Question 9.22: It is required to design a rigid type of flange coupling to ...

Given kW = 37.5 n = 180 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 shaft. The factor of safety for the shafts is assumed to be 2.5.
(ii) The keys and bolts are subjected to shear and compressive 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 bolts. It is assumed that the compressive yield strength is 150% of the tensile yield strength. The factor of safety for the keys and the bolts is taken as 2.5.
(iii) Flanges have complex shape and the easiest method to make the flanges is casting. Grey cast iron FG 200 \left(S_{u t}=200 N / mm ^{2}\right) is manufacturing considerations. It is assumed that 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)}{(2.5)}=76 N / mm ^{2} .

(ii) Keys and bolts

\tau=\frac{S_{s y}}{(f s)}=\frac{0.5 S_{y t}}{(f s)}=\frac{0.5(400)}{(2.5)}=80 N / mm ^{2} .

\sigma_{c}=\frac{S_{y c}}{(f s)}=\frac{1.5 S_{y t}}{(f s)}=\frac{1.5(400)}{(2.5)}=240 N / mm ^{2} .

(3) 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
Taking into consideration the service factor of 1.5, the design torque is given by,

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

=\frac{60 \times 10^{6}(37.5)(1.5)}{2 \pi(180)} .

= 2 984 155.18 N-mm.

\tau=\frac{16 M_{t}}{\pi d^{3}} \quad \text { or } \quad 76=\frac{16(2984155.18)}{\pi d^{3}} .

∴          d = 58.48 or 60 mm
Step IV Dimensions of flanges
The dimensions of the flanges are as follows,

d_{h}=2 d=2(60)=120 mm .

l_{h}=1.5 d=1.5(60)=90 mm .

D = 3d = 3(60) = 180 mm.
t = 0.5d = 0.5(60) = 30 mm.

t_{1}=0.25 d=0.25(60)=15 mm .

d_{r}=1.5 d=1.5(60)=90 mm .

D_{0}=\left(4 d+2 t_{1}\right)=4(60)+2(15)=270 mm .

The above dimensions of the flange are shown in Fig. 9.37. The thickness of recess is assumed as 5 mm. The hub is treated as a hollow shaft 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(120^{4}-60^{4}\right)}{32} .

=19085175.37 mm ^{4} .

r=\frac{d_{h}}{2}=\frac{120}{2}=60 mm .

The torsional shear stress in the hub is given by,

\tau=\frac{M_{t} r}{J}=\frac{(2984155.18)(60)}{(19085175.37)}=9.38 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).

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(2984155.18)}{\pi(120)^{2}(30)} .

=4.40 N / mm ^{2} .

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

The stresses in the flange are within limits.

Step V Diameter of bolts
The diameter of the shaft is 60 mm.

\therefore \quad 40<d<100 mm .

The number of bolts is 4.
From Eq. (9.46),

d_{1}^{2}=\frac{8 M_{t}}{\pi D N \tau}                   (9.46).

d_{1}^{2}=\frac{8 M_{t}}{\pi D N \tau}=\frac{8(2984155.18)}{\pi(180)(4)(80)} .

The compressive stress in the bolt is determined by Eq. (9.47).

\sigma_{c}=\frac{2 M_{t}}{N d_{1} t D}                (9.47).

\sigma_{c}=\frac{2 M_{t}}{N d_{1} t D}=\frac{2(2984155.18)}{(4)(12)(30)(180)} .

=23.03 N / mm ^{2} .

\therefore \quad \sigma_{c}<240 N / mm ^{2} .

Step VI Dimensions of keys
From Table 9.3, the standard cross-section of the flat key for a 60-mm diameter shaft is 18 × 11 mm.

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

The length of the key is equal to l_{h} . Or,

l=l_{h}=90 mm .

The dimensions of the flat key are 18 × 11 × 90 mm.

From Eq. (9.27),

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

\tau=\frac{2 M_{t}}{d b l}=\frac{2(2984155.18)}{(60)(18)(90)}=61.4 N / mm ^{2} .

\therefore \quad \tau<80 N / mm ^{2} .

From Eq. (9.28),

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

\sigma_{c}=\frac{4 M_{t}}{d h l}=\frac{4(2984155.18)}{(60)(11)(90)} .

=200.95 N / mm ^{2} .

\sigma_{c}<240 N / mm ^{2} .

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