Simple Stresses and Strains Questions and Answers

Answer: B

The elongation is \(\frac{\gamma l^{2}}{6E}\).

For a cone hanging from its base, the weight below any section is one third of a cylinder of the same section, because the volume of a cone is one third of base times height. So the stress at depth y from the tip is \(\gamma y/3\), and integrating the strain over the length:

\[ \begin{aligned} \delta &= \int_{0}^{l} \frac{\gamma y}{3E}\,dy \\ &= \frac{\gamma l^{2}}{6E} \end{aligned} \]

\(\gamma l^{2}/2E\) is the result for a prismatic bar, three times larger. \(wl/AE\) is the formula for an external load, and \(\gamma l^{3}/6E\) has the wrong dimensions for a length.

Answer: A

Stress and strain are directly proportional.

Hooke's law states that within the elastic limit, stress is proportional to strain, \(\sigma = E\varepsilon\), where the constant of proportionality E is the modulus of elasticity. On a stress-strain graph this is the straight-line portion that starts at the origin.

An inverse relation would mean more stress gives less strain, which is impossible. The relationship is predictable and linear within the elastic limit, so the 'unpredictable' and 'curvilinear' options describe the material only beyond the limit of proportionality, where Hooke's law no longer applies.

Answer: C

It is called the modulus of elasticity.

Since \(E = \sigma/\varepsilon\), putting \(\varepsilon = 1\) gives \(E = \sigma\). So the modulus of elasticity can be read as the stress that would produce unit strain if the material stayed elastic, which is why it has the units of stress.

Unit stress just means stress per unit area. Principal stress is the normal stress on a plane carrying no shear, and normal stress is stress perpendicular to a plane. None of them is defined by unit strain.

Answer: C

Volumetric strain is \(\frac{\delta V}{V}\).

Strain is always a change divided by the original value. Volumetric strain is the change in volume divided by the original volume, a dimensionless ratio. For small strains it equals the sum of the three linear strains, \(e_v = e_x + e_y + e_z\).

\(\frac{V}{\delta V}\) is the reciprocal, and the forms with m (mass) are not strains at all, since they mix mass with volume.

Answer: B

The yield strength is found by drawing 0.2% offset of the strain.

For steels such as high-strength and cold-worked bars that show no sharp yield point, a line is drawn parallel to the initial straight part of the stress-strain curve, starting at 0.002 (0.2 percent) strain. The stress where it cuts the curve is the 0.2 percent proof stress, taken as the yield strength.

0.5 percent is not the standard offset, and initial tangent or secant moduli describe stiffness, not yield.

Answer: C

Poisson's ratio is 1/3.

Young's modulus and bulk modulus are related by \(E = 3K(1 - 2\mu)\). Setting E = K:

\[ \begin{aligned} 1 &= 3(1 - 2\mu) \\ 1 - 2\mu &= \tfrac{1}{3} \Rightarrow \mu = \tfrac{1}{3} \end{aligned} \]

\(\mu = 1/2\) gives K infinite (an incompressible material), not K = E. 1/4 and 3/4 do not satisfy the relation, and 3/4 exceeds the theoretical maximum of 1/2.

Answer: C

Poisson's ratio is (3K - 2G) / (6K + 2G).

Start from the two standard relations between the elastic constants:

\[ \begin{aligned} E &= 3K(1 - 2\mu) \\ E &= 2G(1 + \mu) \end{aligned} \]

Equating them and solving for \(\mu\):

\[ \begin{aligned} 3K - 6K\mu &= 2G + 2G\mu \\ \mu &= \frac{3K - 2G}{6K + 2G} \end{aligned} \]

Check: the formula gives \(\mu = 0.5\) only when K is infinitely large compared with G, that is, for an incompressible material. The options with 4G do not follow from the two relations.

Answer: A

The elastic modulus is three times its shear modulus.

Young's modulus and the shear modulus are linked by Poisson's ratio \(\mu\):

\[ \begin{aligned} E &= 2G(1 + \mu) \\ &= 2G(1 + 0.5) = 3G \end{aligned} \]

A Poisson's ratio of 0.5 describes an incompressible material, such as rubber, for which this result is used. Four times would need \(\mu = 1\), which is impossible for a stable isotropic material, while 'equal to' and 'half of' would need \(\mu\) of \(-0.5\) and \(-0.75\).

Answer: C

The volumetric strain is \(\frac{P}{AE}\left[1-\frac{2}{m}\right]\).

Under axial tension the longitudinal strain is \(\varepsilon = \frac{P}{AE}\), and each lateral strain is \(-\frac{1}{m}\varepsilon\). Volumetric strain is the sum of the three linear strains:

\[ \begin{aligned} e_v &= \varepsilon - \frac{\varepsilon}{m} - \frac{\varepsilon}{m} \\ &= \frac{P}{AE}\left(1 - \frac{2}{m}\right) \end{aligned} \]

The lateral contraction reduces the volume increase, so the bracket must contain a minus sign, ruling out the two '+' options. \(1 - \frac{1}{2m}\) counts only part of one lateral strain instead of two full ones.

Answer: D

The bolt is under tension but the tube is under compression.

Tightening the nut pulls the bolt head and the nut towards each other. The tube, trapped between them, is squeezed, so it is in compression. The bolt, held at both ends while being pulled, is stretched, so it is in tension.

The tensile force in the bolt equals the compressive force in the tube, since there is no external load. The tube cannot be in tension because nothing pulls on its ends, and the bolt cannot be in compression because the nut pulls on it.