Moment-area Triangle: Can I Redraw the -1000Nm Part?

[foo9008]

in this question , can i redraw the -1000Nm moment area triangle ? (red line part) is it wrong to do so ?

 

[foo9008]

If I do so, then my answer would be the same as the author's working, but is my concept correct?

 

[foo9008]

can i do the same thing for the moment -4800Nm??

 

[David Lewis]

The middle diagram with the red line, what is it? A shear diagram? A moment diagram? What does the horizontal black line represent?

 

[foo9008]

The middle diagram with the red line, what is it? A shear diagram? A moment diagram? What does the horizontal black line represent?

I am not sure what does the horizontal line represent
I think it's distance , and I rhink it's a moment -area diagram , where the vertical axis represent value of different moment , am i right ?

 

[David Lewis]

Good question. The horizontal line might be the beam itself? You always want the load, shear & moment curves to be a different line weight, line type, or color than the structural member in your drawing. And each diagram needs to be labelled (SFD = shear force diagram, BMD = bending moment diagram) so you know what you're looking at.

 

[foo9008]

Good question. The horizontal line might be the beam itself? You always want the load, shear & moment curves to be a different line weight, line type, or color than the structural member in your drawing. And each diagram needs to be labelled (SFD = shear force diagram, BMD = bending moment diagram) so you know what you're looking at.

Sorry, this is from my notes, the author didnt indicate whether the graph is bending moment diagram or shear force diagram, so you th ink what the graph represent? I think it represent the bending moment graph, am I right?

 

[David Lewis]

It's not a shear diagram. The SFD would look like this:
(x,V)
0, 1000 N
1, 600 N
2, 200 N
2, -300 N
3, -700 N
4, -700 N

 

[foo9008]

It's not a shear diagram. The SFD would look like this:
(x,V)
0, 1000 N
1, 600 N
2, 200 N
2, -300 N
3, -700 N
4, -700 N

Then, do u have any idea what the graph is? It's not moment diagram?

 

[PhanthomJay]

The graphs are moment diagrams by parts. You solve for the reactions first, then draw separate moment diagrams about a reference axis for each applied load or couple and each reaction. The total moment at any point in the beam is then the sum of the moments of each part. The reference point chosen in this example is the left end of the beam. There are 3 separate moment diagrams (one for the applied concentrated load, one for the reaction load, and one for the distributed load) and the author has opted to show 2 of then on the same graph. Now if instead you chose the right end of the beam as your reference axis, your red line would be correct for the concentrated load but then you would have to determine the moment of the left reaction and distributed load about the right end which are not the same as when moments are summed about the left end. Moment diagrams by parts is useful when calculating moment areas for deflection calculations, but otherwise i do not recommend this method if you just want to determine shears and moments in the beam, instead, use the conventional shear and moment diagram approach that you should become familiar with.

 

[foo9008]

The graphs are moment diagrams by parts. You solve for the reactions first, then draw separate moment diagrams about a reference axis for each applied load or couple and each reaction. The total moment at any point in the beam is then the sum of the moments of each part. The reference point chosen in this example is the left end of the beam. There are 3 separate moment diagrams (one for the applied concentrated load, one for the reaction load, and one for the distributed load) and the author has opted to show 2 of then on the same graph. Now if instead you chose the right end of the beam as your reference axis, your red line would be correct for the concentrated load but then you would have to determine the moment of the left reaction and distributed load about the right end which are not the same as when moments are summed about the left end. Moment diagrams by parts is useful when calculating moment areas for deflection calculations, but otherwise i do not recommend this method if you just want to determine shears and moments in the beam, instead, use the conventional shear and moment diagram approach that you should become familiar with.

so , the three moment area diagram are drawn using A as reference point ?

 

[foo9008]

The graphs are moment diagrams by parts. You solve for the reactions first, then draw separate moment diagrams about a reference axis for each applied load or couple and each reaction. The total moment at any point in the beam is then the sum of the moments of each part. The reference point chosen in this example is the left end of the beam. There are 3 separate moment diagrams (one for the applied concentrated load, one for the reaction load, and one for the distributed load) and the author has opted to show 2 of then on the same graph. Now if instead you chose the right end of the beam as your reference axis, your red line would be correct for the concentrated load but then you would have to determine the moment of the left reaction and distributed load about the right end which are not the same as when moments are summed about the left end. Moment diagrams by parts is useful when calculating moment areas for deflection calculations, but otherwise i do not recommend this method if you just want to determine shears and moments in the beam, instead, use the conventional shear and moment diagram approach that you should become familiar with.

if i take the reference point at B , the moment diagram will look like this ?

Ans , i also couldn't understand what does (Area AB) x A bar and (Area AB) x B bar mean ...it's not bending moment , am i right ?
in the notes , the author has chosen to draw the moment diagram by taking A as refrence point , so , he should only calculate it using (Area AB) x A bar , am i right ?

 

[PhanthomJay]

Example uses A as ref point. Using B as ref point, your moment diagrams are generally correct except the last one will be part quadratic and part linear. Deflection analysis requires summing moments of the areas under the moment diagram, using the centroid of each area to determine moment arm. Since it was 50 years ago when I last did one of these tedious calcs, I can't comment much further.

 

[foo9008]

Example uses A as ref point. Using B as ref point, your moment diagrams are generally correct except the last one will be part quadratic and part linear. Deflection analysis requires summing moments of the areas under the moment diagram, using the centroid of each area to determine moment arm. Since it was 50 years ago when I last did one of these tedious calcs, I can't comment much further.

OK, can you explain what does (area AB x A bar) and (area AB x B bar mean)??

 

[PhanthomJay]

These are the 'moments' of the bending moment areas about a given point and are equal to the areas of the moment graph times the distance of centroid of those areas to that point. For example, where the moment graph is a right triangle, it's area is 1/2 the base (b) times the height and the centroid is located at 1/3 (b) from fat end, as shown. This area of the moment graph part times the distance from Centroid to the given point A or B is called the Moment area .

 

[foo9008]

These are the 'moments' of the bending moment areas about a given point and are equal to the areas of the moment graph times the distance of centroid of those areas to that point. For example, where the moment graph is a right triangle, it's area is 1/2 the base (b) times the height and the centroid is located at 1/3 (b) from fat end, as shown. This area of the moment graph part times the distance from Centroid to the given point A or B is called the Moment area .

why the moment area doesn't mean area under the graph only? why we must multiply the area of the diagram with position of centroid measured from a specific point?

 

[foo9008]

These are the 'moments' of the bending moment areas about a given point and are equal to the areas of the moment graph times the distance of centroid of those areas to that point. For example, where the moment graph is a right triangle, it's area is 1/2 the base (b) times the height and the centroid is located at 1/3 (b) from fat end, as shown. This area of the moment graph part times the distance from Centroid to the given point A or B is called the Moment area .

moments' of the bending moment areas about a given point mean taking moment about different point in a beam,right? why there's a need to multiply the area under graph with position of centroid measured from a specific point?
http://www.mathalino.com/reviewer/m...ution-to-problem-624-moment-diagrams-by-parts
in this example, the author taking moment about different points , so he used 3 different solutions to solve the problem...
in the first solution,he take the moment about 2m from A,
in the second solution, he took the moment about B,
in the third solution, he took the moment about A...

 

[PhanthomJay]

Don't lose sight of the fact that the ultimate goal of these examples is to determine deflections. Determining moments using moment diagrams by parts is only the first step. If all you were concerned with was bending moments, you would stop there, and probably not use moment diagrams by parts, but rather, conventional shear and moment diagrams and free body diagrams which I discussed earlier.. The author gives 3 examples of determining moments by parts, by summing moments about different points. They all lead to the same result. The moments are in units of N-m. But now since you are looking for deflections, you must continue to the next step , which is to determine the areas of the moment diagram graphs (in N-m^2, which, when divided by EI, gives slopes), and then determining the centroids of those areas, and then determining the moments of those moment areas from tje centroid about a chosen point (in N-m^3, which, when divided by EI, gives deflections) in order to go on to determine deflections. This all stems from the calculus of beam theory. Determining deflections by hand calcs is often a tedious task (except for the simpler cases), no matter what method you use.

 

[foo9008]

Don't lose sight of the fact that the ultimate goal of these examples is to determine deflections. Determining moments using moment diagrams by parts is only the first step. If all you were concerned with was bending moments, you would stop there, and probably not use moment diagrams by parts, but rather, conventional shear and moment diagrams and free body diagrams which I discussed earlier.. The author gives 3 examples of determining moments by parts, by summing moments about different points. They all lead to the same result. The moments are in units of N-m. But now since you are looking for deflections, you must continue to the next step , which is to determine the areas of the moment diagram graphs (in N-m^2, which, when divided by EI, gives slopes), and then determining the centroids of those areas, and then determining the moments of those moment areas from tje centroid about a chosen point (in N-m^3, which, when divided by EI, gives deflections) in order to go on to determine deflections. This all stems from the calculus of beam theory. Determining deflections by hand calcs is often a tedious task (except for the simpler cases), no matter what method you use.

But the author want us to compute the area moment diagram only, no deflection is mentioned in question...

 

[PhanthomJay]

But the author want us to compute the area moment diagram only, no deflection is mentioned in question...

I don't know why you would want to compute area moment diagrams if you were not ultimately interested in deflections. Maybe just for practice.