Page 299 - Marks Calculation for Machine Design

P1: Shashi
January 4, 2005
Brown˙C07
Brown.cls
For bending or torsion, use the following relationships for the range of sizes indicated in
Eq. (7.10). 15:4 FATIGUE AND DYNAMIC DESIGN 281
−0.1133


d

 0.11 in ≤ d ≤ 2in

0.3

k b = (7.10)
d
−0.1133


 2.79 mm ≤ d ≤ 51 mm
7.62

For bending and torsion of larger sizes, the size factor (k b ) varies between 0.60 and 0.75.
For machine elements that are round but not rotating, or shapes that are not round, an
effective diameter, denoted (d e ), must be used in Eq. (7.10). For a nonrotating round or
hollow cross section, the effective diameter (d e ) is given in Eq. (7.11) as
d e = 0.370 D (7.11)
where the diameter (D) is the outside diameter of either the solid or hollow cross section.
For a rectangular cross section (b × h), the effective diameter (d e ) is given in
Eq. (7.12) as
d e = 0.808 (bh) 1/2 (7.12)
Consider the following example using the above size factor (k b ) equations.
U.S. Customary SI/Metric
Example 1. Compare the size factor (k b ) for Example 1. Compare the size factor (k b ) for
a 2-in diameter solid shaft in torsion to a 2-in a 51-mm diameter solid shaft in torsion to a
diameter but hollow nonrotating shaft. 51-mm diameter but hollow nonrotating shaft.
solution solution
Step 1. Using Eq. (7.10) calculate the size Step 1. Using Eq. (7.10) calculate the size
factor (k b ) as factor (k b ) as
−0.1133 −0.1133 −0.1133 −0.1133

d 2 d 51
k b = = k b = =
0.3 0.3 7.62 7.62
= (6.67) −0.1133 = 0.81 = (6.69) −0.1133 = 0.81
Step 2. Using Eq. (7.11), calculate the effec- Step 2. Using Eq. (7.11), calculate the effec-
tive diameter (d e ) as tive diameter (d e ) as
d e = 0.370 D = 0.370 (2in) d e = 0.370 D = 0.370 (51 mm)
= 0.74 in = 18.87 mm
Step 3. Using the effective diameter (d e ) from Step 3. Using the effective diameter (d e ) from
step 2 in Eq. (7.10) gives the size factor (k b ) for step 2 in Eq. (7.10) gives the size factor (k b ) for
the hollow cross section as the hollow cross section as
−0.1133 −0.1133 −0.1133 −0.1133
d 0.74 d 18.87

k b = = k b = =
0.3 0.3 0.3 7.62
= (2.47) −0.1133 = 0.90 = (2.48) −0.1133 = 0.90
Notice that there is almost a 10 percent dif- Notice that there is almost a 10 percent dif-
ference between these two size factors. ference between these two size factors.