BIRD-WATCHING BOARDWALK ON DUDDINGSTON LOCH
VIRIDIANA AMARAL GURGEL ANTON ANIKEEV LACHLAN ANDERSON FRANK ALEXANDRA ZERVUDACHI
SITING STRATEGY We decided to site our bird watching boardwalk on the southeast side of Duddingstone Loch, away from the more busy north west side close to the village. Access is from the nearby running track. We felt that this more secluded site gives birdwatchers the opportunity for both long range birdwatching across the lake, as well as close observation of birds within the wetlands, which are currently innacessible. The design leads birdwatchers from the shore onto an island, which they can explore freely.
view to Arthurs Seat view toward main bird landing
platform faces away from the sun path to avoid being blinded when looking up at the sky natural camouflage among the reeds
view of village
DESIGN STRATEGY Our design strategy focused on creating a structure for birdwatchers, going beyond a simple viewing platform. We applied variations to a simple modular geogetry to create a more elaborate and exciting shape. The zig-zaging shape of the broadwalk provides wider angles of views across the loch as well as offering space for wheel chair manoevering. Structurally it also provides bracing for the broadwalk.
AESTHETIC STRATEGY The aesthetic strategy of our broadwalk was to integrate it within its surroundings by immitating the slender verstical shape of the reeds that grow out of the water. The variations in height of the balustrade allow for bird watchers to look out across the landscape at the lower points and conceal themselves behind the heigher points in order to observe birds without scaring them off.
STRUCTURE _ PLANS Columns
Variation in pattern
Primary beams
Secondary beams
Decking 1:100
1:50
STRUCTURE _ SECTION
1:50 section
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B
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1:20 section A
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1:20 section B PRODUCED BY AN AUTODESK EDUCATIONAL PRODUCT
1:5 section A
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A
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STRUCTURE _ CONNECTION DETAILS
STRUCTURE _ CONNECTION DETAILS
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TCUDORP LANOITACUDE KSEDOTUA NA YB DECUDORP
TCUDORP LANOITACUDE KSEDOTUA NA YB DECUDORP
1:5 section C
1:5 plan
TCUDORP LANOITACUDE KSEDOTUA NA YB DECUDORP
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C
N AUTODESK EDUCATIONAL PRODUCT PRODUCED BY AN AUTODESK EDUCATIONAL PRODUCT
STRUCTURE _ BALUSTRADE DETAIL
Sculpture made of larch wood Nicholas Pope,1980 Venice Bienale Discrete steel cable fulfils need for structural bracing while not breaking the vertical aesthetic of the broadwalk
1:5 section
1:10 elevation
TIMBER PROPERTIES Choice of wood
- British Larch
European Larch is known for its natural strength, durability and wateprood property, ideal for the outside without the use of treatement. Aesthetically, its warm reddish brown or terracota colour with golden streaks, which fade to silver after prolonged exposure to sunlight perfectly matches the subtle colour scheme of the loch and its surroungings.
Mechanical Properties -
strength class
C24
-
Bending
fm,k = 24
-
Parallel compression
fc,0,k = 21
-
Perpendicular compression
fc,90,k = 2,5
-
Shear
fv,k = 2,5
-
Mean elasticity modulus
E = 11103
Modification factors Assuming that: -
Service class
3 (external use, fully exposed)
-
Load duration
long term
-
Material
solid timber
Therefore, strength modification factor
Kmod = 0,55
height factor (assuming d > 150mm)
Kh = 1,0
Instability factor (full torsial constraint)
Kcrit = 1,0
Load sharing factor (span < 6m)
Kls = 1,1
Moisture factor (solid timber, class 3)
Kdef = 2,0
Material factor (soid untreated timber
YM = 1,3
CALCULATIONS _ COLUMNS Assume column dimension 100 x 100 mm (100 x 97 in Table 12) x 3000mm E0.005 = 7.4 kN/mm
Rx-x= 28 mm
Kc,y = 0.2793
Total area of floor carried by column (worst case) A= 2 x 1.8 = 3.6 m2
2000
(column carries half of the adjacent spans)
1800
Beam supporting largest area
Total load carried by each column
P = 3.6 m x 4.5 kN/m2 = 16.2 kN
For stress class C24, Compressive strength parallel to the grain
fc,0,k = 21 N/mm2
E 0.05 7400 = = 352.58 fc,0,k 21
Slenderness ratio
ly=
L e 3000 = = 107.14 rxx 28
Maximum permissible stress in the column f c,0,d =
k mod ⋅ kc,90 ⋅ k ls ⋅ f c,0,k 0.60⋅ 1.0⋅ 1.1⋅ 21 = = 10.66 N /mm 2 gM 1.3
kc,90 = 1 as there is no increase in the bearing strength because the applied length ℓ of the uniformly distributed load q is 3 m > 100 mm Actual compressive stress
€
P 16.2⋅ 10 3 sc = = = 1.62 N /mm 2 A 100⋅ 100
Check for buckling strength:
Compressive stress ( s c ) <
Maximum allowable stress ( kc,y ⋅ f c,0,d )
1,62 N/m2
0,28．10,66 = 2,97 N/m2 €
Therefore the Column is safe against buckling
€
CALCULATIONS _ PRIMARY BEAMS Assume rectangular section 50mmx220mm -
Area 2nd moment of inertia Section modulus
A = 50．220 = 11．103 mm2 Ixx = 44,4．106 mm4 Zxx = 403．103 mm3
Bending strength Maximum bending moment For uniformly distributed load (UDL),
For a point load,
w = surface load．span = (imposed load + dead load)．span = (5,0 + 0,5)． 2,0 = 5,5． 2,0 = 11,1 kN/m
P = point load / 2 = 4,5 / 2 = 2,25 kN Therefore an appropriate section modulus Zxx must be greater than 490
→ Total Mmax
= Mmax for UDL + Mmax for point load = 4,46 + 1,01 = 5,47 kNm
13,58 N/mm2
10,94 N/mm
2
>
Bending strength
11,17 N/mm2
Therefore section is NOT satisfactory in bending, sizing must be reconsidered Bending strength must be greater than or equal to the maximum bending strength →
New section size is 50mm x 245mm
Bending stress
Maximum bending stress
Bending stress
Considering Zxx = 500,2
→
<
Bending strength
11,17 N/mm2
Therefore section is satisfactory in bending
Now assume new rectangular section 50mmx245mm Shear strength
f v,d =
k mod ⋅ k ls ⋅ f v,k 0,55⋅ 1,1⋅ 2,5 = = 1,16 N /mm 2 gm 1.3
Maximum shear force for UDL, €
V = surface load．span．length 2 =
5,5⋅ 1,8 = 4,95kN 2
Maximum shear force for point load, V = P/2 = 2,25 / 2 = 1,13 kN € → Total maximum shear force V for UDL + V for point load = 9,9 + 1,13 = 11,03 kNm
Maximum shear stress in rectangular section
Shear stress
<
1,35 N/mm2
td =
3V 3⋅ 6,08⋅ 10 3 = = 0,74N /mm 2 2bd 2⋅ 50⋅ 245
Shear strength
1,51 € N/mm2
Therefore section is satisfactory in shearing
Deflection of beam -
2nd moment of inertia Section modulus
Ixx = 61,3．106 mm4 Zxx = 500,2．103 mm3 4
Max deflection for UDL, w max
3 5 w⋅ L4 5 11,1⋅ (1,8⋅ 10 ) 5,83⋅ 1014 = ⋅ = ⋅ = = 2,25mm 384 E⋅ Ixx 384 11⋅ 61,3⋅ 10 9 2,59⋅ 1014 3
2,25⋅ (1,8⋅ 10 3 ) P⋅ L3 1,31⋅ 1010 −4 Max deflection for point load, w ins = (negligable) = 9 = 13 = 4,04⋅ 10 mm 48⋅ E⋅ Ixx 48⋅ 11⋅ 61,3⋅ 10 3,24⋅ 10 €
→ Total max deflection €
= Wmax + Wins
= 3,1 + 4,04．10-4 = 3,1 kNm
Final deflection Wfin = W (1+kdef) = 3,1(1+2,0) = 3,1．3 = 9,3 mm Recommended limit of final deflection for a member of span between two supports is150 Maximum allowable deflection = L / 150 = 1800 / 150 = 12 mm Final deflection 9,3 mm
<
Maximum allowable deflection 12 mm
CALCULATIONS _ SECONDARY BEAMS Assume rectangular section 75mmx147mm -
Area 2nd moment of inertia Section modulus
A = 75．147 = 11．103 mm2 Ixx = 19,7．106 mm4 Zxx = 270.1．103 mm3
Bending strength Maximum bending moment For uniformly distributed load (UDL),
M max =
w⋅ L2 3,69⋅ 1,9 2 3,69⋅ 3,61 13,32 = = = = 1,67kNm 8 8 8 8
For a point load,
M max =
w = surface load．span = (imposed load + dead load)．span = (5,0 + 0,5)． 0,67 = 5,5． 0,67 = 3,69 kN/m
P = point load / 2 = 1,67 / 2 = 0,84 kN/m
P⋅ L 0,84⋅ 1,8 1,595 = = = 0,4kNm 4 4 4
→ Total Mmax
= Mmax for UDL + Mmax for point load = 1,67 + 0,4 = 2,07 kNm
Maximum bending stress Mmax/ Zxx = 2,07 x 106/ 11,17 = 185,3 mm
3
270,1
For beam 75 x 147 mm
3V 3⋅ 3,92⋅ 10 3 td = = = 0,53N /mm 2 2⋅ 75⋅ 147 2bd
For beam 150 x 147 mm €
td =
therefore ,
f v,d
k ⋅k ⋅ f 0,55⋅ 1,1⋅ 2,5 = mod ls v,k = = 1,16 N /mm 2 gm 1.3
5,5⋅ 0,67⋅ 1,9 = 3,5kN Maximum shear force for UDL, V = surface load．span．length = 2 € 2
3V 3⋅ 3,92⋅ 10 3 = = 0,27N /mm 2 2bd 2⋅ 150⋅ 147
T1 (0, 53)> 1,93
T2(0,27) > 1,93
€
Therefore section is satisfactory in shearing Deflection of beam 4
Max deflection for UDL, w max
> 185,3
Shear strength
Maximum shear stress in rectangular section
3 5 w⋅ L4 5 3,69⋅ (1,9⋅ 10 ) = ⋅ = ⋅ = 1,43mm 384 E⋅ Ixx 384 11⋅ 39,7⋅ 10 9 3
0,42⋅ (1,9⋅ 10 3 ) P⋅ L3 −4 Max deflection for point load, w ins = = 9 = 1,37⋅ 10 mm € 48⋅ E⋅ Ixx 48⋅ 11⋅ 39,7⋅ 10
→ Total max deflection
= Wmax + Wins = 1,43 + 1,37．10-4 = 1,43 kNm € Final deflection Wfin = W (1+kdef) = 1,43(1+2,0) = 1,43．3 = 4,29 mm
Maximum shear force for point load, V = P/2 = 0,84 / 2 = 0,42kN €
Maximum allowable deflection = L / 150 = 1900 / 150 = 12,6 mm
→ Total max shear force
Final deflection 4,3 mm
= V for UDL + V for point load = 3.5 + 0,42 = 3,92 kNm
<
Maximum allowable deflection 12,6 mm
SECOND FLOOR Assume column dimension 150 x 150 mm (150 x 147 in Table 12) x 3000mm E0.005 = 7.4 kN/mm Rx-x= 42.4 mm
Kc,y = 0.5536
Total area of floor carried by column (worst case) A= 2 x 1.8 = 3.6 m2 (column carries half of the adjacent spans) Total load carried by each column: P = 3.6 m x (5.5x2) kN/m2 = 36.4 kN For stress class C24, the compressive strength parallel to the grain: fc,0,k = 21 N/mm2 E 0.05 7400 = = 352.58 f c,0,k 21
Slenderness ratio
ly=
Le 3000 = = 70.75 rxx 42.4
k ⋅k ⋅k ⋅ f 0.60⋅ 1.0⋅ 1.1⋅ 21 € = 10.66 N /mm 2 Permissible stress in the column f c,0,d = mod c,90 ls c,0,k = gM 1.3 Actual compressive stress €
sc =
P 36⋅ 10 3 = = 1. 6N /mm 2 A 150⋅ 150
Check for buckling strength: €
Compressive stress ( s c ) <
Maximum allowable stress ( kc,y ⋅
1,62 N/m2
0,55．10,66 = 5,90 N/m2
< €
Therefore the column is safe against buckling
€
f c,0,d
)
WORK DIVISION CALCULATIONS
DRAWINGS
CONCEPTION / RESEARCH -Design/structural strategy
Group decision
-Plans & Sections
Alexandra
-Columns
Lachlan
-Aesthetic form
Lachlan, Anton
-Connection details
Anton
-Primars beams
Alexandra
-Siting decision
Alexandra
-Axonometrics
Viridiana
-Secondary beams
Anton
-Choice of Timber
Viridiana
-Balustrade detail
Alexandra
-2nd floor
Lachlan
-2nd floor
Anton, Viridiana