What Equations Are Needed for Calculating Beam Stress and Curvature?

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SUMMARY

The discussion focuses on calculating beam stress and curvature for a rectangular hollow beam subjected to a uniformly distributed load of 2 tonnes/m and a point load of 200N at its center. Key equations include the bending stress formula (σ = M/I * y) and the calculation of the bending moment, which is essential for determining maximum stress and radius of curvature. Participants emphasize the importance of calculating support reactions and constructing shear force diagrams to accurately assess the bending moment. The maximum allowable stress for the material is noted as 100MPa, which is critical for evaluating the factor of safety.

PREREQUISITES
  • Understanding of beam mechanics and bending stress calculations
  • Familiarity with shear force and bending moment diagrams
  • Knowledge of static equilibrium principles
  • Ability to perform unit conversions (e.g., tonnes to Newtons)
NEXT STEPS
  • Study the derivation and application of the bending stress formula (σ = M/I * y)
  • Learn how to construct shear force and bending moment diagrams for various loading conditions
  • Research the concept of the radius of curvature in beam theory
  • Explore the factor of safety calculations in structural engineering
USEFUL FOR

Engineering students, structural engineers, and anyone involved in mechanical design or analysis of beam structures will benefit from this discussion.

  • #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
 
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  • #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
 
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  • #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?
 
  • #36
andrewh21 said:
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.
 
  • #37
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?
 

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  • #38
andrewh21 said:
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?
 
  • #39
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?
 
  • #40
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
 
  • #41
andrewh21 said:
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.
 
  • #42
how do I do this?
 
  • #43
andrewh21 said:
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.
 
  • #44
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?
 
  • #45
andrewh21 said:
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.
 
  • #46
using consistent units? KN?
 
  • #47
andrewh21 said:
using consistent units? KN?
Yes. Reactions after all are forces.
 
  • #48
so combined reactions would be 59.0399KN
59.0399/2=29.51995
so a raction of 29.5 KN
 
  • #49
andrewh21 said:
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?
 
  • #50
29.5*1.5/2=22.125KN
 
  • #51
andrewh21 said:
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 = ?
 
  • #52
29.5KN
 
  • #53
andrewh21 said:
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.
 
  • #54
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?
 
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  • #55
andrewh21 said:
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?
 
  • #56
please ignore that
 
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  • #57
how do I calculate thart?
 
  • #58
andrewh21 said:
how do I calculate thart?
Come on. Re-read Post #51.
 
  • #59
therefore it should be 29,530N
is that correct?
 
  • #60
andrewh21 said:
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.
 

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