A 0.75-in.-thick rectangular alloy bar is subjected to a tensile load P by pins at A and B as shown in Figure P3.4/5. The width of the bar is w = 3.0 in. Strain gages bonded to the specimen measure the following strains in the longitudinal (x) and transverse (y) directions: εx = 840 με and

Question

A 0.75-in.-thick rectangular alloy bar is subjected to a tensile load P by pins at A and B as shown in Figure P3.4/5. The width of the bar is w = 3.0 in. Strain gages bonded to the specimen measure the following strains in the longitudinal (x) and transverse (y) directions: [latex]\varepsilon_{x}=840 \mu \varepsilon ext { and } \varepsilon_{y}=-250 \mu \varepsilon[/latex].
(a) Determine Poisson’s ratio for this specimen.
(b) If the measured strains were produced by an axial load of P = 32 kips, what is the modulus of elasticity for this specimen?

Accepted Answer

(a) Poisson’s ratio for this specimen is

[latex]v=-\frac{\varepsilon_{ ext {lat }}}{\varepsilon_{ ext {long }}}=-\frac{\varepsilon_{y}}{\varepsilon_{x}}=-\frac{-250 \mu \varepsilon}{840 \mu \varepsilon}=0.298[/latex]

(b) The bar cross-sectional area is

[latex]A=(3.0 ext { in. })(0.75 ext { in. })=2.25 ext { in. }^{2}[/latex]

and so the normal stress for an axial load of P = 32 kips is

[latex]\sigma=\frac{32 kips }{2.25 in .^{2}}=14.222222 ksi[/latex]

The modulus of elasticity is thus

[latex]E=\frac{\sigma}{\varepsilon_{ ext {long }}}=\frac{14.222222 ksi }{(840 \mu \varepsilon)\left(\frac{1 in . / in .}{1,000,000 \mu \varepsilon} ight)}=16,931.2 ksi =16,930 ksi[/latex]

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