+ All Categories
Home > Documents > Static Model Bra

Static Model Bra

Date post: 01-Jun-2018
Category:
Upload: patran
View: 230 times
Download: 0 times
Share this document with a friend

of 27

Transcript
  • 8/9/2019 Static Model Bra

    1/27

    Suspension structures Prof Schierle 1

    S

    uspens ion t ruc t u res

  • 8/9/2019 Static Model Bra

    2/27

    Suspension structures Prof Schierle 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

  • 8/9/2019 Static Model Bra

    3/27

  • 8/9/2019 Static Model Bra

    4/27

    Suspension structures Prof Schierle 4

    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

  • 8/9/2019 Static Model Bra

    5/27

    Suspension structures Prof Schierle 5

    Mast / cable details

    The mast detail demonstrates typical use of

    cable or strand sockets. Steel gusset platesusually provide the anchor for sockets.

    Equal anglesA 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

  • 8/9/2019 Static Model Bra

    6/27Suspension structures Prof Schierle 6

    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

  • 8/9/2019 Static Model Bra

    7/27Suspension structures Prof Schierle 7

    Uniform load

    w= 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 in2

    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 =(270 2+72 2)1/2 T=279 k

    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

    Horizontal reaction

    H= M/f= 4320/16 H= 270 k

  • 8/9/2019 Static Model Bra

    8/27Suspension structures Prof Schierle 8

    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 Eo/Em ration

    Try model cross sectionAm = Sg

    2Ao Eo/EmAm = 0.0001x 4.16 x 0.759 Am = 0.0003157

    Model wire size

    = 2(Am/)1/2 = 2(0.0003157/)1/2 = 0.0200Use available wire size = 0.02

    Am = 0.012 Am = 0.0003142

    Force scale

    Sf= AmEm/(AoEo)= 0.0003142 x 29000/(4.16x22000)

    Sf= 0.0001 Sf= 1:10,000

    Model load

    Original load Po = w L = 0.6klf x240 Po = 144 k

    Sf= Pm/Po Pm = Pc SfPm = Po Sf= 144k x 1000 # / 10,000 Pm = 14.4 #

    Load per cup

    Assume 12 load cups (one cup per stay cable)

    Pcup = Pm/12 = 14.4#/12 Pcup = 1.2#

    Original strand Eo = 22,000ksiPiano wire Em = 29,000ksi

    Eo/Em = 22/29 Eo/Em = 0.759

    Original cross section area

    Strand 2 3/4 (70% metallic)

    Ao = 0.7 r2 = 0.7 (2.75/2)2 Ao = 4.16in

    2

  • 8/9/2019 Static Model Bra

    9/27Suspension structures Prof Schierle 9

    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

  • 8/9/2019 Static Model Bra

    10/27Suspension structures Prof Schierle 10

    Uniform suspender 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+438 2)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)

  • 8/9/2019 Static Model Bra

    11/27

  • 8/9/2019 Static Model Bra

    12/27Suspension structures Prof Schierle 12

    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

  • 8/9/2019 Static Model Bra

    13/27Suspension structures Prof Schierle 13

    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

  • 8/9/2019 Static Model Bra

    14/27Suspension structures Prof Schierle 14

    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 strandsto avoid buckling under roof deflection

  • 8/9/2019 Static Model Bra

    15/27Suspension structures Prof Schierle 15

    Assume

    All. strand stress Fy/3 = 210/3 Fa = 70 ksi

    DL = 5 psf 5 psf on arch 5 psf

    LL = 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 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

  • 8/9/2019 Static Model Bra

    16/27Suspension structures Prof Schierle 16

    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 stress

    f = T/Am= 12 k / 0.28 f = 43 ksi

    43 < 70, ok

  • 8/9/2019 Static Model Bra

    17/27Suspension structures Prof Schierle 17

    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) KL = 7

    3 bars, P ~ C / 3 ~ 786 / 3 ~ 262 k

    Try 10 extra strong pipe Pall = 328 > 262

    3xP 10 ok

    (244/25.4 = 9.6)

    (267/25.4 = 10.5)

  • 8/9/2019 Static Model Bra

    18/27Suspension structures Prof Schierle 18

    Oakland Coliseum

  • 8/9/2019 Static Model Bra

    19/27Suspension structures Prof Schierle 19

    Oakland Coliseum

    Architect/Engineer:

    Skidmore Owings and Merrill

    Radial cables, suspended from a concretecompression 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)

  • 8/9/2019 Static Model Bra

    20/27Suspension structures Prof Schierle 20

    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 kCable tension (max.)

    T = (H2 + R2)1/2 = (216 2 + 80 2 )1/2 T = 230 k

    Metallic cross section required

    Am = T/Fa= 230/70ksi Am = 3.3 in2

    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)

  • 8/9/2019 Static Model Bra

    21/27Suspension structures Prof Schierle 21

    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, designedto resist asymmetrical erection loads and lateral wind load.

    The peripheral pylons are triangular concrete walls.

    1 Cross section

    2 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 joists

    D Steel tension ring

    E Steel anchor bracket

  • 8/9/2019 Static Model Bra

    22/27

    Suspension structures Prof Schierle 22

    Clifton Suspension Bridge, UK, 214 m span, 1831

  • 8/9/2019 Static Model Bra

    23/27

    Suspension structures Prof Schierle 23

    Danube Bridge, Budapest, 220 m span, 1840

  • 8/9/2019 Static Model Bra

    24/27

    Suspension structures Prof Schierle 24

    Golden Gate Bridge, San Francisco, 4200 m main span, 1934

  • 8/9/2019 Static Model Bra

    25/27

    Suspension structures Prof Schierle 25

    Tacoma Narrows Bridge, 853 m span, 7, 1, 1940 - 11, 7, 1940

    http://www.youtube.com/watch?v=3mclp9QmCGs

  • 8/9/2019 Static Model Bra

    26/27

    Suspension structures Prof Schierle 26

    Exhibit Hall 8/9, Hanover, Germany, 245x345m, 1998

    Architect: GMP

    Engineer: Schlaich Bergermann

  • 8/9/2019 Static Model Bra

    27/27

    https://www.youtube.com/watch?v=N9fbRcRJY34


Recommended