# Finding the Moment of Point B: Is it Largest?

• freshbox
In summary, the conversation is about finding the largest moment at a specific point on a beam. The original poster is asking for advice on their calculations, which involve moments at points A, B, and C. They are struggling to understand the concept of moments and how to properly calculate them, and are looking for clarification on the correct approach to solving the problem. The conversation also includes some confusion about the direction of moments and equations of equilibrium.
freshbox
Hi can anybody tell me if my working for finding the moment of a b c is correct?

Taking clockwise as +ve

Ma+1.2k x 3 +0.7 x 7 = 0
Ma= -8.5Knm

Mb-8.5+1.9k x 3=0
Mb=2.8kNm

Mc=1.2k x 4 +1.9k x 7 -8.5k=0
Mc=-117.85Nm

Hence Point B has the biggest moment am I right? However the book says is Point A. Please advice, thank you.

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What is the full question?I don't understand what you are trying to do.Find the largest moment?

Taking clockwise as +ve

Ma+1.2k x 3 +0.7 x 7 = 0
Ma= -8.5Knm

Can you explain to me why my working is wrong for Moment A, thank you.

This is the real question. For flexture stress to be maximum at the point, the value of "M' must be largest in the beam

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freshbox said:
Taking clockwise as +ve

Ma+1.2k x 3 +0.7 x 7 = 0
Ma= -8.5Knm

Can you explain to me why my working is wrong for Moment A, thank you.
Isn't that the total moment??
It's also wrong.
M=F*perpendicular distance.
M=1.2k x 3 + 0.7 x 7
=8.5KNm
Why -8.5kNm?
If clockwise is +,and all the forces are acting on clockwise direction,Why should there be any negative value?

EDIT:That tensile flexural state thing is beyond my knowledge.Maybe someone else can help you from here onwards. Sorry

Ma+1.2k x 3 +0.7 x 7 = 0
Ma+8.5=0
Ma=-8.5
I put a = 0 behind my equation and I switched it over, hence I got a negative 8.5

Formula for question

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All my 3 moments are wrong?

freshbox said:
All my 3 moments are wrong?
Moment about A is correct...what does the minus sign mean?
I have no idea what you are doing when summing moments about B and C. The moment at A is an external reaction moment. Elsewhere , moments ate internal in the beam and can be calculated using free body diagrams or shear and moment diagrams.
EDIT: o I think I see now what you are doing...you are messing up your plus and minus signs. These moments are more easily calculated using right hand sections.

Last edited:
For my moment about A my minus means that it is -8.5Nm CCW

whereas for moment B and C, let's start with B, I am trying to take the FBD LHS.

I have attached a picture showing the example of what I am trying to do. The question in this picture is different. I am actually "copying" the steps to help me solve my #1 problem.

I am confused right now

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How I work out C post #1 is based on this question.Taking clockwise as +ve

Ma-5k x 1.2 = 0
Ma=-6kN CCW

Mb+5 x1.2 -6k =0 (I added -6kN which is Moment A into this equation)
Mb =0

So for post #1 working
Mc=1.2k x 4 +1.9k x 7 -8.5k=0 (I added the -8.5k Moment A which I found earlier into this equation)
Mc=-117.85Nm

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You are hopelessly confusing yourself by first not drawing a free-body diagram of your beam. Currently, you are just calculating random moments, adding or subtracting them at will, and hoping you get the right answer somehow.

Draw the FBD of the cantilever and find the reacting force and moment at A. If you can do that, then further help will be forthcoming. Right now, you are just spinning your wheels, and mixing these other problems in with the original problem is not helpful.

I refer to other example because I tried to solve the problem as I study alone and have nobody to ask

This is my working.. All moment I got are the same. So referring to post #4, which point has the largest moment?

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That's just the point. If you follow the hint given in the problem and draw the bending moment diagram of the beam, the location of the maximum moment is immediately obvious.

The beam is restrained at point A. Write your equations of static equilibrium using point A as the reference for the moments. Calculating moments about points B and C is a waste of time because the beam is not restrained at those points, and this is keeping you from proceeding with a solution.

I would like to ask can I just calculate the moment for each point A,B,C manually without drawing the BMD? As shown in post #10.

And what do you mean by "Write your equations of static equilibrium using point A as the reference for the moments."

Thank you.

In order to determine the reaction force and moment applied at A which keeps the free body of the beam in static equilibrium, you write equations which contain the unknown reactions. The beam remains in equilibrium as long as the following equations are satisfied:

∑F = 0

For this particular beam, let's say there is an unknown reaction R$_{A}$ and unknown moment M$_{A}$ at point A. The equations of equilibrium for the beam will be:

[+ forces are pointing up; + moments are counterclockwise]

∑F = R$_{A}$ - 1.2 - 0.7 = 0 [kN]

∑M about A = M$_{A}$ -1.2 * 3 - 0.7 * 7 = 0 [kN-m]

Now, it should be easy to solve these two equations for R$_{A}$ and M$_{A}$.

Once you have done that, you can then construct the shear force and bending moment diagram for the beam.

Yes..I already calculated Ra=1900N and based on the SFD and BMD, I have found that
Ma=-8500Nm
Mb=-2800Nm
Mc=0Nm

I would like to ask without drawing the BMD, can I calculate each Moment individually?

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Your shear force diagram looks OK, but the value of the bending moment at B is not -2800 Nm, and the bending moment does not go to zero between points B and C.

You use the value of the moment at A and the area under the shear force diagram to calculate the value of the bending moment at points B and C.

For example, the BM at point B = MA + Area of the shear force diagram from A to B

You should have obtained MA = 8500 Nm from the equilibrium equations, rather than -8500 Nm.
The reaction moment at A is CCW, given the direction of the the forces on the beam.

Can you calculate the bending moment values at B and C now and correct the bending moment diagram?

I am really confused on why Moment A is not - 8500Nm CW.. Can you please take a look at post #13 again? I set my orientation as +ve CW

This is the new BMD, but it don't make any sense to me. I thought for BMD you always start at 0 and end at 0. For this BMD, I start at 0 but end at 17240. Whereas for the old BMD, I start and end at 0.

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freshbox said:
I am really confused on why Moment A is not - 8500Nm CW.. Can you please take a look at post #13 again? I set my orientation as +ve CW

This is where we differ. In Post #16, I used CCW as +moments for the equilibrium equations.

freshbox said:
This is the new BMD, but it don't make any sense to me. I thought for BMD you always start at 0 and end at 0. For this BMD, I start at 0 but end at 17240. Whereas for the old BMD, I start and end at 0.

Your new bending moment diagram is incorrect because you have taken all moments as having the same sign. The moment at A must have the opposite direction of the moments at B and C, otherwise the beam cannot be in equilibrium.

So I must follow your orientation as CCW +ve, CW as -ve for all the questions if not I'm wrong? I thought we can set out own orientation but must follow accordingly...

And I thought you told me that my Ma should be +ve 8500Nm..

Thank you.

freshbox said:
So I must follow your orientation as CCW +ve, CW as -ve for all the questions if not I'm wrong? I thought we can set out own orientation but must follow accordingly...

And I thought you told me that my Ma should be +ve 8500Nm..

Thank you.

No, you don't have to follow my sign convention, but you must follow a consistent convention if you hope to solve the problem. The equilibrium equations I included in Post #16 are based on the sign convention as disclosed in that post. If another sign convention is chosen, then the signs of the forces and moments in those equations must be adjusted accordingly.

The reaction moment must be the opposite sign of the applied moments, otherwise, the beam cannot be in static equilibrium. That's why your last attempt at drawing the bending moment diagram was unsuccessful.

But for post #13 I did follow my sign convention consistently and why my answer is still wrong? Can you tell me which part is wrong?

Can you tell me what is the correct value for the BMD?

Thank you..

freshbox said:
But for post #13 I did follow my sign convention consistently and why my answer is still wrong? Can you tell me which part is wrong?

Can you tell me what is the correct value for the BMD?

Thank you..

I showed you how to set up the equilibrium equations in Post #16, and I explained how you obtain the bending moment from the shear force diagram in Post #18. You made additional calculation mistakes in the posts after #18 (like having the moments due to the applied forces and the reaction moment at A all having the same sign), which is why you haven't progressed in your solution.

I've gone about as far as I can under the HW help rules at PF. You must show me some additional attempts at correcting your mistakes, and I will be happy to review them and provide further comment.

I still don't understand how I got my calculation wrong for moment A

Taking positive as +ve

Ma+(1200x3)+(700x7)=0
Ma+8500=0
Ma=-8500Nm CCW

freshbox said:
I still don't understand how I got my calculation wrong for moment A

Taking positive as +ve

Ma+(1200x3)+(700x7)=0
Ma+8500=0
Ma=-8500Nm CCW
I told you that was right many posts ago, but i guess you were not paying attention. Continue. The moments at B and C are more easily calculated using free body diagrams of the right hand section. Sorry to be repetitive. Please watch your signage and your math! And your BMD is wrong.

I would like to extend my sincere thanks to PhantomJay for taking his time to explained where I gone wrong.

## 1. What is the significance of finding the moment of point B?

The moment of point B is important because it helps us understand the rotational motion and stability of an object. It tells us the magnitude and direction of the force necessary to cause an object to rotate around point B.

## 2. How is the moment of point B calculated?

The moment of point B is calculated by multiplying the magnitude of the force applied to the object by the perpendicular distance from the point of rotation (point B) to the line of action of the force. This can be expressed mathematically as M = F x d, where M is the moment, F is the force, and d is the distance.

## 3. Is the moment of point B always the largest?

No, the moment of point B is not always the largest. It depends on the location of the force and the distance from the point of rotation. If the force is applied closer to the point of rotation, the moment will be smaller. The largest moment occurs when the force is applied perpendicular to the line of action at the maximum distance from the point of rotation.

## 4. How does the moment of point B affect the stability of an object?

The moment of point B is directly related to the stability of an object. A larger moment of point B means the object is less stable and more prone to tipping over. On the other hand, a smaller moment of point B indicates greater stability and resistance to tipping.

## 5. Can the moment of point B be negative?

Yes, the moment of point B can be negative. This occurs when the force is applied in the opposite direction of the rotation. A negative moment means that the object is more likely to rotate in the opposite direction and is not stable. It is important to consider the sign of the moment in analyzing the stability of an object.

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