1 static safety factor for number 4 and 2 2 fatigue safety f

1) static safety factor for number (4 and 2)?

2) fatigue safety factor for number (4 and 2)?

force acting on arms= 1414lbs

cross-sectional area = 3 in^2

stress=20000 psi

Solution

Here it is given that the force acting on the arms is 1414 lbs

Given area of cross section of the arm is 3 square inches.

Hence, the stress acting on the arm = Force/ Area of cross section

= 1414/3 = 471.33 lb/in2 = 471.33 psi

Given the maximum stress on the component = 20000 psi

We know that design stress = actual stress* Factor of safety

=> 20000 = 471.33 psi*Factor of safety

=> Factor of safety = 42.43

Usually, the endurance limit of the material is half the actual stress.

SO, assuming maximum design stress of 10000 psi, we get

Factor of safety = 10000/471.33 = 21.22