Design a rigid flange coupling to transmit a torque of 257N-m between two coaxial shafts. The shaft is made of alloy steel, flanges out of cast iron and bolts out of steel. Four bolts are used to couple the flanges. The shafts are keyed to the flange hub. The permissible stresses are given below: Shear stress on shaft =100MPa Bearing or crushing stress on shaft =250MPa Shear stress on keys =100MPa Bearing stress on keys = 250MPa Shearing stress on cast iron = 200Mpa Shear stress on bolts =100MPa  Answer in Word please

Mechanics of Materials (MindTap Course List)
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ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter3: Torsion
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Design a rigid flange coupling to transmit a torque of 257N-m between two coaxial shafts. The shaft

is made of alloy steel, flanges out of cast iron and bolts out of steel. Four bolts are used to couple the

flanges. The shafts are keyed to the flange hub. The permissible stresses are given below:

Shear stress on shaft =100MPa

Bearing or crushing stress on shaft =250MPa

Shear stress on keys =100MPa

Bearing stress on keys = 250MPa

Shearing stress on cast iron = 200Mpa

Shear stress on bolts =100MPa 

Answer in Word please 

(iii) Diameter of Pins The number of pins (iv) Dimensions of Bushes Calculate the outer
is usually 4 or 6. The diameter of the pins is diameter of the rubber bush by Eq. (9.50). It is
calculated by the following empirical equation,
given by,
0.5d
%3D
M, =-D; DN
d =
(9.52)
N.
Calculate the effective length of the rubber bush
Determine the shear stress in the pins by Eq. by the following relationship,
(9.51).
8M,
T= -
1, =D,
n df DN
(v) Dimensions of Keys Detemine the standard
The shear stress calculated by the above equation cross-section of flat key from Table 9.3. The length
should be less than 35 N/mm?. Also, determine the of the key in each shaft is I, . Therefore,
bending stresses in the pins and confirm that it is
With the above dimensions of the key, check
the shear and compressive stresses in the key by
Eqs (9.27) and (9.28) respectively.
within limit.
Transcribed Image Text:(iii) Diameter of Pins The number of pins (iv) Dimensions of Bushes Calculate the outer is usually 4 or 6. The diameter of the pins is diameter of the rubber bush by Eq. (9.50). It is calculated by the following empirical equation, given by, 0.5d %3D M, =-D; DN d = (9.52) N. Calculate the effective length of the rubber bush Determine the shear stress in the pins by Eq. by the following relationship, (9.51). 8M, T= - 1, =D, n df DN (v) Dimensions of Keys Detemine the standard The shear stress calculated by the above equation cross-section of flat key from Table 9.3. The length should be less than 35 N/mm?. Also, determine the of the key in each shaft is I, . Therefore, bending stresses in the pins and confirm that it is With the above dimensions of the key, check the shear and compressive stresses in the key by Eqs (9.27) and (9.28) respectively. within limit.
DESIGN PROCEDURE FOR FLEXIBLE COUPLING
The basic procedure for finding out the dimensions
of bushed pin type flexible coupling consists of the
following steps:
The torsional shear stress in the hub can be
calculated by considering it as a hollow shaft
subjected to torsional moment M. The inner and
outer diameters of the hub are d and d, respectively.
The torsional shear stress in the hub is given by,
(i) Shaft Diameter Calculate the shaft diameter
by using the following two equations:
60x10 (kW)
16 M,
and T=-
M,r
M, =
2πη
J
n(d-dt)
|
(ii) Dimensions
of Flanges Calculate
the
J =
32
dimensions of the flanges by the following
empirical relationships:
2d
%3D
The shear stress in the flange at the junction with
the hub is calculated by Eq. (9.45). It is given by,
1.5d
%3D
D = 3d to 4d
t = 0.5d
%3D
1
%3D
%3D
t = 0.25d
%3D
2
Transcribed Image Text:DESIGN PROCEDURE FOR FLEXIBLE COUPLING The basic procedure for finding out the dimensions of bushed pin type flexible coupling consists of the following steps: The torsional shear stress in the hub can be calculated by considering it as a hollow shaft subjected to torsional moment M. The inner and outer diameters of the hub are d and d, respectively. The torsional shear stress in the hub is given by, (i) Shaft Diameter Calculate the shaft diameter by using the following two equations: 60x10 (kW) 16 M, and T=- M,r M, = 2πη J n(d-dt) | (ii) Dimensions of Flanges Calculate the J = 32 dimensions of the flanges by the following empirical relationships: 2d %3D The shear stress in the flange at the junction with the hub is calculated by Eq. (9.45). It is given by, 1.5d %3D D = 3d to 4d t = 0.5d %3D 1 %3D %3D t = 0.25d %3D 2
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