At the proportional limit, a 2-inch gage length of a 0.375-in.-diameter alloy bar has elongated 0.0083 in. and the diameter has been reduced 0.0005 in. The total tension force on the bar was 4.75 kips. Determine the following properties of the material:
(a) the modulus of elasticity.
(b) Poisson’s ratio.
(c) the proportional limit.

Question

Accepted Answer

(a) The bar cross-sectional area is

[latex]A=\frac{\pi}{4} d^{2}=\frac{\pi}{4}(0.375 in .)^{2}=0.110447 in .^{2}[/latex]

and thus, the normal stress corresponding to the 4.75-kip force is

[latex]\sigma=\frac{4.75 kips }{0.110447 in .^{2}}=43.007204 ksi[/latex]

The longitudinal strain in the bar is

[latex]\varepsilon_{ ext {long }}=\frac{\delta}{L}=\frac{0.0083 in .}{2 in .}=0.004150 in . / in .[/latex]

The modulus of elasticity is therefore

[latex]E=\frac{\sigma}{\varepsilon_{ ext {long }}}=\frac{43.007204 ksi }{0.004150 ext { in./in. }}=10,363.2 ksi =10,360 ksi[/latex]

(b) The longitudinal strain in the bar was calculated previously as

[latex]\varepsilon_{ ext {long }}=0.004150 ext { in./in. }[/latex]

The lateral strain can be determined from the reduction of the diameter:

[latex]\varepsilon_{ lat }=\frac{\Delta D}{D}=\frac{-0.0005 in .}{0.375 in .}=-0.001333 in . / in.[/latex]

Poisson’s ratio for this specimen is therefore

[latex]v=-\frac{\varepsilon_{ ext {lat }}}{\varepsilon_{ ext {long }}}=-\frac{-0.001333 ext { in./in. }}{0.004150 ext { in./in. }}=0.321285=0.321[/latex]

(c) Based on the problem statement, the stress in the bar is equal to the proportional limit; therefore,

[latex]\sigma_{P L}=43.0 ksi[/latex]

Question 3.1: At the proportional limit, a 2-inch gage length of a 0.375-i...

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