Page 296 - Marks Calculation for Machine Design

P1: Shashi
January 4, 2005
15:4
Brown.cls
Brown˙C07
STRENGTH OF MACHINES
278
The equation of the straight line from (N = 1) where the fatigue strength (S f ) is the
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ultimate tensile strength (S ut ) to the knee at (N =10 ) cycles has the form
−0.01525
S f = S ut N (7.2)
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where the fatigue strength (S f ) has the value (0.9 S ut ) at (N =10 ) cycles.
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The equation of the straight line from the knee at (N = 10 ) cycles where the fatigue
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strength (S f ) is (0.9 S ut ) to the knee at (N = 10 ) cycles where the fatigue strength (S f ) is
(S e ) has the form in Eq. (7.3)
b
S f = aN (7.3)
where the coefficient (a) has units of stress and is given by Eq. (7.4)
(0.9 S ut ) 2
a = (7.4)
S e
and the exponent (b) is dimensionless and given by Eq. (7.5)
1 0.9 S ut
b =− log (7.5)
3 S e
If the amplitude stress (σ a ) is known, then substitute this value in Eq. (7.5) and solve for
the number of cycles (N), which is given in Eq. (7.6) as
1/b

σ a
N = (7.6)
a
Consider the following example using the above equations for finite life.
U.S. Customary SI/Metric
Example 2. Estimate the following design Example 2. Estimate the following design
information for a machine element made of a information for a machine element made of a
particular steel: particular steel:

a. Rotating-beam endurance limit (S ) a. Rotating-beam endurance limit (S )
e
e
5
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b. Fatigue strength (S f ) at 10 cycles b. Fatigue strength (S f ) at 10 cycles
c. Expected life for a stress level of 60 kpsi c. Expected life for a stress level of 420 MPa
where S ut is 90 kpsi and S y is 70 kpsi. where S ut is 630 MPa and S y is 490 MPa.
solution solution
Step 1. Using the guidelines in Eq. (7.1) where Step1. UsingtheguidelinesinEq.(7.1)where
the ultimate tensile strength (S ut ) is less than the ultimate tensile strength (S ut ) is less than
100 kpsi, the rotating-beam endurance limit 1400 MPa, the rotating-beam endurance limit
(S ) is (S ) is

e e

S = 0.504 S ut = 0.504 (90 kpsi) S = 0.504 S ut = 0.504 (630 MPa)
e
e
= 45.4 kpsi = 317.5MPa
Step 2. Using Eq. (7.4), calculate the coeffi- Step 2. Using Eq. (7.4), calculate the coeffi-
cient (a) as cient (a) as
(0.9 S ut ) 2 (0.9 (90 kpsi)) 2 (0.9 S ut ) 2 (0.9 (630 MPa)) 2
a = = a = =
S e 45.4 kpsi S e 317.5MPa
6,561 kpsi 2 321,489 MPa 2
= = 144.5 kpsi = = 1,013 MPa
45.4 kpsi 317.5MPa