Torsion Rod Formula Fundamentals

The torque ($T$) on a solid circular shaft is related to its length ($L$), polar moment of inertia ($J$), shear modulus ($G$), and angle of twist ($\theta$) using the formula:

$ T = \frac{GJ\theta}{L} $

The polar moment of inertia ($J$) for a solid circular shaft depends on the diameter ($d$) as follows:

$ J = \frac{\pi d^4}{32} $

This means $J$ is directly proportional to $d^4$. Therefore, $T$ is proportional to $\frac{d^4}{L}$. We can write $T = C \frac{d^4}{L}$, where $C$ is a constant incorporating $G$, $\theta$, and $\frac{\pi}{32}$.

Calculating Changes in Torque

We analyze the initial and final states:

• Initial State: Diameter $d_1$, Length $L_1$. Torque $T_1 = C \frac{d_1^4}{L_1}$.
• Final State:
  • Diameter increases by 5%: $d_2 = d_1 \times (1 + 0.05) = 1.05 d_1$.
  • Length increases by 20%: $L_2 = L_1 \times (1 + 0.20) = 1.20 L_1$.
  • Final Torque: $T_2 = C \frac{d_2^4}{L_2}$.

Determining the Percentage Increase (X%)

To find the percentage increase in torque (X%), we first calculate the ratio $\frac{T_2}{T_1}$:

$ \frac{T_2}{T_1} = \frac{C \frac{d_2^4}{L_2}}{C \frac{d_1^4}{L_1}} = \frac{d_2^4}{d_1^4} \cdot \frac{L_1}{L_2} $

Substitute the expressions for $d_2$ and $L_2$:

$ \frac{T_2}{T_1} = \frac{(1.05 d_1)^4}{d_1^4} \cdot \frac{L_1}{1.20 L_1} = (1.05)^4 \cdot \frac{1}{1.20} $

Now, perform the calculation:

$ (1.05)^4 \approx 1.215506 $

$ \frac{T_2}{T_1} \approx \frac{1.215506}{1.20} \approx 1.01292 $

The percentage increase X% is given by:

$ X\% = \left( \frac{T_2}{T_1} - 1 \right) \times 100 $

$ X = (1.01292 - 1) \times 100 = 0.01292 \times 100 \approx 1.29\% $

Therefore, the value of X, accurate to two decimal places, is 1.29.