Spring design problem - use of safety factor

[BrianC]

Ref: Lindberg's Practice Problems, 12 ed, Ch. 52 problem 2 - Spring design.

Can someone clarify why the safety factor is not used in part F when calculating the maximum shear stress? Is it because it was used in part A to calculate the wire diameter, and if the safety factor was applied again it would be redundant? Thanks for any help.

 

[Firefly]

My take on it is that you are asked to find the Maximum shear stress. As the designer, you will only allow the part to see a certain shear stress (thus the reason for the factor of safety and we determined this in part A), however we need to know what the maximum could be.

Knowing this will enable us to either stay with our previous Factor of Safety or select a new one.

There's my 2 cents.

 

[buick455]

It would be very helpfully to post the problem as my 52 problem 2 (second printing) is not what you are describing but in general you apply the safety factor once. I would like to see the problem though.

 

[BrianC]

Sorry about that. I have the 3rd printing. Here is the problem:

A severe service valve spring is to be manufactured from unpeened ASTM A230 steel wire in standard W&M sizes operating continuously between 20 - 30 lbf. The valve lift is 0.30 inch. The spring index is 10. The factor of safety is 1.5.

a. What is the wire diameter? (0.148 in)

b. What is the spring constant? (33.3 lbf/in)

c. What is the number of active coils if the ends are squared and ground? (6.38)

d. What is the total number of coils? (8.28)

e. What is the solid height? (1.24 in)

f. What is the spring force at solid height? (46.2 lbf)

g. What is the deflection at solid height? (1.39 in)

h. What is the minimum free height? (2.63 in)

The solution uses the safety factor for calculating the maximum shear stress in part a), which is used with the Wahl factor to calculate the wire diameter using equation 52.14 of the MERM 12ed. That I understand.

In part f), the maximum shear stress is again calculated, but this time the safety factor is not used. Equation 52.14 is again used, but this time to solve for the force. However, the spring diameter is used from part a) in equation 52.14 to calculate the force, which makes me think the safety factor is inherent in the calculation.

 

[gaidox]

In part a, you are designing the wire diameter thus using "max.allowable shear stress".max. allowable shear stress for design is max shear stress/ FS.

(a lower value of shear stress equals large diameter)

In part f, your already analyzing your designed spring using your designed wire diameter thus you have to compare it with maximum shear stress (without FS).

It seems confusing but my take is to use FS for designing/sizing to have a safe spring size.