Question 7.21: A shaft made of AISI 1030 cold drawn steel transmits 50 kW a......

Given P = 50 kW, speed = 300 rpm, material = AISI 1030 CD steel

\sigma_{\mathrm{ult}}=520 \mathrm{~MPa}, \sigma_y=440 \mathrm{~MPa}

Assumption: Equal strength in tension and compression

Allowable strength for shaft design

Permissible shear stress is

30 \% \text { of } \sigma_y=0.3 \times 440=132.0 \mathrm{~MPa},

18 \% \text { of } \sigma_{\mathrm{ult}}=0.18 \times 520=93.6 \mathrm{~MPa}

Hence the allowable strength is 93.6 MPa.

Because of keyways, the allowable strength is further reduced by 25%

\tau_{\mathrm{all}}=0.75 \times 93.6=70.2 \mathrm{~MPa}

Torque due to power transmission is

T=\frac{9550 \times \mathrm{kW}}{\mathrm{rpm}}=\frac{9550 \times 50}{900}=530.56 \mathrm{~N} \cdot \mathrm{m}

Shear stress induced due to T is

\tau=\frac{16 T}{\pi d^3}=\frac{16 \times 530.56 \times 10^3}{\pi \times d^3} (7.85)

There is no other stress induced in the shaft as weight of the gear and shaft are not considered in this example.

Now equating Eq. (7 .85) to allowable strength, we get

\frac{16 \times 530.56 \times 10^3}{\pi \times d^3}=70.2

Solving, we get

d = 33.76 mm

d = 35 mm

Design of the square key

Width of the key b=d / 4=\frac{35}{4}=8.75 \mathrm{~mm}

Height or thickness of the key h = 8.75 mm

Length of the key is obtained from the shear stress equation,

\tau=\frac{8 T}{d^2 l}=\frac{8 \times 530.56 \times 10^3}{(35)^2 \times l} (7.86)

Equating (7.86) to allowable strength

\frac{8 \times 530.56 \times 10^3}{(35)^2 \times l}=70.2

Solving, we get

l = 49.35 mm

= 50.0 mm

From crushing failure consideration,

\sigma_c=\frac{16 T}{d^2 l}=\frac{16 \times 530.56 \times 10^3}{(35)^2 \times 50}=138.59 \mathrm{~MPa}

Now the factor of safety is

f o s=\frac{440}{138.59}=3.17

Therefore, the design is safe. Hence, the dimensions of the key are 10 × 10 × 50 mm