Question 9.24: It is required to design a bushed-pin type flexible coupling...

Given kW = 20 n = 720 rpm
design torque = 1.5 (rated torque)
Step I Selection of materials
(i) The shafts are subjected to torsional shear stress. On the basis of strength, plain carbon steel of grade 40C8 \left(S_{y t}=380 N / mm ^{2}\right) is used for the shafts. The factor of safety for the shaft is assumed as 2.
(ii) The keys are subjected to shear and compressive stresses. The pins are subjected
to shear and bending stresses. On the basis of strength criterion, plain carbon steel of
grade 30C8 \left(S_{y t}=400 N / mm ^{2}\right) is selected for the keys and the pins. The factor of safety for the keys and the pins is taken as 2. It is assumed that the compressive yield strength is 150% of the tensile yield strength.

(iii) Flanges have complicated shape and the economic method to make the flanges is casting. Grey cast iron of grade FG 200 \left(S_{u t}=200 N / mm ^{2}\right) is selected as the material for the flanges from manufacturing considerations. It is assumed that the ultimate shear strength is one-half of the ultimate tensile strength. The factor of safety for the flanges is assumed as 6, since the permissible stress is based on the ultimate strength and not on the yield strength.
Step II Permissible stresses
(i) Shaft

\tau=\frac{S_{s y}}{(f s)}=\frac{0.5 S_{y t}}{(f s)}=\frac{0.5(380)}{(2)}=95 N / mm ^{2} .

(ii) Keys

\tau=\frac{S_{s y}}{(f s)}=\frac{0.5 S_{y t}}{(f s)}=\frac{0.5(400)}{(2)}=100 N / mm ^{2} .

\sigma_{c}=\frac{S_{y c}}{(f s)}=\frac{1.5 S_{y t}}{(f s)}=\frac{1.5(400)}{(2)}=300 N / mm ^{2} .

(iii) Pins

\tau=35 N / mm ^{2} \text { (As per IS 2693-1980) } .

\sigma_{t}=\frac{S_{y t}}{(f s)}=\frac{400}{2}=200 N / mm ^{2} .

(iv) Flanges

\tau=\frac{S_{s u}}{(f s)}=\frac{0.5 S_{u t}}{(f s)}=\frac{0.5(200)}{6}=16.67 N / mm ^{2} .

Step III Diameter of shafts
The starting torque of the motor is 150% of the rated torque. Therefore,

M_{t}=\frac{60 \times 10^{6}( kW )}{2 \pi n} \times(1.5) .

=\frac{60 \times 10^{6}(20)(1.5)}{2 \pi(720)}=397887.36 N – mm .

\tau=\frac{16 M_{t}}{\pi d^{3}} \quad \text { or } \quad 95=\frac{16(397887.36)}{\pi d^{3}} .

\therefore \quad d=27.73 \text { or } 30 mm .

Step IV Dimensions of flanges
The dimensions of flanges are as follows:

d_{h}=2 d=2(30)=60 mm .

l_{h}=1.5 d=1.5(30)=45 mm .

D = 4d = 4(30) = 120 mm
t = 0.5d = 0.5(30) = 15 mm.

t_{1}=0.25 d=0.25(30)=7.5 \text { or } 8 mm .

The hub is treated as a hollow cylinder subjected to torsional moment.
From Eq. (9.44),

J=\frac{\pi\left(d_{h}^{4}-d^{4}\right)}{32}                  (9.44).

J=\frac{\pi\left(d_{h}^{4}-d^{4}\right)}{32}=\frac{\pi\left(60^{4}-30^{4}\right)}{32} .

=1192823.46 mm ^{4} .

r=\frac{d_{h}}{2}=\frac{60}{2}=30 mm .

The torsional shear stress in the hub is given by,

\tau=\frac{M_{t} r}{J}=\frac{(397887.36)(30)}{(1192823.46)}=10.01 N / mm ^{2} .

\therefore \quad \tau<16.67 N / mm ^{2} .

The shear stress in the flange at the junction of the hub is determined by Eq. (9.45). It is given by,

M_{t}=\frac{1}{2} \pi d_{h}^{2} t \tau             (9.45).

\tau=\frac{2 M_{t}}{\pi d_{h}^{2} t}=\frac{2(397887.36)}{\pi(60)^{2}(15)}=4.69 N / mm ^{2} .

\therefore \quad \tau<16.67 N / mm ^{2} .

The stresses in the flange are within limit.
Step V Diameter of pins
The number of pins is selected as 6.
N = 6
From Eq. (9.52), the diameter of the pin is given by,

d_{1}=\frac{0.5 d}{\sqrt{N}}                 (9.52).

d_{1}=\frac{0.5 d}{\sqrt{N}}=\frac{0.5(30)}{\sqrt{6}}=6.12 \text { or } 7 mm .

The shear stress in the pin is calculated by Eq. (9.51).

\tau=\frac{8 M_{t}}{\pi d_{1}^{2} D N}                   (9.51).

\tau=\frac{8 M_{t}}{\pi d_{1}^{2} D N}=\frac{8(397887.36)}{\pi(7)^{2}(120)(6)}=28.72 N / mm ^{2} .

\therefore \quad \tau<35 N / mm ^{2} .

The design is safe on the basis of shear strength.
The bending stresses in the pin will be determined at a later stage, after deciding the dimensions of the rubber bushes.
Step VI Dimensions of bushes
The permissible intensity of pressure between the rubber bush and the cast iron flange is taken as 1 N/mm². The ratio of effective length to diameter for the rubber bush is assumed as 1.
From Eq. (9.50),

M_{t}=\frac{1}{2} D_{b}^{2} D N               (9.50).

D_{b}^{2}=\frac{2 M_{t}}{D N}=\frac{2(397887.36)}{(120)(6)} .

\text { or } D_{b}=33.25 \text { or } 35 mm .

\therefore \quad D_{b}=l_{b}=35 mm .

Step VII Bending stresses in pins
As shown in Fig. 9.40, the force P acting on each pin depends upon the torque and the relationship is given by,

M_{t}=P \times \frac{D}{2} \times N .

Therefore,

P=\frac{2 M_{t}}{D N}=\frac{2(397887.36)}{(120)(6)}=1105.24 N .

The subassembly of the bush and the pin is shown in Fig. 9.43. It is assumed that the force P is uniformly distributed over 35 mm of effective length \left(l_{b}\right) of the bush. At the section-XX,

M_{b}=P\left(5+\frac{35}{2}\right)=1105.24\left(5+\frac{35}{2}\right) .

=(1105.24 \times 22.5) N – mm .

\sigma_{b}=\frac{32 M_{b}}{\pi d_{1}^{3}} \quad \text { or } \quad 200=\frac{32(1105.24 \times 22.5)}{\pi d_{1}^{3}} .

\therefore \quad d_{1}=10.82 \text { or } 12 mm .

In the initial stages, the diameter of the pin was calculated as 7 mm by using empirical formula.
It was sufficient to keep the maximum shear stress within limit. The shear stress induced in the pin was 28.72 N/mm² and the limiting stress was 35 N/mm². However, it is observed that a 7 mm diameter pin is not sufficient to withstand bending stresses and the minimum diameter of the pin should be 12 mm. Therefore, bending becomes the criterion of design.
As shown in Fig. 9.43, the diameter of the pin should be enlarged at the section-XX in order to fix the pin in the driven flange. This enlarged diameter is taken as (12 + 6) or 18 mm. Other dimensions of the pin and the bush are shown in Fig. 9.44. The thickness of the brass lining is 2 mm. Therefore, the inner diameter of the rubber bush is (18 + 4) or 22 mm. The minimum thickness of the rubber bush is usually 10 mm. Therefore, the outside diameter of the rubber bush is (22 + 20) or 42 mm. The dimensions of the driving flange are shown in Fig. 9.45.
Step VIII Dimensions of keys
From Table 9.3, the standard cross-section of a flat sunk key for 30 mm diameter is 8 × 7 mm. The length of the key is equal to l_{h} .

Table 9.3 Dimensions of square and rectangular sunk keys ( in mm )

\therefore \quad l=l_{h}=45 mm .

The dimensions of the flat key are 8 × 7 × 45 mm.
From Eq. (9.27),

\tau=\frac{2 M_{t}}{d b l}                    (9.27)

\tau=\frac{2 M_{t}}{d h l}=\frac{2(397887.36)}{(30)(7)(45)}=168.42 N / mm ^{2} .

\therefore \quad \tau<100 N / mm ^{2} .

From Eq. (9.28),

\sigma_{c}=\frac{4 M_{t}}{d h l}               (9.28)

\sigma_{c}=\frac{2 M_{t}}{d h l}=\frac{2(397887.36)}{(30)(7)(45)}=168.42 N / mm ^{2} .

\therefore \quad \sigma_{c}<300 N / mm ^{2} .

The shear and compressive stresses induced in the key are within permissible limits and the design is safe.