Strength of Materials Questions and Answers

Answer: D

The aluminium bar needs 22.86 cm².

The elastic strain under axial load is \(\varepsilon = F/(AE)\). Equal strain under the same load means equal axial stiffness per unit length, AE:

\[ \begin{aligned} A_{Al}E_{Al} &= A_{s}E_{s} \\ A_{Al} &= 8 \times \frac{200}{70} = 22.86\ \text{cm}^{2} \end{aligned} \]

Aluminium is less stiff than steel, so it needs a larger area, which rules out 11.43 cm² (which halves the answer) and points to the largest option. Poisson's ratio and the length play no part here.

Answer: B

The elastic constants are connected by \(E = 2G(1 + \nu)\), where E is Young's modulus, G the modulus of rigidity and \(\nu\) Poisson's ratio. So

The value 0.2 is just Poisson's ratio. The values 2.8 and 3.2 would come from using a Poisson's ratio of 0.4 or 0.6 and are not consistent with the data. The ratio of E to G for any real material lies between 2 and 3.

Answer: B

The elongation of an axially loaded bar is \(\delta = PL/(AE)\). The area is \(20 \times 20 = 400\) mm\(^2\), the load is 40,000 N, the length is 1000 mm and \(E = 200{,}000\) N/mm\(^2\):

The options 0.25, 0.75 and 0.95 mm correspond to a wrong load, area or modulus (for example 0.25 mm would be found with an area of 800 mm\(^2\)).

Answer: D

Hooke's law says that strain is proportional to the applied stress up to the proportional limit. The constant of proportionality is the modulus of elasticity. The law does not link strain to temperature, pressure or the volume of the solid in this way, so the options A, B and C are wrong.

Answer: A

The correct match is a-ii, b-iii, c-i. Young's modulus is the ratio of linear stress to linear strain (ii). The bulk or volume modulus is the ratio of bulk stress to bulk strain (iii). The shearing or rigidity modulus is the ratio of the shearing stress applied to the body to the shearing strain produced (i). \[ Y = \frac{\text{linear stress}}{\text{linear strain}} \] Each of the other options gives at least one modulus the definition of a different one.

Answer: A

The correct relation is \(R = \frac{P - Q}{2}\).

For an isotropic material, Hooke's law in Lamé form, with similar equations for the other components, is

\[ \begin{aligned} \sigma_{xx} &= (\lambda + 2\mu)\varepsilon_{xx} \\ &\quad + \lambda(\varepsilon_{yy} + \varepsilon_{zz}) \\ \tau_{yz} &= \mu\gamma_{yz} \end{aligned} \]

Comparing with the matrix, \(P = \lambda + 2\mu\), \(Q = \lambda\) and \(R = \mu\),

so \(P - Q = 2\mu = 2R\), giving \(R = (P - Q)/2\). This is the isotropy condition: of the three constants only two are independent.

\((Q - P)/2\) would make the shear modulus negative, since \(P \gt Q\) for any real material. The two options for \(Q\) rearrange the relation incorrectly; the correct rearrangement is \(Q = P - 2R\).

Answer: C

The relation is \(E = 2G(1 + \nu)\).

For an isotropic linear elastic material only two of the elastic constants are independent. Analysing pure shear as equal tension and compression at 45° gives the link between them:

\[ G = \frac{E}{2(1 + \nu)} \]

For steel, with E about 200 GPa and \(\nu \approx 0.3\), this gives G about 77 GPa, the familiar value. With \(\nu = 0.3\), the option \(2G = E(1 + \nu)\) gives G = 0.65E instead of about 0.38E, and the options \(G = 2E(1 + \nu)\) and \(2E = G(1 + \nu)\) make G larger than E, which no real material shows.

Answer: A

The engineering shear strain is 0.

The x and y axes run along the diagonals AC and BD. In the deformation, AC shortens and BD lengthens symmetrically about O, so points on the x axis stay on the x axis and points on the y axis stay on the y axis. The angle between the x and y directions remains 90°, so \(\gamma_{xy} = 0\), and the strains are purely normal:

\[ \begin{aligned} \varepsilon_{xx} &= \frac{19.96 - 20}{20} = -0.002 \\ \varepsilon_{yy} &= \frac{20.04 - 20}{20} = +0.002 \end{aligned} \]

The x and y axes are principal directions here. The sides of the square do change their angle: the shear strain measured along axes at 45° to x and y is \(\varepsilon_{yy} - \varepsilon_{xx} = 0.004\), the trap in options C and D. The value 0.002 is a normal strain, not a shear strain.

Answer: B

The lightest rod is made from Material β.

For a strength-based design the cross-section must carry P at the yield stress, \(A = P/\sigma_{y}\). The mass of a rod of length l is then

\[ m = \rho A l = P\,l\,\frac{\rho}{\sigma_{y}} \]

so the best material has the highest specific strength \(\sigma_{y}/\rho\). In MPa per (kg/m³): α 270/2700 = 0.100, β 900/4500 = 0.200, γ 520/7800 = 0.067, δ 540/9000 = 0.060. Material β gives half the mass of α and about a third that of γ or δ.

Young's modulus would matter for a stiffness-based design (limiting the extension), not for a strength-based one, which is why γ, with the highest modulus, is not the answer.

Answer: B

A fully anisotropic linear elastic material has 21 independent elastic constants.

Hooke's law relates the six stress components to the six strain components through a 6 x 6 stiffness matrix of 36 entries. The existence of a strain energy function makes this matrix symmetric, which leaves

\[ \frac{6 \times 7}{2} = 21 \]

independent constants.

36 counts every entry and ignores that symmetry. 2 is the number for an isotropic material (for example E and \(\nu\), or the two Lame constants), and 9 is the orthotropic case; 10 does not correspond to the general anisotropic material.