Answer:
[tex]R_{min} = 4.84\times 10^{-3} m[/tex]
Explanation:
Given data:
Applied force 750 N
Flexural strength is 105 MPa
separation is 50 mm = 0.05 m
flexural strength is given as
[tex]\sigma_f = \frac{FL}{\pi R_{min}^3}[/tex]
solving for R so we have
[tex]R_{min} = [\frac{FL}{\pi \sigma_f}]^{1/3}[/tex]
plugging all value to get minimum radius
[tex]R_{min} = [\frac{750 \times 0.05}{\pi 105 \times 10^6}]^{1/3}[/tex]
[tex]R_{min} = 4.84\times 10^{-3} m[/tex]
