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Structures load calculations   shelters
Structures load calculations   shelters
Structures load calculations   shelters
Structures load calculations   shelters
Structures load calculations   shelters
Structures load calculations   shelters
Structures load calculations   shelters
Structures load calculations   shelters
Structures load calculations   shelters
Structures load calculations   shelters
Structures load calculations   shelters
Structures load calculations   shelters
Structures load calculations   shelters
Structures load calculations   shelters
Structures load calculations   shelters
Structures load calculations   shelters
Structures load calculations   shelters
Structures load calculations   shelters
Structures load calculations   shelters
Structures load calculations   shelters
Structures load calculations   shelters
Structures load calculations   shelters
Structures load calculations   shelters
Structures load calculations   shelters
Structures load calculations   shelters
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Structures load calculations shelters

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This document briefs on calculating loads and forces applied on structures

This document briefs on calculating loads and forces applied on structures

Published in: Education, Business, Technology
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  • Tell the students a story: Imagine you are a workshop owner. A customer comes to you asking for a tractor shed (or shelves, or trailer…), and says that he has been shopping around local businesses as he has not much money and needs a quality shed for a low price. You have never built a tractor shed before, so cannot simply give him a quote based on experience. However, if you understand forces, you can work out the minimum amount of material required to make the tractor shed stand up. If you can save on material, you spend less on manufacture and can provide a quality product for lower cost. This customer will continue to use your services above those who offer heavier, more costly designs. Gandhi once said that the most important person in any business was the customer. To satisfy this customers needs, you need to be confident at working out how to build cheap, quality designs. It is worth the effort, as this will make your business preferable to competing businesses, and make the business more prosperous.
  • The first step in working out the minimum materials a structure needs is to look at the overall loads on the building. Ask the students what kind of loads they might expect on a building. After a time, show them the slide in full, explaining the difference between dead and dynamic loading. Dead load remains still. Examples include self weight, objects on a shelf, permanent furniture in a building etc. Dynamic loads include wind and rain, people walking on the structure, objects being loaded onto and off the structure. Dynamic loads often have a much more significant effect than dead load.
  • Tell the students that they are going to pretend they are the workshop owner attempting to produce a quote for the man requesting a tractor shed. The dimensions are as above (these are obviously quite large, but it makes for a more interesting exercise).
  • The important features are the trusses, the columns and the foundations, as these are the sections which will carry the load. Work top to bottom when working out the weight that each section must carry. The trusses must carry the weight of the roof plus their own weight. The columns must carry the weight of the roof, the weight of the trusses, and their own weight. The foundations must carry the weight of the roof, trusses, columns and their own weight.
  • To start, it is important to look at the self weight each part of the structure need to carry. Ask the students to identify the dead load that the trusses must carry. Talk through the calculation, asking the students for answers all along. It is the aim that by the end of the course students should be able to understand how to create such a calculation, rather than merely compute the numbers. Give the students space to try and work out what steps are needed to work out the load, but a lot of coaxing may be required at this early stage.
  • Calculation continued.
  • Calculation completed.
  • Ask the students to identify what loads need to be considered when working out the load on the columns. Work out the load on a column using the combined weight of the trusses, roof and columns.
  • Show the students the video of the tornado. Point out that the building fails by the roof being ripped off like the lid from a tin can. Wind on buildings can cause either a downwards pressure on the roof, or a suction which can rip the roof off a building in strong winds, depending on the direction of the wind. In a tornado, suction is cause by low air pressure towards the centre of the tornado. The low pressure is caused by the wind rotating in a circle, the air flowing along curved paths. Pressure drops towards the centre of the circle.
  • The diagram illustrates how wind follows a curved path as it flows over a building. The effect is the same as in a tornado; towards the centre of curvature, the pressure drops. The wind causes high pressure on the wall it hits, and low pressure on the wall behind the building.
  • The greater the curvature, the lower the pressure and the greater the suction. The less streamlined the building in the more suction the wind will cause on the roof. As can be seen a square building experiences large forces, a pitched roof experiences less suction, and a domed roof experiences least force. Dome shapes have started being used in hurricane prone regions of America.
  • Ask the students what factors may affect wind load. Factors include; size of building, height of building, shape of building, area of roof, region (some regions are more prone to wind than others), landscape (wind accelerates over hills, so a building on a hill will experience higher loads than those in valleys).
  • This is a chart of the maximum expected windspeed in regions across India. You simply look where you are on the map, and use the key to find the maximum expected windspeed. Ask the students why maximum windspeed is used, not average windspeed. This is because you always want to design for worst case scenario. You want a building which can survive stormy nights and monsoons, not just everyday weather. You may get bad weather anytime, and you do not want your building to fail in the first gale.
  • Safety factors are used to make the wind speed bigger or smaller depending on whether the wind is likely to be bigger or smaller at your specific location. Safety factors are based on worst case scenarios, and are useful for ensuring your building is strong enough without having to do long and difficult calculations. In the diagram, watch how the windspeed changes after it blows in from the coast. If you are in a wooded area, the windspeed will drop. On top of a hill, it will be very windy. If sheltered in a valley the windspeed will be low again.
  • Wind pressure is the force per square metre on a surface. It is calculated using the formula above. Once you know the wind pressure, you can work out the force on any particular surface of the building.
  • The wind pressure can be used using the formula above to work out the wind force on any surface of the building. There is another safety factor that needs considering. Wind blowing through the building can cause internal pressure changes. If the pressure outside drops, and the pressure inside rises, then the overall pressure change on a surface is larger. A safety factor of 2 is required to account for this.
  • Get the students to work out the forces on the roof of their designs!
  • Transcript

    • 1. Vigyan Ashram, Pabal
    • 2. Why should you know how to design aWhy should you know how to design a structure?structure?
    • 3. Step 1 – What is the load on the structure?Step 1 – What is the load on the structure? 1) Dead Load  Self weight  Static loads 2) Dynamic Load  Wind  Rain  People  Loading/Unloading
    • 4. Tractor ShedTractor Shed 10m 4m 9m 5m Truss Column Foundations
    • 5. Dead LoadDead Load Self weight Golden Rule: Work top to bottom 1. Truss 2. Column 3. Foundations
    • 6. 1. Load on Truss1. Load on Truss 10m 4m Steel Sheet – 10kg/m2 Mass 1 Truss – 100kg Area of roof (m2 ) = x x10 m 4 m 2 = 80 m2 Mass of roof (kg) = x Weight of roof (N) = x = = 80 m2 10 kg/m2 800 kg 800 kg 10 8000 N
    • 7. Weight of 1 truss (N) = x 1. Load on Truss1. Load on Truss 10m 4m Steel Sheet – 1kg/m2 Mass 1 Truss – 100kg 100 kg 10 8000N = 1000 N
    • 8. Total weight on trusses (N) = + 1. Load on Truss1. Load on Truss 10m 4m Steel Sheet – 1kg/m2 Mass 1 Truss – 100kg 3000 N 8000N = 11,000 N 1000 N 1000 N 1000 N 8000 N
    • 9. ActivityActivity How much roof weight must your structure support? Plastic Sheet – 4.5 kg/m2 Mass 2’ PVC Pipe – 0.75 kg/m Mass bamboo 1’ – 0.2 kg/m Mass steel pipe 0.5’ – 0.36 kg/m Mass steel pipe 0.75’ – 0.5 kg/m Mass steel pipe 1’ – 0.75 kg/m Area of roof (m2 ) = x x Mass of roof (kg) = x Weight of roof (N) = x 10 m 4 m 2 = = = 80 m2 80 m2 10 kg/m2 800 kg 800 kg 10 8000 N Weight of 1 truss (N) = x100 kg 10 = 1000 N Total weight on trusses (N) = + 3000 N = 11,000 N 8000 N
    • 10. 2. Load on Column. Load on Column 5m Mass 1 Column – 10 kg/m 8000N 1000 N 1000 N 1000 N Mass 1 pole = x Weight 1 pole = x = = 50 kg 5m 10kg/m 50kg 500 N 10
    • 11. 2. Load on Column. Load on Column 5m Mass 1 Column – 10 kg/m 11,000N 500 N 500 N 500 N 500 N 500 N 500 N Force on each column = +11,000 N 3000 N = 14,000 N
    • 12. Try it youself!Try it youself! How much weight does each lower support hold? Mass 1 pole = x Weight 1 pole = x = = 50 kg 5m 10kg/m 50kg 500 N 10 Force on columns = +11,000 N 3000 N = 14,000 N Mass 2’ PVC Pipe – 0.75 kg/m Mass bamboo 1’ – 0.2 kg/m Mass steel pipe 0.5’ – 0.36 kg/m Mass steel pipe 0.75’ – 0.5 kg/m Mass steel pipe 1’ – 0.75 kg/m
    • 13. Weight of Foundations = x 3. Load on Foundations. Load on Foundations 14,000N Force on Foundations = + 2000kg 10 = 34,000N Mass Foundations – 2000kg 20,000N 14,000N 20,000N =
    • 14. Try it yourself:Try it yourself: How much weight do the bottom supports hold? Mass 2’ PVC Pipe – 0.75 kg/m Mass bamboo 1’ – 0.2 kg/m Mass steel pipe 0.5’ – 0.36 kg/m Mass steel pipe 0.75’ – 0.5 kg/m Mass steel pipe 1’ – 0.75 kg/m Force on Foundations = +11,000 N 20,000 N = 34,000N
    • 15. Extreme WindExtreme Wind normal air pressure low air pressure Tornado
    • 16. Windflow over buildingsWindflow over buildings high pressure low pressure low pressure
    • 17. So how do I calculate wind load?So how do I calculate wind load? What things do you think affect wind load?
    • 18. Zone Windspeed (m/s) 55 50 47 44 39 33 Step 1 – Highest WindspeedStep 1 – Highest Windspeed Pabal – 39 m/s
    • 19. Wind speed is affected by local terrain If you are on a hill, increase the wind speed by multiplying by 1.1 Safety FactorSafety Factor Design wind speed = 1.1 x 39 m/s = 43 m/s
    • 20. Pressure is the force per square metre on the building Wind PressureWind Pressure Pressure (N/m2 ) = 0.6 x windspeed2 = x x430.6 1109 N/m2= 43 1m 1m
    • 21. Wind force is the specific force on a particular part of the building Safety Factor: account for internal pressure changes with a safety factor of 2 Wind ForceWind Force Force (N) = Pressure (N/m2 ) x Area (m2 ) x Safety Factor
    • 22. Wind ForceWind Force Force (N) = Pressure (N/m2 ) x Area (m2 ) x Safety Factor Force On Wall (N) = x x Area Of Wall (m2 ) = x = = 9m 5m 45 m2 21109 N/m2 9,9810 N 9m 5m Wind Pressure = 1325 N/m2 Safety Factor = 2 45 m2 Wind Force (N)
    • 23. Wind ForceWind Force Force (N) = Pressure (N/m2 ) x Area (m2 ) x Safety Factor Force On Roof (N) = x x Area Of Roof (m2 ) = x = = 10m 4m 40 m2 21109 N/m2 8,8720 N 10m 4mWind Pressure = 1325 N/m2 Safety Factor = 2 Wind Force (N) 40 m2
    • 24. Activity – Own designs!Activity – Own designs! Force (N) = Pressure (N/m2 ) x Area (m2 ) x Safety Factor 1 2 Safety Factor of 1.2 Pressure (N/m2 ) = 0.6 x windspeed2 3 4

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