Question

(b) If the rod is a rotating cantilever of 20 cm length with a transverse force F at its free end. Find the maximum force Fmax that will never break the rod due to fatigue.

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Answer 1

The assumed loading configuration is given below:

Assumed diameter = d m

Moment of Inertia

$\small I=\frac{\pi d^4}{64}$

The max bending moment is(at the base of cantilever)

$\small M=FL=0.2F\: Nm$

Therefore max bending stress is

$\small \sigma=\frac{My}{I}=\frac{0.2F(d/2)}{\pi d^4/64}=\frac{6.4F}{\pi d^3}=\frac{2.0372F}{d^3}\: Pa$

For a point located at the base of cantilever at surface of rod , the above stress will vary in sign from positive to negative.

Hence maximum stress is

$\small \sigma_{max}=\frac{2.0372F}{d^3}, \sigma_{min}=-\frac{2.0372F}{d^3}\: Pa$

Therefore mean stress is

$\small \sigma_m=\frac{\sigma_{max}+\sigma_{min}}{2}=0$

Stress amplitude is

$\small \sigma_a=\frac{\sigma_{max}-\sigma_{min}}{2}=\frac{2.0372F}{d^3}\: Pa$

Calculation of endurance limit for infinite cycles(Ref Shigley Chap-6)

Since material is not given we assume AISI 1040 steel(machined)

Tensile strength S_ut= 620 MPa

Yield strength S_y= 415 MPa

The rotating beam beam specimen(base) endurance limit is

$\small S'_e=0.5S_u_t=0.5(620)=310\: MPa$

The surface condition factor is (for machined surface)

$\small k_a=aS_u_t^b=(4.51)(620^{-0.265})=0.8207$

The size factor is(assuming d <51 mm)

$\small k_b=1.24(1000d)^{-0.107}=0.6065d^{-0.107}$

Loading factor k_c= 1 (for bending)

Temperature factor k_d= 1 (assumed room temperature)

Reliability factor k_e= 0.897 (assumed 90% reliability)

Hence the actual endurance limit is

$\small S_e=k_ak_bk_ck_dk_eS'_e=(0.8207)(0.6065d^{-0.107})(1)(1)(0.897)(310)=138.41d^{-0.107}\: MPa$

Applying modified Goodman criterion

$\small \frac{\sigma_a}{S_e}&plus;\frac{\sigma_m}{S_u_t}=\frac{1}{n}$

n= fatigue factor of safety=1 (for max F)

Hence

$\small \frac{2.0372F/d^3}{138.41d^{-0.107}\times10^6}&plus;\frac{0}{620}=\frac{1}{1}$

Or

$\small 1.4718\times10^{-8}d^{-2.893}F=1$

Hence

$\small F_{max}=67.94\times10^6d^{2.893}=826\: N$

For example if d= 20 mm= 0.02 m

$\small F_{max}=67.94\times10^6(0.02^{2.893})=826\: N$