Question 2.7: The cylindrical rod in Fig. 1 is made of steel with E = 30 ×......

Plan the Solution From the load P and the cross-sectional area A we can determine the stress σ. Then, if σ ≤ σ_Y, we can use Hooke’s Law to relate the stress σ to the strain ε. Then we can relate the uniform strain,
ε, to the elongation ΔL. The change in diameter is due to the Poisson’sr-atio
effect.

Equilibrium: From the free-body diagram in Fig. 2,

∑F_x = 0:                                  F(x) = P = const

From Eq. 2.5,

σ_x = \frac{P} {A} = \frac{10  kips} {π(0.5  in.)^2} = 12.73 ksi < 50 ksi            (1)

Therefore, linearly elastic behavior occurs when the load P = 10 kips is applied to the bar.

Material Behavior: From Eq. 2.14,

σ_x = Eε_x              Hooke’s Law         (2.14)

ε_x = \frac{σ_x}{E} = \frac{P}{AE}                  (2a)

and, from Eq. 2.15,

ε_{\text{transv}} = -νε_{\text{longit}}          Poisson’s Ratio       (2.15)

ε_{\text{radial}} = -vε_x = – \frac{vP} {AE}                   (2b)

Strain-Displacement: From Eq. 2.7,

ε = ε _{avg} = \frac{ΔL}{L}    Axial Strain    (2.7)

ΔL = Lε_x = \frac{PL}{AE}                            (3a)

and

Δd = dε_{\text{radial}} = \frac{vPd}{AE}            (3b)

Substituting numerical values into Eqs. (3), we get

ΔL = Lε_{x} = \frac{(10  Kips)(48  in.)}{\pi (0.5 in.)^2(30\times 10^3  ksi)} = 20.4(10^{-3}) in.                 (4a)

and

Δd = dε_{\text{radial}} = \frac{0.3(10  Kips)(1  in.)}{\pi (0.5 in.)^2(30\times 10^3  ksi)}              (4b)

= -127(10^{-6}) in.

Review the Solution The changes in length and diameter are quite small in comparison with the original length and the original diameter, respectively, as they should be. Δd should definitely be smaller than ΔL, and the signs should be different, which is the case. The extensometer in Fig. 2.8a is capable of measuring both ΔL and Δd.