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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
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
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)
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
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