Page 160 - Marks Calculation for Machine Design

P1: Rakesh
14:16
January 4, 2005
Brown˙C03
Brown.cls
142
STRENGTH OF MACHINES
strength in compression that was found in the previous section to be half the yield stress
(S y ). Converting this statement into a factor-of-safety expression is given in Eq. (3.28) as
S y τ max 1
τ max < S sy = → = (3.28)
2 S y n
2
For the next example, consider a Mars rover with solid spherical titanium alloy wheels,
which during testing on Earth, is driven over flat granite rock to simulate the Mars landscape.
The rover has four wheels that carry the total load evenly.
U.S. Customary SI/Metric
Example 1. Determine whether the design Example 1. Determine whether the design
of the titanium wheels for the rover described of the titanium wheels for the rover described
above will be safe during the test on granite above will be safe during the test on granite
rock, where rock, where
W = 8,000 lb = 4F W = 36,000 N = 4F
d wheel = 3ft = 36 in, d rock =∞ d wheel = 1m, d rock =∞
S y = 110 kpsi (titanium) S y = 770 MPa (titanium)
ν 1 = 0.33 (titanium) ν 1 = 0.33 (titanium)
3
E 1 = 15 × 10 kpsi = 15 Mpsi (titanium) E 1 = 105 GPa (titanium)
ν 2 = 0.3 (granite) ν 2 = 0.3 (granite)
3
E 2 = 10 × 10 kpsi = 10 Mpsi (granite) E 2 = 70 GPa (granite)
solution solution
Step 1. Using Eq. (3.19), calculate the radius Step 1. Using Eq. (3.19), calculate the radius
of the contact area (a) as of the contact area (a) as

2 2 2 2
1 − ν 1 − ν 1 − ν 1 − ν
1 2 1 2
3F + 3F +
E 1 E 2 E 1 E 2
a = 3 a = 3
8 1 + 1 8 1 + 1
d 1 d 2 d 1 d 2

2 2 2 2
1 − 0.33 1 − 0.3 1 − 0.33 1 − 0.3

+
+
3(2,000 lb) 15 Mpsi 10 Mpsi 3(9,000 N) 105 GPa 70 GPa
= = 1 1
8 1 1 8
3 + 3 +
36 in ∞ 1m ∞

0.89 0.91 0.89 0.91

3 3 3 3
= (750)(36) + in = (3,375) + m
15 × 10 6 10 × 10 6 105 × 10 9 70 × 10 9
$ $
3 −7 3 3 −11 3
= (750)(36) 1.5 × 10 in = (3,375) 2.15 × 10 m
$ %
3
3 −3 3 = (7.25 × 10 −8 ) m 3
= (4.06 × 10 ) in
= 0.16 in = 0.0042 m = 0.42 cm
Step 2. Using Eq. (3.21) calculate the maxi- Step 2. Using Eq. (3.21) calculate the maxi-
mum pressure (p max ) mum pressure (p max ).
3F 3F
p max = p max =
2πa 2 2πa 2
3(2,000 lb) 6,000 lb 3(9,000 N) 27,000 N
= = = =
2π(0.16 in) 2 0.16 in 2 2π(0.0042 m) 2 0.00011 m 2
= 37,302 lb/in 2 = 243,600,000 N/m 2
= 37.3 kpsi = 243.6MPa