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A close-coiled helical spring has a mean diameter of 80 mm, a spring constant of 100 kN/m, and 10 coils. The maximum shear stress is 200 MPa. What is the diameter of the spring wire? What is the maximum load the spring can carry? Take G = 80 GPa.

(a) Diameter: 8 mm, Maximum Load: 160 N
(b) Diameter: 10 mm, Maximum Load: 200 N
(c) Diameter: 12 mm, Maximum Load: 240 N
(d) Diameter: 14 mm, Maximum Load: 280 N

User Jgibson
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Final answer:

The diameter of the spring wire and the maximum load the spring can carry are determined by using formulas from spring design and material strength theories, considering the maximum shear stress, spring constant, and mean diameter of the coils.

Step-by-step explanation:

To determine the diameter of the spring wire and the maximum load the spring can carry for a close-coiled helical spring given the mean diameter, spring constant, number of coils, and the maximum shear stress, we can use relevant formulas from spring design and material strength theories.

The wire diameter (d) can be found using the torsional shear stress formula for springs: τ = (8 * D * F) / (π * d^3), where τ is the shear stress, D is the mean diameter of the coils, and F is the force. By rearranging the formula, we can solve for d.

The maximum load (Fmax) that the spring can carry is determined using the spring constant (k = F/δ), where δ is the deflection. The maximum shear stress sets a limit to the maximum force, so Fmax = τ * (π * d^3) / (8 * D).

With the given numeric values, these formulas can be solved to obtain the wire diameter and the maximum load. Keeping in mind the provided options (a) to (d), only one of these will correctly match the calculated values.

User Dmitry Sokurenko
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