Calculating Stresses at Points A, B, & C

  • Thread starter teng125
  • Start date
  • Tags
    Points
In summary: As I have mentioned in the previous post, when a bending moment is applied to a material, each layer of atom in the material experiences different stress. There exists a location where the stress is maximum and we call this the maximum stress. It has nothing to do with limit stress or allowable stress. It depends on the force applied on the material, which yields a bending moment and hence the stress. So we need to know the bending moment in order to find the maximum stress. In summary, the conversation discusses finding the stresses at points A, B, and C when a maximum bending moment is applied. The bending stress distribution is determined by the distance from the neutral axis to the outermost fiber and can be calculated using the formula \sigma(y)=
  • #1
teng125
416
0
http://www.sendspace.com/file/r2469z

attached file

find the stresses at points A;B and C when the maximum moment is applied.

should i consider the total three beams or only one beam if i want to calculate for the moment of inertia in order to use the formula (bending stress = moment / moment of inertia) to answer the ques above at point A,B and C ??

pls help

thanx
 
Physics news on Phys.org
  • #2
I think you might want to put your file on another site.
 
  • #3
pls try again because i have just try it and it can be downloaded

anyway,this is another link http://files.filefront.com//;5404421;;/
 
  • #4
Given the maximum bending stress has been found, then the bending stress distribution is given by

[tex]\sigma(y)=\frac{\sigma_{max}}{C}y[/tex]

where C is the distance between the neutral axis (where the layer of atoms experience zero resultant stress) to the outermost fiber.

The bending stress distribution varies with the position y. So by substituting the distance y into [tex]\sigma(y)[/tex] gives the corresponding bending stress at the designated location.
 
  • #5
what is the difference between allowable,limit and max stress??

i think that allowable and limit stress is the same while how can i calculate max stress if allowable and limit stress is given??
 
  • #6
so here,C is the same for all points but y is the distance from neutral axis to the respective points??

am i right??
 
  • #7
what is the difference between allowable,limit and max stress??

i think that allowable and limit stress is the same while how can i calculate max stress if allowable and limit stress is given??
Good question.

The limit stress, by definition is the stress a particular material can withstand before it
a) exceeds the yield strength (yield point) OR
b) exceeds the tensile strength (also known as breaking point, ultimate tensile strength: UTS)
However, for most of the machines, we do not want the stress experienced by its construction material (steel etc) to exceeds the yield strength for they will fail to work as desired or designed within the range of plastic deformation. Thus b) is out of interest. Hence, the limit stress [tex]\sigma_{lim}[/tex] is practically the yield stress, [tex]R_{p\ 0.2}[/tex], that is

[tex]\sigma_{lim}=R_{p\ 0.2}[/tex]

according to DIN-EN-10002-1, 1991


The allowable stress, is the stress that is permitted to be loaded on a given machine. It is always less than the limit stress. The reason is that while a particular material is theoritically able to withstand its limit stress, practically this is not the case. Most material experience plastic deformation under a stress of not even one third of the limit stress. It depends on many factors, such as the quality of the material, operation environment and control, risk and cost, just to name a few. We call this the safety factor. There should be a complete list stating these factor in your study material (you are a mechanical engineering student, aren't you hmm?). Hence, the allowable stress [tex]\sigma_{all}[/tex] is also known as the actual stress [tex]\sigma_{act}[/tex]. This applies to all three types of stress: tension, compression and shear.

[tex]\sigma_{all}=\frac{\sigma_{lim}}{n_{s}}[/tex]

where [tex]n_{s}[/tex] is the safety factor.


When a bending moment is applied to a material, each layer of atom in the material experiences different stress. There exists a location where the stress is maximum and we call this the maximum stress. It has nothing to do with limit stress or allowable stress. It depends on the force applied on the material, which yields a bending moment and hence the stress.

[tex]\sigma_{max}=\frac{M_{b}}{W_{by}}[/tex]

where [tex]M_{b}[/tex] is the bending moment applied and [tex]W_{by}[/tex] is the Widerstandsmoment.



so here,C is the same for all points but y is the distance from neutral axis to the respective points??

am i right??
Bingo.
 
Last edited:
  • #8
Also worth mentioned that the bending stress distribution is best to be written in the form

[tex]\sigma=\frac{y}{C}\sigma_{max}[/tex] where the ratio [tex]\frac{y}{C}[/tex] is emphasized. Thus y can takes the following range of values:

-C < y < C

with + indicating tensile, - indicating compression stress


EDIT:
I think the explanation about limit stress, allowable stress and maximum stress given above might somehow gives you a hint in finding the maximum stress given limit stress and/or allowable stress.

[tex]\sigma_{max}=\sigma_{all}=\frac{\sigma_{lim}}{n_{s}}[/tex]

Yes, there isn't much useful information we can obtain from this relation as the question isn't meaningful. However, usually we are interested in the maximum MOMENT given the limit stress and/or allowable stress for we will know the maximum load the material of interest can withstand.
 
Last edited:

1. How do you calculate stresses at points A, B, and C?

To calculate stresses at these points, you will need to use the formulas for stress and strain. The stress formula is stress = force/area, and the strain formula is strain = change in length/original length. First, determine the force acting on the point and the area of the material where the force is applied. Then, measure the change in length of the material. Plug these values into the formulas to calculate the stress and strain at each point.

2. What units are used to measure stress?

Stress is typically measured in units of force per unit area, such as pounds per square inch (psi) or newtons per square meter (N/m2). Other common units include pascals (Pa) and megapascals (MPa).

3. What factors can affect the stress at a particular point?

The stress at a point can be affected by several factors, including the magnitude and direction of the applied force, the area of the material where the force is applied, and the material's properties such as elasticity and yield strength. Other factors such as temperature and external loads can also impact stress at a point.

4. How can the stresses at points A, B, and C be used in engineering applications?

The stresses at these points can be used to determine the structural integrity and stability of a material or structure. Engineers can use stress calculations to design and analyze various components, such as bridges, buildings, and machines, to ensure they can withstand the expected loads and forces without failure.

5. Can stress calculations be used for all types of materials?

Stress calculations can be used for most materials, including metals, plastics, and composites. However, different materials may have different properties and behaviors under stress, so it is essential to consider these factors when performing stress calculations. Additionally, certain materials, such as brittle materials, may require more specialized methods for stress analysis.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
11
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
7
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
586
  • Engineering and Comp Sci Homework Help
Replies
4
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
11
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
22
Views
2K
Replies
3
Views
587
Back
Top