Awning and Canopy

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Both awnings and canopy offer you the chance to create a calm, protected outdoor space to enjoy, whatever the weather. Therefore, finding a professional contractor to install awnings and canopies for your beloved home is very important.

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What is the Difference Between an Awning and a Canopy?

When purchasing an outdoor covering for their property, some people are unclear as to what the difference is between an awning and a canopy. First, let’s outline what’s the same about them.

Awnings and canopies are made from similar materials to each other, that are designed to be durable and weatherproof, but there are several material options to choose from to suit your individual needs.

Despite their similarities, awnings and canopies also have their differences too.


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Awnings are sheets of material attached to the exterior of your home, fixed above a window or a door, and overhanging a porch, patio or walkway. They can offer shade and protection from the sun both outdoors and indoors, blocking the light from entering through windows, thus preventing artwork and furniture from fading, and also keeping it cool indoors.

Unlike canopies, awnings are a permanent fixture to the outside of your home and do not require any disassembly, though this does reduce flexibility. There are two main types of awnings: window awnings, and retractable awnings, which can fold away when they are not needed.

Because of their permanent state, awnings can act as a dry storage space for wet or muddy shoes and bikes; a preferable location over a hallway.

Awnings vary in length, but are are generally smaller than canopies. However, it’s important to choose the right size for your requirements. If you require a space to hold parties and BBQs, then a bigger awning may be better, or possibly a canopy.

A range of colours, patterns and styles are available to suit the architecture of any house. Remember that awnings are a permanent structure and therefore it’s advisable to choose some that complement the current style and decor of your garden, taking into account your garden furniture and window dressings.

Awning are also widely used as a fixture for motorhomes and campervans, offering a relaxed dining area.


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Canopies are often large, free-standing coverings, used to provide shade and protection from the elements for a seated, outside area. They are made from a large canvas of a durable fabric, strung over a metal frame and between supporting posts. However, they can sometimes be placed directly next to your windows and doors, which is perhaps why people sometimes confuse them with awnings.

Canopies tend to be bigger than awnings, making them more suitable for gatherings, parties and BBQs.

They are portable, and are easily disassembled and stored away when not in use. Because they are transportable, they are also ideal for outdoor events, festivals or trips to the park.

Because they are not anchored in one place, canopies offer more flexibility than awnings, allowing you to choose exactly where you want to set up your shaded space.

Usage and Benefits of Awning and Canopy

Awning and Canopy have 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

Our method and technology for awing and canopy service

We offer installation services of awnings for a wide variety of uses.

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    Polycarbonate Awning

    Glass Awning

    Glass Awning

    Aluminium Composite Panel

    Aluminium Composite Panel


    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

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    The Problem

    Design a mounting solution for an awning covering a window in a home.

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    Understand the problem

    Analyse to understand the problem in the best way possible

    Understand the design of a generic awning there are certain elements to the design

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    Extract the Input

    How to extract Input ?

    • Reverse engineering
    • Images

    Reverse engineering -Measurements

    Taking physical measurements of the problem space.

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    Form of Input Data.

    Measurements should be converted into a CAD model

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    • Well defined model is not required
    • Extracting data which is most important.


    • 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

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    Concept of mounting 1

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    Awning is connected to the Wall by a Link with pin joints at either end

    Link is Aluminium channel

    Concept of mounting 2

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    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

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    • 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

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    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

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    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


    • 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.

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    Free body diagram

    Kinematically it is a structure -Triangle

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    Free body diagram –Using Method of sections

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    Resolving the load into vertical and horizontal components

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    Resolving the Fce force into vertical and horizontal components

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    Free Body Diagram

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    Equating Forces

    Equating all Forces in X direction to 0 due to static equilibrium

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    Equating moments about C

    Σ M = 0

    Sign convention for moment

    Clock wise –negative

    Counter clockwise –positive


    + (731.7)(487.7-337) + (164.9)(109.8-76) = 0

    -(Rva)(487.7) -(Rha)(109.8) +115840 = 0  3

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    Finding Fce

    1. Rha–(164.9) +(0.857*Fce)=0  >>  Rha= (164.9) -(0.857*Fce)
    2. va–(731.7)+(0.515*Fce)=0  >>  Rva= (731.7) -(0.515*Fce)
    3. -(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

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    Finding Rhe, Rce

    Rve= 332.2 N

    Rhe= 553.2

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    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

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    Net load = 645.5 N

    Calculation for Compressive stress

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    Hence design is safe in Compression


    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 =?2??/?2

    E –Youngs modulus

    I –Moment of Inertia of section

    L –Length of Column

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    Calculation for buckling stability

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    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

    ???= 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.

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    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.

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    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

    ???= 3.142∗69∗109∗7.253∗10−10 / 0.572

    ???= 1518 N

    Safety factor = ???/ 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

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    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

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    Force acting on the Pin

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    Stress area of pin.

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    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

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    This analysis can be done by FEA analysis of the Awning frame with shown boundary conditions.

    Setting up the Analysis

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    Material –Steel with Yield strength 240 Mpa

    Boundary conditions

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    Load case

    Loading : 1500 N

    Area : Across the top surface of the full frame

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    Max Von mises Stress at centre = 100 MPa

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    Well within the Acceptable limit of yield strength of generic steel



    Max Displacement at centre –2.2 mm

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    1. Formulated the Load on the awning
    2. Used Static analysis to find the Reactions at the Pins
    3. Analyse the Link for compressive strength
    4. Analysed the Link for buckling
    5. Analysed the Pin for Shear failure.
    6. FEA analysis of Awning Frame

    Formulation for Link angle

    Angle of link

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    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

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    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

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    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

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    Formulation –Buckling Critical Load

    ???= 3.142∗69∗109∗7.253∗10−10 / (?2

    ???= 493.4 / (?2)

    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

    ???= ???.? / (?)?

    Shear Stress on Pin = Fce/ 28.26

    Now only with input as angle , all the above can be calculated

    Case studies

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    At θ= 30 deg

    Fce= 500

    Buckling critical load = 518

    L = 0.975 m long

    Not desirable , Link is too long

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    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.

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    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

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    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

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    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.

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