Shear force and bending moments graph

[physea]

Hello

I am a bit confused how to calculate the shear force and bending moment graph for a beam under point loads, distributed loads and bending moments.

Is the discontinuous functions the best method to solve such problems?

Can anyone outline an approach?

I don't understand for example how to divide the loads in regions etc.

Thanks

 

[Chestermiller]

Hello

I am a bit confused how to calculate the shear force and bending moment graph for a beam under point loads, distributed loads and bending moments.

Is the discontinuous functions the best method to solve such problems?

Can anyone outline an approach?

I don't understand for example how to divide the loads in regions etc.

Thanks

Can you please provide a specific problem that we can focus on, so it will become clearer what your issues are?

 

[physea]

In this example, the equations are as below. How do I make the graphs out of these equations ie how do I find the right function for each zone?

 

[Chestermiller]

You start out by determining the reaction force at A. What is that equal to, and what is its direction (upwards or downward)?

Next, you make a cut in the beam at location x between A and B. You then remove the portion of the beam to the right of x, and apply an equivalent shear force on the cross section at x so that the portion to the left of the cut is still in force equilibrium. What is the magnitude and direction of that shear force?

 

[physea]

You start out by determining the reaction force at A. What is that equal to, and what is its direction (upwards or downward)?

Next, you make a cut in the beam at location x between A and B. You then remove the portion of the beam to the right of x, and apply an equivalent shear force on the cross section at x so that the portion to the left of the cut is still in force equilibrium. What is the magnitude and direction of that shear force?

I want to know how to use the discontinuity functions in the second image to create the shear and bending moment graphs.

I know the first step to do that is to divide the beam into regions. I know the regions are:
0-2.5
2.5-5
5-10
But what's next?

 

[Chestermiller]

I want to know how to use the discontinuity functions in the second image to create the shear and bending moment graphs.

I know the first step to do that is to divide the beam into regions. I know the regions are:
0-2.5
2.5-5
5-10
But what's next?

The next step is to answer the handful of questions I asked. Do you think you can do that?

 

[physea]

The next step is to answer the handful of questions I asked. Do you think you can do that?

I am not stuck at that point, but further down, but anyway.

Something else: Can you tell me what are the boundary conditions for the bending moment function between B and C?
What is the moment in B and C? I know the moment with centre A, is 170Nm. I know the bending equation in B<x<C is in the form of ax+b but I don't know a and b.
I am having difficulty on how to find the moments in the section B-C. Any hint?

FYI so that you won't ask me again, these are the reactions forces and reaction moments (28kN and 170Nm):

 

[Chestermiller]

I am not stuck at that point, but further down, but anyway.

Something else: Can you tell me what are the boundary conditions for the bending moment function between B and C?
What is the moment in B and C? I know the moment with centre A, is 170Nm. I know the bending equation in B<x<C is in the form of ax+b but I don't know a and b.
I am having difficulty on how to find the moments in the section B-C. Any hint?

FYI so that you won't ask me again, these are the reactions forces and reaction moments (28kN and 170Nm):

You are being very uncooperative, so I'm going to give you the shear force variation:
$V=28\ kN$ for $(0<x<2.5)$
$V=20\ kN$ for $(2.5<x<5)$
$V=20-4(x-5)\ kN$ for $(5<x<10)$

 

[physea]

You are being very uncooperative, so I'm going to give you the shear force variation:
$V=28\ kN$ for $(0<x<2.5)$
$V=20\ kN$ for $(2.5<x<5)$
$V=20-4(x-5)\ kN$ for $(5<x<10)$

Thanks but I asked for the moments in 2.5<x<5?

 

[Chestermiller]

Thanks but I asked for the moments in 2.5<x<5?

Well, now that you know V vs x, and you know the moment at A, you should be able to determine this. So, first, what would you get for the moment variation with x in the region (0<2.5<x)?

 

[physea]

Well, now that you know V vs x, and you know the moment at A, you should be able to determine this. So, first, what would you get for the moment variation with x in the region (0<2.5<x)?

Following the method of graphs, we integrate the V(x) so for 0<x<2.5, it is:
M(x) = 28x+C
At x=0, we have M=170, so the above becomes:
M(x)=28x+170

But what about 2.5<x<5?
Again we integrate V(x), so it is:
M(x) = 20x+D
But what are the boundary conditions for M in 2.5<x<5? (that was my initial question)

 

[physea]

Something else that I am a bit confused is the signs conventions, so can you correct the below appropriately please:

To calculate ∑F and ∑M, positive are the downwards point forces and distributed loads and the anticlockwise moments (signs are reversed when the beam is right supported)

To use Graphs Method, positive is upwards distributed loads, downwards point forces and anticlockwise moments (reverse in right supported beam)

To use Discontinuity Functions Method, positive is upwards distributed loads, upwards point forces and clockwise moments (reverse in right supported beam)

This is the mess I read on my hideous lecture notes and I wonder if this is accurate?

 

[Chestermiller]

Following the method of graphs, we integrate the V(x) so for 0<x<2.5, it is:
M(x) = 28x+C
At x=0, we have M=170, so the above becomes:
M(x)=28x+170

Actually, the way you have drawn the moment in post #7, the equation for the moment should read $M(x)=170-28x$. At x = 2.5, the moment is M(2.5)=100

But what about 2.5<x<5?
Again we integrate V(x), so it is:
M(x) = 20x+D
But what are the boundary conditions for M in 2.5<x<5? (that was my initial question)

The moment is continuous across 2.5 (since there is not a concentrated moment applied at 2.5). So, in this region, $M(x)=100-20(x-2.5)=150-20x$

 

[Chestermiller]

Something else that I am a bit confused is the signs conventions, so can you correct the below appropriately please:

To calculate ∑F and ∑M, positive are the downwards point forces and distributed loads and the anticlockwise moments (signs are reversed when the beam is right supported)

To use Graphs Method, positive is upwards distributed loads, downwards point forces and anticlockwise moments (reverse in right supported beam)

To use Discontinuity Functions Method, positive is upwards distributed loads, upwards point forces and clockwise moments (reverse in right supported beam)

This is the mess I read on my hideous lecture notes and I wonder if this is accurate?

Rules like these drive me crazy. They are too much to remember. So, in each individual problem, I re-derive the relationships that apply to that particular situation. I have very little value for rules like these.

 

[physea]

Rules like these drive me crazy. They are too much to remember. So, in each individual problem, I re-derive the relationships that apply to that particular situation. I have very little value for rules like these.

Same with me, but how do you re-derive them since they should be the starting point of reference?

 

[Chestermiller]

Same with me, but how do you re-derive them since they should be the starting point of reference?

They are not the starting points. They were derived from force and moment balances. So I just do the differential force and moment balances for the specific system I'm considering.

 

[physea]

They are not the starting points. They were derived from force and moment balances. So I just do the differential force and moment balances for the specific system I'm considering.

So what steps you take? You choose arbitrarily a sign for eg downwards point forces?
Then you use the same sign for downwards distributed loads?
That way, you can expect their sum to be opposite sign of the total reaction.

But what about moments? How do you sign moments so that they have the correct sign?

 

[Chestermiller]

So what steps you take? You choose arbitrarily a sign for eg downwards point forces?
Then you use the same sign for downwards distributed loads?
That way, you can expect their sum to be opposite sign of the total reaction.

But what about moments? How do you sign moments so that they have the correct sign?

Draw a free body diagram of the section of the beam between x = 0 and arbitrary x (<2.5) showing the shear forces and moments acting on the section. Let's see what your diagram looks like.

 

[physea]

Can I ask something please, you said it doesn't make any difference which sign you use for forces and moments, as long as you are consistent and you give opposite sign to opposite directions, right?

If you choose to use + for forces going upwards, do you need to use a specific sign for clockwise or anticlockwise moments? Or you choose them arbitrarily?

Thanks

 

[Chestermiller]

Can I ask something please, you said it doesn't make any difference which sign you use for forces and moments, as long as you are consistent and you give opposite sign to opposite directions, right?

If you choose to use + for forces going upwards, do you need to use a specific sign for clockwise or anticlockwise moments? Or you choose them arbitrarily?

Thanks

I don't know how to answer this. I always derive the equations for the shear forces and moments from scratch (as it pleases me aesthetically on that day), and don't follow any specific conventions. Maybe this isn't very efficient, but it helps me avoid making mistakes.

 

[physea]

Also, something else please (this topic drives me mad as there is no COMPREHENSIVE approach).

In this graph, you have Sum(M at A) = 0 so 3*20 + Rc*8 = 0.

In this graph, you have again Sum(M at B) = 0 but Mb + 3*1 + 2*4 = 0.

Why in the first instance we don't say Ma, while in the second instance we do say Mb?

 

[Chestermiller]

Also, something else please (this topic drives me mad as there is no COMPREHENSIVE approach).

In this graph, you have Sum(M at A) = 0 so 3*20 + Rc*8 = 0.
View attachment 225031

In this graph, you have again Sum(M at B) = 0 but Mb + 3*1 + 2*4 = 0.
View attachment 225032

Why in the first instance we don't say Ma, while in the second instance we do say Mb?

Are you asking why, in A, there is no moment at A?

 

[physea]

Are you asking why, in A, there is no moment at A?

Exactly.

 

[Chestermiller]

Exactly.

In A, the beam is mounted on a swivel which is incapable of applying a moment. In B, the beam is built into the wall (cantilevered), which can apply a moment to the beam.

 

[physea]

In A, the beam is mounted on a swivel which is incapable of applying a moment. In B, the beam is built into the wall (cantilevered), which can apply a moment to the beam.

1) So only in fixed/built-in supports we have a reaction moment?
2) I am not sure why in swivel you cannot get a moment, isn't it that the beam can still rotate around the swivel?

 

[Chestermiller]

1) So only in fixed/built-in supports we have a reaction moment?
2) I am not sure why in swivel you cannot get a moment, isn't it that the beam can still rotate around the swivel?

1. If anything constrains the beam to prevent it from freely rotating at a given location, it can exert a moment on the beam at that location.

2. A beam being able to rotate is not a criterion for a moment to exist (where did you get that idea?); a moment will exist if the beam is constrained from rotating.