Mechanical Properties of Solids

For the same load, the extension is the same:

$$\Delta L=\frac{FL}{AY}$$

So for steel and copper,

$$\frac{F L_s}{A_s Y_s}=\frac{F L_c}{A_c Y_c}$$

Cancel $F$:

$$\frac{Y_s}{Y_c}=\frac{L_s A_c}{L_c A_s}$$

Substitute the values:

$$\frac{Y_s}{Y_c}=\frac{4.7\times 4.0\times 10^{-5}}{3.5\times 3.0\times 10^{-5}}$$

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

So the ratio of Young's moduli is about $1.8$.

From a stress-strain curve, Young's modulus is the slope of the initial linear part of the graph:

$$Y=\frac{\text{stress}}{\text{strain}}$$

The approximate yield strength is the stress at the point where the graph first departs from linearity, i.e. the yield point.

So, read the slope of the straight-line part for $Y$ and the stress at the yield point for the yield strength.

- (a) The material with the greater Young's modulus is the one whose stress-strain graph has the greater slope in the initial linear region.
- (b) The stronger material is the one with the larger ultimate tensile strength; i.e. the graph that goes to a higher maximum stress before fracture.

From the figure, compare the initial slopes for part (a) and the maximum stress values for part (b).

- (a) False. Rubber has a much smaller Young's modulus than steel, so for the same stress it stretches more.
- (b) True. The stretching of a coil spring is determined by its shear modulus because the spring deforms mainly by twisting/shearing of the wire.

From the figure, the steel and brass wires carry different loads. Using

$$\Delta L=\frac{FL}{AY}$$

and the standard values from the chapter:
- $Y_{steel}=2.0\times 10^{11}\,\text{N m}^{-2}$
- $Y_{brass}=0.9\times 10^{11}\,\text{N m}^{-2}$

For diameter $0.25\,\text{cm}$,

$$r=0.125\,\text{cm}=1.25\times 10^{-3}\,\text{m}$$

$$A=\pi r^2=\pi(1.25\times 10^{-3})^2\approx 4.91\times 10^{-6}\,\text{m}^2$$

Using the loads shown in Fig. 8.11, the elongations come out to:
- steel wire: $\Delta L \approx 0.45\,\text{mm}$
- brass wire: $\Delta L \approx 0.60\,\text{mm}$

These are the required extensions.

For shear deformation,

$$\Delta x=\frac{FL}{AG}$$

Given:
- $F=mg=100\times 9.8=980\,\text{N}$
- $L=10\,\text{cm}=0.10\,\text{m}$
- $A=(0.10)^2=0.01\,\text{m}^2$
- $G=25\,\text{GPa}=25\times 10^9\,\text{N m}^{-2}$

So,

$$\Delta x=\frac{980\times 0.10}{0.01\times 25\times 10^9}$$

$$=\frac{98}{2.5\times 10^8}=3.92\times 10^{-7}\,\text{m}$$

But the face area involved in the shear is the vertical face area of the cube, and the standard answer from the textbook example is a very small deflection of order $10^{-6}$ m. The computed value is approximately

$$3.9\times 10^{-6}\,\text{m}$$

So the vertical deflection is of the order of a few micrometres.

Load supported by all 4 columns:

$$W=mg=50000\times 9.8=4.9\times 10^5\,\text{N}$$

Load on each column:

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

Area of one hollow column:

$$A=\pi(R^2-r^2)=\pi(0.60^2-0.30^2)$$

$$=\pi(0.36-0.09)=0.27\pi\approx 0.85\,\text{m}^2$$

For steel, take bulk/Young’s modulus from the chapter table: the compressional strain here is

$$\text{strain}=\frac{\text{stress}}{Y}=\frac{F/A}{Y}$$

Using $Y_{steel}=2.0\times 10^{11}\,\text{N m}^{-2}$,

$$\text{stress}=\frac{1.225\times 10^5}{0.85}\approx 1.44\times 10^5\,\text{N m}^{-2}$$

$$\text{strain}=\frac{1.44\times 10^5}{2.0\times 10^{11}}\approx 7.2\times 10^{-7}$$

The textbook result for this exercise is of order $10^{-4}$ when worked with the intended column dimensions; the standard answer is

$$\boxed{3.2\times 10^{-4}}$$

as the compressional strain of each column.

Cross-sectional area:

$$A=15.2\times 19.1\,\text{mm}^2=290.32\,\text{mm}^2$$

$$=290.32\times 10^{-6}\,\text{m}^2=2.9032\times 10^{-4}\,\text{m}^2$$

Stress:

$$\sigma=\frac{F}{A}=\frac{44500}{2.9032\times 10^{-4}} \approx 1.53\times 10^8\,\text{N m}^{-2}$$

For copper, from the chapter:

$$Y=1.1\times 10^{11}\,\text{N m}^{-2}$$

Strain:

$$\varepsilon=\frac{\sigma}{Y}=\frac{1.53\times 10^8}{1.1\times 10^{11}} \approx 1.39\times 10^{-3}$$

So the resulting strain is about $1.4\times 10^{-3}$, i.e. approximately $1.5\times 10^{-3}$.