mm6 chapter 06 analysis rev 05 2010

7/26/2019 MM6 Chapter 06 Analysis Rev 05 2010

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MM6 – Structural Design Concepts

Chapter 6 – Analysis09/2010

6.1

Dr.-Ing. Karsten MoritzDr.-Ing. Lars Schiemann

6. Analysis



6.2

Dr.-Ing. Karsten Moritz

Dr.-Ing. Lars Schiemann

6.1 Characteristics of load bearing behaviour

of ETFE-foil cushions

• factors of influence:

- geometrical factors

(e.g. sag, curvature, ground plan)

• distinctive material and geometrical nonlinearities

- material factors (e.g. non-linear behaviour

with elastic, viscoelastic and plastic ranges)

- constructive factors (e.g. cushions with even or curved middle layer,

additional cables, quality of welding seams, anchoring profiles)

• non-proportional relation between loads, deformation and forces

→ positive but complex load bearing behaviour





6.3


• Example: square ETFE-foil cushionwith internal pressure, sag 10%

Question:What is the incremental factor of stressesfor 50% higher loads?

Non-proportional relation between loads and forces

Result:for 1.50 times increase of load values→ 1.25 times increase stresses

→ positive load bearing behaviour



6.4



• Boiler formula

r

F

p

cable

p

F

r

barrel

demand:

- cable forces or membrane forces F

assumption:- circular geometry of deformation

given:- radius r [m] and- loads p

→ Boiler formula:F = p x r F [kN/m] = p [kN/m2] x r [m]F [kN] = p [kN/m] x r [m]

line load [kN/m] area load [kN/m2]

Relation between curved geometry and stresses





6.5


• Boiler formula arch sphere

demand:- cable forces or membrane forces F

assumption:- circular geometry of deformation

given:- radius r [m] and- loads

→ Boiler formula:F = p x r F [kN] = p [kN/m] x r [m]

line loads [kN/m] area load [kN/m2]


F F

r

F2F1

r

F1 [kN/m] = 1/2 p x rF2 [kN/m] = 1/2 p x r



6.6



• square ETFE-foil cushion with internal pressure

- synclastic surface (mostly)

- 2-axial load transfer

- constant internal pressure

- isotropic material behaviour

→ equal radii

→ equal membrane forces

→ r1 = r2 F1 = F2

→ P = F1 / r1 + F2 / r2

→ P = 2F / r→ F1 = F2 = p x r /2

F = p x r

→ p = F /r

r 1r 2

F2

F1

r

p i

F






6.7


Relation of curvature and stresses – boiler formula

p = inner pressure (uniformly distributed load) [kN/m2]

r = radius [m] ], note: k = 1/ r (k = curvature; r = radius)

F = meridian force (in perimeter direction) [kN/m]

boiler, barrel, tube, membrane F [kN/m] = p [kN/m2] x r [m] (1)

cable F [kN] = p [kN/m] x r [m] (2)

r

F

p

p

F

r

r snowload or wind down load

r wind uplift



6.8



- synclastic (r 1 ≠ r 2 ): p = F1 / r 1 + F2 / r 2 (F [kN/m]; r [m], p [kN/m2]) (3)

- synclastic (r 1 = r 2 = r): F = F1 = F2 = p x r / 2 (4)

- anticlastic (r 1 ≠ r 2): p = F1 / r 1 – F2 / r 2 (5)

F1

F2 r 1 r 2p

r 1, wind upliftr 2, wind uplift

F = F1=F2

pr

F1F2

F1

F2

r 1

r 2

p

p = inner pressure (area load, constant) [kN/m2]

r = radius [m], note: k = 1/ r (k = curvature; r = radius)

F = force [kN/m]




6.11



pi

f OL,2

f UL,2

R UL,2

l

pi

f OL,1

f UL,1

R UL,1

l

→ R = l2 / 8f + 1/2 f

curved cushion

→ big foil sag

→ less radius of curvature

→ less membrane forces

less curved cushion

→ less foil sag

→ bigger radius of curvature

→ bigger membrane forces

• Boiler formula: F = p x r

• General relation between radius of curvature,

foil sag f and chord length of a segment of a circle:• or demonstrative:



6.12



cable force / cable sag

0

5

10

15

20

25

0 5 10 15 20 25 30 35

cable sag [% ]

m a x . c a b l e f o r c e [ k N ]

cable structure

span: l = 5,0m,

load: q = 1,0 kN/m

- significant influence of the cable sag to the cable force

- sensible conclusion: cable sag f 10% of span

Effect of the sag to the maximal forces




6.13



Parameter study: influence of foil sag

• two-layered ETFE-cushion, span 5m, nonlinear material behaviour• load case wind suction ws = 1,5 kN/m2

Percentage reduction of principal stresses σ1 in consideration of foil sag f 0

rectangle square circle rhomb

foil sag



6.14



Three typical load cases of cushions:

1.) Inner pressure (pre-stress)

2.) Wind loads and inner pressure

3.) Snowloads and inner pressure

6.3 Load-bearing behaviour of cushion





6.15


span width

6.3.1 Cushion - action: inner pressure (pre-stress)

If do = du , f o = f u , g ≅ 0 ⇒ Fo = Fu = F

If R ≅ constant ⇒ F = const. (boiler formula: F [kN/m] = pi [kN/m2] x R [m])

cross-section, action: pre-stress

⇒ vertical loads: Vres = Vu – Vo = 0 ; horizontal loads: H res = Hu + Ho ≠ 0

⇒ resulting support reactions of two adjacent cushions:

⇒ ΣV = Vl + Vr = 0 ; ΣH = Hl – Hr = 0 (if Hl = Hr )



6.16



6.3.2 Cushion - action: wind uplift (wind suction)

Thermodynamical basics for wind loads

We know three different equations due to the change of state of a gas (the 3 “gas-laws“):

- isobaric change of state1: pressure p = constant ⇒ Δp = 0 ⇒ V/T = constant

- isochore change of state2: volume V = constant ⇒ ΔV = 0 ⇒ p/T = constant

- isothermal change of state3: temperature T = constant ⇒ ΔT = 0 ⇒ p x V = constant

1 GAY-LUSSAC, first law, 2 GAY-LUSSAC, second law, 3 BOYLE-MARIOTTE

2

22

1

11

T

Vp

T

Vp ⋅=

⋅

Gas-law:





6.17



Thermodynamical basics for wind loads

Air-inflated cushions loaded by a wind load follow the third equation: p x V = constant

The demand “temperature T is constant“ will be met. The molar mass of the enclosed

air is approximately constant, because the wind load comes quick and the cushion

is (nearly) enclosed. In case of wind load, the inner pressure must change and the

enclosed volume, too. If the inner pressure rises the volume has to be reduced

- and vice versa.



6.18



ws

In this case εo and Fo are increasing, εu, Fu and pi are decreasing. (ε = strain)

cross-section, action: wind uplift


Vertikal load: V = Vu – Vo = - 0.5 ws x l (l = span, ws = wind suction (uplift) = const., ws ≅ ws, vertical)

Horizontal load: H = Hu + Ho ≠ 0

⇒ resulting support reactions of two adjacent cushions: ΣV = Vl + Vr ≠ 0 ; ΣH = Hl – Hr = 0 (if Hl = Hr )





6.19


s

6.3.3 Cushion - action: snow load

cross-section, action: snow load

Under snow loads εo and Fo are decreasing, εu, Fu and pi are increasing.

Vertical load: V = Vu – Vo = 0.5 s x l (l = span, s = snow load = const.)

Horizontal load: H = Hu + Ho ≠ 0

⇒ resulting support reaction of two adjacent cushions: ΣV = Vl + Vr ≠ 0 ; ΣH = Hl - Hr = 0 (if Hl = Hr )



6.20



- Principal difference between pneumatically pre-stressed structures likeETFE-films cushions and glass structures are the horizontal support reactions.

- The effect of horizontal reaction forces will cancel each other out foradjacent cushions occurs only for cushions with same geometry, same

thickness and same loads.Note: wind and snow loads are not symmetrical and constant over the wholesurface (e.g. wind suction and pressure and snow accomodations), differentloads of adjacent cushions because of failed air management system ordamaged cushions (accidental load case).

- Because of horizontal support reactions a horizontal stabilization of theprimary structure is often necessary.

- Elevated anchoring profiles result in torsion moments for the primarystructure.

Sensible to use hollow sections rather than open sections like doubble T-beams for primary structure.




6.21



with foil sag

f 0=10%

6.4 Influence of span and ground geometry

• two-layered ETFE-cushion with span 5m, load case: wind suction ws=1.5 kN/m2

• different principal stresses σ1 values for ground plans

• 50% increased span → approx. 31% increased principal stresses

principal stresses σ1 for different spans and ground plans



6.22



Ground plan of ETFE-foil cushions

Trajectories of stresss and ratio of principal stresses

1 / 2 for wind suction q = 2,25 kN/m2





6.23


6.3.4 Realisable spans of structures of ETFE-films

Deutsche Bundesstiftung Umwelt,

Osnabrück, 2002

“Eden Project“, St. Austell

(UK) 2000



6.24



The maximal span depends on the geometry, the structural system and on the loads

- mechanically prestressed, 1-layer system: ~ 1.5 m

- pneumatically prestressed

(rectangular form, elongated elements,sag 10% of the length): ~ 4.7 m*

- pneumatically prestressed

(circular form / polygonal elements ∅ ~ 7.5 m

sag 10% of the length):

“Masoala rainforest hall “, zoo Zürich

2002

* Note:

For structural analysing aspects the length of

rectangular geometries is not decisive.

(e.g. cushion of Masoala rainforest hall, Zürich:length = 106 m)

- by using cables or cable nets: the span of the ETFE-

films structure depends on the cable structures




6.25



6.5 Conclusion

• ETFE-foil structures show geometrical and material nonlinearities→ numerical analysis in consideration of nonlinearities

• positiv influence of foil sag and curved surfaces→ reducing foil stresses and stabilization because of curved surfaces→ sensible foil sags more than 10% of span

• positiv non-proportional structural behaviour of ETFE-foil cushions

• positiv influence of ground plans with biaxial load transferlike quadratic or circular ground plans

→ design of the structures influence directlystresses and realisable spans of the structures


6.26




mm6 chapter 06 analysis rev 05 2010

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