Stress Mechanics Link AB: -20 ksi, 12 ksi Pins: Diameter + Bearing Stress

  • Thread starter Thread starter eaon21
  • Start date Start date
  • Tags Tags
    Mechanics Stress
Click For Summary
SUMMARY

The discussion focuses on the mechanics of Link AB, which has a width of 2 inches and a thickness of 1/4 inch, subjected to an average normal stress of -20 ksi and average shearing stress of 12 ksi in the pins. Participants aim to determine the diameter of the pins and the average bearing stress in the link. The approach involves equating the formulas for shear and normal stress, utilizing the known values to solve for the unknowns. The conversation emphasizes the importance of understanding the relationship between stress areas and the corresponding forces.

PREREQUISITES
  • Understanding of normal and shear stress concepts
  • Familiarity with stress equations and formulas
  • Basic knowledge of mechanics of materials
  • Ability to perform calculations involving area and force
NEXT STEPS
  • Research the calculation of pin diameter using shear stress formulas
  • Study the relationship between bearing stress and contact area
  • Learn about the mechanics of materials, focusing on stress distribution
  • Explore examples of similar problems in structural engineering
USEFUL FOR

Mechanical engineers, structural engineers, and students preparing for exams in mechanics of materials will benefit from this discussion, particularly those focusing on stress analysis in structural components.

eaon21
Messages
2
Reaction score
0
Link AB, of width b = 2 in. and thickness t = 1/4 in., is used to support the end of a horizontal beam. Knowing that the average normal stress in the link is -20 ksi, and that the average shearing stress in each of the two pins iS 12 ksi, determine (a) the diameter d of the pins, (b) the average bearing stress in the link.

dff.jpg
 
Engineering news on Phys.org
This is not for homework questions. Even so, you need to show some effort on your part. Show us what you think and have tried first.
 
im sorry..i was just answering questions in my book for my upcoming exam...
actually..my biggest concern right now is the area under the normal and shear stress..im still confused... though i think that this problem will be solved by equating the formulas for shear and normal stress since they will have the same P, right? i have the normal stress and shear stress average..i can equate the two P's..how about the areas in each stresses?
(that is for question letter a, still not thinking about letter b)
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
3K
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
18
Views
3K
  • · Replies 2 ·
Replies
2
Views
17K
  • · Replies 11 ·
Replies
11
Views
17K
Replies
3
Views
4K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
2
Views
4K