Correct Answer - Option 2 : 72.75 N/mm
2 and 54.54 N/mm
2
Concept:
Shear stress-induced in the shaft at any location (r) from the centre is given by,
\(\frac{\tau }{r} = \frac{T}{{{I_P}}}\)
Where, r = Radial distance from centre to the location at which shear stress is calculated (m), IP = Polar moment of inertia (m4) and τ = Shear stress N/m2
Polar moment of inertia for hollow circular shaft,
\({I_p} = \frac{\pi }{{32}} \times \left( {D_o^4 - D_i^4} \right)\)
Where, Do = External diameter and Di = Internal diameter of shaft
Calculation:
External diameter (do) = 16 mm
Internal Diameter (di) = 12 mm
Torque Applied (T) = 40 N-m
Polar moment of inertia of hollow circular shaft –
\({I_{P}} = \frac{\pi }{{32}}\left( {{{16}^4} - {{12}^4}} \right) = 4396\;m{m^4}\)
Shear stress at inside of material,
\({\tau _{inside}} = \frac{T}{{{I_P}}} \times {r_i}\)
\( = \frac{{40}}{{4396}} \times \frac{{0.012}}{2}\)
\(= 54.595\; \times {10^{ - 6}}\;N/{m^2} \approx 54.595\;MPa\)
Shear stress at outside of material,
\({\tau _{inside}} = \frac{T}{{{I_P}}} \times {r_o}\)
\( = \frac{{40}}{{4396}} \times \frac{{0.016}}{2}\)
\( = 72.793 \times {10^{ - 6}} \approx 72.793\;MPa\)