Determine the position and magnitude of the max bending moment

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  • #1
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Homework Statement


upload_2018-9-26_20-25-54.png

For my question, I am following a previous method I used in my HNC where I had to determine the position and magnitude of the maximum bending moment. I am now working on my project and I have revisited this question but I am having some issues in my method. The above diagram is of a simply supported structure with the weight of an average cat in the centre. The structure is 500mm wide and simply supported at either end as shown.
The cat weighs 4.5kg which translates to a force of 44.2N and as it is stood in the centre of the structure (250mm), R1 and R2 must react with 22.1N to maintain the structure.
Therefore my first answer is below in FIG 1.


Next I introduced Snow to the structure giving s uniform distributed load across.
I worked out that, typically snow weighs between 70kg and 150kg per cubic meter. I have a surface area of 500mm x 500mm with a possible depth of 150mm snow on the structure. Therefore, the cubic area would be

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Which leads me to my next diagram as shown below.
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This time due to the snow fall on the roof, the reactive force from R1 and R2 must compensate for the additional weight. This method is shown in FIG 2.

I cannot see how the maximum bending moment in both situations much less than the force from the cat alone. I have tried this several ways and ended up with different answers each time. I believe my error is in the units of measurment as I am showing my answer in N m and the structure is only 500mm or 0.5m wide. Should I show my answer in N mm? am I making another error somewhere along the way?


Homework Equations


All my methods are below.

The Attempt at a Solution


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FIG 1.

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FIG 2.
 

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Answers and Replies

  • #2
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I would convert the uniformly distributed load to a single equivalent concentrated load.
 
  • #3
PhanthomJay
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The total snow load on the full width and length of the beam is 55 N. Since the beam is 0.5 m in length, that load is distributed at 55/0.5 or 110 N/m. You can use the 55N total load as a concentrated load at the cg of the distributed load to calculate end reactions, but when determining internal bending moments along the beam, you must use the distributed 110 N/m load.
You should prove to yourself that the max moment in a simply supported beam with a concentrated load P at the center occurs at the midpoint of the beam length and is equal to PL/4, whereas the max moment in a simply supported beam with a uniformly distributed load P/L occurs at the midpoint of the beam length and is equal to PL/8, that is, the max moment is halved when you distribute a load over the beam length versus concentrating it at a single point in the center.
 
  • #4
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I understand it would be easier to define the UDL as a single point concentrated load however the reason I am working it out in the way I am is due to the next step, which I have not yest posted. This is confirming my calculations using macaulays method. A method I previously used on an assignment which I also seem to be getting an unusual result. I need to know if my calculations are correct to understand what I am doing wrong in the following method
 
  • #5
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I will post my results for Macaulays method to show you what I mean..

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  • #6
PhanthomJay
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I understand it would be easier to define the UDL as a single point concentrated load however the reason I am working it out in the way I am is due to the next step, which I have not yest posted. This is confirming my calculations using macaulays method. A method I previously used on an assignment which I also seem to be getting an unusual result. I need to know if my calculations are correct to understand what I am doing wrong in the following method
I haven'tooked at Macaulay, but for starters, the uniform distributive load is 110 N/m, per post #3.
 
  • #7
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I haven'tooked at Macaulay, but for starters, the uniform distributive load is 110 N/m, per post #3.

Reference https://www.physicsforums.com/threa...d-magnitude-of-the-max-bending-moment.956272/
Thanks, I will update my calculations with that information. I remember doing that early on in my lessons now but it has been a while since I have needed to use it.
I think there is another error somewhere in my method. If my original method for obtaining the bending moment is correct ( apart from the actual value of the UDL) then there is something wrong in my work using macaulays method.
 
  • #8
haruspex
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Perhaps you have already corrected this, but your first eqn in Fig 2 is wrong. You have already calculated the total snow load as 55N. It makes no sense to multiply that by the beam length. Indeed, it would have the wrong dimensions.

However, what you posted in Fig 2 is at least consistent in, effectively, using 55N/m as the distributed load, so there must be another error to get a lower answer than with the cat alone.
Check your arithmetic for 0.54.

The trouble with posting your work as images is that it makes it hard to comment on specific steps. Please number your equations.
 
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  • #9
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Check your arithmetic for 0.54.
I get 0.0625 for this which is obviously much less than 0.5. So this is why my answers are showing as a lesser force although the force is initially heavier.

The trouble with posting your work as images is that it makes it hard to comment on specific steps. Please number your equations.
I will work through and correct any issues tonight when I get a bit of time, for now I have labelled each step. I cannot copy and paste from microsoft word into the forum and I am not fully up to speed with typing out long equations on here, so I apologise for the inconvenience in quoting back the work.

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  • #10
PhanthomJay
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Where did the 10 kN point load come from if the cat weighs just 44.2 N? And why are you even looking at Macaulay? The point of max moment occurs at the point of 0 shear. Are you familiar with shear and moment diagrams? Once you know the point of max moment, just sum moments about that point in a free body diagram of the section of beam cut at that point. And if you don’t use the distributed load as 110 N/m, you will never arrive at the correct answer.
 
  • #11
haruspex
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I get 0.0625 for this which is obviously much less than 0.5. So this is why my answers are showing as a lesser force although the force is initially heavier.


I will work through and correct any issues tonight when I get a bit of time, for now I have labelled each step. I cannot copy and paste from microsoft word into the forum and I am not fully up to speed with typing out long equations on here, so I apologise for the inconvenience in quoting back the work.

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You have failed to incorporate corrections to errors that have already been pointed out. The total load is 44.2+55=99.2N, or 49.6N per support, not 35.85.
The distributed load is 55N/0.5m=110N/m.
 
  • #12
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Where did the 10 kN point load come from
Good Spot on that one, that was me being sloppy as I have used this method before and I have tried to follow the same method to suit my project. I will rework everything now and repost my answers.
 
  • #13
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Does this look better? If so I will continue my work into the macaulays method and repost my answers from there. Thank you all for your help! I apaolgise about the delay in responding to your advise as I am working 7 day weeks at the moment and trying to fit this in during all the chaos.
 

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  • #14
haruspex
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Does this look better?
Yes, that certainly fixes the earlier errors.
Of course, assessing the moment at intervals and picking the largest does not necessarily identify the greatest moment, but it is fairly clear in such a simple case.
 
  • #15
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Check your arithmetic for 0.5^4.
I have just realised this error in macaulays method. Thanks
 
  • #16
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I am not entirely certain I am following the correct method but where the brackets are highlighted in red, it indidicates the value of the sum within is a negative value and shouldnt be included in the following sum. therefore I have been following this method until I inevitably get to my answer in 1e, where I must include (0.3-0.25)^3 in the sum and this is throwing my answer way off. Obviously I am doing something wrong?
 

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  • #17
haruspex
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I am not entirely certain I am following the correct method but where the brackets are highlighted in red, it indidicates the value of the sum within is a negative value and shouldnt be included in the following sum. therefore I have been following this method until I inevitably get to my answer in 1e, where I must include (0.3-0.25)^3 in the sum and this is throwing my answer way off. Obviously I am doing something wrong?
I am not following line 4b. I get A more like -1.2.

Where does the 4.583 come from in 2e?
 
  • #18
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I dont know how I managed to get -0.79 on my first calculation
I have corrected that to be -1.25 and as that is the value of A it transfers throughout the rest of the method so I have reworked it. I get 4.583 from 2e from 110/24 as I am not including the (0.3-0.25)^4 because the answer is a negative value. There is still something wrong which I cant put my finger on? The values should plot a nice negative curve on a graph.
 

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  • #19
haruspex
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I get 4.583 from 2e from 110/24 as I am not including the (0.3-0.25)^4 because the answer is a negative value.
Not following.
You correctly exclude the 0.3-0.5 term (I don't see a 0.3-0.25 term). The term in 1e with 110/24 in it also has a factor 0.34. This whole term you correctly evaluate as the 0.037... in 2e. There is no other term with 110/24 in it.
 
  • #20
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Not following.
You correctly exclude the 0.3-0.5 term (I don't see a 0.3-0.25 term). The term in 1e with 110/24 in it also has a factor 0.34. This whole term you correctly evaluate as the 0.037... in 2e. There is no other term with 110/24 in it.
upload_2018-10-30_21-49-25.png

the 0.3-0.25 term is highlighted in orange, I include this because the value was not a negative, therefore I think it should be included?
For your comment 'there is no other term with 110/24 in it.' I have highlighted my value in blue, I didnt include the (0.3-0.5)^4 therefore I was left with 110/24 giving the 4.583. I dont remember following any other unique method than, not including the negative values so I cannot put my finger on the problem
Thanks for your help with this one, I really appreciate your time.
 

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  • #21
haruspex
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I didnt include the (0.3-0.5)^4 therefore I was left with 110/24
Not including a term with a negative factor means setting it to zero. You are effectively setting the factor to 1.
 
  • #22
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Im sorry I dont think I am following, I can still include the 110 as it is not part of the method to eliminate anything in the nominator if one value is a negative?
If the ()^4 is raised to a factor and effectively becomes zero e.i (negative value)^4 and zero^1 is still zero, I am only left with 110/24 = 4.583 or I could exclude the whole term and set the value to 1/24 to arrive at 0.0416?
 
  • #23
haruspex
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Im sorry I dont think I am following, I can still include the 110 as it is not part of the method to eliminate anything in the nominator if one value is a negative?
If the ()^4 is raised to a factor and effectively becomes zero e.i (negative value)^4 and zero^1 is still zero, I am only left with 110/24 = 4.583 or I could exclude the whole term and set the value to 1/24 to arrive at 0.0416?
You seem to misunderstand an essential part of the method.
If you have a term like ##\frac{(load factor)(a-b)^n}{n!}## and a<b then you must throw away the whole term. It becomes zero, not ##\frac{(load factor)}{n!}##
 
  • #24
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I have attempted to rework the method following your advice and in the first instance at step 2d, I get a positive value in the results of 0.3102. I thought there was still something I am missunderstanding so I attempted it again, this time excluding the term if the entire term was negative (not just the term within the brackets). Following this I get a set of all negative results however, they do not form a smooth curve as I would have expected when plotting in a graph. I have posted this method example below.
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Below is an example of how we got through this problem the first time round and successfully produced a graph to match the situation. I have posted it as a reference to find out what is going on in this situation.

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  • #25
haruspex
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this time excluding the term if the entire term was negative (not just the term within the brackets)
No, no. It is only the sign of the displacement that determines whether to include it.
The principle is this: at any point along the beam all of the torques and forces on one side exert a net torque about that point, and this is balanced by the net torque from the other side. It is the magnitude of these two opposing torques that determines the curvature at the point. So to find that curvature, you add up all the torques from one side only.
Forces on one side will have a positive displacement from the point, while forces on the other will have a negative displacement. So that determines whether to include it in the sum: only include those with positive displacement. The force itself may be up or down, so the term as a whole may be positive or negative.
Some authors use a special notation for this. E.g. you could put angle brackets <> around displacements. The interpretation is <x>=max{x, 0}, i.e. if x is negative substitute zero.
 
  • #26
haruspex
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I have attempted to rework the method following your advice and in the first instance at step 2d, I get a positive value in the results of 0.3102.
Are you saying that is wrong?
Please post your updated working.
 
  • #27
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I somehow miscalculated step 2e and step 2f leading me to think I had unusual results. These results look good, and are what I was expecting. Does this look ok to you? I have been sloppy with my calculations all the way through. If I have learnt something from this, it is to check my calculations more than twice if I still dont have the correct answer.


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  • #28
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I wonder why there is a delay from 0 to 0.1 before the natural curve begins to take place? is there still something wrong with my method?
 

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  • #29
haruspex
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I somehow miscalculated step 2e and step 2f leading me to think I had unusual results. These results look good, and are what I was expecting. Does this look ok to you? I have been sloppy with my calculations all the way through. If I have learnt something from this, it is to check my calculations more than twice if I still dont have the correct answer.


View attachment 233543
Reread my post #23 and have another go at line 2c. If the (a-b) is negative you must throw away the whole term. It is not a question of whether (a-b)n is negative.
 
  • #30
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This is what I expected to find! Thank you so much for taking the time to help me again... The world needs more people like yourself!
 

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  • #31
haruspex
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This is what I expected to find! Thank you so much for taking the time to help me again... The world needs more people like yourself!
Glad to have helped you understand how to apply the method.
However, from a purist perspective, finding the maximum moment by checking the moment at various points along the line is not satisfactory.
A neater method is to consider each segment for which the distributed load is constant (maybe zero) and there are no point loads. In the present question that would be just two segments. For each segment you can throw away the negative displacement terms from the torque equation (they will be the same terms for the whole segment) and differentiate to look for a local maximum torque within the segment. If there is no local maximum within the segment then the maximum for the segment occurs at one of its endpoints.
Having found the maximum for each segment you would then compare them to find the maximum for the whole beam.
 
  • #32
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That sounds like an interesting method which I am unfamiliar with. I am interested to learn a bit more about calculus as I have not used it a great deal throughout my course. I am going to set myself a further challenge to complete which you may be able to assist me with using alternative methods. I am looking to design a automated system to open an close the lid of the greenhouse using a counterweight and a microservo to tip the balance given by the weight.
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I have put together this diagram of how it would look and I have made a diagram for the lid and mechanism to show the calculations required. As this is a fictional project I am going to say the net weight of the lid is 1kg or 1000g to simplify the calculations. The micro servo attached to the pulley will transmit the torque in favour of the counterweight to open or to the lid to close. I have produced a basic diagram of how this would look.
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I was going to follow my previous method showing the maximum bending moment required to tip the balance over to the counterweight. Again I can say the maximum angular torque available from the servo is 10g as I am not using genuine parts.
Is there anything you can offer me to get started with this problem? Thank you.
 

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  • #33
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Here is a stripped down version with the facts I have set out to find the maximum force required at the big red arrow. I will do some calculations to try and get some results myself.
 

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  • #34
haruspex
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That sounds like an interesting method which I am unfamiliar with. I am interested to learn a bit more about calculus as I have not used it a great deal throughout my course. I am going to set myself a further challenge to complete which you may be able to assist me with using alternative methods. I am looking to design a automated system to open an close the lid of the greenhouse using a counterweight and a microservo to tip the balance given by the weight.
View attachment 233880
I have put together this diagram of how it would look and I have made a diagram for the lid and mechanism to show the calculations required. As this is a fictional project I am going to say the net weight of the lid is 1kg or 1000g to simplify the calculations. The micro servo attached to the pulley will transmit the torque in favour of the counterweight to open or to the lid to close. I have produced a basic diagram of how this would look.
View attachment 233885
I was going to follow my previous method showing the maximum bending moment required to tip the balance over to the counterweight. Again I can say the maximum angular torque available from the servo is 10g as I am not using genuine parts.
Is there anything you can offer me to get started with this problem? Thank you.
If you want a real workable design, this is not it. The problem is that the torque exerted by the counterweight is roughly constant, whereas the torque load from the lid will reduce as it rises.
To fix that, hang the counterweight from that stub of lid that sticks out to the left. Now it is just a seesaw.
 
  • #35
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I have redrawn the diagram to show the idea. I have started with working out the net weight of the lid as a UDL and worked out 1000g into Newtons = 9.81N. as a UDL this would be 9.81/0.6m = 16.35N.
 

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