A cylindrical specimen of this alloy 12 mm in diameter and 188 mm long is to be pulled in tension. Assume a value of 0.34 for Poisson's ratio. Calculate the stress (in MPa) necessary to cause a 0.0105 mm reduction in diameter.

This question is incomplete, the missing image in uploaded along this answer below.

The required stress is 200 Mpa

Explanation:

Given the data in the question;

diameter D = 12 mm = 12 × 10⁻³ m

Length L = 188 mm = 188 × 10⁻³ m

Poisson's ratio v = 0.34

Reduction in diameter Δd = 0.0105 mm = 0.0105 × 10⁻³ m

The transverse strain will;

εˣ = Δd / D

εˣ = -0.0105 × 10⁻³ /  12 × 10⁻³ m

εˣ = -0.00088

The longitudinal strain will be;

= - ( εˣ  / v )

= - ( -0.00088  / 0.34 )

= - ( - 0.002588 )

= 0.0026

Now, Using the values for strain, we get the value of stress from the graph provided in the question, ( first image uploaded below.

From the graph, in the Second image;

The stress is 200 Mpa

Therefore, The required stress is 200 Mpa

unlike the threaded fasteners which are used to clamp the machine members, power screws are used to transmit power. common applications of power screws are: screw jacks: to lift weight. lead screw of lathe: for axial movement of tool and its precise positioning.

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