Overhanging Beam Shear Force and Bending Moment Diagrams Tutorial

In summary, the conversation is about a student seeking help with creating a shear force and bending moment diagram for an overhanging beam problem. The student discusses their approach and equations used, but also expresses confusion and asks for assistance with understanding the problem.
  • #1
Hati
10
1

Homework Statement



So I have a beam here, I'll post a Photoshop diagram of the problem at the bottom of this section, but its an overhanging beam simply supported at one end, and mid way down another, points A and B. I've changed up the values so I can do my work myself, I don't want someone blurting the answer to me. I need to learn rather than cruise through my class.

The beam is carrying a continuous uniformly distributed load that represents it's own weight of 4kN/m and two vertical loads of 16kN and 10kN

The setup looks like this

6eQNQ.png


I'm to create a shear force and bending moment diagram for this beam.

Homework Equations



So, my relevant equations are very simple, the sum of the moments = 0 because I'm in equilibrium, and so are the forces. I have these equations to work with;

∑M=0
∑F=0

The Attempt at a Solution



Well, to create my shear force diagram I decided to turn the UDL into a point load in the middle, considering I had a load of 4kN/m over 16 meters I ended up with 64kN in the middle of the beam at 8 meters. In hindsight I can see how this might not be right, considering it doesn't take into account support B but I don't know how to take it into account. It isn't at the end of the beam.

Continuing on and hoping for the best I calculated my moments and did the following

∑MA=0
0= 0+(16*4)+(64*8)-(MB*5)+(10*16)
0= 64+512-(MB*5)+160

MB*5=736

∴MB=736/5

=147.2kN


Ok, so I have now the reaction force at support B. I can now use a different calculation to get ahold of A.

if ∑F=0
then A= 64+512-147.2-160
=268.8

and from that I can set up a shear force diagram I'm not so sure is correct but I'll have to go with it.

Homework Statement



so now the bending moment diagram. Time to work out my bending moments.

Homework Equations



I'll take point A, and my equation is this

MA=0

The Attempt at a Solution



so what I have in this beam are a number of forces to work out bending moments from.

I have support A
4 meters down the beam is force a at 16kN
then the translated UDL is force b at 64kN, 8 meters away.
support B exerting a force of 147.2kN upwards
and force c exerting a force of 5kN 10m from A

(Yes, I'm using capitals to represent supports and lowercase to represent force)

so what I did is this

MA=0
a= (16*4)+(64*8)-(147.2*10)-(10*16)
=-1056

b=(64*4)-(147.2*6)-(10*12)
=-747.2

B=-(147.2*2)-(10*8)
=-374.4

c=-(10*6)
=-60

and I get a simple weird bending moment diagram that doesn't fit my situation, you can see it'd be kind of this shape √.

Means I don't really get what's going on unfortunately, I need some help with this. What am I doing wrong?
 
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  • #2
Hi, Hati, welcome to PF!
Hati said:

Homework Statement



So I have a beam here, I'll post a Photoshop diagram of the problem at the bottom of this section, but its an overhanging beam simply supported at one end, and mid way down another, points A and B. I've changed up the values so I can do my work myself, I don't want someone blurting the answer to me. I need to learn rather than cruise through my class.

The beam is carrying a continuous uniformly distributed load that represents it's own weight of 4kN/m and two vertical loads of 16kN and 10kN

The setup looks like this

6eQNQ.png


I'm to create a shear force and bending moment diagram for this beam.

Homework Equations



So, my relevant equations are very simple, the sum of the moments = 0 because I'm in equilibrium, and so are the forces. I have these equations to work with;

∑M=0
∑F=0

The Attempt at a Solution



Well, to create my shear force diagram I decided to turn the UDL into a point load in the middle, considering I had a load of 4kN/m over 16 meters I ended up with 64kN in the middle of the beam at 8 meters. In hindsight I can see how this might not be right, considering it doesn't take into account support B but I don't know how to take it into account. It isn't at the end of the beam.
When determining support reactions, you can always represent the uniformly distributed load as a resultant point load acting the cg. But when drawing shear and bending moment diagrams, you must keep the load as a distributed load, not as a point load at the cg.
Continuing on and hoping for the best I calculated my moments and did the following

∑MA=0
0= 0+(16*4)+(64*8)-(MB*5)+(10*16)
0= 64+512-(MB*5)+160

MB*5=736

∴MB=736/5

=147.2kN
You should call the reaction force at B as RB or something else instead of MB , to avoid using the letter M which usually stands for moment, not force. Also, you must have mixed uo some numbers somewhere, because the distance from B to A is 10, not 5.
Ok, so I have now the reaction force at support B. I can now use a different calculation to get ahold of A.

if ∑F=0
then A= 64+512-147.2-160
=268.8
you are summing forces, not moments, you need to be careful with your values.
and from that I can set up a shear force diagram I'm not so sure is correct but I'll have to go with it.
please google on 'Shear and bending moment diagrams' for help with drawing them

Homework Statement



so now the bending moment diagram. Time to work out my bending moments.

Homework Equations



I'll take point A, and my equation is this

MA=0

The Attempt at a Solution



so what I have in this beam are a number of forces to work out bending moments from.

I have support A
4 meters down the beam is force a at 16kN
then the translated UDL is force b at 64kN, 8 meters away.
support B exerting a force of 147.2kN upwards
and force c exerting a force of 5kN 10m from A

(Yes, I'm using capitals to represent supports and lowercase to represent force)

so what I did is this

MA=0
a= (16*4)+(64*8)-(147.2*10)-(10*16)
=-1056

b=(64*4)-(147.2*6)-(10*12)
=-747.2

B=-(147.2*2)-(10*8)
=-374.4

c=-(10*6)
=-60

and I get a simple weird bending moment diagram that doesn't fit my situation, you can see it'd be kind of this shape √.

Means I don't really get what's going on unfortunately, I need some help with this. What am I doing wrong?
you need to corect your numbers and throw that uniform load back in there and watch signs etc. and draw good FBD's of cut sections as well as the diagram.
 
  • #3
PhanthomJay said:
Hi, Hati, welcome to PF!
When determining support reactions, you can always represent the uniformly distributed load as a resultant point load acting the cg. But when drawing shear and bending moment diagrams, you must keep the load as a distributed load, not as a point load at the cg.
Thanks, nice to meet you to. Anyway, the thing that complicates matters is I know how to deal with a UDL but its a bit of a problem when it spans the entire beam, because its on both sides of support B. What do I do with it? I can deal with it if it only spanned between A and B, but it runs the full length of the beam here. Do I do anything different with it?

You should call the reaction force at B as RB or something else instead of MB , to avoid using the letter M which usually stands for moment, not force.

Thanks, I'll amend that.

Also, you must have mixed uo some numbers somewhere, because the distance from B to A is 10, not 5. you are summing forces, not moments, you need to be careful with your values.

Yeah, the confusion is because I changed the values so I could learn the method here and do the real question on my own. I got confused between the two, that 5 is meant to be 10. Shouldn't matter, really. As long as somehow I get the right idea.

please google on 'Shear and bending moment diagrams' for help with drawing them

Already know how, the issue is I don't know if it's right but I'll have to go with it.

you need to corect your numbers and throw that uniform load back in there and watch signs etc. and draw good FBD's of cut sections as well as the diagram.
 
  • #4
Hati said:
Thanks, nice to meet you to. Anyway, the thing that complicates matters is I know how to deal with a UDL but its a bit of a problem when it spans the entire beam, because its on both sides of support B. What do I do with it? I can deal with it if it only spanned between A and B, but it runs the full length of the beam here. Do I do anything different with it?
No, when determining support reactions, just represent it (as you have done) as a single concentrated load of 64 kN acting 8 m from A. Then in determining the support reactions , you sum moments about A = 0 to solve for the support reaction at B, then sum moments about B = 0 to determine the support reaction at A, then check your results to be sure that the sum of the applied and support reaction forces in the y direction add to 0.
 
  • #5
PhanthomJay said:
No, when determining support reactions, just represent it (as you have done) as a single concentrated load of 64 kN acting 8 m from A. Then in determining the support reactions , you sum moments about A = 0 to solve for the support reaction at B, then sum moments about B = 0 to determine the support reaction at A, then check your results to be sure that the sum of the applied and support reaction forces in the y direction add to 0.

Yes, thankyou. I've gotten as far as the shear force diagram, and it seems about right. I cannot turn the UDL into a point load for my bending moments which is giving me a headache. I have to do something different here.

There's no way I know of to account for a UDL when I'm creating bending moments with the method above, which works fantastic for point loads. I heard somewhere that the area of shapes I have on each span of the beam I have on my shear force diagram gives me the right values to plot on my bending moment diagram, is this the case?

I have a place in my shear force diagram, right after the 16kn load where the slope in the line crosses 0 and gives me two triangles. It crosses over at exactly 4 meters giving me one triangle, and where support B offers it's reaction force it crosses back over 0 and gives me that second triangle. Do I measure the areas of both triangles and add them?
 
  • #6
Hati said:
Yes, thankyou. I've gotten as far as the shear force diagram, and it seems about right. I cannot turn the UDL into a point load for my bending moments which is giving me a headache. I have to do something different here.
you can't turn the UDL into a point load when determining shear forces either. If you are getting triangles in your shear diagram, I guess you understand that correctly.
There's no way I know of to account for a UDL when I'm creating bending moments with the method above, which works fantastic for point loads. I heard somewhere that the area of shapes I have on each span of the beam I have on my shear force diagram gives me the right values to plot on my bending moment diagram, is this the case?

I have a place in my shear force diagram, right after the 16kn load where the slope in the line crosses 0 and gives me two triangles. It crosses over at exactly 4 meters giving me one triangle, and where support B offers it's reaction force it crosses back over 0 and gives me that second triangle. Do I measure the areas of both triangles and add them?
Without seeing your diagram, I am not sure of the triangles of which you speak...nonetheless, when drawing the moment diagram, note the following, which comes from the calculus:
-The area under the Shear Diagram between 2 points is the change in moment between those points...(with consideration given to positive and negative areas)...and
-The slope of the Moment Diagram at any point is equal to the shear at that point.

Also don't forget your friend; Free Body Diagrams of cut sections!
 
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1. What is an overhanging beam?

An overhanging beam is a type of structural element used in construction that extends beyond its supports on one or both ends. This allows for more versatile and complex designs, but also requires careful consideration of shear force and bending moment.

2. What is shear force in a beam?

Shear force is a type of internal force that acts parallel to the cross-section of a beam. It is caused by external loads and can result in the beam bending or breaking if it exceeds the beam's strength. Shear force is represented by a diagram that shows the magnitude and direction of the force at different points along the beam.

3. What is a bending moment diagram?

A bending moment diagram is a graphical representation of the internal bending forces acting on a beam at different points along its length. It is created by calculating the moment (or torque) at each point and plotting it on a graph. The resulting diagram helps engineers and designers understand how the beam will behave under different loading conditions.

4. How do you calculate shear force and bending moment?

Shear force and bending moment can be calculated using equations based on the principles of statics. The magnitude of shear force can be calculated by taking the derivative of the bending moment equation, and the bending moment can be calculated by integrating the shear force equation. Alternatively, they can be calculated numerically using a shear and moment diagram or with the help of computer software.

5. How do you interpret a shear force and bending moment diagram?

A shear force and bending moment diagram shows the distribution of internal forces along the length of a beam. The magnitude and direction of these forces can be interpreted to determine how the beam will behave under different loads. For example, a large shear force at a certain point may indicate that the beam is likely to fail at that point. Engineers use these diagrams to design safe and efficient structures.

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