Page 195 - Marks Calculation for Machine Design

P1: Shibu
January 4, 2005
14:25
Brown.cls
Brown˙C04
U.S. Customary COMBINED LOADINGS SI/Metric 177

2

2
Eδ R Eδ R
p = 1 − p = 1 −
2R r o 2R r o
9
2
6
2
(30 × 10 lb/in )(0.0005 in) (207 × 10 N/m )(0.00001 m)
= =
2 (0.75 in) 2 (0.02 m)
2

2
0.75 in 0.02 m

× 1 − × 1 −
4in 0.1m
15,000 lb/in 2,070,000 N/m
= (1 − 0.035) = (1 − 0.04)
1.5in 0.04 m
2
2
= (10,000 lb/in )(0.965) = (51,750,000 N/m )(0.96)
2
2
= 9,650 lb/in = 9.65 kpsi = 49,680,000 N/m = 49.68 MPa
Location of the Maximum Stress Element. Figure 4.23 shows a press fitted gear (no teeth
shown) and solid shaft assembly with the relative dimensions of the assembly in Example 8.
Stress elements T
r o Gear
R
Shaft
FIGURE 4.23 Element for maximum stress.
Two stress elements are identified in Fig. 4.23, and the determination of which one has
the maximum stress state is related to how the individual stresses due to the combination
of loads vary with respect to the radius (r) from the center of the assembly. Also, the two
elements shown are not specific to a particular angular location around the assembly. What
follows is a discussion of how the maximum stress element is chosen.
First, the counterclockwise torque (T ) is caused by a mating spur gear not shown. This
torque produces a shear stress (τ xy ) in the body of the spur gear that is maximum at the
outside radius (r o ), usually taken as the radius to the root of the teeth of the gear, and
minimum at the the inside radius of the gear that is, the interface radius (R). (The stresses
on the gear teeth themselves is a topic in itself, not covered in this book.)
The shear stress (τ xy ) due to the torque (T ) is given by Eq. (4.20)
Tr
τ xy = (4.20)
J
where the polar moment of inertia (J) for the gear is given by Eq. (4.21) as
1 4 4
J = π r − R (4.21)
o
2