Calculating direct stress safety factor

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The discussion revolves around a self-assessment question related to calculating the safety factor for steel rods connected by a clevis joint. The user, gpsza, successfully completed part 1 but found their safety factor for part 2 to be 12.5, differing from the provided answer of 7. They detailed their calculations for both parts, including the working shear stress and direct stress experienced by the rods. Another user confirmed that gpsza's calculations appear correct, and they also provided feedback on proper unit notation and formatting. The conversation emphasizes the importance of accurate calculations in determining safety factors in engineering applications.
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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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