What Equations Are Needed for Calculating Beam Stress and Curvature?

In summary, In this problem, you need to find the maximum stress due to bending (occurring at the center point), the value of the radius of the curvature of the neutral layer, and the factor of safety if the maximum allowable stress of the material is 100MPa. If the beam is hollow, then you're stuck.
  • #1
andrewh21
35
0
Help needed In my course paperwork I can't seem to find the relevant equations needed to complete the following in if someone could help point me in the right direction of equations needed I would be very grateful

1. Homework Statement

A rectangular hollow beam length of 3m is simply supported at its ends it has width of 10 mm and depth of 200 mm the beam is subjected to a uniformly disturbed load of 2 tonnes/m and a point load of 200N at its centre position
I) determine the maximum stress due to bending (occurring at the centre point)
ii) determine the value of the radius of the curvature of the neutral layer
iii) the factor of safety if the maximum allowable stress of the material is 100MPa

 
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  • #2
If the beam is hollow, what are the dimensions of the interior portion? That's kinda necessary to work this problem.

Also, I can't believe that you have allowed your notes to become so disorganized that you can't write down anything pertaining to solving this problem. No textbook? Google? Bueller?
 
  • #3
SteamKing thanks for your reply this is the problem all the info on dimensions is as above this is why I am struggling with this I do have textbooks etc but cannot find anything relevant to solve the above is this possible with the above information?
 
  • #4
andrewh21 said:
SteamKing thanks for your reply this is the problem all the info on dimensions is as above this is why I am struggling with this I do have textbooks etc but cannot find anything relevant to solve the above is this possible with the above information?

If the beam is hollow and the internal dimensions are not provided, then you're stuck. A clarification from the instructor of the course is in order.

It's not clear what textbooks you have, but searching for 'bending of beams', 'bending stress', or 'radius of curvature' would be a good place to start, if you have a textbook on strength of materials. Even if your notes and books have been eaten by zombies, you can Google these terms and come up with something.
 
  • #5
thanks very much I have been told to treat it as a solid so from that do I first calculate the second moment area then (stress/y=M/I)?
 
  • #6
andrewh21 said:
thanks very much I have been told to treat it as a solid so from that do I first calculate the second moment area then (stress/y=M/I)?
Yes, but only after you have determined the bending moment at the center of the beam.
 
  • #7
how do I calculate the bending moment in the centre of the beam?
 
  • #8
Can anyone help I am struggling as the point load in the centre of the beam keeps throwing me can anyone help point me in the right direction
 
  • #9
Have you calculated the support reactions for the beam? Can you draw the shear force curve for this beam?

Simply saying 'I don't know.' or 'I lost all of my notes.' is not sufficient to receive help at PF. You must show some effort, either at working your problem directly or at least researching how to solve it.

You were given some search terms in Post #4. Did you try them?
 
  • #10
hi andrewh21.

You have so far shown no effort on this problem. SteamKing has been trying to elicit some response from you, but none has been forthcoming, other than "I don't have a clue." This is unacceptable. We are not here to do your homework for you or to teach you the entire subject. I have given you a 1 point warning.

The least you could have done was to articulate how you would begin to approach a problem like this. Tell us your thinking. What do you think is happening physically?

At one point, you indicated that you could handle the case of a distributed load, but not the point load case. OK. Tell us what you would do if you had a distributed load.

What textbook are you using?

I am about to close this thread if I don't see some effort on your part.

Chet
 
  • #11
As a first step, calculate the reaction forces at the supports. That is just a statics problem, and has nothing to do with beam bending. What do you get for the reaction forces?

Chet
 
  • #12
Ra=Rb
Ra*3m=200tonnes/m*3*1.5
=900Tonnes/m
from which Ra=900/3=300 tonnes/m =Rb

anywhere near?
 
  • #13
andrewh21 said:

Homework Statement


A rectangular hollow beam length of 3m is simply supported at its ends [and] ... is subjected to a uniformly disturbed load of 2 tonnes/m and a point load of 200N at its centre position

Ra=Rb
Ra*3m=200tonnes/m*3*1.5
=900Tonnes/m
from which Ra=900/3=300 tonnes/m =Rb

anywhere near?

Where do you get 200 tonnes and 300 tonnes from?
 
  • #14
What is the total of the distributed and the point load on the beam?

Chet
 
  • #15
sorry should have been
Ra=Rb
Ra*3m=19613nm/m*3*1.5
=88258.5nm/m
from which Ra=88258.5/3=29419.5Nm/m =Rb

any better?
 
  • #16
andrewh21 said:
sorry should have been
Ra=Rb
Ra*3m=19613nm/m*3*1.5
=88258.5nm/m
from which Ra=88258.5/3=29419.5Nm/m =Rb

any better?

And what is a N-m/m ?
 
  • #17
Nm
 
  • #18
andrewh21 said:
Nm
Since when are the units of force Nm?

Chet
 
  • #19
this is where I am falling down odiously I have UDL of 2 tonnes/m but also this point load I have viewed dozens of worked examples trying to find something similar but nothing I have looked at contains a UDL and a point load it seems to be one or the other any advice on where to look etc would be greatly appreciated

Andrew
 
  • #20
andrewh21 said:
this is where I am falling down odiously I have UDL of 2 tonnes/m but also this point load I have viewed dozens of worked examples trying to find something similar but nothing I have looked at contains a UDL and a point load it seems to be one or the other any advice on where to look etc would be greatly appreciated

Andrew
What if the distributed load wasn't there? Would you then be able to determine the reaction forces at the two ends resulting from the point load? This is a simple freshman physics problem involving static equilibrium.

Chet
 
  • #21
so Ra=Rb
Ra*3=200N*3*1.5=
=900N
from which Ra=900/3=300N=Rb

Is this more like it?
 
  • #22
andrewh21 said:
so Ra=Rb
Ra*3=200N*3*1.5=
=900N
from which Ra=900/3=300N=Rb

Is this more like it?

If you have a 200 N point load in the center of the beam, how can the sum of the reactions be greater than the magnitude of the load?
 
  • #23
If you have units of Newtons multiplied by meters, making N-m, then dividing by meters gives what units? [N*m/m = ?]
 
  • #24
guys thanks for bearing with me I know I am a total newbie
W=2*3=6
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
assuming this is correct and it is as simple as combining both of the forces would I be correct in assuming that
the reaction forces at each support would be 29.15KN
 
  • #25
andrewh21 said:
guys thanks for bearing with me I know I am a total newbie
W=2*3=6
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
assuming this is correct and it is as simple as combining both of the forces would I be correct in assuming that
the reaction forces at each support would be 29.15KN

Yes.
 
  • #26
thanks now I need to calculate the bending moment in the centre of the beam
from what I know to do this I need to effectively cut the beam into 3 sections A,B,C,D and E each section 0.6m for this UDL per section will be 11.77kn
Mb=(29.15*0.6)-11.77*1=5.72
Mc=(29.15*1.2)-(11.77+0.2*1)=23.01
Md=(29.15*0.6)-11.77*1=5.72

Ma=0
Mb=5.72
Mc=23.01
Md=5.72
Me=0

therefore the bending moment in the canter of the beam would be 23.01 KN
 
  • #27
andrewh21 said:
thanks now I need to calculate the bending moment in the centre of the beam
from what I know to do this I need to effectively cut the beam into 3 sections A,B,C,D and E each section 0.6m for this UDL per section will be 11.77kn
Mb=(29.15*0.6)-11.77*1=5.72
Mc=(29.15*1.2)-(11.77+0.2*1)=23.01
Md=(29.15*0.6)-11.77*1=5.72

Ma=0
Mb=5.72
Mc=23.01
Md=5.72
Me=0

therefore the bending moment in the canter of the beam would be 23.01 KN

I'm afraid I don't follow these calculations for the bending moment.

A more universal method would be to construct the shear force diagram, and then calculate the area under the shear force curve. This method works for UDLs, point loads, whatever.
 
  • #28
for the sheer moment diagram including the point load i took the moment at point a and 29.15
i then *this load by the distance to the center and divided it by 2 29.15KN*1.5m/2=21.9KN as it is equal at both ends the sheer moment value in the center of the beam 21.9KN
 
Last edited:
  • #29
22.5 for the point load
 
  • #30
andrewh21 said:
so I have drawn the sheer moment diagram for the UDL alone and also with the point load am I correct in calculating the sheer moment for the UDL alone would be 22.05 KN and with the point load

andrewh21 said:
22.5 for the point load

It's not clear what you have calculated here. What is the moment you obtained at the center of the beam?

Remember, if you have omitted something for your post, or you need to clarify something you can always edit your post.
 
  • #31
edited please advice
although my above work may not be correct how to i get the maximum stress due to bending from the sheer moment diagram?

Thanks Andrew
 
Last edited:
  • #32
Do you know how bending stress is calculated?

The shear force diagram (not sheer force), is a tool used to calculate the bending moment.
 
  • #33
so assuming that my shaer force diagram is correct what do I do next
 
Last edited:
  • #34
andrewh21 said:
so assuming that my shaer force diagram is correct what do I do next

Going back a couple of posts, what value of the bending moment did you calculate at the center of the beam? You never clarified what you had posted previously.
 
  • #35
the bending moment the figure I came to was 23.01KN
sorry steam king what would you like me to clarify?
 

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