Answer:
24.87 mm
Explanation:
The area of the bolt is given by
[tex]A_b=\pi r^{2}[/tex]
Since diameter is 16mm, the radius is 16/2= 8 mm= 0.008 m
Area, [tex]A_b=\pi\times 0.008^{2}=0.000201062 m^{2}[/tex]
For safe design of the bolt, we use stress of 201 Mpa
[tex]\sigma_b=\frac {P}{A_b}[/tex] where P is the load and [tex]\sigma_b[/tex] is normal stress.
Making P the subject then
[tex]P=A_b \sigma_b[/tex]
Substituting the figures given and already calculated area of bolt
[tex]P=201\times 10^{6}\times 0.000201062 m^{2}=40413.4479 N[/tex]
The area of spacer is given by
[tex]\pi (r_o^{2}- r_i^{2})[/tex] where r is radius and the subscripts o and I denote inner and outer respectively
The value of 142 Mpa by default becomes the stress on spacer hence
[tex]\sigma_s=\frac {P}{A_s}[/tex] and making [tex]A_s[/tex] the subject then
[tex]A_s=\frac {P}{\sigma_s}[/tex]
[tex]\pi (r_o^{2}- r_i^{2})=\frac {P}{\sigma_s}[/tex]
Since we already calculated the value of P as 40413.4479 N and inner radius is 8mm= 0.008 m then
[tex]\pi (r_o^{2}- 0.008^{2})=\frac {40413.4479}{142\times 10^{6}}[/tex]
[tex]r_o^{2}=0.000154592[/tex]
[tex]r=\sqrt{0.000154592}=0.012433 m[/tex]
Therefore, [tex]d=2r=2*0.012433=0.024867 m=24.86697 mm\approx 24.87 mm[/tex]
