/AgAg= T/ σ=172.12 kips/15 ksi=11.47 in2
Required Area for 1 Truss (A)A= 1/2 Ag=5.74 in2=3703.22 mm2
A= πr2
r=1.35 in=34.29 mmRequired Area for Secondary Cables (As)As= 1/2 A= πrs
2
rs=0.95 in= 24.13 mm
Maximum Moment in Truss(Mmax) Mmax=wL2/64r=0.95 in= 24.13 mmrs=0.68 in= 17.27 mm
Vassar Skybridge
d=0.91m (3 ft)
T=607.23 kN (136.51 kips)
C=607.23 kN (136.51 kips)
Mmax=555.25 kNm (409.53 klbft)
Structural Design[4.463]Building Structural Systems IIProfessor: John Ochsendorf Andrea LoveTA: Caitlin Tobin Mueller
Laura Renee Schmitz Enid Xuezhu Tian
32.54 m
4.57m
9.14m
3.05m
0.91m
r=35.81mm
rs=20.06 mm
PLAN
SECTION A
SECTION B
Primary Cable Section
Secondary Cable Section
Primary Truss Member Section
Secondary member
MODULUS OF ELASTICITY (E)Steel: 29000 ksi
DEFLECTION
d1=1.5 ft
d2=1.5 ft
y’ y
A3
A4
A1
A2
d3
d4
Dimensions and Sectional Properties of Round HSS
for hollow steel tube, d=6.625 in
Δmax= 5wL4/384EI L=106.75 ft/2=53.375 ft We calculated for the deflection of half of the bridge for asymmetrical loadingΔmax ≤ L/360=53.375 ft/360=640.5 in/360=1.779 inwLL=(80 psf)(10 ft)+(45psf)(19.5 ft)=1677.5 plf =1.6775 k/ft=0.139 k/inE=29000 k/in2
I ≥5 wLLL4/384E Δmax= 5(0.139 k/in)(640.5 in)4/384(29000 k/in2)(1.779 in)=5094.1 in4
I= ΣIi=ΣAidi2
A1=A2=A3=A4= 5.74 in2
d1=d2=d3=d4=1.5 ft=18 in
I=4(5.74 in2)(18 in)2=7439.04 in4
7439.04 in4> 5094.1 in4
43
Bldg 36
cable anchored to �oor platecable cable
roof
20*
deck3’
10’
15’
cable anchored to �oor plate
Bldg 46
Deck
Roof
Cable
Closing line
(30’)
A B C D E F G H I J K L
a
b
c
d
e
f
g
h
i
j
k
l
29.4 k
29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k
point load
245.7 kMaximum Force
Force Polygon
9.7ft
o
2.96 m
130.86 kN
1092.93 kN
9.14 m
θ=20o
ah
θ=20o
PPh
Bldg 36
cable anchored to �oor platecable cable
roof
20*
deck3’
10’
15’
cable anchored to �oor plate
Bldg 46
Deck
Roof
Cable
Closing line
(30’)
A B C D E F G H I J K L
a
b
c
d
e
f
g
h
i
j
k
l
29.4 k
29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k29.4 k
point load
245.7 kMaximum Force
Force Polygon
9.7ft
o
2.96 m
130.86 kN
1092.93 kN
9.14 m
ESTIMATED LOADSFloor Dead Load: 3.11 kN/m2 (65 psf)Floor Live Load: 3.83 kN/m2 (80 psf)Roof Dead Load: 1.44 kN/m2 (30 psf)Roof Live Load: 2.15 kN/m2 (45 psf)Cladding Load: 0.48 kN/m2 (10 psf)
CONCEPTThe Vassar Skybridge will allow for visibility down the street and by its slanted geometry it will allow for people on the bridge to lean over and see the street life below.
DESIGN ASSUMPTIONSBridge length: 32.54 m (106.75 ft)Bridge Width: 3.05 m (10 ft)Bridge Height: 4.57 m (15 ft)
ALLOWABLE STRESSES (σ)Steel: 103 N/mm2 (15 ksi)
POINT LOADS
Tributary Area - Floor(65 psf+80 psf )(10 ft)(9.7 ft)=14.065 kips
Tributary Area - Roof(30 psf+45 psf )(10 ft)(9.7 ft)=10.67 kips
Tributary Area - Cladding(10 psf)(15 ft)(9.7 ft)(2)=2.91 kips
Vertical Point Loads (P)27.645 kips =122.97 kN
Point Loads for Cable (Ph)Ph =P/cos(20o)=29.419 kips=130.86 kN
Maximum Force in Cable (Pm)Pm =245.7 kips=1092.93 kN
Gross Required Area for 2 Cables (Ag)σ=Pm/AgAg= Pm/ σ=245.7 kips/15 ksi=16.4 in2
Required Area for 1 Cable (A)A= 1/2 Ag=8.2 in2=5290.31 mm2
A= πr2
r=1.41 in=35.81 mm
SIZING THE CABLES
Required Area for Secondary Cables (As)As= Ph/ σ= πrs
2
rs=0.79 in= 20.06 mm
DISTRIBUTED LOADS
w=(80 psf+45 psf)(10 ft) =1250 plf=21161.155 N/m
SIZING THE TRUSSES
Maximum Moment in Truss(Mmax)Mmax=wL2/32Mmax=(1250 plf)(106.75 ft)2/32=516.36 klbft=1110.50 kNm
Maximum Tension and Compression Forces in Truss(T/C)T=C=Mmax/dT=C=516.36 klbft/3 ft=172.12 kips=765.63 kN
Gross Required Area for 2 Trusses (Ag)σ=T/AgAg= T/ σ=172.12 kips/15 ksi=11.47 in2
Required Area for 1 Truss (A)A= 1/2 Ag=5.74 in2=3703.22 mm2
A= πr2
r=1.35 in=34.29 mmRequired Area for Secondary Cables (As)As= 1/2 A= πrs
2
rs=0.95 in= 24.13 mm
Maximum Moment in Truss(Mmax) Mmax=wL2/64r=0.95 in= 24.13 mmrs=0.68 in= 17.27 mm
Vassar Skybridge
d=0.91m (3 ft)
T=607.23 kN (136.51 kips)
C=607.23 kN (136.51 kips)
Mmax=555.25 kNm (409.53 klbft)
Structural Design[4.463]Building Structural Systems IIProfessor: John Ochsendorf Andrea LoveTA: Caitlin Tobin Mueller
Laura Renee Schmitz Enid Xuezhu Tian
32.54 m
4.57m
9.14m
3.05m
0.91m
r=35.81mm
rs=20.06 mm
PLAN
SECTION A
SECTION B
Primary Cable Section
Secondary Cable Section
Primary Truss Member Section
Secondary member
MODULUS OF ELASTICITY (E)Steel: 29000 ksi
DEFLECTION
d1=1.5 ft
d2=1.5 ft
y’ y
A3
A4
A1
A2
d3
d4
Dimensions and Sectional Properties of Round HSS
for hollow steel tube, d=6.625 in
Δmax= 5wL4/384EI L=106.75 ft/2=53.375 ft We calculated for the deflection of half of the bridge for asymmetrical loadingΔmax ≤ L/360=53.375 ft/360=640.5 in/360=1.779 inwLL=(80 psf)(10 ft)+(45psf)(19.5 ft)=1677.5 plf =1.6775 k/ft=0.139 k/inE=29000 k/in2
I ≥5 wLLL4/384E Δmax= 5(0.139 k/in)(640.5 in)4/384(29000 k/in2)(1.779 in)=5094.1 in4
I= ΣIi=ΣAidi2
A1=A2=A3=A4= 5.74 in2
d1=d2=d3=d4=1.5 ft=18 in
I=4(5.74 in2)(18 in)2=7439.04 in4
7439.04 in4> 5094.1 in4
44
Load Calculations:
ωds 957.6 Pa:= Beam Depth: d 1.4 m:=Superimposed Dead Load on Deck:
ωll 4.8 103× Pa:= Beam Length: L 34.1 m:=Live Load on Deck:
Deck Width: w 3 m:=
Factored_Load 1.2ωds 1.6 ωll⋅+ 8.8 103× Pa⋅=:= Momentmax
Factored_Load w⋅ L2⋅
83.8 106
× N m⋅⋅=:=
TensionmaxMomentmax
d2.7 106
× N⋅=:= CompressionmaxTensionmax
21.4 106
× N⋅=:=
Size Tension Member:
Material Strength Reduction: ϕt 0.9:= Material Yeild Strength: Fy 3.4 108× Pa:=
AgTensionmax
Fy80.9 cm2
⋅=:=
HSS 12.750 x 0.375 gross area = 87.1 cm2
Pipe Diameter: 32.4 cm
Pipe Thickness: 1 cm
Size Compression Members:
Material Strength Reduction: ϕc 0.9:=Effective Length Factor: K 0.7:=Member Radius: r 32.4 cm:=Material Young's Modulus: E 2 1011
× Pa:=Length of Compression Member: Lc 5.2 m:=
Feπ
2 E⋅
K L⋅r
23.6 108
× Pa⋅=:= AgcCompressionmax
Fe37.8 cm2
⋅=:=
HSS 12.750 x 0.375 gross area = 87.1 cm2
Pipe Diameter: 32.4 cm
Pipe Thickness: 1 cm
Vassar Street Skywalk
• A unique feature of MIT’s campus is its system of interconnected buildings which allow students and faculty to travel across campus indoors.
• The Brain and Sciences Complex was completed in 2005, and has no connection to this interconnected network. This requires students and faculty to cross Vassar Street if they wish to go to buildings on the main campus.
• MIT would like to connect the recently built Brain and Cognitive Sciences Building to the Main Campus via a Skywalk over Vassar Street
Schematic Plan
Load Path
Prepared By: Emidio Piermarini – PM Tzoni Tzonev – SE Congyi Qian – AIA
Project Background Calculations
17' - 0"
A B
A BK = 0.7
ISOMETRIC VIEW PLAN VIEW
ELEVATION VIEW
TYP. SECTION
4'-9"(1.5 m)
10'-0"(3 m)
BLDG 46COG. SCI.
BLDG 36
34° TYP.
10'-0"(3 m)
112'-0"(34 m)
17'-6", TYP. SPAN(5.3 m)
33°
45
Load Calculations:
ωds 957.6 Pa:= Beam Depth: d 1.4 m:=Superimposed Dead Load on Deck:
ωll 4.8 103× Pa:= Beam Length: L 34.1 m:=Live Load on Deck:
Deck Width: w 3 m:=
Factored_Load 1.2ωds 1.6 ωll⋅+ 8.8 103× Pa⋅=:= Momentmax
Factored_Load w⋅ L2⋅
83.8 106
× N m⋅⋅=:=
TensionmaxMomentmax
d2.7 106
× N⋅=:= CompressionmaxTensionmax
21.4 106
× N⋅=:=
Size Tension Member:
Material Strength Reduction: ϕt 0.9:= Material Yeild Strength: Fy 3.4 108× Pa:=
AgTensionmax
Fy80.9 cm2
⋅=:=
HSS 12.750 x 0.375 gross area = 87.1 cm2
Pipe Diameter: 32.4 cm
Pipe Thickness: 1 cm
Size Compression Members:
Material Strength Reduction: ϕc 0.9:=Effective Length Factor: K 0.7:=Member Radius: r 32.4 cm:=Material Young's Modulus: E 2 1011
× Pa:=Length of Compression Member: Lc 5.2 m:=
Feπ
2 E⋅
K L⋅r
23.6 108
× Pa⋅=:= AgcCompressionmax
Fe37.8 cm2
⋅=:=
HSS 12.750 x 0.375 gross area = 87.1 cm2
Pipe Diameter: 32.4 cm
Pipe Thickness: 1 cm
Vassar Street Skywalk
• A unique feature of MIT’s campus is its system of interconnected buildings which allow students and faculty to travel across campus indoors.
• The Brain and Sciences Complex was completed in 2005, and has no connection to this interconnected network. This requires students and faculty to cross Vassar Street if they wish to go to buildings on the main campus.
• MIT would like to connect the recently built Brain and Cognitive Sciences Building to the Main Campus via a Skywalk over Vassar Street
Schematic Plan
Load Path
Prepared By: Emidio Piermarini – PM Tzoni Tzonev – SE Congyi Qian – AIA
Project Background Calculations
17' - 0"
A B
A BK = 0.7
ISOMETRIC VIEW PLAN VIEW
ELEVATION VIEW
TYP. SECTION
4'-9"(1.5 m)
10'-0"(3 m)
BLDG 46COG. SCI.
BLDG 36
34° TYP.
10'-0"(3 m)
112'-0"(34 m)
17'-6", TYP. SPAN(5.3 m)
33°
46
9.14 m max.
2.74 mmin.
STRUCTUREPRECEDENTS
CONNECTION DETAIL PRECENTS
SCHLAICH BERGERMANN UND PARTNERRipshorst Bridge
SCHLAICH BERGERMANN UND PARTNERRipshorst Bridge
BIRDS PORTCHMOUTH RUSSUMPlashet School Footbridge
ELEVATION 1:100
PLAN 1:200SECTION 1:200
VCATHERINE DE WOLFTYLER STEVERMER
BRIDGE
LOADS
PRIMARY STRUCTURE SIZING
SECONDARY STRUCTURE SIZING
1.52 m typ.
6.2 m
6.2 m
33 m
DEFLECTION
y
47
9.14 m max.
2.74 mmin.
STRUCTUREPRECEDENTS
CONNECTION DETAIL PRECENTS
SCHLAICH BERGERMANN UND PARTNERRipshorst Bridge
SCHLAICH BERGERMANN UND PARTNERRipshorst Bridge
BIRDS PORTCHMOUTH RUSSUMPlashet School Footbridge
ELEVATION 1:100
PLAN 1:200SECTION 1:200
VCATHERINE DE WOLFTYLER STEVERMER
BRIDGE
LOADS
PRIMARY STRUCTURE SIZING
SECONDARY STRUCTURE SIZING
1.52 m typ.
6.2 m
6.2 m
33 m
DEFLECTION
y
School of archiTecTure + Planning
Design by Caitlin Mueller, 2013Photos by Ana Vargas and Caitlin Mueller, 2012
“The experience of having Jorg Schlaich in our class was particularly great: not only was he was able to pinpoint the flaws in the structural logic, but he also gave very good advice in the program and aesthetics of the design, which I appreciated very much as an architect. It was really great to learn from him; thanks very much!”
– Shiyu Wei, Master of Architecture Student