1.44 Consider two helical springs with the following characteristics: Spring 1: material steel; number of turns -10; mean coil diameter - 12 in.; wire diameter - 2 in.; free length - 15 in.; shear modulus - 12 x10^6 psi Spring 2: material aluminum; number of turns - 10; mean coil diameter - 10 in.; wire diameter -1 in.; free length - 15 in.; shear modulus - 4 x10^6 psi Determine the equivalent spring constant when (a) spring 2 is placed inside spring 1, and (b) spring 2 is placed on top of spring 1.

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
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1.44 Consider two helical springs with the following characteristics:
Spring 1: material steel; number of turns -10; mean coil diameter - 12 in.; wire diameter - 2 in.; free length - 15 in.; shear modulus - 12 x10^6 psi
Spring 2: material aluminum; number of turns - 10; mean coil diameter - 10 in.; wire diameter - 1 in.; free length - 15 in.; shear modulus - 4 x10^6 psi
Determine the equivalent spring constant when (a) spring 2 is placed inside spring 1, and (b) spring 2 is placed on top of spring 1.
Transcribed Image Text:1.44 Consider two helical springs with the following characteristics: Spring 1: material steel; number of turns -10; mean coil diameter - 12 in.; wire diameter - 2 in.; free length - 15 in.; shear modulus - 12 x10^6 psi Spring 2: material aluminum; number of turns - 10; mean coil diameter - 10 in.; wire diameter - 1 in.; free length - 15 in.; shear modulus - 4 x10^6 psi Determine the equivalent spring constant when (a) spring 2 is placed inside spring 1, and (b) spring 2 is placed on top of spring 1.
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