Mechanical Properties of Solids

Given:
Steel: $L_s = 4.7 \text{ m}, A_s = 3.0 \times 10^{-5} \text{ m}^2$.
Copper: $L_c = 3.5 \text{ m}, A_c = 4.0 \times 10^{-5} \text{ m}^2$.
Condition: Same Load ($F$) and same Extension ($\Delta L$).

Formula: $Y = \frac{F L}{A \Delta L}$.

Q8.2 Analysis:
(a) Young’s Modulus ($Y$): It is the slope of the linear part of the graph.
Let’s take a point from the graph (example values): Stress $= 150 \times 10^6 \text{ N m}^{-2}$, Strain $= 0.002$.
$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{150 \times 10^6}{0.002} = \mathbf{7.5 \times 10^{10} \text{ N m}^{-2}}$.
(b) Yield Strength ($S_y$): The stress at which the graph stops being linear (elastic limit).
From graph reading: $\approx \mathbf{300 \times 10^6 \text{ N m}^{-2}}$.

Q8.3 Comparison:
(a) Greater Young’s Modulus: Material A. (The graph for A is steeper; a steeper slope indicates a higher modulus).
(b) Stronger Material: Material A. (It has a higher breaking stress/fracture point on the y-axis, meaning it can withstand more stress before failing).

(a) Young’s modulus of rubber > steel: FALSE.
For the same applied force, rubber stretches much more (larger strain). Since $Y = \text{Stress}/\text{Strain}$, a larger strain implies a smaller $Y$. Steel resists deformation more strongly, making it more elastic in physics terms.

(b) Stretching of a coil is determined by its shear modulus: TRUE.
When a helical spring stretches, the wire itself undergoes torsion (twisting). Twisting involves shear strain, not tensile strain of the wire material itself. Therefore, the Shear Modulus is the determining factor.

Data: $d=0.25 \text{ cm} \Rightarrow r=0.125 \text{ cm}$.
Area $A = \pi r^2 = \pi (1.25 \times 10^{-3})^2 \approx 4.9 \times 10^{-6} \text{ m}^2$.
$Y_{steel} = 2.0 \times 10^{11} \text{ Pa}$, $Y_{brass} = 0.91 \times 10^{11} \text{ Pa}$.

1. Brass Wire (Bottom):
Load = Mass of block M1 = $6 \text{ kg}$.
Force $F_B = 6 \times 9.8 = 58.8 \text{ N}$. Length $L_B = 1.0 \text{ m}$.
$\Delta L_B = \frac{F_B L_B}{A Y_B} = \frac{58.8 \times 1.0}{4.9 \times 10^{-6} \times 0.91 \times 10^{11}} \approx \mathbf{1.3 \times 10^{-4} \text{ m}}$.

2. Steel Wire (Top):
Load = Total Mass ($4 \text{ kg} + 6 \text{ kg}$) = $10 \text{ kg}$.
Force $F_S = 10 \times 9.8 = 98 \text{ N}$. Length $L_S = 1.5 \text{ m}$.
$\Delta L_S = \frac{F_S L_S}{A Y_S} = \frac{98 \times 1.5}{4.9 \times 10^{-6} \times 2.0 \times 10^{11}} \approx \mathbf{1.5 \times 10^{-4} \text{ m}}$.

Given: Cube side $L = 10 \text{ cm} = 0.1 \text{ m}$. Face Area $A = 0.1 \times 0.1 = 0.01 \text{ m}^2$.
Load Mass $M = 100 \text{ kg}$. Shear Force $F = Mg = 100 \times 9.8 = 980 \text{ N}$.
Shear Modulus $\eta = 25 \text{ GPa} = 25 \times 10^9 \text{ Pa}$.

Total Mass $M = 50,000 \text{ kg}$. Weight $W = Mg = 50000 \times 9.8 = 4.9 \times 10^5 \text{ N}$.
Since there are 4 columns, load per column $F = W/4 = 1.225 \times 10^5 \text{ N}$.
Area $A = \pi (R^2 – r^2)$ where $R=0.6 \text{ m}, r=0.3 \text{ m}$.
$A = \pi (0.36 – 0.09) = \pi(0.27) \approx 0.848 \text{ m}^2$.
$Y_{steel} = 2.0 \times 10^{11} \text{ Pa}$.

Compressional Strain:
$\text{Strain} = \frac{\text{Stress}}{Y} = \frac{F/A}{Y} = \frac{1.225 \times 10^5}{0.848 \times 2.0 \times 10^{11}}$
$= \mathbf{7.22 \times 10^{-7}}$.

Given: Force $F = 44,500 \text{ N}$.
Dimensions: $15.2 \text{ mm} \times 19.1 \text{ mm} = (15.2 \times 10^{-3}) \times (19.1 \times 10^{-3}) \text{ m}^2$.
Area $A \approx 2.9 \times 10^{-4} \text{ m}^2$.
$Y_{copper} \approx 1.1 \times 10^{11} \text{ Pa}$.

Strain Calculation:
Stress $\sigma = F/A = \frac{44500}{2.9 \times 10^{-4}} \approx 1.53 \times 10^8 \text{ Pa}$.
Strain $\varepsilon = \sigma/Y = \frac{1.53 \times 10^8}{1.1 \times 10^{11}} \approx \mathbf{1.39 \times 10^{-3}}$.

Max Stress $\sigma_{max} = 10^8 \text{ N m}^{-2}$.
Radius $r = 1.5 \text{ cm} = 1.5 \times 10^{-2} \text{ m}$.
Area $A = \pi r^2 = \pi (2.25 \times 10^{-4}) \approx 7.07 \times 10^{-4} \text{ m}^2$.

Condition: Same Tension ($F$) and Same Length ($L$).
For the rigid bar to remain horizontal, the extension ($\Delta L$) must be the same for all wires.
Formula: $\Delta L = \frac{F L}{A Y} = \frac{F L}{(\frac{\pi d^2}{4}) Y}$.

Since $\Delta L, F, L$ are constant across wires:
$d^2 Y = \text{constant} \implies d \propto \frac{1}{\sqrt{Y}}$.

Ratio: $\frac{d_{Cu}}{d_{Fe}} = \sqrt{\frac{Y_{Fe}}{Y_{Cu}}} = \sqrt{\frac{1.9 \times 10^{11}}{1.1 \times 10^{11}}} \approx \sqrt{1.727} \approx \mathbf{1.31}$.

Mass $m = 14.5 \text{ kg}$. Radius $l = 1.0 \text{ m}$. Angular velocity $\omega = 2 \text{ rev/s} = 4\pi \text{ rad/s}$.
Area $A = 0.065 \text{ cm}^2 = 6.5 \times 10^{-6} \text{ m}^2$. $Y_{steel} = 2.0 \times 10^{11} \text{ Pa}$.

Tension at Lowest Point:
Net force provides centripetal acceleration: $T – mg = m\omega^2 l$.
$T = m(g + \omega^2 l) = 14.5(9.8 + (4\pi)^2 \times 1)$
$T = 14.5(9.8 + 157.9) = 14.5(167.7) \approx 2431.65 \text{ N}$.

Elongation:
$\Delta l = \frac{T l}{A Y} = \frac{2431.65 \times 1.0}{6.5 \times 10^{-6} \times 2.0 \times 10^{11}}$
$\Delta l \approx 1.87 \times 10^{-3} \text{ m} = \mathbf{1.87 \text{ mm}}$.

Data: $V = 100.0 \text{ L}$. $\Delta V = 100.5 – 100.0 = 0.5 \text{ L}$.
$\Delta P = 100 \text{ atm} = 100 \times 1.013 \times 10^5 \text{ Pa} = 1.013 \times 10^7 \text{ Pa}$.

Comparison: $B_{air} \approx 1.0 \times 10^5 \text{ Pa}$ (at STP).
Ratio $\frac{B_{water}}{B_{air}} \approx 2 \times 10^4$.
Reason: Liquids are much less compressible than gases because their molecules are closer together, resulting in stronger intermolecular forces resisting compression.

Density at surface $\rho = 1.03 \times 10^3 \text{ kg m}^{-3}$.
Pressure $\Delta P = 80 \text{ atm} = 80 \times 1.013 \times 10^5 \text{ Pa}$.
Bulk Modulus of water $B \approx 2.2 \times 10^9 \text{ Pa}$.

Formula: $\rho’ = \frac{\rho}{1 – \frac{\Delta V}{V}} = \frac{\rho}{1 – \frac{\Delta P}{B}}$.
$\frac{\Delta P}{B} = \frac{8.104 \times 10^6}{2.2 \times 10^9} \approx 0.00368$.
$\rho’ = \frac{1030}{1 – 0.00368} \approx 1030(1 + 0.00368) \approx \mathbf{1034 \text{ kg m}^{-3}}$.

$\Delta P = 10 \text{ atm} = 10 \times 1.013 \times 10^5 \text{ Pa}$.
Bulk modulus of glass $B \approx 37 \times 10^9 \text{ Pa}$.

$\frac{\Delta V}{V} = \frac{\Delta P}{B} = \frac{1.013 \times 10^6}{37 \times 10^9} \approx \mathbf{2.74 \times 10^{-5}}$.

Side $L = 10 \text{ cm} = 0.1 \text{ m}$. Volume $V = L^3 = 10^{-3} \text{ m}^3$.
Pressure $P = 7.0 \times 10^6 \text{ Pa}$.
Bulk Modulus of Copper $B \approx 140 \times 10^9 \text{ Pa}$.

Percentage compression $\frac{\Delta V}{V} \times 100 = 0.1\% \implies \frac{\Delta V}{V} = 0.001$.
Bulk Modulus of Water $B \approx 2.2 \times 10^9 \text{ Pa}$.

$\Delta P = B \times \frac{\Delta V}{V} = 2.2 \times 10^9 \times 0.001 = \mathbf{2.2 \times 10^6 \text{ Pa}}$.