Page 324 - Marks Calculation for Machine Design

P1: Shashi
15:4
January 4, 2005
Brown.cls
Brown˙C07
STRENGTH OF MACHINES
306
solution
Step 1. Using Eq. (7.8) and the values for the coefficient (a) and exponent (b) from
Table 7.1, calculate the surface finish factor (k a ) as
b −0.995
k a = aS ut = (39.9 kpsi)(90 kpsi) = (39.9)(0.0114) = 0.45
Step 2. Using Eq. (7.10) and the given diameter, calculate the size factor (k b ) as
−0.1133 −0.1133

d 1.5 −0.1133
k b = = = (5) = 0.83
0.3 0.3
Step 3. The shaft is in torsion so the load type factor (k c ) from Eq. (7.14) is
k c = 0.577
Step 4. As the shaft is operating at room temperature, the temperature factor (k d ) from
Eq. (7.15) and Table 7.2 is
k d = 1
Step 5. Using the given geometric shear stress concentration factor (K ts ) and the notch
sensitivity (q), calculate the reduced concentration factor (K f ) from Eq. (6.23) as
K f = 1 + q(K ts − 1) = 1 + (0.9)(1.65 − 1) = 1 + 0.585 = 1.585
Step 6. Using the reduced stress concentration factor (K f ) found in step 5, calculate the
miscellaneous effect factor (k e ) using Eq. (7.16) as
1 1
k e = = = 0.63
K f 1.585
Step 7. Using the given ultimate tensile stress (S ut ) and the guidelines in Eq. (7.1), calculate

the test specimen endurance limit (S ) as
e

S = 0.504 S ut = (0.504)(90 kpsi) = 45.4 kpsi
e
Step 8. Using the test specimen endurance limit (S ) found in step 7, and the modifying

e
factors found in steps 1 through 6, calculate the endurance limit (S e ) for the solid shaft
using the Marin equation in Eq. (7.7) as

S e = k a k b k c k d k e S = (0.45)(0.83)(0.577)(1)(0.63)(45.4 kpsi)
e
= (0.136)(45.4 kpsi) = 6.2 kpsi
Step 9. Calculate the mean torque (T m ) and the alternating torque (T a ) as
T max + T min (2,200 ft · lb) + (1,800 ft · lb)
T m = =
2 2
4,000 ft · lb
= = 2,000 ft · lb = 24,000 in · lb
2
T max − T min (2,200 ft · lb) − (1,800 ft · lb)
T a = =
2 2
400 ft · lb
= = 200 ft · lb = 2,400 in · lb
2