Problem statement
Solution video1
Solution video2
DISCUSSION THREAD
Please post questions here on the homework, and take time to answer questions posted by others. You can learn both ways.
HOMEWORK 31.B - SUM 24
28 thoughts on “HOMEWORK 31.B - SUM 24”
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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.
We should just use the parameters as provided.
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.
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.
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.