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Ray Of Light [21]

3 years ago

9

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 part

icular 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 .

1 answer:

Norma-Jean [14]3 years ago

7 0

Answer:

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

Explanation:

<u>Polar moment of Inertia</u>

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

= 0.14374 in 4

<u>Maximum sustainable torque on the solid circular shaft</u>

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

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

= 3658.836 lb.in

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

= 304.9 lb.ft

<u>Maximum sustainable torque on the tubular shaft</u>

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

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

= 3334.8 lb.in

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

= 277.9 lb.ft

<u>Maximum sustainable power in the solid circular shaft</u>

$P_{max} = 2 \pi f_T$

= $2\pi(2.1) \times 304.9$

= 4023.061 lb. ft/s

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

= 7.315 hp

<u>Maximum sustainable power in the tubular shaft</u>

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

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

= 3666.804 lb.ft /s

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

= 6.667 hp

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<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> docx </span>

<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> pdf </span>

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