Question #172803

Two concentric springs, the central spring being 5 mm shorter than the outer spring, are to be loaded as shown. The outer spring consists of 40 turns of 15 mm diameter wire coiled on a mean diameter of 100 mm. The central spring, on the other hand, has 80 turns of 20 mm diameter wire coiled on a mean diameter of 80 mm. Use 83 GPa and 76 GPa for the shear modulus of the outer and the central springs, respectively. If the load P = 2 kN, determine the stress in the central spring and the displacement of the plate.

Expert's answer

A force of *2 kN* is acting on a circular rod with diameter *20 mm*. The stress in the rod can be

*σ = (20 10*^{3}* N)* */ (π ((20 10*^{-3}* m) / 2)*^{2}*)*

* = 127388535*2 (N/m*^{2}*) *

* = 127*2 (MPa)*

*= 381(MPa)*

The change of length can be calculated by:

* dl = σ l*_{o }*/ E*

* = (381 10*^{6}* Pa) (2 m) / (83 10*^{9}* Pa) *

* = 0.00381 m*

* = 3.81 mm*

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