Question 7.25: A solid shaft is required to transmit a torque of 125 N·m. A......

The dimensions of the Woodruff key are shown in Figure 7.31.

As per notation, refer to Figure 7.30.

D = 16.0 mm

h = 4.0 mm

Diameter of the shaft is obtained as

d=\left(\frac{16 T}{\pi \tau_s}\right)^{1 / 3}=\left(\frac{16 \times 125 \times 1000}{\pi \times 87.075}\right)^{1 / 3}=19.4 \mathrm{~mm}

d = 20 mm

where

\tau_s=0.18 \times 645=116.1 \mathrm{~MPa}

or =0.3 \times 532=159.6 \mathrm{~MPa}

Hence the allowable stress is

\tau_s=0.75 \times 116.1=87.075 \mathrm{~MPa}

Force acting on the key is obtained from the relation

F=\frac{2 T}{d}

=\frac{2 \times 125 \times 1000}{20}=12.5 \mathrm{~kN}

Torque transmitted by shear of the key is

T=\tau_{\mathrm{key}} \times A_s \times \frac{d}{2}

125 \times 1000=204.64 \times A_s \times \frac{20}{2}

Solving for shear area A_s

A_s=\frac{125 \times 1000 \times 2}{204.64 \times 20}=61.08 \mathrm{~mm}^2

where \tau_{\text {key }}=0.577 \times \frac{\sigma_y}{2}=\frac{0.577 \times 532}{2}=204.64 \mathrm{~MPa}

Shear area A_s=\mathrm{CD} \times h

61.08 = CD × 4

CD = 15.27 mm

Torque capacity of the key in compression (inside the shaft)

T=\sigma_c \times A_c \times \frac{d}{2}

125 \times 1000=\frac{532}{1.5} \times A_c \times \frac{20}{2}

A_c=35.24 \mathrm{~mm}^2

Area of the key inside the hub

A_{c(h)}= Area of the key – area of the key in contact with the shaft

=A-A_{c(s)}

The total area of the key, A=\frac{\pi}{8} D^2

=\frac{\pi}{8} \times 16^2=100.53 \mathrm{~mm}^2

Considering the area A_c=A_{c(h)}, i.e. crushing area in the hub

35.24=100.53-A_{c(s)}

Solving, we get

A_{c(s)}=66.29 \mathrm{~mm}^2

\sin \theta=\frac{\mathrm{DE}}{\mathrm{OD}}=\frac{\mathrm{CD}}{2 \times \mathrm{OD}}

\sin \theta=\frac{15.27}{2 \times 8}=0.954

\theta=72.55^{\circ}

Now,

A_{c(s)}=\frac{2 \theta}{180} A-\frac{1}{2} \mathrm{CD} \times \mathrm{OE}

Substituting the values, we get

66.29=\frac{2 \times 72.55}{180} \times 100.53-\frac{1}{2} \times 15.27 \times \mathrm{OE}

Solving, we get

OE = 1.93 mm

Hence, the key should be at least 1.93 mm inside the hub.