Calculate The Moments about the Point A, [Guidance needed]

[Dellis]

Hi, I need some guidance regarding this exercise,

Homework Statement

Calculate The Moments about the Point A

I attached the visual representation of the exercise, it has common A(PIN)---------B(ROLLER) element, then two forces going down( 5 Kips & 2 kips ), then the distances about point A and B( 6 ft-8ft-10ft )

Homework Equations

M= F x D

The Attempt at a Solution

I need guidance, I tried to do it myself and got up to here, if is wrong please guide me to the right route.


\SigmaM= 5(6)-7(14) + 2( 24 )= 30-98-48= -20

 

[ronaldlover]

as far as i am concerned, you are off to the right start.
I, as well, only got up to this far.
Meassage me when you make a break through. i would love to hear about it.

 

[Dellis]

Really?, then we need to see others input on the exercise, that would be a good thing.

I am here wondering what if that is the actual answer to the exercise?, it is asking for the Moment right?, and basically that's all I got up to.

Lets wait for somebody else to give their input, clarify the situation.

 

[rl.bhat]

Reaction on A and B is, say, R1 and R2.
What is R1 + R2?
Reaction on B is not equal to (5 + 2).
Whether the rod is in equilibrium? Whether the roller is free to move?

 

[Redbelly98]

Since the question asks for the moments (note plural), it seems you are to calculate the individual moments for the loads of 5 kips, 2 kips, and the roller separately.

Is this supposed to be in equilibrium, as rl.bhat asked? What can you say about the sum of the moments in that case?

Also, the roller force is not necessarily 7 kips. There is another force at point A, that you seem to have ignored when balancing all the forces.

 

[Dellis]

Since the question asks for the moments (note plural), it seems you are to calculate the individual moments for the loads of 5 kips, 2 kips, and the roller separately.

Is this supposed to be in equilibrium, as rl.bhat asked? What can you say about the sum of the moments in that case?

Also, the roller force is not necessarily 7 kips. There is another force at point A, that you seem to have ignored when balancing all the forces.

But you forget it says ABOUT POINT A, it says nothing about Point B?


I don't think its supposed to be in equilibrium, it just want the moments about Point A


You think their is a force at point A?, how can that even be found?, 5 - 1?. I don't understand.

 

[Dellis]

Reaction on A and B is, say, R1 and R2.
What is R1 + R2?
Reaction on B is not equal to (5 + 2).
Whether the rod is in equilibrium? Whether the roller is free to move?

But is asking for the moments ABOUT POINT A, why would point B come into the equation? but since you asked there then

Reaction on B would be 10' + 2 kips then?

 

[rl.bhat]

But you forget it says ABOUT POINT A, it says nothing about Point B?
I don't think its supposed to be in equilibrium, it just want the moments about Point A
You think their is a force at point A?, how can that even be found?, 5 - 1?. I don't understand.

It the system is not in equilibrium, what is the role of roller?
Two forces are acting on the rod in the downward direction. There must be reaction for these forces. According to your calculations, it is acting on B only. What about A? Whether the rod is not attached to the pin at A? If the rod is not in equilibrium, in which direction it is moving? If you take the moments about B due to 5 kips and 2 kips, you can see the moment due to 5 kips is greater that due to 2 kips. Do you mean that the roller roles in either direction? How can you find the moment ignoring B?

 

[Dellis]

It the system is not in equilibrium, what is the role of roller?
Two forces are acting on the rod in the downward direction. There must be reaction for these forces. According to your calculations, it is acting on B only. What about A? Whether the rod is not attached to the pin at A? If the rod is not in equilibrium, in which direction it is moving? If you take the moments about B due to 5 kips and 2 kips, you can see the moment due to 5 kips is greater that due to 2 kips. Do you mean that the roller roles in either direction? How can you find the moment ignoring B?

I think you're confused by the fact this exercise is not like other common ones where the roller would be all way at the very end?. I never ignored B but you said I ignored A?, how so if 5 kips is part of A?, I didn't ignore B either since 2 kips is part of Point B.

In all honesty I just need some guidance here, I know you're trying to do that and I appreciate it. I think I am on the right path but I only found a moment and as stated above it actually says MOMENTS ABOUT POINT A.

This type of problem is clearly a basic status one but we're thinking too much making it complex.

 

[stewartcs]

I think you're confused by the fact this exercise is not like other common ones where the roller would be all way at the very end?. I never ignored B but you said I ignored A?, how so if 5 kips is part of A?, I didn't ignore B either since 2 kips is part of Point B.

In all honesty I just need some guidance here, I know you're trying to do that and I appreciate it. I think I am on the right path but I only found a moment and as stated above it actually says MOMENTS ABOUT POINT A.

This type of problem is clearly a basic status one but we're thinking too much making it complex.

The rod is in equilibrium since no other forces are shown acting on it that would cause it to move.

Hence, the sum of the moments about A must equal zero. The sum of the moments about B must equal zero. The reactions at A (since it is pinned it has a horizontal and vertical reaction) and B (since it is a roller it only has a vertical reaction) will be found by summing the forces in the x and y direction. All of your unknowns can be found by doing this.

The problem appears to just want you to calculate each moment about the point A. So just pick a convention (+/-, CW/CCW or whatever you want) and do the math (FxD) noting that the sum of the moments about A is zero.

Also, note that the drawing you attached seems to show that point A is at 6-ft, point B is at 8-ft and the end of the rod is at 10-ft. Hence, the 5 kip load is applied at 7-ft relative to the 6-ft origin. Thus the moment about a resulting from the 5 kip load would be 5 kips times 1 foot = 5 kip-ft (in the direction you chose based on your selected sign convention).

Hope this helps.

CS

 

[Dellis]

The rod is in equilibrium since no other forces are shown acting on it that would cause it to move.

Hence, the sum of the moments about A must equal zero. The sum of the moments about B must equal zero. The reactions at A (since it is pinned it has a horizontal and vertical reaction) and B (since it is a roller it only has a vertical reaction) will be found by summing the forces in the x and y direction. All of your unknowns can be found by doing this.

The problem appears to just want you to calculate each moment about the point A. So just pick a convention (+/-, CW/CCW or whatever you want) and do the math (FxD) noting that the sum of the moments about A is zero.

Also, note that the drawing you attached seems to show that point A is at 6-ft, point B is at 8-ft and the end of the rod is at 10-ft. Hence, the 5 kip load is applied at 7-ft relative to the 6-ft origin. Thus the moment about a resulting from the 5 kip load would be 5 kips times 1 foot = 5 kip-ft (in the direction you chose based on your selected sign convention).

Hope this helps.

CS

Thanks for the very in depth post, I went counter clock wise for the calculations I did.So you're saying keep on that same route but add the multiplication of 5 x 1 to the

calculation?, I figured when the others said I was ignoring Point A's force that they wanted

me to find it by doing 5 x 1 but didn't tell me, I threw a guess and said 5 + 1 clearly not the

right way.

 

[stewartcs]

Thanks for the very in depth post, I went counter clock wise for the calculations I did.


So you're saying keep on that same route but add the multiplication of 5 x 1 to the

calculation?, I figured when the others said I was ignoring Point A's force that they wanted

me to find it by doing 5 x 1 but didn't tell me, I threw a guess and said 5 + 1 clearly not the

right way.

I'm saying that you need to find each moment separately and that their sum should be zero.

The 5 kip-ft I gave you was the first moment due to the 5 kip load applied 1 foot from point A (presuming I've interpreted the distances on your drawing correctly). You need to do the same for the reaction at B and the 2 kip load at the end of the beam. Their sum should be zero since the beam is in static equilibrium.

So, there are three moments about point A. The first one was just given to you. Now use the same approach and find the other two.

Hint: Find the reaction at point B (vertical direction) first.

CS

 

[ronaldlover]

I see.
well i agree since it is asking ony for a moment not moments. so therefore there are 3 moments in point A and one was given to you.

CS is on the correct track as far as i can see.

 

[Dellis]

I'm saying that you need to find each moment separately and that their sum should be zero.

The 5 kip-ft I gave you was the first moment due to the 5 kip load applied 1 foot from point A (presuming I've interpreted the distances on your drawing correctly). You need to do the same for the reaction at B and the 2 kip load at the end of the beam. Their sum should be zero since the beam is in static equilibrium.

So, there are three moments about point A. The first one was just given to you. Now use the same approach and find the other two.

Hint: Find the reaction at point B (vertical direction) first.

CS

Oh alright, thanks for clearing that up

 

[Redbelly98]

Also, note that the drawing you attached seems to show that point A is at 6-ft, point B is at 8-ft and the end of the rod is at 10-ft. Hence, the 5 kip load is applied at 7-ft relative to the 6-ft origin. Thus the moment about a resulting from the 5 kip load would be 5 kips times 1 foot = 5 kip-ft (in the direction you chose based on your selected sign convention).

Thanks for chiming in and helping!

Are you sure about this interpretation of the distances given in the figure? To me, it seemed pretty clear that the values were distances in between the different load points. I'm uncomfortable with just assuming the 5 kips point is halfway between points A and B.

 

[stewartcs]

Thanks for chiming in and helping!

Are you sure about this interpretation of the distances given in the figure? To me, it seemed pretty clear that the values were distances in between the different load points. I'm uncomfortable with just assuming the 5 kips point is halfway between points A and B.

Hi Red,

No I'm not sure at all that's why I said "seems to show". Nevertheless, the OP should be able to tell from the book drawing (which hopefully is more clear and standardized). I can certainly see your point though. Thanks for clarifying.

CS

 

[Dellis]

Ok I tried doing the exercise, but I need somebody to check it for me. please! this is due tomorrow morning, check it for me, I think its correct.

\SigmaF=

Reaction A - 5 k + 5.57 k - 2 k = 0

Reaction A = 1.43 k\SigmaM= 5k x 6ft- Reaction B x 14ft + 2k x 24ft= 0

Reaction B = [30 k ft + 48 k ft]/14 ft

Reaction B = 5.57 k

 

[stewartcs]

Ok I tried doing the exercise, but I need somebody to check it for me. please! this is due tomorrow morning, check it for me, I think its correct.

\SigmaF=

Reaction A - 5 k + 5.57 k - 2 k = 0

Reaction A = 1.43 k\SigmaM= 5k x 6ft- Reaction B x 14ft + 2k x 24ft= 0

Reaction B = [30 k ft + 48 k ft]/14 ft

Reaction B = 5.57 k

Yes, the vertical reactions are equal to 1.43 kips at A and 5.57 kips at B (presuming the distances given are those between the points and not at the points).

However, is this what the problem asks you to find (initially you asked about the moments about A)?

CS

 

[Dellis]

Yes, the vertical reactions are equal to 1.43 kips at A and 5.57 kips at B (presuming the distances given are those between the points and not at the points).

However, is this what the problem asks you to find (initially you asked about the moments about A)?

CS

Well yea but I figured maybe doing this part would open up the process of finding the moments, in a easier manner?

So if the reaction is 1.43kips at POINT A, then if it is asking for the moments at POINT A

that means I need to do what I learned while doing that other exercise, going from left to

right looking at the distances ABOUT POINT A?, then calculate.

 

[stewartcs]

Well yea but I figured maybe doing this part would open up the process of finding the moments, in a easier manner?

So if the reaction is 1.43kips at POINT A, then if it is asking for the moments they mean

I need to do what I learned while doing that other exercise, going from left to right

looking at the distances ABOUT POINT A?

How did you find the reaction at B? It would have required you to sum the moments about A in order to find the reaction at B. So you probably have already done the work!

Look back a few posts where we talked about starting at point A and calculating each individual moment using FxD.

CS

 

[Dellis]

How did you find the reaction at B? It would have required you to sum the moments about A in order to find the reaction at B. So you probably have already done the work!

Look back a few posts where we talked about starting at point A and calculating each individual moment using FxD.

CS

I thought it was 7?, I guess not then but let me look back at page 1

even though you just told me how to do it there, perhaps I already did do the work here.

 

[Dellis]

Damn I don't get it, when I looked at this problem again today it reminded me of the
other one I just did, so I attacked it again with the mindset of knowing how to approach it but clearly I a running into issues understanding the distances and all that again.

 

[stewartcs]

I thought it was 7?, I guess not then but let me look back at page 1

even though you just told me how to do it there, perhaps I already did do the work here.

OK, here is a summary of what you should have done:

\Sigma{F_y} = R_{Ay} -5 + R_{By} -2 = 0

which reduces to:

R_{Ay} + R_{By} = 7 kips

Now, since there are two unknowns, we need another equation in order to solve them simultaneously right?...So summing the moments about A we have:

\Sigma{M_A} = 5(6) - R_{By}(14) + 2(24) = 0

Solving for the reaction at B we get:

14R_{By} = 78

R_{By} = 5.57 kips

Now plug that into the first equation and you get your 1.43 kips for the vertical reaction at A.

So, to find the moments about point A (now that you know the missing reaction at B)...just use FxD...you'll end up with three moments that will sum to zero...

Note that we have found the moments already in the process of finding the reactions, so just pick them out from the calculations above.

Does this help?

CS

 

[Dellis]

Yeah you broke it down in a more calculated manner, where I can see what is going on, I will be here trying to figure things out and post my solution.

 

[stewartcs]

Yeah you broke it down in a more calculated manner, where I can see what is going on, I will be here trying to figure things out and post my solution.

Always try to approach these problems systematically like this...it will greatly simplify them for you.

CS

 

[Dellis]

Yeah I need to get familiar with the processes or ways to approach the exercises, like I said before that is what I need to get better at. Don't go to far I will be here trying to get this done, if any other contributor/mentor reading wants to help feel welcome to.

 

[Dellis]

Ok here is my attempt, again not to sure, but I am trying by best to learn, with what you said there I believe reaction at B is 5.57 right?, which I then plugged into the procedure where before we didn't have that reaction, so then I used F X D and got this, I hope I did the right thing...\Sigma{M_A} = 5(6) - 5.57(14) - 2(24) = -95.98 lb/ftwhat do you think stewartcs?

 

[stewartcs]

Ok here is my attempt, again not to sure, but I am trying by best to learn, with what you said there I believe reaction at B is 5.57 right?, which I then plugged into the procedure where before we didn't have that reaction, so then I used F X D and got this, I hope I did the right thing...


\Sigma{M_A} = 5(6) - 5.57(14) - 2(24) = -95.98 lb/ft

No.

Take a look at the post above where I gave you the summary. It shows the reactions at A and B.

It also show the moments about A, AND, that they sum to zero.

The three moments about point A are:

5(6)      =  30 ft-lb
-5.57(14) = -78 ft-lb
2(24)     =  48 ft-lb
---------------------
Sum       =   0 ft-lb

(Note the negative sign in the second one since it produces a ccw rotation - based on my assumed sign convention)

Adding those up you get zero.

CS

 

[Dellis]

Damn I knew it wasn't that simple, so you basically did the exercise right there right?,

Ok going by that, then you're basically saying do this?..\Sigma{M_A} = 5(6) - 5.57(14) - 2(24) = 0

\Sigma{M_A}= 30 ft-lb-78 ft-lb + 48 ft-lb= 0

 

[stewartcs]

Damn I knew it wasn't that simple, so you basically did the exercise right there right?,

so basically...

\Sigma{M_A} = 5(6) - 5.57(14) = 78 2(24) = 0

The exercise was done in post #24...The last post just showed you how to pick the answers from it...

CS

 

[Dellis]

The exercise was done in post #24...The last post just showed you how to pick the answers from it...

CS

What?, I though you were just showing mostly how to calculate B in there

my previous post you quoted is updated.

 

[Dellis]

You said the reaction of AB= 7, so I tried it out and got this.

\Sigma{M_A} = 5(6) - 5.57(7) - 2(24) = 0

 

[stewartcs]

What?, I though you were just showing mostly how to calculate B in there

my previous post I updated it.

Your original question was to find the moments about point A right?

In order to find them, you needed to know the reaction at B. This was found in post #24.

While finding the reaction at B (in post #24) you had to sum the moments about A as shown below:

\Sigma{M_A} = 5(6) - 5.57(14) + 2(24) = 0

From that part (summing the moments about A), all you needed to do was pick out the moments and list them as I did in post #29.

5(6)      =  30 ft-lb
-5.57(14) = -78 ft-lb
2(24)     =  48 ft-lb
---------------------
Sum       =   0 ft-lb

Hence the moments about A are listed above. This is your answer based on the question you first asked in this thread.

Does this help?

CS

 

[stewartcs]

You said the reaction of AB= 7, so I tried it out and got this.

\Sigma{M_A} = 5(6) - 5.57(7) - 2(24) = 0

That does not sum to zero.

The negative sign in front of 2(24) should not be there based on your sign convention.

CS

 

[Dellis]

Your original question was to find the moments about point A right?

In order to find them, you needed to know the reaction at B. This was found in post #24.

While finding the reaction at B (in post #24) you had to sum the moments about A as shown below:

\Sigma{M_A} = 5(6) - 5.57(14) + 2(24) = 0

From that part (summing the moments about A), all you needed to do was pick out the moments and list them as I did in post #29.

5(6)      =  30 ft-lb
-5.57(14) = -78 ft-lb
2(24)     =  48 ft-lb
---------------------
Sum       =   0 ft-lb

Hence the moments about A are listed above. This is your answer based on the question you first asked in this thread.

Does this help?

CS

So that is the answer to the problem?, showing all the calculated moments, justthat?

 

[stewartcs]

So that is the answer to the problem?, showing all the calculated moments, you don't need to add them or anything like that?, just show that?

Based on the question you asked...yes. Those are the moments about A.

CS

 

[Dellis]

That does not sum to zero.

The negative sign in front of 2(24) should not be there based on your sign convention.

CS

Ok so this below and the "Moments about A" calculation equal to= 0 as stated above?, that is the answer to all of this?

\Sigma{M_A} = 5(6) - 5.57(7) + 2(24) = 72.43 kips/ft


5(6) = 30 ft-lb
-5.57(14) = -78 ft-lb
2(24) = 48 ft-lb
---------------------
Sum = 0 ft-lb

 

[stewartcs]

Ok so this is? and the "Moments about A" equal to 0 as stated above?, that is the answer to all of this?

\Sigma{M_A} = 5(6) - 5.57(7) + 2(24) = 72.43 kips/ft


5(6) = 30 ft-lb
-5.57(14) = -78 ft-lb
2(24) = 48 ft-lb
---------------------
Sum = 0 ft-lb

No.

-5.57(7) is not correct...it should be, as stated numerous times above, -5.57(14). The 14 comes from 6 + 8 which is the distance between point A and reaction B.

CS

 

[stewartcs]

The answer is in post #34.

Answer: 30 ft-lb, -78 ft-lb, and 48 ft-lb

CS

 

[Dellis]

No.

-5.57(7) is not correct...it should be, as stated numerous times above, -5.57(14). The 14 comes from 6 + 8 which is the distance between point A and reaction B.

CS

Ok sorry about that\Sigma{M_A} = 5(6) - 5.57(14) + 2(24) = .025(6) = 30 ft-lb
-5.57(14) = -78 ft-lb
2(24) = 48 ft-lb
---------------------
Sum = 0 ft-lb

 

[Dellis]

The answer is in post #34.

Answer: 30 ft-lb, -78 ft-lb, and 48 ft-lb

CS

I know but I need to show how I got them though, I need to know how to properly display the answer to the whole exercise.

Does that calculation I did in my previous post even part of it?, I don't see it asking for that right. It is just asking to calculate the Moments about A, I can just show how I get those and that's it right?.

 

[stewartcs]

Ok sorry about that

\Sigma{M_A} = 5(6) - 5.57(14) + 2(24) = 72.43 kips/ft

I want you to multiply each term in this equation and right it down without adding them up and post the results.

CS

 

[stewartcs]

I know but I need to show how I got them though, I need to know how to properly display the answer to the whole exercise.

Does that calculation I did in my previous post even part of it?, I don't see it asking for that right, I can just show how I get those and that's it?.

Combine post #24 and #34 and that is the "work" that you should show.

CS

 

[stewartcs]

Ok sorry about that


\Sigma{M_A} = 5(6) - 5.57(14) + 2(24) = .02


5(6) = 30 ft-lb
-5.57(14) = -78 ft-lb
2(24) = 48 ft-lb
---------------------
Sum = 0 ft-lb

The 0.02 is due to rounding errors...it should be 0.

CS

 

[Dellis]

The 0.02 is due to rounding errors...it should be 0.

CS

Ah ok, so other then that(which I will fix below) that is the end of this exercise?

\Sigma{M_A} = 5(6) - 5.57(14) + 2(24) = 05(6) = 30 ft-lb
-5.57(14) = -78 ft-lb
2(24) = 48 ft-lb
---------------------
Sum = 0 ft-lb

 

[stewartcs]

Ah ok, so other then that(which I will fix below) that is the end of this exercise?




\Sigma{M_A} = 5(6) - 5.57(14) + 2(24) = 0


5(6) = 30 ft-lb
-5.57(14) = -78 ft-lb
2(24) = 48 ft-lb
---------------------
Sum = 0 ft-lb

Yes.

CS

 

[Dellis]

Yes.

CS

Thank you 1 million times, I really appreciate all the help and guidance.


As you seen on the forum I got a new thread, dealing with Free Body Diagram though and I got it done in a picture, just want to see what forces I missed, can you help me with that please?.