A small question regarding force/stress/area

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When applying a force to a curved surface, the calculation of stress becomes more complex than with a flat surface. The stress can be approximated using the formula F/A, where A is the projected area, but this assumes uniform distribution, which may not be accurate. The interaction between the pin and the hole complicates the stress distribution, especially if the pin is significantly smaller than the hole, leading to a line load scenario. Engineers often rely on experimental data and nominal assumptions about projected area despite known inaccuracies. It is crucial to consider different types of stress, including tensile, bearing, and shear stress, to ensure proper assessment of the material's strength.
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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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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?
 
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...
 
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.
 
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.
 
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.
 
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