Average Shear Stress on a Plate

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SUMMARY

The discussion focuses on calculating average shear stress on a plate with a long bolt passing through it. The force in the bolt shank is 8 kN, leading to an average shear stress of 4.716 MPa along the cylindrical area of the plate. The initial miscalculation of 18.52 MPa was corrected by recognizing that the shear force should be divided by the appropriate area, considering double shear conditions. The final calculations confirm the accuracy of the textbook answer, reinforcing the importance of precise force distribution in shear stress calculations.

PREREQUISITES
  • Understanding of shear stress calculations using the formula τ = V / A
  • Knowledge of normal stress calculations using the formula σ = N / A
  • Familiarity with concepts of double shear in structural analysis
  • Basic proficiency in geometry related to cylindrical areas
NEXT STEPS
  • Study advanced shear stress analysis techniques in mechanical engineering
  • Learn about the implications of double shear in bolted connections
  • Explore the use of finite element analysis (FEA) for stress distribution
  • Investigate the effects of different materials on shear strength
USEFUL FOR

Mechanical engineers, structural analysts, and students studying materials science who are involved in stress analysis and bolted joint design.

Colt 45 J
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1. Problem Statement
The long bolt passes through the 30-mm-thick plate. If the force in the bolt shank is 8kN, determine the average normal stress in the shank, the average shear stress along the cylindrical area of the plate defined by the section lines a-a, and the average shear stress in the bolt head along the cylindrical area defined by the section lines b-b.

http://img340.imageshack.us/img340/698/solidsproblem1102.jpg

Note: I have solved the shear on the bolt head, as well as the average normal stress in the shank, it's the average shear along the plate that I am having issues with.

Homework Equations


τ = V / A
o= N / A

The Attempt at a Solution



Distribution of forces so the equivalent is 4kN

τ = 4000/(pi*(.009^2-.0035^2))
τ = 18.52MPa

The answer key (which has yet to be wrong in my experience for this textbook) says that the average shear should be 4.72MPa

Edit:
Figured out my error

τ= 8000/(pi*(.018*.03))
τ=4.716MPa
 
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I'm pleased you are happy with your answer, but I didn't understand why it is 4 kN, and not 8kN. If you have a moment to explain I would be grateful.
 
Ah originally it was because with a rectangular element it's usually double shear instead of single, in the case of the answer thank you, as I had put a typo and meant to put 8000N in instead of the 4000, the answer was calculated using the 8000 however and is correct.
 

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