Confused on this deformation problem (statics)

[Patdon10]

Homework Statement

2. The attempt at a solution

Ok. so I'm a little bit confused as to how to approach this problem. I would know how to do this problem if it was laying right to left, but because it's vertical, that makes it a lot more confusing for me.

I know A = (1/4)(pi)(d^2) = 7.0686*10^-4 m^2

I can think of two ways to do this problem, and I need to know which approach is the right one. The first one is straight forward...

I know the combined forces of 10kN + 20kN = 30 kN tensile stress going into the upper wall and out the bottom of the bar at C.

tensile stress = P/A ----> 30kN / (7.0686*10^-4 m^2) = 43441.22 kPa

Using hooke's law: deformation = PL/AE
= (30kN) (3.5m)/ (7.0686*10^-4) (70 * 10^6 kN/m^2) = 0.002122m

is that right? I would be happy to do the next method, just let me know if I need to because I don't want to to do the work if I don't need to. I would use the method of sections, and break the two tensions apart then add them up.

 

[PhanthomJay]

Homework Statement

2. The attempt at a solution

Ok. so I'm a little bit confused as to how to approach this problem. I would know how to do this problem if it was laying right to left, but because it's vertical, that makes it a lot more confusing for me.

You can ignore the weight of the aluminum bar , which is not given anyway, so whether the rod is horizontal or vertical doesn't matter.

I know A = (1/4)(pi)(d^2) = 7.0686*10^-4 m^2

I can think of two ways to do this problem, and I need to know which approach is the right one. The first one is straight forward...

I know the combined forces of 10kN + 20kN = 30 kN tensile stress going into the upper wall and out the bottom of the bar at C.

No, you have 30 kN tensile force at the upper wall, but at C, you only have 20 kN applied force

tensile stress = P/A ----> 30kN / (7.0686*10^-4 m^2) = 43441.22 kPa

Using hooke's law: deformation = PL/AE
= (30kN) (3.5m)/ (7.0686*10^-4) (70 * 10^6 kN/m^2) = 0.002122m

is that right?

No, the tensile stress is not the same throughout the bar, it varies in the sections.

I would be happy to do the next method, just let me know if I need to because I don't want to to do the work if I don't need to. I would use the method of sections, and break the two tensions apart then add them up.

Yes, try this.

 

[Patdon10]

Thanks for your response, Jay. So let's try this again...

I know how to do the method of sections with multiple forces pushing/pulling in different directions. I was a little hesitant to attempt it because I would have to pretend it was horizontal and I'd have to figure out what the overall tensile force was at what part, and it just wasn't working out. However, it's a lot easier than that. Can I not just think of it as 2 bars each with a force then just add the two components? It would be like this:

A = (1/4)(pi)(d^2) = 7.0686*10^-4 m^2
Using hooke's law:
Deformation_Section1 = PL/AE
= (10kN)(2m)/(7.0686*10^-4 m^2)(70*10^6 kN/m^2)
= 0.0004042 m
Deformation_Section2 = PL/AE
= (20kN)(3.5m)/(7.0686*10^-4 m^2)(70*10^6 kN/m^2)
= 0.001415 m

Deformation_Section1 + Deformation_Section2 = Total Deformation
0.0004042 m + 0.001415 m = 0.0018189 m

I feel better about this. Does that look right?

 

[PhanthomJay]

Thanks for your response, Jay. So let's try this again...

I know how to do the method of sections with multiple forces pushing/pulling in different directions. I was a little hesitant to attempt it because I would have to pretend it was horizontal and I'd have to figure out what the overall tensile force was at what part, and it just wasn't working out. However, it's a lot easier than that. Can I not just think of it as 2 bars each with a force then just add the two components? It would be like this:

A = (1/4)(pi)(d^2) = 7.0686*10^-4 m^2
Using hooke's law:
Deformation_Section1 = PL/AE
= (10kN)(2m)/(7.0686*10^-4 m^2)(70*10^6 kN/m^2)
= 0.0004042 m
Deformation_Section2 = PL/AE
= (20kN)(3.5m)/(7.0686*10^-4 m^2)(70*10^6 kN/m^2)
= 0.001415 m

Deformation_Section1 + Deformation_Section2 = Total Deformation
0.0004042 m + 0.001415 m = 0.0018189 m

I feel better about this. Does that look right?

No, but it's getting better. You should first determine the reaction at the top support A. What is it? Then cut a section just below B, and determine the force in the top section (it is not 10). Then calculate the deformation of that section. Then look at the bottom section, and deternine the force and deformation in that piece...this part you got the correct force but the wrong length. Then add up the deformations. So your only errors are in determining the force in the top section, it's not 10 , but rather ____??___? And the length of the bottom section is not 3.5, it's ___??___?
Note another way to do this is to look at each applied force separately and apply the superposition principle. You'll get the same result. EDIT: Oh, shoot, that's what you did, the superposition method, using the full length, not the section length, for the 20 kN force, so your answer is correct, nice...but your explanation is a bit off. .

 

[Patdon10]

The force at section A (ignoring any mass) would have to be 30 k/N upwards. This would mean a section just below B would be 20 k/N upwards? So the first section is actually a compressive force? :/

 

[PhanthomJay]

The force at section A (ignoring any mass) would have to be 30 k/N upwards. This would mean a section just below B would be 20 k/N upwards? So the first section is actually a compressive force? :/

I edited my previous response, you got the correct answer, but you correctly used the full length of the bar when calculating the deformation under the 20 kN laod, not the bottom section length. Using the other method, if you have 30 kN at A, then cutting a section below B must also give you a 30 kN force down at that section, per Newton 1, so you have 30 kN in tension in that section and calculate that section deformation using that top section length. Then in the bottom section, you have 20 kN tension, and calculate that section deformation, using the length of the bottom section, and you get the same answer when you add them up. Use the method you are most comfortable with, as long as you understand why.

 

[Patdon10]

Ok. As long as I got the right answer, I won't edit anything.
I'm trying to make sense of what you said was the method of sections. I'm going to draw this out to help myself figure it out. I'm going to remove the point where the bar is attached, and draw a 30 kN force upward. I understand from from A to B I have an 30 kN tensile force, but what happens when I go from A to C? (I'm calling C halfway between B and C, or if you prefer we'll call it B^+)
Wouldn't that give me a tensile force of 20 Kn and a length of 2.75m?

nvm. I think I get it. A to B is 30 kN tensile force for that first section. A to C^- is a 20 kN force for the full length of the bar. A to C^+ is a balanced bar, so you can't really use that. Did that make sense?

Anyways, thanks again for the help!

 

[p21bass]

It would be easiest, when cutting the bar, to use the following free body diagrams, considering the full FBD and calculations determined the correct 30kN upward force at A...

For section A-B, cut the bar between A and B and use the FBD that includes the fewest external forces to solve for T<sub>AB[/AB] - in this case, the section above the cut. Since you only have one other force (30kN at A), you'll know that the tension throughout section AB* is 30kN.

For B-C, cut the bar between B and C, and use the FBD below the cut. Again - you'll have one known force (the downward 20kN at C), so you'll know that TBC* will be 20kN.

Calculate the deformations and add them all together. This method works no matter how many sections there are. It's the same procedure, whether the beam is horizontal, vertical, or on an angle.* - Note that the tension is throughout the entire section of the beam under consideration. It doesn't matter how close to the points of interest you cut. Therefore, you consider the length of the entire section of interest. Therefore, "Wouldn't that give me a tensile force of 20 Kn and a length of 2.75m?" is not correct. The length of the section that sees 20kN of tension is only BC - 1.5m.</sub>