Question

Consider the brass alloy for which the stress-strain behavior is shown in the Animated Figure 7.12. A cylindrical specimen of this material 10.0 mm (0.3937 in.) in diameter and 100.6 mm (3.961 in.) long is pulled in tension with a force of 9970 N (2241 lbf). If it is known that this alloy has a value for Poisson's ratio of 0.35, compute
(a) the specimen elongation,
(b) the reduction in specimen diameter.

because the diameter decreases, enter a minus sign in your answer.

Answer 1

Answer:

(a) 0.1509 mm

(b) 0.00525 mm

Explanation:

Stress, $\sigma$ is given by

$\sigma=\frac {F}{A}$ where F is force and A is area and area is given by $\frac {\pi d^{2}}{4}$ hence

$\sigma=\frac {4F}{\pi d^{2}}$ where d is the diameter. Substituting 9970 N for F and 10mm=0.01 m for d hence

$\sigma=\frac {4*9970 N}{\pi 0.01m^{2}}=126941982.6 N/m^{2}$

$\sigma \approx 127 Mpa$

From the attached stress-strain diagram, the stress of 127 Mpa corresponds to strain of 0.0015 and since strain is given by

$\epsilon=\frac {\triangle l}{l}$ where$\epsilon$ is the strain, $\triangle l$ is elongation and l is original length and making elongation the subject

$\triangle l= \epsilon \times l$ and substituting strain with 0.0015 and length l with 100.6 mm then

$\triangle l=0.0015\times 100.6=0.1509 mm$

(b)

Lateral strain is given by

$\epsilon_{lat}=\frac {\triangle d}{d}$ and substituting $-v\epsilon$ for $\epsilon_{lat}$ where v is poisson ratio then

$-v\epsilon=\frac {\triangle d}{d}$ and making $\triangle d$ the subject then

$\triangle d=-vd\epsilon$ and substituting 0.35 for v, 0.0015 for strain and 10 mm for d

$\triangle d=-(0.35)*10*0.0015=-0.00525 mm$ and the negative sign indicates decrease in diameter