Nice to meet you. I am an engineer who designs parts. Could you please tell me about the strength of the crossed-over hooks of extension springs? I calculated it under the following conditions using the formula for the machine half hooks on your company’s website.
Wire drawing material = SWP-B; d = 0.32φ; D = 2.1; r1 = 0.54; r2 = 0.45 (calculated by force) P = 4.3N
① Is it okay to calculate the strength of the crossed-over hooks using the formula for the machine half hooks?
② Under the above conditions, the calculation result of the machine half hooks was as follows. Could you check it?
Note σ = 1852.6; τ = 701.74
③ Could you please tell me about the maximum tensile stress and maximum shear stress of SWP-B? I know that you have to answer many other questions in this forum, but I would appreciate it if you could let me know. How should we judge the calculation result σ and τ? Thank you in advance for your reply.
Thank you for writing to us! We will answer your questions below.
①: There is no fixed formula for the crossed over hooks, so I think it is okay to evaluate the strength calculation using the formula for the half-machine hook.
② I think that the calculation results are calculated correctly!
③ The maximum tensile stress and maximum shear stress of SWP-B change according to the wire diameter as shown in the stress graph for each wire diameter in the spring’s technical information page here. https://www.tokaibane.com/en/spring-design/extension-springs-formulas
The stress value when the wire diameter is 0.32 mm is shown below.
・ Maximum working tensile stress: 2000N/mm^2 x 0.8 ＝ 1600N/mm^2
・ Maximum working shear stress: 1190N/mm^2 x 0.64 ＝ 762N/mm^2
* The above stress value is considered as the maximum working stress value, It is necessary to not exceed this value when designing.”
Thank you for your reply. Regarding the safety factor of the maximum shear stress used, how was 0.64 derived?
Hello! The maximum shear stress used is based on JIS B 2704. The allowable shear stress of the tensile spring is 80% of the allowable shear stress of the compression coil spring, and the maximum shear stress used is 80% or less, so 0.8 x 0.8 = 0.64.
