Question

A cylindrical shaft is subjected to an alternating stress of 100 mpa

Answer 1

Explanation:

• Given data

• Here given shaft is subjected to alternating stress. Magnitude of alternating stress is 100 MPa.

• Means it is subjected to maximum and minimum stress.But here it is not clear maximum and minimum value of stress.So we take

        $\mathbf{\sigma _{max}= 100 \ MPa}$

        $\mathbf{\sigma _{min}= -100 \ MPa}$

• Magnitude of mean stress is average of maximum and minimum stress.

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

        $\mathbf{\sigma _{m}=\frac{100-100}{2}}$

        $\mathbf{\sigma _{m}=0 \ MPa}$

• Amplitude of stress is difference between maximum stress and mean stress.

        $\mathbf{\sigma _{a}=\sigma _{max}-\sigma _{m}}$

        $\mathbf{\sigma _{a}=100-0=100 \ MPa}$

Answer 2

Complete question:

A cylindrical shaft is subjected to an alternating stress of 100 Mpa. Fatigue strength to sustain 1000 cycles is 490 Mpa if the corrected endurance  strength is 70 Mpa, estimated shaft life will be

Answer:

The estimated shaft life is 281914 cycles.

Explanation:

From Basquin's equation,

A = $S_fL^B$

Where,

$S_f$ = Fatigue strength = 490 Mpa

$L^B$ = Number of cycles = 1000 cycles

Now, on substituting the values, we get,

A = 490 × $10^{3B}$ → (equation 1)

For 10⁶ cycles, $S_f = S_e =$ 70 Mpa

A = 70 × $10^{6B}$ → (equation 2)

On dividing equation (1) and (2), we get,

1 = 490/70 × $10^{3B}$/$10^{6B}$

$10^{3B}$ = 490/70

On taking ln on both sides, we get,

3B ln 10 = ln 490/70

∴ B = 0.281

Now, we need to find the endurance for 70 Mpa, we get,

A = 70 × $10^{(6 \times 0.281)}$ = 3397

Now,

$3397 \cdot 7=100 \cdot L^{0.281}$

∴ L = 281914 cycles