What Equations Are Needed for Calculating Beam Stress and Curvature?

[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.