Calculating direct stress safety factor

In summary, the conversation discusses a self-assessment question about direct and shear stress and the calculation of a safety factor. The question is in two parts and involves determining the maximum allowed force and the safety factor for steel rods connected with a clevis joint. The conversation includes variables and equations used in the calculations, as well as a description of the approach used. The final answer for the safety factor is 12.5, which is confirmed by another participant in the conversation. Some formatting and notation tips are also provided.
  • #1
gpsza
2
0
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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  • #2
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.
 
  • #3
Thank you nvn for your reponse to the question and the additional help it is really appreciated.

gpsza
 

Related to Calculating direct stress safety factor

1. What is direct stress safety factor?

Direct stress safety factor is a measure of the strength and stability of a material or structure under direct stress. It is calculated by dividing the maximum allowable stress by the actual stress applied to the material or structure.

2. How is direct stress safety factor calculated?

The direct stress safety factor is calculated by dividing the maximum allowable stress (or yield strength) of the material by the actual stress applied to it. The resulting value is the safety factor, which indicates how much stronger the material is compared to the stress it is subjected to.

3. Why is direct stress safety factor important?

The direct stress safety factor is important because it helps engineers and designers determine the strength and reliability of a material or structure. It ensures that the material or structure can withstand the expected load without failure or collapse, providing a margin of safety.

4. What is a good direct stress safety factor?

The ideal direct stress safety factor varies depending on the specific application and industry standards. Generally, a safety factor of 2 or higher is considered acceptable, as it provides a significant margin of safety and accounts for any potential uncertainties in the calculation.

5. How can the direct stress safety factor be improved?

The direct stress safety factor can be improved by using stronger materials, increasing the thickness or size of the structural element, or redesigning the structure to distribute the load more evenly. Conducting thorough stress analysis and testing can also help identify potential weak points and improve the safety factor.

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