Calculation Procedure:- Compute the temperature rise required
Use the relation AT = AZ)/(c£>), where Ar = temperature rise required,^0 F (^0 C); AD =
change in cylinder diameter, in (mm); c = coefficient of expansion of the cylinder = 6.5 x
10-6/^0 F (1.17 x IQ-V^0 C); D = cylinder internal diameter before heating, in (mm). Thus
AJ- (3/32)/[6.5 x 10-^6 (48)1 = 30O^0 F (167^0 C). - Compute the hoop stress in the cylinder
Upon cooling, the cylinder has a diameter Ys2 in (0.8 mm) larger than originally. Compute
the hoop stress from s = EkDID -3Ox 1O^6 CX 32 )MS = 19,500 lb/in^2 (134,452.5 kPa). - Compute the associated radial pressure
Use the relationp = 2tsfD, where/? = radial pressure, lb/in^2 (kPa), with the other symbols
as given earlier. Thus/? = 2(5/16)(19,500)/48 = 254 lb/in^2 (1751.3 kPa).
TORSION OFA CYLINDRICAL SHAFTA torque of 8000 lb-ft (10,840 N-m) is applied at the ends of a 14-ft (4.3-m) long cylindri-
cal shaft having an external diameter of 5 in (127 mm) and an internal diameter of 3 in
(76.2 mm). What are the maximum shearing stress and the angle of twist of the shaft if the
modulus of rigidity of the shaft is 6 x 106 lb/in^2 (4.1 x 104 MPa)?
Calculation Procedure:
- Compute the polar moment of inertia of the shaft
For a hollow circular shaft, J = (7r/32)(Z)
4
- d
4
), where J = polar moment of inertia of a
transverse section of the shaft with respect to the longitudinal axis, in^4 (cm^4 ); D — external
diameter of shaft, in (mm); d - internal diameter of shaft, in (mm). Substituting gives
J= (7T/32)(5^4 - 34 ) = 53.4 in^4 (2222.6 cm^4 ).
- Compute the shearing stress in the shaft
Use the relation ss = TRIJ, where ss = shearing stress, lb/in^2 (MPa); T = applied torque,
Ib-in (N-m); H = radius of shaft, in (mm). Thus ss = [(8000)(12)(2.5)]/53.4 = 4500 lb/in^2
(31,027.5 kPa). - Compute the angle of twist of the shaft
Use the relation 6 = TLIJG 9 where 6 = angle of twist, rad; L = shaft length, in (mm);
G = modulus of rigidity, lb/in^2 (GPa). Thus 0 = (8000)(12)(14)(12)/[53.4(6,000,000)] =
0.050 rad, or 2.9°.
ANALYSIS OFA COMPOUND SHAFT
The compound shaft in Fig. 24 was formed by rigidly joining two solid segments. What
torque may be applied at B if the shearing stress is not to exceed 15,000 lb/in^2 (103.4
MPa) in the steel and 10,000 lb/in^2 (69.0 MPa) in the bronze? Here G 5 = 12 x io^6 lb/in^2
(82.7 GPa); Gb = 6 x IO^6 lb/in^2 (41.4 GPa).