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Awnings Singapore
Le Fong offers installation services for awning, canopies and metal roofs for your everyday use. Each of them are used for different purposes and we can recommend a suitable kind of overhead covers if you let us know what functions your overhead cover should do.
What is an Awning?
Awnings are fixed structures that are either mounted above an outside window or patio to prevent harsh light and shadows from entering the house. The main functions are to keep the temperature low in the home and to protect furniture from sun damage. Retractable awnings mounted on patios are used to keep the pouch cool and comfortable even in the heat.
Usage and Benefits of an Awning
An awning has a relatively fixed usage. It is commonly used for:
- preventing harsh light and shadow from entering the house
- keeping the temperature of the house low
- protecting furniture from sun damage
- keeping a pouch cool and comfortable
Le Fong offers the following types of Awning services:
- Polycarbonate Awning
- Glass Awning
- Aluminium Composite Panel
We offer installation services of awnings for a wide variety of uses.
Polycarbonate Awning
Glass Awning
Aluminium Composite Panel
Aluminium Composite Panel
RETRACTABLE FOLDING ARM AWNINGS AND RETRACTABLE PATIO COVER SYSTEMS SIDE BY SIDE COMPARISON
Category | RETRACTABLE FOLDING ARM AWNINGS | RETRACTABLE PATIO COVER SYSTEMS |
Maximum width | Maximum width for a single unit with one motor and one piece of fabric 52 1/2 feet (1600cm) in one piece (Palermo Plus model) | Maximum width with using one piece of fabric and one motor 42’7″ (1300cm) for all models |
Adjacent units | Multiple units can be butted together with a 5″-6″ (12.7cm – 15.24cm) gap between units | Multiple systems can be butted together and caulked with no gap so no rain or snow enters between systems |
Maximum projection | Maximum projection 16 foot (Roma model). Largest in the world | Maximum projection 29’6″ (Siracusa, Sassari and Rimini models) largest in the world |
Wind load | Maximum wind load tested to Beaufort scale 5 (Palermo/Palermo Plus/Bologna models) | Maximum wind load tested (Beaufort scale 10). Most models |
Drainage | Front water drains off front bar and can splash on hard surface | Front water drains to front and then down through internal, invisible gutters and downspouts (most models) OR off the side(s) of the fabric using single sided or alternating sides |
Rainfall | For LIGHT rainfall only – less than .30″ (7.6mm) per hour per the American Metrorological Society | For HEAVY rainfall – more than .30″ (7.6mm) per hour per the American Meteorological Society |
Water | Fabrics are water RESISTANT and include Para, Sunbrella, Dickson & Sattler | Fabrics are water PROOF (Ferrari 502 & Ferrari 602) or water RESISTANT (Para only) |
Hail & Snow | Not hail or snow load approved | Tested for hail. Not snow load approved |
Powder coating | Qualital® powder coated frames & hood available in white, ivory, sand and brown with sand and brown having a $200 surcharge | Qualital® powder coated frames available in all RAL colors with each model available in a standard color or colors and other RAL colors available for a surcharge |
Fire retardancy | Non fire retardant fabrics are standard with fire retardant fabrics available for a surcharge | Fire retardant fabric is standard on all models |
Posts, Arms and Guides | Spring loaded arms with no posts | Belt driven system with posts (no posts on the Rimini and Firenze models) and guides |
Enclosed | Sides and front of lateral folding arm awnings cannot be enclosed as doing so would damage the system when opened and closed | Sides and front of retractable patio cover systems can be completely enclosed so the area can be air conditioned in the summer and heated in the winter and also allowing for a bug and pollen free area |
Drainage options | Front water drainage only | Front or single sided or alternating side water drainage available |
Free standing or attached | Cannot be free standing unless attached to a custom made structure | Any model can be free standing. Standard free standing models are the Forli, Trento and Vicenza |
Model names | Models include Genova, Palermo, Palermo PLUS, Bologna, Venezia and Roma | System examples include Ferrari, Firenze, Firenze PLUS, Forli, Monza, Monza PLUS, Ravenna, Rimini, Sassari, Sassari, Salerno, Siracusa, Trento and Vicenza models |
Before and After
Façade of a Building
The Problem
Design a mounting solution for an awning covering a window in a home.
Understand the problem
What is an awning ?
An awning or overhang is a covering attached to the exterior wall of a building to provide shelter from the elements like Rain , sunshine.
Example !
Analyse to understand the problem in the best way possible
Understand the design of a generic awning there are certain elements to the design
Extract the Input
How to extract Input ?
- Reverse engineering
- Images
Reverse engineering -Measurements
Taking physical measurements of the problem space.
Form of Input Data.
Measurements should be converted into a CAD model
- Well defined model is not required
- Extracting data which is most important.
Requirements:
- Functional
- Structural
- Assembly
- Maintenance
- Environmental
- Aesthetic
- Safety
- Reliability
- Ergonomics
Structural requirements:
- To sustain the loads on awning from wind and self-weight and impact loads due to debris falling on awning.
- To maintain stability of structure.
Environmental requirements:
To withstand environmental conditions of rain, heat and cold without deterioration in material property.
Assembly requirements
An assembly mechanism
- to mount the frame of awning to the wall
- Be able to dismantle the assembly if required
Aesthetic requirements
To maintain the look of the façade of the building and not look out of place..
Concept development:
Now that the requirements are understood. Next step is to generate concepts.
How to Generate Concepts ?
- Learning From Research and Gather Ideas
- Brainstorm and Ideate
- Synthesis of Ideas into concepts
- Should meet requirements
Awning -Concept
Concept of mounting 1
Awning is connected to the Wall by a Link with pin joints at either end
Link is Aluminium channel
Concept of mounting 2
Awning is connected to the Wall rigidly through the Red frame on either side of Awning
Bracket is Wooden
Comparison of Concepts
Concept 1 | Concept 2 |
Uses a link to connect the Awning to the wall | Uses a structural bracket which connects the awning to the wall |
Uses Pin joints -temporary | Uses Permanent joints |
Easier to dismantle | Cannot be dismantled |
Lighter | Heavier |
Additional brackets for mounting pins | No additional brackets for pin mountings |
Inferior in Aesthetics | Superior in Aesthetics |
Can take lesser loads | Can take larger loads |
Cheaper | Costlier |
Concept Selection
- Concept 1 is inferior to concept 2 in aesthetics but superior in assembly requirements.
- Concept 2 is stronger but that extra strength may not be required for the conditions.
- Concept 1 is cheaper and light weight.
Selecting a concept and the reasoning behind it ?
- Ease of installation
- Simple and light
- Cheaper
Although in this case Concept 1 has been considered superior but this might not be the case in all other instances of design
Engineering Analysis
Analysis of the Design with respect to Engineering principles
1.Engineering mechanics
2.Strength of materials
3.Machine design
Converting the design problem to an Engineering problem
Formulating the engineering problem
Goal
- Get the reaction forces
- Use reaction forces to calculate stress in the member
Arriving at the Design load
- Wind loads
- Self weight
Wind Load direction
Wind velocity
Wind pressure calculations
150 km / h =42 m / s
Generic formula for wind pressure , P = 0.613 V^2
Source : Wikihow
Considering the area of application faces a year round probability of storms then the maximum velocity of wind
For a sever storm is maximum 150 km/h
Calculated Pressure (N/m^2) –
= 0.613 * 42^2
= 1081 N/m^2
Calculate the Load from pressure
Projected area of the Awning is 1 metre X 0.7 metre = 0.7 m^2
Load acting = Pressure X Area
Load = 1081 * 0.7 = 756.7 N or 76 kg
Self Weight calculation
Self weight , Sheet = 11 kg
Frame = 10 kg
Total Design load = Wind load + Self weight load
= 76 +11+10 = 100 kg
Assumed :
Although the Weight will act vertically downwards, the Load due to self weight is a fraction of the wind load.
Considering Overload factor as 1.5
Cause of Overloads?
- Debris falling onto the awning
- Extreme gale force winds
- Material variations
- Dimensional variations
Load to be considered = 150 kg = 1500 N
Static Force and Moment analysis
Procedure
- Find the Force and Moment equilibrium equations
- Solve to Find the Unknowns using Simultaneous equations method.
Free body diagram for system
Load acting is 1500/ 2 (at one side) at centre of the link.
Free body diagram
Kinematically it is a structure -Triangle
Geometry
Free body diagram –Using Method of sections
Resolving the load into vertical and horizontal components
Resolving the Fce force into vertical and horizontal components
Free Body Diagram
Equating Forces
Equating all Forces in X direction to 0 due to static equilibrium
Dimension
Equating moments about C
Σ M = 0
Sign convention for moment
Clock wise –negative
Counter clockwise –positive
-(Rva)(487.7)-(Rha)(109.8)
+ (731.7)(487.7-337) + (164.9)(109.8-76) = 0
-(Rva)(487.7) -(Rha)(109.8) +115840 = 0 3
Finding Fce
- Rha–(164.9) +(0.857*Fce)=0 >> Rha= (164.9) -(0.857*Fce)
- va–(731.7)+(0.515*Fce)=0 >> Rva= (731.7) -(0.515*Fce)
- -(Rva)(487.7) -(Rha)(109.8 +115840 = 0
Substituting Rhaand Rvain equation 3 to find Fce
-((731.7 -0.515*Fce)(487.7)) –((164.9 -0.857*Fce)(109.8)) +115840 = 0
-356850 + 251.2 Fce –18106 + 94.1Fce + 115840 = 0
-222904 + 345.3 Fce = 0
Fce = 222904 / 345.3
Fce = 645.5 N
Finding Rva, Rha
Rha= (164.9) -(0.857 * 645.5) Rha= -388.29 N (opposite direction to what was assumed)
Rva= (731.7) -(0.515 * 645.5 Rva= 399.3 N
Final FBD of Link AD
Link AD
Finding Rhe, Rce
Rve= 332.2 N
Rhe= 553.2
Axial Force on link
From the reactions already calculated find the component of force in the axis of the link
This will be an input to calculate the strength of the link.
Calculation for Compression loading
Fce = 645.5 N
Next step is to calculate the Compressive stress and Buckling stability and design the section of the link and see if it exceeds the Tensile strength of the material
Net load = 645.5 N
Calculation for Compressive stress
Hence design is safe in Compression
Buckling
What needs to be calculated?
Whether the load which is acting on the link exceed the Critical load which would buckle the link?
Formula :
Critical Load =𝜋2𝐸𝐼/𝐿2
E –Youngs modulus
I –Moment of Inertia of section
L –Length of Column
Calculation for buckling stability
Inputs :
1.E -Elastic modulus –Aluminium -69 GPa
2.I –Moment of Inertia – I = 1.232e-008 m^4
3.L –Length of Link –0.57 m
Calculating using formula
𝑃𝑐𝑟= 3.142∗69∗109∗1.232∗10−8/0.572 = 25 kN
Critical Load is much larger than 645.5 N , hence the link is safe from Buckling.
The Link is Strong in buckling and compression with very large margins.
This means that the section of the link can be reduced to reduce weight and maintain strength and stability.
Re-calculating Compressive stress
Stress = Load / Area
Stress = 645.5 N / 6.4e-005 m^2
Compressive Stress = 10 Mpa
Re-calculating Buckling critical Load
𝑃𝑐𝑟= 3.142∗69∗109∗7.253∗10−10 / 0.572
𝑃𝑐𝑟= 1518 N
Safety factor = 𝑃𝑐𝑟/ Load = 1518/654.4 = 2.3
Design of Pin -Shearing vs Bending
Mode of failure of joint:
- Shearing off ,
- bending
Which is more likely to happen?
In this case it would be shearing of as the length of the pin is not long
Design of pin joint for strength
Probable failure mode of Pin :
Failure due to Shear failure
The Pin has to be designed keeping in mind this Shearing action
Force acting on the Pin
Stress area of pin.
Shear Stress calculation
Stress = Force / Area
Stress = Force acting on the Pin / Shear Area
The Pin diameter is 6 mm
Area = 28.26 mm^2
Force = 654.5
Stress = 654.5 / 28.26 = 25 N/mm^2 or MPa
Steel Shear strength = 240 N/mm^2 (MPa)
Design is safe in Pin Shear
FEA Analysis of Awning
This analysis can be done by FEA analysis of the Awning frame with shown boundary conditions.
Setting up the Analysis
Material –Steel with Yield strength 240 Mpa
Boundary conditions
Load case
Loading : 1500 N
Area : Across the top surface of the full frame
Results
Stress:
Max Von mises Stress at centre = 100 MPa
Well within the Acceptable limit of yield strength of generic steel
Results
Displacement:
Max Displacement at centre –2.2 mm
Summary
- Formulated the Load on the awning
- Used Static analysis to find the Reactions at the Pins
- Analyse the Link for compressive strength
- Analysed the Link for buckling
- Analysed the Pin for Shear failure.
- FEA analysis of Awning Frame
Formulation for Link angle
Angle of link
The same equations can be used to generate simple FORMULATION to find the optimal link length and inclination considering the Awning remains the same
Resolving the Fce force into vertical and horizontal components
Equating Forces
Equating all Forces in X direction to 0 due to static equilibrium
Σ Fx= 0
Rha= (164.9) -(Fce.sinθ) A
Equating all Forces in Y direction to 0 due to static equilibrium
Σ Fy= 0
Rva= (731.7) -(Fce.cosθ ) B
Rewriting Equation 3 from earlier section
-(Rva)(487.7) -(Rha)(109.8) +115840 = 0 3
Substituting A and B in equation 3
-((731.7 -Fce.cosθ )(487.7)) –((164.9 -Fce.sinθ)(109.8)) +115840 = 0
-(356850-487.7. Fce.cosθ) –(18106 -Fce.sinθ.109.8) +115840 = 0
-259116+487.7. Fce.cosθ+Fce.sinθ.109.8 = 0
487.7. Fce.cosθ+Fce.sinθ.109.8 = 259116
Fce(487.7.cosθ+109.8.sinθ)= 259116
Fce= 259116 / ((487.7.cosθ+109.8.sinθ))
Formulation –Length
Formulation –Buckling Critical Load
𝑃𝑐𝑟= 3.142∗69∗109∗7.253∗10−10 / (𝐿2
𝑃𝑐𝑟= 493.4 / (𝐿2)
Formulation –All equations
Fce= 259116 / ((487.7.cosθ+109.8.sinθ))
Rha= (164.9) -(Fce.sinθ)
Rva= (731.7) -(Fce.cosθ )
L= 487.7/ sinθ
𝑷𝒄𝒓= 𝟒𝟗𝟑.𝟒 / (𝑳)𝟐
Shear Stress on Pin = Fce/ 28.26
Now only with input as angle , all the above can be calculated
Case studies
At θ= 30 deg
Fce= 500
Buckling critical load = 518
L = 0.975 m long
Not desirable , Link is too long
At θ= 70 deg
Fce= 750
Buckling critical load = 1800
L = 0.519 m long
Shear Stress on Pin = Fce/ 28.26
= 26 MPa
Desirable , Link is short.
At θ= 90 deg
Fce= 1355.3
Buckling critical load = 2074.5
L = 0.487 m long
Shear Stress on Pin = Fce/ 28.26
= 48 MPa
Link is Shortest
Detail Design
- This is the stage when the Design is refined and specifications are laid out.
- Preparation of Bill of Materials
- Each and every dimensions of each part is finalized at this point.
Bill of materials
Next Steps :
- Preparation of Drawings for the Parts for Fabrication and manufacturing
- Refining the Details to suit manufacturing requirements
- Tweaking Dimensions for Fit and clearances
- Dimensional Engineering –GD& T
Procedure to find reactions -Solving Link AD
Reactions to be found Rha, Rva, Rhc, Rvc,
1.Force equilibrium of Force in X and Y
1.Output two equations in terms of Rha, Rva, Rhc, Rvc
2.Moment Equilibrium of Force about Point D
1.Output one equation in terms of Rha, Rva, Rhc, Rvc
3.Moment equilibrium of Force about Point B (load point )
1.Output one equation in terms of Rha, Rva, Rhc, Rvc
4.Replace values of Rha and Rhc in equations of moment balance
5.Arrive at two equations containing Rva and Rvc
6.Solve to find the remaining terms.