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Suspension structures Copyright Prof Schierle 2012 1
SSuspension Structures
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Suspension structures Copyright Prof Schierle 2012 2
Suspension Structures
Effect of:
Support Form
Stability
1. Circular support to balancelateral thrust
2. Bleachers to resist lateral thrust
3. Self weight: catenary funicular4. Uniform load: parabolic funicular
5. Point loads: polygonal funicular
6. Point load distortion7. Asymmetrical load distortion
8. Wind uplift distortion
9. Convex stabilizing cable10.Dead load to provide stability
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Suspension structures Copyright Prof Schierle 2012 3
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:
small sag = large force
Large sag = small force
Large sag requires costly tallsupport
Optimal span/sag is usually ~10
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Cable details
1 Strand (good stiffness, low flexibility)E=22,000 to 24,000 ksi, 70% metallic
2 Wire rope (good flexibility, low stiffness)E = 14,000 to 20,000 ksi, 60% matallic
3 Bridge Socket (adjustable)
4 Open Socket (non-adjustable)5 Wedged Socket (adjustable)
6 Anchor Stud (adjustable)
A Support elements
B Socket / studC Strand or wire rope
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Mast / cable details
The mast detail demonstrates typical use of
cable or strand sockets. Steel gusset platesusually 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
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Loyola University Pavil ion
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 gravityStabilizing cables:
resist wind uplift
resist non-uniform load
provide prestress
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Uniform loadw= 30 psf x 20 / 1000 w= 0.60 klf
Global momentM= wL2/8= 0.60x2402/8 M= 4320 k
Vertical reactionR= wL/2= 0.60x240/2 R= 72 k
Gross cross section (70% metallic)Ag=Am/0.70=3.99/0.7 0 Ag=5.70 in
2
Cable size=2(Ag/)1/2=2(5.70/)1/2=2.69 in use 2
Metallic cross section requiredAm=T/Fa=279/70 ksi Am=3.99 in
2Graphic method
Draw vector of vertical reaction Draw equilibrium vectors at support
Length of vectors give cable forceand horizontal reaction
Cable tension (max.)T=(H2+R2)1/2=(2702+722)1/2 T=279 k
Assume: Suspension cables spaced 20 ftAllowable cable stress Fa = Fy/3 Fa = 70 ksLL = 12 psf (60% of 20 psf for trib. area>600 ft2
DL = 18 psf = 30 psf
Horizontal reactionH= M/f= 4320/16 H= 270 k
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Static model
Assume:Piano wire as model cables
Geometric scale Sg = 1:100
Strain scale Ss = 1 (due to large deflections)
Force scale
Sf= Am Em / (Ao Eo)
To keep Ss = 1 adjust Am by Em / Eo ration
Try model cross section
Am = Sg2Ao Em/Eo
Am = 4.16 x 1.32/10,000 Am = 0.000549
Model wire size
= 2(Am/)1/2 = 2(0.000549/)1/2 = 0.026
Consider available size use 0.025
Am = r2 = (0.025/2)2 Am = 0.00049in
2
Force scale
Sf= AmEm/(AoEo)= 0.00049x29000/(4.16x22000)Sf= 0.0001553 Sf= 1 : 6441
Model load
Original load Po
= w L = 0.6klf x240 Po
= 144k
Sf= Pm/Po Pm = Pc SfPm = Po Sf= 144k x 1000# / 6441 Pm = 22#
Load per cup
Assume 12 load cups (one cup per stay cable)Pcup = Pm/12 = 22#/12 Pcup = 1.83#
Original strand Eo = 22,000ksiPiano wire Em = 29,000ksi
Em/Eo = 29/22 Em/Eo = 1.32
Original cross section area
Strand 23/4 (70% metallic)
Ao = 0.7 r2 = 0.7 (2.75/2)2 Ao = 4.16in
2
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Exhibit Hall Hanover
Architect: Thomas Herzog
Engineer: Schlaich Bergermann
Suspended steel bands of 3x40 cm (1.2x16 inch) supportprefab 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 andfooting 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
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Uniformsuspender load
w= 1.7 kN/m2
x 5.5m w = 9.35 kN/mGlobal moment
M=wL2/8= 9.35 x 642 / 8 M= 4787 kN-m
Horizontal reaction
H= M/f= 4787/7 H = 684 kN
Vertical reaction R (max.)
Reactions are unequal; use R/H ratio
(similar triangles) to compute max. RR / H= (2f+h/2) / (L/2), hence
R= H (2f+h/2) / (L/2)
R= 684 (2x7+13/2)/(64/2) R= 438 kN
Exhibit Hall Hanover
Suspender tension (max.)T= (H2+R2)1/2= (6842+4382)1/2 T= 812 kN
Given LL = 0.5 kN/m2 (10 psf)
DL = 1.2 kN/m
2
(25 psf) = 1.7 kN/m2 (35 psf)
Suspenders 3x40 cm (~1x16), spaced at 5.5 m (18)
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Suspender tension
(from previous page) T = 812 kN
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
Suspender cross section area
A= 0.03 x 0.4 m) A = 0.012 m2
Draw equilibrium vectors parallel to cable tangents
Draw equilibrium vectors for right support
Draw equilibrium vectors for left support
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Dulles Airport Terminal (1963)
Architect: Ero Saarinen
Engineer: Ammann and Whitney
150x600, 40-65 high Concrete/suspension cable roof
Support piers spaced 40
D ll Ai t T i l
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Dulles Airport Terminal
Span L = 150
Sag f = 15
Support height differential h = 25
Strand spacing e = 5
Allowable strand stress Fa = 70 ksi
DL = 38 psf
LL = 12 psf
= 50 psfUniform strand load
w = 50psf x5/1000 w = 0.25 klf
Horizontal reaction H = wL2/(8f)
H = 0.25x1502
/(8x15) H = 46.9 kMax. vertical reaction
R=H(2f+h/2)/(L/2)
R = 46.9(30+12.5)/(75) R = 26.6 k
Strand tension T =(H2+R2)1/2
T =(46.92+ 26.62)1/2 T = 53.9 k
Cross section required (70% metallic)
A = 53.9/(0.7x70) ksi A = 1.1 in2
Strand diameter = 2(A/ )1/2
=2(1.1/3.14)1/2 = 1.18Use = 1 3/16
L
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Skating rink MunichArchitect: AckermannEngineer: Schlaich / Bergermann
A prismatic steel truss arch of 100 m span, risingfrom 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 strandsto avoid buckling under roof deflection
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AssumeAll. strand stress Fy/3 = 210/3 Fa = 70 ksiDL = 5 psf 5 psf on arch 5 psfLL = 20 psf 12 psf on arch uplift 21 psf
= 25 psf 17 psf on arch 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 momentM= w L2/8= 0.0625 x 1102/8 M = 95 k
Horizontal reactionH = 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 (8.62+ 7.62 )1/2 T = 11.5 k
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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 k12 > 11.5 Wind governs
Metallic cross section area(assume twin net cables, 70% metallic)Am = 0.7x2r2= 0.7x2(0.5/2)2 Am= 0.28 in2
Cable stressf = T/Am= 12 k / 0.28 f = 43 ksi
43 < 70, ok
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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 reactionH= M/d = wL2/(8d)= 2.6x3282/(8x53) H= 660k
Vertical reactionR= w L/2 = 2.6 x 328 / 2 R = 426k
Arch forceC= (H2+ R2)1/2 = (6602+ 4262)1/2 C = 786k
Panel bar length (K=1) KL = 7
3 bars, P ~ C / 3 ~ 786 / 3 ~ 262 kTry 10 extra strong pipe Pall = 328 > 262
3xP 10 ok
(244/25.4 = 9.6)
(267/25.4 = 10.5)
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Oakland ColiseumArchitect/Engineer:
Skidmore Owings and Merrill
Radial cables, suspended from a concretecompression ring and tied to a steel tensionring, are stabilized against wind uplift andnon-uniform load by prefab concrete ribs.
AssumeAllowable 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)
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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
Horizontal reaction
H = M/f= 6,468/30 H = 216 k
Vertical reaction
R = wL/2 = (0.52/2+0.12) x 420/2 R = 80 k
Cable tension (max.)
T = (H2 + R2)1/2 = (216 2 + 80 2 )1/2 T = 230 k
Metallic cross section requiredAm = T/Fa= 230/70 ksi Am = 3.3 in
2
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
Recycling Center Vienna (1981)
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Recycling Center, Vienna (1981)
Architect: L. M. LangEngineer: 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, designedto resist asymmetrical erection loads and lateral wind load.The peripheral pylons are triangular concrete walls.
1 Cross section2 Roof plan
3 Top of central support mast4 Tension rib base support
A Laminated wood ribs, 20x80-110 cm (7.8x32-43 in)B Laminated wood ring beams, 12x39 cm (5x15 in)C Wood joistsD Steel tension ringE Steel anchor bracket
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Clifton Suspension Bridge, UK, 214 m span, 1831
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Danube Bridge, Budapest, 220 m span, 1840
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Golden Gate Bridge, San Francisco, 4200 m main span, 1934
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Tacoma Narrows Bridge, 853 m span, 7, 1, 1940 - 11, 7, 1940http://www.youtube.com/watch?v=3mclp9QmCGs
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Exhibit Hall 8/9, Hanover, Germany, 245x345m, 1998
Architect: GMP
Engineer: Schlaich Bergermann
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suspended cable car
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