Chapter 8: Mechanical Properties of Solids

Given: radius $$r = 10 \, \mathrm{mm} = 1.0 \times 10^{-2} \, \mathrm{m}$$, length $$L = 1.0 \, \mathrm{m}$$, force $$F = 100 \, \mathrm{kN} = 1.0 \times 10^{5} \, \mathrm{N}$$, Young's modulus $$Y = 2.0 \times 10^{11} \, \mathrm{N\,m^{-2}}$$.

Cross-sectional area of the rod:

$$A = \pi r^{2} = \pi (1.0 \times 10^{-2})^{2} \approx 3.14 \times 10^{-4} \, \mathrm{m^{2}}$$

(a) Stress. Stress is the force per unit area:

$$\sigma = \frac{F}{A} = \frac{1.0 \times 10^{5}}{3.14 \times 10^{-4}} \approx 3.18 \times 10^{8} \, \mathrm{N\,m^{-2}}$$

(b) Elongation. From the definition of Young's modulus $$Y = \sigma / \varepsilon = (F/A)/(\Delta L / L)$$, we get

$$\Delta L = \frac{F L}{A Y} = \frac{(1.0 \times 10^{5})(1.0)}{(3.14 \times 10^{-4})(2.0 \times 10^{11})} \approx 1.59 \times 10^{-3} \, \mathrm{m}$$

(c) Strain.

$$\varepsilon = \frac{\Delta L}{L} = \frac{1.59 \times 10^{-3}}{1.0} = 1.59 \times 10^{-3}$$

Equivalently, the strain equals the stress divided by Young's modulus, $$\sigma / Y = (3.18 \times 10^{8})/(2.0 \times 10^{11}) \approx 1.59 \times 10^{-3}$$, which checks our result.

Both wires share the same diameter $$d = 3.0 \, \mathrm{mm}$$, so they have the same cross-sectional area:

$$A = \pi r^{2} = \pi (1.5 \times 10^{-3})^{2} \approx 7.07 \times 10^{-6} \, \mathrm{m^{2}}$$

Because the wires are joined end-to-end, the same tension $$F$$ (= the applied load) acts in both. The total elongation is the sum of the individual elongations:

$$\Delta L = \Delta L_{\mathrm{Cu}} + \Delta L_{\mathrm{steel}} = \frac{F L_{\mathrm{Cu}}}{A Y_{\mathrm{Cu}}} + \frac{F L_{\mathrm{steel}}}{A Y_{\mathrm{steel}}} = \frac{F}{A}\left(\frac{L_{\mathrm{Cu}}}{Y_{\mathrm{Cu}}} + \frac{L_{\mathrm{steel}}}{Y_{\mathrm{steel}}}\right)$$

Using $$Y_{\mathrm{Cu}} = 1.1 \times 10^{11} \, \mathrm{N\,m^{-2}}$$ and $$Y_{\mathrm{steel}} = 2.0 \times 10^{11} \, \mathrm{N\,m^{-2}}$$:

$$\frac{L_{\mathrm{Cu}}}{Y_{\mathrm{Cu}}} + \frac{L_{\mathrm{steel}}}{Y_{\mathrm{steel}}} = \frac{2.2}{1.1 \times 10^{11}} + \frac{1.6}{2.0 \times 10^{11}} = (2.0 + 0.8) \times 10^{-11} = 2.8 \times 10^{-11} \, \mathrm{m^{3}\,N^{-1}}$$

Setting $$\Delta L = 7.0 \times 10^{-4} \, \mathrm{m}$$:

$$\frac{F}{A} = \frac{7.0 \times 10^{-4}}{2.8 \times 10^{-11}} = 2.5 \times 10^{7} \, \mathrm{N\,m^{-2}}$$

Therefore the load applied is

$$F = (2.5 \times 10^{7})(7.07 \times 10^{-6}) \approx 1.77 \times 10^{2} \, \mathrm{N} \approx 180 \, \mathrm{N}$$

The performer's own body lies on the ground, so his weight does not pass through the femurs. The extra mass loaded on his thighbones is

$$m = 280 - 60 = 220 \, \mathrm{kg}$$

This load is shared equally by his two thighbones, so each carries

$$F = \frac{m g}{2} = \frac{(220)(9.8)}{2} = 1078 \, \mathrm{N}$$

The cross-sectional area of a thighbone is

$$A = \pi r^{2} = \pi (2.0 \times 10^{-2})^{2} \approx 1.26 \times 10^{-3} \, \mathrm{m^{2}}$$

The compressive stress in the bone is

$$\sigma = \frac{F}{A} = \frac{1078}{1.26 \times 10^{-3}} \approx 8.57 \times 10^{5} \, \mathrm{N\,m^{-2}}$$

Taking Young's modulus of compact bone as $$Y = 9.4 \times 10^{9} \, \mathrm{N\,m^{-2}}$$, the compressive strain is $$\sigma / Y$$, so the shortening of one femur (length $$L = 0.50 \, \mathrm{m}$$) is

$$\Delta L = \frac{\sigma L}{Y} = \frac{(8.57 \times 10^{5})(0.50)}{9.4 \times 10^{9}} \approx 4.56 \times 10^{-5} \, \mathrm{m}$$

That is, each thighbone shortens by only about $$45.6 \, \mathrm{\mu m}$$ — very small, which is why human bones can support such loads without obvious deformation.

The narrow face of the slab is a rectangle of dimensions $$50 \, \mathrm{cm} \times 10 \, \mathrm{cm}$$, so the area over which the shear force acts is

$$A = (0.50)(0.10) = 5.0 \times 10^{-2} \, \mathrm{m^{2}}$$

The shearing stress on this face is

$$\sigma_{s} = \frac{F}{A} = \frac{9.0 \times 10^{4}}{5.0 \times 10^{-2}} = 1.8 \times 10^{6} \, \mathrm{N\,m^{-2}}$$

Using shear modulus of lead $$G = 5.6 \times 10^{9} \, \mathrm{N\,m^{-2}}$$, the shear strain is

$$\theta = \frac{\sigma_{s}}{G} = \frac{1.8 \times 10^{6}}{5.6 \times 10^{9}} \approx 3.21 \times 10^{-4}$$

For a shear deformation, the displacement $$\Delta x$$ of the upper edge is the shear strain times the height $$L$$ over which the deformation occurs. Here the slab is sheared along its full height $$L = 0.50 \, \mathrm{m}$$, so

$$\Delta x = \theta \, L = (3.21 \times 10^{-4})(0.50) \approx 1.6 \times 10^{-4} \, \mathrm{m} = 0.16 \, \mathrm{mm}$$

The hydrostatic pressure at a depth $$h = 3000 \, \mathrm{m}$$ below the surface (taking water's density $$\rho = 1.0 \times 10^{3} \, \mathrm{kg\,m^{-3}}$$) is

$$p = \rho g h = (1.0 \times 10^{3})(10)(3000) = 3.0 \times 10^{7} \, \mathrm{N\,m^{-2}}$$

The bulk modulus $$B$$ relates the volumetric stress to the volumetric strain:

$$B = \frac{p}{|\Delta V / V|} \quad\Longrightarrow\quad \frac{\Delta V}{V} = \frac{p}{B}$$

Substituting,

$$\frac{\Delta V}{V} = \frac{3.0 \times 10^{7}}{2.2 \times 10^{9}} \approx 1.36 \times 10^{-2}$$

So water at the bottom of the Indian Ocean is compressed by about $$1.36\%$$ of its surface volume.

Let the common load be $$F$$ and the common elongation be $$\Delta L$$. From the definition of Young's modulus,

$$Y = \frac{F L}{A \, \Delta L}$$

Applying this to each wire,

$$Y_{s} = \frac{F L_{s}}{A_{s} \, \Delta L}, \qquad Y_{c} = \frac{F L_{c}}{A_{c} \, \Delta L}$$

Since $$F$$ and $$\Delta L$$ cancel when we take the ratio,

$$\frac{Y_{s}}{Y_{c}} = \frac{L_{s}}{L_{c}} \cdot \frac{A_{c}}{A_{s}} = \frac{4.7}{3.5} \cdot \frac{4.0 \times 10^{-5}}{3.0 \times 10^{-5}}$$

$$\frac{Y_{s}}{Y_{c}} = \frac{4.7 \times 4.0}{3.5 \times 3.0} = \frac{18.8}{10.5} \approx 1.79$$

Hence Young's modulus of steel is about $$1.79$$ times that of copper, consistent with steel being the stiffer material.

(a) Young's modulus. Young's modulus equals the slope of the stress-strain curve in its initial (linear, proportional) region. Reading from Fig. 8.9, a representative point on the straight-line portion corresponds to a stress of about $$1.5 \times 10^{8} \, \mathrm{N\,m^{-2}}$$ at a strain of about $$2 \times 10^{-3}$$. Therefore

$$Y = \frac{\mathrm{stress}}{\mathrm{strain}} = \frac{1.5 \times 10^{8}}{2 \times 10^{-3}} \approx 7.5 \times 10^{10} \, \mathrm{N\,m^{-2}}$$

So Young's modulus is of the order of $$7 \times 10^{10} \, \mathrm{N\,m^{-2}}$$.

(b) Yield strength. The yield strength is the stress at which the curve departs noticeably from the straight line (plastic deformation begins). From the graph this occurs at approximately

$$\sigma_{y} \approx 3 \times 10^{8} \, \mathrm{N\,m^{-2}} = 300 \, \mathrm{MPa}$$

From Fig. 8.11 the arrangement is: the steel wire (length $$1.5 \, \mathrm{m}$$) hangs from the ceiling and supports a $$4 \, \mathrm{kg}$$ mass; the brass wire (length $$1.0 \, \mathrm{m}$$) hangs from that mass and supports an additional $$6 \, \mathrm{kg}$$ at its lower end.

Therefore the tension in each wire is

$$F_{\mathrm{steel}} = (4 + 6) \, g = 10 \times 9.8 = 98 \, \mathrm{N}, \qquad F_{\mathrm{brass}} = 6 \, g = 6 \times 9.8 = 58.8 \, \mathrm{N}$$

Both wires have the same diameter $$d = 0.25 \, \mathrm{cm}$$, so radius $$r = 0.125 \, \mathrm{cm} = 1.25 \times 10^{-3} \, \mathrm{m}$$, and

$$A = \pi r^{2} = \pi (1.25 \times 10^{-3})^{2} \approx 4.909 \times 10^{-6} \, \mathrm{m^{2}}$$

Take $$Y_{\mathrm{steel}} = 2.0 \times 10^{11} \, \mathrm{N\,m^{-2}}$$ and $$Y_{\mathrm{brass}} = 0.91 \times 10^{11} \, \mathrm{N\,m^{-2}}$$.

Elongation of the steel wire:

$$\Delta L_{\mathrm{steel}} = \frac{F_{\mathrm{steel}} L_{\mathrm{steel}}}{A \, Y_{\mathrm{steel}}} = \frac{(98)(1.5)}{(4.909 \times 10^{-6})(2.0 \times 10^{11})} = \frac{147}{9.818 \times 10^{5}} \approx 1.5 \times 10^{-4} \, \mathrm{m}$$

Elongation of the brass wire:

$$\Delta L_{\mathrm{brass}} = \frac{F_{\mathrm{brass}} L_{\mathrm{brass}}}{A \, Y_{\mathrm{brass}}} = \frac{(58.8)(1.0)}{(4.909 \times 10^{-6})(0.91 \times 10^{11})} = \frac{58.8}{4.467 \times 10^{5}} \approx 1.3 \times 10^{-4} \, \mathrm{m}$$

The opposite face of the cube is pulled downward (parallel to itself) by the hanging weight, while the face glued to the wall is held fixed. This is a pure shear deformation.

Side of cube $$L = 0.10 \, \mathrm{m}$$. Area of the sheared face:

$$A = L^{2} = (0.10)^{2} = 1.0 \times 10^{-2} \, \mathrm{m^{2}}$$

Tangential (shearing) force:

$$F = m g = (100)(9.8) = 980 \, \mathrm{N}$$

Shearing stress:

$$\sigma_{s} = \frac{F}{A} = \frac{980}{1.0 \times 10^{-2}} = 9.8 \times 10^{4} \, \mathrm{N\,m^{-2}}$$

With shear modulus $$G = 25 \, \mathrm{GPa} = 25 \times 10^{9} \, \mathrm{N\,m^{-2}}$$, the shear strain is

$$\theta = \frac{\sigma_{s}}{G} = \frac{9.8 \times 10^{4}}{25 \times 10^{9}} = 3.92 \times 10^{-6}$$

The vertical deflection of the loaded face equals the shear strain times the distance from the fixed face, $$L$$:

$$\Delta x = \theta \, L = (3.92 \times 10^{-6})(0.10) = 3.92 \times 10^{-7} \, \mathrm{m}$$

Total weight to be supported:

$$W = M g = (5.0 \times 10^{4})(9.8) = 4.9 \times 10^{5} \, \mathrm{N}$$

With four columns sharing the load equally, the compressive force on one column is

$$F = \frac{W}{4} = \frac{4.9 \times 10^{5}}{4} = 1.225 \times 10^{5} \, \mathrm{N}$$

Cross-sectional area of one hollow column (with inner radius $$r_{1} = 0.30 \, \mathrm{m}$$ and outer radius $$r_{2} = 0.60 \, \mathrm{m}$$):

$$A = \pi (r_{2}^{2} - r_{1}^{2}) = \pi (0.60^{2} - 0.30^{2}) = \pi (0.36 - 0.09) = 0.27 \pi \approx 0.848 \, \mathrm{m^{2}}$$

Compressive stress in a column:

$$\sigma = \frac{F}{A} = \frac{1.225 \times 10^{5}}{0.848} \approx 1.444 \times 10^{5} \, \mathrm{N\,m^{-2}}$$

Taking Young's modulus of steel $$Y = 2.0 \times 10^{11} \, \mathrm{N\,m^{-2}}$$, the compressional strain is

$$\varepsilon = \frac{\sigma}{Y} = \frac{1.444 \times 10^{5}}{2.0 \times 10^{11}} \approx 7.22 \times 10^{-7}$$

Cross-sectional area of the rod:

$$A = (15.2 \times 10^{-3})(19.1 \times 10^{-3}) = 2.9032 \times 10^{-4} \, \mathrm{m^{2}}$$

Tensile stress:

$$\sigma = \frac{F}{A} = \frac{4.45 \times 10^{4}}{2.9032 \times 10^{-4}} \approx 1.533 \times 10^{8} \, \mathrm{N\,m^{-2}}$$

Within the elastic limit, strain follows Hooke's law $$\varepsilon = \sigma / Y$$. Using Young's modulus of copper $$Y = 1.1 \times 10^{11} \, \mathrm{N\,m^{-2}}$$:

$$\varepsilon = \frac{\sigma}{Y} = \frac{1.533 \times 10^{8}}{1.1 \times 10^{11}} \approx 1.39 \times 10^{-3}$$

Cross-sectional area of the cable:

$$A = \pi r^{2} = \pi (1.5 \times 10^{-2})^{2} = \pi (2.25 \times 10^{-4}) \approx 7.069 \times 10^{-4} \, \mathrm{m^{2}}$$

The maximum load is obtained when the stress reaches the permitted limit $$\sigma_{\max} = 10^{8} \, \mathrm{N\,m^{-2}}$$:

$$F_{\max} = \sigma_{\max} \cdot A = (10^{8})(7.069 \times 10^{-4}) \approx 7.07 \times 10^{4} \, \mathrm{N}$$

Converted to a mass (using $$g = 9.8 \, \mathrm{m\,s^{-2}}$$), this corresponds to

$$m_{\max} = \frac{F_{\max}}{g} = \frac{7.07 \times 10^{4}}{9.8} \approx 7.2 \times 10^{3} \, \mathrm{kg}$$

The bar is rigid and is held symmetrically by three wires of the same length $$L$$ from the same horizontal ceiling. As the bar hangs, all three lower attachment points must remain in the same horizontal line, so all three wires must elongate by the same amount $$\Delta L$$.

Using $$\Delta L = F L / (A Y)$$ and setting the elongations equal for the (equal-tension) wires,

$$\frac{T L}{A_{\mathrm{Cu}} Y_{\mathrm{Cu}}} = \frac{T L}{A_{\mathrm{Fe}} Y_{\mathrm{Fe}}}$$

The tension $$T$$ and length $$L$$ cancel, giving

$$A_{\mathrm{Cu}} \, Y_{\mathrm{Cu}} = A_{\mathrm{Fe}} \, Y_{\mathrm{Fe}}$$

For wires of circular cross-section $$A = \pi d^{2}/4$$, so

$$\frac{d_{\mathrm{Cu}}^{2}}{d_{\mathrm{Fe}}^{2}} = \frac{Y_{\mathrm{Fe}}}{Y_{\mathrm{Cu}}} \quad\Longrightarrow\quad \frac{d_{\mathrm{Cu}}}{d_{\mathrm{Fe}}} = \sqrt{\frac{Y_{\mathrm{Fe}}}{Y_{\mathrm{Cu}}}}$$

Using $$Y_{\mathrm{Fe}} = 1.9 \times 10^{11} \, \mathrm{N\,m^{-2}}$$ and $$Y_{\mathrm{Cu}} = 1.1 \times 10^{11} \, \mathrm{N\,m^{-2}}$$:

$$\frac{d_{\mathrm{Cu}}}{d_{\mathrm{Fe}}} = \sqrt{\frac{1.9 \times 10^{11}}{1.1 \times 10^{11}}} = \sqrt{1.727} \approx 1.31$$

Hence the copper wires need to be slightly thicker than the iron wire in the ratio above.

Angular velocity:

$$\omega = 2 \, \mathrm{rev\,s^{-1}} = 2 \times 2\pi \, \mathrm{rad\,s^{-1}} = 4\pi \, \mathrm{rad\,s^{-1}}$$

At the lowest point of the vertical circle the tension in the wire must both support the weight and provide the centripetal force:

$$T = m g + m \omega^{2} L = m\,(g + \omega^{2} L)$$

With $$m = 14.5 \, \mathrm{kg}$$, $$L = 1.0 \, \mathrm{m}$$, $$\omega^{2} = (4\pi)^{2} = 16\pi^{2} \approx 157.9 \, \mathrm{rad^{2}\,s^{-2}}$$:

$$T = 14.5 \,(9.8 + 157.9 \times 1.0) = 14.5 \times 167.7 \approx 2.43 \times 10^{3} \, \mathrm{N}$$

The cross-section is $$A = 0.065 \, \mathrm{cm^{2}} = 6.5 \times 10^{-6} \, \mathrm{m^{2}}$$ and $$Y_{\mathrm{steel}} = 2.0 \times 10^{11} \, \mathrm{N\,m^{-2}}$$. The elongation is

$$\Delta L = \frac{T L}{A Y} = \frac{(2.43 \times 10^{3})(1.0)}{(6.5 \times 10^{-6})(2.0 \times 10^{11})} = \frac{2430}{1.3 \times 10^{6}} \approx 1.87 \times 10^{-3} \, \mathrm{m}$$

So the wire stretches by roughly $$1.87 \, \mathrm{mm}$$ at the lowest point.

The magnitude of the volume change is

$$|\Delta V| = 0.5 \, \mathrm{L} = 0.5 \times 10^{-3} \, \mathrm{m^{3}}$$

and the pressure increase is

$$\Delta p = 100 \times 1.013 \times 10^{5} = 1.013 \times 10^{7} \, \mathrm{Pa}$$

With initial volume $$V = 100 \, \mathrm{L} = 0.1 \, \mathrm{m^{3}}$$, the bulk modulus of water is

$$B_{\mathrm{water}} = \frac{\Delta p}{|\Delta V| / V} = \frac{\Delta p \, V}{|\Delta V|} = \frac{(1.013 \times 10^{7})(0.1)}{0.5 \times 10^{-3}} \approx 2.026 \times 10^{9} \, \mathrm{N\,m^{-2}}$$

For air at constant temperature, $$p V = \mathrm{constant}$$, so the isothermal bulk modulus equals the pressure itself; at atmospheric pressure

$$B_{\mathrm{air}} \approx 1.0 \times 10^{5} \, \mathrm{N\,m^{-2}}$$

Hence

$$\frac{B_{\mathrm{water}}}{B_{\mathrm{air}}} \approx \frac{2.026 \times 10^{9}}{1.0 \times 10^{5}} \approx 2 \times 10^{4}$$

Why is the ratio so large? In water (a liquid) the molecules are already packed almost as closely as possible; compressing them further requires overcoming strong intermolecular repulsion, so water resists volume change very strongly. In a gas the molecules are far apart with mostly empty space between them, and only weak interactions, so they can be squeezed together easily with a small pressure increase.

Treat the extra (gauge) pressure $$\Delta p = 80 \, \mathrm{atm} = 80 \times 1.013 \times 10^{5} \approx 8.10 \times 10^{6} \, \mathrm{Pa}$$ as compressing a fixed mass of water by a small fractional volume change. With bulk modulus $$B = 2.2 \times 10^{9} \, \mathrm{Pa}$$,

$$\frac{|\Delta V|}{V} = \frac{\Delta p}{B} = \frac{8.10 \times 10^{6}}{2.2 \times 10^{9}} \approx 3.68 \times 10^{-3}$$

Since the mass $$m$$ is conserved, $$\rho \propto 1/V$$. Writing the new volume as $$V' = V - |\Delta V|$$,

$$\rho' = \frac{m}{V'} = \rho \cdot \frac{V}{V - |\Delta V|} = \frac{\rho}{1 - |\Delta V|/V}$$

For small fractional changes, $$1/(1-x) \approx 1 + x$$, so

$$\rho' \approx \rho \,\bigl(1 + \tfrac{|\Delta V|}{V}\bigr) = (1.03 \times 10^{3})(1 + 3.68 \times 10^{-3})$$

$$\rho' \approx 1.03 \times 10^{3} \times 1.00368 \approx 1.034 \times 10^{3} \, \mathrm{kg\,m^{-3}}$$

The density increases by only about $$0.37\%$$, confirming that water is nearly incompressible.

The applied pressure is

$$p = 10 \, \mathrm{atm} = 10 \times 1.013 \times 10^{5} \approx 1.013 \times 10^{6} \, \mathrm{Pa}$$

For a hydraulic (isotropic) compression, the fractional change in volume is

$$\frac{\Delta V}{V} = \frac{p}{B}$$

Using the bulk modulus of glass $$B = 3.7 \times 10^{10} \, \mathrm{N\,m^{-2}}$$:

$$\frac{\Delta V}{V} = \frac{1.013 \times 10^{6}}{3.7 \times 10^{10}} \approx 2.74 \times 10^{-5}$$

So the glass slab is compressed by only about 27 parts per million of its volume.

Initial volume of the cube:

$$V = (0.10)^{3} = 1.0 \times 10^{-3} \, \mathrm{m^{3}}$$

For a hydraulic compression, $$\Delta V / V = p / B$$. Using the bulk modulus of copper $$B = 1.4 \times 10^{11} \, \mathrm{N\,m^{-2}}$$:

$$\frac{\Delta V}{V} = \frac{p}{B} = \frac{7.0 \times 10^{6}}{1.4 \times 10^{11}} = 5.0 \times 10^{-5}$$

Therefore the volume contraction is

$$\Delta V = \left(\frac{\Delta V}{V}\right) V = (5.0 \times 10^{-5})(1.0 \times 10^{-3}) = 5.0 \times 10^{-8} \, \mathrm{m^{3}}$$

or equivalently $$5.0 \times 10^{-2} \, \mathrm{cm^{3}} = 0.05 \, \mathrm{cm^{3}}$$.

The required fractional change in volume is

$$\frac{\Delta V}{V} = 0.10\% = 1.0 \times 10^{-3}$$

From the definition of bulk modulus $$B = \Delta p / (\Delta V / V)$$, the required pressure change is

$$\Delta p = B \cdot \frac{\Delta V}{V}$$

Using the bulk modulus of water $$B = 2.2 \times 10^{9} \, \mathrm{N\,m^{-2}}$$:

$$\Delta p = (2.2 \times 10^{9})(1.0 \times 10^{-3}) = 2.2 \times 10^{6} \, \mathrm{N\,m^{-2}}$$

In terms of atmospheres,

$$\Delta p = \frac{2.2 \times 10^{6}}{1.013 \times 10^{5}} \approx 21.7 \, \mathrm{atm}$$

The initial volume of $$1 \, \mathrm{L}$$ does not enter the answer because the required quantity is the pressure increase, which depends only on the fractional compression and the bulk modulus.