homework 6 compression answers

ME 270: Basic Mechanics I

Homework 31.b - sum 24.

Problem statement Solution video1 Solution video2

DISCUSSION THREAD

34 thoughts on “HOMEWORK 31.B - SUM 24”

From the picture, d1 diameter is bigger than d2. Do we need to swap the values between d1 and d2? Also, when checking the factory of safety, do we calculate both rods' safety?

I’m assuming that we calculate the factor of safety for both members as long as neither member fails. Also I’m assuming that we just keep the diameters that are described for each member. I don’t think we switch them.

Hey, yeah need to calculate the factor of safety for both members but that can only be done after calculating the stress in each member. Then yes the diameters are the same as given.

We should just use the parameters as provided.

Agreed, images aren't always to scale in these types of problems

What is a factor of safety?

Sigma yp divided by the stress in each member gives the factor of safety I think. A factor of safety is how strong a member in this case needs to be to support its load, and if it is just as strong then its factor of safety is 1.

I googled it and found the factor of safety to be equal to the "Yield Stress / Working Stress" for ductile materials like steel, as per testbook.com . So based on that, in our case I think the factor of safety is the given yield strength divided by the average stress in each member of the rod. I'm also guessing that we don't need to calculate the factor of safety for members that fail.

I think that you are correct, except I would use the highest stress experienced by either member as the denominator instead of the average. This ratio would show how close the overall rod is to yielding/failing which I think makes more sense.

Are there two factors of safety (one for each member)? If it's one for the entire rod, would we use the greatest stress or the average stress?

There are two factors of stress I believe because one could fail while the other might not and they are more or less two rods fixed to each other rather than just one giant rod. I also believe that we would use the instant of greatest stress as in general a material will fail at a specific location, rather than just throughout the entire object.

Yes, I believe you would use the instant of highest stress to get the most accurate number to the failure point.

Yes, I think that there are indeed two factors of safety, which would be one for each member of the rod. Also, I think that it would be based on their respective stresses. I think that since we need a single factor of safety for the entire rod, we should probably then use the one with the greatest stress since I feel this is how it is done in real life to ensure the safest design.

You can use the equation (oYP)/(stress of rod) to determine the value. Then, you can compare it to 1 to determine if it will fail or not.

The Young's modulus given is extraneous information, correct?

I did not use Young's modulus, since we only need to calculate factor of safety.

I also did not use Young's Modulus but it's probably good to know how to solve a problem where it requires using Young's modulus on the final.

Same, I also did not use it, as it asks us for the stress, not strain.

I also did not need to use the given Young's Modulus for this problem.

I realize this might be a very dumb question, but in our free body diagrams for the individual members of the rod, does the normal force always point to the left? Asking because in an example video with a problem similar to this one, the free body diagrams for each member showed the normal force on both members pointing to the left even though the applied forces on both members were opposite to each other.

I pointed the normal force to the right, I think it should still be opposite to the direction of the applied force.

Pointing both ways works, the only difference is when pointing left, if the force is positive, it is a tension, if it is negative, it is a compression. But if pointing right, if the force is positive, it is a compression, if it is negative, it is a tension.

Not a dumb question at all, the direction used and sign of answer can get very confusing. When picking the direction of the normal force, you just need to remember that pointing away from the member assumes tension and pointing into the member assumes compression. This factors into your answer when dealing with the sign of the stress.

What I would do and I've seen a lot of help youtube videos do is always draw the FBD in tension even if you know its compression. You can base all the reaction force directions of of this. Then if you need to solve for the truss and get a negative number you know it is in compression meaning the reaction force is going in the opposite direction you drew it. At least that's how I think about it, I struggled a little bit earlier with FBD conventions but found it was easier to standardize it and work around the negative numbers.

Just to make sure, the force of 2P is acting on rod 1 right? Not rod 2

That’s the way I did it.

Yeah, I did this way and it should have worked.

They are calling sigma yp the yield of strength on the homework. I assume they refer to the yield of stress since the strain is usually epsilon, but I am not completely sure.

Yes, I agree that sigma yp is the yield strength and epsilon refers to strain. I plan on using sigma yp to determine if the rod will fail and if not calculate the safety factor.

Even though the force of P is applied on member 2, it still should show up in free body diagram for member 1, correct? This way I got a force of P pointing left and a net force of P pointing inwards from the right side.

If the factor of safety is equal to or above 1 can we assume that the member would not fail or is there any other way to determine if a member would fail or not?

Yea, you can assume it will not fail as long as the factor is above 1.

If the factor of Safety is more than 1, then it does not fail. In this homework, both were more than 1 and that's why nothing failed.

I'm under the impression that the factor of safety is determined for each member individually, provided that both remain intact. Additionally, I believe we retain the specified diameters for each member as outlined, without interchanging them.

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Compression Q&A: Answers to Common Questions

• Thread starter hofner
• Start date Oct 31, 2021
• Tags Compression
• Oct 31, 2021

• Physicists report new insights into exotic particles key to magnetism
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A PF Universe

Welcome to PF. Can you post a better image of your work? And if you could type your math equations into the Edit window (preferably using LaTeX), that would be a big help. I'm not able to read your attachment so far...  

That helps some, thanks. Can you post the figure that was given with the problem? It seems like it's two inclined beams at different angles supporting a pail of water in the middle? And as I mentioned, it is a lot easier for us to help if you post your work using LaTeX. Check out the "LaTeX Guide" link below the Edit window.  

I hope that helps! Thank again  

Wow! Ok thanks for that!  

You're welcome. The weird thing is that my LaTeX is not rendering correctly for me (after I told you to use it, sheesh). I'll keep trying to figure out why, but hopefully you can decode what I posted.  

Lol. We shall try thanks again.  

A PF Multiverse

The problem is a little easier if you solve it in the other order. The vertical component of the compression is obvious, so you can find the horizontal component by taking moments about the end point of a beam; no trig required. Then you just need Pythagoras to find the total compressive force. In my experience, the great majority of mechanics problems in which angles are implied by lengths can be solved without ever calculating the angles. You usually just need the trig functions of the angles, but here you do not even need those.  

Related to Compression Q&A: Answers to Common Questions

1. what is compression and why is it important.

Compression is the process of reducing the size of a file or data without losing any important information. It is important because it allows for more efficient storage and transmission of data, saving time and resources.

2. How does compression work?

Compression works by using algorithms to identify and remove redundant or unnecessary data from a file. This can include repeating patterns, empty spaces, or unused characters. The remaining data is then encoded in a more efficient way.

3. What are the different types of compression?

There are two main types of compression: lossless and lossy. Lossless compression preserves all data from the original file, while lossy compression sacrifices some data in order to achieve a higher level of compression. Examples of lossless compression include ZIP and PNG, while MP3 and JPEG are examples of lossy compression.

4. What are the benefits of using compression?

The main benefit of using compression is that it reduces the size of files, making them easier and faster to transfer and store. This is especially useful for large files, such as videos or images. Compression also helps to save storage space and can improve the overall performance of a system.

5. Are there any downsides to compression?

While compression offers many benefits, there are some downsides to consider. Lossy compression can result in a lower quality of the compressed file compared to the original. Additionally, compressing and decompressing files can take up processing power and time. It is also important to note that some file types, such as already compressed files, may not benefit from further compression.

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