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.

Aluminium Composite Panel

RETRACTABLE FOLDING ARM AWNINGS AND RETRACTABLE PATIO COVER SYSTEMS SIDE BY SIDE COMPARISON

Before and After

Façade of a Building

The Problem

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

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 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

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

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

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

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.