#### Categories

- Waterproofing
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- Concrete Repair
- Wall Crack Repair
- Structural Repairs
- Grouting and Injection
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- Thermal Imaging
- Thermal Insulation
- Awning and Canopy
- Rope Access Works
- Anti Slip
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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**

**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.