Question: Determine the diameter of a solid steel shaft which will transmit 90 kW at 160 rpm. Also find the length of the shaft if the twist must not exceed $1^{o}$ over the entire length.
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The maximum shear stress is limited to 60 N/mm2. Take the value of modulus of rigidity = 8 × 104 N/mm2.

somd-2 • 27 views
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modified 8 days ago  • written 8 days ago by gravatar for bharathchippa49 bharathchippa490
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Given:

For Solid circular steel shaft-

p = 90 kW = $90 \times 10^{3} watts$,

N = 160 rpm , $q_{max} = 60 N/mm^{2}$,

$G = 8 \times 10^{4} N/mm^{2}$

$\theta = a^{o} = 1 \times \frac{\pi}{180} = 0.0175 rad$

Solution:

$P = \frac{2 \pi N T_{avg}}{60}$

$90 \times 10^{3} = 3 \pi \times 160 \times T_{avg}/60$

$\therefore T_{avg} = 5.371 \times 10^{3} N-m = 5.371 \times 10^{6} N-mm$

Student may assume $T_{max} = T_{avg} = 5.371 \times 10^{6} N-mm$

Using the relation,

$\frac{T}{i_{P}} = \frac{q_{max}}{R}$

$\therefore \frac{5.371 \times 10^{6} \times 32}{\pi \times 60 \times 2} = \frac{60 \times 2}{d}$

$\therefore d^{3} = \frac{5.371 \times 10^{6} \times 32}{\pi \times 60 \times 2} = 455.90 \times 10^{3}$

$\therefore$ d = 76.96 mm

Now, $I_{p} = \frac{\pi}{32} \times d^{4} = \frac{\pi}{32} \times 76.96^{4} = 3.444 \times 10^{6} mm^{4}$

Using the relation,

$\frac{T}{I_{p}} = \frac{G_{o}}{L}$

$\therefore L = \frac{G_{o} - I_{p}}{T} = \frac{8 \times 10^{4} \times 0.0175 \times 3.444 \times 10^{6}}{5.371 \times 10^{6}}$

L = 897.71 mm

$\therefore$ Length of shaft = L = 0.897m say 0.9m.

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modified 8 days ago  • written 8 days ago by gravatar for bharathchippa49 bharathchippa490
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