A small question regarding force/stress/area

  • Thread starter tina-duncan
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In summary: Tensile and shear stresses can cause failure. Bearing stress is important, but shear stress is more important overall.
  • #1
tina-duncan
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Hello friends, I have a small query regarding a force being applied to a surface...

I have a force of 0.1MN which is being applied in the -y direction onto a surface.
Now I know that if said surface was flat, I would divide the force by the area to give me the stress.

However the force is being applied to a curved surface (a u shape).
Is the process the same?
Do I just simply calculate the area of that surface or is there another procedure involved?

Many thanks,
Tina
 
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  • #2
This needs a sketch and a description of how the force is applied. Does 'flat' mean horizontal, vertical or sloping? Does 'u-shape' mean 2-dimensional or 3-dimensional?
 
  • #3
My apologies. Here is a rather crude sketch of the problem.
2lbiyxz.png


The load acting on the hole is supplied by a pin of diameter 0.0195m.

This is all the information that is given to me, so assumptions may have to be made I would imagine...
 
  • #4
The stress on the contact area between pin and u hole is complex. If the pin and hole are of approximately equal size, an approximation can be used you could calculate your compressive stress as F/A with the area being .02x.01
If the pin is much smaller than the hole, I'll see if I can find the equation.
 
  • #5
The hole has to be larger than the pin if it is to be fitted cold. Therefore the actual contact is theoretically a line load with no thickness. However, if you know the yield strength of the pin you can work out the lower bound on the contact area from that. However, this is a very common problem and most engineers calculate the stress as load/ projected area (as DTM has suggested). In assessing strength they will be using experimental data using the same nominal assumption of projected area, even though it is known to be false. There are other complications: To get the load into the pin from outside the object illustrated, the pin has to bend a little, throwing most of the stress to the edges of the hole (in a way that is indeterminate) rather than uniformly distributed as is usually assumed.
 
  • #6
There are 3 diffrent types of stress you should consider for the plate: tensile, bearing, and shear stress (tear out). DTM describes the bearing stress of the bolt acting on the plate. Tear out should be fine if you follow the standards, but it's good to check anyway.
 

1. What is the relationship between force, stress, and area?

The relationship between force, stress, and area is described by the equation: stress = force/area. This means that as the force applied to an object increases, the stress on the object also increases. However, if the area of the object increases, the stress decreases.

2. How does force affect an object's stress?

Force has a direct impact on an object's stress. As the force applied to an object increases, the stress on the object also increases. This is because the force acts over a certain area, causing the object to experience a greater amount of internal resistance or strain.

3. How does area affect an object's stress?

The area of an object has an inverse relationship with its stress. As the area of an object increases, the stress on the object decreases. This is because the force applied to the object is spread out over a larger surface area, resulting in a lower amount of internal resistance or strain.

4. What is the difference between force and stress?

Force is a physical quantity that describes the push or pull on an object, while stress is a measure of the internal resistance or strain experienced by the object. In other words, force is the external cause of stress on an object.

5. How does stress affect an object's strength?

Stress can have a significant impact on an object's strength. When an object is subjected to high levels of stress, it can lead to deformation, damage, or failure. However, some materials can withstand higher levels of stress than others, making them stronger and more suitable for certain applications.

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