The steel bar with the geometry shown below has an axial fluctuating load applied from $-5000 \, \text{lbf}$ to $12000 \, \text{lbf}$. The bar has a hole with a $0.25 \, \text{inch}$ diameter, is $0.25 \, \text{inches}$ thick, has a fillet radius of $0.175 \, \text{inches}$, and a notch sensitivity factor $q = 0.95$.

The fully adjusted endurance limit of the bar is $40 \, \text{ksi}$, the bar has an ultimate strength of $185 \, \text{ksi}$, and a fatigue strength fraction $f = 0.78$.

Using the Goodman criteria, determine the most conservative $N$ to design to, where $N$ is the life of the bar in terms of cycles to failure (worst case scenario).

- Theoretical Stress Concentration Factor $K_f$
- Less than 1 cycle
- 1 to 5 cycles
- 17000 to 29000 cycles
- 40000 to 60000 cycles
- 63000 to 80000 cycles
- 5 to 6 million cycles
- Infinite life

The most conservative N to design using the Goodman criteria can be determined by finding the maximum allowable fluctuating stress. In this case, the most conservative N to design to is approximately 3.3552 cycles to failure.

Explanation:

The most conservative N to design using the Goodman criteria can be determined by finding the maximum allowable fluctuating stress. According to the Goodman criteria, the maximum allowable stress is given by:

Sa = Se/ (1 + (Sm/Su))

Where:

Sa is the maximum allowable fluctuating stress.

Se is the fully adjusted endurance limit of the bar.

Sm is the mean stress (average of maximum and minimum stress)

Su is the ultimate strength of the bar.

In this case, Se = 40 ksi, Sm = (12000 - 5000)/2 = 8500 lbf, and Su = 185 ksi. Converting the values to kips, we have Se = 0.04-kip, Sm = 0.0085 kip, and Su = 0.185 kip. Substituting these values into the equation:

Sa = 0.04/ (1 + (0.0085/0.185))

Simplifying this expression gives:

Sa = 0.029 kip

The most conservative N to design to can then be found using the fatigue strength fraction:

N = f * (Sa/σa)

Where:

N is the life of the bar in terms of cycles to failure

f is the fatigue strength fraction (0.78 in this case)

Sa is the maximum allowable fluctuating stress

σa is the alternating stress (range of maximum and minimum stress)

In this case, σa is given by:

σa = (12000 - (-5000))/2 = 8500 lbf

Converting this value to kips, we have σa = 0.0085 kip. Substituting the values into the equation:

N = 0.78 * (0.029/0.0085)

Simplifying this expression gives:

N ≈ 3.3552

Therefore, the most conservative N to design to is approximately 3.3552.