Question
Download Solution PDFA helical spring of 10 N/mm rating is mounted on top of another helical spring of 8 N/mm rating. The force required for a total combined deflection of 45 mm through the two springs is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Spring on top of one another = Spring in series
When ‘n’ springs are connected in series
Equivalent stiffness is given by:
Calculation:
Given:
k1 = 10 N/mm, k2 = 8 N/mm, and δ = 45 mm
Equivalent stiffness
Equivalent stiffness
Load = Stiffness × Deflection
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