Respuesta :
Answer:
[tex] d_s = 16 mm \sqrt{1+\frac{216 Mpa}{143 Mpa}} =25.35 mm[/tex]
So then the outer diameter for the spaces should be 25.35 in order to have the most economical and safe design.
Explanation:
Let's define some notation first:
[tex] \sigma_b =216 Mpa[/tex] represent the tensile stress for the bolt
[tex] \sigma_s = 143Mpa[/tex] represent the tensile stress for the spacer
[tex] d_b = 16 mm[/tex] the bolt diameter
[tex] d_s[/tex] the value that we want to find.
For this case we need to remember the definition of tensile stress given by:
[tex] \sigma = \frac{F}{A} = \frac{P}{ \frac{\pi}{4} d^2}[/tex]
And if we solve for P we got:
[tex] P = \frac{\pi}{4} d^2 \sigma_b[/tex] where P represent the normal stress
For this case we want that the most economical space and we can do this setting equal the both values of P for this case like this:
[tex] P_b =P_s[/tex]
And if we replace we got:
[tex] \frac{\pi}{4} \sigma_s (d^2_s - d^2_b) = \frac{\pi}{4} \sigma_b d^2_b[/tex]
And if we cancel the constant terms we got:
[tex] d^2_s - d^2_b = \frac{\sigma_b}{\sigma_s} d^2_b [/tex]
And if we solve for the variable of interest we got:
[tex] d^2_s = d^2_b +\frac{\sigma_b}{\sigma_s} d^2_b [/tex]
And taking common factor we got:
[tex] d^2_s = d^2_b (1+\frac{\sigma_b}{\sigma_s})[/tex]
And if we take the square root on both sides we got:
[tex] d_s = d_b \sqrt{1+\frac{\sigma_b}{\sigma_s}}[/tex]
And if we replace we got:
[tex] d_s = 16 mm \sqrt{1+\frac{216 Mpa}{143 Mpa}} =25.35 mm[/tex]
So then the outer diameter for the spaces should be 25.35 in order to have the most economical and safe design.
