Suspension Structures
Suspension structures
Prof Schierle
Suspension Structures
Effect of:
Support
Form
Stability
1. Circular support to balance
lateral thrust
2. Bleachers to resist lateral thrust
3. Self weight: catenary funicular
4. Uniform load: parabolic funicular
5. Point loads: polygonal funicular
6. Point load distortion
7. Asymmetrical load distortion
8. Wind uplift distortion
9. Convex stabilizing cable
10. Dead load to provide stability
Suspension structures
Prof Schierle
Sag/span vs. force
Considering force vectors at support;
H = horizontal reaction
V = vertical reaction
T = tension of cable at support
Reveal, for a given vertical reaction:
Suspension structures
small sag = large force
Large sag = small force
Large sag requires costly tall
support
Optimal span/sag is usually ~10
Prof Schierle
Cable details
1
2
Strand (good stiffness, low flexibility)
E=22,000 to 24,000 ksi, 70% metallic
Wire rope (good flexibility, low stiffness)
E = 14,000 to 20,000 ksi, 60% matallic
3
4
5
6
Bridge Socket (adjustable)
Open Socket (non-adjustable)
Wedged Socket (adjustable)
Anchor Stud (adjustable)
A
B
C
Support elements
Socket / stud
Strand or wire rope
Suspension structures
Prof Schierle
Mast / cable details
The mast detail demonstrates typical use of
cable or strand sockets. Steel gusset plates
usually provide the anchor for sockets.
Equal angles A and B result in equal forces
in strand and guy, respectively.
A Mast / strand angle
B Mast / guy angle
C Strand
D Guy
E Sockets
F Gusset plates
G Bridge socket (to adjust prestress)
H Foundation gusset (at strand and mast)
I
Mast
Suspension structures
Prof Schierle
Loyola University Pavilion
Architect: Kahn, Kappe, Lottery, Boccato
Engineer: Reiss and Brown
Consultant: Dr. Schierle
Roof spans the long way to provide open view for
outdoor seating for occasional large events
Lateral wind and seismic loads are resisted by:
Roof diaphragm
In width direction by concrete shear walls
In length direction by guy cables and
Handball court walls
Guy cables resist lateral trust
Suspension cables resist gravity
Stabilizing cables:
resist wind uplift
resist non-uniform load
provide prestress
Suspension structures
Prof Schierle
Assume:
Suspension cables spaced 20 ft
Allowable cable stress Fa = Fy/3
Fa = 70 ks
LL = 12 psf (60% of 20 psf for trib. area>600 ft2
DL = 18 psf
= 30 psf
Uniform load
w= 30 psf x 20 / 1000
Global moment
M= wL2/8= 0.60x2402/8
Horizontal reaction
H= M/f= 4320/16
Vertical reaction
R= wL/2= 0.60x240/2
Cable tension (max.)
T=(H2+R2)1/2 =(270 2+72 2)1/2
Graphic method
Draw vector of vertical reaction
Draw equilibrium vectors at support
Length of vectors give cable force
and horizontal reaction
Suspension structures
Metallic cross section required
Am=T/Fa=279/70 ksi
w= 0.60 klf
M= 4320 k
H= 270 k
R= 72 k
T=279 k
Am=3.99 in2
Gross cross section (70% metallic)
Ag=5.70 in2
Ag=Am/0.70=3.99/0.7 0
Cable size
=2(Ag/)1/2=2(5.70/)1/2=2.69 in
Prof Schierle
use 2
7
Static model
Assume:
Piano wire as model cables
Geometric scale Sg = 1:100
Strain scale Ss = 1 (due to large deflections)
Original strand
Eo = 22,000ksi
Piano wire
Em = 29,000ksi
Eo/Em = 22/29
Eo/Em = 0.759
Original cross section area
Strand 2 3/4 (70% metallic)
Ao = 4.16in2
Ao = 0.7 r2 = 0.7 (2.75/2)2
Suspension structures
Force scale
Sf = Am Em / (Ao Eo)
To keep Ss = 1 adjust Am by Eo/Em ration
Try model cross section
Am = Sg2 Ao Eo/Em
Am = 0.0001x 4.16 x 0.759
Am = 0.0003157
Model wire size
= 0.0200
= 2(Am/)1/2 = 2(0.0003157/)1/2
Use available wire size
= 0.02
Am = 0.0003142
Am = 0.012
Force scale
Sf = AmEm/(AoEo)= 0.0003142 x 29000/(4.16x22000)
Sf = 1:10,000
Sf = 0.0001
Model load
Po = 144 k
Original load Po = w L = 0.6klf x240
Sf = Pm/Po Pm = Pc Sf
Pm = Po Sf = 144k x 1000 # / 10,000
Pm = 14.4 #
Load per cup
Assume 12 load cups (one cup per stay cable)
Pcup = 1.2#
Pcup = Pm/12 = 14.4#/12
Prof Schierle
Exhibit Hall Hanover
Architect: Thomas Herzog
Engineer: Schlaich Bergermann
Suspended steel bands of 3x40 cm (1.2x16 inch) support
prefab wood panels, filled with gravel to resist wind uplift.
In width direction the roof is slightly convex for drainage;
which also provides an elegant interior spatial form.
Curtain wall mullions are pre-stressed between roof and
footing to prevent buckling under roof deflection.
Unequal support height is a structural disadvantage since
horizontal reactions of adjacent bays dont balance; but it
provides natural lighting and ventilation for sustainability
Suspension structures
Prof Schierle
Exhibit Hall Hanover
(10 psf)
LL =
0.5 kN/m2
(25 psf)
DL =
1.2 kN/m2
=
1.7 kN/m2
(35 psf)
Suspenders 3x40 cm (~1x16), spaced at 5.5 m (18)
Given
Uniform suspender load
w= 1.7 kN/m2 x 5.5m
w = 9.35 kN/m
Global moment
M=wL2/8= 9.35 x 642 / 8
M= 4787 kN-m
Horizontal reaction
H= M/f= 4787/7
Suspension structures
H = 684 kN
Vertical reaction R (max.)
Reactions are unequal; use R/H ratio
(similar triangles) to compute max. R
R / H= (2f+h/2) / (L/2), hence
R= H (2f+h/2) / (L/2)
R= 684 (2x7+13/2)/(64/2)
R= 438 kN
Suspender tension (max.)
T= (H2+R2)1/2= (6842+438 2)1/2
T= 812 kN
Prof Schierle
10
Suspender tension
(from previous page)
T = 812 kN
Suspender cross section area
A= 0.03 x 0.4 m)
A = 0.012 m2
Suspender stress
f=T/A= 812/0.012= 67,667 kPa
f = 68 MPa
US units equivalent
68 MPa x 0.145 =
f = 9.9 ksi
9.9 < 22 ksi, OK
Graphic method
Draw vector of total vertical load W = w L
Draw equilibrium vectors parallel to cable tangents
Draw equilibrium vectors for right support
Draw equilibrium vectors for left support
Suspension structures
Prof Schierle
11
Dulles Airport Terminal (1963)
Architect: Ero Saarinen
Engineer: Ammann and Whitney
150x600, 40-65 high
Concrete/suspension cable roof
Support piers spaced 40
Suspension structures
Prof Schierle
12
Dulles Airport Terminal
Span
Sag
Support height differential
Strand spacing
Allowable strand stress
Uniform strand load
w = 50psf x5/1000
Horizontal reaction H = wL2/(8f)
H = 0.25x1502/(8x15)
Max. vertical reaction
R=H(2f+h/2)/(L/2)
R = 46.9(30+12.5)/(75)
Strand tension T =(H2+R2)1/2
T =(46.92+ 26.62)1/2
Cross section required (70% metallic)
A = 53.9/(0.7x70) ksi
Strand diameter = 2(A/ )1/2
=2(1.1/3.14)1/2 = 1.18
Use
Suspension structures
Prof Schierle
L = 150
f = 15
h = 25
e = 5
Fa = 70 ksi
DL = 38 psf
LL = 12 psf
= 50 psf
w = 0.25 klf
H = 46.9 k
R = 26.6 k
T = 53.9 k
A = 1.1 in2
= 1 3/16
13
Skating rink Munich
Architect: Ackermann
Engineer: Schlaich / Bergermann
A prismatic steel truss arch of 100 m span, rising
from concrete piers, support anticlastic cable nets
A translucent PVC membrane is attached to wood
slats that rest on the cable net
Glass walls are supported by pre-stressed strands
to avoid buckling under roof deflection
Suspension structures
Prof Schierle
14
Assume
All. strand stress Fy/3 = 210/3
DL = 5 psf 5 psf on arch
LL = 20 psf 12 psf on arch
=
25 psf 17 psf on arch
Fa = 70 ksi
5 psf
uplift 21 psf
16 psf
Cable net
Uniform load (cable spacing 75 cm = 2.5)
Gravity w= 25 psf x2.5/1000
w = 0.0625 klf
Wind p= 16 psf x 2.5/1000
p = 0.040 klf
Global moment
M= w L2/8= 0.0625 x 1102/8
M = 95 k
Horizontal reaction
H = M / f = 95 / 11
H = 8.6 k
Vertical reaction
R/H= (2f+h/2 ) / (L/2); R= H (2f+h/2 ) / (L/2)
R= 8.6 (2x11+53/2)/(110/2)
R = 7.6 k
Gravity tension (add 10% residual prestress)
T = 1.1 (H2 + R2)1/2
T = 1.1 (8.6 2 + 7.6 2 )1/2
T = 11.5 k
Suspension structures
Prof Schierle
15
Gravity tension (from previous slide)
T = 11.5 k
Wind tension (10% residual prestress)
Wind suction is normal to surface, hence
T= 1.1 p r= 1.1 x 0.04 x 262
Wind T = 12 k
12 > 11.5
Wind governs
Metallic cross section area
(assume twin net cables, 70% metallic)
Am= 0.28 in2
Am = 0.7x2r2= 0.7x2(0.5/2)2
Cable stress
f = T/Am= 12 k / 0.28
Suspension structures
Prof Schierle
f = 43 ksi
43 < 70, ok
16
Truss arch design (prismatic truss of 3 steel pipes)
Floor area 4,200 m2 / 0.30482
45,208ft2
Arch load
w = (45,208x17psf/328)/1000+0.26klf arch DL)
w = 2.6klf
Horizontal reaction
H= M/d = wL2/(8d)= 2.6x3282/(8x53)
H= 660k
Vertical reaction
R= w L/2 = 2.6 x 328 / 2
R = 426k
Arch force
C= (H2 + R2)1/2 = (6602 + 4262)1/2
C = 786k
Panel bar length (K=1)
3 bars, P ~ C / 3 ~ 786 / 3 ~ 262 k
Try 10 extra strong pipe
KL = 7
Pall = 328 > 262
3xP 10 ok
(244/25.4 =
9.6)
(267/25.4 = 10.5)
Suspension structures
Prof Schierle
17
Oakland Coliseum
Suspension structures
Prof Schierle
18
Oakland Coliseum
Architect/Engineer:
Skidmore Owings and Merrill
Radial cables, suspended from a concrete
compression ring and tied to a steel tension
ring, are stabilized against wind uplift and
non-uniform load by prefab concrete ribs.
Assume
Allowable cable stress
Fa= 70 ksi
(1/3 of 210 ksi breaking strength)
Cables spaced 13 @ outer compression ring
LL reduced to 60% of 20 psf per UBC
for tributary area > 600 sq. ft.)
LL = 12 psf (60% of 20 psf)
DL = 28 psf (estimate)
= 40 psf
+ 0.12 klf for concrete ribs
(0.15kcf x 4 x 29/144 uniform load)
Suspension structures
Prof Schierle
19
Distributed load
w = 40 psf x 13/1000
w = 0 to 0.52 klf
Global moment (due to triangular roof load)
(cubic parabola with origin at mid-span)
Mx = w L2/24 (1 8 X3 / L3 )
For max. M at mid-span, X=0, hence
M = wL2 / 24= 0.52 x 4202 / 24
M = 3,822 k
Global moment (due to uniform rib load)
M = w L2/8 = 0.12 klf x 4202/8
M = 2,646 k
Moments = 3,822 + 2,646
M = 6,468 k
Horizontal reaction
H = M/f= 6,468/30
Vertical reaction
R = wL/2 = (0.52/2+0.12) x 420/2
Cable tension (max.)
T = (H2 + R2)1/2 = (216 2 + 80 2 )1/2
Metallic cross section required
Am = T/Fa= 230/70ksi
Suspension structures
Prof Schierle
H = 216 k
R = 80 k
T = 230 k
Am = 3.3 in2
20
Recycling Center, Vienna (1981)
Architect: L. M. Lang
Engineer: Natterer and Diettrich
This recycling center is a wood structure of 170 m (560 ft)
diameter of 67 m (220 ft) height concrete mast.
The roof consists of 48 radial laminated wood ribs that
carry uniform roof load in tension but asymmetrical loads
may cause bending in the semi-rigid tension bands.
The mast cantilevers from a central foundation, designed
to resist asymmetrical erection loads and lateral wind load.
The peripheral pylons are triangular concrete walls.
Suspension structures
1
2
3
4
Cross section
Roof plan
Top of central support mast
Tension rib base support
A
B
C
D
E
Laminated wood ribs, 20x80-110 cm (7.8x32-43 in)
Laminated wood ring beams, 12x39 cm (5x15 in)
Wood joists
Steel tension ring
Steel anchor bracket
Prof Schierle
21
Clifton Suspension Bridge, UK, 214 m span, 1831
Suspension structures
Prof Schierle
22
Danube Bridge, Budapest, 220 m span, 1840
Suspension structures
Prof Schierle
23
Golden Gate Bridge, San Francisco, 4200 m main span, 1934
Suspension structures
Prof Schierle
24
Tacoma Narrows Bridge, 853 m span, 7, 1, 1940 - 11, 7, 1940
http://www.youtube.com/watch?v=3mclp9QmCGs
Suspension structures
Prof Schierle
25
Exhibit Hall 8/9, Hanover, Germany, 245x345m, 1998
Architect: GMP
Engineer: Schlaich Bergermann
Suspension structures
Prof Schierle
26
https://www.youtube.com/watch?v=N9fbRcRJY34
Suspension structures
Prof Schierle
27
