0
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.

The maximum shear stress is limited to 60 N/mm2. Take the value of modulus of rigidity = 8 × 104 N/mm2.

somd-2 • 297  views
0  upvotes
0

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.

0  upvotes
Please log in to add an answer.

Next up

Read More Questions

If you are looking for answer to specific questions, you can search them here. We'll find the best answer for you.

Search

Study Full Subject

If you are looking for good study material, you can checkout our subjects. Hundreds of important topics are covered in them.

Know More