What Equations Are Needed for Calculating Beam Stress and Curvature?

[SteamKing]

the bending moment the figure I came to was 23.01KN
sorry steam king what would you like me to clarify?

Yeah, I didn't understand this calculation from Post #26, which is why I mentioned finding the area under the shear force diagram. Then, your Posts #28 and #29 were still confusing me.

 

[andrewh21]

sorry please ignore the bending moment post
as for the shear force diagram basically I followed a tutorial for this first I found the moments and I have attached a rough sketch
I then took the load at one end 29.5kn multiplied it by 1.5 m length of the triangle and / 2 (area of the triangle)
therefore 29.15*1.5/2=21.9KN
as the opposite side is equal do these then want added together giving a moment of 42.8kn?

is this the correct approach?

 

[SteamKing]

sorry please ignore the bending moment post
as for the shear force diagram basically I followed a tutorial for this first I found the moments and I have attached a rough sketch
I then took the load at one end 29.5kn multiplied it by 1.5 m length of the triangle and / 2 (area of the triangle)
therefore 29.15*1.5/2=21.9KN
as the opposite side is equal do these then want added together giving a moment of 42.8kn?

is this the correct approach?

No. Remember, the bending moment at each end of this beam is zero and is a maximum at the center.

You've found the BM as a result of the UDL. What about the BM due to the force applied at the center of the beam?

 

[SteamKing]

You are using 29.15 kN as the value of the shear at each end of the beam. This is different from the reaction of 29.42 kN which you previously calculated. Why is that?

 

[andrewh21]

as the reaction I think I have got confused
the reaction at each end is at 2 tonnes /m is 6 tonnes in total so the UDL reactions would be 3 tonnes at each end
and for the point load toal200 n so either end would be 100N
3tonees converted to Newtons is 29419.95008592 +200=29619.9500859N converted to KN=29.619950086
so reaction at each end would be 29.6 KN

therefore
29.6n multiplied it by 1.5 m length of the triangle and / 2 (area of the triangle)
therefore 29.6*1.5/2=22.2KN
the moment in the centre of the beam would be 22.2KN

Am I on the right track I am very sorry if I seem slow to pick this up I assure you that I am trying very hard and putting the effort in thank you for your help

 

[SteamKing]

as the reaction I think I have got confused
the reaction at each end is at 2 tonnes /m is 6 tonnes in total so the UDL reactions would be 3 tonnes at each end
and for the point load toal200 n so either end would be 100N
3tonees converted to Newtons is 29419.95008592 +200=29619.9500859N converted to KN=29.619950086
so reaction at each end would be 29.6 KN

You can't add 200 N to the reaction due to the UDL at each end of the beam. You must add the reaction due to the 200 N point load instead.

 

[andrewh21]

how do I do this?

 

[SteamKing]

how do I do this?

Figure out what the reactions are for a 200 N load placed in the center of the beam.

Look, you've already done this in Post #24. You can do it by inspection.

 

[andrewh21]

W= 6 tones
so the reaction forces at each support would be 3 tonnes
an for the point load
W=200N
as this is central the reaction forces at the supports would be 100N

is this the one you mean steamking?

 

[SteamKing]

W= 6 tones
so the reaction forces at each support would be 3 tonnes
an for the point load
W=200N
as this is central the reaction forces at the supports would be 100N

is this the one you mean steamking?

Yes. These reactions can be combined, using consistent units, of course.

 

[andrewh21]

using consistent units? KN?

 

[SteamKing]

using consistent units? KN?

Yes. Reactions after all are forces.

 

[andrewh21]

so combined reactions would be 59.0399KN
59.0399/2=29.51995
so a raction of 29.5 KN

 

[SteamKing]

so combined reactions would be 59.0399KN
59.0399/2=29.51995
so a raction of 29.5 KN

OK. A reaction of 29.5 kN means that the shear force curve starts at a value of 29.5 kN. What is the value of the shear force in the middle of the beam?

 

[andrewh21]

29.5*1.5/2=22.125KN

 

[SteamKing]

29.5*1.5/2=22.125KN

That's not the value of the shear force in the middle of the beam.

At x = 0 m, which is where the left support is located, the shear force = reaction at the support = 29.5 kN.

In order to calculate the value of the shear force at x = 1.5 m, which is the middle of the beam, you've got to calculate how much load is supported by the beam
between x = 0 m and x = 1.5 m. That load = ?

 

[andrewh21]

29.5KN

 

[SteamKing]

29.5KN

According to your previous posts, the load over 1.5 m was 3 tonnes, which when converted to Newtons was 29,419.95 N.

 

[andrewh21]

so very sorry please forgive me I am getting confused again I have tried that many calcs trying to get this I am in a bit of a muddle
1.5m at 2 tonnes/m = 3 tonnes
3 tonnes converted to KN =29.892049152000002KN
so 29.9KN from x=o to x=1.5
not forgetting the point load in the centre of 200N am I correct to assume that the load from x=0m to x=1.5m
=30.1KN
or does the point load not get took into consideration for this?

 

[SteamKing]

so very sorry please forgive me I am getting confused again I have tried that many calcs trying to get this I am in a bit of a muddle
1.5m at 2 tonnes/m = 3 tonnes
3 tonnes converted to KN =29.892049152000002KN
so 29.9KN in the centre
not forgetting the point load in the centre of 200N am I correct to assume that the load from x=0m to x=1.5m
=30.1KN
or does the point load not get took into consideration for this?

Let's try to simplify things a bit.

1. There's no need to carry calculations to 50 decimal places.
2. We can use a value of g = 9.81 m/s² or even 10 m/s², without significant loss of accuracy.

For a UDL of 2 tonnes/meter, the total load on the beam will be 2 tonne/m * 3m = 6 tonnes, half of which is 3 tonnes.

3 tonnes = 3000 kg = 3000 * 9.81 = 29,430 N

The reactions on the beam, using g = 9.81 m/s², will be 29,530 N at the left and right supports.

Now, what is the value of the shear force at x = 1.5 m from the left support?

 

[andrewh21]

please ignore that

 

[andrewh21]

how do I calculate thart?

 

[SteamKing]

how do I calculate thart?

Come on. Re-read Post #51.

 

[andrewh21]

therefore it should be 29,530N
is that correct?

 

[SteamKing]

therefore it should be 29,530N
is that correct?

Concentrate. If the reaction at the left support is 29,530 N, how can the shear force be 29,530 N at the center of the beam? Like I said, re-read Post #51. You also must account for the direction of the reaction force and the forces due to the loads applied to the beam.

 

[andrewh21]

is the shear force in the centre of the beam calculated by calculating the area of the triangle so 29,530N*1.5m/2= 22,147N

 

[SteamKing]

is the shear force in the centre of the beam calculated by calculating the area of the triangle so 29,530N*1.5m/2= 22,147N

No. That's how the bending moment would be calculated from considering just the UDL by itself.

Doesn't any of this ring a bell from taking notes? Haven't you done any research online about beam bending?

 

[andrewh21]

so would that be considered the pure bending moment if it included the point load?
and including the point load would it be 22,347N?

the reason i ask this is i have found some notes that will help me a whole bunch if i correctly find the pure bending moment.

My next question is there is no indication of E so i will just consider that it is mild steel with a 210 gpa

i have done a lot of research and spent a heck of a long time trying to crack this and i am grateful for your time and help along the way if this is the pure bending moment i will go away and answer the whole of the question and come back to you (if that is ok) to review and give me your feed back

Many thanks Andrew

 

[np86]

Hello all,

I'm struggling with a beam problem and hoping someone can help. I have a beam with a point load in the centre (Beam A), one end of the beam is treated as simply supported and the other is attached to a metal plate (Beam B) which is fixed at both end (welded). I am trying to find the max deflection and max stresses in each beam but am unsure of how to approach such a problem. Any guidance would be greatly appreciated. I have looked in Rourke and searched the internet but can't find a relevant example or formula. I presume it will require a combination of beam formulas but do not know where to start!
I am thinking the the load on Beam A will create a moment in Beam B (Ma). Then this moment could be could be split into two forces, one positive and one negative, then use fixed beam formulas and method of superposition to combine the 2 deflections due to the 2 resulting forces on Beam B?

 

[SteamKing]

np86:
Please don't hijack other member's HW threads.

If you have a HW problem to post, please follow the Rules and post it using the HW Template.

What you have here is a frame of some sort and must be analyzed as such. There is no simple formula for it, nor do I expect will you find one in a handbook like Roark's. Most engineers will use software to model such a frame to find the deflections, if it were a simple frame. According to your description, beam B is a plate of some sort, which usually cannot be modeled as a simple beam. More information, including a better sketch of the components of this structure, would be required to analyze it.

 

[np86]

didn't mean to "hijack",

just registered today so I was unaware of the rules sorry.