Question

Consider the fatigue properties of tool steel as shown below. Assume the data and endurance limits represent the fatigue properties when R = 0. Find the endurance limits when R = 0.3 and R = -0.3.

APTER 7 Mechanical Properties: Part Two 120 100 80 100,000 cycle fatigue life at 90,000 psi applied stress Tool steel Endurance limit 60,000 psi 3 60 40 20 0 Aluminum alloy 104 105 106 107 Number of cycles Figure 7-17 The stress-number of cycles to failure (S-N) curves for a tool steel and an lminum allov. (Credit:Cengage Learning 2014)

Answer 1

SOLUTION: We have to analyse the fatigue properties of tool steel given above in the figure and find the endurance limits for different R values.

Analysis: We have plotted a S-N curve based on the property diagram shown above:

logio Sf logio 110 logio 90 Endurance Limit logio 60 logio N 10 4

From above S-N curve we can derive the number of cycles as a function of $\frac{\log 110 - \log 60 }{7-4}=\frac{\log 110 - \log S{_{f}}}{\log N-4 }$ .

$\Rightarrow \log N=27.26449 - \frac{\log S_{f}}{0.087747}$

With the help of above formula we can plot a chart or graph of N as a function of S_f.

Sf (in ksi)	log N	N (no of cycles)
10	15.868	7380874246000000
20	12	1000000000000
40	9	1000000000
60	7	10000000
80	5.576	376704
100	4.47	29512
120	3.56	3630.78

We are given three cases for the values of R. First we will find out the relation between R, mean stress and amplitude stress then we will analyse the effect of mean and amplitude stresses on the failure stress as a function of R.

mean stress,

$\sigma _{m}=\frac{\sigma _{max}+\sigma _{min}}{2}$

stress amplitude,

$\sigma _{a}=\frac{\sigma _{max}-\sigma _{min}}{2}$

Stress Ratio,

$R=\sigma _{min}/\sigma _{max}$

Amplitude ratio,

$A=\sigma _{a}/\sigma _{m}=\frac{1-R}{1+R}$

Case (1): R=0

$R=\sigma _{min}/\sigma _{max}$

$\Rightarrow \sigma _{min}=0$

$\Rightarrow \sigma _{m}=\sigma _{a}$

The tool steel undergoes cyclic of loading and unloading in tensile or only compressive manner.

Case (2): R=0.3

$\Rightarrow \sigma _{min}/\sigma _{max}=0.3$

$\Rightarrow \sigma _{min}=0.3\sigma _{max}$

The tool steel undergoes Fluctuating stresses where minimum stress is 0.3 times the maximum stress. These both stress limits may lie in tensile zone or in compressive zone.

Case (3):R=-0.3

$\Rightarrow \sigma _{min}/\sigma _{max}=-0.3$

$\Rightarrow \sigma _{min}=-0.3\sigma _{max}$

In this case, tool steel will be subjected to alternate tensile and compressive stresses and here also the magnitude of minimum stress will be 0.3 times that of maximum stress.  

To develop a relationship between mean stress, stress amplitude and the number of cycles we draw Haigh diagram as one shown below:

Finite Life Region N-103 2 N-104 N-10 N-106 Lines of Constant Life 5 Infinite Life Region Mean Stress. σ

A very substantial amount of testing is required to generate a Haigh diagram, and it is usually
impractical to develop curves for all combinations of mean and alternating stresses.There are few methods to find out the endurance limit and yield strength based on the given values of mean stress and stress amplitude which are given below:

Goodman (England, 1899): -a + =1 tu Gerber (Germany, 1874): Sa Sm Se Sv Soderberg (USA, 1930): Sa Sm Morrow (USA, 1960s):

Other observations related to the mean stress equations include:

• All methods should only be used for tensile mean stress values. For cases where the mean stress is small relative to the alternating stress (R << 1), there is little difference in the methods.
• The Soderberg method is very conservative. It is used in applications where neither fatigue failure nor yielding should occur.
• For hard steels (brittle), where the ultimate strength approaches the true fracture stress, the Morrow and Goodman curves are essentially equivalent.

Effect of Load ratio on fatigue life time:

1. Fatigue cracks are initiated at an early fatigue stage, i.e., 5–10% of fatigue life, indicating that total of fatigue lives N_f can be approximated as the crack propagation lives, N_p.
2. The effect of mean stress on the fatigue strength was given approximately by the Gerber relationship.
   0 Gerber line Goodman line Soderberg line 0 0 0 0 0 0 0 Mean stress, σ Compressive stress Tensile stress
   $\frac{\sigma _{a}}{S_{e}}=\frac{\sigma _{m}}{S_{ut}}^{2}=1$
3. For the case of the negative R ratios such as R =-1 and R = -2, compressive yield strength should be used in analyzing short fatigue crack propagation and in predicting the S–N curves based on the M parameter.