Mechanics of Materials Practice Exam — Free

The angle of twist of a uniform circular shaft is $\phi = TL/(GJ)$, where $J$ is the polar moment of area of the cross-section. For a solid circle of diameter $d$:

$$J = \frac{\pi d^4}{32} = \frac{\pi (0.050)^4}{32} = 6.136 \times 10^{-7}\,\text{m}^4$$

Convert everything to base SI units before substituting: $T = 2000\,\text{N}\cdot\text{m}$, $L = 1.2$ m, $G = 80 \times 10^9$ Pa.

$$\phi = \frac{TL}{GJ} = \frac{2000(1.2)}{(80 \times 10^9)(6.136 \times 10^{-7})} = \frac{2400}{4.909 \times 10^4} = 0.0489\,\text{rad}$$

The formula gives radians, so convert: $\phi = 0.0489 \times (180/\pi) = 2.80^\circ$.

As a consistency check, the maximum shear stress is $\tau = Tc/J = 2000(0.025)/(6.136 \times 10^{-7}) = 81.5$ MPa, and $\phi = \tau L/(Gc) = (81.5 \times 10^6)(1.2)/[(80 \times 10^9)(0.025)] = 0.0489$ rad, the same value.

The tempting answer 5.60° comes from using the second moment of area $I = \pi d^4/64$ instead of the polar moment $J = \pi d^4/32$; torsion uses $J$, which is twice $I$ for a circle. 0.175° results from substituting the diameter into $J = \pi r^4/2$, and 0.049° is the radian value read as degrees.

$\phi = 2.80^\circ$ (0.0489 rad)

The load is axial and the cross-section is uniform, so the normal stress is simply force over area. Cross-sectional area:

$$A = \frac{\pi d^2}{4} = \frac{\pi (0.020)^2}{4} = 3.142 \times 10^{-4}\,\text{m}^2$$

Normal stress (tensile, taken positive):

$$\sigma = \frac{P}{A} = \frac{30 \times 10^3}{3.142 \times 10^{-4}} = 95.5 \times 10^6\,\text{Pa} = 95.5\,\text{MPa}$$

This is below the yield strength of common structural aluminum alloys, consistent with the stated elastic behaviour. Hooke's law gives the strain, $\varepsilon = \sigma/E = (95.5 \times 10^6)/(70 \times 10^9) = 1.364 \times 10^{-3}$, and the elongation is the strain times the original length:

$$\delta = \varepsilon L = \frac{PL}{AE} = \frac{(30 \times 10^3)(2.0)}{(3.142 \times 10^{-4})(70 \times 10^9)} = 2.73 \times 10^{-3}\,\text{m}$$

Equivalently $\delta = (1.364 \times 10^{-3})(2000\,\text{mm}) = 2.73$ mm. A strain of about 0.14% stretching a 2 m rod by a few millimetres is the right order of magnitude for a metal loaded elastically.

$\sigma = 95.5\,\text{MPa}$ (tension), $\delta = 2.73\,\text{mm}$

Reactions: by symmetry each support carries $wL/2 = 6.0(4.0)/2 = 12$ kN.

Internal forces: the shear force is largest at the supports, $V_{max} = 12$ kN, and the bending moment is largest at midspan, where the shear passes through zero:

$$M_{max} = \frac{wL^2}{8} = \frac{6.0(4.0)^2}{8} = 12\,\text{kN}\cdot\text{m}$$

Section properties for the 100 mm by 200 mm rectangle (neutral axis at mid-depth, so $c = 100$ mm):

$$I = \frac{bh^3}{12} = \frac{(0.100)(0.200)^3}{12} = 6.667 \times 10^{-5}\,\text{m}^4,\qquad A = 0.020\,\text{m}^2$$

(a) Flexure formula at the extreme fibres:

$$\sigma_{max} = \frac{M_{max}c}{I} = \frac{(12 \times 10^3)(0.100)}{6.667 \times 10^{-5}} = 18.0 \times 10^6\,\text{Pa} = 18.0\,\text{MPa}$$

A positive (sagging) moment puts the top fibre in compression and the bottom fibre in tension, both at midspan.

(b) For a rectangular section the transverse shear stress is parabolic through the depth, zero at the top and bottom and maximum at the neutral axis, where $\tau_{max} = VQ/(Ib) = 1.5\,V/A$:

$$\tau_{max} = \frac{1.5(12 \times 10^3)}{0.020} = 0.900 \times 10^6\,\text{Pa} = 0.900\,\text{MPa}$$

This occurs at the neutral axis of the sections adjacent to the supports, where $V$ is greatest. Using the first moment of the area above the neutral axis directly, $Q = (0.100)(0.100)(0.050) = 5.0 \times 10^{-4}\,\text{m}^3$ and $VQ/(Ib) = 12000(5.0 \times 10^{-4})/[(6.667 \times 10^{-5})(0.100)] = 0.900$ MPa, the same value. The shear stress is only 5% of the bending stress, which is typical of a slender beam.

(a) $\sigma_{max} = 18.0\,\text{MPa}$ at midspan (top in compression, bottom in tension); (b) $\tau_{max} = 0.900\,\text{MPa}$ at the neutral axis next to each support