Stress in a hole due to pin force

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SUMMARY

The discussion focuses on calculating the compression stress in a plate with a pin-hole configuration where the hole diameter exceeds the pin diameter by three times. The maximum compressive stress is defined by the equation σc,max = 0.798 * √(p / (KD * CE)), where p is the load per unit length, KD and CE are derived from the geometrical and material properties of the pin and socket. Additionally, for a circular hole, the actual stress is calculated as σA = 3σc, indicating that the stress concentration increases due to the hole's geometry.

PREREQUISITES
  • Understanding of compression stress calculations
  • Familiarity with material properties such as modulus of elasticity and Poisson's ratio
  • Knowledge of stress concentration factors in mechanical design
  • Basic principles of finite element analysis (FEA)
NEXT STEPS
  • Research the application of finite element analysis (FEA) for stress analysis in mechanical components
  • Study the effects of stress concentration in circular and elliptical holes in materials
  • Learn about the derivation and application of bearing stress equations
  • Explore the relationship between load distribution and material deformation in engineering contexts
USEFUL FOR

Mechanical engineers, structural analysts, and anyone involved in the design and analysis of load-bearing components with pin-hole configurations will benefit from this discussion.

guideonl
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Hi everyone,

What is the area should be taken to determine the compression stress of pin-hole in tention force (i.e plate with known thickness having a hole for lifting shakle pin) where the hole diameter is greater than the pin diameter (> x3) (theoretically, it is not an area just line of contact between the hole and the pin, which is the thickness of the plate).
Thank you, Guideon
 
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guideonl said:
Hi everyone,

What is the area should be taken to determine the compression stress of pin-hole in tention force (i.e plate with known thickness having a hole for lifting shakle pin) where the hole diameter is greater than the pin diameter (> x3) (theoretically, it is not an area just line of contact between the hole and the pin, which is the thickness of the plate).
Thank you, Guideon

Hey man, I would like to help you, but I don't quite understand your description. A diagram would be really helpful here.
 
Thank you,
Hereby a sketch attached to clarify the question, I hope it would explain it better.

Guideon
 

Attachments

  • pin acting in a hole 001.jpg
    pin acting in a hole 001.jpg
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guideonl said:
Thank you,
Hereby a sketch attached to clarify the question, I hope it would explain it better.

Guideon

Oh wow, that's great, thanks for the clarification. When I get home from work today I'll hook you up. Do you have numbers? I could also run a FE solution for you.
 
guideonl said:
Hi everyone,

What is the area should be taken to determine the compression stress of pin-hole in tention force (i.e plate with known thickness having a hole for lifting shakle pin) where the hole diameter is greater than the pin diameter (> x3) (theoretically, it is not an area just line of contact between the hole and the pin, which is the thickness of the plate).
Thank you, Guideon

Your attachment calcs look like you're trying to solve for bearing stress. However, since the pin is much smaller than the socket, the max compressive stress is actually given by:

\sigma c_{max} = 0.798 \cdot \sqrt{\frac{p}{K_D \cdot C_E}}

Where,

p = load per unit length
E1 = modulus of elasticity of bottom socket
E2 = modulus of elasticity of pin
v1 = poisson's ratio of bottom socket
v2 = poisson's ratio of pin

K_D = \frac{D_2 \cdot D_1}{D_1 - D_2}

C_E = \frac{1- \nu_1^2}{E_1} + \frac{1- \nu_2^2}{E_2}

CS
 

Attachments

  • CompressiveStressCyl.JPG
    CompressiveStressCyl.JPG
    2.1 KB · Views: 1,012
I agree with Stewart, though I would also add that the stress in your plate increases due to the hole.

For an elliptical center hole,

\sigma_{A} = \sigma_{c}\left(1+\frac{2a}{b}\right)

Since your hole is circular, a = b = r, thus

\sigma_{A} = 3\sigma_{c}

where

\sigma_{c} is compressive stress
\sigma_{A} is the actual stress, taking into account the geometry of the hole
2a is the major axis
2b is the minor axis
r is the radius of your circular hole

Source: Anderson, T.L., Fracture Mechanics Fundamentals and Applications, 3rd Ed., CRC / Taylor and Francis, 2005, p27.
 
Unit91Actual said:
I agree with Stewart, though I would also add that the stress in your plate increases due to the hole.

For an elliptical center hole,

\sigma_{A} = \sigma_{c}\left(1+\frac{2a}{b}\right)

Since your hole is circular, a = b = r, thus

\sigma_{A} = 3\sigma_{c}

where

\sigma_{c} is compressive stress
\sigma_{A} is the actual stress, taking into account the geometry of the hole
2a is the major axis
2b is the minor axis
r is the radius of your circular hole

Source: Anderson, T.L., Fracture Mechanics Fundamentals and Applications, 3rd Ed., CRC / Taylor and Francis, 2005, p27.

The stress concentration formula you're talking about was derived for uniaxial tension. I don't think those boundary conditions apply in this case. Also, since this is a contact stress problem, the stresses will be altered because traction forces are applied to the hole.
 
Thank you all, you are great
I just want to be sure that I undersdand your answers:
stewartcs,
a) Is the stress equation you sent depends on the units you use in the equation (SI/american)?
b) The 0.798 factor is dimensionless?
Unit91Actual
a) Is the compressive stress you used in the equation for an elliptical center hole is the result of the max compressive stress equation in stewartcs reply? If not, how can I calculate this sress? that was my original question.
b) According to your important note, actually, the major stress is not where the pin force acts, but at 90 degrees right/left from the hole center (3 times the compressive stress)?

Thank you all again, Guideon
 
guideonl said:
Thank you all, you are great
I just want to be sure that I undersdand your answers:
stewartcs,
a) Is the stress equation you sent depends on the units you use in the equation (SI/american)?
b) The 0.798 factor is dimensionless?
Unit91Actual

As with all engineering formulas, the equation should be dimensionally consistent.

The 0.798 is a dimensionless factor.

CS
 

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