A solid circular rod with a diameter of d = 16 mm is shown in Figure P3.2. The rod is made of an aluminum alloy that has an elastic modulus of E = 72 GPa and Poisson’s ratio of ν = 0.33. When subjected to the axial load P, the diameter of the rod decreases by 0.024 mm. Determine the magnitude of

Question

A solid circular rod with a diameter of d = 16 mm is shown in Figure P3.2. The rod is made of an aluminum alloy that has an elastic modulus of E = 72 GPa and Poisson’s ratio of ν = 0.33. When subjected to the axial load P, the diameter of the rod decreases by 0.024 mm. Determine the magnitude of load P.

Accepted Answer

(a) The lateral strain in the rod is

[latex]\varepsilon_{ lat }=\frac{\Delta d}{d}=\frac{-0.024 mm }{16 mm }=-1,500 imes 10^{-6} mm / mm[/latex]

Using Poisson’s ratio, compute the corresponding longitudinal strain:

[latex]\varepsilon_{ ext {long }}=-\frac{\varepsilon_{ ext {lat }}}{v}=-\frac{-1,500 imes 10^{-6} mm / mm }{0.33}=4,545.455 imes 10^{-6} mm / mm[/latex]

Use Hooke’s law to calculate the stress in the rod:

[latex]\sigma=E \varepsilon_{ ext {long }}=(72,000 MPa )\left(4,545.455 imes 10^{-6} mm / mm ight)=327.273 MPa[/latex]

The cross-sectional area of the rod is:

[latex]A=\frac{\pi}{4} d^{2}=\frac{\pi}{4}(16 mm )^{2}=201.062 mm ^{2}[/latex]

Consequently, the force P that acts on the rod must be

[latex]P=\sigma A=(327.273 MPa )\left(201.062 mm ^{2} ight)=65,802.086 N =65.8 kN[/latex]

Question 3.2: A solid circular rod with a diameter of d = 16 mm is shown i...

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