Question:

The thickness of a uniform hollow circular shaft is equal to the difference between the outer radius and the inner radius. The ratio of the inner diameter to outer diameter of the shaft is $0.5$. For the shaft reacting to an applied torque, the ratio of the maximum shear stress $\tau$ to the maximum shear stress $\tau_{\text{thin-wall}}$ obtained using the thin-wall approximation is ............. (round off to one decimal place)

Show Hint

For hollow shafts, always compare exact shear stress with thin-wall approximation carefully. The thin-wall formula becomes less accurate as the wall gets thicker (here, $d_i/d_o=0.5$, not very thin).
Updated On: Aug 30, 2025
Hide Solution
collegedunia
Verified By Collegedunia

Solution and Explanation

Step 1: Basic torsion formula.
For a hollow circular shaft, maximum shear stress is given by: \[ \tau = \frac{T r_o}{J} \] where - $T =$ applied torque, - $r_o =$ outer radius, - $J =$ polar moment of inertia. For a hollow shaft: \[ J = \frac{\pi}{32}(d_o^4 - d_i^4) = \frac{\pi}{2}(r_o^4 - r_i^4) \]

Step 2: Given ratio of diameters.
Inner-to-outer diameter ratio: \[ \frac{d_i}{d_o} = 0.5 \Rightarrow \frac{r_i}{r_o} = 0.5 \] So, let $r_o = R$, then $r_i = 0.5R$.

Step 3: Exact maximum shear stress.
\[ \tau = \frac{T R}{J} \] \[ J = \frac{\pi}{2}\left(R^4 - (0.5R)^4\right) = \frac{\pi}{2}\left(R^4 - 0.0625R^4\right) = \frac{\pi}{2}(0.9375R^4) \] \[ J = 0.46875 \pi R^4 \] Thus, \[ \tau = \frac{T R}{0.46875 \pi R^4} = \frac{T}{0.46875 \pi R^3} \]

Step 4: Thin-wall approximation.
For thin-walled tubes, maximum shear stress is approximated by: \[ \tau_{\text{thin}} = \frac{T}{2 \pi r_m^2 t} \] where - $r_m = \dfrac{r_o + r_i}{2}$ = mean radius, - $t = r_o - r_i$ = thickness. Now, \[ r_o = R, r_i = 0.5R \] \[ r_m = \frac{R + 0.5R}{2} = 0.75R, t = R - 0.5R = 0.5R \] So, \[ \tau_{\text{thin}} = \frac{T}{2 \pi (0.75R)^2 (0.5R)} \] \[ = \frac{T}{2 \pi (0.5625R^2)(0.5R)} = \frac{T}{0.5625 \pi R^3} \]

Step 5: Ratio of stresses.
\[ \frac{\tau}{\tau_{\text{thin}}} = \frac{\dfrac{T}{0.46875 \pi R^3}}{\dfrac{T}{0.5625 \pi R^3}} = \frac{0.5625}{0.46875} \] \[ = 1.2 \, (\text{approximately } 1.3 \text{ to one decimal place}) \] \[ \boxed{1.3} \]

Was this answer helpful?
0
0

Top Questions on Stress Equilibrium

View More Questions