Projected Loads and Snow Loads - Intro to Structural Analysis
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- čas přidán 7. 08. 2024
- This video defines projects loads and presents an example problem using the most common type of projected load, snow loading on a roof truss.
In this example, we will:
1) Compute the reactions forces for combined dead plus snow loading
2) Convert projected loads to distributed loads in the axial and shear directions
3) Compute the axial, shear, and moment diagrams.
For a projected load S, with an angle A between the loading plane and the element to which the load is applied, there are a few shortcut equations:
~ Distributed Shear Load = S*cos(A)*cos(A)
~ Distributed Axial Load = S*cos(A)*sin(A)
Try these out, and you'll see they match exactly the results shown in the video! In this example, cos(A) = 12/13 and sin(A) = 5/13. - Věda a technologie
Professor, thanks for the video. Was hoping you could address a question if time allows. Per ASCE, for roof exposure factor Ce, having a hard time understanding why a fully exposed roof has a reduced factor VS a sheltered roof. Wouldn't a fully exposed roof see the more snow load when compared to the sheltered roof?
You are a blessing sir thank you
Thank you sir❤️
@StructuresProfH
How would I calculate the dead and dynamic load capacity on a ridge beam for the purpose of hanging substantial weight from it?
dumb question, but around minute 7:48 dividing the forces into their components, f/13 and 12/13, can this be explained more? That was the issue I was having was projecting the loads and snow loads onto a beam. Thank you very much for the video.
Very well explained thank you, I have a question tho, if we had the wind load as well which would be perpendicular to the roof how could we calculate it as well?
The wind load, applied perpendicular to the surface of the roof, does not need to be converted like a projected load. It is already applied per unit surface area (not projected area), and it is already in the "shear" direction with no "axial" direction component. So no conversion necessary.
@@StructuresProfH how we use the perpendicular wind load in the load combinations?
The reality is that snow does not fall straight down and with a symmetric roof will build up more on one side than on the other. Also with asphalt shingles, as I know all too well in New England, heavy, wet snow does not slide off the roof by itself
All very true. Chapter 7: Snow Loads in ASCE/SEI 7 considers these factors. In particular, section 7.4 describes the sloped roof snow load - the "slope factor" is often 1, particularly for rough surfaces like asphalt shingles, meaning snow is generally not sliding off that roof. Section 7.6 describes unbalanced load cases where more snow accumulates on one side versus the other.
Regardless, the design snow load (whether balanced, unbalanced, or reduced due to sliding) is always intended to be applied as a projected load acting vertically. This video was just meant to introduce the concept of how to handle that projected load to find shear and moment diagrams. I think a follow-up video on how to compute snow loads in more detail would be a good idea though!
So the Ax and Cx reactions would dictate the required thrust force to be resisted. Would you do an example where a collar tie is installed and no ceiling tie as in many wood framed homes?
Yes, the Ax and Cx reactions are the thrust forces. Normally these could be resisted by a ceiling or rafter tie. A collar tie near the peak of roof does not effectively reduce these thrust forces. I believe the primary purpose for collar ties is to better secure the rafters to the ridge beam.
Anyway, if you don't have a ceiling tie, that thrust force needs to be resisted by the walls that support the roof. The other option is to have a structural ridge beam. In structural analysis terms, that means adding a third vertical support at the roof peak - this effectively eliminates the horizontal thrust forces because the truss is no longer being "squashed" out by the vertical load.
Thanks for a well explained video. In spite of that, I do have a couple of questions. I have learned that a truss that is statically indeterminate can't be solved. Is that just for internal forces(you obviously did solve it). secondly could you make a video on how to use these figures to design a wooden truss? How many members and what size they need to be.
G,day from Sydney Australia.
Once you have calculated the forces applied in compression and tension, what is the easiest way to select materials that can resist that load per foot: manufacturers specifications or actually testing?
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That's a MUCH bigger question than can be possibly tackled in a CZcams comment. First, design can't be done just knowing the load per foot - we need to know the flexural moment, shear, and axial force (tension or compression) demands in the members we are designing (and torsion if it comes up), plus we need to make sure it works for other concerns like serviceability (limiting deflections or vibrations, etc.). These will all depend on the span lengths, how things are connected together, and so on. Different materials will have different design standards, but in general there are accepted standards or specifications that will describe how to compute the nominal design capacities.
To get back to your question, for pre-fabricated systems, it's likely that the manufacturer has already done a lot of this for you. They may even present the results as a nice "load per foot" value, but there is a lot of background to those numbers. Actual testing is typically only done for unique circumstances that might be pushing the envelope in some way, or if the project budget and stakes are really high and can afford some proof-of-concept tests. For most typical applications, the specifications (as long as they are properly checked and used in the proper context) are sufficient.
@@StructuresProfH Thank you for your time and explanation: assessment of materials used in building construction, snow covered. I only have experience selecting pre-determined stress graded timber and steel. This was in residential building surveying qualifications.
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Thanks for the video. Is it correct to state that the only reason the reactions have an X component is due to the hinge at the peak of the roof?
The hinge certainly doesn't help, but you will still get horizontal (X-component) reactions even without the hinge. As the load presses down on the structure, it tends to "flatten out" which will be resisted as horizontal reactions. The hinge means it tends to "flatten" more under vertical load than an equivalent system without a hinge, but the phenomena is present for either case.
Why is it then with truss analysis vertical loads never transfer lateral force to the roller/pin? Due to a 3 force member in equilibrium? And in the roof case, the bottom chord would resist the tension?
@@WG-ft6tz For truss analysis, we typically assume that one support is a pin and the other is a roller (meaning, it allows for horizontal movement). In this configuration, any horizontal force is resisted by tension in the bottom chord. However, if both supports are pins (meaning both resist horizontal motion), then you will once again get horizontal reactions even if you have a bottom chord. This is because the deformation of the truss will push these two supports outward, which is resisted by inward horizontal reactions.
Ah thank you! So used to analyzing pin/roller trusses in examples. Subscribed!
How do you calculate the dead load.
In this particular example, I just made up convenient numbers for the loads. In reality, you would need to know the weights of your materials for your roof/floor/whatever you are designing. You can find typical material densities online, but the Chapter C3 commentary of ASCE/SEI 7 (Minimum Design Loads for Buildings and Other Structures) provides a lot of useful values for structural engineers.