Calculating direct stress safety factor

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SUMMARY

The discussion focuses on calculating the direct stress safety factor for two steel rods connected by a clevis joint. The user, gpsza, calculated a safety factor of 12.5, while the study guide indicated a solution of 7. Key variables include a pin diameter of 8 mm, an ultimate shear stress of 80 MPa, and a rod yield stress of 160 MPa. The calculations involved determining the working shear stress, force, direct stress, and safety factor using established formulas.

PREREQUISITES
  • Understanding of shear stress and direct stress concepts
  • Familiarity with safety factor calculations in engineering
  • Knowledge of basic geometry for area calculations
  • Proficiency in using units such as MPa and mm
NEXT STEPS
  • Review the principles of shear and direct stress in mechanical engineering
  • Study the calculation of safety factors in structural analysis
  • Learn about the properties of materials, specifically yield stress and ultimate shear stress
  • Explore advanced topics in stress analysis, such as finite element analysis (FEA)
USEFUL FOR

Mechanical engineers, structural engineers, and students studying material mechanics will benefit from this discussion, particularly those involved in stress analysis and safety factor calculations for structural components.

gpsza
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Hello Everyone,
Below is a self assesment question I found in a study guide related direct and shear stress, it is in 2 parts. The guide only gives the answers. I managed part 1 but for part 2 my answer differs from the given safety factor solution of 7 mine is 12.5. In my attempt below i have tried to give a description of the approach I used.

I would appreciate if someone could point out the error in my approach.

Thanks in advance
gpsza

Homework Statement



We have 2 steel rods connected using a clevis joint.
Part 1. Determine maximum allowed force using given shear stress safety factor
Part 2. Based on answer of part 1 determine the safety factor for the rods

variables:
pin diameter = 8 mm
ultimate shear stress of pin material = 80 Mpa or Alternatively 80 N/(mm*mm)
shear stress Safety Factor = 2

rod material yield stress = 160 Mpa or Alternatively 160 N/(mm*mm)
rod diameter = 20 mm


Homework Equations



Part 1:
(working shear stress)=(Maximum Shear Stress)/(Safety Factor)
Force = (shear stress)*(2*(Cross sectional area of pin))
(Circle Area)=(pi*diameter*diameter)/4

Part 2:
(direct stress)=Force/Area
(safety factor)=(Stress at Failure)/(Maximum Working Stress)

The Attempt at a Solution



Part 1:
Calculate working stress:
(working shear stress) = 80/2 = 40 Mpa or Alternatively 40 N/(mm*mm)

Calculate force keeping in mind the need to double the area of the pin:
Area = (pi*8*8)/4 = 50.2654824574 (mm*mm)
2*Area = 100.5309649149 (mm*mm)
Force = 40 * 100.5309649149 = 4021.2385965949 = 4.02 kN

Part 2:
Approach: use force calculated in part 1 to calculate the direct stress the rods would experience, use the direct stress and given yield stress to calculate the safety factor
Area = (pi*20*20)/4 = 314.159265359 (mm*mm)
(direct stress) = 4021.2385965949 / 314.159265359 = 12.8 Mpa
(safety factor)=(Stress at Failure)/(Maximum Working Stress)
(safety factor)= 160/12.8 = 12.5
 
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gpsza: Your answer currently looks correct, I think.

By the way, the unit symbol for megapascal is spelled MPa, not Mpa. Only write MPa; do not write its derivation (N/mm^2) each time. Also, for exponentiation, write 8^2 or mm^3, not 8*8 nor mm*mm*mm. Also, usually do not list more than four to six decimal places for numerical values.
 
Thank you nvn for your reponse to the question and the additional help it is really appreciated.

gpsza
 

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