Question

A solid circular shaft and a tubular shaft, both with the same outer radius of c=co = 0.550 in , are being considered for a particular design. The tubular shaft has an inner radius of ci = 0.300 in The shafts are to be powered by a motor operating at a frequency of f = 2.70 Hz and are to handle an attached load. Assume the cross sections are uniform throughout the lengths of the shafts and that the materials have an allowable shear stress of τallow = 14.5 ksi .

Answer 1

Answer:

The power for circular shaft is 7.315 hp and tubular shaft is 6.667 hp

Explanation:

Polar moment of Inertia

$(I_p)s = \frac{\pi(0.55)4}2$

      = 0.14374 in 4

Maximum sustainable torque on the solid circular shaft

$T_{max} = T_{allow} \frac{I_p}{r}$

         =$(14 \times 10^3) \times (\frac{0.14374}{0.55})$

         = 3658.836 lb.in

         = $\frac{3658.836}{12}$ lb.ft

        = 304.9 lb.ft

Maximum sustainable torque on the tubular shaft

$T_{max} = T_{allow}( \frac{Ip}{r})$

          = $(14 \times10^3) \times ( \frac{0.13101}{0.55})$

          = 3334.8 lb.in

          = $(\frac{3334.8}{12} )$ lb.ft

          = 277.9 lb.ft

Maximum sustainable power in the solid circular shaft

$P_{max} = 2 \pi f_T$

          = $2\pi(2.1) \times 304.9$

          = 4023.061 lb. ft/s

          = $(\frac{4023.061}{550})$ hp

          = 7.315 hp

Maximum sustainable power in the tubular shaft

$P _{max,t} = 2\pi f_T$

            = $2\pi(2.1) \times 277.9$

            = 3666.804 lb.ft /s

            = $(\frac{3666.804}{550})$hp

            = 6.667 hp