Answer :

Correct answer is : 2

Given b = 12 mm, d = 20 mm; σ_ut = 600 MPa, σ_e = 200 MPa;

Load varies from 800 N to 1600 N

⇒ Moment varies from 800 × 0.1 N-m to 1600 × 0.1 N-m

⇒ M_max = 160 N-m, M_min = 80 N-m;

Now

${\sigma }_{max}=\frac{6\times 160}{0.012\times {20}^2}=200MPa$

${\sigma }_{min}=\frac{6\times 80}{0.012\times {20}^2}=100MPa$

Now

σ_m = 150 MPa, σ_a = 50 MPa;

Substituting in modified goodman line,

$\Rightarrow \frac{150}{600}+\frac{50}{200}=\frac{1}{N}$

⇒ N = 2;