Respuesta :
Answer:
P = 0.490 kip
Explanation:
given data
allowable bearing stress = 2 ksi
allowable tensile stress = 18 ksi
diameter = 0.31 in
outer diameter = 0.75 in
inner diameter (hole) = 0.50 in
solution
we find here cross section area of shank that is express as
Area = [tex]\frac{\pi }{4} \times d^2[/tex] ..................1
area = [tex]\frac{\pi }{4} \times 0.31^2[/tex]
area = 0.0754 in²
and
now we get here allowable load in bolt will be
[tex]\sigma = \frac{P}{A}[/tex] ...................2
P = [tex]\sigma \times A[/tex]
P = 18 × 10³ × 0.0754
P = 1357.2 = 1.357 kip
and
now find here area of washer is
Area = [tex]\frac{\pi }{4} \times (d^2-d1^2)[/tex] .......................3
put here value
Area = [tex]\frac{\pi }{4} \times (0.75^2-0.5^2)[/tex]
area = 0.2454 in²
so now we get here allowable load of washer will be
[tex]\sigma = \frac{P}{Area}[/tex] .....................4
P = 2 × 10³ × 0.245
P = 490 = 0.490 kip
