Shaft Dimensions Analysis

We are given two solid shafts, P and Q, transmitting the same torque ($T$). They are made of the same material.

• Shaft P has cross-sectional area $A_P = A$.
• Shaft Q has cross-sectional area $A_Q = 4A$.

The cross-section is circular. Let $r_P$ and $r_Q$ be the radii of shafts P and Q, respectively.

The area of a circular cross-section is given by $A = \pi r^2$. Therefore:

For shaft P: $A_P = \pi r_P^2 = A$

For shaft Q: $A_Q = \pi r_Q^2 = 4A$

Comparing the areas:

$\pi r_Q^2 = 4 (\pi r_P^2)$

$r_Q^2 = 4 r_P^2$

$r_Q = 2 r_P$

This means the radius of shaft Q is twice the radius of shaft P.

Polar Moment of Inertia Comparison

The polar moment of inertia ($J$) for a solid circular shaft is given by $J = \frac{\pi r^4}{2}$.

For shaft P: $J_P = \frac{\pi r_P^4}{2}$

For shaft Q: $J_Q = \frac{\pi r_Q^4}{2}$

Substituting $r_Q = 2 r_P$ into the expression for $J_Q$:

$J_Q = \frac{\pi (2 r_P)^4}{2} = \frac{\pi (16 r_P^4)}{2} = 16 \left( \frac{\pi r_P^4}{2} \right)$

$J_Q = 16 J_P$

The polar moment of inertia of shaft Q is 16 times that of shaft P.

Torsional Shear Stress Calculation

The maximum torsional shear stress ($\tau$) in a solid circular shaft is related to the applied torque ($T$), the outer radius ($r$), and the polar moment of inertia ($J$) by the formula:

$\tau = \frac{T r}{J}$

For shaft P, the maximum shear stress is given:

$\tau_P = \frac{T r_P}{J_P} = 160 \text{ MPa}$

For shaft Q, the maximum shear stress is:

$\tau_Q = \frac{T r_Q}{J_Q}$

Substitute $r_Q = 2 r_P$ and $J_Q = 16 J_P$ into the equation for $\tau_Q$:

$\tau_Q = \frac{T (2 r_P)}{16 J_P} = \frac{2}{16} \frac{T r_P}{J_P} = \frac{1}{8} \frac{T r_P}{J_P}$

Since $\tau_P = \frac{T r_P}{J_P}$, we can write:

$\tau_Q = \frac{1}{8} \tau_P$

Now, substitute the value of $\tau_P$:

$\tau_Q = \frac{1}{8} \times 160 \text{ MPa}$

$\tau_Q = 20 \text{ MPa}$

The maximum torsional shear stress developed in shaft Q is 20 MPa.