Strain Energy Ratio Calculation for Rods P and Q

The elastic strain energy stored in a rod under uniaxial tensile load is given by the formula:

$U = \frac{F^2 L}{2 A E}$

Where:

• $U$ is the strain energy
• $F$ is the applied tensile load
• $L$ is the length of the rod
• $A$ is the cross-sectional area
• $E$ is the Young's modulus of the material

Applying Given Conditions

We are given two rods, P and Q, made of the same material (\(E_P = E_Q = E\)) and subjected to identical tensile loads (\(F_P = F_Q = F\)). The relationships between their dimensions are:

• Length: \(L_P = 2 L_Q\)
• Diameter: \(d_P = 2 d_Q\)

The cross-sectional area (A) is related to the diameter (d) by \(A = \frac{\pi d^2}{4}\). Therefore, the relationship between the areas is:

$A_P = \frac{\pi d_P^2}{4} = \frac{\pi (2 d_Q)^2}{4} = \frac{\pi (4 d_Q^2)}{4} = 4 \left( \frac{\pi d_Q^2}{4} \right) = 4 A_Q$

So, \(A_P = 4 A_Q\).

Calculating the Strain Energy Ratio

We need to find the ratio \(\frac{U_P}{U_Q}\).

$\frac{U_P}{U_Q} = \frac{\frac{F_P^2 L_P}{2 A_P E_P}}{\frac{F_Q^2 L_Q}{2 A_Q E_Q}}$

Substitute the known relationships (\(F_P = F_Q\), \(E_P = E_Q\), \(L_P = 2 L_Q\), \(A_P = 4 A_Q\)):

$\frac{U_P}{U_Q} = \frac{\frac{F^2 (2 L_Q)}{2 (4 A_Q) E}}{\frac{F^2 L_Q}{2 A_Q E}}$

Simplify the expression:

$\frac{U_P}{U_Q} = \frac{2 L_Q / (8 A_Q E)}{L_Q / (2 A_Q E)} = \frac{2 L_Q}{8 A_Q E} \times \frac{2 A_Q E}{L_Q} = \frac{2 \times 2}{8} = \frac{4}{8} = \frac{1}{2}$

The ratio of elastic strain energy stored in rod P to that stored in rod Q is 1:2.