Moments -- sign convention in beam

[fonseh]

Homework Statement

In this question , we can see that in teh second picture , the 12.86kNm is in clockwise direction at the right of span BC .

Homework Equations

The Attempt at a Solution

In meachnic of materials , i learned that when the beam curve upwards ( U shape) , then it has positive bending moment ... So , I agree that the 1.54kNm clockwise rotation on the left of span AB has POSITIVE moment ... But , i don't agee that the 12.86kNm is in clockwise direction at the right of span BC has POSTIVE value ( it's -12.86kNm on the left , so POSITIVE 12.86kNm is added to bring the moment to 0 at C) ... I think the -12.86kNm on the right of span BC will lead to negative moment

 

[PhanthomJay]

Homework Statement

In this question , we can see that in teh second picture , the 12.86kNm is in clockwise direction at the right of span BC .

Homework Equations

The Attempt at a Solution

In meachnic of materials , i learned that when the beam curve upwards ( U shape) , then it has positive bending moment ... So , I agree that the 1.54kNm clockwise rotation on the left of span AB has POSITIVE moment ... But , i don't agee that the 12.86kNm is in clockwise direction at the right of span BC has POSTIVE value ( it's -12.86kNm on the left , so POSITIVE 12.86kNm is added to bring the moment to 0 at C) ... I think the -12.86kNm on the right of span BC will lead to negative moment

Yes, the ends are both fixed and at the left end the deflected curve is upwards, or positive moment by convention, and at the right end the deflected curve is downward so the moment is negative. So the moment is -12.86 at the right end.
Note when drawing the moment diagram, fixed end moments and applied couples , when acting clockwise , add to the moment diagram as a positive value. That is why at C the cw moment of 12.86 is added as a plus to bring the moment to 0 a hair to the right of C. But by convention, the fixed end moment at C is indeed a negative 12.86. It can get confusing, for sure.

 

[fonseh]

Note when drawing the moment diagram, fixed end moments and applied couples , when acting clockwise , add to the moment diagram as a positive value. That is why at C the cw moment of 12.86 is added as a plus to bring the moment to 0 a hair to the right of C.

Do you mean for the clockwise moment 12.86kNm has to become counterclockwise 12.86kNm moment when drawing moment diagram ? This will lead to positive 12086kNm ?

If it's so , why we have to do so ? Why we have to make the 12.86kNm clockwise moment to become couterclockwise when drawing moment diagram ?

 

[PhanthomJay]

Do you mean for the clockwise moment 12.86kNm has to become counterclockwise 12.86kNm moment when drawing moment diagram ? This will lead to positive 12086kNm ?

No.

If it's so , why we have to do so ? Why we have to make the 12.86kNm clockwise moment to become counterclockwise when drawing moment diagram ?

By convention, clockwise moments are negative when acting on a section to the left of it, and clockwise moments are positive when acting on a section to the right of it. Similarly, counterclockwise moments are positive when acting on a section to the left of it, and counterclockwise moments are negative when acting on a section to the right of it. so in your case, the 12.86 moment acts clockwise on a section to the left of it, and is thus negative. Rather confusing I know, so it is always a good idea to roughly sketch in the deflected shape of the beam, and if it faces upward it is a positive moment in that region, and when it faces downward, it is a negative moment in that region.
Now when drawing your moment diagram from left to right, the fixed end moments point upward when clockwise, and downward when counterclockwise.
If you are still confused, don't be, because your last sketch showing positive and negative moments is correct.

 

[fonseh]

No.
By convention, clockwise moments are negative when acting on a section to the left of it, and clockwise moments are positive when acting on a section to the right of it. Similarly, counterclockwise moments are positive when acting on a section to the left of it, and counterclockwise moments are negative when acting on a section to the right of it. so in your case, the 12.86 moment acts clockwise on a section to the left of it, and is thus negative. Rather confusing I know, so it is always a good idea to roughly sketch in the deflected shape of the beam, and if it faces upward it is a positive moment in that region, and when it faces downward, it is a negative moment in that region.
Now when drawing your moment diagram from left to right, the fixed end moments point upward when clockwise, and downward when counterclockwise.
If you are still confused, don't be, because your last sketch showing positive and negative moments is correct.

As we draw the BENDING MOMENT DIAGRAM from left to right , for FIXED END moment , when it's in clockwise direction , it will bend the beam upwards , so , it's positive ?

Conversely , if the FIXED END moment is in anticlockwise direction , it will bend the beam downwards , so , it's negative?

 

[fonseh]

By convention, clockwise moments are negative when acting on a section to the left of it, and clockwise moments are positive when acting on a section to the right of it. Similarly, counterclockwise moments are positive when acting on a section to the left of it, and counterclockwise moments are negative when acting on a section to the right of it.

This applied to the moment 12.84kNm only because it's the couple as you mentioned in post#2 ?

 

[fonseh]

No.
By convention, clockwise moments are negative when acting on a section to the left of it, and clockwise moments are positive when acting on a section to the right of it. Similarly, counterclockwise moments are positive when acting on a section to the left of it, and counterclockwise moments are negative when acting on a section to the right of it. so in your case, the 12.86 moment acts clockwise on a section to the left of it, and is thus negative. Rather confusing I know, so it is always a good idea to roughly sketch in the deflected shape of the beam, and if it faces upward it is a positive moment in that region, and when it faces downward, it is a negative moment in that region.
Now when drawing your moment diagram from left to right, the fixed end moments point upward when clockwise, and downward when counterclockwise.
If you are still confused, don't be, because your last sketch showing positive and negative moments is correct.

Now when drawing your moment diagram from left to right, the fixed end moments point upward when clockwise, and downward when counterclockwise.

This is same as the when drawing the moment diagram, fixed end moments and applied couples , when acting clockwise , add to the moment diagram as a positive value?

 

[PhanthomJay]

As we draw the BENDING MOMENT DIAGRAM from left to right , for FIXED END moment , when it's in clockwise direction , it will bend the beam upwards , so , it's positive ?

The Fixed End Moments do not bend the beam; the loading on the beam between supports causes the bending.

This is same as the when drawing the moment diagram, fixed end moments and applied couples , when acting clockwise , add to the moment diagram as a positive value?

What I mean is that when drawing the moment diagram starting at the left, as in the example bending moment diagram, at the left support there is a clockwise fixed end moment , 1.54, so you start at 0 and add 1.54 upward to the diagram, then as you proceed to the end you get 12.86 negative moment (- 12.86), and to close the diagram so to speak you add the FEM of 12.86 upward because it is clockwise.

 

[fonseh]

FEM of 12.86 upward because it is clockwise.

it is positive 12.86kNm because the clockwise moment will make the left portion of he beam to curve upwards , while the right end in fixed ? Can i explain in this way ??

 

[PhanthomJay]

it is positive 12.86kNm because the clockwise moment will make the left portion of he beam to curve upwards , while the right end in fixed ? Can i explain in this way ??

No you cannot. The 12,86 moment is NEGATIVE if you read my prior posts. The section immediately to the left of it curves downward.

 

[fonseh]

By convention, clockwise moments are negative when acting on a section to the left of it, and clockwise moments are positive when acting on a section to the right of it. Similarly, counterclockwise moments are positive when acting on a section to the left of it, and counterclockwise moments are negative when acting on a section to the right of it. so in your case, the 12.86 moment acts clockwise on a section to the left of it, and is thus negative.

in the previous post , you said my sign convention curve u shape upwards is positive is correct (refer to photo) ? And your current comment stated that when the beam curve 'n' shape is positive ?

Can you clarify it ?

 

[PhanthomJay]

Where did I say that? Upwards U shape is pos mom and upside downU shape is neg mom

 

[fonseh]

Upwards U shape is pos mom and upside downU shape is neg mom

what are you saying ?

 

[fonseh]

By convention, clockwise moments are negative when acting on a section to the left of it, and clockwise moments are positive when acting on a section to the right of it. Similarly, counterclockwise moments are positive when acting on a section to the left of it, and counterclockwise moments are negative when acting on a section to the right of it.

What you mean is like this ? The red part has positive sign convention ?

But , i also remembered you said that my sign convention positive is correct when the beam bent upwrads ( green part) in post 2 ?

So , which beam has positive sign convention ? bent upwards or bend downwards ?

 

[PhanthomJay]

What you mean is like this ? The red part has positive sign convention ?

But , i also remembered you said that my sign convention positive is correct when the beam bent upwrads ( green part) in post 2 ?

So , which beam has positive sign convention ? bent upwards or bend downwards ?

As you noted in your very first post attachment 621.png, bent upwards is positive and bent downwards is negative. I thought that was already established. You will also note in that attachment that clockwise moments acting on a section to the right of it, and counterclockwise moments acting on a section to the left of it, are positive moments. You also showed this correctly. The sketch is 100 percent correct.
You can confirm this by looking at figure 11-10 (b). Notice at the extreme left , the beam bends upwards initially, and the moment M_AB is clockwise acting on he section to the right of it, thus, this agrees fully that the moment M_AB is positive. And notice at the extreme right, the beam bends downward initially, and the moment M_CB is clockwise acting on the section to the left of it, thus, this agrees fully that the moment M_CB is negative.

Is this clear now?

 

[fonseh]

You will also note in that attachment that clockwise moments acting on a section to the right of it, and counterclockwise moments acting on a section to the left of it, are positive moments

No , i show that in that attachment that clockwise moments acting on a section to the right of it, and counterclockwise moments acting on a section to the left of it, are NEGATIVE moments , (curve downwards)

 

[fonseh]

As you noted in your very first post attachment 621.png, bent upwards is positive and bent downwards is negative. I thought that was already established. You will also note in that attachment that clockwise moments acting on a section to the right of it, and counterclockwise moments acting on a section to the left of it, are positive moments. You also showed this correctly. The sketch is 100 percent correct.
You can confirm this by looking at figure 11-10 (b). Notice at the extreme left , the beam bends upwards initially, and the moment M_AB is clockwise acting on he section to the right of it, thus, this agrees fully that the moment M_AB is positive. And notice at the extreme right, the beam bends downward initially, and the moment M_CB is clockwise acting on the section to the left of it, thus, this agrees fully that the moment M_CB is negative.

Is this clear now?

Can you explain how the negative moment -12.86kNm is brought to 0 at the extreme right end ?

 

[PhanthomJay]

No , i show that in that attachment that clockwise moments acting on a section to the right of it, and counterclockwise moments acting on a section to the left of it, are NEGATIVE moments , (curve downwards)

What we have here is a failure to communicate. The attachment shows that clockwise moments acting on a section to the LEFT of that moment is NEGATIVE. Please look at the lower sketch in your attachment, the one showing the upside down U deflected shape of the beam. We know that moments are negative in this section. And the clockwise moment at the far right acts on a section to its left. I guess you mean that the clockwise moment lies to the right of the section. Whatever is the way to describe it in words, the sketch is correct.

Can you explain how the negative moment -12.86kNm is brought to 0 at the extreme right end ?

Well when drawing moment diagrams, first off, the fixed end (concentrated couple) moments (FEM) at either end don't in theory go to zero, they are what they are, that is, in this example, at the right end, the FEM is negative 12.86 kN-m. In actuality, concentrated moments , just like concentrated point loads, don't exist, because they are applied over zero width, which leads to the infinities that nature despises. But anyway, let's assume the moments start and end at 0. As you draw the moment diagram (after first drawing the shear diagram) from left to right, when you get to the right end , the moment in the beam is -12.86, then in order to close it to 0, you add in the clockwise FEM of 12.86 to get -12.86 + 12.86 = 0. Notice here that when drawing moment diagrams , clockwise moments are always considered positive, and this is unrelated to our previous discussion on moment signage .
Consider a simple cantilever beam 2 m in length and subject to a downward point load of 10 N at the free end. The free end is at the left of the beam and the fixed end is at the right of the beam. The FEM is 20 N-m clockwise and the beam section it acts on is to its left, thus a negative moment, and of course the beam deflects downward as an upside down 'U' shape of sorts. Now in drawing the moment diagram. starting from the left, the moment in the beam is 0 at the free end, and continues sloping downward to -20 N-m at the fixed end, then since the FEM of 20 N-m is clockwise, you draw a straight line upward back to 0.

Okay?

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[fonseh]

What we have here is a failure to communicate. The attachment shows that clockwise moments acting on a section to the LEFT of that moment is NEGATIVE. Please look at the lower sketch in your attachment, the one showing the upside down U deflected shape of the beam. We know that moments are negative in this section. And the clockwise moment at the far right acts on a section to its left. I guess you mean that the clockwise moment lies to the right of the section. Whatever is the way to describe it in words, the sketch is correct.

Can you explain how the negative moment -12.86kNm is brought to 0 at the extreme right end ?

 

[fonseh]

so in your case, the 12.86 moment acts clockwise on a section to the left of it, and is thus negative.

Can you explain this one ? I noticed that the clockwise 12.86kNm clockwise moment acting to the right of span ...

 

[PhanthomJay]

Can you explain how the negative moment -12.86kNm is brought to 0 at the extreme right end ?

see post 18

 

[PhanthomJay]

Can you explain this one ? I noticed that the clockwise 12.86kNm clockwise moment acting to the right of span ...

You are posting faster than I can respond, so we're getting messed up here.

Yep, the clockwise moment acts to the right of the span, therefore, the span is to the left of the moment, same thing, moment is negative, sir.

 

[fonseh]

As you draw the moment diagram (after first drawing the shear diagram) from left to right, when you get to the right end , the moment in the beam is -12.86, then in order to close it to 0, you add in the clockwise FEM of 12.86 to get -12.86 + 12.86 = 0. Notice here that when drawing moment diagrams , clockwise moments are always considered positive, and this is unrelated to our previous discussion on moment signage .

Do you mean that the moment clockwise 12.86kNm is acting to the left of span ? So , to become 0 , anticlockwise moment 12.86kNm is required to bring it to 0 ?

 

[fonseh]

Consider a simple cantilever beam 2 m in length and subject to a downward point load of 10 N at the free end. The free end is at the left of the beam and the fixed end is at the right of the beam. The FEM is 20 N-m clockwise and the beam section it acts on is to its left, thus a negative moment, and of course the beam deflects downward as an upside down 'U' shape of sorts. Now in drawing the moment diagram. starting from the left, the moment in the beam is 0 at the free end, and continues sloping downward to -20 N-m at the fixed end, then since the FEM of 20 N-m is clockwise, you draw a straight line upward back to 0.

Okay?

sorry , why for clockwise fixed end moment is positive ?

 

[fonseh]

Notice here that when drawing moment diagrams , clockwise moments are always considered positive, and this is unrelated to our previous discussion on moment signage .

I'm confused of the sign convention of fixed end moment , can you explain it here ?

 

[PhanthomJay]

I'm confused of the sign convention of fixed end moment , can you explain it here ?

I am sorry you are still confused. I'm probably making it worse for you.
The issue may be this:
There are 2 different sign conventions,
-- one for external moments, and
-- one for internal moments.

Let's look at the internal moments first. Internal moments exist within the beam itself and are caused by the applied loadings.. The sign of those moments have been I hope well established already in the prior posts, that is to say, internal moments that cause the beam to deflect concave upward (like a smiley face) are positive, while internal moments that cause the beam to deflect concave downward (upside down U) are negative. The cw or ccw direction of these moments are as established in your famous third attachment on post 1. These signs by the way...plus or minus...will come directly from your moment diagram without having to look at the curved shape of the deflected beam...provided you draw the moment diagram correctly, however.

So speaking of the moment diagram , now let's look at the external moments. External moments are either directly applied moments (couples) or fixed end support reaction moments (FEM). Both types of these external moments have the convention of positive if they are clockwise, or negative if they are counterclockwise. This is unlike the internal moments where clockwise moments can be plus or minus depending on which side of the beam section you are looking at. But again, clockwise external moments are always positive.

So in conclusion, looking your original problem, the internal moment in the beam at the right end is clockwise on a left hand section and thus a negative 12.86, and since the FEM is an external clockwise moment, it is a plus 12.86, which closes your moment diagram to 0.

Well , whaddya think?

 

[fonseh]

I am sorry you are still confused. I'm probably making it worse for you.
The issue may be this:
There are 2 different sign conventions,
-- one for external moments, and
-- one for internal moments.

Let's look at the internal moments first. Internal moments exist within the beam itself and are caused by the applied loadings.. The sign of those moments have been I hope well established already in the prior posts, that is to say, internal moments that cause the beam to deflect concave upward (like a smiley face) are positive, while internal moments that cause the beam to deflect concave downward (upside down U) are negative. The cw or ccw direction of these moments are as established in your famous third attachment on post 1. These signs by the way...plus or minus...will come directly from your moment diagram without having to look at the curved shape of the deflected beam...provided you draw the moment diagram correctly, however.

So speaking of the moment diagram , now let's look at the external moments. External moments are either directly applied moments (couples) or fixed end support reaction moments (FEM). Both types of these external moments have the convention of positive if they are clockwise, or negative if they are counterclockwise. This is unlike the internal moments where clockwise moments can be plus or minus depending on which side of the beam section you are looking at. But again, clockwise external moments are always positive.

So in conclusion, looking your original problem, the internal moment in the beam at the right end is clockwise on a left hand section and thus a negative 12.86, and since the FEM is an external clockwise moment, it is a plus 12.86, which closes your moment diagram to 0.

Well , whaddya think?

ok , how do we know that whether MAB , MBA , MBC and MCB are internal or extrenal moment ?

 

[fonseh]

I am sorry you are still confused. I'm probably making it worse for you.
The issue may be this:
There are 2 different sign conventions,
-- one for external moments, and
-- one for internal moments.

Let's look at the internal moments first. Internal moments exist within the beam itself and are caused by the applied loadings.. The sign of those moments have been I hope well established already in the prior posts, that is to say, internal moments that cause the beam to deflect concave upward (like a smiley face) are positive, while internal moments that cause the beam to deflect concave downward (upside down U) are negative. The cw or ccw direction of these moments are as established in your famous third attachment on post 1. These signs by the way...plus or minus...will come directly from your moment diagram without having to look at the curved shape of the deflected beam...provided you draw the moment diagram correctly, however.

So speaking of the moment diagram , now let's look at the external moments. External moments are either directly applied moments (couples) or fixed end support reaction moments (FEM). Both types of these external moments have the convention of positive if they are clockwise, or negative if they are counterclockwise. This is unlike the internal moments where clockwise moments can be plus or minus depending on which side of the beam section you are looking at. But again, clockwise external moments are always positive.

So in conclusion, looking your original problem, the internal moment in the beam at the right end is clockwise on a left hand section and thus a negative 12.86, and since the FEM is an external clockwise moment, it is a plus 12.86, which closes your moment diagram to 0.

Well , whaddya think?

i have an example below .The beam has UDL load of 15kN/m , conc couple of 30.71kNm clockwise . for moment about A and moment about B , i let the concentrated couple clockwise couple 30.71KNm as positive , but i didnt get the ans . So , for the sign convention of the clockwise concentrated couple is positive only applicable in drawing bending moment diagram only ? But , not when calculating the forces using equilibrium equation ?

Here's actually a part of working in my book . So , i just take the certain part only . Not the whole question .
When i made another attempt , i gt moment about A = RB(5) -30.71 - 15(5)(2.5) = 0 , RB = 43.5kN upwards , i gt the ans

 

[PhanthomJay]

ok , how do we know that whether MAB , MBA , MBC and MCB are internal or extrenal moment ?

External moments are the applied moments or the reaction at support moments, So only MAB and MCB are external, everything else is internal within the beam.

i have an example below .The beam has UDL load of 15kN/m , conc couple of 30.71kNm clockwise . for moment about A and moment about B , i let the concentrated couple clockwise couple 30.71KNm as positive , but i didnt get the ans . So , for the sign convention of the clockwise concentrated couple is positive only applicable in drawing bending moment diagram only ? But , not when calculating the forces using equilibrium equation ?

The positive sign convention of the clockwise concentrated couple is applicable in drawing bending moment diagram. When calculating the forces using equilibrium equation, you can choose clockwise or counterclockwise as plus or minus, but be consistent. I always recommend using clockwise moments as positive when determining force reactions.

Here's actually a part of working in my book . So , i just take the certain part only . Not the whole question .
When i made another attempt , i gt moment about A = RB(5) -30.71 - 15(5)(2.5) = 0 , RB = 43.5kN upwards , i gt the ans

Sure, your first attempt messed up the signage. Use clockwise moments as plus, whether from a 'force times distance moment' or applied moment or couple, and you'll get the right answer. Your equation summing moments about B is good! But when summing moments about A, it should be RB(-5) + 30.71 + 15(5)(2.5) = 0. RB = 43.64 up.

 

[fonseh]

Sure, your first attempt messed up the signage. Use clockwise moments as plus, whether from a 'force times distance moment' or applied moment or couple, and you'll get the right answer. Your equation summing moments about B is good! But when summing moments about A, it should be RB(-5) + 30.71 + 15(5)(2.5) = 0. RB = 43.64 up.

is my working of RB(5) -30.71 - 15(5)(2.5) = 0 incorrect ?

For moment about A , i have M(A) -RB(5) +30.71 + 15(5)(2.5) = 0 , thus , moment about A = RB(5) -30.71 - 15(5)(2.5) = 0 , RB = 43.64 up

I did in this way because for the left end of span , clockwise moment is positive ( cause the beam to bend upwards) , similarly , when finding moment about B , I assume anticlockwise as positive .(because anticlockwise moment causing the right span to bend upwards) Moment = M(B) -RA(5) + 30.71-15(5)(2.5) = 0 , 31.36 up

Or should I be consistent , keeping clockwise moment as positive and anticlockwise as negative or vice versa?

 

[PhanthomJay]

is my working of RB(5) -30.71 - 15(5)(2.5) = 0 incorrect ?

It is correct. Whenever you are summing moments about a point to find reaction forces, you can assume applied clockwise moments, and applied clockwise moments from forces, all as plus, or all as minus, and you get the same answer. I suggest however to assume clockwise moments as plus.

For moment about A , i have M(A) -RB(5) +30.71 + 15(5)(2.5) = 0 , thus , moment about A = RB(5) -30.71 - 15(5)(2.5) = 0 , RB = 43.64 up

I did in this way because for the left end of span , clockwise moment is positive ( cause the beam to bend upwards) ,

Forget about upwards or downwards in this step when determining reaction forces. Just use cw as plus and ccw as minus.

similarly , when finding moment about B , I assume anticlockwise as positive .(because anticlockwise moment causing the right span to bend upwards) Moment = M(B) -RA(5) + 30.71-15(5)(2.5) = 0 , 31.36 up

Or should I be consistent , keeping clockwise moment as positive and anticlockwise as negative or vice versa?

Once again, when determining reaction forces, keep it a simple set of rules: Consider clockwise moments, whether an applied moment couple or an applied 'force times distance moment about a point', as PLUS. And counterclockwise as MINUS.

I will try to sum up the signage rules later with a sketch, when I get a moment.

 

[fonseh]

It is correct. Whenever you are summing moments about a point to find reaction forces, you can assume applied clockwise moments, and applied clockwise moments from forces, all as plus, or all as minus, and you get the same answer. I suggest however to assume clockwise moments as plus. Forget about upwards or downwards in this step when determining reaction forces. Just use cw as plus and ccw as minus.Once again, when determining reaction forces, keep it a simple set of rules: Consider clockwise moments, whether an applied moment couple or an applied 'force times distance moment about a point', as PLUS. And counterclockwise as MINUS.

I will try to sum up the signage rules later with a sketch, when I get a moment.

https://www.physicsforums.com/threads/pinned-end-supported-span.912159/https://www.physicsforums.com/threads/moment-in-beam.912156/

Can you help me in this 2 threads ? It seems like you are the only one who is familiar with the structural engineering problem